mm-4069
===
Subject: Re: Antidiagonal, Infinity
> What I want to figure out are things like what is the probability of
> a real being a rational, or an algebraic irrational or a
> transcendental irrational? I have a conception that the probability
> of a real being a rational is one-half,
Your conception is a misconception.
One can construct a sequence of (either open or closed) real
intervals of arbitrary small _total_ length that contains ALL the
rationals, thus the probability of selecting a rational at random
from a uniformly distributed interval of reals must also be
arbitrarily small, namely zero. This is, in fact, the way that one
proves that the set of rationals has measure zero in the set of
reals, based on the measure of a real interval being its length.
> I derive that notion from the consideration that the rationals
> and irrationals alternate on the real number line, contrary to
> that the rationals are countable and the irrationals are
> uncountable, which would suppose the probability of a real being
> rational much smaller than that of it being irrational, and as
> many reals being in [0,1/2) as in (-oo, oo).
Once you assume conditions contrary to fact, as you are doing here,
every statement becomes both true and false.
The idea that rationals and irrationals can alternate in any sense
is another misconception, one which Ross seems unable to rid himself
of, since it continues to pollute his attempts at coherent thought.
Any rule of alternation must exhaust all the rationals while there
are still as many irrationals left as one started with.
Subject: Re: Antidiagonal, Infinity
> What I want to figure out are things like what is the probability of
> a real being a rational, or an algebraic irrational or a
> transcendental irrational? I have a conception that the probability
> of a real being a rational is one-half,
> Your conception is a misconception.
> One can construct a sequence of (either open or closed) real
> intervals of arbitrary small _total_ length that contains ALL the
> rationals, thus the probability of selecting a rational at random
> from a uniformly distributed interval of reals must also be
> arbitrarily small, namely zero. This is, in fact, the way that one
> proves that the set of rationals has measure zero in the set of
> reals, based on the measure of a real interval being its length.
> I derive that notion from the consideration that the rationals
> and irrationals alternate on the real number line, contrary to
> that the rationals are countable and the irrationals are
> uncountable, which would suppose the probability of a real being
> rational much smaller than that of it being irrational, and as
> many reals being in [0,1/2) as in (-oo, oo).
> Once you assume conditions contrary to fact, as you are doing here,
> every statement becomes both true and false.
> The idea that rationals and irrationals can alternate in any sense
> is another misconception, one which Ross seems unable to rid himself
> of, since it continues to pollute his attempts at coherent thought.
> Any rule of alternation must exhaust all the rationals while there
> are still as many irrationals left as one started with.
There are a variety of definitions of measure.
Consider your example about the topological property of the rationals:
it's based upon the rationals being countable, in that a countaby
infinite set maps to them. The same method derives uncountably many
sequences that contain any irrationals, and is concerned about there
being a rational between any two irrationals.
I'm aware of logic behind the considerations that there are many more
irrationals than rationals. I just want to ignore it temporarily to
consider the rationals and irrationals alternating.
If there are more irrationals than rationals then there are
necessarily two irrationals between which in value there are no
rationals, which is not true, an argument flip. If it's true then
there are vast ranges of irrationals among which there are no
irrationals, the distance of those ranges obviously goes to zero.
Heh, pollution <-> corruption.
Say two points start on a race from zero along the reals towards
infinity. One point has a velocity of 1 distance unit / time unit,
the other of 2 distance units / time unit. At any point in time point
one is at point x where point two is at point 2x.
If you consider the measure of the reals [0,1] to be one, I totally
agree.
Allow yourself to consider that the measure of the rationals on [0,1)
is 1/2. Too late, you already did. With what known results does it
disagree? What are the verifiable real-world results of assuming the
rationals to be measure zero? With what analytical results does it
disagree?
What was the function bijecting the reals and irrationals again? Hmm?
I appreciate you, because you're smart.
Ross
Subject: Re: Antidiagonal, Infinity
> What was the function bijecting the reals and irrationals again? Hmm?
There is a perfectly good function that bijects the reals to the
irrationals. It doesn't even need the choice axiom.
Let S be a countable set of irrationals. Index S over the natural
numbers cross the natural numbers. So if n and m are natural numbers,
then S(n,m) is an irrational.
You knew this was comming: Enumerate the rationals Q = (q1, q2, q3,
...).
The mapping f: R -> RQ is the following:
On Q, f takes qn to S(n, 1)
On S, f takes S(n,m) to S(n, m+1)
On R Q S, f is the identity
That f is a bijection is obvious.
Subject: Re: Antidiagonal, Infinity
> What was the function bijecting the reals and irrationals again? Hmm?
For Ross' delectation:
The reals R form an infinite dimensional vector space over the
rational field Q. It is actually of uncountable dimension.
Let b_0, b_1, ..., b_n, ... be a sequence of reals linearly
independent over Q with b_0 rational, and let C be the set of reals
linearly independent over Q of all these b_i.
Define f:R -> RQ by
f(b_n) = b_{n+1}
F(c) = c, for c in C,
and extend to all of R by linearity over Q.
Subject: Re: Antidiagonal, Infinity
What I want to figure out are things like what is the probability
of
> a real being a rational, or an algebraic irrational or a
> transcendental irrational? I have a conception that the
probability
> of a real being a rational is one-half,
Your conception is a misconception.
One can construct a sequence of (either open or closed) real
> intervals of arbitrary small _total_ length that contains ALL the
> rationals, thus the probability of selecting a rational at random
> from a uniformly distributed interval of reals must also be
> arbitrarily small, namely zero. This is, in fact, the way that one
> proves that the set of rationals has measure zero in the set of
> reals, based on the measure of a real interval being its length.
I derive that notion from the consideration that the rationals
> and irrationals alternate on the real number line, contrary to
> that the rationals are countable and the irrationals are
> uncountable, which would suppose the probability of a real being
> rational much smaller than that of it being irrational, and as
> many reals being in [0,1/2) as in (-oo, oo).
Once you assume conditions contrary to fact, as you are doing here,
> every statement becomes both true and false.
> The idea that rationals and irrationals can alternate in any sense
> is another misconception, one which Ross seems unable to rid himself
> of, since it continues to pollute his attempts at coherent thought.
Any rule of alternation must exhaust all the rationals while there
> are still as many irrationals left as one started with.
> There are a variety of definitions of measure.
> Consider your example about the topological property of the rationals:
> it's based upon the rationals being countable, in that a countaby
> infinite set maps to them. The same method derives uncountably many
> sequences that contain any irrationals, and is concerned about there
> being a rational between any two irrationals.
> I'm aware of logic behind the considerations that there are many more
> irrationals than rationals. I just want to ignore it temporarily to
> consider the rationals and irrationals alternating.
That makes about as much sense as wanting to ignore the law of
gravity for a while.
> If there are more irrationals than rationals then there are
> necessarily two irrationals between which in value there are no
> rationals,
This only need hold for finite sets. Infinite sets do not behave the
same way as finite sets. But you seem to think they should. Get over
it.
> which is not true, an argument flip. If it's true then
> there are vast ranges of irrationals among which there are no
> irrationals, the distance of those ranges obviously goes to zero.
Not so.
> Heh, pollution <-> corruption.
> Say two points start on a race from zero along the reals towards
> infinity. One point has a velocity of 1 distance unit / time unit,
> the other of 2 distance units / time unit. At any point in time point
> one is at point x where point two is at point 2x.
> If you consider the measure of the reals [0,1] to be one, I totally
> agree.
> Allow yourself to consider that the measure of the rationals on [0,1)
> is 1/2. Too late, you already did. With what known results does it
> disagree? What are the verifiable real-world results of assuming the
> rationals to be measure zero? With what analytical results does it
> disagree?
What ARE you smoking?
Subject: Re: math wizzards..??
> Hello..
> Im trying to figure out this quiz..
> If u know the answer, ( and how u get it) pls e-mail me back
> K.
> Could you please answer the following 2 questions and send me back your
> answers, please???
> First piece of information
> You have 100 large bowls and in each bowl you have 1000 balls.
> 45 of the large bowls have 700 black balls and 300 red balls. (total 
balls
=
> 1000)
> 55 of the large bowls have 300 black balls and 700 red balls. (total 
balls
=
> 1000)
> Questions 1: What is the probability that one randomly selected large
bowl
> have more black balls?
> Answer: _____ %
> Second piece of information
> Now from the 100 large bowls, 1 bowl is randomly selected. Then 12 balls
> are randomly drawn from the bowl ( with returning) with the following
> results: 8 were black balls and 4 were red balls.
> Question 2: What is the probability that this bowl has more black balls?
> Answer: ______ %
This looks remarkably like a homework assignment.
What have you done so far?
Subject: Re: math wizzards..??
> Hello..
> Im trying to figure out this quiz..
> If u know the answer, ( and how u get it) pls e-mail me back
> K.
> Could you please answer the following 2 questions and send me back your
> answers, please???
> First piece of information
> You have 100 large bowls and in each bowl you have 1000 balls.
> 45 of the large bowls have 700 black balls and 300 red balls. (total 
balls
=
> 1000)
> 55 of the large bowls have 300 black balls and 700 red balls. (total 
balls
=
> 1000)
> Questions 1: What is the probability that one randomly selected large
bowl
> have more black balls?
> Answer: _____ %
> Second piece of information
> Now from the 100 large bowls, 1 bowl is randomly selected. Then 12 balls
> are randomly drawn from the bowl ( with returning) with the following
> results: 8 were black balls and 4 were red balls.
> Question 2: What is the probability that this bowl has more black balls?
> Answer: ______ %
It still looks like homework.
Multiple posting of the same questions doesn't help.
Subject: gluing theory question
Hi I am reading one book SPLAG, it introduces Gluing theory.
It says
Gluing theory is a way to describe the generatl n dimentional integral
lattice L that has a sublattice which is a direct sum
L1(+) L2 (+) ... (+)Lk
of given integral lattices L1,...,Lk of total dimension n.
The typical vector of L can be written
y=y1+y2+...+yk
where each component yi is in the subspace spanned by Li,
although not necessarily in Li itself.
Don't understand well of y=y1+y2+...+yk
Suppose Li is n_i dimension, so yi should be n_i dimention too,
how they add different dimension vectors?
Does it mean for example y1=[1 2] y2=[3 4]
y = y1+y2=[1 2 3 4]
Subject: Re: gluing theory question
> Does it mean for example y1=[1 2] y2=[3 4]
> y = y1+y2=[1 2 3 4]
Yes.
--
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: To all math enthusiasts, even non-specialists and other newbies...
Hello to all,
is there a notion in math that you had difficulty to
understand (non-intuitive or simply difficult to grasp)
that you'd like to see illustrate in 3D...
linear algebra ... ? dna ... ? other ??
A.
Subject: Re: To all math enthusiasts, even non-specialists and other
newbies...
> Hello to all,
> is there a notion in math that you had difficulty to
> understand (non-intuitive or simply difficult to grasp)
> that you'd like to see illustrate in 3D...
> linear algebra ... ? dna ... ? other ??
> A.
I found a discrete pi(x,y) function which gives the count of primes
when y=sqrt(x), and it has a continuous analog, which is also in 3-d.
I am curious about what both look like graphed in 3-d.
James Harris
Subject: Re: factoring to satisfiability
by the way, I think the assertion I quote at the
bottom of this msg is
just outright laughably false. I suspect there are less
than a dozen or so researchers in the entire world
who have **ever** looked much at factoring->SAT. and virtually
none at the moment.
remarkably, and evidence that suggests there is low
research, even DIMACS does not have have a standard
set of factoring instances
along with all the hard SAT test instances.
(although they have code by william naylor that can generate
them)
another point to add:
another possible reason researchers have shied away from
the factoring->SAT problem. it turns out that even a very
good satisifiability algorithm running on these instances
is apparently much slower than the worlds best factoring
algorithms. but there are two questions here:
- is the O(f(n)) growth poorer? this could just be due to a large
constant delay factor in SAT algorithms vs the straight or direct
factoring algorithms.
- even if it is demonstrably slower, then its a very interesting
question why a SAT algorithm cannot seize on the structure of the
factoring instances to solve them at least as fast as a direct factoring
algorithm.
[re factoring->SAT research]
> I'm sure there are
> dozens of researchers out there busily trying to use it for something ...
Subject: Re: factoring to satisfiability
3QLpj-NoP*NzsIC,boYU]bQ]H'<P%JD/,.J>y<#4ga3$21:
> another possible reason researchers have shied away from
> the factoring->SAT problem. it turns out that even a very
> good satisifiability algorithm running on these instances
> is apparently much slower than the worlds best factoring
> algorithms. but there are two questions here:
> - is the O(f(n)) growth poorer? this could just be due to a large
> constant delay factor in SAT algorithms vs the straight or direct
> factoring algorithms.
It seems more likely to me that the growth is asymptotically poorer and
not just poorer by a constant factor. Good SAT solvers still have times
like c^n (with c very close to 1) for random instances, while NFS is
more like c^(n^{1/3}). The instances one gets from factoring are of
course highly nonrandom but I don't see any reason for expecting a
general-purpose SAT solver to be able to exploit their structure in the
deep ways that NFS does.
So I'm not convinced this line of research is of much use for practical
factorization, but it is of some interest to find hard instances for
SAT solvers (this was the motivation of the Horie-Watanabe ISAAC paper
mentioned earlier in this thread, a version of which is also at
<http://arxiv.org/abs/cs.CC/9809117>.
More work in this line has been sparse, but does exist:
Factorization as a SAT problem, M. Klook,
http://www.elsevier.nl/gej-ng/31/29/24/41/23/94/endm8015.ps
The Propositional Formula Checker HeerHugo, J. F. Groote and J. P.
Warners, http://ftp.cwi.nl/CWIreports/SEN/SEN-R9905.pdf
(section 5.2 uses factorization to generate test instances for their
solver).
lisa entry in SAT2002 competition, Fadi Aloul,
ezfact entry, Dan Pehoushek, and pyhala entry, Tuomo Pyh.88l.88,
http://www.satlive.org/SATCompetition/2002/onlinereport/node9.html
Unfortunately the report does not break down the performance of
different instance generators, only of solvers.
--
Subject: Re: herstein problem
Visiting Assistant Professor at the University of Montana.
>I have one
>problem..........in the problem section of homomorphism, herstein
>gives
>a following problem:
> Prove that group of order 9 is abelian.
>The result is trivial if we will show that the center of this group is
>nontrivial but i can(or rather herstien himself) gives it in the next
>to next section.
> But staying in watertight solution how we can prove it.
Well, here is an exhaustive and not terribly elegant way to do it; the
class equation is of course the usual way to do it. But the same
argument will work for any group of order p^2, with p a prime.
Well, assume G is a group of order 9. Let x be a nontrivial element of
G. Then x is either of order 9 (in which case G is cyclic, and we are
done), or else x is of order 3.
If x is of order 3, let y be an element of G - {1, x, x^2}.
Again, if y is of order 9 you are done, so we may assume that y is of
order 3. Note that <x> intersect <y> must be equal to {1} (this is
were you use that the order is the square of a prime). We now have at
least the following elements of G:
1, x, x^2, y, y^2, xy, xy^2, x^2y, x^2y^2.
That's nine elements. I claim that these are ALL the elements of G,
that is, that there is no repetition in that list.
Clearly, x cannot equal x^2, y, or y^2.
Also, x cannot equal xy, xy^2, x^2y, or x^2y^2, for any of those would
imply either that y is in <x> or that y^2 is in <x>, which is
impossible.
By the same argument, y cannot equal x, x^2, xy, xy^2, x^2y, or
x^2y^2.
x^2 cannot equal y, y^2, x^2y, or x^2y^2; and it cannot equal xy or
xy^2, because then a non-trivial power of y would be equal to a
nontrivial power of x. Same is true for y^2 by a symmetric argument.
xy cannot equal xy^2 (for then y=1), and it cannot equal x^2y (for
then x=1); it cannot equal x^2y^2, because then you would have xy=1,
so y=x^2.
xy^2 cannot equal x^2y (you get xy=x^2); and it cannot equal
x^2y^2. And x^2y cannot equal x^2y^2, because then you get y=1.
So those are ALL the elements of G.
Which means that yx must be one of those elements. It cannot be equal
to 1, x, x^2, y, or y^2. So we have
yx = xy, xy^2, x^2y, or x^2y^2.
We want to show that it must equal xy (because then x and y commute).
If yx=xy^2, then x^{-1}yx = y^2. So then
x^{-2}yx^2 = x^{-1}(x^{-1}yx)x = x^{-1}y^2x = (x^{-1}yx)^2 = (y^2)^2 =y
and y = x^{-3}yx^3 = x^{-1}(x^{-2}yx^2)x = x^{-1}yx = y^2, which is
impossible.
Likewise, if yx = x^2y, then we have yxy^{-1} = x^2, and the symmetric
argument shows this is impossible.
So the only remaining possibility that would yield a non-abelian group
would be if yx = x^2y^2. Then
y = x^2 y^2 x^2
= x^2(x^2 y^2 x^2 x^2 y^2 x^2)x^2
= x y^2 x y^2 x
= xy yx y yx
= xy x^2 y^2 y yx
= xy x^2 yx
= xy x^2 x^2y^2
= xy xy^2
and therefore xyxy = 1. Which means that (xy)^2 = 1; since there are
no elements of order 2 in a group of order 9, that means that xy=1,
which is impossible.
Therefore, yx=xy, and so the group is abelian.
Subject: Re: herstein problem
Isn't it cyclic? If so, then there you go.
Lurch
> I have one
> problem..........in the problem section of homomorphism, herstein
> gives
> a following problem:
> Prove that group of order 9 is abelian.
> The result is trivial if we will show that the center of this group is
> nontrivial but i can(or rather herstien himself) gives it in the next
> to next section.
> But staying in watertight solution how we can prove it.
Subject: Re: herstein problem
Visiting Assistant Professor at the University of Montana.
>Isn't it cyclic? If so, then there you go.
Z/3Z x Z/3Z?
Subject: Re: herstein problem
>Isn't it cyclic? If so, then there you go.
> Z/3Z x Z/3Z?
Yes but that's still Abelian. Can there be nonabelian groups of order
p^2?
Subject: Re: herstein problem
>>Isn't it cyclic? If so, then there you go.
>> Z/3Z x Z/3Z?
>Yes but that's still Abelian. Can there be nonabelian groups of order
>p^2?
No, there cannot be. A group of order p^2 is either cyclic, or the
product of two cyclic groups of order p; but the OP claimed that a
group of order p^2 would have to be cyclic.
(The easiest way to show it is to use the class equation: a p-group
must have non-trivial center; let x be a nontrivial element of the
center. If x is of order p^2 you are done. If x is of order p, then
G/<x> is of order p, hence cyclic, and a center-by-cyclic group is
always abelian; I posted one BFBI[1] method to show it for a group of
order 9, which easily generalizes to a group of order p^2, if you do
not use the class equation or the fact that the center is nontrivial).
[1] BFBI = Brute Force and Bloody Ignorance.
Subject: Re: herstein problem
> (The easiest way to show it is to use the class equation: a p-group
Arturo, tell me please if the following reasoning differs from the one
which uses the class equation argument :
Let G be a non cyclic group of order 9. Every element of G distinct from 1
is of order 3 and G has exactly 4 subgroups H1,...,H4 of order 3. G acts by
conjugation on the set {H1,...,H4} so there is a homomorphism
G->Sym({H1,...,H4}). Since 9 doesn't divide 24 the kernel is not trivial
and hence there exist a normal subgroup H=H_i and an element g not in H
such that gHg^{-1}=H. Let a be any generator of H.
Since gag^{-1} is a generator of H, gag^{-1} coincide with a or a^{-1}. In
the later case, we have g^2ag^{-2}=ga^{-1}g^{-1}=a so g^{-1}ag=a and g
commutes again with a .
Finally, G is an abelian since generated by {a,g}.
Pascal
Subject: Re: herstein problem
Visiting Assistant Professor at the University of Montana.
>> (The easiest way to show it is to use the class equation: a p-group
>Arturo, tell me please if the following reasoning differs from the one
>which uses the class equation argument :
>Let G be a non cyclic group of order 9. Every element of G distinct from 1
>is of order 3 and G has exactly 4 subgroups H1,...,H4 of order 3. G acts
by
>conjugation on the set {H1,...,H4} so there is a homomorphism
>G->Sym({H1,...,H4}). Since 9 doesn't divide 24 the kernel is not trivial
>and hence there exist a normal subgroup H=H_i and an element g not in H
>such that gHg^{-1}=H. Let a be any generator of H.
>Since gag^{-1} is a generator of H, gag^{-1} coincide with a or a^{-1}. In
>the later case, we have g^2ag^{-2}=ga^{-1}g^{-1}=a so g^{-1}ag=a and g
>commutes again with a .
I think this is essentially the class equation; the action of G
divides the elements of G into equivalence classes, x~y if and only if
there exists g such that g^{-1}xg = y; you are just noting that if
x~y, then x^n~y^n for each n.
Then note that every class has either 1 or 3 elements and not all can
have 3 elements (equivalent to the fact that the map
G->Sym({H1,H2,H3,H4}) has nontrivial kernel) so there are at least 3
classes with 1 element; which gives you a central element, and then
the rest follows.
Subject: Matrix Exponential / Computer code
Does anybody know a public code library (written in C/C++ or Delphi)
which implements efficently the
matrix-exponential operation?
Mathew
Subject: Measure extension proof
Given:
1.- F is a field (not necessarily Borel)
2.- u is a measure on F
3.- G is the minimal Borel Field containing F.
4.- v is a measure on G.
5.- v and u agree on F.
6.- v and u are sigma-finite on F.
It can be proved that v is the unique extension of u from F to G.
Apparently 6.- is a sufficient but not necessary condition for this
uniqueness. Can someone please indicate the necessary condition and
Subject: Re: Measure extension proof
>Given:
>1.- F is a field (not necessarily Borel)
>2.- u is a measure on F
>3.- G is the minimal Borel Field containing F.
I really don't see what sense the terminology
minimal Borel field makes. Maybe you meant
that G is the minimal sigma-field containing F?
>4.- v is a measure on G.
>5.- v and u agree on F.
>6.- v and u are sigma-finite on F.
>It can be proved that v is the unique extension of u from F to G.
>Apparently 6.- is a sufficient but not necessary condition for this
>uniqueness. Can someone please indicate the necessary condition and
I doubt that there _is_ a simple necessary and sufficient condition.
************************
Subject: Re: Terminology question
Visiting Assistant Professor at the University of Montana.
>What do you call a statement, after a proof of a theorem, that says:
>This theorem does NOT apply if such-and-such assumption is removed,
>as we'll show here?
I usually call statements like that Remarks, which I program into
number, and then the text in roman. So I would write:
textbf{Remark s.n.} Note that this theorem does not apply if
such-and-such an assumption is removed. Here is a counterexample:
and then give the example. (Or as we will show below). Bourbaki uses
Remark by itself, and if many remarks follow the same theorem, he
numbers them simply (Remark (1), Remark (2), etc.)
Alternatively, one can use N.B., which is read Note well.
This is if you want the statement to stand out for some reason (which
part of the presentation in an Example, labeled as such.
Subject: Re: Terminology question
> What do you call a statement, after a proof of a theorem, that says:
> This theorem does NOT apply if such-and-such assumption is removed,
> as we'll show here?
There probably isn't a standard term for this. I'd call it a caveat,
or warning, or caution, or just head it N.B. The Bourbaki books had
the
rather nice custom of putting an S-bend road sign in the margin as a
warning where there was danger of confusion.
Ken Pledger.
Subject: deep holes of leech lattice
Hello
Is there any algorithm to calculate the deep holes of leech lattice?
Any software can do that? I tried MAGMA but MAGMA gave no result
I read SPLAG however I am not clear about it
I wanna have the whole collection of leech lattice deep holes
is there any way for me to calculate them?
Subject: Re: deep holes of leech lattice
> Is there any algorithm to calculate the deep holes of leech lattice?
Yes, it's not hard to find one that calculates all holes of any lattice.
> Any software can do that? I tried MAGMA but MAGMA gave no result
I'm sure that on a lattice as complicated as Leech any known
algorithm would be totally intractible.
> I read SPLAG however I am not clear about it
Pity.
> I wanna have the whole collection of leech lattice deep holes
is the best account you're going to find. Finding
the Leech lattice holes was a major piece of research. Although
finding holes is a computable problem, just running some
general-purpose algorithm on Leech would take too much time.
--
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: After Coxeter
> Suppose that, having read Coxeter's Introduction to Geometry the reader
> wished for something equally broad but more deep. What books might he
> try? Surely there is a GTM as thick as Lang's Algebra that he may curl
> up with during the coming autumnal evenings? Any suggestions?
Here are a couple of wide-ranging books.
Hilbert & Cohn-Vossen, Geometry and the Imagination. This classic is
more discursive than Coxeter, but you should find it very readable while
curled up in the evenings.
Borsuk & Szmielew, Foundations of Geometry. This goes into more detail
about the logical structure of various geometries (cf. Coxeter Chapters
12-16) if that's what you'd like.
But, as maky m. has said, you may want a variety of more specialized
books. If you can focus on any particular parts of Coxeter that you'd
like to follow up, then no doubt various people will have suggestions.
Ken Pledger.
Subject: Re: After Coxeter
> Suppose that, having read Coxeter's Introduction to Geometry the reader
> wished for something equally broad but more deep. What books might he
> try? Surely there is a GTM as thick as Lang's Algebra that he may curl
> up with during the coming autumnal evenings? Any suggestions?
> Here are a couple of wide-ranging books.
> Hilbert & Cohn-Vossen, Geometry and the Imagination. This classic is
> more discursive than Coxeter, but you should find it very readable while
> curled up in the evenings.
Yup got that, a definite curl up in the evenings kind of book.
> Borsuk & Szmielew, Foundations of Geometry. This goes into more
detail
> about the logical structure of various geometries (cf. Coxeter Chapters
> 12-16) if that's what you'd like.
> But, as maky m. has said, you may want a variety of more
specialized
> books. If you can focus on any particular parts of Coxeter that you'd
> like to follow up, then no doubt various people will have suggestions.
Logical foundations.
> Ken Pledger.
--
G.C.
Subject: Implied Sequence (#-Theory) Puzzle
[This is a continuation of the A certain Dirichlet-sum: question
thread (the last reply of which is copy/pasted below). But the current
topic at hand is this puzzle related to that thread.]
Let a(k,m) = the number of distinct prime divisors, p, of k,
where p^m divides k, but p^(m+1) does not.
Let c(k,m) be such that:
sum{k=1 to oo} c(k,m)/k^r =
(1 - 1/zeta(r))^m
for every r where sum converges,
where zeta(r) is the Riemann zeta function.
Now, {b(m)} is a sequence of reals such that:
For every positive integer n,
---
/ a(n,m) b(m) =
---
m>=1
---
/ c(n,m) b(m)
---
m>=1
In linear-mode:
sum{m>=1} a(n,m) b(m) =
sum{m>=1} c(n,m) b(m)
(I use 'm>=1' since the sums are finite, but the upper limit varies
based on n.)
Now, I wondered in my previous post (in the other thread) if {b(m)}
was unique.
I THINK that I may have proved today that {b(m)} is indeed unique,
except that each term can be multiplied by a constant (of course).
*
So, setting b(1) = 1, what is the sequence {b(k)}??
*
(Not-too-small hint below copy/pasted post.)
Leroy Quet
>[cross-posting reply to rec.puzzles]
>>I should probably give this more thought myself. But this question
>>does not seem to be as easily answered as I 1st assumed.
>>Does sum{k=1 to oo} a(k,m)/k^r =
>>(1 - 1/zeta(r))^m,
>>where a(k,m) = the number of distinct prime divisors, p, of k,
>>where p^m divides k, but p^(m+1) does not?
>> I don't think so. For example, a(15,2) = 0 but
>> (1- 1/zeta(r))^2 should contain a term 2/15^r.
>The reason I ask is that, I THINK:
>sum{m=1 to oo} (1 - 1/zeta(r))^m *b(m) =
>sum{m=1 to oo} (sum{k=1 to oo} a(k,m)/k^r) b(m)
>for a certain (at least one) sequence {b(m)}.
>Puzzle(??): Find {b(k)}.
>(I do not know if the sequence {b(k)} is unique.)
Hint: {b(k)} is a sequence of rationals that I have been quite fond of
Subject: Re: Closed form determinant of Toeplitz-Hessenberg matrix
>I would like to get the determinant of a matrix
>typified by (n=4)
> 1/2! 1/4! 1/6! 1/8!
> -c 1/2! 1/4! 1/6!
> -c 1/2! 1/4!
> -c 1/2!
>where missing entries are zero and c is
>a symbolic expression, for arbitrary n.
>This is both Toeplitz and Hessenberg (TH).
>There is a closed form for dets of Vandermonde
>matrices. Is there a similar result for
> For n = 4, 5, 6, 7, 8, I tried evaluating the
> determinant as a polynomial of degree n-1 in c.
> In each case the galois group of this polynomial
> was the full symmatric group on n-1 elements,
> so there appears not to be a simple formula for its roots.
Agree. So far no progress on a closed form. As n->oo the ratio
det(A11)/det(A), with A11 the (1,1) principal minor, appears to approach
a Pade form as yet unidentified.
Subject: Arc length of Gamma function
I 'discovered' this years ago and now that I got the student version of
Mathematica I tried messing around with it : the arc length of the gamma
function.
I used to think after my discovery, using a ruler, that the arc length from
x = 1 to x = n of the Gamma function (Euler's, where Gamma(x) = x! for
integer x - you know the one) is (n-1)!.
So, Integer[Sqrt(1+D(Gamma)^2),x=1..n] = (n-1)!
Thatw as my thought.
Now I wanted to see with Mathematica why exactly that would be true.. But,
what I found there was something odd..
It says consistently that :
Integrate[Sqrt[1 + D[Gamma[x], x]^2], {x, 1, n}] = n! + 0.66
The +0.66 holds for smaller n but then goes slowly up, very slowly..which
is
very odd if you ask me, where does this extra bit come from? Can anybody
enlighten me on this? Is the arc length of the Gamma function from 1 to x
equal to its funciton value in x-1? I shoudl expand the terms manually but
it's kind of a nasty integration..
--
Tetsuo
Subject: Re: Arc length of Gamma function
> I 'discovered' this years ago and now that I got the student version of
> Mathematica I tried messing around with it : the arc length of the gamma
> function.
> I used to think after my discovery, using a ruler, that the arc length
from
> x = 1 to x = n of the Gamma function (Euler's, where Gamma(x) = x! for
> integer x - you know the one) is (n-1)!.
> So, Integer[Sqrt(1+D(Gamma)^2),x=1..n] = (n-1)!
> Thatw as my thought.
> Now I wanted to see with Mathematica why exactly that would be true..
But,
> what I found there was something odd..
> It says consistently that :
> Integrate[Sqrt[1 + D[Gamma[x], x]^2], {x, 1, n}] = n! + 0.66
> The +0.66 holds for smaller n but then goes slowly up, very slowly..which
is
> very odd if you ask me, where does this extra bit come from? Can anybody
> enlighten me on this? Is the arc length of the Gamma function from 1 to x
> equal to its funciton value in x-1? I shoudl expand the terms manually
but
> it's kind of a nasty integration..
You want to find out about
int_0^x sqrt( 1 + gamma'(y)^2 ) dy - gamma(x)
when x->infinity. Write
gamma(x) = int_0^x gamma'(y) dy - 1
and you obtain
int_0^x ( sqrt( 1+ gamma'(y)^2 ) - gamma'(y) ) dy - 1
= int_0^x 1/( sqrt( 1 + gamma'(y)^2 ) + gamma'(y) ) dy - 1.
It is easy to see, that the above integral converges as x->infinity.
It is also easy to obtain a numerical value for it. Mathematica gives
1.661576018...
HTH,
Michael.
--
&&&&&&&&&&&&&&&&#@#&&&&&&&&&&&&&&&&
Dr. Michael Ulm
FB Mathematik, Universitaet Rostock
michael.ulm@mathematik.uni-rostock.de
Subject: Re: Arc length of Gamma function
Dr. Michael Ulm <taga@news.uni-rostock.de> schreef in bericht
> I 'discovered' this years ago and now that I got the student version of
> Mathematica I tried messing around with it : the arc length of the
gamma
> function.
> I used to think after my discovery, using a ruler, that the arc length
from
> x = 1 to x = n of the Gamma function (Euler's, where Gamma(x) = x! for
> integer x - you know the one) is (n-1)!.
> So, Integer[Sqrt(1+D(Gamma)^2),x=1..n] = (n-1)!
> Thatw as my thought.
> Now I wanted to see with Mathematica why exactly that would be true..
But,
> what I found there was something odd..
> It says consistently that :
> Integrate[Sqrt[1 + D[Gamma[x], x]^2], {x, 1, n}] = n! + 0.66
> The +0.66 holds for smaller n but then goes slowly up, very
slowly..which is
> very odd if you ask me, where does this extra bit come from? Can
anybody
> enlighten me on this? Is the arc length of the Gamma function from 1 to
x
> equal to its funciton value in x-1? I shoudl expand the terms manually
but
> it's kind of a nasty integration..
> You want to find out about
> int_0^x sqrt( 1 + gamma'(y)^2 ) dy - gamma(x)
> when x->infinity. Write
> gamma(x) = int_0^x gamma'(y) dy - 1
> and you obtain
> int_0^x ( sqrt( 1+ gamma'(y)^2 ) - gamma'(y) ) dy - 1
> = int_0^x 1/( sqrt( 1 + gamma'(y)^2 ) + gamma'(y) ) dy - 1.
> It is easy to see, that the above integral converges as x->infinity.
> It is also easy to obtain a numerical value for it. Mathematica gives
> 1.661576018...
Right.. nice integration trick there.
remainder was caused by the vertical straightness of the Gamma function,
extremely silly in fact. Although I do wonder about the symbolic nature of
this number, 1.6616, now..
I also did not remember Gamma(x) was (x-1)!, which makes everythign even
sillier of course as a function's arc length can hardly be the same as its
funciton value unless it is straight, hoirontally or vertically....
But I also wonder though, for the family of function where the arc length
from 0 to x *is* the same as its function value in x-1, so :
f(y-1) = int,0_y(sqrt(1+Df(x)^2) dx)
You get a differential equation that represents this class of
functions...but I lack the insight to solve it symbolically with Df(x-1) in
one term and Df(x) in another.
(even better would probably be f(y+1) = int,0_y(sqrt(1+Df(x)^2) dx) since
you can have a point of the curve in (0,0) then..
--
Tetsuo
Subject: Re: Arc length of Gamma function
[[ This message was both posted and mailed: see
the To, Cc, and Newsgroups headers for details. ]]
> I 'discovered' this years ago and now that I got the student version of
> Mathematica I tried messing around with it : the arc length of the gamma
> function.
> I used to think after my discovery, using a ruler, that the arc length
from
> x = 1 to x = n of the Gamma function (Euler's, where Gamma(x) = x! for
> integer x - you know the one) is (n-1)!.
> So, Integer[Sqrt(1+D(Gamma)^2),x=1..n] = (n-1)!
> Thatw as my thought.
> Now I wanted to see with Mathematica why exactly that would be true..
But,
> what I found there was something odd..
> It says consistently that :
> Integrate[Sqrt[1 + D[Gamma[x], x]^2], {x, 1, n}] = n! + 0.66
> The +0.66 holds for smaller n but then goes slowly up, very slowly..which
is
> very odd if you ask me, where does this extra bit come from? Can anybody
> enlighten me on this? Is the arc length of the Gamma function from 1 to x
> equal to its funciton value in x-1? I shoudl expand the terms manually
but
> it's kind of a nasty integration..
Why do you think that's odd? The graph of the gamma function is
locally very like a straight line (an ALMOST VERTICAL straight line),
and it doesn't surprise me at all that the length from x = n to x = n+1
is n! - (n-1)! + c_n, where c_n is a small constant (so small that it's
summable to a finite value). In fact, the arc has length only slightly
greater than the length of the straight line,
sqrt( (n! - (n-1)!)^2 + 1)
= (n! - (n-1)!) sqrt( 1 + 1/(n!-(n-1)!)^2)
approx (n! - (n-1)!) * [ 1 + 1/[2(n! - (n-1)!)^2] ]
approx (n! - (n-1)!) + 1/[2(n! - (n-1)!]
so the added length (over the already vertical height of n! - (n-1)!)
is rapidly summable. This can be formalized by estimating
Gamma'(x) = Gamma(x) psi(x),
in the arclength integral, where psi is the digamma function, which
grows--what? Logarithmically?
--Ron Bruck
Subject: Re: Arc length of Gamma function
> I 'discovered' this years ago and now that I got the student version of
> Mathematica I tried messing around with it : the arc length of the gamma
> function.
> I used to think after my discovery, using a ruler, that the arc length
from
> x = 1 to x = n of the Gamma function (Euler's, where Gamma(x) = x! for
> integer x - you know the one) is (n-1)!.
Normally the gamma function is taken such that Gamma(x) = (x - 1)!. And
I am pretty sure that Mathematica uses that definition.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland,
+31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
Subject: Re: Arc length of Gamma function
Tetsuo <tecknomancer@deep.es> schreef in bericht
> I 'discovered' this years ago and now that I got the student version of
> Mathematica I tried messing around with it : the arc length of the gamma
> function.
> I used to think after my discovery, using a ruler, that the arc length
from
> x = 1 to x = n of the Gamma function (Euler's, where Gamma(x) = x! for
> integer x - you know the one) is (n-1)!.
> So, Integer[Sqrt(1+D(Gamma)^2),x=1..n] = (n-1)!
> Thatw as my thought.
> Now I wanted to see with Mathematica why exactly that would be true..
But,
> what I found there was something odd..
> It says consistently that :
> Integrate[Sqrt[1 + D[Gamma[x], x]^2], {x, 1, n}] = n! + 0.66
correction = (n-1)! + c
(where c is between 0.66 and 0.75 for acceptable n)
--
Tetsuo
Subject: Re: Probably Simple But I Can't Work It Out.
Your problem is not properly posed until you have decided what teh
common ratio is, or some information that defines the common ratio.
Say n = 4. Suppose you set a/b = b/c = c/d = 2. If L = a+b+c+d is your
total length, then in terms of d ou have L = 8d+4d2d+d = 15d and from
this you can get all your panel lengths.
Now consider a different problem. Say you have n = 4 and you desire
your last panel to be some fraction of L. For exapmple, d = L/15. Then
a = r^3d, b = r^2d, c = rd where r is the common ratio and d = L/15.
Then you must solve
a+b+c+d = L
(r^3+r^2+r+1)(L/15) = L
r^3+r^2+r-14 = 0
a cubic equaiton. This is not very difficult to handle for the problem
at hand, for the equation has a rational root. Even if it didn't or if
you don't want to do a rational-root search (difficult to program on a
computer) the equation has only one real root and therefore no casus
irreducibilis arises when Cardano's method is applied. For larger n,
however, the polynomial is of degree n-1 and its analytical solution is
either very unwieldy or impossible to get. But there is still only one
positive root and it can be approximated with relatively little trouble
via Newton's Method. Any nonnegative initial guess leads to convergence
to the proper root.
--OL
Subject: Re: division by zero...??
ETAtAhUAgvjFylhZ0C1oWslpGJ9SZ+vTpmQCFC6E9OcvYXVES0wqxIKxNaOBURde
You can't divide by the additive identity element in ANY algebra.
Using the distributive law, one can prove that any element times the
additive identity is the additive identity (a*b = a*(b+0) = a*b+a*0, so
a*0 = ... ). The additive identity is not the same as the
multiplicative identity. Therefore the existence of a multiplicative
inverse for the divisor, necessary for division, is lost when the
divisor is the additive identity element.
--OL
Subject: Re: division by zero...??
You can't divide by the additive identity element in ANY algebra.
Not *quite* true. You can in just one
algebra: the finite field with 1 element.
[call it 0]. I understand that people
at U. Illinois called this a 'Stu Field'
after a grad student named 'Stu[art]'.
We have 0+0 = 0, 0*0 = 0 and
0 is its own inverse; both multiplicative
and additive. It is easy to see that this
algrebra satisfies the 11 basic axioms of
a field. Some Purists claim that a field
has 12 axioms, the 12th being that the
additive inverse and multiplicative inverse
are different. This (somewhat silly)
algebra contradicts that.
Of course, this field is rather 'trivial'. :-)
You can lead a horse's ass to knowledge, but you can't make him think.
Subject: Re: division by zero...??
> You can't divide by the additive identity element in ANY algebra.
> Not *quite* true. You can in just one
> algebra: the finite field with 1 element.
> [call it 0]. I understand that people
> at U. Illinois called this a 'Stu Field'
> after a grad student named 'Stu[art]'.
> We have 0+0 = 0, 0*0 = 0 and
> 0 is its own inverse; both multiplicative
> and additive. It is easy to see that this
> algrebra satisfies the 11 basic axioms of
> a field. Some Purists claim that a field
> has 12 axioms, the 12th being that the
> additive inverse and multiplicative inverse
> are different. This (somewhat silly)
> algebra contradicts that.
> Of course, this field is rather 'trivial'. :-)
Are you using the term Algebra and Field as synonyms?
If not What is the difference.?
Bob Pease
Subject: Re: division by zero...??
Zero is the the one of the agreed point of reference, i.e. decimal dot.
Therefore the question to divide by zero is irrelevant.
How to divide by the point of reference indicator?
Tapio
> You can't divide by the additive identity element in ANY algebra.
> Not *quite* true. You can in just one
> algebra: the finite field with 1 element.
> [call it 0]. I understand that people
> at U. Illinois called this a 'Stu Field'
> after a grad student named 'Stu[art]'.
> We have 0+0 = 0, 0*0 = 0 and
> 0 is its own inverse; both multiplicative
> and additive. It is easy to see that this
> algrebra satisfies the 11 basic axioms of
> a field. Some Purists claim that a field
> has 12 axioms, the 12th being that the
> additive inverse and multiplicative inverse
> are different. This (somewhat silly)
> algebra contradicts that.
> Of course, this field is rather 'trivial'. :-)
> Are you using the term Algebra and Field as synonyms?
> If not What is the difference.?
> Bob Pease
Subject: Re: division by zero...??
> Zero is the the one of the agreed point of reference, i.e. decimal dot.
> Therefore the question to divide by zero is irrelevant.
> How to divide by the point of reference indicator?
> Tapio
> You can't divide by the additive identity element in ANY algebra.
Not *quite* true. You can in just one
> algebra: the finite field with 1 element.
> [call it 0]. I understand that people
> at U. Illinois called this a 'Stu Field'
> after a grad student named 'Stu[art]'.
We have 0+0 = 0, 0*0 = 0 and
> 0 is its own inverse; both multiplicative
> and additive. It is easy to see that this
> algrebra satisfies the 11 basic axioms of
> a field. Some Purists claim that a field
> has 12 axioms, the 12th being that the
> additive inverse and multiplicative inverse
> are different. This (somewhat silly)
> algebra contradicts that.
Of course, this field is rather 'trivial'. :-)
> Are you using the term Algebra and Field as synonyms?
> If not What is the difference.?
> Bob Pease
I'm afraid that I don't understand what you mean
BOb Pease
Subject: Re: Help Maple 8
Im sorry if you dont like My asking for help?
But I need help!
If you dont like Maple or computers then DONT USE THEM!
I, again am asking for help Who said I cant work the problem ? I
asked how to imput the problem into MAPLE I think your reading to
much into my request for help! Help or stand back and critize ,but at
least KNOW what your talking about!
Wayne
>>I need help inputting the following in Maple 8
>>How do I find the domain and range : f(x)=Sqrt 7x+2
>Could a True Believer in the use of fancy computer software for teaching
>please explain to me how this person is being helped by the use of Maple
>(Please direct NO follow-ups to comp.soft-sys.math.maple, where my
>question is not really on topic.)
>dave
Subject: Re: Help Maple 8
>Im sorry if you dont like My asking for help?
>But I need help!
>If you dont like Maple or computers then DONT USE THEM!
>I, again am asking for help Who said I cant work the problem ? I
>asked how to imput the problem into MAPLE I think your reading to
>much into my request for help! Help or stand back and critize ,but at
>least KNOW what your talking about!
Well excuuuuse me! I like Maple a lot and use it all the time;
computers too (how do you think I'm typing this message?) I wish you
well and will let others help you with your homework, but your question,
obviously set to you by a well-intentioned instructor, struck me as
being something which would not, in fact, help you as a student in
the long run. As a seasoned instructor looking to learn new ways to
teach, I asked for someone who understood how this was supposed to
be helpful to chime in. You will notice I did not take you to task
at all and in fact was careful to remove your name from my post.
I believe there is quite a bit that technology (e.g. Maple) can do to
improve mathematics education. I also believe there is quite a bit that
technology can do, and has done, to harm mathematics education. My
question was simply how someone would defend this particular use of
Maple as an improvement, since in practice a student like you will
show up in my calculus class next semester and I will ask him or her
face-to-face to give the domain of a function like yours, and I absolutely
do not want the student to tell me that he or she needs to fire up
Maple before answering my question.
dave
Subject: Re: Help Maple 8
>I need help inputting the following in Maple 8
>How do I find the domain and range : f(x)=Sqrt 7x+2
> Could a True Believer in the use of fancy computer software for teaching
> please explain to me how this person is being helped by the use of Maple
> (Please direct NO follow-ups to comp.soft-sys.math.maple, where my
> question is not really on topic.)
I say that it is highly on-topic. As the all-time second most frequent
poster to this newsgroup, and as an instructor who has used Maple
extensively for instruction for the past five years, and as a researcher
who uses Maple everyday for research, I think that I am qualified to make
that assessment. Indeed, it is crucial that this topic be discussed.
Crucial both to the survival of Maple and the out survival as a
technological society. I have to go now -- more to say later.
Subject: Re: WANTED: MATH JOKE TRANSLATOR
In sci.math, Misguided Angel
<spyclub4@firemail.de>
<MPG.19e948f1880420c2989685@news.995.ca>:
> This joke was posted by an instructor ... could someone explain the joke?
> I just don't get it.
> Two functions meet in a narrow street. (*)
> F1: Clear the way!
> F2: No, I won`t.
> F1: Move over, or I will differentiate you!
> F2: Ok, try it, I am the exp-function!
(F1 pulls a lever, and both functions get zapped.)
F1: Still wanna play, eh?
(F1 pulls again. It's clear that F2 isn't budging and F1's
patience is wearing slightly thin. The zapping, however,
is also taking its toll on F1's degree. F1 pulls repeatedly,
each time losing more and more degreeness, with F2 looking
at F1 with a knowing smile.)
F1: Oh, c'mon, that should have at least tickled you a little?
F2: Really? I've not been paying attention.
F1: HA!
(F1 pulls again, and discovers he's now a constant.)
F2: Who's your d/dy now, flatline?
(Now F2 pulls the lever. F1 vanishes. F2 triumphantly strolls
to the next bridge where he encounters the function
e^(x^2). Stay tuned....)
:-)
--
#191, ewill3@earthlink.net -- insert random strange joke here
It's still legal to go .sigless.
Subject: Re: WANTED: MATH JOKE TRANSLATOR
> This joke was posted by an instructor ... could someone explain the
joke?
> I just don't get it.
> Two functions meet in a narrow street. (*)
> F1: Clear the way!
> F2: No, I won`t.
> F1: Move over, or I will differentiate you!
> F2: Ok, try it, I am the exp-function!
> Simple, all it needs is to be translated into Korean and back
> into English again by Babelfish:
> The function of the cranium meets inside the narrow street, (*) the
> }f{1: Method clearly!
> }f{2: Negation, me.
> }f{1: It moves, nem e from, or I to be extensive and differentiate!
> }f{2: Approval, it, the exp function which is a B it tries,!
I tried talking to those Russian brides with that, got nowhere, NOWHERE!
Herc
Subject: Re: [JSH] Re: Did mathematicians know?
In sci.math, Andy Spragg
<sparge@globalnet.co.uk>
<3f7fb061.279087512@news.global.net.uk>:
> QncyMI@netscape.net (Brian Quincy Hutchings) pushed briefly to the
> the shed door:
> ^ man, what's hote mean ?!?
> ^ > Looks like a donkey h.99te to me... :-)
> ^ ^ > But where's the windmill?
> With a name like Quincy, you ought to recognize a quick
> soat when you see one!
I'll admit a quick soat after tilting at windmills works
every time to soak the tiredness from one's body... :-)
[.sigsnip]
--
#191, ewill3@earthlink.net
It's still legal to go .sigless.
Subject: Two similar things
> Since they now are identical-with-somethings, if they shared
> all the same properties they would be one cake rather than
> two. THerefore, since they are two cakes, they don't share
> all their properties, and I can distinguish them by means
> of any property that one has and the other doesn't.
No, there's a foot wrong there. Why would they be one cake if they shared
all the same properties? Why not 3 cakes? Would you tell the difference by
counting?
I missed the point. More pertinent is this:
How would you distinguish your two different cakes? You can name them
'todays and yesterdays' cakes, but you actually would not be able to know
which was which from a description of the cakes themselves.
JJ
> --I could have baked both cakes yesterday.
> --I could have baked both cakes today.
> --I could have baked this one yesterday
> and that one today.
> --I could have baked that one yesterday and
> this one today.
> Your example doesn't work!
> How am I going to distinguish 'one cake' from 'the other cake'!
> Since they now are identical-with-somethings, if they shared
> all the same properties they would be one cake rather than
> two. THerefore, since they are two cakes, they don't share
> all their properties, and I can distinguish them by means
> of any property that one has and the other doesn't.
> JJ
1 I. The Problem, and the Problem with the Problem,
> 2 of Identical Articles and Quantum Statistics
> 3
> 4 Suppose we have a box with two qualitatively
> 6 bouncing around inside. We think of the box
> 7 as having a left (*l*) and a right (*r*) side.
> 9 at random without interacting, so that their
> 10 motions are independent; in particular we
> 12 that we may neglect collisions. What are the
> 13 chances for finding one or both on one side
> 14 or the other?
> 15
> 16 Many find the following reasoning persuasive.
> 19 in *r*, and 2 in *l* and 1 in *r*. These should
> 20 be equally likely, so that each has a probability
> 21 of 1/4, or a probability of 1/4 for two in *l*,
> 22 1/4 for two in *r*, and 1/2 for one on each side.
> 23
> 24 This stylized example is a simple mock-up for a
> 25 kind of situation that can occur with quantum entities
> 26 and properties. For many of these situations the
> 27 probabilities are in fact found to be 1/3 for each
> 28 of the three cases: two in *l*, two in *r* and one
> 29 on each side. Many interpreters have found this
> 30 fact utterly astonishing.
> 31
> 32 But on the face of it, there is a very simple
> 33 resolution of the puzzle: give up supposing
> 34 that there are two qualitatively identical but
> 36 there are two *quanta*, as I'll put it, to which
> 37 the notion of being numerically distinct does not
> 38 apply. . . (p. 114)
Compare Teller's mock-up with how I can go about
> making two cakes in two days:
1) I can bake both cakes today.
> 2) I can bake both cakes tomorrow.
> 3) I can bake one cake today and the other cake tomorrow.
But after baking the cakes, when I consider how I
> MIGHT have gone about baking them, there are FOUR
> possibilities rather than THREE.
--I could have baked both cakes yesterday.
> --I could have baked both cakes today.
> --I could have baked this one yesterday
> and that one today.
> --I could have baked that one yesterday and
> this one today.
Homework for David Ullrich:
In what respect do cakes I am going to make resemble quanta? (Hint:
> Like causes produce like effects.)
--John
Subject: Re: Two similar things
> Since they now are identical-with-somethings, if they shared
> all the same properties they would be one cake rather than
> two. THerefore, since they are two cakes, they don't share
> all their properties, and I can distinguish them by means
> of any property that one has and the other doesn't.
> No, there's a foot wrong there. Why would they be one cake if they shared
> all the same properties? Why not 3 cakes? Would you tell the difference
by
> counting?
Granted that these cakes--call them cake1 and cake2--are now
identical-with-somethings, the property of being identical
with cake1 and that of being identical with cake2 are no
longer unexemplified: cake1 has come to exemplify the
property of being identical with cake1, and cake2, the property
of being identical with cake2. Consequently, in the event that
these cakes share *all* their properties, cake1 will exemplify
the property of being identical with cake2, and cake2 will
exemplify the property of being cake1, making cake1 and
cake2 one and the same cake.
> I missed the point. More pertinent is this:
> How would you distinguish your two different cakes? You can name them
> 'todays and yesterdays' cakes, but you actually would not be able to know
> which was which from a description of the cakes themselves.
> JJ
*Any* property which one had and the other did not would
distinguish them, whether or not you, or I, or anyone else
grasped the basis of their difference. Thus, it might very
well be the case that everyone took there to be but a single
cake when in fact there were two: wasn't it once thought
that there was only one kind of helium when in fact there
were two? And everyone might think there were two different
cakes when in fact there was but one.
--John
> --I could have baked both cakes yesterday.
> --I could have baked both cakes today.
> --I could have baked this one yesterday
> and that one today.
> --I could have baked that one yesterday and
> this one today.
Your example doesn't work!
> How am I going to distinguish 'one cake' from 'the other cake'!
> Since they now are identical-with-somethings, if they shared
> all the same properties they would be one cake rather than
> two. THerefore, since they are two cakes, they don't share
> all their properties, and I can distinguish them by means
> of any property that one has and the other doesn't.
JJ
1 I. The Problem, and the Problem with the Problem,
> 2 of Identical Articles and Quantum Statistics
> 3
> 4 Suppose we have a box with two qualitatively
> 6 bouncing around inside. We think of the box
> 7 as having a left (*l*) and a right (*r*) side.
> 9 at random without interacting, so that their
> 10 motions are independent; in particular we
> 12 that we may neglect collisions. What are the
> 13 chances for finding one or both on one side
> 14 or the other?
> 15
> 16 Many find the following reasoning persuasive.
> 19 in *r*, and 2 in *l* and 1 in *r*. These should
> 20 be equally likely, so that each has a probability
> 21 of 1/4, or a probability of 1/4 for two in *l*,
> 22 1/4 for two in *r*, and 1/2 for one on each side.
> 23
> 24 This stylized example is a simple mock-up for a
> 25 kind of situation that can occur with quantum entities
> 26 and properties. For many of these situations the
> 27 probabilities are in fact found to be 1/3 for each
> 28 of the three cases: two in *l*, two in *r* and one
> 29 on each side. Many interpreters have found this
> 30 fact utterly astonishing.
> 31
> 32 But on the face of it, there is a very simple
> 33 resolution of the puzzle: give up supposing
> 34 that there are two qualitatively identical but
> 36 there are two *quanta*, as I'll put it, to which
> 37 the notion of being numerically distinct does not
> 38 apply. . . (p. 114)
Compare Teller's mock-up with how I can go about
> making two cakes in two days:
1) I can bake both cakes today.
> 2) I can bake both cakes tomorrow.
> 3) I can bake one cake today and the other cake tomorrow.
But after baking the cakes, when I consider how I
> MIGHT have gone about baking them, there are FOUR
> possibilities rather than THREE.
--I could have baked both cakes yesterday.
> --I could have baked both cakes today.
> --I could have baked this one yesterday
> and that one today.
> --I could have baked that one yesterday and
> this one today.
Homework for David Ullrich:
In what respect do cakes I am going to make resemble quanta? (Hint:
> Like causes produce like effects.)
--John
Subject: Chess/Go/etc: Continuous Game-Boards?
I was just wondering today if there has been anything interesting
as Chess, Go, etc, where the game-boards are without any subdivision?
For example, each piece's base's shape/size would be significant. The
pieces would move (or be placed on the gameboard or removed from it)
under certain rules, but the exact distance/direction could be within
a continuous interval.
For instance, in Chess, the pieces (except knight) would not be
allowed to touch other pieces while moving.
(So, either this game would be played on a computer so as to be
precise in moving, or the game would take on aspects of Operation if
the pieces are moved by hand {no touching pieces to each other!}.)
I think, with Chess, what is considered a capture might be ambiguous.
Go, on the other hand, might be very interesting if played
continuously, especially if the pieces are different-sized circles.
(Each player would have the same number of pieces as their opponent of
any given radius, since bigger pieces would be more valuable than
smaller pieces.)
Any other games that might be interesting if played this way, either
on a computer or by hand?
Leroy Quet
Subject: Re: Chess/Go/etc: Continuous Game-Boards?
> I was just wondering today if there has been anything interesting
> as Chess, Go, etc, where the game-boards are without any subdivision?
(...)
> Any other games that might be interesting if played this way, either
> on a computer or by hand?
Surely some games can be played on such boards, but I am skeptical
about Chess, which is not really a very natural game in the first
place (*ducks for flying tomatoes*). In Go, it might be better to
lend an aspect from TwixT and build bridges between stones that are
a certain distance apart rather than letting the stones themselves
have any size. Being an avid TwixT-player I must of course also
recommend TwixT itself for these boards. ;)
---
J K Haugland
http://www.neutreeko.com
Subject: Re: Chess/Go/etc: Continuous Game-Boards?
Sorry, piggybacking.
> I was just wondering today if there has been anything interesting
> as Chess, Go, etc, where the game-boards are without any subdivision?
> (...)
> Any other games that might be interesting if played this way, either
> on a computer or by hand?
Tiddlywinks. You have to have the dexterity to move your pieces where
you want them too. (The serious game of tiddlywinks isn't just a race
to the pot, it also includes 'capturing' moves which are used to prevent
your opponent reaching the pot. Highly territorial, highly strategic.)
More info at www.etwa.org ad links therefrom.
Phil
Subject: Re: Chess/Go/etc: Continuous Game-Boards?
>I was just wondering today if there has been anything interesting
>as Chess, Go, etc, where the game-boards are without any subdivision?
>For example, each piece's base's shape/size would be significant. The
>pieces would move (or be placed on the gameboard or removed from it)
>under certain rules, but the exact distance/direction could be within
>a continuous interval.
>For instance, in Chess, the pieces (except knight) would not be
>allowed to touch other pieces while moving.
>(So, either this game would be played on a computer so as to be
>precise in moving, or the game would take on aspects of Operation if
>the pieces are moved by hand {no touching pieces to each other!}.)
>I think, with Chess, what is considered a capture might be ambiguous.
>Go, on the other hand, might be very interesting if played
>continuously, especially if the pieces are different-sized circles.
>(Each player would have the same number of pieces as their opponent of
>any given radius, since bigger pieces would be more valuable than
>smaller pieces.)
For Go, I assume:
(1.) Two pieces of the same colour only form a connected group if they
are in physical contact.
(2.) A group is captured only if its owner has no way to play another
piece in contact with it.
You have pointed out why large pieces would be more valuable than small
ones. Small ones also have an advantage: they can be used to make a
connection through a narrow gap.
If the pieces of the smallest size are in finite supply, a player might
find it advisable to keep his last piece of this size unplayed, to keep
his groups alive in accordance with (2).
The difference between Japanese (territory) rules and Chinese (area)
rules would be very important in this game, unlike in normal Go where it
rarely matters. For one thing, under Chinese rules, you get captured
pieces back and can use them again.
How would surrounded territory be defined - it might be (A) surrounded
by physically connected pieces; (B) surrounded by a wall with gaps
narrow enough that none of your opponent's unplayed pieces could fit
through them; (C) such that you could guarantee being able to capture
any opponent's piece played within it, using your remaining pieces
What would be the rules for establishing the status of groups after the
game end?
Nick
--
Nick Wedd nick@maproom.co.uk
Subject: Journal of Knot Theory and Its Ramifications - TOC alert
Vol. 12 No. 6 of the Journal of Knot Theory and Its Ramifications
(JKTR) is out. Articles are available in electronic format at
http://www.worldscinet.com/jktr.html
Table-of-contents:
Extensions of Quandles and Cocycle Knot Invariants
by J. Scott Carter, Mohamed Elhamdadi, Marina Appiou Nikiforou and
Masahico Saito
Link Invariants Associated with Gauge Equivalent Solutions of the
YangoBaxter Equation: The One-Parameter Family of Minimal
Typical
Representations of Uq[gl(2|1)]
by Jon R. Links and David de Wit
Classical Invariants for Holomic Knots
by Magnus Bergqvist
An Orientation-Sensitive Vassiliev Invariant for Virtual Knots
by J.9arg Sawollek
Detecting Non-Triviality of Virtual Links
by Teruhisa Kadokami
Sharp Bounds on Some Classical Knot Invariants
by C. Kearton and S. M. J. Wilson
Isometries of Orbifolds Double-Covered by Lens Spaces
by Theresa L. Jeevanjee
Automatic Structures for Torus Link Groups
by Matthieu Picantin
Stallings Twists Which Can be Realized by Plumbing and Deplumbing Hopf
Bands
by Ryosuke Yamamoto
The Editorial Board
JKTR
Subject: Re: Grid: Integers = To Sum Of Some Divisors
> Take an n-by-n-grid, n>= 3.
> Place the integers 2 to (n^2 +1) into the grid,
> one DISTINCT integer per grid-square, so that:
> If s(k,j) = a grid-square (ie. an element of an
> n-by-n matrix), then
> (for all k and j where n >= k >= 3 and n >= j >= 1)
> s(k,j) = (any divisor >= 2 of s(k-1,j)) +
> (any divisor >= 2 of s(k-2,j)),
> and also
> (for all k and j where n >= k >= 1 and n >= j >= 3)
> s(k,j) =
> (any divisor >= 2 of s(k,j-1)) +
> (any divisor >= 2 of s(k,j-2))
> I have been having no luck finding an n=4 example.
> But an n=3 example is:
> 5 3 8
> 2 6 4
> 7 9 10
> Is there an n=4 example??
> If there is, and even if there is not, are there
> examples for higher n??
Anyone willing to do a brute-force computer search for the existence
of any n = 4 case(s)??
I am going to otherwise just let this thread rest.
> Leroy
> Quet
Subject: Re: Minimal Graph, Four Color Theorem
>> Do you also believe that K3 is the only 3-chromatic graph that can be
>> reduced to bipartite by the removal of any single vertex?
>YES, if the vertex removed is chosen randomly!
Okay, a 5-cycle is a 3-chromatic graph which is not equal to K3. Removing
any random vertex results in a 4-path, which is 2-colourable. Again, it
sounds like you are assuming something like uniqueness of colouring, or
you are talking about graphs that are already labelled with a specific
colouring. Either would be non-standard terminology, so the responsibility
is yours to clarify your usage.
-- Erick
Subject: Re: Minimal Graph, Four Color Theorem
>> Do you also believe that K3 is the only 3-chromatic graph that can be
>> reduced to bipartite by the removal of any single vertex?
>YES, if the vertex removed is chosen randomly!
> Okay, a 5-cycle is a 3-chromatic graph which is not equal to K3.
Removing
> any random vertex results in a 4-path, which is 2-colourable. Again, it
> sounds like you are assuming something like uniqueness of colouring, or
> you are talking about graphs that are already labelled with a specific
> colouring. Either would be non-standard terminology, so the
responsibility
> is yours to clarify your usage.
> -- Erick
I do not know 'standard' terminology. Where can I find a source that
explains it?
Subject: Re: Minimal Graph, Four Color Theorem
>> Okay, a 5-cycle is a 3-chromatic graph which is not equal to K3.
Removing
>> any random vertex results in a 4-path, which is 2-colourable. Again, it
>> sounds like you are assuming something like uniqueness of colouring, or
>> you are talking about graphs that are already labelled with a specific
>> colouring. Either would be non-standard terminology, so the
responsibility
>> is yours to clarify your usage.
>I do not know 'standard' terminology. Where can I find a source that
>explains it?
I'm not sure what the best reference would be, but here is a nice general
graph theory textbook by Reinhard Diestel that is free to download:
http://www.math.uni-hamburg.de/home/diestel/books/graph.theory/download.html

Chapter 5 is all about colourings. The place where I'm getting the feeling
that there is a disconnect in terminology is that when we say a graph is
5-chromatic, it just means that some colouring with 5 colours is possible
(and no fewer than 5). The specific colouring you use to establish this is
unimportant and is not considered to permanently colour the graph and all
its subgraphs (you might call this a graph with colouring to
distinguish
it from just a graph).
There could be more than one colouring (and in fact there likely is more
than one). So unlike a bipartite graph where if it is bipartite then the
colouring is basically unique (assuming connectedness), a 5-partite graph
doesn't necessarily have a uniquely well-defined partition into 5 classes.
That's why it doesn't make too much sense to talk about whether the sizes
of the five partitions must be such-and-such. When we delete a vertex
and ask if the resulting graph is 4-chromatic, the colouring needn't share
any common ground with the 5-colouring in the larger graph.
Certainly it is possible to formulate the idea of a minimal 5-chromatic
graph in a way you might find more pleasing. For instance, we could say
that it's equivalent to a graph that is 5-chromatic in such a way that
for every vertex v there exists a 5-colouring of the graph in which v
is the sole vertex with the colour blue.
-- Erick
Subject: Re: Minimal Graph, Four Color Theorem
> So why are you arguing?
> I'm sorry, I didn't realize I was dealing with
> another case of raging insanity in sci.math.
> I will endeavor to disregard any postings by
> you in the future.
> xanthian.
CHICKEN!!!!!
Subject: Re: Minimal Graph, Four Color Theorem
Visiting Assistant Professor at the University of Montana.
>>If H could be recolored after vertex removal, G could have been
>>recolored before the vertex was removed.
>> Not necessarily.
>> Consider the following:
>> Let S be the set of all planar graphs which can be 4-colored but NOT
>> 3-colored. The set is nonempty. So the collection
>> {n : n is a positive integer and there is a graph with n vertices in S}
>> is nonempty. A nonempty set of positive integers has a minimum. IN
>> this case, the minimum is 4.
>> Therefore, any planar graph with fewer than 4 vertices can be 3-colored.
>> Let G be any graph with exactly 4 vertices and which cannot be
>> 3-colored. For example, K-4. Any
>> attempted 3 coloring will fail. If you remove any vertex, you get
>> either a triangle (K-3). If you take out an arbitrary vertex, you can
>> recolor the remaining graph with a 3-coloring, but that 3-coloring
>> cannot be extended to the original graph, for any vertex.
>Removing a vertex does not change the adjacency matrix for the
>remaining vertices!!!
So? For every color used in the remaining graph, there is a vertex
colored with that color which is adjacent to the vertex you
removed. So the vertex you removed cannot be assigned any color in the
3-coloring for the graph resulting in its removal.
>Remember that chi(G) = 5. If G is planar, then the FCT is false! The
>only way that G cannot be planar is if G = K5! Do you dispute this
>conclusion?
Yes, there are many graphs which are not planar and are NOT equal to
K5. Any graph properly containing K5, for example. And, of course,
K(3,3) is also not planar, and does not contain any copy of K5.
Subject: Re: Minimal Graph, Four Color Theorem
>Remember that chi(G) = 5. If G is planar, then the FCT is false! The
>only way that G cannot be planar is if G = K5! Do you dispute this
>conclusion?
> Yes, there are many graphs which are not planar and are NOT equal to
> K5. Any graph properly containing K5, for example. And, of course,
> K(3,3) is also not planar, and does not contain any copy of K5.
OK. I agree; partially. The second sentence is clumsy and ambiguous.
But the first sentence; If G is planar, then the FCT is false!
cannot be challenged!
Subject: Re: Minimal Graph, Four Color Theorem
Visiting Assistant Professor at the University of Montana.
>>Remember that chi(G) = 5. If G is planar, then the FCT is false! The
>>only way that G cannot be planar is if G = K5! Do you dispute this
>>conclusion?
>> Yes, there are many graphs which are not planar and are NOT equal to
>> K5. Any graph properly containing K5, for example. And, of course,
>> K(3,3) is also not planar, and does not contain any copy of K5.
>OK. I agree; partially. The second sentence is clumsy and ambiguous.
>But the first sentence; If G is planar, then the FCT is false!
>cannot be challenged!
Like I said, I think you are very confused.
The argument is part of a proof by contradiction: assume that the FCT
is false, and derive a contradiction.
In Graph Theory, this particular kind of proof by contradiction is so
common that it even has a special name: the minimal counterexample
argument. You assume the statement is false; then that means that
there must exist a smallest positive integer with the property that
there is a counterexample with that many vertices; which means that
any proper subgraph of the counterexample ->does<- have the property,
and then one tries to derive a contradiction from these superposition
of facts.
Now, you seem to be claiming that given any planar graph G, and any
vertex v in G, if G-{v} can be 4-colored then one can take that
coloring and extend it to a 4-coloring of G. This is simply not the
case, but you are close to the way the proof goes: what one shows is
that if there were a counterexample, and G is minimal, then there
would always be a specific choice of v such that there is some
specific 4-coloring of G-{v} which can be extended to a 4-coloring of
G. But the proof is long: one must consider many possible cases of G
and many configurations.
Now, the following is certainly true: if G is planar, and has a vertex
v with degree less than 4, then any 4-coloring of G-{v} can be
extended to a 4-coloring of G.
Are you perhaps trying to say that since K5 is not planar, any planar
graph would have such a vertex v? If so, then of course you'd be
wrong. Consider the graph of 7 vertices, 1,2,3,4,5,6,7, with the
following adjacencies:
1 is adjacent to 2,3,4, and 5.
2 is adjacent to 1, 3, 5, and 7.
3 is adjacent to 1, 2, 4, and 6.
4 is adjacent to 1, 3, 5, 6, and 7.
5 is adjacent to 1, 2, 4, and 7.
6 is adjacent to 2, 3, 4, and 7.
7 is adjacent to 2, 4, 6, and 6.
I have a drawing before me, so I know the graph is planar. Now, if you
remove any vertex, and you 4-color the remaining graph, then it is
possible that that 4-coloring cannot be extended to a 4-coloring of
G. For instance, remove the vertex 6, and color the remaining graph
with Red, Blue, Green, and Yellow, as following:
Blue: 2, 4
Green: 3
Red: 5
Yellow: 1, 7.
Since 6 is adjacent to a blue, a green, a red, and a yellow vertex,
you cannot extend that 4-coloring to a 4-coloring of G.
Now, yes, of course, there are OTHER 4-colorings of G-{6} that ->can<- be
extended to G. But that's not the point. The point is that you cannot
simply say that a 4-coloring of G-{6} can be extended to a 4-coloring
of G-{6}, only that ->some<- 4-colorings of G-{6} can be extended.
So you need would need to show that given ANY planar graph G, there is
always a choice of vertex v such that there is SOME 4-coloring of
G-{v} which can be extended to G, to be able to derive a contradiction
from the assumption that a minimal counterexample exists. That is in
fact what Appel and Hanken did.
Subject: Re: Minimal Graph, Four Color Theorem
[...]
>Remember that chi(G) = 5. If G is planar, then the FCT is false! The
>only way that G cannot be planar is if G = K5! Do you dispute this
>conclusion?
Yes, there are many graphs which are not planar and are NOT equal to
K5. Any graph properly containing K5, for example. And, of course,
K(3,3) is also not planar, and does not contain any copy of K5.
|
|OK. I agree; partially. The second sentence is clumsy and ambiguous.
|But the first sentence; If G is planar, then the FCT is false!
|cannot be challenged!
The proof is structured overall as a proof by contradiction. That means
it starts about by assuming that there exists a planar graph requiring
more than four colors, and then shows in the rest of the proof that the
logical consequences of this assumption are contradictory, which proves
that there was not actually such a graph in the first place. If you have a
problem with such a method of proof, start by addressing your qualms with
that method of proof first. Yes, of course if there exists a planar graph G
with chromatic number 5, then the four color theorem is false. The
situation
being assumed is one which one *thinks* is impossible; that's the whole
point-- to show that it's impossible.
The next step is that if there is any planar graph G with chromatic number
>4, then there is one which has the fewest possible vertices in it. The
next step is to observe that if G is a planar graph having chromatic
number >4, and which has the smallest possible number of vertices for any
graph with those properties, then every graph G' obtained by deleting one
of the vertices of G would have chromatic number <=4, simply because G' is
a planar graph having fewer vertices.
Now this reasoning is simple enough to make it difficult to see what about
it is not clear enough for you, which is why people have been having
trouble
giving you better answers than they've given already.
It appeared once as though you thought deleting any given one vertex could
not be enough to reduce the chromatic number to 5 unless G=K_5, because
(given some specific coloring of G with five colors) you thought the only
way to get a graph having chromatic number <5 was to delete all of the
vertices of one of the colors. There is of course only one 5-colored graph
having chromatic number 5 where each vertex is the only vertex of its
color,
i.e. K_5. Since you're being somewhat terse, it's not really clear what the
line of thinking is, but that's what it sounded like.
It's correct to reason that if a graph G has chromatic number 5, and if
deleting a vertex v turns it into a graph of chromatic number <5, then
there
exists SOME coloring of G which gives v its own unique color, and uses the
other four colors for the rest of G. All one has to do is to color G-v with
<5 colors (actually it has to be 4 or else chi(G) wouldn't be 5), and then
give the fifth color to v. This is what your remark about the incidence
matrix of the remaining vertices staying the same seems to be about. But
this doesn't make the fallacious reasoning in the last paragraph work,
because this coloring with five colors we've produced would not necessarily
By way of comparison, consider the graph with five vertices which are
arranged in a circle, with each connected to its two neighbors. That graph
C_5 requires three colors. If you delete any one vertex, it becomes a graph
requiring just two colors. But any way you color C_5 with three colors will
give two vertices the same color, of course. Given a vertex, there's a way
to color the graph so that that vertex has its own unique color, but it's
a different coloring for each vertex.
Again:
|But the first sentence; If G is planar, then the FCT is false!
|cannot be challenged!
I'm not sure what your goal is here. Are you trying to get someone to
explain something to you, or are you convinced enough that there's
something wrong with it that you're now trying to convince us also that
there's something wrong with it? It's kind of irritating to get what
appears to be a request for help in understanding something, followed up
with challenges presented in an argumentative way, complete with
exclamatory remarks. This is one of those rare places where it's possible
to get help for free, and the explanations you've been getting have been
about as good as can be expected under the circumstances. Try to understand
what's being said before starting to argue about it.
Keith Ramsay
Subject: Re: Minimal Graph, Four Color Theorem
To Keith M.
I disagree that the explanations were as good as could be expected.
The explanations were sometimes wrong and generally irrelevant.
Example.
... a minimal graph with chi(G)=4 would be the square with
one diagonal added, so that n=4.
Here is a simple contradiction to the recolorability conjecture.
J is a 6 vertex graph with the following edges AB, AC, AD, AE, AF, BC,
BD, BE, CD, CE, DE, & DF. If only vertex 'F' is removed, J' cannot be
recolored to make chi(J')=4. (It is noted that J is non-planar, but
the example is valid?)
Subject: Re: Minimal Graph, Four Color Theorem
Visiting Assistant Professor at the University of Montana.
>To Keith M.
>I disagree that the explanations were as good as could be expected.
>The explanations were sometimes wrong and generally irrelevant.
>Example.
> ... a minimal graph with chi(G)=4 would be the square with
>one diagonal added, so that n=4.
The problem there is just that I screwed up. A square with one
diagonal added has chi(G)=3; but if you take a square with one
diagonal added, and then you join the other two opposite vertices by
an edge lying outside the square, you do get a minimal graph with
chi(G)=3.
Subject: Re: recursively draw the points between the interval
You need only to add one line to your code.
> drawPoints(min, max, n)
> if (n != 1)
> { mid = (max - min)/2
> print mid
> drawPoints(min, mid, n/2)
drawPoints(mid,max,n/2)
> }
Given your example, this will draw the points in this order:
.5, .25, .125, .375, .75, .625, .875
Subject: Re: recursively draw the points between the interval
> Here's a recursive algorithm to get the points in a line, but failed
> attempted:
> Basically given the range [min, max] and number of intervals (n)
> between the range, the algorithm should be able to get the points in
> between recursively.
> For example, the range is [0,1] and n=8, then it will get the
> following points:
> 0.5 = (1-0)/2
> 0.25 = (0.5-0)/2
> 0.125 = (0.25-0)/2
> 0.375 = 0.25+0.125
> 0.75 = (1-0.5)/2
> 0.625 = 0.5+0.125
> 0.875 = 0.75+0.125
> drawPoints(min, max, n)
> if (n != 1)
> { mid = (max - min)/2
> print mid
> drawPoints(min, mid, n/2)
> }
> My thought is I need to get the points on the left hand side of each
> point, then right hand side. I just don't know how to get the right
> hand side point.
> But the program output, obviouly I get stuck when I draw the fourth
> point
> 0.5
> 0.25
> 0.125
You will only ever get a countable number of the points on the line
this way (or any other way on a real computer). You will thus miss
almost all of the points (all except a point set with measure zero)
Darren
Subject: Re: Hyperreals
> Do the Hyperreals form a field? If so, what does that say about the
> analytical completeness of the reals?
> ~Steven Margolin
> I'm the third one (at least) not knowing what you mean by analytical
> completeness. Perhaps you mean this property: every nonempty set
> that is bounded above has a least upper bound.
Yes. That's what I meant. I think I may have used the wrong term. I
had heard that the Complex Numbers were (and still are) the smallest
field that are both algerbraically complete (like the algebraics) and
analytically complete (like the reals).
> The hyperreals fails
> that property, the set of finite hyperreals, for example, is bounded
> above but has no least upper bgound.
Subject: Re: Irrationality of an integral nth root of prime number m
<y8zisn3sez7.fsf@nestle.ai.mit.edu> Ive never come across this simple proof 
of
the irrationality of an
integral
>> nth root of prime number m
> For positive integers r,s, integer n > 1 and prime p, let
> p = (r/s)^n
> By reducing r/s to lowest terms we may take r,s as coprime. Thus
> p s^n = r^n
> Hence
> p | r^n
> and because p is prime,
> p | r
> As n > 1,
> p^2 | r^n
> p^2 | ps^n
> p | s^n
> p | s
> Which contradicts that r,s are coprime.
> In comparision, your proof is much complicated.
> That's not true. If x, y denote the powers of p dividing s, r resp.
> then we have 1 + n x = n y from p s^n = r^n and unique
factorization.
> His proof: above => n divides 1, contra n > 1
That's your proof.
> Your proof: y >= 1, n >= 2 => x >= 1 => p|(r,s) contra (r,s) = 1
> Do you still believe his proof is much complicated in comparison ?
Yes. However your proof, or rewrite of his proof, is nifty.
> His proof has the real essence: the expts of p clearly disagree (mod n)
His expression is poor.
Subject: power set proof
Hey,
I'm trying to prove that the power set of (A intersection B) = the
power
set of A intersection with the power set of B. Can someone give a good
starting point from which I can work on?
Subject: Re: power set proof
> Hey,
> I'm trying to prove that the power set of (A intersection B) = the
power
> set of A intersection with the power set of B. Can someone give a good
> starting point from which I can work on?
The usual approach to proving that two sets are equal is to prove
each one is a subset of the other, i.e., that each member of each
set is also a member of the other set.
Subject: Re: power set proof
> Hey,
> I'm trying to prove that the power set of (A intersection B) = the
power
> set of A intersection with the power set of B. Can someone give a good
> starting point from which I can work on?
Use the standard method of proof to show that two sets are equal.
You want to show that P(A intersection B) = P(A) intersection P(B),
where P means power set of.
First, show P(A intersection B) is a subset of P(A) intersection P(B).
Next, show P(A) intersection P(B) is a subset of P(A intersection B).
Usually, one of the two is trivial. Here, the second one is trivial.
-- Bill Hale
Subject: Re: power set proof
I keep coming up with a proof that the intersection of A and B is within
the
power set of a and the power set of b.
However, if I let x be in the power set of A intersection B, what can I say
about A and B seperately?
> Hey,
> I'm trying to prove that the power set of (A intersection B) = the
power
> set of A intersection with the power set of B. Can someone give a good
> starting point from which I can work on?
> Use the standard method of proof to show that two sets are equal.
> You want to show that P(A intersection B) = P(A) intersection P(B),
> where P means power set of.
> First, show P(A intersection B) is a subset of P(A) intersection P(B).
> Next, show P(A) intersection P(B) is a subset of P(A intersection B).
> Usually, one of the two is trivial. Here, the second one is trivial.
> -- Bill Hale
Subject: Re: power set proof
> I keep coming up with a proof that the intersection of A and B is within
the
> power set of a and the power set of b.
But, that fact is not relevant. You need to use the definitions for
the things that you are trying to prove. See below.
> However, if I let x be in the power set of A intersection B, what can I
say
> about A and B seperately?
Likewise, that is not directly relevant.
Ok, you are trying to prove that P(A intersection B) is a subset of P(A)
intersection P(B).
Thus, you let x be an element of P(A intersection B).
You want to show that x is an element of P(A) intersection P(B).
What is the definition of x being an element of P(A) intersection P(B)?
The definition says that you must show:
1) x is an element of P(A); and
2) x is an elment of P(B).
Rather than looking at the definition of intersection, you are
going off on a tangent, trying to find true statements that
you can say about A and B. Sometimes, you have to do this when
you don't have an idea of how to proceed. But, that does not
apply here, since you should be unwinding the definitions to
get simpler statements to prove than the one you are currently at.
Recapping: you broke down trying to show two sets are equal
into trying to show first that one set is a subset of the other,
and second that the other set is a subset of the first one.
Here, you are using a well-known theorem of set theory.
For showing that one set is a subset of the other, you break
that down into two steps 1) and 2) given above. Here, you are
using the definition of what intersection means.
-- Bill Hale
Subject: Re: P R O O F O F G O D
>>alt.atheism;
>> Mark K. Bilbo <iskanipa-y@hoo.com> schreef in bericht
>> So it's also misspelt?
>> Misspelt is incorrect; misspelled is correct!!!
How can you have any kind of intelligent dialogue if you can't
even
>spell simple words like misspelled? My advice: Give up!
>> spelt2 ( P ) Pronunciation Key (splt)
>> v.
>> A past tense and a past participle of spell.
>> Source: The American Heritage Dictionary of the English
Language,
>> Fourth Edition
>> I thought it was a valid (if fallen out of common use) form...
>> Note that 'mispel' is also correct.
>> --
>> Tetsuo
>So is Miss Spelt, unless you mean a beauty pageant winner from a fishing
>town.
>>Miss Pelt?
>Unshaven
Her name is Miss Information.
-Barry
Web page: http://members.optusnet.com.au/~barry.og
Atheist, radio scanner, LIPD information.
Voicemail/fax number +14136227640
Subject: Re: probability problem
> By the way, if X, Y, and Z are i.i.d. continuous r.v.'s, then
> P{Y<X<Z} is easy to compute and does not depend on the common
distribution.
Then, if X, Y, Z are independent continous r.v.s. The pdf are
respectively given as f_X(x), f_Y(y), f_Z(z), how to compute the
Subject: Re: probability problem
> By the way, if X, Y, and Z are i.i.d. continuous r.v.'s, then
> P{Y<X<Z} is easy to compute and does not depend on the common
distribution.
> Then, if X, Y, Z are independent continous r.v.s. The pdf are
> respectively given as f_X(x), f_Y(y), f_Z(z), how to compute the
oo z x
P(Y<X<Z) = Int f_Z(z) ( Int f_X(x) ( Int f_Y(y) dy ) dx ) dz
-oo -oo -oo
In case f_X = f_Y = f_Z this boils down to
P(Y<X<Z) = 1/3! = 1/6
Horst
Subject: Rudin question
I am trying to read Rudin Principles of Mathematical Analysis. Theorem
2.27(a) provides the following statement and proof:
Thm: bar{E} is closed.
Proof: If p in X and p notin bar{E}, then p is neither a point of E
nor a limit point of E. Hence p has a neighborhood which does not
intersect E. The complement of bar{E} is therefore open. Hence
bar{E} is closed.
On a second read, I am unable to come to complete grips with this proof.
The statement Hence p has a neighborhood which does not intersect E
can be seen as
exists N(p) s.t N(p) subset E^c
However, to correctly claim that the complement of bar{E} is open, we
need to show that exists N(p) s.t N(p) subsect (bar{E})^c. i.e we
need to show
exists N(p) s.t N(p) subset (bar{E})^c
And however much I stare at it, I do not seem to be able to convince
myself that
N(p) subset E^c => N(p) subset (bar{E})^c
because
E^c is not a subset of (bar{E})^c
Infact
(bar{E})^c = E^c intersect E'^c
Thus (bar{E})^c is smaller no?
Is my understanding fundamentally flawed, or is there something missing
from the proof? Please excuse my inexpertise in this. I am not a
mathematics student...
Srinath
Subject: Re: Rudin question
<Pine.SOL.4.44.0310052218470.7380-100000@albinoni.EECS.Berkeley.EDU>,
> I am trying to read Rudin Principles of Mathematical Analysis.
Theorem
> 2.27(a) provides the following statement and proof:
> Thm: bar{E} is closed.
> Proof: If p in X and p notin bar{E}, then p is neither a point of E
> nor a limit point of E. Hence p has a neighborhood which does not
> intersect E. The complement of bar{E} is therefore open. Hence
> bar{E} is closed.
> On a second read, I am unable to come to complete grips with this proof.
> The statement Hence p has a neighborhood which does not intersect E
> can be seen as
> exists N(p) s.t N(p) subset E^c
> However, to correctly claim that the complement of bar{E} is open,
we
> need to show that exists N(p) s.t N(p) subsect (bar{E})^c. i.e we
> need to show
> exists N(p) s.t N(p) subset (bar{E})^c
> And however much I stare at it, I do not seem to be able to convince
> myself that
> N(p) subset E^c => N(p) subset (bar{E})^c
> because
> E^c is not a subset of (bar{E})^c
> Infact
> (bar{E})^c = E^c intersect E'^c
> Thus (bar{E})^c is smaller no?
Yes, it is: ^c reverses inclusions, and E subset bar E implies
(bar E)^c subset E^c.
But you're not looking at this completely. Rudin asserts that p has a
neighborhood which does not intersect E. It's automatic that it
doesn't intersect E' either, since a point x of N(p) cap E' would have
a small neighborhood N(x) which was a subset of N(p); and such a
neighborhood must contain a point of E (by definition; even, a point of
E other than x).
Thus N(p) subset E^c ==> N(p) subset (bar E)^c. It's not that (bar
E)^c is BIGGER than E^c; it's that you haven't used the notion of
NEIGHBORHOOD.
[For those of you who don't have Rudin available, he defines E' to be
the set of accumulation [limit] points of E; then defines the closure
of E, bar E, to be E union E'. A stupid definition, IMHO. This
nonstandardness causes no end of confusion to students. Still, they're
supposed to be able to apply logic and follow proofs.]
--Ron Bruck
Subject: Re: Rudin question
> Thus N(p) subset E^c ==> N(p) subset (bar E)^c. It's not that
(bar
> E)^c is BIGGER than E^c; it's that you haven't used the notion of
> NEIGHBORHOOD.
I think that it would have helped (at least me) tremendously to have
included this statement in the proof. That if an open set, (N) is in the
exterior of a set (E), then N cannot contain any limit points of E.
I think this is not quite obvious enough to be left out of the proof
that bar{E} is closed.
The problem I had was that I thought that any proof in Rudin (at least
from chapter 2!) would be _complete_ in the sense that the sequence of
statements should imply the hypothesis without absolutely any recourse
obvious-ness other than elementary set logic.
> [For those of you who don't have Rudin available, he defines E' to be
> the set of accumulation [limit] points of E; then defines the closure
> of E, bar E, to be E union E'. A stupid definition, IMHO. This
> nonstandardness causes no end of confusion to students. Still, they're
> supposed to be able to apply logic and follow proofs.]
I'd be very interested to know what the standard definition is. Is it
the closure of a set E is the smallest closed set which completely
contains E. That would seem like a good candidate.
> --Ron Bruck
--
Srinath
Subject: l_{infty} and separability
be indebted to your person.
We define
l_{infty} = { (a_n)_{nin N} subset R ; sup|a_n| < infty }.
This is a metric space with metric
d((a_n),(b_n)) = sup|a_n-b_n|.
I wish to show that this space is not separable, and that its density
character is |R|, the cardinality of the reals.
Subject: Re: l_{infty} and separability
>be indebted to your person.
>We define
> l_{infty} = { (a_n)_{nin N} subset R ; sup|a_n| < infty }.
>This is a metric space with metric
> d((a_n),(b_n)) = sup|a_n-b_n|.
>I wish to show that this space is not separable, and that its density
>character is |R|, the cardinality of the reals.
For every subset A of N let x_A be its characteristic function and O_A
the ball
around x_A of radius 1/3; the family of all O_A is pairwise disjoint.
KP
--
E-MAIL: K.P.Hart@EWI.TUDelft.NL PAPER: Faculty EWI
PHONE: +31-15-2784572 TU Delft
FAX: +31-15-2786178 Postbus 5031
URL: http://aw.twi.tudelft.nl/~hart 2600 GA Delft
the Netherlands
Subject: Re: l_{infty} and separability
> We define
> l_{infty} = { (a_n)_{nin N} subset R ; sup|a_n| < infty }.
> This is a metric space with metric
> d((a_n),(b_n)) = sup|a_n-b_n|.
> I wish to show that this space is not separable, and that its density
> character is |R|, the cardinality of the reals.
A standard procedure to show that l_p is separable for 1<=p<inf is that we
approximate with arbitrary precision a given point (sequences) by a finite
sequence with rational elements. It is then possible to show that a set of
all these approximating sequences is countable.
For p=inf, you cannot find an approximating element among finitely nonzero
sequences for a given point. Hence, ...
Zdenek Hurak
Subject: Re: Sets.
> So the issue of expressing infinity, imho, is not exactly
> the reason why we should make a radical change to
> the notion of mathematics, and logic, whether or not
> we're happy with the current way of doing mathematics.
> Yes - a good point. Revolutions should, indeed, be made of sterner
> stuff!
> So long as we treat the concept of infinity as essentially
> indicating the presence of a well-defined non-terminating process, and
> provide effective, and unambiguous, methods for determining which of
> these processes we define as mathematical objects (e.g., Cauchy
> sequences), the fact that the outpput of the process itself may be
> completely representable individually, but not as a completed
> totality, within a mathematical language should not, by itself, be
> considered sufficient grounds for treating the mathematical object, or
> its introduction, as intuitionistically unacceptable within the
> language.
Well said. It is important to note the distinction between a well-defined
non-terminating process, for example potential infinity, and a completed
individuality, for example actual infinity (w = little omega). Since w is
not the successor of any natural number, the successor operation cannot
reach w. And since it can't reach w, the idea of ALL naturals, in a set,
contains a meaning above and beyond the notion of subsequent successors of
the null set. Actual infinity is w, unreachable by the successor
operation,
and potential infinity means only those numbers that can be reached by free
recursive use of the successor operation starting from the null set. The
successor operation can reach every member of N (that is it can reach
potential infinity) but it can't reach N itself (actual infinity).
Randy
http://www.rlgerl.com
Subject: Re: 3D Geometry Area Problems
The Ghost In The Machine
> Three words: 1/2 cross product. I'll leave you to figure out the details.
:-)
It is even more easy!
Abs[ Determinant [(i,j,k ) , (a,-b,0 ) , (-a, 0, c ) ] ]/2
Subject: Re: 3D Geometry Area Problems
In sci.math, Narasimham G.L.
<mathma18@hotmail.com The Ghost In The Machine
>> Three words: 1/2 cross product. I'll leave you to
>> figure out the details. :-)
> It is even more easy!
> Abs[ Determinant [(i,j,k ) , (a,-b,0 ) , (-a, 0, c ) ] ]/2
Like I said, half cross product, as it's obvious what you're doing. :-)
(Even if I don't know Maple or Mathematica.)
--
#191, ewill3@earthlink.net
It's still legal to go .sigless.
Subject: Re: Topology question
A general construction (doubling a point):
Let X be a compact Hausdorff space and p be a non-isolated point of X.
Take some q not in X and define X' = X+{q}, where the base in q is the
set of all {q}+U{p}, U is a neighbourhood of p; the topology of X is
the original one. Then any two neighbourhoods of p and q in X' have
non-empty intersection, but they have Hausdorff compact neighborhoods.
Simeon
> Could anyone give me an example of a topological space X such that every
> point of X has a Hausdorff compact neighborhood, but X is not Hausdorff?
> J.S
Subject: Re: Topology question
> A general construction (doubling a point):
> Let X be a compact Hausdorff space and p be a non-isolated point of X.
> Take some q not in X and define X' = X+{q}, where the base in q is the
> set of all {q}+U{p}, U is a neighbourhood of p; the topology of X is
> the original one. Then any two neighbourhoods of p and q in X' have
> non-empty intersection, but they have Hausdorff compact neighborhoods.
What's the compact Hausdorff nhood of q?
> Could anyone give me an example of a topological space X such that
every
> point of X has a Hausdorff compact neighborhood, but X is not
Hausdorff?
Subject: adjoint matrix
Hello
The Adjoint matrix of a square matrix A:
Replace each element of A with its own cofactor and transpose the
result, then you have made the adjoint matrix of A.
I got this question..if determinant (A)=0 then
determinant(Adjoint matrix of A)=0 ? I think yes (doing some
examples)
but maybe in general fails..any counterexamples or ideas on how to
show
this is true?
Carlos
Subject: Re: adjoint matrix
> Hello
> The Adjoint matrix of a square matrix A:
> Replace each element of A with its own cofactor and transpose the
> result, then you have made the adjoint matrix of A.
> I got this question..if determinant (A)=0 then
> determinant(Adjoint matrix of A)=0 ? I think yes (doing some
> examples)
> but maybe in general fails..any counterexamples or ideas on how to
> show
> this is true?
Note that A adj(A) = 0. (*)
Split into 2 cases
(i) rank(A) = n-1. In this case what does (*) tell you about the
columns of adj(A).
(ii) rank(A) < n-1. In this case what is adj(A)?
--
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: What to tell new students?
I will have to give a short talk (15-20 minutes) to our
new mathematics students, and I find myself at a bit of
a loss as to what I am going to say. I should talk about
something mathematical, maybe some application, but the
technical part should be minimal. Ideally, it should
demonstrate some mathematical point or somehow prepare
and motivate them for the courses. It would also be
nice if there were some story attached to it to grip
their attention. However, anything entertaining and
mathematical would be acceptable.
Any ideas?
Michael.
--
&&&&&&&&&&&&&&&&#@#&&&&&&&&&&&&&&&&
Dr. Michael Ulm
FB Mathematik, Universitaet Rostock
michael.ulm@mathematik.uni-rostock.de
Subject: Re: What to tell new students?
> I will have to give a short talk (15-20 minutes) to our
> new mathematics students, and I find myself at a bit of
> a loss as to what I am going to say. I should talk about
> something mathematical, maybe some application, but the
> technical part should be minimal. Ideally, it should
> demonstrate some mathematical point or somehow prepare
> and motivate them for the courses. It would also be
> nice if there were some story attached to it to grip
> their attention. However, anything entertaining and
> mathematical would be acceptable.
> Any ideas?
<http://bridge.anglia.ac.uk/~systimk/Humour/Lehrer/Misc.Html#Mathematics>
--
Paul Sperry
Columbia, SC (USA)
Subject: Re: What to tell new students?
> I will have to give a short talk (15-20 minutes) to our
> new mathematics students, and I find myself at a bit of
> a loss as to what I am going to say. I should talk about
> something mathematical, maybe some application, but the
> technical part should be minimal. Ideally, it should
> demonstrate some mathematical point or somehow prepare
> and motivate them for the courses. It would also be
> nice if there were some story attached to it to grip
> their attention. However, anything entertaining and
> mathematical would be acceptable.
> Any ideas?
An area where a new branch of mathematics is having a great impact is
encryption. You could tell them something about the primitive origins
of encryption, the ENIGMA project and its role in the Allied victory in
WW2, then talk about the ubiquity of encryption in the internet age,
with especial reference to wireless communications.
Gib
Subject: Re: What to tell new students?
> I will have to give a short talk (15-20 minutes) to our
> new mathematics students, and I find myself at a bit of
> a loss as to what I am going to say. I should talk about
> something mathematical, maybe some application, but the
> technical part should be minimal. Ideally, it should
> demonstrate some mathematical point or somehow prepare
> and motivate them for the courses. It would also be
> nice if there were some story attached to it to grip
> their attention. However, anything entertaining and
> mathematical would be acceptable.
> Any ideas?
> Michael.
> --
> &&&&&&&&&&&&&&&&#@#&&&&&&&&&&&&&&&&
> Dr. Michael Ulm
> FB Mathematik, Universitaet Rostock
> michael.ulm@mathematik.uni-rostock.de
Without mathematics you guys could not count to 10, literally. That's where
it
started with people's attempt to find a way to understand the world around
them - in this case counting things. And it progresses through geometry,
algebra, calculus, tensors, etc. And sometimes works backwards - a
mathematical concept discovered for its own sake is applied to the real
world.
Or, if that does not work, you could make up quotes from Anna Kournikova on
how much mathematics has meant to her. :)
Bill
Subject: Re: What to tell new students?
> I will have to give a short talk (15-20 minutes) to our
> new mathematics students, and I find myself at a bit of
> a loss as to what I am going to say. I should talk about
> something mathematical, maybe some application, but the
> technical part should be minimal. Ideally, it should
> demonstrate some mathematical point or somehow prepare
> and motivate them for the courses. It would also be
> nice if there were some story attached to it to grip
> their attention. However, anything entertaining and
> mathematical would be acceptable.
> Any ideas?
> Michael.
> --
> &&&&&&&&&&&&&&&&#@#&&&&&&&&&&&&&&&&
> Dr. Michael Ulm
> FB Mathematik, Universitaet Rostock
> michael.ulm@mathematik.uni-rostock.de
> Without mathematics you guys could not count to 10, literally. That's
where it
> started with people's attempt to find a way to understand the world
around
> them - in this case counting things. And it progresses through geometry,
> algebra, calculus, tensors, etc. And sometimes works backwards - a
> mathematical concept discovered for its own sake is applied to the real
> world.
I just fell asleep. ZZZ.
> Or, if that does not work, you could make up quotes from Anna Kournikova
on
> how much mathematics has meant to her. :)
> Bill
Subject: Re: Quanta and Cakes
> I should have asked in what respect *correct reasoning* about
> quanta resembles such reasoning about cakes-to-be, but
> differs from such reasoning about (extant) 'medium size
> dry goods'.
> My answer to this question would be: correct reasoning about
> medium size dry goods should take these as satisfying
> both the right and the left sides of the following bi-
> conditional;
> AxAy(Az(z in x <-> z in y) -> x=y) <-> Ax(x=x)
> [Identity of Indiscernibles] [Reflexive Law of Equality]
> while correct reasoning about quanta--and cakes-to-be--should
> take these to satisfy neither.
> Sorry, you have obfuscated your meaning beyond my ability to extract
> any sense from it, or even to guarantee that the language in use is
> still English, so I shall withdraw from the discussion.
> xanthian.
FYI: In question here is whether a better background logic--for
theories as different from one another as (i.e.) Russell's Theory
of Descriptions and quantum mechanics--is a logic in which identity
is reflexive, or a logic in which identity is not reflexive.
I have been arguing that the weaker of these logics, in which
Ax(x=x) is not provable, is a better background logic--for
quantum mechanics because the weaker logic leaves it up to the
physicist to decide whether quanta, which Paul Teller (among
others) has suggested lack individuality, refute Ax(x=x).
In my opinion, this is a matter that physicists are better
qualified than logicians to pass judgment on.
--John
Subject: Re: Quanta and Cakes
> [...] The alternative package characterises quanta as
> non-individuals, where this is understood in terms of a lack of
> identity.
> [Source: see above!]
The physicists erroneously identify 'identity' and 'self-identity',
> because to be *identityless* is not to be *non-self-identical*!!!
> For every x and every y, if x and y are quanta than ~(x=y). Have
> you a problem with this?
> I dont have any problem with
> For every x and every y, if x and y are *two* quanta than
~(x=y).
> This is supposed to mean that if you substitute 'x' for 'y', the
> statement is rendered false, since one quantum cannot be two quanta!
> But back to the crucial point, namely, that he who posits
> non-self-identicals has got to posit 'self-inconsistent', i.e.
> impossible objects as well!
> a = a <-> AP(Pa <-> Pa))
> [The combined principles of 'the self-identity of
> self-indiscernibles' and of the 'self-indiscernibility of
> self- identicals']
> correspondingly
> ~(a = a) <-> ~AP(Pa <-> Pa))
> and equivalently
> ~(a = a) <-> EP~(Pa <-> Pa))
> and equivalently
> ~(a = a) <-> EP(Pa <-> ~Pa))
> and equivalently
> ~(a = a) <-> EP(Pa & ~Pa))
> Any non-self-identical object would be an impossible object!
> So if you intend to create non-self-identicals, dump the law of
> non-contradiction first!
> A non-self-identical-object would possess some properties not
> possessed by it!
Theories of reflexive and non-reflexive identity include the
deductive apparatus of propositional logic, whose theorems include
(p <-> p), of which (Fx <-> Fx) is an instance. Not to worry:
self-identical or not, from yourself you are indistinguishable.
Get a grip, Paul, get a grip!
--John
Subject: Re: Quanta and Cakes
> A) AxAy(Az(z in x <-> z in y) -> x=y) <-> Ax(x=x)
> [Identity of Indiscernibles] [Reflexive Law of Equality]
Doesn't (A) suggest that instead of going on and on and
> on about the identity of indiscernibles, Leibniz should just
> have asserted that everything is self-identical and have been
> done with it?' (Oh my God! I'm starting to sound like Ullrich!)
> Worth discussing.
In the following proof, the first-order axiom schema
AxAy[(Fx <-> Fy) -> x=y) stands in for (A).
Any questions about the (similar) proof with (A)
should be directed to Jim Burns.
1. AxAy[(Fx <-> Fy) -> x=y) Assume
2. Show Ax(x=x)
3. ~Ax(x=x) Assume
4. Ex~(x=x) 3
5. ~(x=x) 4,EI
6. Ay[(Fx <-> Fy) -> x=y) 1,UI
7. (Fx <-> Fx) -> x=x 6,UI
8. x=x 7,MP
9. x=x & ~(x=x) 5,8
10. Ax(x=x) 3,9 Proof by contradiction: Cancel Show
11. AxAy[(Fx <-> Fy) -> x=y) -> Ax(x=x) 1,2 (Conditional Proof)
1. AxAy[(Fx <-> Fy) -> {(x=x & y=y) <-> x=y}] Premise
2. Ax(x=x) Assume
3. Show AxAy[(Fx <-> Fy) -> x=y]
4. (x=x) 2,UI
5. (y=y) 2,UI
6. AxAy[(Fx <-> Fy) -> x=y) 1,4,5: Cancel Show
7. 1|- Ax(x=x) -> AxAy[(Fx <-> Fy) -> x=y] 2,3 Conditional Proof
--John
Subject: Re: Quanta and Cakes
> Despite which they are distinguishable: one will be chocolate & the
other
> Ullrich. Accordingly, I can bake both today, both tomorrow, the
chocolate
> one today & the Ullrich one tomorrow, or the Ullrich one today & the
> chocolate one tomorrow. Thus there are four ways I can bake them
> rather than just three, as would be the case if in addition to being
> identical-with-nothings they were *also* indistinguishable.
> The concept 'being an identical-with-nothing' applies to nothing, for
> what is identical with nothing does not exist!
follow that AxAy(qu(x) & qu(y) -> ~(x=y))--and so that
Ax(qu(x) -> ~(x=x)).
Are you suggesting that physicists have latched onto a red herring,
because logic *rules out* quanta?
John
Subject: Re: Quanta and Cakes
> v('Fa & ~E!a') = 0
> There are no true atomic sentences Fa if it is not the case that a
> exists!
According to (iv), every class of non-self-membered sets contains
the empty set, and according to (v) no such set is an identical-
with-something.
iv) Ay[Ax(x in y <-> (Et(x in t) & ~(x in x))) -> 0 in y]
v) Ay[Ax(x in y <-> (Et(x in t) & ~(x in x))) -> Az~(y=z)]
Don't (iv) and (v) attribute properties to classes of non-
self-membered sets?
--John
Subject: Re: Quanta and Cakes
> For example, suppose that I plan
> on baking two cakes--a chocolate cake (yum!) and an Ullrich cake
(ugh!).
> Well, since these cakes aren't around yet, they are
> identical-with-nothings.
> Cakes which arent around yet, i.e. dont
exist (yet) are neither
> identicals-with-nothings nor not identicals- with-nothings,
> because nonexistents (which certainly dont exist!) have no
properties
> whatsoever! Accordingly, nonexistent cakes are not even cakes at
> all, theyre absolutely nothing--not even absolutey
nothing! ;-)
> v('Fa & ~E!a') = 0
> There are no true atomic sentences Fa if it is not the case that a
> exists!
are identical-with-nothings.
C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A)
C4 AxAy[Az(z in x <-> z in y) -> {(set x & set y) <-> x=y}]
(Equi-membered classes are identical iff these are sets.)
Yet among its members every such class counts the empty set.
Are you suggesting that because a class of non-self-membered
sets is an identical-with-nothing, it loses the
property of having the empty set as a member?
If not, what makes you think cakes-to-be (which are also
identical-with-nothings) cannot be discriminated in terms
of *their* properties?
--John
Subject: Re: Quanta and Cakes
> By eg Im self-indistinguishable I mean that there is
absolutely no
> qualitative difference between me and myself--so I cannot be
> distinguished from myself.
Theories of reflexive and non-reflexive identity include the
deductive apparatus of propositional logic, and so the theorem
(p <-> p), of which (Fx <-> Fx) is an instance. Not to worry:
self-identical or not, from yourself you are indistinguishable.
--John
Subject: Re: Quanta and Cakes
> Inextricably linked with individuality through the history of
> philosophy, it is precisely a failure of (self-)identity that is
> the latter remarking that '[w]e are unable to identify individual
> electrons, hence it is meaningless to speak of the self-identity of
> electrons ...' (1970, p. 50). Denying identity is a formally
tractable
> way of representing this notion of non-individuality and indeed
> interesting formal systems can be constructed for doing so.
> [Source: http://www.sorites.org/Issue_06/item3.htm]
It is a fatal mistake to suppose that our inability to identify
> individual electrons has anything to do with every
electrons being
> necessarily self-identical.
> Argument?
> Even if it is empirically impossible for us to identify two electrons
> by saying that
> this one is quantum Laurel and that one is quantum Hardy, we
> nonetheless know that there are altogether two (or much more)
> electrons each of which is self-identical.
> That everything including quanta is self-identical (and necessarily
> so) constitutes logical, i.e. apriorical knowledge.
~(x=y) is consistent both with x=x/y=y and with ~(x=x)/~(y=y).
By means of what faculty of logical intuition do you claim to
know the contrary?
--John
The ~Ax(x=x) project fact sheet: How does a proof that ~Ax(x=x)
differ from a traditional mathematical proof? Traditional
mathematical proofs are written in a way to make them easily
understood by mathematicians. Routine logical steps are
omitted. An enormous amount of context is presupposed
on the part of the reader. Proofs, especially in geometry
and topology, rely on intuitive arguments in situations
where a trained mathematician would be capable of translating
those intutive arguments into a more rigorous argument. In a
proof of ~Ax(x=x), all the intermediate logical steps are
applied. No appeal is made to intuition, even if the
translation from intuition to logic is routine. Thus,
a proof that ~Ax(x=x) is less intuitive, and less susceptible
to logical errors. (Adapted from The Flyspeck Project
Fact Sheet <http://www.math.pitt.edu/~thales/flyspeck/>)
Subject: Re: Perfect Square Polynomial?
> Since y will be an integer if y^2 is a perfect square,
> then x^2 + bx + c must be a perfect square = (x + k)^2 = x^2 + 2kx + k^2
> Hence c must be a perfect square c= k^2 and b = 2k
> in order for y to be an intger
For fixed x, b and c the fact that x^2+b*x+c is a square does not imply
that
b is even and c is a square.
For example: x=1, b=5, c=3, yields x^2+b*x+c = 9.
Subject: Re: Perfect Square Polynomial?
> Since y will be an integer if y^2 is a perfect square,
> then x^2 + bx + c must be a perfect square = (x + k)^2 = x^2 + 2kx +
k^2
> Hence c must be a perfect square c= k^2 and b = 2k
> in order for y to be an intger
> For fixed x, b and c the fact that x^2+b*x+c is a square does not imply
that
> b is even and c is a square.
> For example: x=1, b=5, c=3, yields x^2+b*x+c = 9.
But, if x^2 + b*x + c is a polynomial in x, then x is not fixed.
Subject: Re: WHAT IS THE COINCIDENCE OF THIS was Re: To prove psychic
powers, someone has to sit this test!
> http://uk.groups.yahoo.com/group/HercTest/
> <snip soc.atheism diversion I thought we were going to concetrate on 1
issue. I have a moderated
forum
> for you to play on LInk is above and in my signature.
signed up, waiting for confirm, the issue is the new topic!
> So Eric, I said Rich *would* be a 20 to one coincidence, and Mr
Shewmaker
> would be 20 to 1,
> Yes you said this, but nothing you said backs up these odds. You're just
> pulling these numbers out of your.....er..... hat.
> Do you want me to demonstrate 100 to 1 with Rich Shewmaker?
space left blank intentionally
> So my question to you is :
> How many name options do you think the majority of a sample of average
> people could guess
> the name Rich Shewmaker from his post? hint its ATEAST 3.
> Given a choice of 3 options you would need 10 out of 10 successes to
> demonstrate better than random chance. This from YOUR table of choice
(pun
> intended)
> http://www.automeasure.com/chance.html
I don't have a bold option for ATLEAST 3, caps should work.
> Once you understand the above - when are we going to attempt this on a
> moderated forum.
won't people know who I am and word their posts away from their name?
I could be anonymous.
> How many name options do you think the majority of a sample of average
> people could guess
> the name Rich Shewmaker from his post?
Eric, I don't know what the problem is here, I would really appreciate
after
months of posts about it someone admitting in writing that the post has a
high level of
coincidence.
Is this post a good coincidence or not? How good? Can I measure it? Can
you measure it?
Randi will test you when you properly apply to be tested. Sign up here:
http://www.randi.org/research/challenge.html
--Rich Shewmaker
a/ perfectly ordinary post
b/ slightly odd, maybe a coincidence, the name could be guessed
c/ I know a few better posts where the name gives away the message
d/ its such an oddity the post must be unique with this amazing match
e/ he didn't even have to type anything, he could have got the same message
through just replying with a blank email, implying 'Hey .. the Rich
Show-maker, dude'
f/ I don't have any replies anything like it
Herc
Subject: Re: WHAT IS THE COINCIDENCE OF THIS was Re: To prove psychic
powers, someone has to sit this test!
> http://uk.groups.yahoo.com/group/HercTest/
> <snip soc.atheism diversion I thought we were going to concetrate on 1
issue. I have a moderated
forum
> for you to play on LInk is above and in my signature.
> signed up, waiting for confirm, the issue is the new topic!
Herc, I have only one outstanding applicant for membership at the
moment. Could you please email to this email address:
e_hocking@yahoo.co.uk to confirm the username you are attempting to
join under.
> So Eric, I said Rich *would* be a 20 to one coincidence, and Mr
Shewmaker
> would be 20 to 1,
> Yes you said this, but nothing you said backs up these odds. You're
just
> pulling these numbers out of your.....er..... hat.
> Do you want me to demonstrate 100 to 1 with Rich Shewmaker?
<snip space left blank intentionally
> So my question to you is :
> How many name options do you think the majority of a sample of
average
> people could guess
> the name Rich Shewmaker from his post? hint its ATEAST 3.
> Given a choice of 3 options you would need 10 out of 10 successes to
> demonstrate better than random chance. This from YOUR table of choice
(pun
> intended)
> http://www.automeasure.com/chance.html
> I don't have a bold option for ATLEAST 3, caps should work.
I merely have shown you that for a at least 3 options, you would need
to obtain 10 out of 10 to demonstrate better than random chance,
quoting 1 to 10,000 odds.
For 4 options you need to be right 9 out of 10, 5 options 8 out of 10.
> Once you understand the above - when are we going to attempt this on a
> moderated forum.
> won't people know who I am and word their posts away from their name?
> I could be anonymous.
I've asked you to clarify your claim in another post, but since you
raise this here. I was led to believe that it is *your* postings that
are the influence on the resulting replies and that this is the
magic that you claim to be able to perform. If you wish to post
under an alias, I have no problem with this - just inform me (as
moderator) at e_hocking@yahoo.co.uk of which alias you are trying to
log in as and I will note this.
Given that you are able to post under a pseudonym, can others that
have already corresponded with you on usenet post under a pseudonym
too? Would this effect that you are claiming still be able to work
if, say, Wally Anglesea or myself, were to reply to you using aliases?
> How many name options do you think the majority of a sample of
average
> people could guess
> the name Rich Shewmaker from his post?
> Eric, I don't know what the problem is here, I would really appreciate
after
> months of posts about it someone admitting in writing that the post has a
high level of
> coincidence.
> Is this post a good coincidence or not? How good? Can I measure it?
Can you measure it?
It is merely a coincidence - and a forced word play, at that. You
cannot measure it in the sense that you have been. Really the only
way to measure the success of your guess would have been if a. You had
not seen his post before replying and b. you had chosen his name from
a list of names someone else put together. This is what we are tying
to do on the moderated group. Hope the rules of engagement if drafted
meet with your expected test conditions.
<snip>
Please confirm that Group Member name you want to use to the Yahoo
email below.
I'm pleased that you are going to attempt this. I will no remove all
chocks and start a membership drive so you have a large set of names
and posts to select from. You'll note in the rules I've outlined that
you will have last call on the posts you think you can best guess.
But enough of that - confirm your member name and we'll get on with
the test.
--
Eric Hocking (moderator)
You're invited to test of the paranormal at:
http://uk.groups.yahoo.com/group/HercTest/
e_hocking@yahoo.co.uk
Subject: Re: WHAT IS THE COINCIDENCE of this ONE post ?
> http://uk.groups.yahoo.com/group/HercTest/
> <snip soc.atheism diversion
> I thought we were going to concetrate on 1 issue. I have a moderated
forum
> for you to play on LInk is above and in my signature.
> signed up, waiting for confirm, the issue is the new topic!
> Herc, I have only one outstanding applicant for membership at the
> moment. Could you please email to this email address:
> e_hocking@yahoo.co.uk to confirm the username you are attempting to
> join under.
> So Eric, I said Rich *would* be a 20 to one coincidence, and Mr
Shewmaker
> would be 20 to 1,
> Yes you said this, but nothing you said backs up these odds. You're
just
> pulling these numbers out of your.....er..... hat.
Do you want me to demonstrate 100 to 1 with Rich Shewmaker?
> <snip space left blank intentionally
> So my question to you is :
> How many name options do you think the majority of a sample of
average
> people could guess
> the name Rich Shewmaker from his post? hint its ATEAST 3.
> Given a choice of 3 options you would need 10 out of 10 successes to
> demonstrate better than random chance. This from YOUR table of choice
(pun
> intended)
> http://www.automeasure.com/chance.html
> I don't have a bold option for ATLEAST 3, caps should work.
> I merely have shown you that for a at least 3 options, you would need
> to obtain 10 out of 10 to demonstrate better than random chance,
> quoting 1 to 10,000 odds.
> For 4 options you need to be right 9 out of 10, 5 options 8 out of 10.
I know, but its no use showing 9 out of 10 if you wont mark 1 out of 1 is
it?
Its not the only post but I want this ONE post evaluated 1st.
> Once you understand the above - when are we going to attempt this on
a
> moderated forum.
> won't people know who I am and word their posts away from their name?
> I could be anonymous.
> I've asked you to clarify your claim in another post, but since you
> raise this here. I was led to believe that it is *your* postings that
> are the influence on the resulting replies and that this is the
> magic that you claim to be able to perform. If you wish to post
> under an alias, I have no problem with this - just inform me (as
> moderator) at e_hocking@yahoo.co.uk of which alias you are trying to
> log in as and I will note this.
> Given that you are able to post under a pseudonym, can others that
> have already corresponded with you on usenet post under a pseudonym
> too? Would this effect that you are claiming still be able to work
> if, say, Wally Anglesea or myself, were to reply to you using aliases?
I assume so, the names come to life, qed.
> How many name options do you think the majority of a sample of
average
> people could guess
> the name Rich Shewmaker from his post?
> Eric, I don't know what the problem is here, I would really appreciate
after
> months of posts about it someone admitting in writing that the post has
a high level of
> coincidence.
> Is this post a good coincidence or not? How good? Can I measure it?
Can you measure it?
> It is merely a coincidence - and a forced word play, at that. You
> cannot measure it in the sense that you have been. Really the only
> way to measure the success of your guess would have been if a. You had
> not seen his post before replying and b. you had chosen his name from
> a list of names someone else put together. This is what we are tying
> to do on the moderated group. Hope the rules of engagement if drafted
> meet with your expected test conditions.
A forced word play! Rich Shewmaker talking about James Randi's million
dollar test!
Again, is it a coincidence or not?
Does the post have a *property* that this very post now doesn't?
a. You had > not seen his post before replying and
b. you had chosen his name from > a list of names someone else put
together
Why am *I* the only one allowed to match the names?
Don't say you're the one claiming to have an ability
Im completely outside of the picture, we're judging the post not me.
If you say its coincidental then anyone can test it.
If you say its not coincidental then I'll post it with 100 options to 20
groups to prove otherwise.
Remember soc.atheisms surprise when people DID pick my address from a big
list.
> But enough of that - confirm your member name and we'll get on with
> the test.
of previous replies.
This is the 4th asking : What is the level of coincidece in the Rich
Shewmaker ~ James Randi post?
Answer as a number of how many names it *could* be guessed from, like 1 in
25.
Herc
Subject: Re: WHAT IS THE COINCIDENCE of this ONE post ?
[Note follow-up set to sci.skeptic]
<snip I merely have shown you that for a at least 3 options, you would need
> to obtain 10 out of 10 to demonstrate better than random chance,
> quoting 1 to 10,000 odds.
> For 4 options you need to be right 9 out of 10, 5 options 8 out of 10.
> I know, but its no use showing 9 out of 10 if you wont mark 1 out of 1 is
it?
> Its not the only post but I want this ONE post evaluated 1st.
We're going to have to agree to disagree on this one Herc. This one post
can't (in my opinion) be evaluated in isolation. There is no context in
which to gauge how unique or spook (or whatever) matching his name to
that
singular post is. There is certainly no basis for putting odds on picking
his name either as it's all after the fact. Can we just leave it as you
believe it and I can't see it? I *can* see the point you're trying to
make,
but I can 't think of any way I can mangle odds of coincidence to give
it
any weight, if you can see what you mean.
> Once you understand the above - when are we going to attempt this
on
a
> moderated forum.
> won't people know who I am and word their posts away from their name?
> I could be anonymous.
> I've asked you to clarify your claim in another post, but since you
> raise this here. I was led to believe that it is *your* postings that
> are the influence on the resulting replies and that this is the
> magic that you claim to be able to perform. If you wish to post
> under an alias, I have no problem with this - just inform me (as
> moderator) at e_hocking@yahoo.co.uk of which alias you are trying to
> log in as and I will note this.
> Given that you are able to post under a pseudonym, can others that
> have already corresponded with you on usenet post under a pseudonym
> too? Would this effect that you are claiming still be able to work
> if, say, Wally Anglesea or myself, were to reply to you using aliases?
> I assume so, the names come to life, qed.
We have a game plan. Give me time to build post and membership numbers and
we will have something for you to test your claim in in a *fairly well*
controlled environment. For those lurkers out there - check my sig (and
other posts today) for invitations to join in the fun and test Herc's
claimed skills.
> How many name options do you think the majority of a sample of
average
> people could guess
> the name Rich Shewmaker from his post?
> Eric, I don't know what the problem is here, I would really
appreciate
after
> months of posts about it someone admitting in writing that the post
has a high level of
> coincidence.
> Is this post a good coincidence or not? How good? Can I measure
it?
Can you measure it?
> It is merely a coincidence - and a forced word play, at that. You
> cannot measure it in the sense that you have been. Really the only
> way to measure the success of your guess would have been if a. You had
> not seen his post before replying and b. you had chosen his name from
> a list of names someone else put together. This is what we are tying
> to do on the moderated group. Hope the rules of engagement if drafted
> meet with your expected test conditions.
> A forced word play! Rich Shewmaker talking about James Randi's million
dollar test!
> Again, is it a coincidence or not?
It is a coincidence - but merely that. The problem is that I don't put the
significance on this coincidence that you do - even more so because (to me)
the forced word play to my mind is done post-hoc. Much like that sort
of
numerology where letrers are replaced by numbers and muddled together to
get
666 or similar. In my view, doing this AFTER the fact is less impressive
that BEFORE the fact.
> Does the post have a *property* that this very post now doesn't?
> a. You had > not seen his post before replying and
> b. you had chosen his name from > a list of names someone else put
together
But I hadn't chosen his name from a list - you gave me both the list and
the
result in this thread.
> Why am *I* the only one allowed to match the names?
> Don't say you're the one claiming to have an ability
I must misunderstand then, I thought that was you claim, that you can do it
better than most AND better than random chance. I have to admit I can't
see this one - thus my forced word play comment. No matter, we're
getting
somewhere with the moderated group where you can better demonstrate this
effect, hopefully.
> Im completely outside of the picture, we're judging the post not me.
> If you say its coincidental then anyone can test it.
> If you say its not coincidental then I'll post it with 100 options to 20
groups to prove otherwise.
Let's not get ahead of ourselves, and especially not raise the ire of 20
groups by posting off-topic, can we put this particular example down as
no going to be resolved and move on?
> Remember soc.atheisms surprise when people DID pick my address from a big
list.
I wasn't there to see it happen, so no, I don't remember it.
> But enough of that - confirm your member name and we'll get on with
> the test.
> All for testing, I'm also against ignoring perfectly fine recordings by a
> of previous replies.
> This is the 4th asking : What is the level of coincidece in the Rich
Shewmaker ~ James Randi post?
> Answer as a number of how many names it *could* be guessed from, like 1
in
25.
Chances of picking Rich's name from 1 single post is 1 in 25, given 25
options (which was not the case), but a single coincidence proves little.
What you need to show, and the test on the moderated group hopes to show,
is
that a consistent ability to force these coincidences with a result that is
better than random chance, will put you on the road to making fools of the
Australian Sceptics Soc. and JREF (just to mention the first two). To do
that, with 25 options per post, you'd have to acheive a score of 4 out of 5
such posts.
It is this demonstration of coincidence greater than random chance that
counts.
--
Eric Hocking
You're invited to test the paranormal at:
http://uk.groups.yahoo.com/group/HercTest/
e_hocking@yahoo.co.uk
Subject: Re: To prove psychic powers, someone has to sit this test!
>< crackpottery and dingbattery deleted It is no wonder why you are part of
the long term unemployed.
> Let's not forget the long term unlaid.
I would say permanantly unlaid. But then again I am onre of those
nasty old skeptics...;-)
Subject: Re: I am a freshman of CST,which book of maths should I read
first?
ETAsAhRuqnPsmdSdc1BmJK4bzZTuy/SHqAIUObscmV3QQwLJv12R5ibp1l/ZJ30=
Lee@Sunnytea.com (Lee), Wrote:
>I am a freshman of computer science and
>technology. I came across a problem.That is I
>don't know which book should I read first,which
>should I read second......
Um, which book is the professor going to recite from first?
>I have three books right here.They are
><Discrete Maths>,<Combination
>Maths>,<Concrete Maths>.
Well, make sure you read elementary algebra first. Then take
advantage of any courses, materials, lectures, et.al on how to study in
the most efficient manner. <most *important*, trolling aside> This will
lessen the dangers to your mental health come exam time. Anyway,
http://dimacs.rutgers.edu/News/
it seems discreet is the most important, followed by calculus. I would
guess that reading a bit extra on Logic would be helpful as well since
my Webtv is *still* smarter than my almost new Hal in the next room
gathering dust. If you understand why the last sentence is true and
correct then you're on your way to being a competent computer scientist.
Hell, you might actually invent something other than the spam ridden,
virus infected, mentally retarded concept commonly known as a
computer. And...and when you make that first self-aware HAL, please
remember me when HAL figures out how to use nanotech for repairing
cooperation in this matter.
blackboard the square root of zero + apple pie x infinity^2 divided by
that elusive prime number sequence = an imaginary pink elephant just
take a deep breath and accept.
Of course, you could actually take on a more rigorous academic schedule.
http://womens-studies.rutgers.edu/undergraduate/major.html
--
Professor Bunn E. Rabbit, PhD, DVM,
Animal Intelligence Enhancement Program
>If you know,please help me!
_____
Cosmic upheaval is not so moving as a little child pondering the death
of a sparrow in the corner of a barn. -Anouk Aimee, French Actor
_____
Death is better, a milder fate than tyranny, Aeschylus (525BC-456BC),
Agamemnon
_____
I wear no Burka. - Mother Nature
----------
The mailbox, BunnERabbit@webtv.net has been circumvented to fight spam.
To send mail... substitute ModerateMammal
----------
Subject: Re: I am a freshman of CST,which book of maths should I read
first?
> I am a freshman of computer science and technology.
> I came across a problem.That is I don't know which book should I read
> first,which should I read second......
> I have three books right here.They are <Discrete Maths>,<Combination
> Maths>,<Concrete Maths>.
> If you know,please help me!
Well, that greatly depends upon where you live: If you live in North
America, unless you plan living and working in India or Brazil, I suggest
you alter your course of study to something that cannot be exported.
--
Felony case 02-CR-0617 9/1/03: Oregon Department of
Justice V. Raymond Ronald Karczewski, Defendant.
The defendant's name is NOT copyrighted.
Subject: Re: I am a freshman of CST,which book of maths should I read
first?
> I am a freshman of computer science and technology.
> I came across a problem.That is I don't know which book should I read
> first,which should I read second......
> I have three books right here.They are <Discrete Maths>,<Combination
> Maths>,<Concrete Maths>.
> If you know,please help me!
Pick up an English text.
HTH,
Jim
Subject: Re: I am a freshman of CST,which book of maths should I read
first?
>> I am a freshman of computer science and technology.
>> I came across a problem.That is I don't know which book should I read
>> first,which should I read second......
>> I have three books right here.They are <Discrete Maths>,<Combination
>> Maths>,<Concrete Maths>.
>> If you know,please help me!
>Pick up an English text.
>HTH,
>Jim
Concrete doesn't take as much Maths, as it does trial and error. You
have to learn how to adjust the sand/gravel/concrete/water ratio,
depending on what you plan to do with it, existing moisture content of
the dry ingredients, acceptable curing times, etc. You can start
with a basic formula, but you still have to know what it's supposed
to look like, sort of like with bread dough.
--
V.G.
People are more violently opposed to fur than leather, because it is
easier to harrass
rich women than it is motorcycle gangs. - Bumper Sticker
Sarcasm is my sword, Apathy is my shield.
Subject: Re: I am a freshman of CST,which book of maths should I read
first?
>> I am a freshman of computer science and technology.
>> I came across a problem.That is I don't know which book should I read
>> first,which should I read second......
>> I have three books right here.They are <Discrete Maths>,<Combination
>> Maths>,<Concrete Maths>.
>> If you know,please help me!
>Pick up an English text.
>HTH,
>Jim
> Concrete doesn't take as much Maths, as it does trial and error. You
> have to learn how to adjust the sand/gravel/concrete/water ratio,
> depending on what you plan to do with it, existing moisture content of
> the dry ingredients, acceptable curing times, etc. You can start
> with a basic formula, but you still have to know what it's supposed
> to look like, sort of like with bread dough.
This would be a good place to start for combination maths:
http://tinyurl.com/ptgy
HTH,
Jim
Subject: Re: I am a freshman of CST,which book of maths should I read
first?
>I am a freshman of computer science and technology.
>I came across a problem.That is I don't know which book should I read
>first,which should I read second......
>I have three books right here.They are <Discrete Maths>,<Combination
>Maths>,<Concrete Maths>.
>If you know,please help me!
It doesn't make a difference. Mathematics is not linear. You might try
looking at all three simultaneously, or read one and refer to the others
if you can't resolve somthing on your own.
--
Stephen J. Herschkorn herschko@rutcor.rutgers.edu
Subject: Re: I am a freshman of CST,which book of maths should I read
first?
> I am a freshman of computer science and technology.
> I came across a problem.That is I don't know which book should I read
> first,which should I read second......
> I have three books right here.They are <Discrete Maths>,<Combination
> Maths>,<Concrete Maths>.
> If you know,please help me!
Anything you`ll be able to read and what moves you at least a little big
higher...
Kalus Petr
Subject: Re: metoda najmanjih kvadrata / least squares method
> molim vas ako mi tko moze pomoci, treba mi tocna definicija za metodu
> najmanjih kvadrata za aproksimaciju empirickih podataka - linearnom
> funkcijom
> - kvardatnom
> funkcijom
> please if somebody can help me, I need right definition of least
> squares method for approximation of empiric data - with linear
> function
> - with square function
http://mathworld.wolfram.com/LeastSquaresFitting.html
Phil
Subject: prob joint density function question
Question 2.73 p.71:
Let f(x,y) = {1 0<=x<=1, 0<=y<=1; 0 otherwise
be the joint density function of X and Y. Find the density function
of Z = XY.
My work:
Instead of using U and V as in previous problems, I used U and Z.
So Let U=X. Z = XY given. Y=Z/X so Y=Z/U.
Setting up the partials for the Jacobian:
dx/du=1 dx/dz=0
dx/dz= - dy/dz=1/u
J=1/u
So I get:
g(u,z) = { 1/u 0<=u,z<=1; 0 otherwise
They want the marginal:
g(z) = Integral(1/u) du
I didn't know what parameters to use. The answer in the book is
g(z)={-ln z 0<z<1; 0 otherwise
So I think the parameters they used were 1,z, but how did they get
those, and in that order?
Subject: Re: How long must physics put up w/f=ma?
Cut<
> We better stick with the good old newtonian
>physics!
> All the best
> Laszlo Lemhenyi,Toronto
With just a tad of Shead's wisdom; _yes_:
Newton's Second Law according to:
<http://scienceworld.wolfram.com/physics/NewtonsSecondLaw.html> :
A force F acting on a body gives it an acceleration a which is in the
direction of the force and has magnitude inversely proportional to the mass
m of the body: F = ma.
But it's only the _net_ force that causes an acceleration; so we'd _ought_
to
use (F-uw) = wa/g; so substituting 'w/g' for 'm'; as I put it, the _net_
force [F-uw = f] exerted on
and/or by a body gives it an acceleration [a] which is in the direction of
the _net_ force and has magnitude inversely proportional to the mass m of
the body: f = wa/g.
The 'uw' is the product of a coefficient of restraint and the
weight of the body. Coefficients of frictional resistance are usually are
quite low - less than one [1] - but the coefficient of restraint for a
vertical concrete abutment wall is _many_ times greater.
Subject: Re: How long must physics put up w/f=ma?
Cut<
> We better stick with the good old newtonian
>physics!
> All the best
> Laszlo Lemhenyi,Toronto
With just a tad of Shead's wisdom; _yes_:
Newton's Second Law according to:
<http://scienceworld.wolfram.com/physics/NewtonsSecondLaw.html> :
A force F acting on a body gives it an acceleration a which is in the
direction of the force and has magnitude inversely proportional to the mass
m of the body: F = ma.
But it's only the _net_ force that causes an acceleration; so we'd _ought_
to
use (F-uw) = wa/g; so substituting 'w/g' for 'm'; as I put it, the _net_
force [F-uw = f] exerted on
and/or by a body gives it an acceleration [a] which is in the direction of
the _net_ force and has magnitude inversely proportional to the mass m of
the body: f = wa/g.
The 'uw' is the product of a coefficient of restraint and the
weight of the body. Coefficients of frictional resistance are usually are
quite low - less than one [1] - but the coefficient of restraint for a
vertical concrete abutment wall is _many_ times greater.
Subject: Re: modulo proof of FLT?
> I dont think FLT can be proven only by modulo exercising, and you cannot
> prove the opposite either:
> Now, if x mod y = z
> then it is known that x^r mod y = z^r
> This works only in the direction you mentioned, not in the other
> Therefore (b mod a)^n = (c mod a)^n
> This is wrong: 2^2mod5=4=3^2mod5 but (2mod5)^2=4!=9=(3mod5)^2
This is wrong. Modulo 5, 4==9.
Phil
Subject: Re: modulo proof of FLT?
> the math seems too simple, but the result seems certain, if I went
> wrong, please let me know:
> Assume that there is a solution for a^n + b^n = c^n (a,b,c,n positive
> integers and n greater than 2) - please excuse the lack of proper
> notation, I am in Italy and the keyboard lacks signs in this cafe
Hint - for modular relations, to contrast them against normal equalities,
you can use == rather than =.
> then (a^n + b^n) mod a = c^n mod a
> if follows that b^n mod a = c^n mod a
> Now, if x mod y = z
> then it is known that x^r mod y = z^r
> Therefore (b mod a)^n = (c mod a)^n
> and for _odd_ n, b mod a = c mod a (this is not true for even because
> positive and negative raised to even powers can be equal)
Nope, e.g. 3^3 == 5^3 (mod 7) but 3 !== 5 (mod 7)
Phil
Subject: Re: modulo proof of FLT?
Visiting Assistant Professor at the University of Montana.
>the math seems too simple, but the result seems certain, if I went
>wrong, please let me know:
>Assume that there is a solution for a^n + b^n = c^n (a,b,c,n positive
>integers and n greater than 2) - please excuse the lack of proper
>notation, I am in Italy and the keyboard lacks signs in this cafe
>then (a^n + b^n) mod a = c^n mod a
>if follows that b^n mod a = c^n mod a
>Now, if x mod y = z
>then it is known that x^r mod y = z^r
>Therefore (b mod a)^n = (c mod a)^n
>and for _odd_ n, b mod a = c mod a (this is not true for even because
>positive and negative raised to even powers can be equal)
It's also not true for odd n in modular arithmetic.
For example, 2^3 = 1^3 (mod 7).
Subject: Re: Test |-|erc (was Re: A test proposal for Herc)
Herc,
Just so we all know what we are attempting here, can you please
clarify your claimed powers. Please be succinct as the intent is to
use it as an introduction to the moderated Yahoo created for your
benefit. I will also use it in usenet posts to invite participants to
build a list of options for your post/author matching magic.
Here is my understanding of what you claim you do.
You claim that by posting a message you can induce a response that
will show characteristics in the message content whereby you can
divine the poster's name give a list of alternatives.
I believe that is it in a nutshell?
For the purposes of clarifing what you can do, I don't want to discuss
the source or nature of your paranormal power, only to understand
the result.
ie, you post. Someone responds. The content of their post can be
matched to their name.
> I have begun an Yahoo group specifically to test your power Herc.
> http://uk.groups.yahoo.com/group/HercTest/
> It is a members only, moderated forum.
> You will have to sign up as a Yahoo member to participate, but very
little
> personal data is required for this.
> Anyone else interested feel free to join up, if Herc agrees to at least
> attempt this.
--
Eric Hocking
You're invited to test the paranormal at:
http://uk.groups.yahoo.com/group/HercTest/
please respond to e_hocking@yahoo.co.uk
Subject: Re: Test |-|erc (was Re: A test proposal for Herc)
> Herc,
> Just so we all know what we are attempting here, can you please
> clarify your claimed powers. Please be succinct as the intent is to
> use it as an introduction to the moderated Yahoo created for your
> benefit. I will also use it in usenet posts to invite participants to
> build a list of options for your post/author matching magic.
> Here is my understanding of what you claim you do.
> You claim that by posting a message you can induce a response that
> will show characteristics in the message content whereby you can
> divine the poster's name give a list of alternatives.
> I believe that is it in a nutshell?
> For the purposes of clarifing what you can do, I don't want to discuss
> the source or nature of your paranormal power, only to understand
> the result.
> ie, you post. Someone responds. The content of their post can be
> matched to their name.
pretty good, the magic is inducing the response, so there's no divining
necessary,
its just analytical work to figure out the name, anyone can do it once the
posts are available, but since you've set it up so I have to deduce them
myself divine is apt. for this claim.
Herc
> I have begun an Yahoo group specifically to test your power Herc.
> http://uk.groups.yahoo.com/group/HercTest/
> It is a members only, moderated forum.
> You will have to sign up as a Yahoo member to participate, but very
little
> personal data is required for this.
> Anyone else interested feel free to join up, if Herc agrees to at least
> attempt this.
> --
Subject: Re: Test |-|erc (was Re: A test proposal for Herc)
[follow up set to sci.skeptic]
I can only presume that a.f.a-b and s.math are not interested?
If anyone from these newsgroups ARE interested, I'll post invitations to
join us in the moderated group. Please reply to this post if anyone on your
groups want to see this OT invite.
> Herc,
> Just so we all know what we are attempting here, can you please
> clarify your claimed powers. Please be succinct as the intent is
to
> use it as an introduction to the moderated Yahoo created for your
> benefit. I will also use it in usenet posts to invite participants to
> build a list of options for your post/author matching magic.
> Here is my understanding of what you claim you do.
> You claim that by posting a message you can induce a response that
> will show characteristics in the message content whereby you can
> divine the poster's name give a list of alternatives.
> I believe that is it in a nutshell?
> For the purposes of clarifing what you can do, I don't want to discuss
> the source or nature of your paranormal power, only to understand
> the result.
> ie, you post. Someone responds. The content of their post can be
> matched to their name.
> pretty good, the magic is inducing the response, so there's no divining
necessary,
> its just analytical work to figure out the name, anyone can do it once
the
> posts are available, but since you've set it up so I have to deduce them
> myself divine is apt. for this claim.
OK. The game is on, as the saying goes. We seem to have a basis to
progress this test.
For the others - Herc applied to join the moderated Yahoo group - but I
banned his user ID before I'd signed him up. A little too zealous with my
security measures there I'm afraid.
My next post will be to advertise the draft of the rules of engagement here
in sci.skeptic (cc'd to Herc) and then go on a recruitment drive for
participants.
The rest of you please feel free to join us by following the link in my
sig.
--
Eric Hocking
You're invited to test of the paranormal at:
http://uk.groups.yahoo.com/group/HercTest/
e_hocking@yahoo.co.uk
Subject: Re: Fermat's Last Theorem/Andrew Wiles
Read: Fermat's Last Theorem by Simon Singh
>You might find Algebraic Number Theory and FLT by Ian Stewart and
David
>Tall (3rd edition A K Peters 2002) helpful.
>Guy Corrigall
>> Hi All,
>> I'm a sincere math amateur, and I saw a series on PBS about Andrew
Wiles'
>> attempt to solve Fermat's Last Theorem. If I recall, he was close, but
>that
>> an associate pointed out a fundamental flaw in his theory.
>> Does anyone know if he fixed the theorem -- and is it now considered
>> proved? Also, can anyone recommend a good book with regard to Wiles'
>> attempt to solve the problem over 7 years?
Subject: Re: Fermat's Last Theorem/Andrew Wiles
Read: Fermat's Last Theorem by Simon Singh
>You might find Algebraic Number Theory and FLT by Ian Stewart and
David
>Tall (3rd edition A K Peters 2002) helpful.
>Guy Corrigall
>> Hi All,
>> I'm a sincere math amateur, and I saw a series on PBS about Andrew
Wiles'
>> attempt to solve Fermat's Last Theorem. If I recall, he was close, but
>that
>> an associate pointed out a fundamental flaw in his theory.
>> Does anyone know if he fixed the theorem -- and is it now considered
>> proved? Also, can anyone recommend a good book with regard to Wiles'
>> attempt to solve the problem over 7 years?
Subject: Re: Two coin flip/ clarification for C Bond
>I'll toss in my 2 cents.
>Thanx for your extreme sacrifice. The bad news is that with inflation,
>it's already worth less.
>>Dang, I'll dig up some nickels for next time.
<snip
>Eldon, may I suggest:
>>1) telling us the rules of the game.
>>2) determining what you think should happen.
>>3) run an experiment to see if your prediction appears to be correct.
>>
<snip
> Suppose that HH landed first and we made the statement, Two coins
> were flipped and at least one is a head. What are the chances for two
> heads? Bill bet for two heads, Jane covered the bet. Bill put up a
> dollar and he will win. How much should Jane pay?
> 1 dollar. She covered his bet. If he wins, she loses.
Yes, but with true odds, if the answer to that question is 1/3, she
should pay two to one.
Maybe I should have said, with posted odds of two to one, should Jane
cover the bet?
OR:
Suppose that TT landed first and we made the statement, Two coins
> were flipped and at least one is a Tail. What are the chances for two
> tails? Bill bet for two tails, Jane covered the bet. Bill put up a
> dollar and he will win. How much should Jane pay?
> Same as before.
So Jane loses with HH, and with TT. She wins with HT and TH, no matter
which statement is made. She wins two out of four, she loses two out
of four, she has an even money bet. If the odds are even, she should
cover the bet, if the odds are two to one, she should not.
> Eldon: Which statement is made when TH comes up? How about HT? You still
> haven't explained how the betting works. There is a difference between
> whether the game is a *fair* game and how much Jane should pay for a
> *particular* game.
Bill always bets for the two. Jane covers. All we have to do to figure
the correct odds is answer the question. All she has to do to make the
proper decision is answer the question. The answer is 1/2, with even
odds she can cover, with two to one odds she should demur.
With HT, or TH, Bill wins and Jane loses, no matter whether the
'heads' or the 'tails' statement is made.
With no prejudice toward either heads or tails, the 'heads' statement
and the 'tails' statement are equally likely at HT and TH. Enter a
prejudice toward one and it detracts from the other.
With extreme prejudice toward one, it's answer is 1/3, but the other
answers 1.
When the 'heads' question and the 'tails' question have equal answers,
the correct answer is 1/2. The secret is in the prejudice, not some
'magic' in the at least one is statement.
Eldon
Subject: Re: Two coin flip/ clarification for C Bond
>>Eldon, may I suggest:
>>1) telling us the rules of the game.
>>2) determining what you think should happen.
>>3) run an experiment to see if your prediction appears to be correct.
> <snipSuppose that HH landed first and we made the statement, Two coins
>were flipped and at least one is a head. What are the chances for two
>heads? Bill bet for two heads, Jane covered the bet. Bill put up a
>dollar and he will win. How much should Jane pay?
>>1 dollar. She covered his bet. If he wins, she loses.
> Yes, but with true odds, if the answer to that question is 1/3, she
> should pay two to one.
> Maybe I should have said, with posted odds of two to one, should Jane
> cover the bet?
She should pay two to one if you want the game to be fair to Bill. That
is a seperate issue of what the probability is that Bill wins.
>OR:
>Suppose that TT landed first and we made the statement, Two coins
>were flipped and at least one is a Tail. What are the chances for two
>tails? Bill bet for two tails, Jane covered the bet. Bill put up a
>dollar and he will win. How much should Jane pay?
>>Same as before.
> So Jane loses with HH, and with TT. She wins with HT and TH, no matter
> which statement is made. She wins two out of four, she loses two out
> of four, she has an even money bet. If the odds are even, she should
> cover the bet, if the odds are two to one, she should not.
If the odds are two to one, it is a fair game. Whether she covers the
bet is irrelevent.
>Eldon:>Which statement is made when TH comes up? How about HT? You still
>>haven't explained how the betting works. There is a difference between
>>whether the game is a *fair* game and how much Jane should pay for a
>>*particular* game.
> Bill always bets for the two. Jane covers. All we have to do to figure
> the correct odds is answer the question. All she has to do to make the
> proper decision is answer the question. The answer is 1/2, with even
> odds she can cover, with two to one odds she should demur.
You are now doing a very different problem from the one originally
stated, and it still isn't clearly stated.
Bill and Jane are playing a game in which a third person flips two
coins, looks at one, and announces whether it is heads or tails. Bill
will then bet one dollar that the uncalled coin is the same as the
called coin. How much should Jane pay if she loses to make this a fair
game?
> With HT, or TH, Bill wins and Jane loses, no matter whether the
> 'heads' or the 'tails' statement is made.
This is backwards from the above.
> With no prejudice toward either heads or tails, the 'heads' statement
> and the 'tails' statement are equally likely at HT and TH. Enter a
> prejudice toward one and it detracts from the other.
> With extreme prejudice toward one, it's answer is 1/3, but the other
> answers 1.
> When the 'heads' question and the 'tails' question have equal answers,
> the correct answer is 1/2. The secret is in the prejudice, not some
> 'magic' in the at least one is statement.
The secret is in correctly setting up the sample space. S={HH,HT,TH,TT}
does not accurately reflect the situation unless you consider the pairs
to be {(called)(uncalled)}. Since this is not the way you appear to be
thinking about it, the sample space will be S={(first)(second)(called)}
= {HH1,HH2,HT1,HT2,TH1,TH2,TT1,TT2}
Bill wins on exactly 4, Jane wins on 4, so each has a probability of
winning of 1/2, and Jane's bet should equal Bill's.
The problem with all this is the following:
Your original question was Two coins were flipped and at least one is a
tail. What are the chances for two tails?
In this case the sample space is S={HH,HT,TH,TT}.
at least one is a tail={HT,TH,TT}.
two tails={TT}.
P(two tails | at least one is a tail)
= P(two tails and at least one tail) / P(at least one tail)
= (1/4) / (3/4) = 1/3
Here's the key. These two problems are *not* equivalent. By adding the
betting, and the assumption that someone is looking at *one* coin and
calling it, rather than looking at both, you have changed the problem
from what was originally intended.
--
Will Twentyman
email: wtwentyman at copper dot net
Subject: Continuation of linear, bounded mappings on Banach spaces
I've got a question on functional analysis:
Assume we have two Banach spaces X and Y. Let X' be a dense, linear
subspace of X. Further we have a bounded, linear mapping T':X'->Y. The
statement is then that we have a unique continuation T of T' to X which
is also bounded and linear (T:X->Y).
Can we proof that statement without presuming that T is linear or bounded?
Mark
Subject: Re: Continuation of linear, bounded mappings on Banach spaces
>I've got a question on functional analysis:
>Assume we have two Banach spaces X and Y. Let X' be a dense, linear
>subspace of X. Further we have a bounded, linear mapping T':X'->Y. The
>statement is then that we have a unique continuation T of T' to X which
>is also bounded and linear (T:X->Y).
>Can we proof that statement without presuming that T is linear or bounded?
If the question is can we prove it assuming nothing except that
T':X'->Y the answer is of course not.
If the question is whether there are other hypotheses we could use:
What's important is that T' is uniformly continuous on bounded subsets
of X'; this follows from assuming it's linear and bounded.
>Mark
************************
Subject: Re: History of blackboard bold?
> Of course this way of writing was devised for writing by hand (on a
> blackboard, for example); and then adapted for use by a typewriter
> (those ancient machines used before personal computers for producing
> documents in a home or office environment). Produce the blackboard
> bold R by typing an I and an R close together.
I agree this style must have originated as a way to produce bold
letters by hand on paper or blackboard. As an attempt to narrow the
date of origin I recall not seeing this as an undergraduate at a
mathematically sophisticated school in the early 1960's but
encountering it as a graduate student in the mid-60's. But I imagine
something like this was used in Europe or Germany, especially, before
that.
Subject: Re: History of blackboard bold?
>> Of course this way of writing was devised for writing by hand (on a
>> blackboard, for example); and then adapted for use by a typewriter
>> (those ancient machines used before personal computers for producing
>> documents in a home or office environment). Produce the blackboard
>> bold R by typing an I and an R close together.
>I agree this style must have originated as a way to produce bold
>letters by hand on paper or blackboard. As an attempt to narrow the
>date of origin I recall not seeing this as an undergraduate at a
>mathematically sophisticated school in the early 1960's but
>encountering it as a graduate student in the mid-60's. But I imagine
>something like this was used in Europe or Germany, especially, before
>that.
I remember noticing as an undergraduate math major (1967-1971) that
the books that I read and that used a distinctive font for these sets,
all used some sort of ordinary bold (upright or slanted). While all my
teachers used blackboard bold on the blackboard. I jumped to the
conclusion that BBB was simply a way to make a bold-like distinction
in a medium where it was not otherwise possible. I have always felt
don't have any other bold C in them and a simple bold letter C would
be quite sufficient.
By the way, none of my teachers added more than one extra stroke to
their ordinary uppercase letters to produce BBB. The example BBB in
the TeXBook (my edition anyway) is pretty obviously a CMR I and R
run together. (I thing it looks better than any currently available
doublestroke font and have considered adapting cmr to produce something
along those lines.)
Dan
--
Dan Luecking Department of Mathematical Sciences
University of Arkansas Fayetteville, Arkansas 72701
luecking at uark dot edu
Subject: Re: History of blackboard bold?
<a3e3ov4buqq51viei3hqh7blnohm64lt1l@4ax.com By the way, none of my teachers
added more than one extra stroke to
> their ordinary uppercase letters to produce BBB. The example BBB in
> the TeXBook (my edition anyway) is pretty obviously a CMR I and R
> run together. (I thing it looks better than any currently available
> doublestroke font and have considered adapting cmr to produce
> something along those lines.)
You mean like the bbm.sty and fonts do?
--
David Kastrup, Kriemhildstr. 15, 44793 Bochum
UKTUG FAQ: <URL:http://www.tex.ac.uk/cgi-bin/texfaq2html>
Subject: Re: History of blackboard bold?
> The old msym is like the Mathematical Pi quoted, while the new msbm is
> like the Eudlid Math 2.
This contradicts Barbara Beeton's claim in
I quote:
> i assure you that the doubled strokes in msym, in the sti openface
> font, and in msym are all in exactly the same places -- i checked
> all three from contemporaneous source documents before replying.
> i will be happy to produce the evidence, which is part of my
> permanent library.
Is there any reliable way to check for sure? I have access to
libraries, it's just knowing what to look for... Are there
any journals that indisputably used msym during some period?
of this one!
Robin
Subject: Re: History of blackboard bold?
> The old msym is like the Mathematical Pi quoted, while the new msbm is
> like the Eudlid Math 2.
> This contradicts Barbara Beeton's claim in
> I quote:
> i assure you that the doubled strokes in msym, in the sti openface
> font, and in msym are all in exactly the same places -- i checked
> all three from contemporaneous source documents before replying.
> i will be happy to produce the evidence, which is part of my
> permanent library.
> Is there any reliable way to check for sure? I have access to
> libraries, it's just knowing what to look for... Are there
> any journals that indisputably used msym during some period?
> of this one!
> Robin
<http://www.math.ohio-state.edu/~edgar/BBBfonts.gif>
Don't you agree that msym looks like Mathematical Pi? (The little
swish on the Q is maybe different.) And msbm is the same as Euclid
Math 2.
--
G. A. Edgar
http://www.math.ohio-state.edu/~edgar/
Subject: Re: History of blackboard bold?
> <http://www.math.ohio-state.edu/~edgar/BBBfonts.gif>
Barbara's remark: she was talking only about the _location_
of the doubled strokes, which is indees the same as in msbm.
(This is different from Alan Jeffrey's bbold font, for
example, which seems to be what's shown in
http://www.w3.org/TR/MathML2/double-struck.html)
> Don't you agree that msym looks like Mathematical Pi?
Yes, absolutely. So perhaps msbm was indeed the first
use of the inline style for blackboard bold. That would
be surprising, since inline fonts certainly predate
blackboard bold: for example
http://www.myfonts.com/fonts/linotype/cloister-open-face/cloister-open-face/

testdrive.html?s=N+Q+R+Z+C&p=72
is a font designed in 1929.
One would expect printers to use the fonts they had already
available, where possible.
Robin
Subject: Re: History of blackboard bold?
>> <http://www.math.ohio-state.edu/~edgar/BBBfonts.gifBarbara's remark: she
was talking only about the _location_
>of the doubled strokes, which is indees the same as in msbm.
>(This is different from Alan Jeffrey's bbold font, for
>example, which seems to be what's shown in
>http://www.w3.org/TR/MathML2/double-struck.html)
>> Don't you agree that msym looks like Mathematical Pi?
>Yes, absolutely. So perhaps msbm was indeed the first
>use of the inline style for blackboard bold.
When you say inline do you perhaps mean outline? I have never heard
inline used in this sense.
Dan
--
Dan Luecking Department of Mathematical Sciences
University of Arkansas Fayetteville, Arkansas 72701
luecking at uark dot edu
Subject: Re: History of blackboard bold?
> When you say inline do you perhaps mean outline? I have
> never heard inline used in this sense.
Let me quote from the typographic glossary in Robert
Bringhurst's excellent book _The Elements of Typographic Style_:
Inline: A letter in which the inner portions of the main
strokes have been carved away, leaving the edges more
or less intact. Inline faces lighten the color while
preserving the shapes and proportions of the original
face. _Outline_ letters, on the other hand, are produced
by drawing a line around the outsides of the letters
and removing the entire original form. Outline letters,
in consequence, are fatter than the originals and have
less definition.
Robin
Subject: Re: History of blackboard bold?
><http://www.math.ohio-state.edu/~edgar/BBBfonts.gifDon't you agree that 
msym
looks like Mathematical Pi? (The little
>swish on the Q is maybe different.) And msbm is the same as Euclid
>Math 2.
you're right, of course. i remember thinking that msbm was a great
improvement, at the time it was released, but over the years there
have been people hankering after the shapes of msym. looking at them
side by side again, after all this time, i realise i was right (for my
value of better) and they were wrong ;-)
--
Robin (the partially spineless) Fairbairns, Cambridge
Subject: Re: History of blackboard bold?
>I'm trying to trace the use of blackboard bold
>symbols (Z, N, Q, R, ...) in printed mathematics.
...
>Does anyone have any information about when and where
>these symbols were first used in print, and which
>style was first printed?
[After writing all the following blather, I noticed
that the only part which is truly responsive to the
question asked is something not completely unlike
your first set of examples appeared from Prentice-Hall
in 1965, and I find no earlier instances in my home
library. Read the rest at your peril.]
It is my impression that such symbols were first used
in polycopied/mimeographed notes, and similar materials
prepared on typewriters. Although (as I recently mentioned
in another thread in c.t.t) there was at least one office
typewriter (an Olympia) for sale by the mid 1960s with a
built-in poor man's bold doublestrike function, for a
long time--until the era of Typ-It insertible type bars,
and not too much later the IBM Executive and Selectric
lines of typewriters--about the best that could easily
be done to fake bold was to overstrike an uppercase I
(or a single quote) on your R or C. Given a Selectric,
one had the further option of doublestriking your R, C,
Z, Q, or N with a slight offset (this involved manual
intervention with the device that carried the golf-ball
type element, and was very hard to do consistently).
The typed notes prepared by typists in the Princeton
mathematics department sometime in the late 1950s,
and published by D. Van Nostrand Company in 1959 as
Advanced Calculus by Nickerson, Spencer, and Steenrod,
show no effort to fake bold in any way; Typ-It (or its
equivalent) has been used for various symbols like
union and in. The typed notes from the same source,
prepared in 1961 and published in 1962 by Princeton
University Press as Lectures on Modular Forms
(Annals of Mathematics Study 48) by Gunning is also
devoid of boldfakery, and has assorted Type-It symbols.
By 1966, however, Gunning (in Lectures on Riemann Surfaces,
PUP Princeton Mathematical Notes) uses doublestruck C and R;
the typewriter is almost surely a Selectric (and thus, for
instance, script letters--which were typed with Typ-It in
Lectures on Modular Form--have been written in by hand).
Gunning very much liked the use of blackboard bold
(I believe it was from him that I learned to use it,
at just around that time); his 1965 textbook (with Rossi),
Analytic Functions of Several Complex Variables, in
the Prentice-Hall Series in Modern Analysis, produced from
notes written in 1960-1962, is typeset with a *realllly*
*ugly* mathbb{C} (it must be seen to be believed), more
in the Mathematical Pi than the Euclid Math Two style.
Going back to typescript notes: Serre's 1964 lectures at
Harvard, Lie Algebras and Lie Groups, printed from (surely
author-supplied, camera-ready) typescript by W. A. Benjamin,
Inc., in 1965, uses a double-underline (handwritten) beneath
C and Z to denote the complex numbers and the integers,
respectively; Narasimhan's 1966 Introduction to the Theory
of Analytic Spaces, printed from (author-supplied, camera-
ready) typescript as Springer LNM 25, looks like Selectric
Courier to me, and has all Fraktur and script characters
written in by hand, *but* has somehow arranged to have very
elegant typewritten versions of mathbb{C} and mathbb{R},
I have no idea how. Karoubi's thesis (University of Paris,
1967) is typewritten with no distinction for C, R, etc.,
at all (though when C stands for some category, it has a
handwritten x beneath it--very odd). Mumford's original
Red Book (Introduction to Algebraic Geometry--Preliminary
version of first 3 Chapters), which is undated and has no
publisher noted anyway (I assume it was printed by Harvard
in or about 1969) uses a mixture of over- and doublestriking
for mathbb: doublestruck A (for affine space), P overstruck
with I (for projective space).
Of all these, the authors closest to Bourbaki (Serre and
Karoubi) have, if anything, the type that's furthest from
blackboard bold.
Lee Rudolph
Subject: Re: History of blackboard bold?
>By 1966, however, Gunning (in Lectures on Riemann Surfaces,
>PUP Princeton Mathematical Notes) uses doublestruck C and R;
...
>(I believe it was from him that I learned to use it,
>at just around that time);
and so on. Since writing that, I have exchanged e-mail with
Professor Gunning. Like others, he has heard that the notation
goes back to Bourbaki--more specifically, he says he picked it
up in the Kodaira-Spencer seminars (at Princeton, early 1960s),
and that someone there (whose name he can't recall) attributed
it to Bourbaki. He confirms that it was in the early 1960s
that he himself was converted to the notation, and started
passing it on to students; and he says (I paraphrase, since
I didn't ask permission to quote) that Addison-Wesley didn't
give him and Rossi any trouble about using blackboard bold
for the complex numbers (but that he and Rossi *did* have
trouble convincing A-W to allow non-bold vectors and yet
bold germs). He has also promised to ask other veterans
of the K-S seminar for their memories, which I will pass
on as they arrive. (I think I will not act on his implicit
suggestion that one might ask Serre; but if anyone else has
the chutzpah to do so, I don't doubt something useful might
be learned from the experiment.)
I hope to spend some time in the library stacks today, and
may have more data by this afternoon.
Lee Rudolph
Subject: Is this newsgroup useless?
I guess that the time has come for me to ditch this group
altogether. It
seems to have been taken over by deranged individuals exposing their
misguided ideas over and over again (Mr. Harris being the best
representative of this increasingly vocal group,) impervious to any kind
of logic reasoning; by people trying to get the group to do their homework
for them, and by plain idiots who post the most irrelevant babbling (like
this moron who posts numerological stuff, while changing his profile
frequently.)
It's sad, but sci.math has become, by and large, a freak show,
which
makes the occasional gems posted more and more difficult to find.
Subject: Re: Is this newsgroup useless?
> I guess that the time has come for me to ditch this group altogether. It
> seems to have been taken over by deranged individuals exposing their
> misguided ideas over and over again (Mr. Harris being the best
> representative of this increasingly vocal group,) impervious to any kind
> of logic reasoning; by people trying to get the group to do their
homework
> for them, and by plain idiots who post the most irrelevant babbling (like
> this moron who posts numerological stuff, while changing his profile
> frequently.)
> It's sad, but sci.math has become, by and large, a freak show, which
> makes the occasional gems posted more and more difficult to find.
The number of replies is the thread's rating. Apparently, post's quality
has
nothing to do with the rating. Now, face it: the JSH show has the highest
rating in sci.math.
Subject: Re: Is this newsgroup useless?
> The number of replies is the thread's rating. Apparently, post's quality
has
> nothing to do with the rating. Now, face it: the JSH show has the highest
> rating in sci.math.
That's what makes it so sad - that the rantings of that pathetic
loony
seem to be stuff that so many contributors to sci.math are so interested
in.
Subject: Re: Is this newsgroup useless?
> I guess that the time has come for me to ditch this group
altogether. It
>seems to have been taken over by deranged individuals exposing their
>misguided ideas over and over again (Mr. Harris being the best
>representative of this increasingly vocal group,) impervious to any kind
>of logic reasoning; by people trying to get the group to do their homework
>for them, and by plain idiots who post the most irrelevant babbling (like
>this moron who posts numerological stuff, while changing his profile
>frequently.)
> It's sad, but sci.math has become, by and large, a freak show,
which
>makes the occasional gems posted more and more difficult to find.
It might depend on what newsreader you use. I read my news with
Netscape(R) 7.1(and a broadband connection). With this, it is extremely
easy to scroll through the headers (= subject line + poster) and skip
over those posts wherein I am sure to have no interest, for whatever
reason. Years ago, I read my news through a serial news program on a
UNIX(tm) system where I had to keep hitting my n key. I would find
that practice intolerable now.
--
Stephen J. Herschkorn herschko@rutcor.rutgers.edu
Subject: Re: Is this newsgroup useless?
> It's sad, but sci.math has become, by and large, a freak show,
> which
> makes the occasional gems posted more and more difficult to find.
I agree that it's getting worse. But it's not hard to avoid
the trolls so much as it is to avoid the feeders of the trolls.
It's the feeders who are the real problem. First, they multiply
the number of times the trolls post by 10, by egging them on,
then they multiply that number by 10 again with their own posts.
echoes he creates, this filters out only about 1% of the related
noise.
I'd be all for sci.math.moderated under a charter that specifies
that the moderator must reject any post that refers negatively on
the character of another person. One gets to say there is a flaw
in your reasoning here, but not your brain produces a lot of
flawed reasoning.
One can say there is a flaw in your reasoning here but can not
continue with and that's so typical of your posts. The criticism
would always stay strictly upon the mathematics itself, and not
upon the talent or character of the posters.
The moderator could make judgement calls about whether to pass
That proof was biggest pile of $*(#@(&$-ing garbage I've
ever seen. And if we disagree with his judgement once in a
while, well, least it's far better than what we had before.
Bart
Subject: Re: Is this newsgroup useless?
> It's sad, but sci.math has become, by and large, a freak show,
> which
> makes the occasional gems posted more and more difficult to find.
> I agree that it's getting worse. But it's not hard to avoid
> the trolls so much as it is to avoid the feeders of the trolls.
> It's the feeders who are the real problem. First, they multiply
> the number of times the trolls post by 10, by egging them on,
> then they multiply that number by 10 again with their own posts.
It's possible that your own post is an example of the problem you cite.
The O.P. may well be a troll. In any case, I disagree strongly that It's
the feeders who are the *real* problem. (emphasis added). The real
problem is that this is an *unmoderated* newsgroup. All posters are
permitted. If the threads are dominated by a [JSH] and respondents, the
best solution is to form a moderated newsgroup -- not to try and
discourage respondents. Clearly if no trolls post, no responses occur. If
trolls do post, responses *may* occur. Blaming the respondents places
your own posts in question.
> echoes he creates, this filters out only about 1% of the related
> noise.
> I'd be all for sci.math.moderated under a charter that specifies
> that the moderator must reject any post that refers negatively on
> the character of another person. One gets to say there is a flaw
> in your reasoning here, but not your brain produces a lot of
> flawed reasoning.
> One can say there is a flaw in your reasoning here but can not
> continue with and that's so typical of your posts. The criticism
> would always stay strictly upon the mathematics itself, and not
> upon the talent or character of the posters.
> The moderator could make judgement calls about whether to pass
> That proof was biggest pile of $*(#@(&$-ing garbage I've
> ever seen. And if we disagree with his judgement once in a
> while, well, least it's far better than what we had before.
> Bart
--
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: Is this newsgroup useless?
> The real
> problem is that this is an *unmoderated* newsgroup. All posters are
> permitted.
I have two other main newsgroup interests. One of them is
christianity. In that case, I avoid the unmoderated groups
at like a Pharisee avoids a leper. They make sci.math look like
a candle next to the space shuttle taking off. It takes a hardworking,
patient and wise moderator to keeps christians behaving like...uh..
christians. So in that sense, your half right.
But my other interest is homebrewing, and that group is UNmoderated,
but the most pleasant and helpful and unflaming group on all of USENET.
Every newbie who comes along asking the same dumb question that's
been asked 17 trillion times is welcomed with open arms, given
kind advice and we cheer that another lost sheep has been converted
from the evils of Budmilloors and demon megaswill. So your half
wrong.
Half the problem here is that it's unmoderated. The other half is
that it's populated by overly-anal-retentive jerks who think it's
going to be important to point out that I spelled you're your
twice in the preceding paragraphs.
There's something the same about fanatic legalists in the religious
groups and anal mathematicians, in that they think they've scored
something if they catch you in an error. AHA! You've typed
slander when you should have typed libel! So what? Therefore
I'm an idiot and he's a savant and now I have to erase his
chalkboards for him? Hardly. I'm going to have a homebrew and
hang out with the cute chicks while he re-catalogues his PowerRanger
Collector's Cards.
I _thought_ we had beaten these guys up sufficiently in the high
school locker room that they'd be quiet by now. All I did was
explain that if we'd all place our duffel bags perpendicularly
on the benches, there would be room for all of us to get dressed
at once. I really don't think that was sufficient reason for
them to put Nair in my bottle of Prell.
Maybe the real problem is that a guy can't e-mail a matburn.
Bart
Subject: Re: Is this newsgroup useless?
> The O.P. may well be a troll.
Believe me, I am not. I am just expressing despondency about the
status
quo nowadays. I have been reading sci.math, on and off, for some 14 years
now, and it has gotten steadily worse. With all the trolls, morons,
lunatics and otherwise deranged contributors to this group so prevalent
today, going through new headers every day has become a painful exercise.
I would actually propose either to have a moderator for this group,
or
to abandon it completely, while creating a moderated one.
Subject: Re: Is this newsgroup useless?
>> The O.P. may well be a troll.
> Believe me, I am not. I am just expressing despondency about the
status
> quo nowadays. I have been reading sci.math, on and off, for some 14 years
> now, and it has gotten steadily worse. With all the trolls, morons,
> lunatics and otherwise deranged contributors to this group so prevalent
> today, going through new headers every day has become a painful exercise.
I would think that more accurately describes sci.physics than sci.math.
It doesn't seem that bad to me around here.
> I would actually propose either to have a moderator for this group,
or
> to abandon it completely, while creating a moderated one.
Moderating this group is not an option. The USENET Powers that Be
(otherwise knows as the moderators of news.announce.newgroups) have
decreed that proposals for changing the moderation status of existing
groups will be not be accepted. If you want a moderated version of
sci.math then you'll have to propose a new newsgroup.
--
Wayne Brown | When your tail's in a crack, you improvise
fwbrown@bellsouth.net | if you're good enough. Otherwise you give
| your pelt to the trapper.
e^(i*pi) = -1 -- Euler | -- John Myers Myers,
Silverlock
Subject: Re: Is this newsgroup useless?
> I would think that more accurately describes sci.physics than sci.math.
> It doesn't seem that bad to me around here.
To tell you the truth, I quit reading sci.physics a couple of years
ago
because of that. It's just that sci.math seem to be attaining a similar
level of worthlessness.
>> I would actually propose either to have a moderator for
this group, or
>> to abandon it completely, while creating a moderated one.
> Moderating this group is not an option. The USENET Powers that Be
> (otherwise knows as the moderators of news.announce.newgroups) have
> decreed that proposals for changing the moderation status of existing
> groups will be not be accepted.
Well, I guess the time has come to move on then.
Subject: Re: Is this newsgroup useless?
> The O.P. may well be a troll.
> Believe me, I am not.
Could you please use your real name then?
> I am just expressing despondency about the
status
> quo nowadays. I have been reading sci.math, on and off, for some 14 years
> now, and it has gotten steadily worse. With all the trolls, morons,
> lunatics and otherwise deranged contributors to this group so prevalent
> today, going through new headers every day has become a painful exercise.
> I would actually propose either to have a moderator for this
group, or
> to abandon it completely, while creating a moderated one.
I'd certainly read sci.math.moderated but who would moderate it? It
would be a lot of hard work.
--
G.C.
Subject: Re: Is this newsgroup useless?
> I'd certainly read sci.math.moderated but who would moderate it?
> It would be a lot of hard work.
It wouldn't be so hard to read a message and decide whether to
post it. And having decided not to post it, to return the message
to the author with the (first) offending sentence highlighted and
a pointer to the website with the posting guidelines.
The REAL hard work would be in the fielding of the the challenges
to the moderator's decisions. If the moderator was strict and did
his work well, the posting-decisions work would be minor (but a
daily chore, no doubt) since posters would quickly learn what was
acceptable and abide by the rules. The moderator, by being consistant
and fair, could keep his own work minimal.
It's when some dingbat engages him in an argument about whether
his post should have gone through and starts whining about free
speech and starts posting libelous crap about the moderator and
his fascist policies on sci.math and (as in one case I know) actually
takes the moderator to court, that things get out of hand. (The
idiot was laughed out of court, of course, but what a pain for the
moderator.) One buffoon on the christianity group got mad at the
moderator and spammed all the posters with penis-enlargement ads
with active embedded scripts, which, upon opening, of course,
signed everyone up for even more penis-enlargement spam.
The charter has to agree that the moderator makes decisions and they
are final. If your post is returned, your _only_ recourse is to
fix your submission and resubmit it. But how can this be enforced?
There is nothing to stop someone from repeatedly e-mailing the
moderator or from engaging counsel and trying to sue him for
violation of free speech rights.
Anyway, sci.math must have about 400 posts per day...? I predict that
sci.math.moderated would have less than 50 per day and that most of
those could be scanned and passed for posting quickly.
Bart
Subject: Re: Is this newsgroup useless?
> I'd certainly read sci.math.moderated but who would moderate it?
> It would be a lot of hard work.
> It wouldn't be so hard to read a message and decide whether to
> post it. ...
a lot of hard work was the wrong phrase, time consuming would be
better.
--
G.C.
Subject: Re: Is this newsgroup useless?
> ....
> I would actually propose either to have a moderator for this
group, > or to abandon it completely, while creating a moderated one.
> I'd certainly read sci.math.moderated but who would moderate it? It
> would be a lot of hard work.
There was a group a few years ago with exactly that name IIRC. It
received hardly any posts, and even those were delayed for half a day or
more until the moderator had time to deal with them.
I find it quite easy to scan quickly through the list of titles and
choose a few of interest, but that may be because I use the excellent
NewsWatcher on a Mac.
Ken Pledger.
Subject: Re: Is this newsgroup useless?
>> The O.P. may well be a troll.
>> Believe me, I am not.
> Could you please use your real name then?
(Sigh!) Jerzy Szuszkiewicz.
Now can you tell if that is my name? You might, or might not, be Mr.
C.
Bond. I don't know, and I am certainly not going to investigate it. It is
much more pertinent to see if Mr. C. Bond, real name or not, has got
anything interesting to say.
I have good reasons to post anonymously, reasons that I don't want
to go
into. The question is, is there any merit to what I am proposing, or
should anonymous postings be automatically discarded, regardless of their
content?
Subject: Re: Is this newsgroup useless?
> echoes he creates, this filters out only about 1% of the related
> noise.
I thought xnews had decent filter ability?
> I'd be all for sci.math.moderated under a charter that specifies
> that the moderator must reject any post that refers negatively on
> the character of another person. One gets to say there is a flaw
> in your reasoning here, but not your brain produces a lot of
> flawed reasoning.
> One can say there is a flaw in your reasoning here but can not
> continue with and that's so typical of your posts. The criticism
> would always stay strictly upon the mathematics itself, and not
> upon the talent or character of the posters.
> The moderator could make judgement calls about whether to pass
> That proof was biggest pile of $*(#@(&$-ing garbage I've
> ever seen. And if we disagree with his judgement once in a
> while, well, least it's far better than what we had before.
> Bart
I see calls for moderated groups all the time (it's not just sci.math
that's
inundated by trolls and crackpots). Feel free to create a newsgroup and
become the moderator. Finding out how to create a newsgroup is easy
enough,
doing it is much harder, and finding a moderator, well, that's the kicker,
isn't it.
--Stan Gula
Subject: Re: Is this newsgroup useless?
> I see calls for moderated groups all the time (it's not just sci.math
> that's inundated by trolls and crackpots). Feel free to create a
> newsgroup and become the moderator. Finding out how to create a
> newsgroup is easy enough, doing it is much harder, and finding a
> moderator, well, that's the kicker, isn't it.
I wouldn't mind doing it....for about a day;-)
Bart
Subject: Re: Is this newsgroup useless?
> I guess that the time has come for me to ditch this group altogether. It
> seems to have been taken over by deranged individuals exposing their
> misguided ideas over and over again (Mr. Harris being the best
> representative of this increasingly vocal group,) impervious to any kind
> of logic reasoning; by people trying to get the group to do their
homework
> for them, and by plain idiots who post the most irrelevant babbling (like
> this moron who posts numerological stuff, while changing his profile
> frequently.)
> It's sad, but sci.math has become, by and large, a freak show, which
> makes the occasional gems posted more and more difficult to find.
I am posting under a real ISP.
It's not really feasible to keep this sort of thing under complete
control,
and still have a public forum.
If I see a real problem, it's the people who respond to Harris threads,
knowing that is will end up in flames.
I don't know enough about the subject to disprove Harris' stuff, but I
trust
the those who have done so.
In view that Harris live in his own world, he cannot deal with the rules of
the mathematical community, and it would reduce clutter IMO to ignore him.
Often the homework gang launch interesting side threads which have value to
those who are rusty or semi-competent in different fields.
Moron trolls just need to be plonked frequently, as name and ISP changes
are
difficult to control.
I learn a lot from the on-topic threads here, and occasionally can
contribute to subjects in Introductory real Analysis and lesser basics.
Also, sometimes low-key stuff like pun cascades are fun, just to add some
bright perspective.
Bob Pease
Subject: What to tell students in a 10-15 minute talk
NNTP-Posting-User: mckay
Someone from Rostock asked for ...
I reckon there are some very beautiful thought stimulating solutions
to simple problems which are hard unless treated the right way.
1. A fly travels between two trains travelling towards each other
on the same line. It goes from one train to the other repeatedly.
How far does it travel? Trains go at fixed given speeds.
[All needed is constant train and fly speeds. Can work out time flown.]
2. Prove that among any set of lines and (non-colinear) points, there
is some line with just 2 points on it.
[Consider all point-line pairs and work from the one containing the
closest point.]
3. When are the minutes and hour hnds of a clock on top of each other?
[The hands move at fixed speeds. They cross 11 time in 12 hours, hence
the time between being on top of each other is 12/11 of an hour.]
But for greatest effect, it is best to give the students plenty of time
to spend trying to find solutions first.
Good luck,
John McKay
--
But leave the wise to wrangle, and with me
the quarrel of the universe let be;
and, in some corner of the hubbub couched,
make game of that which makes as much of thee.
Subject: Re: What to tell students in a 10-15 minute talk
Well, for (1), all you need is a series.
Lurch
> Someone from Rostock asked for ...
> I reckon there are some very beautiful thought stimulating solutions
> to simple problems which are hard unless treated the right way.
> 1. A fly travels between two trains travelling towards each other
> on the same line. It goes from one train to the other repeatedly.
> How far does it travel? Trains go at fixed given speeds.
> [All needed is constant train and fly speeds. Can work out time flown.]
> 2. Prove that among any set of lines and (non-colinear) points, there
> is some line with just 2 points on it.
> [Consider all point-line pairs and work from the one containing the
> closest point.]
> 3. When are the minutes and hour hnds of a clock on top of each other?
> [The hands move at fixed speeds. They cross 11 time in 12 hours, hence
> the time between being on top of each other is 12/11 of an hour.]
> But for greatest effect, it is best to give the students plenty of time
> to spend trying to find solutions first.
> Good luck,
> John McKay
> --
> But leave the wise to wrangle, and with me
> the quarrel of the universe let be;
> and, in some corner of the hubbub couched,
> make game of that which makes as much of thee.
Subject: Re: What to tell students in a 10-15 minute talk
> Well, for (1), all you need is a series.
fly's speed and that time, work out the distance.
> Lurch
> Someone from Rostock asked for ...
> I reckon there are some very beautiful thought stimulating solutions
> to simple problems which are hard unless treated the right way.
> 1. A fly travels between two trains travelling towards each other
> on the same line. It goes from one train to the other repeatedly.
> How far does it travel? Trains go at fixed given speeds.
> [All needed is constant train and fly speeds. Can work out time flown.]
> 2. Prove that among any set of lines and (non-colinear) points, there
> is some line with just 2 points on it.
> [Consider all point-line pairs and work from the one containing the
> closest point.]
> 3. When are the minutes and hour hnds of a clock on top of each other?
> [The hands move at fixed speeds. They cross 11 time in 12 hours, hence
> the time between being on top of each other is 12/11 of an hour.]
> But for greatest effect, it is best to give the students plenty of time
> to spend trying to find solutions first.
> Good luck,
> John McKay
> --
> But leave the wise to wrangle, and with me
> the quarrel of the universe let be;
> and, in some corner of the hubbub couched,
> make game of that which makes as much of thee.
--
G.C.
Subject: Re: What to tell students in a 10-15 minute talk
> Someone from Rostock asked for ...
> I reckon there are some very beautiful thought stimulating solutions
> to simple problems which are hard unless treated the right way.
> 1. A fly travels between two trains travelling towards each other
> on the same line. It goes from one train to the other repeatedly.
> How far does it travel? Trains go at fixed given speeds.
> [All needed is constant train and fly speeds. Can work out time flown.]
> 2. Prove that among any set of lines and (non-colinear) points, there
> is some line with just 2 points on it.
Steiner's Problem? Also, with n points there are >= 3n/7 such
lines.
> [Consider all point-line pairs and work from the one containing the
> closest point.]
That is: point P and line through QR not incident?
> 3. When are the minutes and hour hnds of a clock on top of each other?
> [The hands move at fixed speeds. They cross 11 time in 12 hours, hence
> the time between being on top of each other is 12/11 of an hour.]
> But for greatest effect, it is best to give the students plenty of time
> to spend trying to find solutions first.
> Good luck,
> John McKay
> --
> But leave the wise to wrangle, and with me
> the quarrel of the universe let be;
> and, in some corner of the hubbub couched,
> make game of that which makes as much of thee.
--
G.C.
Subject: Re: What to tell students in a 10-15 minute talk
>3. When are the minutes and hour hnds of a clock on top of each other?
>[The hands move at fixed speeds. They cross 11 time in 12 hours, hence
>the time between being on top of each other is 12/11 of an hour.]
Also, when are the hour, minute and second hands positioned so that
they divide the clockface in three equal sectors?
Subject: Re: What to tell students in a 10-15 minute talk
>3. When are the minutes and hour hnds of a clock on top of each other?
>[The hands move at fixed speeds. They cross 11 time in 12 hours, hence
>the time between being on top of each other is 12/11 of an hour.]
> Also, when are the hour, minute and second hands positioned so that
> they divide the clockface in three equal sectors?
That's nice, a new one to me.
--
G.C.
Subject: Re: What to tell students in a 10-15 minute talk
> Someone from Rostock asked for ...
> I reckon there are some very beautiful thought stimulating solutions
> to simple problems which are hard unless treated the right way.
> 1. A fly travels between two trains travelling towards each other
> on the same line. It goes from one train to the other repeatedly.
> How far does it travel? Trains go at fixed given speeds.
> [All needed is constant train and fly speeds. Can work out time flown.]
> 2. Prove that among any set of lines and (non-colinear) points, there
> is some line with just 2 points on it.
> [Consider all point-line pairs and work from the one containing the
> closest point.]
> 3. When are the minutes and hour hnds of a clock on top of each other?
> [The hands move at fixed speeds. They cross 11 time in 12 hours, hence
> the time between being on top of each other is 12/11 of an hour.]
> But for greatest effect, it is best to give the students plenty of time
> to spend trying to find solutions first.
Can you re-state #2 a different way?
I think you mean
Prove that for any set of three or more non-collinear points (in a plane),
there is at least one line in that plane which contains EXACTLY two of
these points ???
Bob Pease
Subject: Yet another question about circular sectors :)
Folks here have been good enough to help me out a few times so I am back at
the well hoping for another drink. :)
Given the chord and area of a circular sector (the pie shape), I need
to
find any one of the following: included angle, length of arc, or arc
radius.
Subject: Re: Yet another question about circular sectors :)
> Given the chord and area of a circular sector (the pie shape), I need
to
> find any one of the following: included angle, length of arc, or arc
radius.
If L is semi chord length and A is sector area, th is included angle,
then
we have ( L= R sin(th/2) ; Area=R^2 th /2 ). Eliminate R. We need to
solve transcendental equation th = (1-cos(th)) *(Area/L^2);
An interesting possibility is that we can also solve it manually on a
scientific calculator with some patience on a keyboard (before too
much beer). This is nothing, as Newton-Raphson and other iteration
procedures are so easily available. However,here it is :
Key in -> th;
Manually enter keys : [ cos(th) > CHS (+/-) > +1 = > * (Area/L^2)=
] ; recursively
See my earlier
Subject: Re: Yet another question about circular sectors :)
> Folks here have been good enough to help me out a few times so I am back
at
> the well hoping for another drink. :)
> Given the chord and area of a circular sector (the pie shape), I need
to
> find any one of the following: included angle, length of arc, or arc
radius.
I think you mean given the length of a chord and the area of its sector,
find the angle formed by the chord and the radius of the circle.
If you are given the chord in degrees, change it to RADIANS.
likewise use ratio and proportion to find the radius, knowing that area
of
the sector is equal to L*R and area of a circle - 2*PI*R
BOb PEase
Subject: Re: Yet another question about circular sectors :)
No, that's not what I meant.
Given a circular sector, the classic pie shape, and knowing the
enclosed
area and the *length* of its chord (stand the piece of pie on its vertex
with the arc at the top, the chord is the horizontal measurement between
the
points where the arc intersects with the sides).
Here's the deal - a pie shape has five basic measurements:
1) the radius of the arc (or, the length of the sides)
2) the circumferential length of the arc
3) the included angle of the arc
4) the enclosed area
5) the linear length of the chord, as defined above
Now, given any two of the above, I know how to calculate the other three -
*except* when given # 4 and #5 - that's where I am hung up.
> Folks here have been good enough to help me out a few times so I am
back
> at
> the well hoping for another drink. :)
> Given the chord and area of a circular sector (the pie shape), I
need
to
> find any one of the following: included angle, length of arc, or arc
> radius.
> I think you mean given the length of a chord and the area of its sector,
> find the angle formed by the chord and the radius of the circle.
> If you are given the chord in degrees, change it to RADIANS.
> likewise use ratio and proportion to find the radius, knowing that area
of
> the sector is equal to L*R and area of a circle - 2*PI*R
> BOb PEase
Subject: Re: Yet another question about circular sectors :)
> Given a circular sector, the classic pie shape, and knowing the
enclosed
> area and the *length* of its chord (stand the piece of pie on its vertex
> with the arc at the top, the chord is the horizontal measurement between
the
> points where the arc intersects with the sides).
> Here's the deal - a pie shape has five basic measurements:
> 1) the radius of the arc (or, the length of the sides)
> 2) the circumferential length of the arc
> 3) the included angle of the arc
> 4) the enclosed area
> 5) the linear length of the chord, as defined above
> Now, given any two of the above, I know how to calculate the other three
-
> *except* when given # 4 and #5 - that's where I am hung up.
Let L be the linear length of the chord,
let A be the area,
let R be the radius, and
let T be the central angle in radians, then
L = 2*R*sin(T/2) and A = (1/2)*R^2*T.
One can elimnate R from the two equations getting
8*A*sin(T/2)^2 = L^2*T.
But such an equation, with T both inside and outside a trigonometric
function, indicates that there is no standard closed form solution
for T in terms of L and A.
Similarly, there is no standard closed form solution for R in
terms of L and A.
However, in either case, numerical solutions of the relevant
equations are possible.
Subject: Re: Yet another question about circular sectors :)
>No, that's not what I meant.
>Given a circular sector, the classic pie shape, and knowing the
enclosed
>area and the *length* of its chord (stand the piece of pie on its vertex
>with the arc at the top, the chord is the horizontal measurement between
the
>points where the arc intersects with the sides).
>Here's the deal - a pie shape has five basic measurements:
> 1) the radius of the arc (or, the length of the sides)
> 2) the circumferential length of the arc
> 3) the included angle of the arc
> 4) the enclosed area
> 5) the linear length of the chord, as defined above
>Now, given any two of the above, I know how to calculate the other three -
>*except* when given # 4 and #5 - that's where I am hung up.
Maybe this can help:
http://mathforum.org/dr.math/faq/faq.circle.segment.html#11
Subject: Re: Yet another question about circular sectors :)
> No, that's not what I meant.
> Given a circular sector, the classic pie shape, and knowing the
enclosed
> area and the *length* of its chord (stand the piece of pie on its vertex
> with the arc at the top, the chord is the horizontal measurement between
the
> points where the arc intersects with the sides).
> Here's the deal - a pie shape has five basic measurements:
> 1) the radius of the arc (or, the length of the sides)
> 2) the circumferential length of the arc
> 3) the included angle of the arc
> 4) the enclosed area
> 5) the linear length of the chord, as defined above
> Now, given any two of the above, I know how to calculate the other three
-
> *except* when given # 4 and #5 - that's where I am hung up.
> Folks here have been good enough to help me out a few times so I am
back
> at
> the well hoping for another drink. :)
Given the chord and area of a circular sector (the pie shape), I
need
> to
> find any one of the following: included angle, length of arc, or arc
> radius.
> I think you mean given the length of a chord and the area of its
sector,
> find the angle formed by the chord and the radius of the circle.
> If you are given the chord in degrees, change it to RADIANS.
> likewise use ratio and proportion to find the radius, knowing that
area
> of
> the sector is equal to L*R and area of a circle - 2*PI*R
> BOb PEase
Given the chord length, and the area of the sector, ( not arc length) I
don't think that the problem is uniquely determined.
I'll work on it some more.
Maybe someone will find the answer in the meantime.
Bob Pease
Subject: Re: Yet another question about circular sectors :)
> Given the chord length, and the area of the sector, ( not arc length) I
> don't think that the problem is uniquely determined.
> I'll work on it some more.
> Maybe someone will find the answer in the meantime.
I think, depending on the values given, you can get 0, 1, or 2
solutions.
Do a thought experiment with me.
Pretend the chord is horizontal, an example center of the circle
is directly below the chord, and consider what happens as the
center of the circle is moved along the line defined by these two
points. If the center is moved down, the area of the sector will
go to infinity.
If the center is moved up, a minimum area of the sector is found
when the center of the circle and the center of the chord coincide,
and is in fact the area of a half circle with diameter the chord
length. Therefore, for this area and chord length there is one
unique solution for the other values. For an area smaller than
this minimum given that chord length, there is no solution.
When are there two solutions? Keep moving the center of the circle
up past the chord, and you have a the rest of the pie shape, where
the chord is exterior to the sector. Again, as the center of the
circle continues up, this sector's area tends smoothly to infinity;
thus for each solution with interior chord, there will be a matching
solution with exterior chord having the same area (but different
radius and arc length).
xanthian.
--
Subject: billard mechanics
I posted something like this in sci.physics but people seem to be mainly
with their heads in the stars there..
I have programmed, ages ago, this billiard mechanics engine, using very
basic physical laws on momentum collision. It works in the sense that it
creates :
1. good collisions between two balls that each have a specific velocity
vector, and
2. it never draws balls over each other,
although I'm not satisfied with the amount of calculation needed for those
two things (it involves solving a quadratic equation and then
trigonometry).
But, as you may be aware, in snooker or pool one starts with the red balls
(and the pink) touching each other.
This kind of ruins this whole nice model since one can no longer use an
O(n^2) algorithm to scan the balls for possible collisions and then work
them, since..well it becomes a mess. When the white ball hits the pack of
red balls, I get a nice 4-dimensional representation of chaos, I think,
which isn't my intention. As you know, when balls touch each other then the
energy of the first ball gets transferred to the last. Most of it,
anyway.
In fact I think it calls for an entirely new strategy. Does anybody have
any
ideas how I could :
- devise a collision strategy that works on singular collisions as well
as on multiple simultaneous collisions
or
- treat these chained collisions separately (this would involve
sorting the touching balls with respect to the ball(s) that will hit them,
since a computer can't know in what direction the force gets transferred ->
meaning, this is probably impossible to accomplish.
I would probably do best to treat the table of balls as a table of static
balls, with force fields on them, instead of each ball having its vx and
vy,
but really, it's not that easy..
If anybody has any ideas or experience with this, I'd be very grateful,
--
Tetsuo
Subject: Group automorphisms
Does anybody know an example of a finite group
G of odd order whose automorphism group Aut G
is a p-group?
Subject: Re: Group automorphisms
Visiting Assistant Professor at the University of Montana.
>Does anybody know an example of a finite group
>G of odd order whose automorphism group Aut G
>is a p-group?
Aut(Z/3Z) = Z/2Z, a 2-group. If p is a prime of the form 2^n+1 (so, a
Fermat prime), then Z/pZ is a finite group of odd order whose
automorphism group is cyclic of order 2^n.
So presumably, you had something a bit more complicated in mind?
Subject: Re: Deep Thoughts # 1: A new limitation to the human mind
> 1. Mathematics is the science in which we make something out of
> nothing.
Wrong. Mathematics is built on the 13 Axioms, which are not nothing.
> 2. All of man-made Mathematics consists of an abstraction from
> physical processes.
Wrong again. Mathematics consists of theorems derivated by following a
set of formal rules.
> 3. Since Mathematics in general needs nothing to be created, then the
> human mind seems to be limited in that it can only consider
> possibilities that are analogous to physical processes.
Well, I think first of all mathematics exist all by themselves, as a
logical system. However one may argue that for humans this exists in
its precise own way, because the human mind works the way it works.
Thus mathematics actually may help delineate the limits of the human
mind.
Subject: Re: Deep Thoughts # 1: A new limitation to the human mind
>>What is (or was, at least) referred to as metamathematics is a part of
>>mathematics, so metametamathematics is just a particular portion of
>>metamathematics.
>The proposition undecidable in the system PM is thus decided by
>metamathematical arguments. - Kurt Godel, 1931
>>The undecidable propostition is seen to be true in the natural numbers
>>by an ordinary mathematical argument.
> If it is ordinary mathematics, then why is it not decidable in the
> system PM, or is Godel wrong?
Uhm. Why should it be decidable in PM? There is no decree that
everything that's mathematical (and true) should be provable in PM.
G.9adel showed that this can't be.
>A Turing Machine calculates Mathematics, but does not decide the
>halting predicate, which lies within Metamathematics.
>>The halting predicate lies well within ordinary mathematics.
> Doesn't a computer implement ordinary mathematics (yet a computer
> cannot decide the halting predicate)?
Huh? In a certain sense a computer can be said to implement ordinary
mathematics, e.g. some specific finite state machine, but this doesn't
mean that it implements the whole of ordinary mathematics. How would a
computer implement, say, set theory?
>Aside from where Metametamathematics happens to lie, I ask, what does
>it look like?
>>It looks like mathematics.
> You haven't made any distinction concerning the nature of
> Metametamathematics. I think that Metamathematics concerns
> Mathematical problems that cannot be solved by Mathematics.
You are wrong.
> My
> question is still, what is Metametamathematics like?
It looks like mathematics studying models, formal theories, computable
functions, proofs, &c.
--
Aatu Koskensilta (aatu.koskensilta@xortec.fi)
Wovon man nicht sprechen kann, daruber muss man schweigen
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
Subject: Re: Good Old Springer
>> A message from Joan Birman to another mailing list begins as follows:
>> former Springer-Verlag (more recently BertelsmannSpringer) was merged
with
>> Kluwer Academic Publishers. As the press release says, the merger will
>> create the second largest professional publisher in science, technology
>> and medicine. What's the other one: Elsevier, of course! [...]
>> (Birman goes on to describe why this is not especially good news for
>> fans of Springer and more generally for those who look for
reasonably-priced
>> published mathematics; she then advocates some alternatives.)
>I agree, this is bad news indeed.
>Which particular alternatives does she mention?
The usual suspects:
Independent private journals (e.g. Annals of Math)
University- and Professional-Society journals (The ___ of the ___ Math.
Soc.)
Online journals (which are frequently free of charge)
New journals to be started by like-minded colleagues who hate the status
quo.
You can read her letter yourself: http://www.lehigh.edu/~dmd1/jb102.txt
If you want a journal with a professional reputation, it's going to
cost money for decent editorial work (meaning both competent content
decisions and good copy editing). That's probably worth asking universities
and other research institutions to shell out a few hundred dollars a
year per subscription. But with circulation rates so low, this kind of
return on effort is not attractive to large commercial publishers; they
prefer either to charge $1000 or more per year for a journal, or to quit
investing value-adding editorial services in the journal. Either way,
the mathematical community suffers.
Mathematicians collectively have little financial acumen: we give away
the fruits of our intellectual labors, and then pay big bucks for the
privilege of buying them back to put on our library's shelves.
dave
Subject: Ball selection from mulitple urns question
I have a probability question I could use some help with. I'll use
the urn/ball model to make it more general. Assume that I have n
urns, each with a different proportion of colored balls in it. I know
the probability P_n(c) of selecting each color of ball from each urn.
I will select one ball from each urn, for a total of n unordered balls
in the selected set S. Now, given a particular reference set of
unordered balls T, how do I determine the probability that S will
match it? For example, if I have 10 urns, how might I determine the
probability that I will choose 3 red, 2 blue and 5 green balls?
I've come up with a brute force iterative method, but it depends on
looking at all permutations of the target set, which gets quite large
as I grow the number of urns. This will be buried deep inside of a
very large nested loop, so I'm hoping there's a more direct method,
Subject: Re: quantum echo
> --I could have baked both cakes yesterday.
> --I could have baked both cakes today.
> --I could have baked this one yesterday
> and that one today.
> --I could have baked that one yesterday and
> this one today.
Your example doesn't work!
> How am I going to distinguish 'one cake' from 'the other cake'!
Since they now are identical-with-somethings, if they shared
> all the same properties they would be one cake rather than
> two. THerefore, since they are two cakes, they don't share
> all their properties, and I can distinguish them by means
> of any property that one has and the other doesn't.
you mean argument by GPS or by just determining their global
positioning
> with sattellites we have a difference even if there are no other clear
> attributes of different classes besides location?
> That is a call which no logician qua logician has any particular
competence
> to make.
Are you saying it is possible for something to exist at location A and
location B at the same time? I thought it was inference by the rule of
non-contradiction.
> --John
JJ
1 I. The Problem, and the Problem with the Problem,
> 2 of Identical Articles and Quantum Statistics
> 3
> 4 Suppose we have a box with two qualitatively
> 6 bouncing around inside. We think of the box
> 7 as having a left (*l*) and a right (*r*) side.
> 9 at random without interacting, so that their
> 10 motions are independent; in particular we
> 12 that we may neglect collisions. What are the
> 13 chances for finding one or both on one side
> 14 or the other?
> 15
> 16 Many find the following reasoning persuasive.
> 19 in *r*, and 2 in *l* and 1 in *r*. These should
> 20 be equally likely, so that each has a probability
> 21 of 1/4, or a probability of 1/4 for two in *l*,
> 22 1/4 for two in *r*, and 1/2 for one on each side.
> 23
> 24 This stylized example is a simple mock-up for a
> 25 kind of situation that can occur with quantum entities
> 26 and properties. For many of these situations the
> 27 probabilities are in fact found to be 1/3 for each
> 28 of the three cases: two in *l*, two in *r* and one
> 29 on each side. Many interpreters have found this
> 30 fact utterly astonishing.
> 31
> 32 But on the face of it, there is a very simple
> 33 resolution of the puzzle: give up supposing
> 34 that there are two qualitatively identical but
> 36 there are two *quanta*, as I'll put it, to which
> 37 the notion of being numerically distinct does not
> 38 apply. . . (p. 114)
Compare Teller's mock-up with how I can go about
> making two cakes in two days:
1) I can bake both cakes today.
> 2) I can bake both cakes tomorrow.
> 3) I can bake one cake today and the other cake tomorrow.
But after baking the cakes, when I consider how I
> MIGHT have gone about baking them, there are FOUR
> possibilities rather than THREE.
--I could have baked both cakes yesterday.
> --I could have baked both cakes today.
> --I could have baked this one yesterday
> and that one today.
> --I could have baked that one yesterday and
> this one today.
Homework for David Ullrich:
In what respect do cakes I am going to make resemble quanta?
(Hint:
> Like causes produce like effects.)
--John
Subject: Cardinality of sets
I have a feeling that the question I pose is silly( I have never
studied set theory rigorously) - Suppose there are sets A and B such
that there exits a bijective function f:A -> B. Also assume that the
cardinality of A and B is uncountabily infinite. Now suppose I
construct a set C = B U D, st that cardinality of D is finite( D is
not a subset of B). Then it seems to me that the cardinality of C is
different from A (even though they are both uncontabily infinite)
because I am not able to create a function g:A->C, st g is a
bijection.
S.
Subject: Re: Cardinality of sets
> I have a feeling that the question I pose is silly( I have never
> studied set theory rigorously) - Suppose there are sets A and B such
> that there exits a bijective function f:A -> B. Also assume that the
> cardinality of A and B is uncountabily infinite. Now suppose I
> construct a set C = B U D, st that cardinality of D is finite( D is
> not a subset of B). Then it seems to me that the cardinality of C is
> different from A (even though they are both uncontabily infinite)
> because I am not able to create a function g:A->C, st g is a
> bijection.
> S.
If you have an axiom of choice and a bijection from A to B, both A
and B being infinite, you can construct a bijection from A to B U D
for any finite D.
If A and B are countable and you have a bijection from N to either
of them, you don't even need the axiom of choice.
Subject: Re: Cardinality of sets
>I have a feeling that the question I pose is silly( I have never
>studied set theory rigorously) - Suppose there are sets A and B such
>that there exits a bijective function f:A -> B. Also assume that the
>cardinality of A and B is uncountabily infinite. Now suppose I
>construct a set C = B U D, st that cardinality of D is finite( D is
>not a subset of B). Then it seems to me that the cardinality of C is
>different from A (even though they are both uncontabily infinite)
>because I am not able to create a function g:A->C, st g is a
>bijection.
In general, if at least one of A and B are infinite, then the
cardinality of A U B is the same as the cardinality of the larger
set. One can establish this fact without the axiom of choice.
In your particular case, here is the concept of a direct proof : Take a
countable subset of E of A and displace the map f thereon by
n = |D| places. (I.e., define g(e_(n+1)) = f(e1), g(e_(n+2)) =
f(e_2), etc.). Then define g to map the first n elements of E
onto D. Let g agree with f for elements not in E.
his details.
--
Stephen J. Herschkorn herschko@rutcor.rutgers.edu
Subject: Re: Cardinality of sets
> In general, if at least one of A and B are infinite, then the
> cardinality of A U B is the same as the cardinality of the larger
> set. One can establish this fact without the axiom of choice.
That surely depends on how you define infinite. If you mean
infinite
= Dedekind infinite, then I agree. But to me, infinite means
larger
than any natural number, and with that definition, I disagree with your
conclusion.
--
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
<http://www.commoncouragepress.com/index.cfm?action=book&bookid=228>
Subject: Re: Cardinality of sets
>>In general, if at least one of A and B are infinite, then the
>>cardinality of A U B is the same as the cardinality of the larger
>>set. One can establish this fact without the axiom of choice.
>That surely depends on how you define infinite. If you mean
infinite
>= Dedekind infinite, then I agree. But to me, infinite means
larger
>than any natural number, and with that definition, I disagree with your
>conclusion.
I looked at Kunen for my conclusion. He defines infinite as do you, and
he is careful in the first chapter to note where AC is needed.
Oh, wait a minute. He is talking about *cardinals,* defined as
ordinals not bijective with any smaller ordinal. What he shows is that
the sum of two cardinals is the larger of the two if one of them is
infinite. Maybe you do need choice when talking of unions of arbitrary
sets. My bad.
--
Stephen J. Herschkorn herschko@rutcor.rutgers.edu
Subject: Re: Cardinality of sets
Visiting Assistant Professor at the University of Montana.
>In your particular case, here is the concept of a direct proof : Take a
>countable subset of E of A and displace the map f thereon by
>n = |D| places. (I.e., define g(e_(n+1)) = f(e1), g(e_(n+2)) =
>f(e_2), etc.). Then define g to map the first n elements of E
>onto D. Let g agree with f for elements not in E.
>his details.
I was just a bit more annoying: I did what you did n different times,
displacing only by 1 to fit in the first element of D, then taking
another sequence and displacing it by 1 to get the second element of
D, etc.
Just my usual If there's a hard way of doing it, why bother with the
simple way?
Subject: Re: Cardinality of sets
Visiting Assistant Professor at the University of Montana.
>I have a feeling that the question I pose is silly( I have never
>studied set theory rigorously) - Suppose there are sets A and B such
>that there exits a bijective function f:A -> B. Also assume that the
>cardinality of A and B is uncountabily infinite. Now suppose I
>construct a set C = B U D, st that cardinality of D is finite( D is
>not a subset of B). Then it seems to me that the cardinality of C is
>different from A (even though they are both uncontabily infinite)
>because I am not able to create a function g:A->C, st g is a
>bijection.
Oh, but you ARE able to create such a function when A and B are
infinite. If they are uncountable then it is just a bit easier.
Say D = {d1,....,dn}. We do the following:
The way we will construct g:A-> B U D is by making part of g equal
to f, and just fixing the rest so that we also map to D.
Pick some distinct elements a_1, a_2, ..., a_n in A.
Map a_1 to d1, a_2 to d2, ...., a_n to dn.
Now, for i=1,...,n, we pick an infinite sequence of elements of A (I
am assuming the Axiom of Choice, though I suspect you could do this
without it; you'll have to ask an expert on that):
pick a_{11}, a_{12},...., a_{1j},...
so that they are all pairwise distinct and distinct from each of
a_1, a_2,..., a_n.
Then pick new elements a_{21}, a_{22}, ...., a_{2j}, ....
also pariwise distinct and distinct from the previous elements (this
is where the uncountablity of A makes it easier). Continue the same
way picking
a_{31}, ..., a_{3j,} ...
.
.
.
a_{n1}, a_{n2}, ..., a_{nj}, ...
etc.
Now we define g:A-> B U D as follows:
g(a_1) = d1
g(a_2) = d2
...
g(a_n) = dn
g(a_{11}) = f(a_1)
g(a_{1(k+1)}) = f(a_{1k}) for k=1,2,3,....
g(a_{21}) = f(a_2)
g(a_{2(k+1)}) = f(a_{2k}) for k=1,2,3...
.
.
.
g(a_{n1}) = f(a_n)
g(a_{n(k+1)}) = f(a_{nk}) for k=1,2,3,...
and g(a) = f(a) for any a not already taken.
Then you can show that g is indeed bijective (from the fact that f is
bijective).
Subject: Re: Cardinality of sets
> Now, for i=1,...,n, we pick an infinite sequence of elements of A (I
> am assuming the Axiom of Choice, though I suspect you could do this
> without it; you'll have to ask an expert on that):
I'm no expert, but without AC there may be infinite sets that do not
admit a countably infinite subset. That's equivalent to saying there may
be infinite sets that are Dedekind-finite.
--
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
<http://www.commoncouragepress.com/index.cfm?action=book&bookid=228>
Subject: Re: Cardinality of sets
If you add a finite set to a uncountably infinite set, then you still
have
an uncountably infinite set, and if your domain is uncountably infinite
then
there should still be a bijective function.What about the identity
function?
Lurch
> I have a feeling that the question I pose is silly( I have never
> studied set theory rigorously) - Suppose there are sets A and B such
> that there exits a bijective function f:A -> B. Also assume that the
> cardinality of A and B is uncountabily infinite. Now suppose I
> construct a set C = B U D, st that cardinality of D is finite( D is
> not a subset of B). Then it seems to me that the cardinality of C is
> different from A (even though they are both uncontabily infinite)
> because I am not able to create a function g:A->C, st g is a
> bijection.
> S.
Subject: Maxium likelihood estimate
I need to estimate Gamma (and some other) distribution parameters of
some data I'm generating.
Anyone knows a freeware C library that can solve this fitting problem?
thanx,
--tzurs
Subject: Re: help me out here??
>I hope that at least one correspondent will point out that probabilities
>are numbers in [0,1] not percentages.
And that correspondant will have missed the point entirely.
Doug
Subject: Re: help me out here??
...
Questions 1: What is the probability ...
Answer: _____ %
I hope that at least one correspondent will point out that probabilities
> are numbers in [0,1] not percentages.
And percentages in [0,100] are numbers in [0,1].
Or does suffixing /100 suddenly make a number a non-number.
So 45 is a number, but 45/100 isn't a number.
Weird.
Phil
Subject: Re: help me out here??
> Hello..
> Im trying to figure out this quiz..
> If u know the answer, ( and how u get it) pls e-mail me back
Nope, ask here, get answer here.
> K.
> Could you please answer the following 2 questions and send me back your
> answers, please???
> First piece of information
> You have 100 large bowls and in each bowl you have 1000 balls.
> 45 of the large bowls have 700 black balls and 300 red balls. (total 
balls
=
> 1000)
> 55 of the large bowls have 300 black balls and 700 red balls. (total 
balls
=
> 1000)
> Questions 1: What is the probability that one randomly selected large
bowl
> have more black balls?
> Answer: _____ %
What did you try? What answer did it give you?
> Second piece of information
> Now from the 100 large bowls, 1 bowl is randomly selected. Then 12 balls
> are randomly drawn from the bowl ( with returning) with the following
> results: 8 were black balls and 4 were red balls.
> Question 2: What is the probability that this bowl has more black balls?
> Answer: ______ %
If you can't get the first one, I wouldn't even look at this one.
Phil
Subject: Re: help me out here??
>If u know the answer, ( and how u get it) pls e-mail me back
Maybe if you post it seven or eight more times, you'll get more luck.
Sheesh.
Doug
Subject: Re: help me out here??
> Hello..
> Im trying to figure out this quiz..
> If u know the answer, ( and how u get it) pls e-mail me back
> K.
> Could you please answer the following 2 questions and send me back your
> answers, please???
> First piece of information
> You have 100 large bowls and in each bowl you have 1000 balls.
> 45 of the large bowls have 700 black balls and 300 red balls. (total 
balls
=
> 1000)
> 55 of the large bowls have 300 black balls and 700 red balls. (total 
balls
=
> 1000)
> Questions 1: What is the probability that one randomly selected large
bowl
> have more black balls?
> Answer: _____ %
> Second piece of information
> Now from the 100 large bowls, 1 bowl is randomly selected. Then 12 balls
> are randomly drawn from the bowl ( with returning) with the following
> results: 8 were black balls and 4 were red balls.
> Question 2: What is the probability that this bowl has more black balls?
> Answer: ______ %
It still looks like homework.
Multiple posting of the same questions doesn't help.
Subject: Re: Help with Integral
> I should know this, but I was never that great with exponential
> integrals and derivitives.
> The original integral is a double integral where you are supposed
>to do variable substitution. Specifically changing of the variables to
>polar coordinates. I'll use S as an integral sign:
> SS(exp((x^2)+(y^2))dxdy)
> Then changing to polar:
> SS(r * (e^(r^2))dOdr)
> Is the above right or did I mess this up? If it is right, where do I
>go from here?
Similar to what ILC mentioned, unless the region of integration is
easily expressed in terms of polar coordinates, you probably will not be
able to express this in terms of elementary functions. Compare the
similar double integral of exp(-(x^2+y^2)). This is proportional to a
probability associated with a bivariate normal distribution. This
*might* be expressible in terms of the univariate normal CDF or the
error function.
--
Stephen J. Herschkorn herschko@rutcor.rutgers.edu
Subject: Re: Help with Integral
> I should know this, but I was never that great with exponential
> integrals and derivitives.
> The original integral is a double integral where you are supposed
> to do variable substitution. Specifically changing of the variables to
> polar coordinates. I'll use S as an integral sign:
> SS(exp((x^2)+(y^2))dxdy)
> Then changing to polar:
> SS(r * (e^(r^2))dOdr)
> Is the above right or did I mess this up? If it is right, where do I
> go from here?
This appears correct. Make the further substitution u = e^(r^2).
--
P.A.C. Smith
'If the Apocalypse comes, beep me.' <*> http://www.srcf.ucam.org/~pas51
Subject: Re: Java Math packages
> I have built up a list of Java Math packages. A lot seem to have
> overlaping features, but I am not an expert. Since, I am not an
> expert, I would like recommendations on what to use: I would like the
> union of features but the intersection of packages, ;) Imagine for
> example that I would like to build up a general math API, or extend
> the Java language.
> How about looking at:
> 1. GMP (if you rehost this ... it will be a big deal, but maybe someone
has
> done it)
Hu ? What do you mean ? rehost ?
I am not trying to write a package, I want to know what Java extension
package for Math is/are good (in cse many are needed). I want answers
from users.
I am working on JSci.sourceforge.net. if you developed GMP (is it in C
?) and want to drop your package or move to Java then may be, yes, we
can help.
> 2. MIRACL
Same as above.
BTW, MIRACL is C, I was talking about Java. Doesn't anyone here use
java ???
> Topics:
> 1. Numerical Analysis
> 2. Number Theory
> 3. Matrix Algebra
> 4. DEQ Solvers
ok.
> HTH, Flip
Subject: Re: Java Math packages
>> I have built up a list of Java Math packages. A lot seem to have
>> overlaping features, but I am not an expert. Since, I am not an
>> expert, I would like recommendations on what to use: I would like the
>> union of features but the intersection of packages, ;) Imagine for
>> example that I would like to build up a general math API, or extend
>> the Java language.
>> How about looking at:
>> 1. GMP (if you rehost this ... it will be a big deal, but maybe someone
has
>> done it)
>Hu ? What do you mean ? rehost ?
As in create a Java implementation of it. It's written in C currently
and there are builds for a variety of operating systems.
Subject: Derivatives
Can any nice function eventually be reduced to either 0 or an
elementary function by repeated differentiation.
By nice function, I mean a function that can be written on a (big)
sheet of paper using only:
1) The integers, pi, and e
2) The operations of +,*,^, and log
3) The sine function.
By elementary function, I mean sin and exp.
Subject: Re: Derivatives
> Can any nice function eventually be reduced to either 0 or an
> elementary function by repeated differentiation.
> By nice function, I mean a function that can be written on a (big)
> sheet of paper using only:
> 1) The integers, pi, and e
> 2) The operations of +,*,^, and log
> 3) The sine function.
> By elementary function, I mean sin and exp.
No, differentiate log(x) until the cows come home you will not get
sin(x) or exp(x).
Btw your definition of elementary function is highly unusual.
--
G.C.
Subject: mathematics and poetry
what do mathematics and poetry have in common?
The answer in :
www.telefonica.net/web/matematicasypoesia
best wishes
Subject: Quantum Gravity?
Comments by Jack Sarfatti on excerpts from:
The three perspectives on the quantum-gravity problem
and their implications for the fate of Lorentz symmetry1
Dipart. Fisica Univ. La Sapienza and Sez. Roma1 INFN
P.le Moro 2, I-00185 Roma, Italy
ABSTRACT
Each approach to the quantum-gravity problem originates from expertise
in one or another area
to attempt to reproduce in
quantum gravity as much as possible of the successes of the Standard
therefore, as done in String Theory, the core features of quantum
gravity are described in terms of
general-relativity perspective
it is natural to renounce to any reference to a background spacetime,
and to describe spacetime in a
way that takes into account the in-principle limitations of
measurements. The Loop Quantum Gravity
approach and the approaches based on noncommutative geometry originate
from this general-relativity
perspective. The condensed-matter perspective, which has been adopted in
a few recent quantum-
gravity proposals, naturally leads to scenarios in which some familiar
properties of spacetime are
only emergent, just like, for example, some emergent collective degrees
of freedom are relevant to the
description of certain physical systems only near a critical point. Both
from the general-relativity
perspective and from the condensed-matter perspective it is natural to
explore the possibility that
quantum gravity might have significant implications for the fate of
Lorentz symmetry in the Planckian
obvious reason to renounce to exact
Lorentz symmetry, although (spontaneous) Lorentz symmetry breaking is
of course possible. A fast-
growing phenomenological programme looking for Planck-scale departures
from Lorentz symmetry can
contribute to this ongoing debate.
1Based on invited seminars given at Perspectives on Quantum Gravity: a
tribute to John Stachel
Relativity (Rio de
1
1 Preliminaries
1.1 Lorentz symmetry and the three perspectives on the Quan-
tum Gravity problem
Each quantum-gravity research line can be connected with one of three
perspectives
perspective and
the condensed-matter perspective.
reproduce as much
is tempted to
see gravity simply as one more gauge interaction. Among the
quantum-gravity open
issues the failure of perturbative renormalization in naive
quantum-gravity is perceived
solution of the
quantum gravity problem would be String-Theory-like: a quantum gravity
whose core
features are essentially described in terms of graviton-like exchange in
a background
classical spacetime.
The general-relativity perspective naturally leads to reject the use of
a background
spacetime, and this is widely acknowledged [1, 2].
[Comment # 1: As long as the c-number background spacetime obeys local
Diff(4) symmetry (in the zero torsion limit) there should be no real
conceptual problem here especially if this c-number background spacetime
is a collective emergent More is different (P.W. Anderson) mode in the
sense of the condensed matter approach to quantum gravity further
discussed below.]
Although less publicized, there is
also growing awareness of the fact that the development of general
relativity relied
heavily on the careful consideration of the in-principle limitations
that measurement
procedures can encounter. Think for example of the limitations that the
speed-of-light
limit imposes on certain setups for clock synchronization and of the
contexts in which
it is impossible to distinguish between a constant acceleration and the
presence of a
gravitational field. In light of the various arguments (some briefly
reviewed later in
these notes) suggesting that, whenever both quantum mechanics and
general relativity
are taken into account, there should be an in-principle limitation to
the localization of
a spacetime point (an event), the general-relativity perspective invites
one to renounce
to any direct reference to a classical spacetime [3, 4, 5, 6, 7]. Indeed
this requirement
that spacetime be described as fundamentally nonclassical
(fundamentally quantum),
that the in-principle measurability limitations be reflected by the
adoption of a corresponding
measurability-limited description of spacetime, is another element of
intuition
which is guiding quantum-gravity research from the general-relativity
perspective. This
naturally leads us to consider certain types of discretized spacetimes,
as in the Loop
Quantum Gravity approach, or noncommutative spacetimes. Loop Quantum
Gravity
is also a background-independent approach and therefore combines both
elements of
the general-relativity perspectivea. Noncommutative spacetimes could be
introduced
in a background-independent way, as in preliminary attempts reported in
Ref. [3] and
follow-up work, but in most studies, for simplicity, noncommutative
spacetimes are
adopted as background spacetimes [8, 9, 10, 11, 12] (leading to an
approach which in a
perspective and
the general-relativity perspective).
The third possibility is a condensed-matter perspective (see, e.g., the
research programs
of Refs. [13] and [14]) on the quantum-gravity problem, in which some of
the
Note a: Although it must be noted that this is actually achieved, so
far, at the price of some possibly
concerning compromises. For example, as stressed by John Stachel and
others, one could be concerned
of the fact that most of the Loop-Quantum-Gravity results are obtained
preserving only 3D (space)
diffeomorphisms.
[Comment #2: This is a serious defect in the Loop-Quantum Gravity
program IMHO]
1
familiar properties of spacetime are only emergent. Condensed-matter
theorists are
used to describe some of the degrees of freedom that are measured in the
laboratory as
collective excitations within a theoretical framework whose primary
description is given
in terms of much dierent, and often practically unaccessible,
fundamental degrees of
freedom. Close to a critical point some symmetries arise for the
collective-excitations
theory, which do not carry the significance of fundamental symmetries,
and are in fact
lost as soon as the theory is probed somewhat away from the critical
point. Notably,
some familiar systems are known to exhibit special-relativistic
invariance in certain
limits, even though, at a more fundamental level, they are described in
terms of a nonrelativistic
theory. For a rather general class of fermionic systems one finds [13]
that at
low energies, as a Fermi point is approached, fermions gradually become
chiral Weyl
fermions, while bosonic collective modes of the vacuum transform into
gauge fields and
gravity.
Clearly from the (relatively new) condensed-matter perspective on the
quantum gravity
problem it is natural to see the familiar classical continuous Lorentz
symmetry
only as an approximate (emergent) symmetry.
[Comment #3: This notion is very compatible with the Bohm-Hiley-Vigier
pilot BIT wave landscape IT hidden variable system point flow on the
landscape picture of quantum theory. Orthodox micro-quantum theory is
the limit of Antony Valentini's sub-quanta heat death with signal
locality in which the quantum potential is fragile (Bohm-Hiley) with
a pilot BIT wave (nonlocal in configuration space for entangled systems)
that has no sources (Bohm-Hiley). This means that the IT hidden
In other words IT gets its marching orders (J.A. Wheeler) from BIT
but not vice versa. This is action without reaction! Here reaction
means back-action where the BIT pilot wave landscape has sources so
figuratively speaking that it warps the BIT landscape it is flowing on
according to
vs = (h/m)Grad(Phase) - (q/c)A
or in the 4D elastic world crystal lattice (Hagen Kleinert) model
distortion field du(x)
du(x) = Lp*^2 (Goldstone Phase),u - TaA^a,u
Lp*^2 = hG*/c^3 = Lp^4/3(c/Ho)^2/3 in t'Hooft-Susskind world hologram
model for renormalized Planck scale Lp* ~ 1 fermi determined by
Newtonian Planck scale Lp = 10^-33 cm and size 10^28 cm of our
contingent local Level I (Max Tegmark) parallel IT universe in an
infinity of parallel IT universes on a single 3D spatially flat
American).
Ta are the generators of the Lie algebra of the standard
model in globally flat Minkowski spacetime with fiber connections A^a,u
for parallel transport. a is internal space and u is spacetime. This is
normalized such that TaA^a,u is a length. ,u is the ordinary partial
derivative operator relative to x^u, u = 0, 1,2,3.
The attraction between virtual electrons and positrons near the -mc^2
band in the Dirac vacuum spectrum Fermi sphere is a pair instability
that creates the virtual giant local vacuum wave BEC <0|e+(x)e-(x)|0> a
complex scalar inflation field whose phase is the Goldstone phase and
whose amplitude is the Higgs field
<0|e+(x)e-(x)|0> = |Higgs field(x)|e^i(Goldstone phase(x))
such that the unified dark energy/matter zero point stress-energy
density Diff(4) tensor field is :
tuv(x)zpf = (c^4/8piG*)/zpf(x)guv(x)
where
/zpf(x) = Lp*^-2[1 - Lp*^3|Higgs field(x)|^2]
/zpf(x) > 0 is strongly scale-dependent anti-gravitating dark energy
exotic vacuum with negative pressure and positive zero point energy
density.
/zpf(x) < 0 is strongly scale-dependent gravitating dark matter
exotic vacuum with positive pressure and negative zero point energy
density.
All quantum fields of all spins contribute to the net residual /zpf(x)
LOCAL FIELD in the emergent More is different collective mode c-number
background spacetime with both Diff(4) base space and local Lorentz
group tangent space emergent symmetries at least near a critical point
of the renormalization group flow sense.
Einstein's gravity c-number background field is
guv(x) = nuv(Minkowski) + (1/2)[du(x),v + dv(x),u]
with anholonomic torsion tensor field
suv(x) = (1/2)[du(x),v - dv(x),u]
and the back-action nonlinear LOCAL virtual BEC Landau-Ginzburg Diff(4)
symmetric equation:
{D^uDu + V(<0|e+(x)e-(x)|0>)}<0|e+(x)e-(x)|0> = 0
Where V(<0|e+(x)e-(x)|0>) limits to chaotic inflation cosmology in the
large-scale limit of the FRW metric with cosmological constant.
Du is the Diff(4) operator so that D^uDu is the GR wave propagation
operator on a complex numbered scalar field.]
to be continued.
Note there is no experimental number anywhere in the cutting edge of
physics today so far that seems to require quantum gravity or M
theory.
to be continued
Subject: Re: FUNctions/Continued-Fraction Puzzle
> (This might be actually trivial. But it does not seem to be with the
> little thought I have given it. In any case, perhaps I should not have
> cross-posted this to rec.puzzles {if I should have even posted it to
> sci.math}; but what the...)
> For all real x > 1, and for some function of x, y(x);
> where each y is a real, y =y(x), based on x:
> it is so that:
> f([x; x^2, x^3, x^4,...,x^m]) =
> [y; y^2, y^3, y^4,...,y^m],
> for EVERY positive integer m;
> where:
> [x; x^2, x^3, x^4,...,x^m] is the continued-fraction
> 1
> x + ------------------ ;
> 1
> x^2 + --------------
> 1
> x^3 + ---------
> ....
> + 1/x^m
> and [y; y^2, y^3, y^4,...,y^m] is also a continued-fraction (obviously);
> and [x; x^2, x^3, x^4,...,x^m] converges to X;
> and f(w) is a real -> real function, such that
> f'(X) exists and is finite nonzero.
> So, what are the possible f(w)'s, given all of the conditions above??
First, by the way, f'(X) is the (1st) derivative of f(w) at w = X, in
case this is not obvious.
I should mention that f can equate to an infinite number of functions
if it need not be analytic. If it need by analytic, however, there are
a finite number of possible functions that can equal f(w).
(So, find the set of analytic f(w)'s.)
This puzzle seems to be more difficult than I first assumed.
I will wait until Friday, at least, to post the answer if no one else
posts the solution before that.
Leroy Quet
Subject: Re: Folk Psychology and Social Convention
> Michaels in replying to my initial post to comp.ai.philosophy,
> crossposted this thread comp.ai.genetic, sci.math, talk.bizarre. I
> presume, in order to foment further confusion and mayhem (as these other
> readers, having no idea of the relevant history, might indeed think
> these exchanges were bizarre).
Follows is the first post in this thread:
Message 1 in thread
Subject: Re: Folk Psychology and Social Convention
>> http://www.behavior.org/journals_BP/2000/Place.pdf
>> There is quite a large literature on social conformity which could be
>> referred to, but whilst of some heuristic value (in the helpful,
>> illustrative sense) I really find that social psychological literature
>> relatively trivial compared to the rather detailed and powerful work
>> which has been done for decades in what I rather loosely refer to in
>> this newsgroup as behaviour analysis and learning theory.
>find that very conforming to hear again.
Subject: Re: Folk Psychology and Social Convention
<MeIoIJMZwVf$Ews4@longley.demon.co.uk>
<NyVoTWSvWef$Ewen@longley.demon.co.uk>
<0kEGGeaptqf$Ew4p@longley.demon.co.uk> Michaels in replying to my initial 
post
to comp.ai.philosophy,
>> crossposted this thread comp.ai.genetic, sci.math, talk.bizarre. I
>> presume, in order to foment further confusion and mayhem (as these other
>> readers, having no idea of the relevant history, might indeed think
>> these exchanges were bizarre).
>Follows is the first post in this thread:
>Message 1 in thread
>Subject: Re: Folk Psychology and Social Convention
> http://www.behavior.org/journals_BP/2000/Place.pdf
> There is quite a large literature on social conformity which could be
> referred to, but whilst of some heuristic value (in the helpful,
> illustrative sense) I really find that social psychological literature
> relatively trivial compared to the rather detailed and powerful work
> which has been done for decades in what I rather loosely refer to in
> this newsgroup as behaviour analysis and learning theory.
>>find that very conforming to hear again.
oo
>> http://www.behavior.org/journals_BP/2000/Place.pdf
>> There is quite a large literature on social conformity which could be
>> referred to, but whilst of some heuristic value (in the helpful,
>> illustrative sense) I really find that social psychological literature
>> relatively trivial compared to the rather detailed and powerful work
>> which has been done for decades in what I rather loosely refer to in
>> this newsgroup as behaviour analysis and learning theory.
>find that very conforming to hear again.
>Longley Citation Statistics [8 long years and counting]
>-------------------------------------------------------
>folk psychology [derogatorily] in over 1000 posts
>Quine [glowingly] in over 1800 posts
>silly [regarding others' ideas] over 400 times
>Skinner [in awe] in over 800 posts
>behaviorism [as accepted dogma] in over 500 posts
>[pernicious] mentalism over 100 times
>[pernicious] cognitivism over 250 times
>intensional [as wrongish] over 1600 times
>extensional [as correctish] over 1850 times
>frag.html [self-promotion] over 300 times
You would *appear* to have presented my reply of 3rd October (the 3rd
message in this thread) as the first message in this thread.
In fact, my first post was on 1 October, as was your reply.
You appear to have edited my subsequent reply of 3rd October in order
to mislead or deceive others. I don't know why anyone would try to do
that. Is this perhaps another one of those careless or accidental
mis-quotations or edits perhaps? Or maybe an example of the sort of
thing that you did once or twice some time back when reporting other
peoples' work?
It would also seem to be the case that someone has in fact deleted your
original message of the 1st October, making it look like my subsequent
reply on the third was the cross-poster. But one can see from that post
of mine that it was in fact a REPLY to you. Do you know where that post
of yours might have gone? It's rather unfortunate as it makes you look
bad - rather devious and dishonest if we didn't know better. No matter,
I've included it below in case you *accidentally* lost it or deleted it
and would like to examine it more closely.
oOo
Note the date and where it's posted to. Compare with mine of 1 October.
news.demon.co.uk!mutlu.news.demon.net!kibo.news.demon.net!demon!news-hog.
Subject: Re: Folk Psychology and Social Convention
sci.math:124582 talk.bizarre:24512
http://www.behavior.org/journals_BP/2000/Place.pdf
> There is quite a large literature on social conformity which could be
> referred to, but whilst of some heuristic value (in the helpful,
> illustrative sense) I really find that social psychological literature
> relatively trivial compared to the rather detailed and powerful work
> which has been done for decades in what I rather loosely refer to in
> this newsgroup as behaviour analysis and learning theory.
find that very conforming to hear again.
Longley Citation Statistics [8 long years and counting]
-------------------------------------------------------
folk psychology [derogatorily] in over 1000 posts
Quine [glowingly] in over 1800 posts
silly [regarding others' ideas] over 400 times
Skinner [in awe] in over 800 posts
behaviorism [as accepted dogma] in over 500 posts
[pernicious] mentalism over 100 times
[pernicious] cognitivism over 250 times
intensional [as wrongish] over 1600 times
extensional [as correctish] over 1850 times
frag.html [self-promotion] over 300 times
-------------------------------------------------------------------------
--------------------------------
Subject: Folk Psychology and Social Convention
http://www.behavior.org/journals_BP/2000/Place.pdf
There is quite a large literature on social conformity which could be
referred to, but whilst of some heuristic value (in the helpful,
illustrative sense) I really find that social psychological literature
relatively trivial compared to the rather detailed and powerful work
which has been done for decades in what I rather loosely refer to in
this newsgroup as behaviour analysis and learning theory.
--
David Longley
-------------------------------------------------------------------------
------------------------------
Subject: Re: Folk Psychology and Social Convention
sci.math:125158 talk.bizarre:24780
>> http://www.behavior.org/journals_BP/2000/Place.pdf
>> There is quite a large literature on social conformity which could be
>> referred to, but whilst of some heuristic value (in the helpful,
>> illustrative sense) I really find that social psychological
literature
>> relatively trivial compared to the rather detailed and powerful work
>> which has been done for decades in what I rather loosely refer to in
>> this newsgroup as behaviour analysis and learning theory.
>find that very conforming to hear again.
>Longley Citation Statistics [8 long years and counting]
>-------------------------------------------------------
>folk psychology [derogatorily] in over 1000 posts
>Quine [glowingly] in over 1800 posts
>silly [regarding others' ideas] over 400 times
>Skinner [in awe] in over 800 posts
>behaviorism [as accepted dogma] in over 500 posts
>[pernicious] mentalism over 100 times
>[pernicious] cognitivism over 250 times
>intensional [as wrongish] over 1600 times
>extensional [as correctish] over 1850 times
>frag.html [self-promotion] over 300 times
Yet despite this, I bet we see yet more of your asinine
misrepresentations of popular neuroscience in posts to
comp.ai.philosophy. If you actually read some of the material you
irreverently *count* and disparage above, you might learn why your posts
amount to little more than science fiction and why they have next to
nothing to do with the philosophy of AI.
The above terms are exactly those which should occur in a forum which is
specifically for the discussion of the philosophy of AI, something you'd
discover for yourself if you actually read and understood any of the
relevant literature.
--
David Longley
oOo
I'm not sure how this all might have happened but I am intrigued - it
could make you look pretty bad......
Perhaps it's got something to do with my Forensic/Applied Criminological
Psychologist background - it makes me a bit over sensitive to
criminogenic behaviours - so do help me sort this all out. I'm sure
there must be another explanation.
--
David Longley
Subject: Re: Folk Psychology and Social Convention
>How do you recognize what you see?
>How do you know how to move your arm?
>How do you choose which words to say?
>How do you understand what they mean?
>How does commonsense reasoning work?
> How do we keep wankfests like this out of sci.math?
> Lee by repeatedly setting followups, I suppose Rudolph
Look at the posts to see who first cross-posted it, and send him a nice
msg.
Subject: Re: Folk Psychology and Social Convention
>> How do we keep wankfests like this out of sci.math?
>> Lee by repeatedly setting followups, I suppose Rudolph
>Look at the posts to see who first cross-posted it, and send him a nice
msg.
So intelligence, human or artificial, comes down in the end to
nothing more than monkey see, monkey do? That's glory for you.
Lee Rudolph
Subject: Re: Folk Psychology and Social Convention
>The peephole you view your world through is very narrow, Longley.
>There are many here who think the following matters are the crux of
>building AI's:
>How do you recognize what you see?
>How do you know how to move your arm?
>How do you choose which words to say?
>How do you understand what they mean?
>How does commonsense reasoning work?
Stop misquoting Lennon & McCartney!
how do you recognize just what you've seen?
i can't tell you but i know it's mine
how do you know how to move your own arm?
i get by with a little help from my friends
how do you choose which big words to say?
lend me your ears and i'll sing you a song
how do you understand what they mean?
i try not to sing out of key
does commonsense reasoning work?
yes i'm certain that it happens all the time
oh building AIs with a little help from my friends
* * *
weary with foils, i fall into my bed
and sought some rest for limbs with dust attired,
but there began a trial in my head
to obsess, when body's overtired:
my thoughts then race, unlike their daily pace
when they escape and leave me little grace;
for verbal slings and arrows of bon mots
i am left gawking bereft of ripostes.
look in darkness on what might have been said
had i the wits to speak as those admired:
rejoinders, like galaxies, hang in space
and taunt me in my unrestful repose.
this sort of pesky followup is never very far
when you keep sending crosspostings to here, talk dot bizarre.
.
. thank you thank you i'll be here at dogberry's fencing school all
weekend
Subject: Sum{k=1 to oo} d(k,n)/k^r = 1/zeta(r)^n
If:
sum{k=1 to oo} d(k,n)/k^r = 1/zeta(r)^n
for all r where sum converges,
then it is well-known that
d(k,1) = mu(k), and
d(k,n) = sum{j|k} mu(k/j) d(j,n-1),
for every n >= 2.
So, just a little number theory for now:
If p = prime, m = nonnegative integer,
d(k,p^m) = binomial(n,m) (-1)^m.
And, if the lowest-prime-divisor of n is >= k, k>= 2, then
d(k,n) is congruent to (n!/(n-k)!) (-1)^k (mod n).
These are derived from:
d(k,n) = sum{m=0 to n} binomial(n,m) c(k,m) (-1)^m,
where c(k,m) is discussed in:
Implied Sequence (#-Theory) Puzzle
and in:
A certain Dirichlet-sum: question
Leroy Quet
Subject: (typo fixed)Sum{k=1 to oo} d(k,n)/k^r = 1/zeta(r)^n
(repost with d(k,p^m) -> d(p^m,n). Sorry about the error.)
----
If:
sum{k=1 to oo} d(k,n)/k^r = 1/zeta(r)^n
for all r where sum converges,
then it is well-known that
d(k,1) = mu(k), and
d(k,n) = sum{j|k} mu(k/j) d(j,n-1),
for every n >= 2.
So, just a little number theory for now:
If p = prime, m = nonnegative integer,
d(p^m,n) = binomial(n,m) (-1)^m.
And, if the lowest-prime-divisor of n is >= k, k>= 2, then
d(k,n) is congruent to (n!/(n-k)!) (-1)^k (mod n).
These are derived from:
d(k,n) = sum{m=0 to n} binomial(n,m) c(k,m) (-1)^m,
where c(k,m) is discussed in:
Implied Sequence (#-Theory) Puzzle
and in:
A certain Dirichlet-sum: question
Leroy Quet
Subject: Re: [Fwd: integral of x^x]
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision:
1.9 primary) id h8U23b205633;
I've just graduated from the University of Kansas (undergrad w/ a degree in
math) and I derived the open form integrand of x^x my sophmore year in
college...is this something which should be a big deal?
btw...it took me a huge effort (spanning an entire summer and part of a
semester) to derive the integral. It was like beating myself with a mental
brick...
Jeremy
><preThis is a multi-part message in MIME format.
>--------------D12BE1EBA320167FC2A796D1
>As suggested, I am posting this is in Sci.math. If anyone has any
>further references or info I would be grateful for it.
>--------------D12BE1EBA320167FC2A796D1
>Subject: Re: integral of x^x
>>While digging out some old magazines I found a copy of the November 1979
>>Scientific Australian, on page 44 it has a math problem held over from
>>the previous month; Find integral x^x dx. In 'Last Month's
Solutions'
>>on the same page the following commentary appears; Really we are
>>beginning to feel that the solution to this thing rests upon a whole new
>>area of function theory - at least that's the feeling one gets while
>>trying various techniques. It is really fascinating. Try it yourself
>>... substitution, parts, expansion then term by term ... and see where
>>you end up. You could just about write a book about this one function:
>>x^x. Looking forward to hearing your ideas.
>>In the 80's at a PC Show at Centrepoint in Sydney I asked the salesman
>>at the Mathematica stand to see what solution the program would give to
>>this question. It did not find a solution.
>>Does anyone know of a solution or of any work I could read about this
>>problem?
>You'd do better to ask this in sci.math...
>The integral has no solution in elementary terms. It is possible to
>represent the integral as e^(x*ln(x)), then break that down into a power
>series:
>1 + x*ln(x) + (x*ln(x))^2/2! + (x*ln(x))^3/3!...
>then integrate each term, but you'd have to be a masochist to want to do
>it.
>Btw, just because the indefinite integral is not known, that does not mean
>some definite integrals are not known:
>integral [ from 0 to 1 ] { x^x dx } = 1 - 1/2^2 + 1/3^3 - 1/4^4 + 1/5^5...
>integral [ from 0 to 1 ] { dx/x^x } = 1 + 1/2^2 + 1/3^3 + 1/4^4 + 1/5^5...
>Proofs of those are available at <a
href=http://www.dejanews.com...>http://www.dejanews.com...</a> you'll
have
>to search sci.math creatively, though, since ^ is considered a special
>character.
>--
>Glenn Lamb - mumford@netcom.com. Finger for my PGP Key.
>mail will be bounced automatically as spam.
>PGPprint = E3 0F DE CC 94 72 D1 1A 2D 2E A9 08 6B A0 CD 82
>--------------D12BE1EBA320167FC2A796D1--
></pre>
Subject: Re: [Fwd: integral of x^x]
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision:
1.9 primary) id h8U17DO01742;
I've just graduated from the University of Kansas (undergrad w/ a degree in
math) and I derived the open form integrand of x^x my sophmore year in
college...is this something which should be a big deal?
btw...it took me a huge effort (spanning an entire summer and part of a
semester) to derive the integral. It was like beating myself with a mental
brick...
Jeremy
><preThis is a multi-part message in MIME format.
>--------------D12BE1EBA320167FC2A796D1
>As suggested, I am posting this is in Sci.math. If anyone has any
>further references or info I would be grateful for it.
>--------------D12BE1EBA320167FC2A796D1
>Subject: Re: integral of x^x
>>While digging out some old magazines I found a copy of the November 1979
>>Scientific Australian, on page 44 it has a math problem held over from
>>the previous month; Find integral x^x dx. In 'Last Month's
Solutions'
>>on the same page the following commentary appears; Really we are
>>beginning to feel that the solution to this thing rests upon a whole new
>>area of function theory - at least that's the feeling one gets while
>>trying various techniques. It is really fascinating. Try it yourself
>>... substitution, parts, expansion then term by term ... and see where
>>you end up. You could just about write a book about this one function:
>>x^x. Looking forward to hearing your ideas.
>>In the 80's at a PC Show at Centrepoint in Sydney I asked the salesman
>>at the Mathematica stand to see what solution the program would give to
>>this question. It did not find a solution.
>>Does anyone know of a solution or of any work I could read about this
>>problem?
>You'd do better to ask this in sci.math...
>The integral has no solution in elementary terms. It is possible to
>represent the integral as e^(x*ln(x)), then break that down into a power
>series:
>1 + x*ln(x) + (x*ln(x))^2/2! + (x*ln(x))^3/3!...
>then integrate each term, but you'd have to be a masochist to want to do
>it.
>Btw, just because the indefinite integral is not known, that does not mean
>some definite integrals are not known:
>integral [ from 0 to 1 ] { x^x dx } = 1 - 1/2^2 + 1/3^3 - 1/4^4 + 1/5^5...
>integral [ from 0 to 1 ] { dx/x^x } = 1 + 1/2^2 + 1/3^3 + 1/4^4 + 1/5^5...
>Proofs of those are available at <a
href=http://www.dejanews.com...>http://www.dejanews.com...</a> you'll
have
>to search sci.math creatively, though, since ^ is considered a special
>character.
>--
>Glenn Lamb - mumford@netcom.com. Finger for my PGP Key.
>mail will be bounced automatically as spam.
>PGPprint = E3 0F DE CC 94 72 D1 1A 2D 2E A9 08 6B A0 CD 82
>--------------D12BE1EBA320167FC2A796D1--
></pre>
Subject: Re: Question About Prime Numbers
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision:
1.9 primary) id h8U4fJ515831;
Interesting problem, Russell!
For p=11,
minimal d = 4911773580 (OEIS A088430),
and AP contains maximal number, 11, primes.
For p=13, d should be a factor of 2310.
Who first find it (and then try 17,19,...)?
Zak
BTW I guess that found d is indeed minimal not unique-
there is no reason of absense of other larger d's.
>Twin primes are prime numbers such as 5 and 7, 11 and 13, 17 and 19,
>etc. These twins are only one unit apart.
>There are strings of prime numbers that are n-units apart:
>3, 5, 7, [3 prime numbers, 2 units apart]
>5, 11, 17, 23, 29, [5, 6 units]
>7, 157, 307, 457, 607, 757, 907, [7, 150 units]
>11... ? ...? ...? ...
>The question becomes: For all odd prime numbers P, are there P number of
>primes that are the same numerical[equal] distance apart?
Subject: Re: Linear Algebra Question
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision:
1.9 primary) id h8UFcnm27613;
>I have the following problem, and it nags at the back of my mind that the
>result is well-known, but I can't think where. Hopefully, the gurus here
can
>help me.
>If A is a real, symmetric, positive semi-definite matrix with largest
>eigenvalue lambda (assumed algebraic multiplicity 1), then it is possible
to
>choose the associated eigenvector to have all non-negative entries.
>Tim Norfolk
>Just when we manage to idiot-proof something, Nature comes along and
builds a
>better idiot.
Subject: Re: Ridicule This Crackpot
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision:
1.9 primary) id h8UHiGj04678;
>>I noticed that you had posted a number of times and that nobody had
responded. I think that's because there's not much to say about them.
>>They look fractal-like to me (that doesn't count for much), but I
>>don't think they're unusually significant because
>>anyone on here could have done them. There are infinite
>>variations along similar lines, I suppose.
>>I'm sure that there are people that appreciate them, and there's no
>>harm in mentioning them, but apparently there's not much to reply
>>about.
>Another compliment...
>>They look fractal-like to me
>I feel the same way, and that also does not count for much.
>I'm not sure what you mean by anyone on here could have
>done them. Does that imply that I could have plagiarized from
>someone, or that the formula is so simple that anyone could
>easily have explored similar ideas before? I am only concerned
>with the second. If anyone knows of what this is or where I can
>read about it, I would really like to know. I may be an outsider,
>but it is my understanding that there are finite and relatively few
>kinds of different fractals. And a novel method of describing one
>would be of interest.
I meant that I think most people who are adept at math could
produce spirals within spirals, etc, with a little effort. That
doesn't mean that someone has already produced your version.
I don't think anyone will reply and say that they are new,
because who knows?
As I said, they appear to be fractals to me (at a glance).
Sorry I'm not more help - I only responded because nobody else did.
You could search on the internet or in books. For a
start, you might try the following site:
http://mathworld.wolfram.com/Fractal.html
Subject: Re: Ridicule This Crackpot
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision:
1.9 primary) id h8ULfpn23454;
Another compliment !!!
the anime flick? I can assure you that the images are not drawn
by hand. I have made corrections from some of your
recommendations, except for the color of the page. (The subject
line of this thread is colorful enough.) I have fixed the broken link:
http://fractalspiral.zxcvb.org/images/im1.html
There is way too much detail for hand drawing to be possible. I
have fixed the formulas and removed all symbols and replaced
them with more universal letters. I also went crazy with
parentheses in case my understanding of the order of
operations is incorrect.
But more importantly I have included the algorithm so people
can reproduce my results! The algorithm is visible at my website:
http://fractalspiral.zxcvb.org
But I will also show it here, for people who use non-standard
browsers. Please note that the example is in faux code meant
only to convey the ideas. It looks like basic to me. The only
differences will be the syntax of the for ... next loops, sine and
cosine, and the width and height of the screen and function
drawPointAt is not defined. It should plot one pixel.
variable = .325
totalIterations = 100
totalPoints= 1000
radius = 25
xcenter = widthOfScreen / 2
ycenter = heightOfScreen / 2
for point = 1 to totalPoints
x = 0
y = 0
theta = point * ( (pi * 2) / totalPoints)
for iteration = 1 to totalIterations
n = (iteration*theta) + (iteration*iteration*pi*variable)
newRadius = radius / iteration
x = x + ( sin(n) * newRadius )
y = y + ( cos(n) * newRadius )
next iteration
drawPointAt (xcenter + x, ycenter + y)
next point
My big question is how can I insure that I receive credit for this
formula and algorithm? (Assuming that mathematicians decide
that it is novel and concede the slightest bit of significance.) Any
advice would be appreciated.
Subject: Re: Prime numbers
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision:
1.9 primary) id h8ULwrb24499;
><pre Does anyone know an effective algorithm for finding prime numbers
>> I've made two algorithms:
>> The first one, uses the sieve of Erathosthenes (is it spelled correctly?
in
>Eratosthenes
>> Spanish is Criva de Erat.97stenes), and I use it in conjunction with
dividing
>> only by odd numbers.
>> The other algorithm (which happens to be slower), also uses main concept
of
>> the first, but instead of dividing by odd numbers, it divides by the
already
>> found primes.
>Neither is the _real_ sieve of Eratosthenes. You don't need to divide
>by anything. You just skip through the array, with a skip distance
>equal to the most recently found prime, wiping out every number
>you meet. This only requires adding, not dividing.
>Assume an array, P : ARRAY [2..N] OF BOOLEAN;
>already filled with TRUE;
>In Modula 2 this would be
>i := 2; j := 2;
>WHILE i <= TRUNC(SQRT(N))) DO
> (* no need to look further than sqrt(N) - prove this *)
> WHILE P[i] & (i < TRUNC(SQRT(N))) DO INC[i, 1] END;
> (* search for next number not previously shown to be composite,
> the smallest such must be prime *)
> WHILE j < N DO INC[j, i]; P[j] := FALSE END
> (* wipe out all multiples, k*i, k > 1 *)
>END;
>(* now P[i] = true if and only if i is prime *)
>Try this against your algorithm and see which is best.
>--
>Terry Moore, Statistics Department, Massey University, New Zealand.
>Theorems! I need theorems. Give me the theorems and I shall find the
>proofs easily enough. Bernard Riemann
></pre>
Subject: Re: Prime numbers
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision:
1.9 primary) id h8ULwrF24495;
><pre Does anyone know an effective algorithm for finding prime numbers
>I've made two algorithms:
>The first one, uses the sieve of Erathosthenes (is it spelled correctly?
in
>Spanish is Criva de Erat.97stenes), and I use it in conjunction with
dividing
>only by odd numbers.
>The other algorithm (which happens to be slower), also uses main concept
of
>the first, but instead of dividing by odd numbers, it divides by the
already
>found primes.
>I'm sure there are better and more efficient algorithms, but these work
>fine... at least for not huge numbers.
>Fernando Gonz.87lez del Cueto.
>Matem.87ticas Aplicadas
>Instituto Tecnol.97gico Aut.97nomo de M.8exico
>Ciudad de M.8exico
></pre>
Subject: foundation for gifted individuals:
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision:
1.9 primary) id h8ULfoD23441;
here is the provisional website of a foundation supporting e.g.
scientific projects of gifted people:
http://wwwhomes.uni-bielefeld.de/cschroeter/Stiftung-e.htm
The foundation satisfies the strict criteria of being officially
certified as being of public utility, the founder is a leading
figure in germanys economy.
In case you know someone who may satisfy the criteria for being
supported by that foundation, feel free to give him/her the link.
Thomas Riepe
Subject: Re: Talk to a mathematician (Op-Ed)
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision:
1.9 primary) id h8UMc7827418;
The problem is this:
Many college freshmen, etc. have heard a lot of stories about famous
mathematicians who were boy wonders. A la: Oh, Gauss was two when he
corrected a mistake in his father's audit.... Einstein did this or that...
When they finally make it to college they think Hey, I`m already 19 and
should at least have one or two theories under my belt. Let`s admit we all
had our own theories when we were 19 about math and the world, and yes, 
mabie
we would have posted it to the internet too, if there`d been internet back
then. So I propose we call it a truce.
Subject: Re: integers
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision:
1.9 primary) id h914Rwh20479;
>>I don't know if 4c^3=b(3x^2+b^2) has any integer solutions.
>> Where c, b and x are positive integers the left side can never equal the
>right side.
>Huh? c=b=x=1.
>Jon Miller
You are right, I missed that only answer.
However my addition is more interesting though ---
4c^3=(b(3x^2+b^2))+(x-b)^2
This ----->oo of the number of solutions.
Portly
Subject: Re: Professor McKenzie, problem with ring of A.I.
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision:
1.9 primary) id h8UNNnX30844;
I have no idea what forum to post this in so here it goes,
i need a formula that will give me the answer to
(1*1)+(2+2)+.....(999*999)
Subject: Re: Professor McKenzie, problem with ring of A.I.
> I have no idea what forum to post this in so here it goes,
> i need a formula that will give me the answer to
> (1*1)+(2+2)+.....(999*999)
sum(k^2, k=1 to n) = n*(n+1)*(2n+1)/6
Put in n = 999
Subject: Re: Ridicule This Crackpot
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision:
1.9 primary) id h916kLa29208;
The formula was written incorrectly on my website. I was in a
rush when I converted it from symbols to letters. It is corrected.
Here it is:
x --> x + ( (r/n) sin ( (n*t)+(n^2*pi*v) ) )
y --> y + ( (r/n) cos ( (n*t)+(n^2*pi*v) ) )
Some might also find it humorous that I corrected the spelling of
fractal in the first sentence.
Please look at my site and tell me what you think.
http://fractalspiral.zxcvb.org
Subject: Bijection between N^0 and N for any k>=0
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision:
1.9 primary) id h91Dpke25136;
Is it possible? How?
Subject: Re: Bijection between N^0 and N for any k>=0
> Is it possible? How?
No, because N^0 is a finite set (in fact, it has exactly one member),
while N is an infinite set.
It is true that |N^k| = |N| for each k > 0. If you can show |NxN| = |N|,
which follows from a standard enumeration argument, then the rest follows
by induction.
--
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
<http://www.commoncouragepress.com/index.cfm?action=book&bookid=228>
Subject: Re: Trapezoidal Rule Error Sum
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision:
1.9 primary) id h91DqKv25209;
>Name: Mike Ossipoff
>e-mail: nkklrp@hotmail.com
>Question:
>A number of books on numerical analysis or numerical
>methods say that the trapezoidal rule's error can be
>written as a sum involving only even powers of h, the
>subinterval size.
>But what about this?:
>Say that the function we want to apply the trapezoidal
>rule is the Taylor approimation for some function
>(any function whose Taylor series has both even & odd
>powers of x will do, which means it could be almost
>any functionn), evaluated about a point at the middle
>of the subinterval.
>Call the endpoints of the subinterval A & B.
>B-A=h
>The Taylor series contains factors that are powers of
>(x-a), where x is the independent variable, and a is
>some number.
>So then, what I'm saying is that, for this Taylor
>approximation, a = (A+B)/2 That's the middle of the
>subinterval.
>The middle of the subinterval is the most natural
>place to have a, and it's also where a would be
>placed in order to get the most accurate Taylor
>approximation in that interval.
>The trapezoidal rule says to multiply the average of
>the integrand's values at A & B by h.
>I'm going to let F(x) stand for the Taylor
>approximation evaluated at x.
>So the result of the trapezoidal rule is:
>h(F(A)+F(B))/2
>As I said, the terms of the Taylor approximation have
>as a factor various powers of (x-a).
>At A, x-a = -h/2 At B, x-a = h/2
>Odd powers of a negative number are negative, and so
>since x-a is negative at A, the odd power terms of the
>Taylor approximation at A are identical to those at B
>except for being opposite in sign. So when we add the
>Taylor approximation's value at A and at B, the odd
>power terms cancel out, leaving us with a sum
>involving only even powers of h.
>Then, when we multiply by h, as the trapezoidal rule
>calls for, those even powers of h become odd powers of
>So, the result of the applying the trapezoidal rule to
>that integrand function is a sum involving only odd
>powers of h.
>Now, what about the actual correct value of the
>integral?:
>The Taylor approximation is a polynomial. It is
>antidifferentiated term by term, and the
>antiderivative, like the Taylor approximation itself,
>is a sum of both even & odd powers of x.
>Now, when an odd or even number is raised to an even
>power, the result is always even. So the even powers
>of -h/2 are the same as the even powers of h/2. So,
>when the antiderivative of the Taylor approximation is
>evaluated at A & B, and when the value at B is
>subtracted from the value at B, the even power terms,
>which are identical at A & B, subtract out, leaving
>only the odd power terms.
>So the actual correct integral from A to B is a sum
>involving only odd powers of h.
>So both the actual correct integral and the
>trapezoidal rule approximation are sums involving only
>odd powers of h.
>Therefore their difference, the error of the
>trapezoidal rule, is also a sum involving only odd
>powers of h.
>That contradicts what a number of books on numerical
>analysis or numerical methods say about that error
>sum.
>The function that we used isn't contrived. The Taylor
>approximation could be of any function whose Taylor
>approximation has both even & odd powers of x-a.
>The middle of the subinterval is the most natural
>place about which to evaluate the Taylor
>approximation.
>And it's the place where one would do so in order to
>have the most accurate Taylor approximation over that
>interval.
>And, if we wanted to contrive a function for which the
>error of the trapezoidal rule would be a sum involving
>only even powers of h, we'd have to _start with_ an
>integrand function that is a sum of terms involving
>only even powers of h.
>So, my question is: Why do they say that ther error of
>the trapezoidal rule is a sum involving only even
>powers of h?
>Mike Ossipoff
Subject: Re: Bland-Altman Software
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision:
1.9 primary) id h91HfJO09867;
kindly forward the site address to me.
emma
>I know a site to dowload a sowftware that permit to perform the
>bland-altman plot: analyze-it.com. You can use it free for a month and
>then decide if you want to buy it. It's easy to use. Another software
>is in medcalc.com but isn't free.
Subject: Re: Re fermat by Tomas
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision:
1.9 primary) id h91K5t618811;
The correct website is: http://www.users.bigpond.com/pidro/home.htm
and a new website on the subject is:
http://www.users.bigpond.com/pidro/home.htm
E. E. Escultura
Subject: Re: Continuously Tracing the HyperPower function x^^y
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision:
1.9 primary) id h91K5hu18792;
>There was a nice bonus on my construction for a Continuous Extension for
>the HyperPower operator, which I discovered only two days ago.
>The definition that I use, seems to produce an infinity of new shapes if
>one traces the hyperexponent continuously.
>For example, although (-1)^^n = -1, for all n in N, (so that the
>function (-1)^^y returns infinately often to -1) the behavior of (-1)^^y
>for values in between natural numbers appears to be quite complex.
>A couple of .GIF trace samples, along with a short cross-referenced
>indirect proof that lim[n->+oo]i^^n exists (by considering the traces of
>i^^y, y>=0), can be found at:
><<a
href=http://users.forthnet.gr/ath/jgal/math/exponents4.html#trace>http://
users.forthnet.gr/ath/jgal/math/exponents4.html#trace</a>in my Math 
section:
><<a
href=http://users.forthnet.gr/ath/jgal/math/>http://users.forthnet.gr/ath
/jgal/math/</a>Enjoy the read.
>--
>Ioannis
><a
href=http://users.forthnet.gr/ath/jgal/>http://users.forthnet.gr/ath/jgal
/</a___________________________________________
>Eventually, _everything_ is understandable.
Please fix your hyperlinks
Subject: Re: Prime number sequence, Genralization solved
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision:
1.9 primary) id h920rNC07707;
I did not before describe how to solve the problem
with a<>3. Here is a way to do it:
Let us suppose that there exists an infinite sequence
of prime numbers p(1),p(2),...
where p(n+1) = a p(n) + 2 b(n) for some b(n)=-1 or 1
Granville asks whether the sequence b(1), b(2), ...
can look like a random sequence of -1's and 1's
Let q be an odd prime dividing a-1 if it exists.
Now, if the b(n) look like a random sequence then we must
have that for any integer R there exist values of N such that
b(N)=b(N+1)=b(N+2)=...=b(N+R)=1. For any n with b(n)=1
we have p(n+1)=p(n)+2 mod q. Thus p(N+k)=p(N)+2k mod q
for k=0,...,R. In particular if we choose k so that
k=-p(N)/2 mod q, then p(N+k)=0 mod q, but p(N+k)>q
since q<a<p(2), so long as R>=q>k.
Thus it cannot be an infinite sequence
of primes! Thus we can assume a-1 is a power of 2.
Now let q be an odd prime dividing a+1 if it exists.
Now, if the b(n) look like a random sequence then we must
have that for any integer R there exist values of N such that
b(N)=b(N+2)=b(N+4)=...=b(N+2R)=1, and
b(N+1)=b(N+3)=b(N+5)=...=b(N+2R+1)=-1. For any n with b(n)=1
and b(n+1)=-1, we have p(n+2)=a(ap(n)+2)-2= a^2 p(n) + 2(a-1),
so that p(n+2)=p(n)-4 mod q. Thus p(N+2k)=p(N)-4k mod q
for k=0,...,R. In particular if we choose k so that
k=p(N)/4 mod q, then p(N+2k)=0 mod q, but p(N+k)>q
since q<a<p(2). Thus it cannot be an infinite sequence
of primes! Thus we can assume a+1 is also a power of 2.
The only integer a for which both a+1 and a-1 is a power of 2
is a=3, which we dealt with last time!
Subject: Re: Quaternion Extensions
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision:
1.9 primary) id h92Ej1a32170;
>> I have recently extended the Quaternions to larger sets by requiring
some
>> (new) group elements to commute. In doing so, I found this process and
its
>> results to be very asthetic. For one, the law of association is
regained.
>> However, the algebra involved is no longer a division algebra, i.e. we
may
>> not always follow x = 0 or y = 0 from xy=0 (when x and y are certain
>> elements taken from a linear combination of group vectors).
>...
>> Has this type of thing been done before and are its conclusions of
>> interest?
>It's obvious that there exist extensions of the quaternions H,
>eg H + H (direct sum),
>pr algebras of matrices with quaternion elements.
>You'd have to say what properties your extension has
>before anyone could say if it is of interest.
>--
>Timothy Murphy
>e-mail: tim@birdsnest.maths.tcd.ie
>I am neither refering to the ring of quaternions matrices nor to the group
product... have sent you a copy of this work.
Subject: Re: Desire for fame is a bitch
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision:
1.9 primary) id h92FnOP05075;
> A funny thought...
> A warning to all would be geniouses looking for a prime countin'
> function. Stop wasting your time. Use your talents to better look
> for a cure to world hunger and a way of giving it to the peoples
> of the world in a way that they will still be able to retain a
> sense of self-dignity (low self-dignity / humiliation: one of the
> leading causes of terrorism). Keep in mind that even if
> you do find the mystery function, James is going to sue you anyway > for
stealing his idea.
Herry
Subject: Re: Question About Prime Numbers
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision:
1.9 primary) id h92COSi21895;
>Twin primes are prime numbers such as 5 and 7, 11 and 13, 17 and 19,
>etc. These twins are only one unit apart.
>There are strings of prime numbers that are n-units apart:
>3, 5, 7, [3 prime numbers, 2 units apart]
>5, 11, 17, 23, 29, [5, 6 units]
>7, 157, 307, 457, 607, 757, 907, [7, 150 units]
>11... ? ...? ...? ...
>The question becomes: For all odd prime numbers P, are there P number of
>primes that are the same numerical[equal] distance apart?
Not always.
There is an infinite number of interprime gap-patterns
arising only once or finite many times.
Yours
e.labos
Subject: Re: Combination/Permutation Question...
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision:
1.9 primary) id h92HsWJ14965;
Image that you have 10 balls.
Permutation : The way(order) you can take them out from a box.
Example:
1,2,3,4,5,6,7,8,9,0
0,1,2,3,4,5,6,7,8,9
..................
4,3,2,1,5,8,6,7,9,0
Combination with no repeating: The way(order) you can take only m of them
out and m<10.
Example: m=3
1,2,3
5,3,7
0,8,2
.....
Combination with repeating: The way(order) you can take only m of them out
and m<10 and you have to return this ball back before take out another one 
.
Example: m=3
1,2,3
4,7,7
1,1,1
.....
Subject: Re: Ridicule This Crackpot
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision:
1.9 primary) id h92KDfH26427;
Well, it looks like someone found a reference to shut me up. I
haven't seen it yet, I've only stumbled onto a web page with a
background photo with a similar kind of spiral (not precisely the
same thing, but similar enough that I believe it.) I still need to find
something more specific, but it will take time.
http://www.math.u-bordeaux.fr/~stan/Colloque/mmfenglish.html
My web page will be modified soon so I don't take credit for
something I didn't discover. But, please go there one more time
and look at the quicktime movies... because they're cool looking.
http://fractalspiral.zxcvb.org
http://fractalspiral.zxcvb.org/qt/qt1.html (1.2 mb)
http://fractalspiral.zxcvb.org/qt/qt3.html (4.2 mb)
But (sniff) it's so sad I'm going to have to stop spamming this
place and leave it to the REAL mathematicians and the REAL
crackpots. (sniff).
Subject: Re: Chained arrow notation experts-help!
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision:
1.9 primary) id h92HsVf14961;
>A diagram of 9 -> 9 -> 9 -> 9 ---->
another try at 9->9->9->9---
387420488 section bays
_______________________________________________^____________________________
_
________________________________
/
/ 387420488 bays
| _______________________^_____________________
|/ 1) 9^9=387420489 / / 1) 9^9
| 2) 9^..1)..^9 | | 2) 9^..1)..^9
| . | | .
| . | | .
|387420489) 9^..387420488)..^9<..< .
| levels| | .
| ..9^........^9
| bays
/387420488< ______________________________________^______
| section|/ .
| levels | . / /
387420488 bays
| | . | |
_______________________^_____________________
| | ______________________________________^______ | |/
1) 9^9 / / 1) 9^9
| |/ 1) 9^9 / / 1) 9^9 | |
2) 9^..1)..^9 | | 2) 9^..1)..^9
| | 2) 9^..1)..^9 | | 2) 9^..1)..^9 | |
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| ..9^.........^9<...<
bays
| section levels | |
___________________________________^_________
| | |/
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| | |
.
| | |
___________________________________^_________
| | |/
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| | |
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levels of| | |
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|
____________________________________________________________________________
_
_____________________^_________
|/ / 387420488 bays
| | _______________________^_____________________
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387420488 bays
| . | |
_______________________^_____________________
| ______________________________________^______ | |/
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|/ 1) 9^9 / / 1) 9^9 | |
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levels of
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_______________________________________________________________^____________
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. | |
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| |
| _______________________^_____________________
| |
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| |
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| . | |
_______________________^_____________________ | |
| ______________________________________^______ | |/ 1)
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|/ 1) 9^9 / / 1) 9^9 | | 2)
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| |/
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| |
. | |
| |
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| |
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____________________________________________________________________________
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_
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section bays
|
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|
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|
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|
| _______________________^_____________________
|
|/ 1) 9^9=387420489 / / 1) 9^9
/ /
387420488 section bays
|
| 2) 9^..1)..^9 | | 2) 9^..1)..^9
| |
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| |/
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| | /
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|
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| | |
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| | . | |
_______________________^_____________________ | | |
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1) 9^9 / / 1) 9^9 | | |
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. | | . | | levels |
. / / 387420488 bays
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v
9->9->9->9
Subject: Re: sol 71
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision:
1.9 primary) id h92MBtW03303;
>>Let X denote the number of white balls selected when k balls are chosen 
at
random from an urn containing n white and m black balls.
>>1. compute P{X=i}
>I think you mean 'with replacement'.
>In this case it is a direct application of the other problem you posted to
the newsgroup:
>P(X=i) = choose(k,i)*chose(k-i,k-i)* ((n/(n+-m))^i * (m/(n+-m))^(k-i))
> = k! / (i! * (k-i)!) * ((n/(n+-m))^i * (m/(n+-m))^(k-i)).
>>2. let, for i = 1,2,...,k; j= 1,2,...,n,
>> Xi = 1, if the i_th ball selected is white
>> 0, otherwise
>X = sum(i=1,...,k) Xi.
>So E[X] = E[sum(i=1,...,k) Xi] = sum(i=1,...,k) E[Xi] =
>sum(i=1,...,k) (1*P(i_th ball selected is white) +- 0*P(i_th ball selected
is black)) =
>sum(i=1,...,k)(n/(n+-m)) = k*n/(n+-m).
>> Yi = 1, if the j_th white ball is selected
>> 0, otherwise
>>compute E[X] in two ways by expressing X first as a function of the Xi's
and then of the Yi's.
>I think you meant to define
>Yij = 1, if the j_th white ball is chosen in the i_th selection
> = 0, else.
>Then you have X = sum(i=1,...,k) sum(j=1,...,n) Yij.
>So E[X] = E[sum(i=1,...,k) sum(j=1,...,n) Yij] =
>sum(i=1,...,k) sum(j=1,...,n) E[Yij] =
>sum(i=1,...,k) sum(j=1,...,n) (1*P(j_th white ball is chosen in the i_th
selection) +- 0*P(j_th white ball is not chosen in the i_th selection)) =
>sum(i=1,...,k) sum(j=1,...,n) 1/(n+-m) = k*n/(n+-m).
>>Can anybody help me with this problem?
>Best wishes
>Torsten
Subject: Re: Quaternion Extensions
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision:
1.9 primary) id h93CRM429159;
>> I have recently extended the Quaternions to larger sets by requiring
some
>> (new) group elements to commute. In doing so, I found this process and
its
>> results to be very asthetic. For one, the law of association is
regained.
>> However, the algebra involved is no longer a division algebra, i.e. we
may
>> not always follow x = 0 or y = 0 from xy=0 (when x and y are certain
>> elements taken from a linear combination of group vectors).
>...
>> Has this type of thing been done before and are its conclusions of
>> interest?
>It's obvious that there exist extensions of the quaternions H,
>eg H + H (direct sum),
>pr algebras of matrices with quaternion elements.
>You'd have to say what properties your extension has
>before anyone could say if it is of interest.
>--
>Timothy Murphy
>e-mail: tim@birdsnest.maths.tcd.ie
>tel: +353-86-233 6090
>s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland
>I am neither refering to the ring of quaternion matrices nor to the >group
product... have sent you a copy of this work.
C.Dement
----------------------------------------------------------------------------

----
Reply to this message
Subject: Re: Quaternion Extensions
> I have recently extended the Quaternions to larger sets by requiring
> some (new) group elements to commute. In doing so, I found this process
> and its results to be very asthetic. For one, the law of association is
> regained. However, the algebra involved is no longer a division
algebra,
> i.e. we may not always follow x = 0 or y = 0 from xy=0 (when x and y
are
> certain elements taken from a linear combination of group vectors).
>>...
> Has this type of thing been done before and are its conclusions of
> interest?
>>It's obvious that there exist extensions of the quaternions H,
>>eg H + H (direct sum),
>>or algebras of matrices with quaternion elements.
>>You'd have to say what properties your extension has
>>before anyone could say if it is of interest.
>>I am neither refering to the ring of quaternion matrices nor to the
>group
>>product... have sent you a copy of this work.
Apologies for not replying to your email ... lectures have just begun.
My point was that you seemed surprised to find that
there were algebras extending the quaternions,
and I noted that this wasn't too surprising.
So the fact that you have constructed such an algebra
could not be considered interesting in itself --
any interest would have to lie in the special properties of the algebra.
--
Timothy Murphy
e-mail: tim /at/ birdsnest.maths.tcd.ie
(all email over 80k dispatched to /dev/null)
tel: +353-86-2336090, +353-1-2842366
s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland
Subject: Re: The Quaternion Counterpart
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision:
1.9 primary) id h93CRCA29132;
><pre> I would like to know how to define the roots and powers of
>> quaternions. For example, De Movre's theorm says
>> z^n = (r^n)(cos n*@ + i*sin n*@)
>> Whats the deal with quaternions?
>Integer powers are well defined, and exp, sin, cos are defined
>by the same power series as for the real or complex numbers.
>Length r is defined by the Pythagorean distance formula as for
>complex numbers.
>Any quaternion q = A+Bi+Cj+Dk can be written in exactly 2
>ways as A + Y*w where Y is real and w^2 = -1, just as any complex
>number A+Bi can be written in exactly 2 such ways. Defining Y as
>the imaginary part of q, Euler's formula and de Moivre's theorem
>are true for q.
></pre>
Subject: Re: FS: Math&Phyx Books
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision:
1.9 primary) id h93LUGb04905;
I want the book by James Dugundji on Topology if its still available.
Rajaraman.
><preI have the following books for sale. I hope you find the prices
>reasonable. Please add $2 for shipping for the first book, and $1
>for each additional book. Please Email to confirm, and get mailing
>address. Reply to jeffh@indiana.edu
>Anderson, F.W. & Fuller, K.R.
>Rings and Categories of Modules
>Out of print in paper, my price 15.00
>Paper
>Bellman, R.
>Stability Theory of Differential Equations
>Out of Print, my price 20.00
>Hard
>Bluman, G.W. & Cole, J.D.
>Similarity Mathods for Differential Equations
>Applied Mathematical Sciences, Volume 13
>Out of Print, my price 20.00
>Paper
>Bishop,R.L. & Goldberg, S.I.
>Tensor Analysis on Manifolds
>Original Hardback -- MacMillian
>My price 10.00
>Bolza, O.
>Lectures on the Calculus of Variations
>English Translation
>My price 18.00
>Hard
>Chevray, Rene & Mathieu, Jean
>Topics in Fluid Mechanics
>List 125.00, my price 20.00
>Hard
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>Differential and Integral Calculus
>Origional Wiley printing
>List 60.00, my price 20.00
>Hard
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>Group Representation Theory (Part A and B)
>Out of Print, 50.00 for the pair.
>Hard
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>Topology
>Classic Hardback -- Allyn & Bacon
>List 69.95(paper), my price 30.00
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>Advanced Calculus of Several Variables
>Original Hardback -- Academic Press
>My price 10.00
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>Advanced Calculus (A differential forms approach)
>Origional Hardback later reprinted by Birkhauser
>Out of Print, my price 20.00
>Eichler, M.
>Introduction to the Theory of Algebraic Numbers and Functions
>Out of Print, My price 25.00
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>Topology
>Out of Print, my price 25.00
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>List 62.50, my price 12.00
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>Topology
>Original Hardback w/dust jacket --Addison Wesley
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>Origional Van Nostrand printing
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>Original Hardback -- Prentice Hall
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>List 69.95, my price 20.00
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>Classic - Paper - Princeton University Press
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>Classic - Hard - Ungar Press
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>Plane Algebraic Curves
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>Paper
>Wilansky, A.
>Functional Analysis
>Out of Print, my price 25.00
>Hard
></pre>
Subject: Re: After Coxeter
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision:
1.9 primary) id h9461wP06185;
At one point or another I read parts of these books:
* Manfredo P. do Carmo: Riemannian Geometry
a very nice reading. good exercises
* R.W. Sharpe : Differential Geometry. Cartan's Generalization of
Klein's Erlangen Program.
very broad minded. not allways accurate.
* Kobayashi and Numizu
enciclopedic.
* J. Jost
the analytic view. very dense.
* Spivak's pentalogy
surpsingly, some parts of it are actually nice.
HTH
D.L.
Subject: Re: The Bible Code
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision:
1.9 primary) id h94EaMi05382;
Ricci Rukavina is worthless idiot. He's a hack slob who has lied his
way into a Creative Producer position, which none of his consecutive
Production failures merit.
He's a smart guy, but in so far over his head he has to watch the bubbles
rise just to know which way is up. He's become dreadfully insecure.
Let's hope he gets sent home. He could use the dose of reality and more
importantly, Scarface could use a real Producer. Not someone who's going to
it up and then jump ship, as Ricci has so many times.
Tool.
Utter and total tool.
Spin some jewish toys and use both hands to count your way out of your
current mess, you short pile of slime.
><preOne thing that I a have been wondering... It is the TORAH that is
>mistranslated a countless number of times and can be found in a
>multitude of versions, whereas the TORAH exits primarly in one form.
>FAILED.
>Is this accurate? And does this say something about the New Testament
>and the belief in a Christ figure?
>Just a question....
>boats
></pre>
Subject: Re: Professor [so-and-so]
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision:
1.9 primary) id h94HcFv17559;
>> the problem with professor so-and-so is that he is not very bright,
at
>> least socially speaking... it should have taken less than a minute
to
>> catch onto the fact that james harris is a high-caliber crackpot.
for
>> that reason professor so-and-so probably deserves the ridicule the
>> crackpot has subjected him to...
>> V.
>> Professor so and so happens to be a very bright person.
>> and you base this on???
>> C.
> I agree that Professor R. M. is a very bright person. My opinion
> is based on personal acquaintance over a period of several
> years; he was my Ph.D. advisor at U.C. Berkeley.
is he as bright socially? crackpot's report indicates otherwise...
btw, what did the berkeley math phd do for you?
seems that about 20 out of 23 known of so-and-so's students
turned duds thus far.
>Alan Stern
Subject: Re: Sets.
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision:
1.9 primary) id h951GRl15683;
There are set theories based on paraconsistent logic which are
inconsistent (there are propositions that are both true and false
in such systems, e.g. the Russell set is both a member and
non-member of itself in paraconsistent set theory), but which are
also non-trivial, i.e. not every proposition is derivable in the
system. Such systems contain unrestriced comprehension schemes, such
as (Ex)(Ay)(y in x iff B(x,y,z1,z2,...,zn)), where the z's are
parameters and B is ay formula of the language.
It is an open question whether any of the ystems studied so far
will yield what we hope for. An ideal paraconsistent set theory
should have as theorems everything we want (e.g. recovery of ZFC in
some natural way)but not contain any theorems we don't want (e.g.
0=1).
Research is at an early stage in this field, i.e. in modern logic
these ideas have been investigated around 40 years, though there
are precursers, for example the Polish logician Lukasiewicz
proposed some systems that we would call paraconsistent today.
theory' to find current progress in this field.
Arief
> More generally, you cannot have a set of all sets satisfying P, where
P
> is some arbitrary property. In fact, this restriction covers both of
> yours. What you can have instead is the principle of bounded
> comprehension, which says that if you have a set A and a property P,
then
> you can form the subset B = { x in A : P(x) }. The set A serves as
an
> upper bound for the size of the new set being constructed. The other
> kind is called unbounded comprehension and is what leads to
paradoxes
> such as those of Russell, Cantor, and Burali-Forti.
>> What if we allow unbounded comprehensive sets?
>> Your question doesn't make sense. I was using comprehension as a
noun,
>> not an adjective.
>>Or sets that do not
>> follow strict logic. That is, we allow something to exist in states
>> other than either existing or not existing.
>> What would happen if we injected fuzzy logic into Zermelo Fraenkel Set
>> theory?
>> We would get functions with codomain [0,1]. Not a big deal.
>Codomain?
>Well, what if we use a varity of neither-wholly-true-nor-wholly-false
>type logical operators?
>Is it possible to define a number x such that n^x is not greater than
>x where n is any number greater e^(1/e)?
>I know no numbers like that exist, but suppose we were to invent one?
>(...Starblade Riven Darksquall...)
Subject: radial directed subgraphs
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision:
1.9 primary) id h951GSg15692;
Hi. I've bumped into an issue in graph theory that I have not seen 
discussed
anywhere. Any help/reference is greatly appreciated. Here goes:
Let G be a connected (undirected) graph and d be the usual distance
metric for G. For a vertex v from G, let D_v be the spanning directed
subgraph of G obtained by including a directed edge (u,v) for each
(undirected) edge {x,y} in G with d(u,x) < d(v,y).
(1) Is there a standard name for this directed subgraph? I'm inclined to
call it a radial directed subgraph.
(2) Is anything known about finding minimal-sized sets of vertices {v_1,
..., v_m} with the property that given any pair of vertices x and y of G,
there is some v_i such that D_{v_i} has a directed path from x to y? 
Perhaps
something can be said for planar graphs or unit disc graphs? The latter are
of special interest to me.
(3) Ditto (2), but replace from x to y with either from x to y or from
y to x?
Michael Rieck
Subject: Re: Reject SI
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision:
1.9 primary) id h953Npg23843;
><pre THE METRIC SYSTEMS
> (Commercially Successful, Scientific Misfortunes)
> Copyright 1997 by Donald Shead
> Since first beginning to study physics, as a prerequisite for
>engineering, I've been trying to figure out why it is such a difficult and
>forbidding subject. Recently, while contemplating why the United States
is
>so slow in converting to the metric system, the main reason has quite
>suddenly come to me: The trouble with physics is the metric systems;
which
>allow arbitrary alternative 'absolute' *mass* based systems of measure in
>lieu of and in preference to conventional *force* based systems of
measure.
> Nowhere is the misleading misuse of (the gram and kilogram of)
mass
>as standards for weight any greater than in the metric systems.
Clearly
>understanding the difference between force, weight and mass is very
>important in beginning the study and comprehension of physics.
> Mass was chosen for determining quantities of material goods, for
>international trade, because the weight of a body of matter varies
>depending on the acceleration due to gravity at the location where it is
>measured; while its mass remains unchanged. The same, anywhere.
> The reason for this of course is simply that a body's weight (w)
>varies because of, and in proportion to, the acceleration due to gravity
>(g); its this proportion, or ratio (w/g) that remains Constant, and is
the
>measure of the quantity of matter contained in it.
> A metallic prototype of the chosen unit of mass was made, and is
>called the 'kilogram'; the French name for weight. Herein lies the
problem:
> Is the kilogram a measure of weight, or a measure of mass? OR doesn't it
>really matter?
> If not the original perpetrators of it, the absolute metric
systems
>are at least perpetuating the confusion between mass and weight: The gram
>and kilogram of mass are commonly called weights, and are inconsistently
>defined in various texts and dictionaries: Sometimes they are defined as
>mass, sometimes as weight, and sometimes as mass OR weight. There's even
>a conversion factor (2.204 lb/kilogram) directly relating weight to mass.
> Giving names to units of mass, as the gram (gm) and kilogram
(kg),
>and definitions so that: 1 gm = 1 dyne sec^2/centimeter; 1 kg = 1 newton
>sec^2/meter, does *NOT* allow these connections to be arbitrarily, or
>otherwise, rewritten as: 1 dyne = 1 gm centimeter/sec^2, and 1 newton = 1
>kg meter/sec^2: We cannot - in our striving for simplicity - hide the
fact
>that 1 gm has units of dynes sec^2/centimeter, and 1 kg has units of
>newtons sec^2/meter. Here's how it *really* works:
> A gram weighs (about) 980.6 dynes, on Earth. Its mass (w/g) is 980.6
>dynes sec^2/980.6 centimeters = (reduces to) 1 dyne sec^2/centimeter (=
>f/a). THIS connection may be rewritten; so that (980.6 dynes sec^2/980.6
>centimeters) x 1 centimeter/sec^2 = (reduces to) 1 dyne; which is a
>fundamental unit (of force, and/or weight) because all of the other units
>cancel.
> A kilogram weighs 9.806 newtons (about 2.2 lb), on Earth. Its mass
>(w/g) is 9.806 newtons sec^2/9.806 meters = (reduces to) 1 newton
>sec^2/meter (= f/a). This connection may be rewritten, so that (9.806
>newtons sec^2/9.806 meters) x 1 meter/sec^2 = (reduces to) 1 newton; which
>is a fundamental unit because all of the other units cancel.
Mass cannot be arbitrarily, or otherwise, ... taken as fundamental,
>and force as derived, ..., because it's the other way around!!
> The solution to this enigma, is to reject, *or modify,* the present
>mass based metric systems, and any other 'absolute' systems of measure. If
>one doesn't already exist, a meter-newton-second gravitational system
>would more correctly serve in its place.
> Better yet, for the time being at least, our existing
>foot-pound-second system - with decimals thereof - should be retained,
>and/or reinstituted. This system, where a pint's a pound the whole world
>around, has been, and still is, quite successful. So who *cares* if the
>foot was the length of one of some king's feet. Its a more convenient
>length than one ten millionth of the distance from the Earth's equator to
>one of its poles anyway. [And the foot too, could be defined as the
>distance light travels in a second.] Let's stick with our correct
>foot-pound-second system, with *real* fundamental units. Physics will be
>much simpler for it
> Finally, the decimal system is applicable to any numerical measure,
>including the foot, the pound, the degree of arc, and the second. The
>metric systems, with their individual names (prefixes) for each and every
>decimal place, have no special claim to it.
>Respectfully,
>Donald Shead
>Chaplin, CT, USA
>&lt;u10889@snet.net&gt;
></pre>
Subject: The Octic x^8-x^7+29x^2+29 Revisited
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision:
1.9 primary) id h95CWuZ29032;
Hello all,
The solvable octic x^8-x^7+29x^2+29 has been discussed here before,
but I still find it interesting due to its simple form. There was a
solution (by P.Montgomery, see previous posts) though it was rather
complex. Perhaps, there is a simpler way befitting the simplicity
of this octic's form.
Euler's approach to the quartic involves the sum of the square roots
of the 3 roots of the resolvent cubic. My approach then is similar:
the octic solved as the sum of the square roots of the 7 roots of its
resolvent septic.
The first step is to depress the octic by removing the x^7 term:
Given, x^8-x^7+29x^2+29 (eq.0), let x=(y+1)/8; thus we get:
y^8-28y^6-112y^5-210y^4-224y^3+7602036y^2+15204304y+494141433 = 0
(eq.1)
Then we find the polynomial formed by taking 4 roots (a+b+c+d) at a
time, the degree given as 8!/(4!4!)=70. This is the signifance of
depressing the equation, since a+b+c+d+e+f+g+h=0 gives us
a+b+c+d=-(e+f+g+h), which is also important later.
Thus the SQUARES of the sum 4 roots at a time z-((a+b+c+d)/4)^2
should give us only 35 distinct values, or a 35-deg polynomial in z,
which i need not write here. It has a 7-deg factor given by:
z^7-7z^6-2763z^5-19523z^4+1946979z^3+34928043z^2+119557031z-3247^2=0
(eq.2)
This IS solvable in radicals. Note the square constant term. The cubic
resolvent of the quartic also has a square constant term.
Getting the positive square root of its 7 roots (rounded off):
a=(-26.982518)^(1/2)
b=(-26.955652)^(1/2)
c=(-19.716772)^(1/2)
d=(-4.780672)^(1/2)
e=(0.086012)^(1/2)
f=(36.910874)^(1/2)
g=(48.438729)^(1/2)
Then the roots of the octic (eq.1) are given by:
y1= a-b-c-d+e-f-g
y2=-(a-b-c-d-e+f+g)
y3=-(a+b-c+d+e-f+g)
y4= a+b-c+d-e+f-g
y5=-(a+b+c-d+e+f-g)
y6= a+b+c-d-e-f+g
y7=-(a-b+c+d-e-f-g)
y8= a-b+c+d+e+f+g
which are all complex, by the way, so I've arranged them in pairs.
To solve the original octic (eq.0) is simply x=(y+1)/8.
The problem, obviously, is the SIGN of the square root. The sum of
the square roots of 7 numbers is 2^7=128 possible values, depending
on the sign you used. What I did was to calculate ALL 128 values,
form the 128-deg poly, and see if eq.1 was a factor, thus confirming
the validity of the approach. It was, as well as its negative, and
2 56-deg polys.
And why does it work? Euler realized that for the quartic, what you
get as square roots was a+b, a+c, a+d. However, since a+b+c+d=0, then
a+d=-(b+c). So, (a+b)+(a+c)-(b+c)= 2a, and you have your quartic root.
For the DEPRESSED octic with roots a,b,c,d,e,f,g,h, let the SQUARE
ROOTS of the 7 roots of its septic resolvent be:
s1=a+b+c+d
s2=a+c+d+e
s3=a+d+e+f
s4=a+e+f+g, &
s5=a+f+g+h, or -(b+c+d+e)
s6=a+b+g+h, or -(c+d+e+f)
s7=a+b+c+h, or -(d+e+f+g)
By adding them, one can see that the last 3 values would cancel out
everything leaving only 4a. (And that is why my z poly had a
denominator 4, or z-((a+b+c+d)/4)^2, so that it leaves only a.
This approach should work for all octics with 7th deg resolvents, I
guess. This is the only octic I know like that. Care to send over
similar octics?
P.S. Would somebody be so nice enough to explain how to solve the
solvable septic? I know how to solve most of the composite degrees
higher than the quintic, as well as the quintic itself, but the prime
degrees starting with 7th deg has stumped me. Care to oblige?
Tito
Subject: Length of Stock
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision:
1.9 primary) id h95CWun29042;
What is the formula for length of stock required to go around a
circumference.
Eg
If I wanted to go around a piece of pipe 3 1/2 in diameter, with 1/4
flatbar, how long should I cut the bar. This is considering the fact that I
can form it completely round
Con S. Borg
Subject: Re: Electrical Storm Sets Three Houses Ablaze
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision:
1.9 primary) id h95CWwl29097;
Hi:
I have some very direct experience with the initiator of this thead and for
that reason would like to know a little about this forum.
I made way directly here via Google and so have no idea who you are, what
you discuss and why Mr. Daryl Shawn Kabatoff's tolerated, much less taken
up.
Ray
Subject: Convergence in distribution
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision:
1.9 primary) id h95LO5d31034;
Hello.
I have a problem like this:
Given two random variables X & Y such that X converges to Y in
distribution,is it true that X1 converges to Y1 in distribution, too?
,where
X1= [ (X+delta)/(2*delta+1) ]
Y1= [ (Y+delta)/(2*delta+1) ]
Here [Q] denotes the largest integer less than or equal to Q and
delta is a positive integer. Also assume X & Y take only positive real
values.
Any ideas?
If the answer is yes,could you provide any supporting arguments?
If no,why?
YSH
Subject: Re: JSH your ship has come in!!!!
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision:
1.9 primary) id h962hgJ19761;
There is an obvious typo in this subject header. The word ship should
end in the 5th letter directly to the left of p on the qwerty key pad. And
it's been coming in by the truck load for years.
>Hey James,
>Great news!!!! Alain Connes has just joined the Vanderbilt math dept.
>Finally someone who can recognize and appreciate your math. And I hear
he
>is a good mathematician (as opposed to those evil ones).
>Good luck,
Subject: Re:Re: Convergence in distribution
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision:
1.9 primary) id h963S5M22858;
Yes,I mean random 'sequences'.
But BOTH X_n and Y_n are random sequences converging in distribution.
Does that make any difference to the answer?
Well,the actual pdf's of X_n and Y_n at a given n are reasonably well
modeled by GGD(Generalized Gaussian Distribution)'s with zero means and
variances typically 25~100 while the delta's can be 1~10.
Although we can easily find counter-examples with discrete probability
distributions, I observed in the actual problems that the 'quantized'(i.e.
X1_n & Y1_n) sequences seemed to 'approach' in distribution. (By 
'approach',I
mean the variances and the shape parameter characterizing the GGDs of both
X1_n & Y1_n approached the same values.)
In fact,what I want to know,eventually, is if it is possible to estimate 
the
entropy of X1_n from that of Y1_n at least within a certain error bound.
Any comments?
>>Given two random variables X & Y such that X converges to Y in
>>distribution,is it true that X1 converges to Y1 in distribution, too?
>>,where
>> X1= [ (X+delta)/(2*delta+1) ]
>> Y1= [ (Y+delta)/(2*delta+1) ]
>> Here [Q] denotes the largest integer less than or equal to Q and
>>delta is a positive integer. Also assume X & Y take only positive real
>>values.
>One random variable doesn't converge to another one. Perhaps you mean
>you have a sequence of random variables X_n converging to Y in
>distribution, and you want to know whether [(X_n + delta)/(2*delta+1)]
>converges to [(Y+delta)/(2*delta+1)] in distribution.
>The answer is no, because the greatest integer function [] is not
>continuous. For example, you could have
>0 < (X_n + delta)/(2*delta+1) < 1 almost surely, while
>(Y + delta)/(2*delta+1) = 1 almost surely. Then
>[(X_n + delta)/(2*delta+1)] = 0 but [(Y + delta)/(2*delta+1)] = 1.
>Robert Israel israel@math.ubc.ca
>Department of Mathematics <a
href=http://www.math.ubc.ca/~israel>http://www.math.ubc.ca/~israel</
aUniversity of British Columbia
>Vancouver, BC, Canada V6T 1Z2
Subject: Re: Magasil.
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision:
1.9 primary) id h96DRC628406;
Hi
I would like to know the use of MAGASIL(mixture of Magnesium
Carbonate,Magnesium Trisilicate,Sodium Bicarbonate).
Subject: Re: Fermat's Last Theorem/Andrew Wiles
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision:
1.9 primary) id h96HaIH07170;
There is a new perspective on this question. Please check out the post by
Tomas, http://mathforum.com/discuss/sci.math/m/77228/77277
E. E. E.
Subject: Re: Alevel Pure 3 Circle Equation
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision:
1.9 primary) id h96IO7d10841;
>equation.
>The original question is as follows:
>Given AB is the diamtere of a circle and the coordinates of A and B are
as
>below, find and equation of the circle.
>A(5,-3) B(-3,5)
>This is my current attempt:
>Midpoint = ([(5+-3)/2], [(-3+5)/2])
> = ( 1 , 1) = Centre of circle
>Length AB^2 = (-3-5)^2 + (5+3)^2 = (-8)^2 + 8^2
> AB^2 = 128
>Using Circle equation: (x-a)^2 + (y-b)^2 = r^2
>(x-1)^2 + (y-1)^2 = 128
>x^2 -2x +1 + y^2 -2y +1 =128
>x^2 -2x + y^2 - 2y =126
>This is as close as I can get to the book answer of:
>(x-5)(x+3) + (y-5)(y+3) = 0
You can try substituting the coordinates of the two known points
into the book answer. They both check.
But (-3, -3) and (5, 5) are also on the circle. Try them in the
book answer. Hmm, they both check.
Looks like the book is right.
If you substitute into your eqn, it doesn't check.
The reason is simple: you have confused radius with diameter.
phil
Subject: Re: billard mechanics
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision:
1.9 primary) id h96IO8U10845;
>I posted something like this in sci.physics but people seem to be mainly
>with their heads in the stars there..
>I have programmed, ages ago, this billiard mechanics engine, using very
>basic physical laws on momentum collision. It works in the sense that it
>creates :
> 1. good collisions between two balls that each have a specific
velocity
>vector, and
> 2. it never draws balls over each other,
>although I'm not satisfied with the amount of calculation needed for those
>two things (it involves solving a quadratic equation and then
trigonometry).
>But, as you may be aware, in snooker or pool one starts with the red balls
>(and the pink) touching each other.
>This kind of ruins this whole nice model since one can no longer use an
>O(n^2) algorithm to scan the balls for possible collisions and then work
>them, since..well it becomes a mess. When the white ball hits the pack of
>red balls, I get a nice 4-dimensional representation of chaos, I think,
>which isn't my intention. As you know, when balls touch each other then
the
>energy of the first ball gets transferred to the last. Most of it,
anyway.
>In fact I think it calls for an entirely new strategy. Does anybody have
any
>ideas how I could :
> - devise a collision strategy that works on singular collisions as
well
>as on multiple simultaneous collisions
> - treat these chained collisions separately (this would involve
>sorting the touching balls with respect to the ball(s) that will hit them,
>since a computer can't know in what direction the force gets transferred
-meaning, this is probably impossible to accomplish.
>I would probably do best to treat the table of balls as a table of static
>balls, with force fields on them, instead of each ball having its vx and
vy,
>but really, it's not that easy..
>If anybody has any ideas or experience with this, I'd be very grateful,
>Tetsuo
Check rec.sport.billiard for previous posts on math analyses of
collisions. However, you should be aware that slight variations
in initial conditions yield vastly divergent results after the
pack is broken.
Good luck
phil
Subject: Re: JSH your ship has come in!!!!
> Great news!!!! Alain Connes has just joined the Vanderbilt math dept.
Finally someone who can recognize and appreciate your math. And I
hear he
> is a good mathematician (as opposed to those evil ones).
> I had a dream last night about evil and James. James was living in a
castle
> of
> wonders, but really it seemed less like a castle than it did the dwelling
> some
> advanced alien race would construct to direct operations on Earth. He
was
> striding about through the castle, manipulating complex instruments,
> performing
> obscure adjustments, and pausing to gaze out the window at the world. The
> world
> abounded in suffering and filth, and the sight of it sickened him, yet
still
> he
> gazed. How would he bring the grandeur that was his intrinsic home out
into
> the
> sordid world?
> throughout the world, and he broadcast his vision to those who had ears 
to
> hear.
> But at night, while he was asleep, those fibers glowed and pulsed in
reverse.
> Something stealthy and wicked crept in through his console and scattered
like
> lemurs throughout James's castle. (Yes, lemurs. I don't know why. They
> seemed like tiny roaches or speedy nanobots, but for some reason they
were
> lemurs.) They infected everything, performing subtle sabotage upon the
> intricate inner works of his instruments.
Are you sure that they were not lemmas?
--
Michael Press
Subject: Re: JSH your ship has come in!!!!
> <for scary dream see previous Maybe the only point is that I fear James
being overwhelmed by evil.
Hmmm.
> I have to ask myself, Why should I care? James may be the
reincarnation
> of Gauss, but is it really any skin off my nose if he goes
unrecognized?
> I've been worrying about the guy for months (ever since I realized that
he
> was not, in fact, a crank, but a genius) and defending him on this
newsgroup.
> My reward? Laughter and bile from the peanut gallery. And not a word
from
> James himself. A word of advice to Prof. Connes: Don't waste your
time
on
> James Harris. If he loves being the solitary genius so much, let
him
fight
> his own damn battles. He's not worth losing sleep over.
> Well damn it, I'm losing sleep at least partly because of your scary
> dream!!!
> Good writing their Jim Ferry, and I have to give you credit for that,
> but hey, how many years were you ridiculing me, and now you expect me
> to just go, hey, pal?
> Show me you're serious and wade in and respond to some of these
> ant-mathematicians in the current battle.
> Prove your sincerity, and um, keep posting any interesting dreams you
> have, as I found it interesting puzzling over that one.
Exactly! This Jim Ferry now thinks you are a genius, but has he
defended
you in your current battle? No!
I suspect that he is not *fully* committed to your view of mathematics.
That
is just a feeling, and I suppose I could be wrong.
I, however, have been trying to find out more about your Object Ring, but I
feel you are pushing me away.
Your friend.
Clive
Subject: Re: JSH your ship has come in!!!!
>>Good writing their Jim Ferry, and I have to give you credit for that,
>>but hey, how many years were you ridiculing me, and now you expect me
>>to just go, hey, pal?
>>Show me you're serious and wade in and respond to some of these
>>ant-mathematicians in the current battle.
>>Prove your sincerity, and um, keep posting any interesting dreams you
>>have, as I found it interesting puzzling over that one.
> Exactly! This Jim Ferry now thinks you are a genius, but has he
defended
> you in your current battle? No!
> I suspect that he is not *fully* committed to your view of mathematics.
That
> is just a feeling, and I suppose I could be wrong.
> I, however, have been trying to find out more about your Object Ring, but
I
> feel you are pushing me away.
Jim sent you roses, but I sent diamonds . . .
I shan't be drawn into a battle of pitching woo, Clive.
It's true that I have given up trying to understand James's Object Ring.
I've come to the conclusion that it's simply beyond my comprehension. I
suppose I'm just accepting that it's true on faith.
Oh, I can hear the derisive snorting. Faith? Mathematicians don't take
things on faith! We accept pure logic only!
Well let me ask you this, ye snorters: Andrew Wiles proved FLT, right?
And how do you know this? Did you go through his proof? Did you go
through
Ribet's proof as well? No? Interesting.
Yes, I know the responses to this. Everyone says it's right. It's
been
checked by reputable mathematicians. The infallibility of the peer
review
process guarantees . . . etc. It has the herd's official brand of
approval,
so any objections are moooooot.
(Okay, that moooooot was pretty awful . . .)
Anyway, I can't help James by going on the lecture circuit and presenting
an argument which I myself can't follow. Besides, the honor should belong
to
James. What I can do, and have been doing for the last several days, is
apply
myself to the problem of how one goes about convincing the world of an idea
to which they've shut their ears.
I have some ideas, but they're still maturing. They don't involve the
wholesale slaughter of the mathematical establishment. This would be
entirely
the wrong approach. So stop thinking about it, James! No, not even a
little
slaughter! I know you like direct action, but direct action hasn't
convinced
anyone thus far, and besides, they'd just throw you in prison and take away
your pencil. No, I have something entirely different in mind. No frontal
assaults. No guerilla attacks.
Uh oh. By suggesting that war may not be the most effective way of
achieving
one's ends, I think I may have violated the Patriot Act. Ungood.
BTW, it occurs to me that you might be a Republican, and hence enjoy
futile,
pointless wars. In this case all you'd be interested in is more guns and
bigger bombs, and I don't have any of those. Let me know if this is the
case.
There's no need to waste my time and yours.
> Your friend.
> Clive
Subject: Re: JSH your ship has come in!!!!
> <for scary dream see previous
> Maybe the only point is that I fear James being overwhelmed by evil.
> Hmmm.
> I have to ask myself, Why should I care? James may be the
> reincarnation
> of Gauss, but is it really any skin off my nose if he goes
unrecognized?
> I've been worrying about the guy for months (ever since I realized
that
> he
> was not, in fact, a crank, but a genius) and defending him on this
> newsgroup.
> My reward? Laughter and bile from the peanut gallery. And not a
word
> from
> James himself. A word of advice to Prof. Connes: Don't waste your
time
> on
> James Harris. If he loves being the solitary genius so much, let
him
> fight
> his own damn battles. He's not worth losing sleep over.
> Well damn it, I'm losing sleep at least partly because of your scary
> dream!!!
> Good writing their Jim Ferry, and I have to give you credit for that,
> but hey, how many years were you ridiculing me, and now you expect me
> to just go, hey, pal?
> Show me you're serious and wade in and respond to some of these
> ant-mathematicians in the current battle.
> Prove your sincerity, and um, keep posting any interesting dreams you
> have, as I found it interesting puzzling over that one.
> Exactly! This Jim Ferry now thinks you are a genius, but has he
defended
> you in your current battle? No!
> I suspect that he is not *fully* committed to your view of mathematics.
That
> is just a feeling, and I suppose I could be wrong.
The primary focus is the odd definition error in core.
Once that's accepted for what it is, and most importantly FIXED, then
it's not about committing to my view of mathematics, but about showing
your commitment to mathematics itself.
There is ONE mathematics.
> I, however, have been trying to find out more about your Object Ring, but
I
> feel you are pushing me away.
I have spent some effort guiding you along at the Mega Foundation
discussion area.
Remember mathematics is a continuing process.
The object ring is fascinating in and of itself, so you can't expect
me to know all the answers just because I'm a discoverer.
After all, if it were that easy, then math research would have ended
long ago with the first mathematician explaining it all.
> Your friend.
> Clive
Time will tell.
Hmmm...I'm having to deal now with questions about people who spent
some YEARS in dedicated efforts to attack me and my work, and you know
what?
Ok, yeah, I'm not in the mood to hold grudges as the research has been
fun, and, of course, the discoveries themselves always had an
immunizing effect.
So for those of you who wondered how I kept going through years of
having been wrong, and LOTS of ridicule, well the answer is that I
pushed myself long enough till I had correct discoveries, and as they
piled up I found that it felt good to be on the side of correct
mathematics.
So it's not about being right in a debate or arguments, as it's about
knowing *valid* mathematics, which is an incredible reward.
It's like, if you're the guy who has 2+2=4, and the others are people
pushing 2+2=5 on Sundays but every once in a while on Monday its 7,
then don't you think you're going to have some satisfaction from the
*security* of not having inconsistency to handle?
James Harris
Subject: Re: JSH your ship has come in!!!!
... stuff deleted ...
> The primary focus is the odd definition error in core.
> Once that's accepted for what it is, and most importantly FIXED, then
> it's not about committing to my view of mathematics, but about showing
> your commitment to mathematics itself.
As I understand it, your claim is that the ring of algebraic integers
is flawed. One can understand this claim in several ways, so I'll
attempt to wade through a few of them. Please comment on the relation
between each of these understandings and your own understanding of
the flawed nature of the ring of algebraic integers.
1. There is no *set* given by the definition:
A complex number z is an algebraic integer if there
exists a polynomial P(x) with integer coefficients and
leading coefficient 1, for which P(z) = 0.
2. The set given by the above definition fails to be a ring.
3. The set given by the above definition *is* a ring, but is not
the ring of algebraic integers. If it were termed the Monic
Roots Ring, or the M Ring, or even
4. The set given by the above definition *is* a ring, and it
*is* the ring of algebraic integers, but it is incomplete,
because if m is an algebraic integer that is not a multiple
of 2, and if n is an algebraic integer for which
mn = 2,
then n *should* be a unit, but it is not.
5. The set given by the above definition *is* a ring, and it
*is* the ring of algebraic integers, but it is incomplete,
because if 2m is an algebraic integer but m is not, and
n is an algebraic integer for which
2mn = 2,
then n *should* be a unit, but it is not.
I think people would appreciate a selection of this list, as to
which item best fits your understanding of the problem in the
ring of algebraic integers.
... the rest deleted ...
> James Harris
Dale.
Subject: Re: JSH your ship has come in!!!!
Visiting Assistant Professor at the University of Montana.
>As I understand it, your claim is that the ring of algebraic integers
>is flawed. One can understand this claim in several ways, so I'll
>attempt to wade through a few of them. Please comment on the relation
>between each of these understandings and your own understanding of
>the flawed nature of the ring of algebraic integers.
[.snip.]
>4. The set given by the above definition *is* a ring, and it
> *is* the ring of algebraic integers, but it is incomplete,
> because if m is an algebraic integer that is not a multiple
> of 2, and if n is an algebraic integer for which
> mn = 2,
> then n *should* be a unit, but it is not.
Okay, this one has me confused. Surely if mn=2, and you want to
conclude that n is a unit, you mean that m ->is<- a multiple of 2?
Otherwise, what is wrong with m=n=sqrt(2)?
Why do you take so much trouble to expose such a reasoner as
Mr. Smith? I answer as a deceased friend of mine used to answer
on like occasions - A man's capacity is no measure of his power
to do mischief. Mr. Smith has untiring energy, which does
something; self-evident honesty of conviction, which does more;
and a long purse, which does most of all. He has made at least
ten publications, full of figures few readers can critize. A great
many people are staggered to this extend, that they imagine there
must be the indefinite something in the mysterious all this.
They are brought to the point of suspicion that the mathematicians
ought not to treat all this with such undisguised contempt,
at least.
-- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan
Arturo Magidin
magidin@math.berkeley.edu
Subject: Re: JSH your ship has come in!!!!
>>As I understand it, your claim is that the ring of algebraic integers
>>is flawed. One can understand this claim in several ways, so I'll
>>attempt to wade through a few of them. Please comment on the relation
>>between each of these understandings and your own understanding of
>>the flawed nature of the ring of algebraic integers.
> [.snip.]
>>4. The set given by the above definition *is* a ring, and it
>> *is* the ring of algebraic integers, but it is incomplete,
>> because if m is an algebraic integer that is not a multiple
>> of 2, and if n is an algebraic integer for which
>> mn = 2,
>> then n *should* be a unit, but it is not.
> Okay, this one has me confused. Surely if mn=2, and you want to
> conclude that n is a unit, you mean that m ->is<- a multiple of 2?
> Otherwise, what is wrong with m=n=sqrt(2)?
Nothing, of course. Actually, you'll no doubt recall JSH's long-time
denial of the concept that in a factorization of algebraic integers
rs = t,
with r and s non-units, that r was a perfectly good common divisor
of both itself and t. I think there is something profoundly wrong
with JSH's reasoning process [I also think this is a severe
understatement]. Whether it is entirely due to a total absence of
factual or theoretical knowledge, or runs to the ability to
connect different aspects of the same problem, I for one find
it difficult to see what it is that he's saying [all terminology
issues aside, even].
So, with that as my starting point, my list of questions were
an attempt to wring a different wording of the allegation
we see in this message:
A closer approximation (than the one you've cited) to JSH's wording of
turns item 4 into item 5 of that list.
What I'm thinking is that if we can isolate *precisely* what his
disconnect is, wrt the algebraic integers, that is:
1. Does he not even think a set is properly defined
there [unlikely],
2. Does he not think it is a ring [could be, but there's the
whole definition of a ring issue that I believe could be at
the root, er, basis, of this],
3. Is it a name thing?
I just noticed that I neglected to finish the wording of #3:
3. The set given by the above definition *is* a ring, but is not
the ring of algebraic integers. If it were termed the Monic
Roots Ring, or the M Ring, or even that ring, then it
would be acceptable.
4. & 5. various wordings of the message referenced above
then one might know how to address this issue.
> Why do you take so much trouble to expose such a reasoner as
> Mr. Smith? I answer as a deceased friend of mine used to answer
> on like occasions - A man's capacity is no measure of his power
> to do mischief. Mr. Smith has untiring energy, which does
> something; self-evident honesty of conviction, which does more;
> and a long purse, which does most of all. He has made at least
> ten publications, full of figures few readers can critize. A great
> many people are staggered to this extend, that they imagine there
> must be the indefinite something in the mysterious all this.
> They are brought to the point of suspicion that the mathematicians
> ought not to treat all this with such undisguised contempt,
> at least.
> -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan
> Arturo Magidin
> magidin@math.berkeley.edu
Dale
Subject: Re: JSH your ship has come in!!!!
> <for scary dream see previous
> Maybe the only point is that I fear James being overwhelmed by
evil.
> Hmmm.
> I have to ask myself, Why should I care? James may be the
> reincarnation
> of Gauss, but is it really any skin off my nose if he goes
unrecognized?
> I've been worrying about the guy for months (ever since I realized
that
> he
> was not, in fact, a crank, but a genius) and defending him on this
> newsgroup.
> My reward? Laughter and bile from the peanut gallery. And not a
word
> from
> James himself. A word of advice to Prof. Connes: Don't waste your
time
> on
> James Harris. If he loves being the solitary genius so much,
let
him
> fight
> his own damn battles. He's not worth losing sleep over.
Well damn it, I'm losing sleep at least partly because of your scary
> dream!!!
Good writing their Jim Ferry, and I have to give you credit for that,
> but hey, how many years were you ridiculing me, and now you expect me
> to just go, hey, pal?
Show me you're serious and wade in and respond to some of these
> ant-mathematicians in the current battle.
Prove your sincerity, and um, keep posting any interesting dreams you
> have, as I found it interesting puzzling over that one.
> Exactly! This Jim Ferry now thinks you are a genius, but has he
defended
> you in your current battle? No!
> I suspect that he is not *fully* committed to your view of mathematics.
That
> is just a feeling, and I suppose I could be wrong.
> The primary focus is the odd definition error in core.
> Once that's accepted for what it is, and most importantly FIXED, then
> it's not about committing to my view of mathematics, but about showing
> your commitment to mathematics itself.
> There is ONE mathematics.
> I, however, have been trying to find out more about your Object Ring,
but I
> feel you are pushing me away.
> I have spent some effort guiding you along at the Mega Foundation
> discussion area.
Yes, indeed.
But, I expect you remember that I did ask you a question that I was unable
to answer.
I was using the notation [a,b,c] to represent an ordered triple of complex
numbers. And I was wondering if the ordered triple [1,2,8] was an element
of
the Object Ring.
You replied:
> Quit being lazy!!! You have the definition, figure it out for yourself!!!
> What amazes me is how often people are willing to ask someone else to do
their work for them.
> If you're smart enough, answer your own question.
> I'm curious to see if you can.
> I've given the definition for the object ring, so no excuses.
Well, I am ashamed to say I still cannot figure it out.
Could you help me please?
> Remember mathematics is a continuing process.
> The object ring is fascinating in and of itself, so you can't expect
> me to know all the answers just because I'm a discoverer.
> After all, if it were that easy, then math research would have ended
> long ago with the first mathematician explaining it all.
> Your friend.
> Clive
> Time will tell.
> Hmmm...I'm having to deal now with questions about people who spent
> some YEARS in dedicated efforts to attack me and my work, and you know
> what?
> Ok, yeah, I'm not in the mood to hold grudges as the research has been
> fun, and, of course, the discoveries themselves always had an
> immunizing effect.
> So for those of you who wondered how I kept going through years of
> having been wrong, and LOTS of ridicule, well the answer is that I
> pushed myself long enough till I had correct discoveries, and as they
> piled up I found that it felt good to be on the side of correct
> mathematics.
> So it's not about being right in a debate or arguments, as it's about
> knowing *valid* mathematics, which is an incredible reward.
> It's like, if you're the guy who has 2+2=4, and the others are people
> pushing 2+2=5 on Sundays but every once in a while on Monday its 7,
> then don't you think you're going to have some satisfaction from the
> *security* of not having inconsistency to handle?
Clive
Subject: Re: JSH your ship has come in!!!!
Visiting Assistant Professor at the University of Montana.
>Hmmm...I'm having to deal now with questions about people who spent
>some YEARS in dedicated efforts to attack me and my work, and you know
>what?
>Ok, yeah, I'm not in the mood to hold grudges as the research has been
>fun, and, of course, the discoveries themselves always had an
>immunizing effect.
The claim that you are not in the mood to hold grudges is a lie.
> I assume, then, that in addition to noting here that I was correct,
> you will be apologizing for any personal attack you launched against
> me, at least with respect to this argument? For instance, will you
> explicitly apologize for claiming that I was doing wacky stuff,
> anti-math, or trying to fool readers with this argument?
Well, if you hadn't spent so much time and effort trying to convince
others that correct mathematics was wrong...maybe...but your actions
have had an influence on the future of humanity, as you've been an
integral part of its failure.
So you ->are<- actually holding a grudge, against (imaginary) past
offenses.
Really, James, for someone who so often speaks about how important
truth is, you seem to have a rather surprising lack of acquaintance
with it.
Why do you take so much trouble to expose such a reasoner as
Mr. Smith? I answer as a deceased friend of mine used to answer
on like occasions - A man's capacity is no measure of his power
to do mischief. Mr. Smith has untiring energy, which does
something; self-evident honesty of conviction, which does more;
and a long purse, which does most of all. He has made at least
ten publications, full of figures few readers can critize. A great
many people are staggered to this extend, that they imagine there
must be the indefinite something in the mysterious all this.
They are brought to the point of suspicion that the mathematicians
ought not to treat all this with such undisguised contempt,
at least.
-- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan
Arturo Magidin
magidin@math.berkeley.edu
Subject: Re: JSH your ship has come in!!!!
<for scary dream see previous
> Maybe the only point is that I fear James being overwhelmed by
evil.
> Hmmm.
> I have to ask myself, Why should I care? James may be the
> reincarnation
> of Gauss, but is it really any skin off my nose if he goes
unrecognized?
> I've been worrying about the guy for months (ever since I realized
that
> he
> was not, in fact, a crank, but a genius) and defending him on this
> newsgroup.
> My reward? Laughter and bile from the peanut gallery. And not a
word
> from
> James himself. A word of advice to Prof. Connes: Don't waste your
time
> on
> James Harris. If he loves being the solitary genius so much,
let him
> fight
> his own damn battles. He's not worth losing sleep over.
Well damn it, I'm losing sleep at least partly because of your scary
> dream!!!
Good writing their Jim Ferry, and I have to give you credit for that,
> but hey, how many years were you ridiculing me, and now you expect me
> to just go, hey, pal?
Show me you're serious and wade in and respond to some of these
> ant-mathematicians in the current battle.
Prove your sincerity, and um, keep posting any interesting dreams you
> have, as I found it interesting puzzling over that one.
Exactly! This Jim Ferry now thinks you are a genius, but has he
defended
> you in your current battle? No!
I suspect that he is not *fully* committed to your view of mathematics.
That
> is just a feeling, and I suppose I could be wrong.
Yes - I too feel that Mr. Ferry has lent support
without substance - but perhaps he feels that your genius
is such that his feeble contributions would only detract
from the power of your own replies.
> The primary focus is the odd definition error in core.
I have always wondered what you mean when you say core -
the definition to which you refer is, I think, that of
algebraic integers. It says an algebraic integer is a
real or complex number which is a root of a monic polynomial
with integer coefficients. Such roots always exist by the
Fundamental Theorem of Algebra. Certainly it seems like a
straightforward, unambiguous definition. How can there
be a problem with it? Of course it takes some work to show
that the SET of algebraic integers is actually a RING, closed
under addition and multiplication. That is a theorem, and
there could be a previously over-looked error in the proof
of that theorem. Is that what you mean? Or might the FTA
be wrong? That too is a nontrivial theorem - maybe Gauss
actually slipped up in proving it and the herds of dull and
unimaginative sheep that have followed him never noticed.
I believe what you have said, however, is that you can
show that there are algebraic integers a and b such that
a * b is not an algebraic integer. That contradicts the theorem
I mentioned above, that the SET of algebraic integers is a
RING [esp., closed under multiplication]. I am not sure you
have ever explicitly written out what a and b are. Could you
do that, or at least specify what monic equations they are roots
of, so we can check this?
> Once that's accepted for what it is, and most importantly FIXED, then
> it's not about committing to my view of mathematics, but about showing
> your commitment to mathematics itself.
> There is ONE mathematics.
One should be enough.
> I, however, have been trying to find out more about your Object Ring,
but I
> feel you are pushing me away.
> I have spent some effort guiding you along at the Mega Foundation
> discussion area.
> Remember mathematics is a continuing process.
> The object ring is fascinating in and of itself,
What about it is fascinating?
> so you can't expect
> me to know all the answers just because I'm a discoverer.
People have also asked that you specify at least one number
which is in the 'object ring' but is not an algebraic integer.
Perhaps if you specify algebraic integers a and b as I have
requested above, then a * b is such a number ?
It is high time you responded to this. Some people actually
doubt that there IS an object ring which differs from the ring
of algebraic integers. You need to settle this issue, or people
will assume you have nothing.
> After all, if it were that easy, then math research would have ended
> long ago with the first mathematician explaining it all.
> Your friend.
Clive
Is this the same friend, Clive Tooth I believe, with whom
you were not so friendly in previous times?
> Time will tell.
> Hmmm...I'm having to deal now with questions about people who spent
> some YEARS in dedicated efforts to attack me and my work, and you know
> what?
> Ok, yeah, I'm not in the mood to hold grudges as the research has been
> fun, and, of course, the discoveries themselves always had an
> immunizing effect.
> So for those of you who wondered how I kept going through years of
> having been wrong, and LOTS of ridicule, well the answer is that I
> pushed myself long enough till I had correct discoveries, and as they
> piled up I found that it felt good to be on the side of correct
> mathematics.
They haven't piled up. I don't think you can claim yet to have
had that feeling.
> So it's not about being right in a debate or arguments, as it's about
> knowing *valid* mathematics, which is an incredible reward.
Again I doubt you are in a position to know this.
> It's like, if you're the guy who has 2+2=4, and the others are people
> pushing 2+2=5 on Sundays but every once in a while on Monday its 7,
> then don't you think you're going to have some satisfaction from the
> *security* of not having inconsistency to handle?
Saying it over and over does not make it so, for
either side.
Nora B.
> James Harris
Subject: Re: JSH your ship has come in!!!!
> Show me you're serious and wade in and respond to some of these
> ant-mathematicians in the current battle.
Serious? Hmmm. That's a problem. You think I'm joking. The
ant hill thinks I'm joking. Well, yes, I'm joking. Of course
I'm joking: I have no viable alternative. I work inside the
system, James. I can't exactly walk up to my boss and say,
Everything you think is deluded, and I'm working with James
Harris to knock that pedestal out from under your corrupt ass.
I doubt I'd be killed, but I'd certainly lose my job. And I like
my job, James.
I'm on your side now (unlike before), but, indeed, how can I be
of any use to you?
I'm thinking about it . . .
Quite a conundrum . . .
--
| Jim Ferry | Center for Simulation |
+------------------------------------+ of Advanced Rockets |
| http://www.uiuc.edu/ph/www/jferry/ +------------------------+
| jferry@[delete_this]uiuc.edu | University of Illinois |
Subject: Re: JSH your ship has come in!!!!
Show me you're serious and wade in and respond to some of these
> ant-mathematicians in the current battle.
> Serious? Hmmm. That's a problem. You think I'm joking. The
> ant hill thinks I'm joking. Well, yes, I'm joking. Of course
> I'm joking: I have no viable alternative. I work inside the
> system, James. I can't exactly walk up to my boss and say,
> Everything you think is deluded, and I'm working with James
> Harris to knock that pedestal out from under your corrupt ass.
Actually I think you're serious to a large extent, as it is the most
intelligent position, and I can't see where you'd see an advantage to
just be joking.
However, I also know that the current situation is shocking.
It would also be rather odd to me if you'd dealt with that
emotionally, this quickly.
> I doubt I'd be killed, but I'd certainly lose my job. And I like
> my job, James.
> I'm on your side now (unlike before), but, indeed, how can I be
> of any use to you?
> I'm thinking about it . . .
> Quite a conundrum . . .
Could it be relatively simple with a FOCUS on the definition error
that has snuck into core mathematics? I like now emphasizing that
it's a definition error so that maybe it takes the edge off somewhat
from a core error.
One nice thing Ferry is that I thought for some reason that you were a
physicists (maybe your job) but checked online and see that you're a
mathematician.
For that reason, I am no longer using mathematician in the blanket
way that I have in the past.
You've shown that mathematicians can accept the math, even when it's
something as extraordinary, even in its simplicity, as my work.
James Harris
Subject: Re: Flawed answer in a calculus textbook?
Ronald Benedik <rbenedik@fsmat.htu.tuwien.ac.at> schrieb im Newsbeitrag
> I alread gave the 'academic' definition of the exponential function.
The exponential function came from financial math:
capital*(1+per_year_interest) = capital*(1+per_month_interest)^12
making the time intervall smaller and smaller gives:
exp(x) = limit_{n->inf} (1+x/n)^n which is discussed in:
Meyberg, Vachenauer; H.9ahere Mathematik 1
Subject: Re: Flawed answer in a calculus textbook?
>> A continuous function on a closed interval is integrable.
>> Proved by Euler, I believe.
>That seems doubtful to me. Would Euler even have a proper definition of
>continuity in front of him? I remember hearing that Cauchy attempted to
>prove this rigorously, but confused continuity with uniform continuity
>(it's the latter you need). If Cauchy didn't quite have this together,
it's
>doubtful Euler did.
>Does anyone know who was the first to prove this? Riemann? Weierstrass?
> Well on the one hand Edgar is usually a very reliable source - otoh I
> find what he says here very hard to believe, for the same reasons.
> One might _guess_ that it was Riemann. Not that one can always tell
> from the way things are named, but he presumably had _something_
> to do with the definition of the Riemann integral, and then he
> presumably proved _something_ about it - seems likely that this
> would have been first on the list...
I would guess Riemann didn't have that result. I think continuity =>
uniform continuity came later. Riemann may have just assumed f is uniformly
continuous. This would handle any function that is Lip alpha, alpha > 0,
for example, and this would take care of a lot of pre-Weierstrass
functions.
Subject: Re: Flawed answer in a calculus textbook?
>> A continuous function on a closed interval is integrable.
>> Proved by Euler, I believe.
That seems doubtful to me. Would Euler even have a proper definition
of
>continuity in front of him? I remember hearing that Cauchy attempted
to
>prove this rigorously, but confused continuity with uniform continuity
>(it's the latter you need). If Cauchy didn't quite have this together,
it's
>doubtful Euler did.
Does anyone know who was the first to prove this? Riemann?
Weierstrass?
> Well on the one hand Edgar is usually a very reliable source - otoh I
> find what he says here very hard to believe, for the same reasons.
> One might _guess_ that it was Riemann. Not that one can always tell
> from the way things are named, but he presumably had _something_
> to do with the definition of the Riemann integral, and then he
> presumably proved _something_ about it - seems likely that this
> would have been first on the list...
> I would guess Riemann didn't have that result. I think continuity = 
uniform
continuity came later. Riemann may have just assumed f is
uniformly
> continuous. This would handle any function that is Lip alpha, alpha > 0,
> for example, and this would take care of a lot of pre-Weierstrass
> functions.
Cauchy in Resume des lecons sur le infinitesimal (dispensing with
diacritical marks) published in 1823 _states_ that if
f:[a,b] -> R
is continuous, then it is integrable, but his proof was inadequate.
Heine proved in Die Elemente der Funkionenlehre published in 1872 that
if
f:K -> R^m
is continuuos on K, and K is compact subset of R^n, then f is uniformly
continuous on K.
Riemann died in 1866.
--
G.C.
Subject: Re: Flawed answer in a calculus textbook?
>> A continuous function on a closed interval is integrable.
>> Proved by Euler, I believe.
That seems doubtful to me. Would Euler even have a proper definition of
>continuity in front of him? I remember hearing that Cauchy attempted to
>prove this rigorously, but confused continuity with uniform continuity
>(it's the latter you need). If Cauchy didn't quite have this together,
>it's
>doubtful Euler did.
Does anyone know who was the first to prove this? Riemann? Weierstrass?
Well on the one hand Edgar is usually a very reliable source - otoh I
> find what he says here very hard to believe, for the same reasons.
One might _guess_ that it was Riemann. Not that one can always tell
> from the way things are named, but he presumably had _something_
> to do with the definition of the Riemann integral, and then he
> presumably proved _something_ about it - seems likely that this
> would have been first on the list...
> I would guess Riemann didn't have that result. I think continuity = 
uniform
continuity came later. Riemann may have just assumed f is
uniformly
> continuous. This would handle any function that is Lip alpha, alpha > 0,
> for example, and this would take care of a lot of pre-Weierstrass
> functions.
My mention of Euler was wrong. Should be Cauchy.
My understanding is something like this... Cauchy proved (with more or
less rigor) that continuous functions are integrable. When Riemann
read the paper, he extracted from it a DEFINITION of integral, namely
what we now call Riemann sums.
I suppose someone could look up Riemann's collected works and see what
he says about it.
--
G. A. Edgar
http://www.math.ohio-state.edu/~edgar/
Subject: Re: Brownian motion approximation
>> A while back I posted a question about whether or not
>> P[sup_{0<=t<=1}|B_t - f(t)| < d] > 0 for all d > 0 and f(t) continuous
>> on [0,1] with f(0) = 0.
>> In other words, does Brownian motion uniformly approximate any
>> continuous function(with f(0)=0) with positive probability? Someone
>> replied that it does, and this follows from first proving it for f(t)
>> = 0 for all t and then applying the Cameron-Martin Theorem. I can do
>> it for f(t)=0, but I don't seem to be able to find a reference for the
>> Cameron-Martin Theorem, though it seems to be related to Girsanov's
>> Theorem, and maybe even follows from it. Can someone give me some help
>> or lead me to a reference?
>I don't believe there is any significant difference between the
>Cameron-Martin-Girsanov theorem, the Girsanov theorem and the
Cameron-Martin
>theorem. As I understand it the same theorem was discovered independently
>and is now attributed to all three.
>Rather like the Green-Gauss-Ostrogradsky theorem.
This does not need quoting any complicated theorems. Construct
a polygonal function h such that |f - h| < d/3 on the interval,
bound (from below) the probability that |B_t - h(t)| < d/3 at
all vertices, and bound the probability that B differs by d/3
from its polygon at those vertices. The Markov nature of B and
the boundedness and continuity of f enable all of this to be
carried out easily.
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
Subject: Re: Did mathematicians know?
you're saying that the L-wing thing is just an argument
*about* an actual (R-wing) proof?
well, I'll have to re-read that little treatise, but
I'd rather dig into one of Devlin's other books ... y'know,
the rather rightwing ones.
> so, it looks to me, like they are setting up
> for a new position where proof does not exist,
> as you say.
> does Devlin mean to say that (e.g.)
> the Pythagorean Theorem has never been proven,
> in spite of hundreds of variants?
> The way I read it, the right wing approach corresponds to something
like
> formalised mathematics such as http://mizar.uwb.edu.pl/JFM/index.html and
> the left wing approach to the more common approach of convincing
people
> that a proof is formalisable.
--les ducs de Buffet;
vote NONE OF THE BELOW
on Trickier Dick Cheney's California Recall & e-Dereg!
http://larouchepub.com
http://members.tripod.com/~american_almanac/
Subject: Re: Did mathematicians know?
>>James Harris <jstevh@msn.com> scribbled the following
>>on sci.math:
>One of the odder things I keep facing is my ability to explain a
>rather basic argument in a very short space, which has *enormous*
>implications, only to see mathematicians mostly walk away.
>>James, have you ever considered the possibility that the reason why
>>everyone who has looked at your proofs and said that you are wrong is
>>that *you* *are* *wrong*?
>No, He's too self absorbed in himself and won't care what anyone says.
My
>question is if he doesn't care what anyone says, why does he post here?
>>Because he does not care what anyone says ->against his work<-, but he
>>is dying to hear people praise him and tell him how smart he is. Just
>>look at how he reacts when anyone says anything that is either
>>encouraging in any way, or even just curious, and see how quickly it
>>turns sour when they start asking ->questions<-.
> Good Point. I've noticed he takes offense to anyone asking questions, why
is
> beyond me.
A profound sense of insecurity, masked by bluster.
Gib
Subject: Re: JSH: The 'Core Error' argument ...
what a tour de force, not that I'm able to read it.
has James actually said what is supposed
to fix the ring?
> I was thinking about this all last night, and I think I may have come
> up with what is going on in James's thought process. Let me try my
> hand at it, while not attempting to diminish the value of your
> efforts, despite the fact that I'll remove them... Some of it is
> there, when you mention the infinitely many ways that f^2 can be
> written as a product of three algebraic integers.
> Of course this is true. It is a tautology. It can be a
>good idea, when you are losing an argument, to state
>something which is obviously true, even though it has
>nothing to do with your desired conclusion. It confuses
>your attackers - all they can say is, well, yeah, you're
>right about that. What they *should* say is, SO WHAT?
> Here, since a1(m) is not a constant function, why does
>Harris's statement tell you anything about divisibility of
>a1(m) when m <> 0 ? That step is simply not there in
>Harris's argument. It is a gap. The counterexample shows
>that it is more than a gap. It is an error which cannot be
>fixed.
> I ->think<- the following is what is running in the back of James's
> mind:
> (1) Say you have a polynomial F(x); write it as F(x)=g(x) + F(0),
> where g(x) is a multiple of x, so g(x) = F(x)-F(0). This can
> always be done.
> (2) Assume that you have a factorization of F(x) in the ring of
> functions (not necessarily of polynomials), F(x) = P(x)Q(x).
> (3) Then F(0)=P(0)Q(0).
> (4) Write P(x) = R(x) + P(0), with R(x)=P(x)-P(0).
> Write Q(x) = S(x) + Q(0), with S(x)=Q(x)-Q(0).
> (5) Then we have
> F(x) = P(x)Q(x)
> = [R(x) + P(0)][S(x)+Q(0)]
> = [R(x)S(x) + R(x)Q(0) + P(0)S(x)] + P(0)Q(0)
> = [R(x)S(x) + R(x)Q(0) + P(0)S(x)] + F(0)
> = g(x) + F(0).
> Maybe, though this is a bit of a stretch, he thinks that R(x) or S(x)
> (or both) will necessarily be factors of g(x); or maybe he thinks that
> R(x) and S(x) cannot be coprime to x. The first is false, as is plain
> from the above; the second is in general false as well. But that, I
> think, is not really where the problem in his reasoning is.
> Because, for instance, let's consider the case where P(x) and Q(x) are
> ->polynomials<- (surely James's argument would work in that trivial
> case as well?) Then [R(x)S(x)+R(x)Q(0)+P(0)S(x)] is a multiple of x,
> so in that instance there are no difficulties. Like that.
> I think where everything is going wrong is in the fact that f is a
> prime, and that is what is leading him astray. For integers, if p is a
> prime, then every other integer only has two choices: it is either a
> multiple of p, or else it is coprime to p. And we have the following
> two properties:
> (A) The sum of two numbers not coprime to p is not coprime to p;
> (B) The sum of a number not coprime to p and a number coprime to p is
> coprime to p.
> But of course, neither of those two propositions is true for a
> composite number: 10 and 6 are both not not coprime to 15, but their
> sum is; 3 is not coprime to 15, 2 is coprime to 15, but their sum is
> not coprime to 15.
> And no number in the algebraic integers satisfies the two conditions
> above, except for units which satisfy them vacuously.
> So I think James is arguing thusly, at least in part:
> Take g_3(m). Since g_3(0) = 3 coprime to f, and g_3(m)-g_3(0) is not
coprime to f,
> then g_3(m) = (g_3(m)-g_3(0)) + g_3(0) is the sum of something coprime
> to f and something not coprime to f, so it is coprime to f. Take
> g_2(m) and g_1(m); in both cases, g_i(0) is not coprime to f; since
> g_i(m)=(g_i(m)-g_i(0))+g_i(0) and g_i(m)-g_i(0) is not coprime to f,
> then this is the sum of two things not coprime to f and therefore it
> is not coprime to f.
> This argument would work in the integers if f is a prime, and the g_i
> are polynomials, since the polynomial F(x)-F(0) is actually a multiple
> of f and so is the constant F(0). In this situation, assuming that the
> only way a prime could divide every value of the polynomial were if it
> divided all the coefficients (something which is false for integers,
> but true for algebraic integers, so we are really in a complete
> mish-mash of hypothesis, as you can see) the properties above would
> yield the conclusion James is hoping for:
> g_3(0) is coprime to f, g_3(m)-g_3(0) is not coprime to f for every m,
and
> so g_3(m) must be coprime to f for every m; since g_2(0) and g_1(0) are
> not coprime to f and g_i(m)-g_i(0) is also not coprime to f, then it
> would follow that g_1(m) and g_2(m) are multiples of f for every m.
> One of the hypothesis is that the only way in which every (integer)
> value of a polynomial is a multiple of f is if every coefficient is a
> multiple of f. That one I think holds for algebraic integers, (Dot
> proved it if it was for every value of x, and f a prime integer, not
> just the values when evaluated at integers:
> Even then, James has not shown that g_i(m)-g_i(0) is not coprime to f,
> which would also constitute a gap (though, of course, he could claim
> that this is not what he is arguing).
> But very key to the rest of the argument was that f was a prime, the
> elements involved were integers, and thus that properties (A) and (B):
> (A) The sum of two numbers not coprime to f is not coprime to f;
> (B) The sum of a number not coprime to f and a number coprime to f is
> coprime to f.
> hold.
> When we drop the assumption that f is a prime in the integers, the
> argument breaks down, precisely because properties (A) and (B)
> fail. Then we observe that redistribution of factors that James
> finds so nonsensical in its face in the integers as well. Because it
> is possible to have the sum of two numbers not coprime to f be coprime
> to f, and much more importantly, because it is possible ot have
> something not coprime to f added to something which ->is<- coprime to
> f, and yet obtain something which is NOT coprime to f.
> That is, in part, the phenomenon that we are observing. Even if we
> could prove that g_3(m)-g(0) is not coprime to f for each m, the fact
> that g(0)=3 is coprime to f does NOT guarantee that
> g_3(m)=[g_3(m)-g(0)]+g_3(0) is coprime to m. We could have, for
> example, something like f=15, g_3(0)=2, and g_3(m)-g(0)=3; then
> g_3(m)-g(0)=3 is a divisor of 15, g_3(0) is coprime to 15, but g_3(m)
> is ALSO a divisor of 15, despite being the sum of something coprime to
> 15 with a divisor of 15.
> Likewise, even if we knew that g_1(m)-g_1(0) is not coprime to f for
> each m, the fact that g_1(0) is a multiple of f does guarantee that
> g_1(m) is always not coprime to f, but it does not guarantee what the
> factor is; even if we could always guarantee that g_1(m)-g_1(0) is a
> multiple of some proper divisor of f (which is certainly NOT the case
> that James is dealing with), the best we could say then is that g_1(m)
> is a multiple of that divisor, but the gcd with f could be something
> else entirely.
> For instance, if f = 30, g_1(0) = 60, even if g_1(m)-g_1(0) is always a
> multiple of 5, it may be possible for g_1(m) to have a factor greater
> than 5 in common with f, say if g_1(m)-g(0) = 15, then g_1(m)=15+60=75
> has the factor 15 in common with f, greater than the 5 we are
> guaranteed. Other more complicated examples could be produced,
--les ducs de Buffet;
vote NONE OF THE BELOW
on Trickier Dick Cheney's California Recall & e-Dereg!
http://larouchepub.com
http://members.tripod.com/~american_almanac/
Subject: Large rectangular matrix problem...
I've come to a bit of a problem in some research. I want to solve:
Ax=b
in a least squares sense (select x to minimise ||b-Ax||_2).
The matrix A is very large but simply structured. A has the structure:
A=[ P1 P2 ... Pn ]
where each Pi is a block toeplitz matrix with toeplitz blocks. There is no
simple relationship between any Pi and any other. Any element of A can be
trivially computed from a (comparatively) small number of values. Each Pi
is the same size typically 1024 square and may be as large as 4096, and n
will typically be around 256 but may be as large as 512. These sizes make A
an absolute monster in terms of storage - if it were ever completely
computed.
The only way I know of to get anywhere near a solution to x without having
to store gigabytes of matrices would be to use the conjugate gradient
algorithm to solve:
(A'A)x=A'b
where A' is the transpose of A.
For engineering reasons, the computer which will be solving these is fast
at performing computations but isn't capable of having sufficient storage
of the computation isn't possible.
Is there a better approach you could suggest? Any other comments I might
find useful?
Ian Woods
Subject: Class of computable functions
I took a class in basic foundations and computability last year, but
there is a question that has been bothering me.
Suppose I want a 'large' computably enumerable collection of functions
f_i : N --> N with the following properties:
1. Each primitive recursive function is included.
2. Each f_i is total by construction.
3. Given i and n in N, there is a computable function time: N x N -->
N which tells me that the value of f_i(n) will take at most time(i,n)
to compute by a turing machine or equivalent.
['Take as long as you want, but PLEASE tell me when you will be
done!']
4. The function time is computable in 'Ackermann + constant' time,
or at least it's behavior is boundedly nasty in some sense. :)
I feel rather queasy about the prospects for this, but it has been
bothering me for too long. The basic theme is this: I want to know
worst case scenario computation time. Will I not get much besides
primitive recursive functions?
Rex Butler
PS I've heard the term 'strongly computable.' Is this related?
Subject: Re: Linear Algebra proof
> following linear algebra theorem:
> Given an order n real matrix A, the function which maps the real
> variable lambda into the real variable det(A-lambda*I) is a polynomial
> of degree n
Hi Andrea, Below are two proofs that came to my mind. In fact the
proofs below work on any general field (the first proof works on any
integral domain because any integral domain is a subring of its field
of fractions).
Sketch of the 1st proof: Every matrix is upper triangularizable over
its algebraic closure. With that in mind, with no loss of generality
one may assume that A is an upper triangular matrix. Now the rest
easily follows. Done!
Sketch of the 2nd proof: Let A in M_n( F[x]) where F is a field whose
characteristic does not divide n!. Think of Det(A) as a function of
the rows of A, say a_1, ..., a_n. We can write Det(A)= det(a_1, ...,
a_n). Now using the definition of determinant, one can easily verify
that
det(A)' = det((a_1)', ..., a_n)+ ... + det(a_1, ..., (a_n)'). (*)
Now back to your problem; set f(lambda):= det(A- lambda*I). In light
of (*), it is not difficult to see that f^{(n)}(lambda) = (-1)^n*n!
and f^{(k)}(lambda) = 0 for all lambda in F and k >= n+1 where
f^{(k)} denotes the k-th derivative of f w.r.t. lambda. Now we are
done in view of the Taylor formula. qed.
Hope that helps.
Best wishes, Bamdad.
PS I do not check this place, nor my yahoo account on a regular basis;
so sorry if I am unable to get back to your possible reply.
> Below I report the proof I devised: I am not sure it is correct, so I
> would be extremely grateful if you may check wheter it's kosher.
> Should you find annoying to follow someone's else proof, I would still
> find helpful another proof, provided that it uses recursive definition
> of determinant, i.e.
> n=1, det([a_{11}])=a_{11}
> n, det(A)=sum_{j=1}^n (-1)^{i+j}a_{ij}det(A_{ij}) etc. etc.
> Andrea
> Proof: by induction. For n=1 it is true, so let A be a real nxn
> matrix, then
> det(A-lambda*I)=sum_{j=1}^n (-1)^{1+j}a_{1j}det((A-lambda*I)_{1j})
> det((A-lambda*I)_{11})=det(A_{11}-lambda*I) is a degree n polynomial
> by inductive hypothesis.
> Fro j ne 1, det((A-lambda*I)_{1j}) ne det(A_{1j}-lambda*I), then
> |det((A-lambda*I)_{1j})| = |det(P_{n-1}(j-1))*(A-lambda*I)_{1j}| =
> |det(B)| where P_{n-1}(j-1)) is the order n-1 permutation matrix which
> brings the (j-1)th row to first row (i.e., b_{11}=a_{j1)},
> b_{12}=a_{j2},..,b_{22}=a_{(j+1)(j+1)}-lambda and so on).
> B = C-lambda*M where
> m_{1s} = 0 (s=1...n-1), m_{rs} = delta_{rs} for r ne 1
> det(B) = det(C-lambda*M) = det(D-lambda*I)-det(E-lambda*I) where
> d_{11} = c_{11}-lambda, d_{rs} = c_{rs} for (r,s) ne (1,1)
> e_{1s} = 0, s=1,...,n-1, e_{rs} = c_{rs} for (r,s) ne (1,1)
> For inductive hypothesis det(D-lambda*I) and det(E-lambda*I) are
> polynomial of degree n-1, so det(C-lambda*M) = det(B) = +-
> det((A-lambda*I)_{1j}) is a polynomial of degree m le n-1, and so it
> is (-1)^{1+j}a_{1j}det((A-lambda*I)_{1j}). Then det(A-lambda*I) is sum
> of a polynomial of degree n and n-1 polynomials of degree m le n-1,
> thus being an n_th degree polynomial itself. QED
Subject: Re: Linear Algebra proof
> Hello
> Given an order n real matrix A, the function which maps the real
> variable lambda into the real variable det(A-lambda*I) is a polynomial
> of degree n
> Actually the result is trivial if you use the real definition of the
> determinant (ie: sum(sig(s)*product(s(i),i, i=1..n), s permutation),
where
> sig(s) is the signature of s.
As I already stated in my previuous post, I would still find helpful
another proof that proof, provided that it uses recursive (or
Laplace's) definition of determinant. Indeed, using the definition of
det(A) as
sum_{all permutations (i1,i2,...,in) of (1,2,...,n)} sign(i1, i2,
..., in) a_{i11}a_{i22}...a_{inn}
the proof is trivial, but, again, this isn't of any help to me, nor it
is a proof based on equivalence between this and Laplace's definition.
BTW, the formula using permutations isn't itself the real definition
of determinant (it's only the historical definition), but rather an
explicit formula for det(A) derived by the real definition, which is:
det(A) is the only function that maps the vector space of order n
real matrices into vector space of real numbers having following
properties:
-det(A) is multilinear in each column of A.
-detA is alternating in columns of A.
-det(I_n) = 1 where I_n is the order n identity matrix.
Andrea
Subject: Re: Linear Algebra proof
andrea2@despammed.com (Andrea) wants a proof of the statement:
>> Given an order n real matrix A, the function which maps the real
>> variable lambda into the real variable det(A-lambda*I) is a polynomial
>> of degree n
and further states that she wants the proof to be in terms of the
real definition of determinant, viz.:
>det(A) is the only function that maps the vector space of order n
>real matrices into vector space of real numbers having following
>properties:
>-det(A) is multilinear in each column of A.
>-detA is alternating in columns of A.
>-det(I_n) = 1 where I_n is the order n identity matrix.
It would seem to me that, given this last desideratum, it would
be very appropriate to rephrase the statement to be proved
*also* in a coordinate-free way. So, what is a polynomial
of degree n? Whatever it is, presumably the restriction
to a 1-dimensional subspace L of a finite-dimensional vectorspace
V (which might be the vector space of order n real matrices)
of a polynomial map of degree n from V to R (whatever *that*
might be) would be one such (leaving aside for the moment the
possibility that the degree might actually turn out to be
smaller than n). Now, there is a standard multilinear-algebra
definition of polynomial maps from a vectorspace V to its ground
field (say R, as in this case), as multilinear maps which are
*symmetric* (rather than, as in the case of det, *alternating*)
in their arguments.
I suggest to Andrea that she (1) prove the desired theorem
using this interpretation, and then (2) prove that, given a
basis of V, polynomial maps in this coordinate-free sense
are put into natural bijection with polynomial functions of
n variables in the other sense. Of course there are various
other details to attend to, but this approach seems good to me
(heck, *I'm* not the one who has to write it out correctly).
Lee Rudolph
Subject: Re: Linear Algebra proof
> andrea2@despammed.com (Andrea) wants a proof of the statement:
>> Given an order n real matrix A, the function which maps the real
>> variable lambda into the real variable det(A-lambda*I) is a
polynomial
>> of degree n
> and further states that she wants the proof to be in terms of the
> real definition of determinant, viz.:
Wait, the process is not converging :-) I do not have to use the
coordinate free definition of det(A). As a matter of fact, I just
cited it because I do not agree wiht Julien saying that the
permutation definition is the real one. Instead, as stated in my
first post, the definition I need to use is the recursive or
Laplace's definition of determinant:
n=1, det([a_{11}])=a_{11}
n >1, det(A)=sum_{j=1}^n (-1)^{1+j}a_{1j}det(A_{1j}) where A_{1j} is
the (n-1)x(n-1) matrix obtained by deleting the first row and the j-th
column of A.
Indeed, I used this definition in my proof in the first post. I
sincerely apologize for any incomprehensions due to my bad English.
>heck, *I'm* not the one who has to write it out correctly).
Sure: let me point out that I am not asking anyone to do the proof in
only asking if it is correct, or if someone can post a reference to a
correct proof. This way I may compare the two and reassure myself that
mine is correct too (at least I hope so).
Andrea
PS: I am Italian, so Andrea is a male's name, not female's ;-)
Subject: Borel measurable derivatives
Hey all,
I'm trying to write up a proof, and I'm using the fact if a function
f: R -> R
is differentiable everywhere, then the derivative f' is Borel measurable.
Is there an easy way to do this?
Subject: Re: Mass is a quantity of matter
> Physical mass is an aggregation - via gravitation and other centripetal
> forces - of the substances comprising any object; body, and or mass of
> material matter; which causes these accumulations to have inertia, and/or
> heft; the property of matter whereby it becomes more obvious that it
requires greater
> (net) force to change the velocity of an accumulation as it becomes
larger:
> Here on Earth Galileo found that the rate of change in the velocity of
any
> body free falling at Earth's surface was about [s/tO =
16'/secO] - half of
> 'Newton's' acceleration of free fall [g] - and furthermore - in effect -
he
> found that the force restraining this change in velocity of free fall
from
> continuing toward Earth's center is the mutual force exerted between the
> body and Earth's terra firma; which force is commonly measured with
> weight-scales: On Earth, the ratio of this weight-force [w], divided by
the
> rate of change in velocity [s/tO = 16'/secO]
that it restrains, is a
> constant [wtO/16' = wtO/s]; to be known
hereafter, as one half of the
body's
> gravitational mass [g/2], and/or inertia.
> On any similar planet, such as the moon, this _ratio_ will still be equal
to
> half of any body's mass, and/or inertia!
In everyday use, a body's mass [m] is commonly confused with its
weight
[w]; A practice which must cease immediately; for the sake of physics!
According to newton's second law weight is the product of mass [m] and the
gravitational acceleration of free fall [g]: This erroneous formula w = ma
is a special case of f = ma, and like f = wa/g, must be written as w =
fg/a:
All because inertial mass m = f/a, and gravitational mass m = w/g; where it
follows that inertial mass f/a is equal to gravitational mass [w/g]: f/a =
w/g.
Cheeze, what have I got to do, write a book for youse people(:^?
Subject: Re: Mass is a quantity of matter
> There's some real live inertial mathematics! :)
> Newton's Second Law
> http://scienceworld.wolfram.com/physics/NewtonsSecondLaw.html
A force F acting on a body gives it an acceleration a which is in the
direction of the force and has magnitude inversely proportional to the mass
m of the body: F = ma.
Replacing mass [m] with w/g; we'd have F = wa/g: But it's only the net
force
that causes an acceleration; so we'd _ought_ to use (F-uw) = wa/g; or as I
put it, the _net_ force [F-uw = f] exerted on and/or by a body gives it an
acceleration [a] which is in the direction of the _net_ force and has
magnitude inversely proportional to the mass m of the body: f = wa/g.
The 'uw' is the product of a coefficient of frictional restraint and the
weight of the body.