mm-70
===
A transaction consists of adding ten balls to a bucket
and removing 1.>(Obviously a transaction is a net increase
of nine balls.) Assume that>in?itely many transactions
somehow occur. Some people here think>there will be no balls
in the bucket afterward!>There are some people who think that
there will only be zero balls>left if we label the balls in a
certain way, otherwise there will be>more than zero!>There
are other people who think that there will be none
left>because the cardinalities of the sets of balls added to
the bucket is>the same as the cardinality of the set of balls
removed!>I tend to say the heck with their theories. I know
that if it were>truly possible for in?itely many
transactions to occur, the bucket>would be VERY full! Where
do theories that say othwise come>from? Does anyone really
disagree that the bucket has more balls>after each
transaction and never has less?> If the balls are labeled and
a labeled ball is removed after each> time it is inserted, is
that ball in the bucket at the end? If it is in the bucket at
the end, how does it get there? If it is not in at the end,
and being not in is true of every ball,> which (allegedly
large number of) balls are in the bucket at the end? Surely
if every transaction takes a ?ite time there is no end.Oh
really, so exactly what does 1/2 sec + 1/4 sec + 1/8 sec
+...+ 1/2^n+...= ? Pesonally I thought that it equals 1 sec.
Gib>
===
[snip]> Surely if every transaction takes a ?ite
time there is no end.<Despair>It seems that some people have
yet to resolve, in their own minds, thesimple paradoxes of
Zeno!Believe it or not, an in?ite series of positive terms
(times) can have a?ite sum. For example,1/2 sec + 1/4 sec +
1/8 sec +...+ 1/2^N sec +... = 1 sec.David
===
Poker Joker
<Poker@wi.rr.com> scribbled the following:> A transaction
consists of adding ten balls to a bucket and removing 1.>
(Obviously a transaction is a net increase of nine balls.)
Assume that> in?itely many transactions somehow occur. Some
people here think> there will be no balls in the bucket
afterward!> There are some people who think that there will
only be zero balls> left if we label the balls in a certain
way, otherwise there will be> more than zero!> There are
other people who think that there will be none left> because
the cardinalities of the sets of balls added to the bucket
is> the same as the cardinality of the set of balls removed!>
I tend to say the heck with their theories. I know that if it
were> truly possible for in?itely many transactions to
occur, the bucket> would be VERY full! Where do theories that
say othwise come> from? Does anyone really disagree that the
bucket has more balls> after each transaction and never has
less?I don't know if I'm responding to a troll here, 
but
given the above,each transaction adds nine balls to the
bucket, thus creating amonotonically increasing function from
the index number of thetransactions to the number of balls in
the bucket. So it is impossiblefor the bucket to be empty *at
any time* after the start of theexperiment.-- /-- Joona
Palaste (palaste@cc.helsinki.? - Finland
- http://www.helsinki.?~palaste
rules! --/As a boy, I often
dreamed of being a baseball, but now we must go forward,
notbackward, upward, not forward, and always whirling,
whirling towards freedom! - Kang
===
> I don't know if I'm
responding to a troll here, but given the above,> each
transaction adds nine balls to the bucket, thus creating a>
monotonically increasing function from the index number of
the> transactions to the number of balls in the bucket. So it
is impossible> for the bucket to be empty *at any time* after
the start of the> experiment.While it is impossible for the
mathematically modeled bucket to be empty after any ?ite
number of transactions, one of the peculiarities of in?ite
mathematical processes is that the bucket CAN be empty after
in?itely many transactions, since each ball put into the
bucket can be, then or later, removed from the bucket in an
in?ite mathematical process completed in ?ite mathematical
time.It is our attempts to interpret the mathematical model
of bucket and balls and transactions in our everyday
non-mathematical world that troubles our understanding.
===
>
Poker Joker <Poker@wi.rr.com> scribbled the following:>A
transaction consists of adding ten balls to a bucket and
removing 1.>(Obviously a transaction is a net increase of
nine balls.) Assume that>in?itely many transactions somehow
occur. Some people here think>there will be no balls in the
bucket afterward!>There are some people who think that
there will only be zero balls>left if we label the balls in
a certain way, otherwise there will be>more than zero!There are other people who think that there will be none
left>because the cardinalities of the sets of balls added to
the bucket is>the same as the cardinality of the set of balls
removed!>I tend to say the heck with their theories. I
know that if it were>truly possible for in?itely many
transactions to occur, the bucket>would be VERY full! Where
do theories that say othwise come>from? Does anyone really
disagree that the bucket has more balls>after each
transaction and never has less?> I don't know if 
I'm
responding to a troll here, but given the above,> each
transaction adds nine balls to the bucket, thus creating a>
monotonically increasing function from the index number of
the> transactions to the number of balls in the bucket. So it
is impossible> for the bucket to be empty *at any time* after
the start of the> experiment.> If after adding 10 balls I
remove the last one added, it is completely equivalent to
adding 9 balls, and so we can say which balls are certainly
going to stay in the bucket. These are the balls numbered
1-9, 11-19, 21-29 etc.In fact, we can map each natural number
to a ball we de?itely know is in the bucket (n -> 10n+1), and
so we know there are in?itely many balls in the bucket after
in?itely many transactions.However, if after adding 10 balls
I remove the *?st* ball (i.e., the ball which was added as
early as possible and is still in the bucket), then we have a
problem. At time t we remove ball number t. In this case, if
you think there are in?itely many balls in the bucket, then
please name one. I know that ball number t was removed in
time t, and so for any natural number t, ball number t is
*not* in the bucket. So the bucket must be empty.To be
honest, I don't think there's any problem. All we can 
say is
that the number of balls depends on the choice of ball to
remove. If I always remove the ?st ball, I'll end up with no
balls. If I always remove ball number t, I'll end up with t-1
balls. If I always remove ball number x_t where x_t is
monotonically increasing, I'll end up with in?itely many
balls.Any set theorist to resolve this?This seems to me like
the paradox of f(x) = 1/x. For every natural number n, you
can ?d a positive number for which f(n) is n+1 times larger
(simply 1/(n*(n+1))). So I say that since, as n increases, we
can ?d more and more positive numbers between 0 and f(n), how
can the limit at in?ity of f(n) be 0? After all, for any
?ite number m, we can ?d at least m numbers, all different,
all positive, between 0 and f(n). Hence, a paradox. The
problem here is that you think of the one step before
in?ity, but there's not such thing. It's a quantum
leap.
===
> To be honest, I don't think there's any problem.
All we can say is> that the number of balls depends on the
choice of ball to remove. If I> always remove the ?st ball,
I'll end up with no balls. If I always> remove ball number t,
I'll end up with t-1 balls. If I always remove> ball number
x_t where x_t is monotonically increasing, I'll end up> with
in?itely many balls.> Any set theorist to resolve
this?This is right.
===
> Poker Joker <Poker@wi.rr.com>
scribbled the following:>A transaction consists of adding ten
balls to a bucket and removing 1.>(Obviously a transaction is
a net increase of nine balls.) Assume that>in?itely many
transactions somehow occur. Some people here think>there
will be no balls in the bucket afterward!>There are some
people who think that there will only be zero balls>left if
we label the balls in a certain way, otherwise there will
be>more than zero!>There are other people who think that
there will be none left>because the cardinalities of the
sets of balls added to the bucket is>the same as the
cardinality of the set of balls removed!>I tend to say the
heck with their theories. I know that if it were>truly
possible for in?itely many transactions to occur, the
bucket>would be VERY full! Where do theories that say
othwise come>from? Does anyone really disagree that the
bucket has more balls>after each transaction and never has
less?> I don't know if I'm responding to a troll here, 
but
given the above,> each transaction adds nine balls to the
bucket, thus creating a> monotonically increasing function
from the index number of the> transactions to the number of
balls in the bucket. So it is impossible> for the bucket to
be empty *at any time* after the start of the> experiment.>
If after adding 10 balls I remove the last one added, it is
completely> equivalent to adding 9 balls, and so we can say
which balls are> certainly going to stay in the bucket. These
are the balls numbered 1-9,> 11-19, 21-29 etc. In fact, we can
map each natural number to a ball we de?itely know is> in the
bucket (n -> 10n+1), and so we know there are in?itely many>
balls in the bucket after in?itely many transactions.
However, if after adding 10 balls I remove the *?st* ball
(i.e., the> ball which was added as early as possible and is
still in the bucket),> then we have a problem. At time t we
remove ball number t. In this case,> if you think there are
in?itely many balls in the bucket, then please> name one. I
know that ball number t was removed in time t, and so for>
any natural number t, ball number t is *not* in the bucket.
So the> bucket must be empty. To be honest, I don't think
there's any problem. All we can say is that> the number of
balls depends on the choice of ball to remove. If I always>
remove the ?st ball, I'll end up with no balls. If I always
remove> ball number t, I'll end up with t-1 balls. If I
always remove ball> number x_t where x_t is monotonically
increasing, I'll end up with> in?itely many balls. Any set
theorist to resolve this?You will not end up with ...,
because there is no such thingas after in?itely many
transactions.Dirk Vdm> This seems to me like the paradox of
f(x) = 1/x. For every natural> number n, you can ?d a
positive number for which f(n) is n+1 times> larger (simply
1/(n*(n+1))). So I say that since, as n increases, we can>
?d more and more positive numbers between 0 and f(n), how
can the> limit at in?ity of f(n) be 0? After all, for any
?ite number m, we> can ?d at least m numbers, all
different, all positive, between 0 and> f(n). Hence, a
paradox. The problem here is that you think of the one> step
before in?ity, but there's not such thing. It's a 
quantum
leap.>
===
> You will not end up with ..., because there is no
such thing> as after in?itely many transactions.Sure there
is; the transactions can be numbered with ordinals andthere
is no problem at all.
===
> Poker Joker <Poker@wi.rr.com>
scribbled the following:> A transaction consists of adding
ten balls to a bucket and removing 1.> (Obviously a
transaction is a net increase of nine balls.) Assume that>
in?itely many transactions somehow occur. Some people here
think> there will be no balls in the bucket afterward!>
There are some people who think that there will only be zero
balls> left if we label the balls in a certain way,
otherwise there will be> more than zero!> There are other
people who think that there will be none left> because the
cardinalities of the sets of balls added to the bucket is>
the same as the cardinality of the set of balls removed!> I
tend to say the heck with their theories. I know that if it
were> truly possible for in?itely many transactions to
occur, the bucket> would be VERY full! Where do theories
that say othwise come> from? Does anyone really disagree
that the bucket has more balls> after each transaction and
never has less?> I don't know if I'm responding to a 
troll
here, but given the above,> each transaction adds nine balls
to the bucket, thus creating a> monotonically increasing
function from the index number of the> transactions to the
number of balls in the bucket. So it is impossible> for the
bucket to be empty *at any time* after the start of the>
experiment.The problem as stated is ambiguous, because it
does not specify how theball to be removed is chosen at each
step.See the concurrent thread Ten Balls In - One Ball Out -
Repeat - HowMany Remain?. Or go to Google groups and search
for the historicalthreads Bucket and Balls and its offshoot
Trans?ite Subway.-- Dave SeamanJudge Yohn's mistakes
revealed in Mumia Abu-Jamal
ruling.<http://www.commoncouragepress.com/index.cfm?action=
book&bookid=228>
===
> A transaction consists of adding ten
balls to a bucket and removing 1.> (Obviously a transaction
is a net increase of nine balls.) Assume that> in?itely many
transactions somehow occur. Some people here think> there will
be no balls in the bucket afterward!> There are some people
who think that there will only be zero balls> left if we
label the balls in a certain way, otherwise there will be>
more than zero!> There are other people who think that
there will be none left> because the cardinalities of the
sets of balls added to the bucket is> the same as the
cardinality of the set of balls removed!> I tend to say the
heck with their theories. I know that if it were> truly
possible for in?itely many transactions to occur, the
bucket> would be VERY full! Where do theories that say
othwise come> from? Does anyone really disagree that the
bucket has more balls> after each transaction and never has
less?> If you say that there's a ball left at the end, which
one is it? It can't be #47, that was removed on the 47th
step. And so forth.
===
A transaction consists of adding ten
balls to a bucket and removing 1.> (Obviously a transaction
is a net increase of nine balls.) Assume that> in?itely many
transactions somehow occur. Some people here think> there will
be no balls in the bucket afterward! There are some people who
think that there will only be zero balls> left if we label the
balls in a certain way, otherwise there will be> more than
zero! There are other people who think that there will be
none left> because the cardinalities of the sets of balls
added to the bucket is> the same as the cardinality of the
set of balls removed! I tend to say the heck with their
theories. I know that if it were> truly possible for
in?itely many transactions to occur, the bucket> would be
VERY full! Where do theories that say othwise come> from?
Does anyone really disagree that the bucket has more balls>
after each transaction and never has less?> If you say that
there's a ball left at the end, which one is it? It> 
can't be
#47, that was removed on the 47th step. And so forth.You don't
know that. If you remove the last ball that was put in,
I'mcertain that balls 1,2,3,4,5,6,7,8,10,... will always stay
in the bucket.By the way, at the end of what?
===
> The length
of the curve f(x) on the continuous curve [a,b] is the>
integral taken from a to b of the squareroot of the sum of
the> derivitive of the function, squared, plus one.> This
formula for arc length is probably in every elementary
calculus text.Lol, not mine, but then again, mine's not all
that great. Woulda savedme a good three hours though.
===
>
Hey, haven't been here in a while...> Anyway, Thursday and
Friday I was working on a problem, basically,> given a
fuciton and a continuous interval of that function, I wanted>
to ?d the length of its curve. I eventually came up with
this:> The length of the curve f(x) on the continuous curve
[a,b] is the> integral taken from a to b of the squareroot of
the sum of the> derivitive of the function, squared, plus
one.> Bobby SimioneThat is a standard result in calculus.
If you discovered iton your own, rather than reading it in a
Calculus text,that is quite good, and shows that you have a
much betterthan average grasp of the material.Kepp trying to
discover things new to you. It does not matterif you are not
the ?st.Martin Cohen
===
>That is a standard result in
calculus. If you discovered it>on your own, rather than
reading it in a Calculus text,>that is quite good, and shows
that you have a much better>than average grasp of the
material.Kepp trying to discover things new to you. It does
not matter>if you are not the ?st.Martin
Cohen~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
~~~
===
R is the reals. That N likely a typo, read R
instead.>You claim A_n ->* A iff>A_n subset A forall n and
for all p > 0, p in R exists q in N>for all a in A for all n
> q exists b in A_n: d(a,b) < pFor all n, let A_n = D be a
dense subset of R.>Let A be any set with D subset A subset
R.>Then A_n ->* A for all such A's.Your uniform convergence,
and hence your convergence doesn't have unique>limits.
Moreover, a sequence can even converge to different sets
with>different cardinalities.-- >Remember the following part
of the ?st >message in this series of posting?No.>You have a
problem here, because a limit is not unique, if A' is a>subset
of A, then A_n -> A implies A_n -> A'.All of this stuff has
topology in it, contary to your title topic.For limits
without topology, for A_n, A subset S, there's:limsup A_n =
/{ { / Aj | j > k } | k in N }liminf A_n = /{ { / Aj | j > k
} | k in N }liminf A_n subset limsup A_n; / / intersect
unionA_n -> A when A = limsup A_n = liminf A_n lim A_n =
liminf A_n = limsup A_ndescending, ascending A_n ==> lim A_n
= /{ A_n | n in N }, /{ A_n | n in N }Whatever math I dream
up is already old hat. -- William's Metatheorem-
===
>
...>am doing. in the end i am trying to show that Aut(Z16) is
isomorphic>to Z2xZ8, and I dont think I want to use the fact
that I know all>groups of order 8.> ...> If Z16 stands for
the cyclic group with 16 elements, then its rather simple to
determine Aut(Z16): choose a generator, for example 1. Then
an automorphism f is uniquely determined by prescribing the
image f(1). This image must be another generator of Z16. So
there are as many automorphisms as generators of Z16. Less
than 16 by the way - so Aut (Z16) cannot be isomorphic to
Z2xZ8.> Hactually, I made a mistake ... that should read:am
doing. in the end i am trying to show that Aut(Z16) is
isomorphicto Z2xZ4, and I dont think I want to use the fact
that I know allgroups of order 8.I already know what Aut(Z16)
is, I am trying to show it by elementarymeans. in fact, i know
that Aut(Z16) is isomorphic to the set of unitsin the ring
(Z16,+,*) or all number relatively prime to 16 equippedwith
multiply. but ?ding a simple proof that Aut(Z16)=Z2xZ4
haseluded me. and by simple I mean not describing all groups
of order 8,since |Aut(Z16)|=8.thanks
===
in message
<83f64ab8.0312132314.5aaab249@posting.google.com>:[...]> I
already know what Aut(Z16) is, I am trying to show it by
elementary> means. in fact, i know that Aut(Z16) is
isomorphic to the set of units> in the ring (Z16,+,*) or all
number relatively prime to 16 equipped> with multiply. but
?ding a simple proof that Aut(Z16)=Z2xZ4 has> eluded me. and
by simple I mean not describing all groups of order 8,> since
|Aut(Z16)|=8.Well, you can get away with describing only the
abelian groupsof order 8 by ?st showing that the
automorphism group of acyclic group is abelian.-- Jim
Heckman
===
> I am given a set of generators (s_1, ...,
s_{n-1}) and relations for a> group G_n:> s_i^2 = 1
(i=1...n-1)> s_i s_j = s_j s_i ( |i-j|> 1)> s_i s_{i+1} s_i =
s_{i+1} s_i s_{i+1} (i=1...n-2)> I have veri?d that these
properties hold for transpositions in S_n, e.g.> for (1,2),
(2,3), (n-1,n), but I don't know that there can't be 
other>
relations for S_n. I want to show that G_n = S_n. Since phi:
G_n -> S_n is> surjective, I was thinking of using
Todd-Coxeter and induction to show> that> |G_n| = |S_n|. I
tried this, but did not get very far. Any hints?You could
show that any word in these generators is equivalentto one in
a canonical form. One possible canonical form is this.De?e
words t(i,j) where 1 <= i <= j <= n. Then t(i,i) is empty
whilet(i,j) = s_i s_{i+1} ... s_{j-1} for i < j. The shw that
each wordin the s_i can be reduced to a wordt(n-1, a_{n-1})
t(n-2, a_{n-2}) .... t(2, a_2)t(1, a_1)by means of the
generators.-- Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the
last laugh. Alan Partridge, _Bouncing Back_ (14
times)
===
Arrrghh.... this calculator is really annoying
me.I've put my fx-991MS calculator into CMPLX mode, and I can
get it to>calculate and show me, for example, the real and
imaginary parts of 1 -3i>(simple I know but it de?itely
works). However I can't get it to showme>the real and
imaginary parts of, for example, 1 - sqrt3i. It says
Math>ERROR. This also happens with *any* other complex
numbers calculation>involving roots or powers. If you mean
(sqrt(3))*i, then you can almost certainly overcome the>
problem by putting in some parentheses.I still can't raise a
complex expression to a power though e.g. (1 + 3i)^3.The only
exception is squaring it - the calulator has a separate
buttonspeci?ally for squaring i.e. I don't have to put in
the ^ then the 3(or any power), I simple press the square
button. Any idea on the correctparenthesis for this?I've
tried (1 + 3i)^3, (1 + 3i)(^3), ((1 + 3i)(^3)) - none seem to
work.> If you mean sqrt(3i) then you are up against a
restriction on the> operations allowed on your calculator.
(Note that sqrt is> multi-valued.)Perhaps this is the reason
I can't square the expression?
===
Arrrghh.... this calculator
is really annoying me.I've put my fx-991MS calculator into
CMPLX mode, and I can get it to>calculate and show me, for
example, the real and imaginary parts of 1 -> 3i>(simple I
know but it de?itely works). However I can't get it to show>
me>the real and imaginary parts of, for example, 1 - sqrt3i.
It says Math>ERROR. This also happens with *any* other
complex numbers calculation>involving roots or powers. If you
mean (sqrt(3))*i, then you can almost certainly overcome the>
problem by putting in some parentheses.> I still can't
raise a complex expression to a power though e.g. (1 +
3i)^3.> The only exception is squaring it - the calulator has
a separate button> speci?ally for squaring i.e. I don't have
to put in the ^ then the 3> (or any power), I simple press
the square button. Any idea on the correct> parenthesis for
this?> I've tried (1 + 3i)^3, (1 + 3i)(^3), ((1 + 3i)(^3))
- none seem to work.> If you mean sqrt(3i) then you are up
against a restriction on the> operations allowed on your
calculator. (Note that sqrt is> multi-valued.)> Perhaps
this is the reason I can't square the expression?> Does
your Casio require explicit multiplication symbols? If so,
then enter (1 + 3*i)^3 or (1 + (3*i))^3 with an explicit
multiplication symbol, and see how it
evaluates.
===
>Arrrghh.... this calculator is really
annoying me.>I've put my fx-991MS calculator into CMPLX
mode, and I can get it to>calculate and show me, for
example, the real and imaginary parts of 1 ->3i>(simple I
know but it de?itely works). However I can't get it to
show>me>the real and imaginary parts of, for example, 1 -
sqrt3i. It says Math>ERROR. This also happens with *any*
other complex numbers calculation>involving roots or
powers.> If you mean (sqrt(3))*i, then you can almost
certainly overcome the> problem by putting in some
parentheses.>I still can't raise a complex expression to a
power though e.g. (1 + 3i)^3.>The only exception is squaring
it - the calulator has a separate button>speci?ally for
squaring i.e. I don't have to put in the ^ then the 3>(or any
power), I simple press the square button. Any idea on the
correct>parenthesis for this?I've tried (1 + 3i)^3, (1 +
3i)(^3), ((1 + 3i)(^3)) - none seem to work.> If you mean
sqrt(3i) then you are up against a restriction on the>
operations allowed on your calculator. (Note that sqrt is>
multi-valued.)Perhaps this is the reason I can't square the
expression?>Well, probably for a similar reason. One approach
to programming the^ operation on a calculator is to take a
logarithm, then do amultiplication, then do an antilogarithm.
I vaguely remember onecalculator I have owned (or it may have
been an early implementationof BASIC) refuse to do things
like (-2)^3, and I assumed that thereason is something to do
with the non-existence (in reals) o?). In complex
numbers the log function is multi-valued. In bothcases,
negative reals and complex numbers, it is possible to
workaround the problem; my current calculator is perfectly
happy with(-2)^3.I don't know the exact algorithms used on
calculators. But they willhave been optimised for real
arithmetic. When programming the complexnumber operations,
the designers will have had to decide when to adaptthe
routines to allow operations with complex numbers, and when
not tobother. It seems that Casio decided that general
exponentiation ofcomplex numbers was not worth the
bother.
===
>Arrrghh.... this calculator is really annoying
me.>I've put my fx-991MS calculator into CMPLX mode, and I
can get it to>calculate and show me, for example, the real
and imaginary parts of1 ->3i>(simple I know but it de?itely
works). However I can't get it toshow>me>the real and
imaginary parts of, for example, 1 - sqrt3i. It
saysMath>ERROR. This also happens with *any* other complex
numbers calculation>involving roots or powers.> If you mean
(sqrt(3))*i, then you can almost certainly overcome the>
problem by putting in some parentheses.>I still can't raise a
complex expression to a power though e.g. (1 +3i)^3.>The only
exception is squaring it - the calulator has a separate
button>speci?ally for squaring i.e. I don't have to put in
the ^ then the3>(or any power), I simple press the square
button. Any idea on the correct>parenthesis for this?I've
tried (1 + 3i)^3, (1 + 3i)(^3), ((1 + 3i)(^3)) - none seem to
work.> If you mean sqrt(3i) then you are up against a
restriction on the> operations allowed on your calculator.
(Note that sqrt is> multi-valued.)Perhaps this is the reason
I can't square the expression?> Well, probably for a similar
reason. One approach to programming the> ^ operation on a
calculator is to take a logarithm, then do a> multiplication,
then do an antilogarithm. I vaguely remember one> calculator I
have owned (or it may have been an early implementation> of
BASIC) refuse to do things like (-2)^3, and I assumed that
the> reason is something to do with the non-existence (in
reals) of> log(-2). In complex numbers the log function is
multi-valued. In both> cases, negative reals and complex
numbers, it is possible to work> around the problem; my
current calculator is perfectly happy with> (-2)^3. I don't
know the exact algorithms used on calculators. But they will>
have been optimised for real arithmetic. When programming the
complex> number operations, the designers will have had to
decide when to adapt> the routines to allow operations with
complex numbers, and when not to> bother. It seems that Casio
decided that general exponentiation of> complex numbers was
not worth the bother.I've got a Casio CFX-9850G, and the
manual speci?aly states that complexoperation is limited to
simple addition, subtraction, multiplication anddivision,
along with the square, square root and reciprocal
functions.Complex-speci? functions for the modulus, argument
and conjugate areprovided, so all other functions (powers,
logs, trig) can be implementedusing these.--Paul V. S.
TownsendInterchange the alphabetic elements to reply
===
>
let I=k[x_1,x_2,..,x_n] is polinomial ring over ?ld of
char=0 and> I_n - subspase gomogenius polinomial of power n.
Let sl_2 - 3 -> dimesional simple lie algebra wich act at I_n
in usual way. How ?d a> irreducible components of
decomposition of this representation? Need> ?d something
like as formulae of (Klebsh-Gordon)for tensor product.>
Try ?ding the highest weight elements, counting dimensions
(?d its> character) and semi-simplicity.At present I ?d as
follows . Let f_i - standart (i+1) -dimensionrepresentation
of sl_2. The decomposition for I_2
isI_2=f_2n+f_(2n-4)+f_(2n-8)+.....I am looking for
generalisation of this for arbitrary I_k
===
> let
I=k[x_1,x_2,..,x_n] is polinomial ring over ?ld of char=0
and> I_n - subspase gomogenius polinomial of power n. Let
sl_2 - 3 -> dimesional simple lie algebra wich act at I_n in
usual way. How ?d a> irreducible components of decomposition
of this representation? Need> ?d something like as formulae
of (Klebsh-Gordon)for tensor product.> Try ?ding the
highest weight elements, counting dimensions (?d its>
character) and semi-simplicity.> At present I ?d as
follows . Let f_i - standart (i+1) -dimension> representation
of sl_2. The decomposition for I_2 is>
I_2=f_2n+f_(2n-4)+f_(2n-8)+.....> I am looking for
generalisation of this for arbitrary I_kSomething struck me
about this: what do you mean act on the degree nhomogeneous
polynomials in the usual way? I know the usual way for
2variables. Incidentally, you're using n twice in the same
statement fordifferent things (I think).
===
>
http://mathworld.wolfram.com/Determinant.html gives the
distribution of |det| if the elements of the matrix are
withinthe> unit disk.> I can't tell of this distribution
depends on the matrix dimension,Of course it does, due to the
mentioned upper bound...> but what> is known about it? What
would be the limit when n increases?That question remains
:)
===
> http://mathworld.wolfram.com/Determinant.html>
gives the distribution of |det| if the elements of the matrix
are within the> unit disk.> I can't tell of this distribution
depends on the matrix dimension, but what> is known about it?
What would be the limit when n increases?> My reading of
this result is that is is an upper bound, not the
distribution.
===
http://mathworld.wolfram.com/Determinant.html gives the
distribution of |det| if the elements of the matrix are
withinthe> unit disk.> I can't tell of this distribution
depends on the matrix dimension, butwhat> is known about it?
What would be the limit when n increases?> My reading of this
result is that is is an upper bound, not the> distribution.The
plots above show the distribution of determinants for random
nxncomplex matrices with entries satisfying |aij|<1 for n =
2, 3, and 4.Try reading more than the ?st line beneath a
plot. How could you plot theupper bound of an 2x2 random
matrix and get a non-constant functionanyway....
===
Find the
values of cos(pi/6 + 3i)Am I correct in making this cos(pi/6)
+ cosh(3)? Probably not because theanswer I've been given is
8.7189 - 5.0089i. So if not, what do I do to getthe
answer?
===
Use the standard formula for the cosine of a
sum:cos (x+y) = cos x cos x - sin x sin yOK now suppose y is
made pure imaginary as iy; then:cos (x+iy) = cos x cos iy -
sin x sin iy.Now the sines and cosines of pure imaginary
numbers are related toHYPERBOLIC functions of the
correspondin real numbers; cos iy = cosh yand sin i = i sinh
y. Plug that into the above and you should get whatyou are
looking for.--OL
===
>Find the values of cos(pi/6 + 3i)Am I
correct in making this cos(pi/6) + cosh(3)? Probably not
because the>answer I've been given is 8.7189 - 5.0089i. So if
not, what do I do to get>the answer?remember that cos(a+b) =
cos(a)cos(b) - sin(a)sin(b)and that cos(ix) = cosh(x) and
sin(ix) = i sinh(x)Rob Johnson <rob@trash.whim.org>take out
the trash before replying
===
> Find the values of cos(pi/6 +
3i)> Am I correct in making this cos(pi/6) + cosh(3)?
Probably not because the> answer I've been given is 8.7189 -
5.0089i. So if not, what do I do to get> the answer?> No,
that's not correct.cos(pi/6 + 3i) = cos(pi/6)cos(3i) -
sin(pi/6)sin(3i)cos(3i) = cosh(3)sin(3i) = i*sinh(3)therefor
cos(pi/6 + 3i) = cos(pi/6)cosh(3) -
i*sin(pi/6)sinh(3)
===
>Find the values of cos(pi/6 + 3i)Am I
correct in making this cos(pi/6) + cosh(3)? No.>Probably not
because the>answer I've been given is 8.7189 - 5.0089i. So if
not, what do I do to get>the answer?Use the de?ition? (cos(z)
= (exp(iz) + exp(-iz))/2.)************************David C.
Ullrich
===
L is a langage containing only one symbol of
binary relation R.T is the theory of equivalence relation
with in?ite many in?ite classesexpressed in L.M and N are
two models of T and M is a sub-structure of N. It has to
beproved that N is an elementary extension of M, with 3
steps:- Every equivalence class of N is represented in M.-
All the equivalence classes of N have the same cardinal.-
General case.Any help appreciated !
===
You know,
_crossposting_ is a better idea than justposting the same
message to two different groups.With a crosspost people
reading one group cansee what people have said about it
elsewhere...sci.math, sci.logic.)>L is a langage containing
only one symbol of binary relation R.T is the theory of
equivalence relation with in?ite many in?ite
classes>expressed in L.M and N are two models of T and M is a
sub-structure of N. It has to be>proved that N is an
elementary extension of M, with 3 steps:You must be leaviing
out part of the problem(?)>- Every equivalence class of N is
represented in M.because it seems clear that this doesn't
follow>- All the equivalence classes of N have the same
cardinal.and while I suppose I could be overlooking something
there,it's _very_ clear that _this_ doesn't follow.>- 
General
case.Any help appreciated !************************David C.
Ullrich
===
Each speci? point doesn't follow the other ones.
They are just easier tobe solve than the general case. May be
my word STEP is wrong, CASE would bebetter.Jean-Pierre.David
C. Ullrich <ullrich@math.okstate.edu> a .8ecrit dans le
message de> You know, _crossposting_ is a better idea than
just> posting the same message to two different groups.> With
a crosspost people reading one group can> see what people have
said about it elsewhere... sci.math, sci.logic.)>L is a
langage containing only one symbol of binary relation R.T is
the theory of equivalence relation with in?ite many
in?iteclasses>expressed in L.M and N are two models of T and
M is a sub-structure of N. It has to be>proved that N is an
elementary extension of M, with 3 steps: You must be leaviing
out part of the problem(?)- Every equivalence class of N is
represented in M. because it seems clear that this doesn't
follow- All the equivalence classes of N have the same
cardinal. and while I suppose I could be overlooking
something there,> it's _very_ clear that _this_ 
doesn't
follow.- General case.Any help appreciated !
************************ David C. Ullrich
===
>Each speci?
point doesn't follow the other ones. They are just easier
to>be solve than the general case. May be my word STEP is
wrong, CASE would be>better.Oh. Yes, the word case would have
been clearer.>Jean-Pierre.David C. Ullrich
<ullrich@math.okstate.edu> a .8ecrit dans le message de> You
know, _crossposting_ is a better idea than just> posting the
same message to two different groups.> With a crosspost
people reading one group can> see what people have said
about it elsewhere...> sci.math, sci.logic.)>L is a langage
containing only one symbol of binary relation R.>T is the
theory of equivalence relation with in?ite many
in?ite>classes>expressed in L.>M and N are two models of T
and M is a sub-structure of N. It has to be>proved that N is
an elementary extension of M, with 3 steps:> You must be
leaviing out part of the problem(?)>- Every equivalence
class of N is represented in M.> because it seems clear that
this doesn't follow>- All the equivalence classes of N have
the same cardinal.> and while I suppose I could be
overlooking something there,> it's _very_ clear that _this_
doesn't follow.>- General case.>Any help appreciated !>
************************> David C.
Ullrich>************************David C. Ullrich
===
KRamsay
schrieb:>Interesting how the center column progress by
12 and the two columns>on either side progress by 3. >
I'm unable really to read Perron's book on continued
fractions> because I don't know any German, but Perron
appears to make> a study of the numbers whose continued
fraction is sort of close> to periodic in this way:
[a1,...,an, b1,...,bn, b1+c1, b2+c2,> b3+c3,..., bn+cn,
b1+2c1, b1+2c2,..., bn+2cn, ...], with each> element of the
cycle increasing in an arithmetic progression.> Gosper has
explained how to compute with continued fraction> numbers,
and you can calculate the expansion of multiples> of e and so
on easily enough, and you'll ?d a bunch of> related numbers
which have the same type of continued fraction.> See for
example>
http://www.inwap.com/pdp10/hbaker/hakmem/cf.html#item101b>
Keith Ramsay> If you allow negative coef?ients, you seem to
get a pretty general schemefor 1/k e for k= ...
-3,-2,-1,{0},1,2,3,...
- 1 3 5 7
9
11--
-cf(e^(1/-2)): [1,-3, 1, 1, -7,
1, 1,-11, 1, 1,-15, 1, 1,-19, 1, 1,-23, 1 ]cf(e^(1/-1)):
[1,-2, 1, 1, -4, 1, 1, -6, 1, 1, -8, 1, 1,-10, 1, 1,-12, 1
]cf(e^(1/ 0)): [1,-1, 1, 1, -1, 1, 1, -1, 1, 1, -1, 1, 1, -1,
1, 1, -1, 1 ] divergent (oscillates on 0 and 1)cf(e^(1/ 1)):
[1, 0, 1, 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, 1, 1, 10, 1
]cf(e^(1/ 2)): [1, 1, 1, 1, 5, 1, 1, 9, 1, 1, 13, 1, 1, 17,
1, 1, 21, 1 ]cf(e^(1/ 3)): [1, 2, 1, 1, 8, 1, 1, 14, 1, 1,
20, 1, 1, 26, 1, 1, 32, 1 ]cf(e^(1/ 4)): [1, 3, 1, 1, 11, 1,
1, 19, 1, 1, 27, 1, 1, 35, 1, 1, 43, 1 ]cf(e^(1/ 5)): [1, 4,
1, 1, 14, 1, 1, 24, 1, 1, 34, 1, 1, 44, 1, 1, 54, 1 ]cf(e^(1/
6)): [1, 5, 1, 1, 17, 1, 1, 29, 1, 1, 41, 1, 1, 53, 1, 1, 65,
1
]
+ 1 3 5 7 9
11--
-for 2/k e for k= ...
-3,-2,-1,{0},1,2,3,...
-
-delta: - 1 12 5 7 36 11 13 60 17 19 84 23
25--
cf(e^(2/-1)); [1,-1, -6, -3, 1, 1, -4, -18, -6, 1, 1, -7,
-30, -9, 1, 1, -10, -42, -12, 1, 1,-23 ]cf(e^(2/1)); [1, 0,
6, 2, 1, 1, 3, 18, 5, 1, 1, 6, 30, 8, 1, 1, 9, 42, 11, 1, 1,
12 ]cf(e^(2/3)); [1, 1, 18, 7, 1, 1, 10, 54, 16, 1, 1, 19,
90, 25, 1, 1, 28, 126, 34, 1, 1, 37 ]cf(e^(2/5)); [1, 2, 30,
12, 1, 1, 17, 90, 27, 1, 1, 32, 150, 42, 1, 1, 47, 210, 57,
1, 1, 62 ]cf(e^(2/7)); [1, 3, 42, 17, 1, 1, 24, 126, 38, 1,
1, 45, 210, 59, 1, 1, 66, 294, 80, 1, 1, 87 ]cf(e^(2/9)); [1,
4, 54, 22, 1, 1, 31, 162, 49, 1, 1, 58, 270, 76, 1, 1, 85,
378, 103, 1, 1, 112 ]cf(e^(2/11)); [1, 5, 66, 27, 1, 1, 38,
198, 60, 1, 1, 71, 330, 93, 1, 1, 104, 462, 126, 1, 1, 95
]
--
delta: + 1 12 5 7 36 11 13 60 17 19 84 23
25--
Also, allowing fractions for coef?ients, the primary
expansion of e = e^(1/1) = e^(2/2) can be inserted in
theprevious
table:-
-
...cf(e^(2/1)): [1, 0, 6, 2, 1, 1, 3, 18, 5, 1, 1, 6, 30,
8, 1, 1, 9, 42, 11, 1 ...cf(e^(2/2)): [1, 0.5, 12, 4.5, 1, 1,
6.5, 36, 10.5, 1, 1, 12.5, 60, 16.5, 1, 1, 18.5, 84, 22.5, 1
...cf(e^(2/3)): [1, 1, 18, 7, 1, 1, 10, 54, 16, 1, 1, 19, 90,
25, 1, 1, 28, 126, 34, 1
......-
-
delta + 0.5 6 2.5 3.5 18 5.5 6.5 60 8.5 9.5 42
11.5
-
-Perhaps this allowing of negative and/or fractional
coef?ients enables also to ?dmore simple regularities for
e^k with
abs(k)>2
===
=====================================
===
=========With the golden-ratio (phi) you can have also
some regularities:Golden-Ratiox = phi ~
1.61803398874989484820...
-?? cf(x^-2)?? cf(x^-1); [ -1, -1,
??????-
--cf(x^1); [ 1, 1, 1, 1, 1, 1, 1, 1, 1 ...cf(x^3); [ 4, 4,
4, 4, 4, 4, 4, 4, 4 ... = a1*3-a_(-1)cf(x^5); [ 11, 11, 11,
11, 11, 11, 11, 11, 11 ... = a3*3-a1cf(x^7); [ 29, 29, 29,
29, 29, 29, 29, 29, 29 ... = a5*3-a3cf(x^9); [ 76, 76, 76,
76, 76, 76, 76, 76, 76 ... = a7*3-a5cf(x^11); [199, 199, 199,
199, 199, 199, 199, 199 ...cf(x^0): [ 1, 0, 0, 0, 0, 0, 0, 0,
0,cf(x^2); [ 2, 1, 1, 1, 1, 1, 1, 1, 1 ... = a0 + a1cf(x^4);
[ 6, 1, 5, 1, 5, 1, 5, 1, 5 ... = a2 + a3cf(x^6); [ 17, 1,
16, 1, 16, 1, 16, 1, 16 ... = a4 + a5cf(x^8); [ 46, 1, 45, 1,
45, 1, 45, 1, 45 ... = a6 +
a7--
===
=======================================
===
=======Maybe one can formulate a regular proof and
systematize this to a greater extend.Gottfried Helms
===
I may
append, that from the e^2/k -table also the e^3/-cfs can
bederived, which seem patternless in their
simple-c-f-representation.When negtive integers and rational
*fractions* as coef?ients ofsimple cfs allowed, seemingly
all rational powers of e can be expressedcompletely
schematically:for 2/k e for k= ...
-3,-2,-1,{0},1,2,3,...
-
-delta: - 1 12 5 7 36 11 13 60 17 19 84 23
25--
cf(e^(2/-1)); [1,-1, -6, -3, 1, 1, -4, -18, -6, 1, 1, -7,
-30, -9, 1, 1, -10, -42, -12, 1, 1,-23 ]cf(e^(2/1)); [1, 0,
6, 2, 1, 1, 3, 18, 5, 1, 1, 6, 30, 8, 1, 1, 9, 42, 11, 1, 1,
12 ]cf(e^(2/3)); [1, 1, 18, 7, 1, 1, 10, 54, 16, 1, 1, 19,
90, 25, 1, 1, 28, 126, 34, 1, 1, 37 ]cf(e^(2/5)); [1, 2, 30,
12, 1, 1, 17, 90, 27, 1, 1, 32, 150, 42, 1, 1, 47, 210, 57,
1, 1, 62 ]cf(e^(2/7)); [1, 3, 42, 17, 1, 1, 24, 126, 38, 1,
1, 45, 210, 59, 1, 1, 66, 294, 80, 1, 1, 87 ]cf(e^(2/9)); [1,
4, 54, 22, 1, 1, 31, 162, 49, 1, 1, 58, 270, 76, 1, 1, 85,
378, 103, 1, 1, 112 ]cf(e^(2/11)); [1, 5, 66, 27, 1, 1, 38,
198, 60, 1, 1, 71, 330, 93, 1, 1, 104, 462, 126, 1, 1, 95
]
--
delta: + 1 12 5 7 36 11 13 60 17 19 84 23
25--
It seems possible to get other rational powers of e simply by
interpolation.For instance, for e^3, which has a seemingly
random pattern in terms of a simplecontinued fraction, a
schematic patter can be found, if we allow negative as wellas
rational coef?ients.First interpolate for the zero-cf, which
diverges, if we try to approximate it.Let's use the symbol #+
for elementwise addition of the coef?ients-lists,
andanaloguously #-,#*,#/ for the appropriate other operation,
then cf(e^(2/0)) = [ cf(e^(2/-1)) #+ cf(e^(2/1)) ] #/ 2Insert
this into the table between the rows of cf(e^(2/-1)) and
cf(e^(2/1))--
-
--...cf(e^(2/-1)); [1,-1, -6, -3, 1, 1, -4, -18, -6, 1,
1, -7, -30, -9, 1, 1, -10, -42, -12, 1 ...cf(e^(2/0)):
[1,-0.5, 0,-0.5, 1, 1,-0.5, 0,-0.5, 1, 1,-0.5, 0,-0.5, 1,
1,-0.5, 0,-0.5, 1 ...cf(e^(2/1)): [1, 0, 6, 2, 1, 1, 3, 18,
5, 1, 1, 6, 30, 8, 1, 1, 9, 42, 11, 1
......-
-
Also let us shorten the cf(...) notation to a c(..)
notation c(-1) = cf(e^(2/-1)) c( 0) = cf(e^(2/ 0)) c( 1) =
cf(e^(2/ 1)) ...and d = c(1) #- c(0)so that d = [0, 0.5, 6,
2.5, 0, 0, 3.5, 18, 5.5, 0, 0, 6.5, 30, 8.5, 0, 0, 0.5, 42,
11.5, 0 ...Then cf(e^3) can simply be interpolated as 2
cf(e^3) = c(2/3) = c(0) #+ d #* 3Thencf(e^3) =
cf(e^(2/(2/3)))= 1 7 11 19 23 31 35 43 =c(2/3) =[1, - -, 4,
-, 1, 1, --, 12, --, 1, 1, --, 20, --, 1, 1, --, 28, --, 1] 6
6 6 6 6 6 6
6
===
==========================================
===
=================And the continued-fraction for any
power of e can be expressed along the schematic ofthat of e^2
cf(e^p) = [ a[5i+0], a[5i+1], a[5i+2], a[5i+3],a[5i+4], {for
i=0..oo} ] where a[5i+0] = 1 a[5i+1] = -0.5 + ( 1 + 6*i)/p
a[5i+2] = 0 + (12 +24*i)/p a[5i+3] = -0.5 + ( 5 + 6*i)/p
a[5i+4] =
1
===
==========================================
===
================ -i=0-
--i=1-- i=2
...cf(e^2)=c(2/2): [1, 0, 6 , 2, 1, 1, 3, 18, 5, 1, 1, 6, 30,
8, 1, ...cf(e^3)=c(2/3): [1,-1/6, 4 , 7/6, 1, 1,11/6, 12 ,
19/6, 1, 1,23/6, 20,31/6, 1, ...cf(e^4)=c(2/4): [1,-2/8, 3 ,
6/8, 1, 1,10/8, 9 , 18/8, 1, 1,22/8, 15,30/8, 1,
...cf(e^5)=c(2/5): [1,-3/10,24/10, 5/10, 1, 1, 9/10,72/10,
17/10, 1, 1,21/10, 12,29/10, 1, ...or even more concise:
-i=0- --i=1
--i=2-- ......cf(e^2)=c(2/2): [ 4, 0, 24 , 8 4,
4, 12 , 72, 20, 1, 4, 24, 120, 32, 4, ... ] #/
4cf(e^3)=c(2/3): [ 6, -1, 24 , 7, 6, 6, 11 , 72, 19, 1, 6,
23, 120, 31, 6, ... ] #/ 6cf(e^4)=c(2/4): [ 8, -2, 24 , 6, 8,
8, 10 , 72, 18, 1, 8, 22, 120, 30, 8, ... ] #/
8cf(e^5)=c(2/5): [10, -3, 24 , 5, 10, 10, 9 , 72, 17, 1,
10,21, 120, 29, 10, ... ] #/10 ...One can see the 5
interleaved arithmetic progressions a+d,a+2d, ... ;Eric
Weisstein at mathworld.wolfram.com mentions some studies on
such progressions
incontinued-fractions-coef?ients.
===
==================
===
==========================================
This ist just by trial&error; a proof would be good...Also I
guess, that this interpolation-method could be applicableto
the more simple representations of cf-s of e^1/p
analoguously.(But I may leave to check that out to another
reader's fun)Gottfried Helms(for computations the program
Maxima (GNU) was used )
===
Gottfried Helms schrieb:> Also I
guess, that this interpolation-method could be applicable> to
the more simple representations of cf-s of e^1/p
analoguously.> (But I may leave to check that out to another
reader's fun)> Well, it was just too simple.So here it
is:with the #/-operator meaning to divide each element of the
list by the divisorcf(e^1) = [1 0 1 1 2 1 1 4 1 1 6 1 1 8 1 1
10 1 ] #/1cf(e^2) = [2 - 1 2 2 1 2 2 3 2 2 5 2 2 7 2 2 9 2 ]
#/2cf(e^3) = [3 - 2 3 3 0 3 3 2 3 3 4 3 3 6 3 3 8 3 ]
#/3cf(e^4) = [4, - 3, 4, 4, - 1, 4, 4, 1, 4, 4, 3, 4, 4, 5,
4, 4, 7, 4 ] #/4cf(e^5) = [5, - 4, 5, 5, - 2, 5, 5, 0, 5, 5,
2, 5, 5, 4, 5, 5, 6, 5 ] #/5...Since these are not the common
simple cf's, their convergence is not optimal;but 
that's
ignored here for the sake of getting a simple
pattern.Gottfried Helms
===
Well, a last addendum:more general
relation: cf(e^x) = [x, 1-x, x, x, 3-x, x x, 5-x, x ... ] #/x
cf(e^(1/x) = [1, x-1, 1 1, 3x-1, 1, 1, 5x-1, 1, ...]Putting
that together, I now assume that cf(e^x) = [1, 1/x-1, 1,
1,3/x-1, 1 1,5/x-1, 1 ... ] for all rational x except x=0and
I think, this is worth to be derived algebraically and to
beproven, if this is not done already.What is especially
interesting, is that a multiplication overall coef?ents of a
continued fraction is an operation, whichnormally modi?s the
?al value in a very complicated way.That with the
euler-constant e such an operation comes outwith such a
regularity is especially astonishing to me.Anyway- now as it
is compressed so much, I think, I've seen thisanywhere? Does
anyone know of any online-reference for that
relation?Gottfried Helms
===
I'd like to get some opinions
about this problem, please.The dependable capacity C
available in an electric system in a monthof the future, say
Dec 2008, is a random variable with densityfunction f de?ed
on [0, Cmax]. We can admit f is continuous on thisinterval.
On the cited month you have to meet a known anddeterministic
demand r<=Cmax. De?e a random variable de?it, D, byD= r-C
if C<r and D=0 if C>=r. Supposing the density function f
isindependent of r, the expected value of D corresponding to
a demand ris given by E(r) = Integral [0,r] (r-c) f(c) dc =
r*Integral [0,r]f(c) dc - Integral [0,r] c f(c) dc. Since f
is continuous, E(r)exists for r<=Cmax and E is differentiable
wrt r. Applying the F.T. OfInt. Calculus, we get E'(r) = r*
f(r) + Integral [0,r] f(c) dc -r*f(r) = Integral [0,r] f(c)
dc = Probability (C<=r) = Probability(D>=0). Though I don't
have f in a closed form, I can estimateProbability (D>=0) by
means of simulation models, using a processsimilar to Monte
Carlo's. Then, for variations on r of about 5%, I canmake the
estimate Delta E(r) = (Delta r) * Probability (D>=0). All
Ineed to know is that f exists and is continuous on [0,
Cmax]. I don'tneed to know how exactly f sends c into f(c).In
some situations this is reasonable, but there are cases when
it'snot admittable at all to suppose f is independent of r.
In such cases,for each r there's a particular density
function f_r. You can stillsuppose r is known, but it affects
the distribuition of C on [0, Cmax](Cmax is always known and
independent of r]. Then I think I havesomething like E(r) =
Integral [0,r] (r-c) g(r,c) dc where g isde?ed on R^2 with
values on R and, for a given r, g(r,c) = f_r(c).Supposing g
is continuous, is that OK if I use Leibiniz formula tocompute
E'(r)? Anyway, if f_r depends on r than that
beautifulconclusion I came to before is no longer true,
right? I'm a bitconfused here.Artur
===
>The dependable
capacity C available in an electric system in a month>of the
future, say Dec 2008, is a random variable with
density>function f de?ed on [0, Cmax]. We can admit f is
continuous on this>interval. On the cited month you have to
meet a known and>deterministic demand r<=Cmax. De?e a random
variable de?it, D, by>D= r-C if C<r and D=0 if C>=r.
Supposing the density function f is>independent of r, the
expected value of D corresponding to a demand r>is given by
E(r) = Integral [0,r] (r-c) f(c) dc = r*Integral [0,r]>f(c)
dc - Integral [0,r] c f(c) dc. Since f is continuous,
E(r)>exists for r<=Cmax and E is differentiable wrt r.
Applying the F.T. Of>Int. Calculus, we get E'(r) = r* f(r) +
Integral [0,r] f(c) dc ->r*f(r) = Integral [0,r] f(c) dc =
Probability (C<=r) = Probability>(D>=0). Though I don't have
f in a closed form, I can estimate>Probability (D>=0) by
means of simulation models, using a process>similar to Monte
Carlo's. Then, for variations on r of about 5%, I can>make
the estimate Delta E(r) = (Delta r) * Probability (D>=0). All
I>need to know is that f exists and is continuous on [0,
Cmax]. I don't>need to know how exactly f sends c into
f(c).>In some situations this is reasonable, but there are
cases when it's>not admittable at all to suppose f is
independent of r. In such cases,>for each r there's a
particular density function f_r. You can still>suppose r is
known, but it affects the distribuition of C on [0,
Cmax]>(Cmax is always known and independent of r]. Then I
think I have>something like E(r) = Integral [0,r] (r-c)
g(r,c) dc where g is>de?ed on R^2 with values on R and, for
a given r, g(r,c) = f_r(c).>Supposing g is continuous, is
that OK if I use Leibiniz formula to>compute E'(r)? Anyway,
if f_r depends on r than that beautiful>conclusion I came to
before is no longer true, right?>Your formulae seem correct.
As long as the partial derivative dg/dr is continuous in some
neigborhood [r0 - e, r0 + e] x [0, r0], you are allowed to use
Leibniz' rule to evaluate the derivative E'(r0). You 
are
correct that the derivative is not as pretty as in your ?st
case.-- Stephen J. Herschkorn
herschko@rutcor.rutgers.edu
===
I've started a thread to go
over some statements by Dik Winter which Isay are crank
statements. I'm also going to outline some crankbehavior by
that person.I don't mind others posting in the thread or in
this one, and I mayreply to people other than Dik Winter, but
I want you to know wherethe focus is.Some of you may know that
I have independent veri?ation of theargument that he attacks,
but I've been puzzled both by hispersistence in making his
claims against those argument, and in theacceptance of his
claims by the sci.math newsgroup.So I'm doing an experiment.
My guess is that despite hearing thatthere's independent
veri?ation of the argument Winter attacks, anddespite the
wackiness of his position the sci.math newsgroup willSTILL
either show support for Winter or fail to correct him.That's
the hypothesis that I'm currently testing.James Harris
===
Some of you may know that I have independent veri?ation of
the> argument that he attacks...>If you don't want to divulge
this independent source, that's ?e. But atleast post this
person's veri?ation. As you say, let the math speak
foritself.l8r, Mike N. Christoff
===
So I'm doing an
experiment. My guess is that despite hearing that> there's
independent veri?ation of the argument Winter attacks, and>
despite the wackiness of his position the sci.math newsgroup
will> STILL either show support for Winter or fail to correct
him.Mr. Harris,great strategy!You have asked Arturo to no
longer comment on your ?math arguments.Dik Winter has
also repeatedly showed you just how ?your
nonsensicaldribble is and now you want him gone too?You have
berated others too like Nora who has also shown the
uncorrectable?n your so-called arguments.Do you believe
that shaming these folks to not replying will make
yourarguments any more acceptable?You should have chosen
comedy as your ?ld of study!What a joke you are. To me, you
have proven that without question!
===
>Mr. Harris,great
strategy!You have asked Arturo to no longer comment on your
?math arguments.I would suggest that asked is not the
correct verb to use to describewhat
transpired.http://groups.google.com/groups?selm=atkubk%24bml%
241%40agate.berkeley.eduhttp://groups.google.com/groups?selm=
atrhku%242cpk%241%40agate.berkeley.edu(towards the end in
both)http://groups.google.com/groups?selm=bo8tqb%241p9o%241%
40agate.berkeley.eduhttp://groups.google.com/groups?selm=
3c65f87.0311041814.5828faa8%40posting.google.com--
===
===========================================
===
====It's not denial. I'm just very selective about 
what
I accept as reality. Calvin (Calvin and
Hobbes)
===
======================================
===
=========Arturo Magidinmagidin@math.berkeley.edu=== >
Some of you may know that I have independent veri?ation of
the> argument that he attacksHow are we supposed to have
gotten this knowledge. This is the ?st mention of the
existence of independent veri?ation made on thenewsgroup. I
for one do not believe that independent veri?ation exists.I
suspect that what we have her is a de?ition
problem.independent veri?ation The rest of the world: A
disinterested third party has veri?d this. James Harris: I
thought up yet another argument.Question: How many legs does
a horse have if James Harris calls a tail a leg.Answer: Four.
The fact that James Harris calls a tail a leg does not make it
one. - William Hughes
===
> So I'm doing an experiment. My
guess is that despite hearing that> there's independent
veri?ation of the argument Winter attacks, and> despite the
wackiness of his position the sci.math newsgroup will> STILL
either show support for Winter or fail to correct him.But we
*haven't* heard there's independent veri?ation, have 
we?
Atleast, we have no means of con?ming or refuting that
someonesomewhere has independently veri?d your claim
(assuming thatsomeone here is certain what, precisely, the
claim is).You've said that there is, but you have given no
means to con?m ordeny this claim, so that hardly counts as
any particular evidence thatthere's been independent
veri?ation.-- Jesse F. HughesI think the burden is on those
people who think he didn't haveweapons of mass destruction to
tell the world where they are. -- White House spokesman Ari
Fleischer
===
>I've started a thread to go over some
statements by Dik Winter which I>say are crank statements.
I'm also going to outline some crank>behavior by that
person.I don't mind others posting in the thread or in this
one, and I may>reply to people other than Dik Winter, but I
want you to know where>the focus is.Some of you may know that
I have independent veri?ation of the>argument that he
attacks, but I've been puzzled both by his>persistence in
making his claims against those argument, and in
the>acceptance of his claims by the sci.math newsgroup.So I'm
doing an experiment. My guess is that despite hearing
that>there's independent veri?ation of the argument Winter
attacks, and>despite the wackiness of his position the
sci.math newsgroup will>STILL either show support for Winter
or fail to correct him.That's the hypothesis that I'm
currently testing.Um, we've only heard about this independent
veri?ation from_you_. GIven your record, nobody is going to
believe anythingyou say just because you say it. Sorry,
that's how that works.Exactly who did this independent
veri?ation?>James Harris************************David C.
Ullrich
===
[snip]> Some of you may know that I have
independent veri?ation of the> argument that he
attacks,Well, *I* haven't seen it posted in this newsgroup,
and your word is noword. Where is the independent veri?ation
of the argument? Who hasveri?d it? Your assertions are
justi?bly suspect, since you rarelypost a single paragraph
which does not contain distortions or lies.> but I've been
puzzled both by his> persistence in making his claims against
those argument, and in the> acceptance of his claims by the
sci.math newsgroup.argument is faulty.> So I'm doing an
experiment.This is ?sci.math', not
?alt.test.your.crackpot.theories'.> My guess is that despite
hearing that> there's independent veri?ation of the argument
Winter attacks, and> despite the wackiness of his position the
sci.math newsgroup will> STILL either show support for Winter
or fail to correct him.Your assurance that independent
veri?ation exists is no assurance atall. You repeatedly
misrepresent or distort everything you post. Yourcredibility
is zero. Why? Because you have consistently defended
yourerrors with the same vigor and passion that you exhibited
when you laterdefended their corrections. You have no
demonstrated ability todistinguish truth from error.> That's
the hypothesis that I'm currently testing.The result is just
in: YOU ARE A CRANK!Wacky, isn't it? But, hey, it's 
just
basic math. Yup, yup, yup.--There are two things you must
never attempt to prove: the unprovable --and the
obvious.--Democracy: The triumph of popularity over
principle.--http://www.crbond.com
===
> [snip]> Some of
you may know that I have independent veri?ation of the>
argument that he attacks,> Well, *I* haven't seen it posted
in this newsgroup, and your word is no> word. Where is the
independent veri?ation of the argument? Who has> veri?d it?
Your assertions are justi?bly suspect, since you rarely> post
a single paragraph which does not contain distortions or
lies.I can tell you what he did: He printed all his arguments
and put the pile of paper on one plate. Then he took a sausage
and put it on the other plate. Then he told his dog: On one
plate there is a printout of a proof of FLT. On the other
plate there is a sausage. Go and eat the sausage. And the dog
ate the sausage, which clearly demonstrates that on the plate
that it didn't touch there was no sausage, but a proof of
FLT.
===
> I can tell you what he did: He printed all his
arguments and put the> pile of paper on one plate. Then he
took a sausage and put it on the> other plate. Then he told
his dog: On one plate there is a printout of> a proof of FLT.
On the other plate there is a sausage. Go and eat the>
sausage. And the dog ate the sausage, which clearly
demonstrates that> on the plate that it didn't touch there
was no sausage, but a proof of> FLT.I think the dog then
crapped on the plate because what was on it was way toostinky
for him to even touch!
===
> I've started a thread to go over
some statements by Dik Winter which I> say are crank
statements. I'm also going to outline some crank> behavior by
that person.> I don't mind others posting in the thread or
in this one, and I may> reply to people other than Dik
Winter, but I want you to know where> the focus is.> Some
of you may know that I have independent veri?ation of the>
argument that he attacks, but I've been puzzled both by his>
persistence in making his claims against those argument, and
in the> acceptance of his claims by the sci.math
newsgroup.>By whom? > So I'm doing an experiment. My guess is
that despite hearing that> there's independent veri?ation of
the argument Winter attacks, and> despite the wackiness of
his position the sci.math newsgroup will> STILL either show
support for Winter or fail to correct him.> Who has
independently veri?d your proof? Where is this veri?ation?>
That's the hypothesis that I'm currently testing.OK, 
that's
H_0, what's H_1?> James Harris
===
>Are you saying we
have no balls left at noon *because* pointwise>convergence
holds, or are you saying that pointwise convergence
holds>*because* it describes the fact (independently
veri?ble via ZF) that>no balls are left at noon?>If the
former, then you have not explained *why* pointwise
convergence>holds.>|De?ition:>|Suppose {f_n} is a
sequence of functions with domain D and that f is
a>|function also with domain D. The sequence {f_n} converges
pointwise to>|f iff for each x in D, f_n(x) converges to
f(x).> Let D be the set of buckets, and f_n(x) be 1 or 0
depending on whether> that bucket has a ball in it or not at
time 12 - 2^{1-n}.> Since f_n(x) can take only two values, 1
or 0, the discrete topology is> about the only topology we
can give to the range of f_n. To converge in> the discrete
topology, all terms must be constant after some point.I do
not *assume* that convergence of any type holds. I assume
that>may *deduce* that pointwise convergence holds, but I do
not need that>fact to solve the problem. I need only set
theory.If Newton's First Law implies pointwise convergence
and you assume that,then you assume even more than pointwise
convergence, but that means youdo assume pointwise
convergence since it is implied by Newton's
FirstLaw.>|De?ition:>|Suppose {a_n} is a sequence in a
space with the discrete topology and a>|is another point in
that same space. The sequence {a_n} converges to a>|iff
there is an N so that a_n = a for all n > N.> This is why I
called discrete convergence monotonous. Perhaps, due to> the
constancy of discrete convergence, it may not appear that
there is> any convergence at all, the terms are all just the
same.You keep thinking that all you need to do is explain
convergence more>fully and completely, and all will become
clear. I know what convergence>means.I was not really sure
why our views differed, but I think that yourpost has ?ally
enlightened me. Hopefully I now understand where
ourapproaches to this problem differ. I apologize if I
insulted you byexplaining these two types of convergence; I
was merely trying to makemy point clearer.> Translating the
de?ition of pointwise convergence to the language of>
buckets and balls and using discrete convergence at each
point, we get>|A bucket is empty at noon iff it has always
been empty since some time>|before noon. A bucket has a ball
in it at noon iff it has always had a>|ball in it since some
time before noon.And this follows from Newton's ?st law,
does it not? Convergence is>merely the language you have
chosen to describe what we observe; it is>not in any way a
*cause* of what we observe.Newton's First Law merely
describes what we observe; it is not in anyway a cause of
what we observe either.> This last quoted statement is
something I believe you take for granted,> but others take
it is an assumption. Whatever, it is just a restatement> of
pointwise convergence. This is why I have said that you are
assuming> pointwise convergence. Perhaps I should say you
are taking pointwise> convergence for granted.I am taking
Newton's ?st law and ZF for granted. Pointwise
convergence>is a *consequence* of those assumptions, not an
assumption in its own>right.As I said, if Newton's First Law
implies pointwise convergence and youassume Newton's First
Law, then you are, in fact, assuming pointwiseconvergence.
Under that assumption, as I have said before, I agreewith you
completely.> Your assumption is that if a bucket is always
empty after a certain time> before noon, then that bucket
is empty at noon. This is the assumption> of pointwise
convergence. >Then you must think the assumption of
pointwise convergence is equally>necessary for solving the
?e apples, take away two problem.> No, because there is no
in?ite sequence in the ?e apples, take away> two problem.
Convergence deals with limits of in?ite sequences.You have
?e apples at 11:55. I take away two at 11:59. How many
do>you have at noon?If f(1) = 5 and f(2) = f(1) - 2, what is
f(3)?>We need to assume that Newton's ?st law applies, just
as we do with the>buckets and balls.In math problems,
especially ones that are physically unrealizable, wedo not
necessarily need to assume physical laws such as Newton's
FirstLaw. In order to explain the seemingly contradicatory
answers that havebeen given for this problem, the way I have
been looking at this problemhas no connection with physical
laws; I have been looking at it as asequence of sets with a
particular de?ition, which is essentiallya_n = {k in Z : n+1
<= k <= 10n}. The problem then is to ?d the limitas n tends
to in?ity of a_n. Different topologies give us
differentlimits and therefore different answers.However, if
you feel we do need to assume Newton's First Law, then thatis
where our approaches differ. Under this added constraint, I
agreewith you that the limit is {}.Truce?Rob Johnson
<rob@trash.whim.org>take out the trash before replying
===
>I
do not *assume* that convergence of any type holds. I assume
that>may *deduce* that pointwise convergence holds, but I do
not need that>fact to solve the problem. I need only set
theory.> If Newton's First Law implies pointwise convergence
and you assume that,> then you assume even more than
pointwise convergence, but that means you> do assume
pointwise convergence since it is implied by Newton's First>
Law.If we assume a few basic axioms, are we assuming that FLT
holds, merelybecause FLT happens to be a consequence of those
axioms?> I was not really sure why our views differed, but I
think that your> post has ?ally enlightened me. Hopefully I
now understand where our> approaches to this problem differ.
I apologize if I insulted you by> explaining these two types
of convergence; I was merely trying to make> my point
clearer.No offense taken. I was merely pointing out that the
de?ition ofconvergence is not where our views differ.>
Translating the de?ition of pointwise convergence to the
language of> buckets and balls and using discrete
convergence at each point, we get>|A bucket is empty at
noon iff it has always been empty since some time>|before
noon. A bucket has a ball in it at noon iff it has always had
a>|ball in it since some time before noon.>And this follows
from Newton's ?st law, does it not? Convergence is>merely
the language you have chosen to describe what we observe; it
is>not in any way a *cause* of what we observe.> Newton's
First Law merely describes what we observe; it is not in any>
way a cause of what we observe either.True. But in solving
word problems it is an unstated assumption thatordinary
physical laws apply insofar as that is possible.> This last
quoted statement is something I believe you take for
granted,> but others take it is an assumption. Whatever, it
is just a restatement> of pointwise convergence. This is why
I have said that you are assuming> pointwise convergence.
Perhaps I should say you are taking pointwise> convergence
for granted.>I am taking Newton's ?st law and ZF for
granted. Pointwise convergence>is a *consequence* of those
assumptions, not an assumption in its own>right.> As I said,
if Newton's First Law implies pointwise convergence and you>
assume Newton's First Law, then you are, in fact, assuming
pointwise> convergence. Under that assumption, as I have said
before, I agree> with you completely.I think the important
point here is that assuming Newton's First Lawimplies
pointwise convergence to the exclusion of other kinds
ofconvergence, at least in the context of this problem.
Assuming Newton'sFirst Law allows us to deduce a unique
answer. If I say outright that Iam assuming pointwise
convergence, then you have every right to proposeother forms
of convergence as equally valid approaches to the
problem.> Your assumption is that if a bucket is always
empty after a certain time> before noon, then that bucket
is empty at noon. This is the assumption> of pointwise
convergence. >Then you must think the assumption of
pointwise convergence is equally>necessary for solving the
?e apples, take away two problem.> No, because there is no
in?ite sequence in the ?e apples, take away> two problem.
Convergence deals with limits of in?ite sequences.>You have
?e apples at 11:55. I take away two at 11:59. How many
do>you have at noon?> If f(1) = 5 and f(2) = f(1) - 2, what
is f(3)?Another common assumption regarding word problems is
that there is nohidden information that would affect the
answer. We are not told of anyother movements of apples, and
therefore none exist.> In math problems, especially ones that
are physically unrealizable, we> do not necessarily need to
assume physical laws such as Newton's First> Law. In order to
explain the seemingly contradicatory answers that have> been
given for this problem, the way I have been looking at this
problem> has no connection with physical laws; I have been
looking at it as a> sequence of sets with a particular
de?ition, which is essentially> a_n = {k in Z : n+1 <= k <=
10n}. The problem then is to ?d the limit> as n tends to
in?ity of a_n. Different topologies give us different>
limits and therefore different answers.I have never denied
that these tools exist and that, if applied, they doindeed
produce a variety of answers. I count that as an argument
againstassuming any particular kind of convergence without
some justi?ation.> However, if you feel we do need to assume
Newton's First Law, then that> is where our approaches differ.
Under this added constraint, I agree> with you that the limit
is {}.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia
Abu-Jamal
ruling.<http://www.commoncouragepress.com/index.cfm?action=
book&bookid=228>
===
>Problem 4. Labeled balls (as in #2 and
#3). At step n, remove both >balls labeled n and 10n. Switch
the labels these balls, then return >the ball newly labeled
10n back in the basket. Discard the ball newly >labeled n.
I think JH will insist there are no balls left in the basket
>at noon. But how does this differ from Problem 3?> Since
this version has the same ball being added and removed an
in?ite> number of times, I don't see how to decide where
that particular ball> might be when we ?ish. Rob Johnson
would probably say the function> fails to converge.> That's
an excellent point, Dave. I lost sight of the fact that we>
must look at the actual set element being reinserted,
regardless of> label. Labels are helpful as a convenience for
most of the other> problems given, but in this case following
the labels does not> follow the balls since each ball can
have a different label at> different times.> Although I had
?st claimed that in Problem #4 the container is> empty, given
Dave's argument, I would now say that it is more like> Problem
#1 and indeterminant.Say we change the problem slightly.
Instead of removing the two ballsand then replacing one, we
switch the labels while both balls arestill in the basket,
then remove only the ball with the smaller label. Now we
don't suffer any more from the problem Dave identi?d.Now do
you agree that the question how many balls are in the
basketis identical to problem #3 and the question how many
labels are inthe basket is identical to problem #2? If so,
the only conclusion isthat there are an in?ite number of
balls remaining in the basket,each without a
label.
===
>Problem 4. Labeled balls (as in #2 and #3). At
step n, remove both >balls labeled n and 10n. Switch the
labels these balls, then return >the ball newly labeled
10n back in the basket. Discard the ball newly >labeled
n. I think JH will insist there are no balls left in the
basket >at noon. But how does this differ from Problem
3?>Since this version has the same ball being added
and removed an in?ite>number of times, I don't see how to
decide where that particular ball>might be when we ?ish.
Rob Johnson would probably say the function>fails to
converge.>That's an excellent point, Dave. I lost sight
of the fact that we>must look at the actual set element being
reinserted, regardless of>label. Labels are helpful as a
convenience for most of the other>problems given, but in this
case following the labels does not>follow the balls since each
ball can have a different label at>different times.Although I
had ?st claimed that in Problem #4 the container is>empty,
given Dave's argument, I would now say that it is more
like>Problem #1 and indeterminant.>I do not understand the
basis for this claim. Your objection to Problem 1 was that it
does not specify which ball we remove. In Problem 4, we
specify precisely which ball we remove.-- Stephen J.
Herschkorn herschko@rutcor.rutgers.edu
===
> I agree that
there are no labels, but I don't agree that there are no>
balls. It seems clear to me that any ball with an initial
label not> divisible by 10 will never be removed. So at noon,
there will still> be an in?ite number of (unlabeled) balls
remaining. It's not> contradictory that the balls are labeled
before noon but not after> noon, just as in the original
problem it's not contradictory that the> bucket is non-empty
before noon but not after noon.> Since by the rules of the
game, no ball is reentered without a label. > I think that no
labels = no balls (please, no jokes on this!) is> therefore
valid. Unless you can show me a natural number n such that>
10n is not a natural number, I think the labeless ball
hypothesis> cannot happen.Here are two problems:Problem 1:
You have a basket and a bunch of balls. At time 12-2^-n,you
put ball n in the basket and remove ball n-1. How many balls
arein the basket at 12?Problem 2: You have a ball and bunch
of labels. At time 12-2^n, youput label n on the ball and
remove label n-1. How many labels are onthe ball at 12?I
don't see how the answers could possibly differ, yet you seem
toclaim that the answer to problem 1 is 0 and the answer to
problem 2 is1.> You have a bucket, initially with no coins in
it. For n>0, at time> 1-2^(-n), you put a coin in the bucket
(say coin n), ?l the> coins in the bucket, and remove
all the coins that show heads. What's> the probability
distribution for the number of coins in the bucket at> time
1?> It feels like the bucket should be empty with
probability 1, but I> haven't been able to put together a
convincing argument.> Hmmm...interesting problem. For any
coin n, the probability that the> coin ?ails is 1/2, and
thus the probability that it will never> go to heads is lim
n->oo (1/2)^n = 0. As this is true for each coin,> I would
think that the ending container is empty happen with>
probability 1.I agree that for any given coin, it will be in
the ending containerwith probability 0. But when you add
together an in?ite number ofprobability 0's, you 
don't
necessarily get another probability 0 back.Here's an example:
Choose a random real number x beetween 0 and 1. What's the
probability that x is a real number between 0 and 1? Forany
speci? real number y, the probability that x = y is 0. But
theprobability that it's ANY real number between 0 and 1 is
1.- Nate
===
> You have a bucket, initially with no coins in
it. For n>0, at time> 1-2^(-n), you put a coin in the bucket
(say coin n), ?l the> coins in the bucket, and remove
all the coins that show heads. What's> the probability
distribution for the number of coins in the bucket at> time
1?> It feels like the bucket should be empty with
probability 1, but I> haven't been able to put together a
convincing argument.> Hmmm...interesting problem. For any
coin n, the probability that the> coin ?ails is 1/2,
and thus the probability that it will never> go to heads is
lim n->oo (1/2)^n = 0. As this is true for each coin,> I
would think that the ending container is empty happen with>
probability 1.> I agree that for any given coin, it will be
in the ending container> with probability 0. But when you add
together an in?ite number of> probability 0's, you 
don't
necessarily get another probability 0 back.> Here's an
example: Choose a random real number x beetween 0 and 1. >
What's the probability that x is a real number between 0 and
1? For> any speci? real number y, the probability that x = y
is 0. But the> probability that it's ANY real number between 0
and 1 is 1.Probability is countably additive. When you add
together a countablenumber of 0's, as in the coin problem,
the result is 0. The real numbersin [0,1] are uncountable,
and therefore countable additivity does notapply.-- Dave
SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal
ruling.<http://www.commoncouragepress.com/index.cfm?action=
book&bookid=228>
===
> You have a bucket, initially with no
coins in it. For n>0, at time> 1-2^(-n), you put a coin in
the bucket (say coin n), ?l the> coins in the bucket,
and remove all the coins that show heads. What's> the
probability distribution for the number of coins in the
bucket at> time 1?> It feels like the bucket should be
empty with probability 1, but I> haven't been able to put
together a convincing argument.> Hmmm...interesting
problem. For any coin n, the probability that the> coin ?ils
is 1/2, and thus the probability that it will never> go
to heads is lim n->oo (1/2)^n = 0. As this is true for each
coin,> I would think that the ending container is empty
happen with> probability 1.> I agree that for any given
coin, it will be in the ending container> with probability 0.
But when you add together an in?ite number of> probability
0's, you don't necessarily get another probability 0 
back.>
Here's an example: Choose a random real number x beetween 0
and 1. > What's the probability that x is a real number
between 0 and 1? For> any speci? real number y, the
probability that x = y is 0. But the> probability that it's
ANY real number between 0 and 1 is 1.> Probability is
countably additive. When you add together a countable> number
of 0's, as in the coin problem, the result is 0. The real
numbers> in [0,1] are uncountable, and therefore countable
additivity does not> apply.On second thought, we're not
adding an in?ite number of probability0's at all -- 
we're
multiplying an in?ite number of probability 1's. Any idea if
probability is countable multiplicative as well?
===
>
Probability is countably additive. When you add together a
countable> number of 0's, as in the coin problem, the result
is 0. The real numbers> in [0,1] are uncountable, and
therefore countable additivity does not> apply.> On second
thought, we're not adding an in?ite number of probability>
0's at all -- we're multiplying an in?ite number of
probability 1's.> Any idea if probability is countable
multiplicative as well?An in?ite product can be converted to
an in?ite sum by taking logarithms.-- Dave SeamanJudge Yohn's
mistakes revealed in Mumia Abu-Jamal
ruling.<http://www.commoncouragepress.com/index.cfm?action=
book&bookid=228>
===
> Probability is countably additive.
When you add together a countable> number of 0's, as in the
coin problem, the result is 0. The real numbers> in [0,1]
are uncountable, and therefore countable additivity does
not> apply.> On second thought, we're not adding an
in?ite number of probability> 0's at all -- we're
multiplying an in?ite number of probability 1's.> Any idea
if probability is countable multiplicative as well?> An
in?ite product can be converted to an in?ite sum by taking
logarithms.Yes, but then you're not dealing with a countable
sum of probabilitiesany more -- you're dealing with a
countable sum of log-probabilities. Are log-probabilities
countably additive?
===
> Probability is countably additive.
When you add together a countable> number of 0's, as in the
coin problem, the result is 0. The real numbers> in [0,1]
are uncountable, and therefore countable additivity does
not> apply.> On second thought, we're not adding an
in?ite number of probability> 0's at all -- we're
multiplying an in?ite number of probability 1's.> Any idea
if probability is countable multiplicative as well?> An
in?ite product can be converted to an in?ite sum by taking
logarithms.> Yes, but then you're not dealing with a
countable sum of probabilities> any more -- you're dealing
with a countable sum of log-probabilities. > Are
log-probabilities countably additive?For each n, we can
compute the probability p_n that a given ball is stillin the
bucket after the n-th step. Since p_n -> 0 as n -> oo, we
canconclude that the probability of that particular ball
being in the bucketat noon is 0.After that, we use countable
additivity over all the balls to show that thebucket is empty
at noon with probability 1.-- Dave SeamanJudge Yohn's mistakes
revealed in Mumia Abu-Jamal
ruling.<http://www.commoncouragepress.com/index.cfm?action=
book&bookid=228>
===
> Probability is countably additive.
When you add together a countable> number of 0's, as in the
coin problem, the result is 0. The real numbers> in [0,1]
are uncountable, and therefore countable additivity does
not> apply.> On second thought, we're not adding an
in?ite number of probability> 0's at all -- we're
multiplying an in?ite number of probability 1's.> Any idea
if probability is countable multiplicative as well?> An
in?ite product can be converted to an in?ite sum by taking
logarithms.> Yes, but then you're not dealing with a
countable sum of probabilities> any more -- you're dealing
with a countable sum of log-probabilities. > Are
log-probabilities countably additive?> For each n, we can
compute the probability p_n that a given ball is still> in
the bucket after the n-th step. Since p_n -> 0 as n -> oo, we
can> conclude that the probability of that particular ball
being in the bucket> at noon is 0.Right.> After that, we use
countable additivity over all the balls to show that the>
bucket is empty at noon with probability 1.Here I don't
follow. If p_k is the probability that ball number k isin the
basket at noon, the probability that the basket is empty
atnoon is prod(1-p_k). I don't immediately see how the fact
thatprobabilities are countably additive helps evaluate that
product.
===
>I think that2^n = sum (nCk) (k ranges from 0 to
n)(by nCk I mean n!/(k!(n-k)!))but can't prove it. does
anyone have any hints? also, is there any standard way to
represent sigma notation in ascii?>the above expression is a
little vague as I've written it.nCk, or as I usually write
it, C(n,k), is usually called a combination.In any case,
remember the recursive formula for Pascal's Triangle:C(n,k) =
C(n-1,k) + C(n-1,k-1). Then try to use induction.Rob Johnson
<rob@trash.whim.org>take out the trash before replying
===
Did
anyone see Amir Aczel on C-SPAN2? He was talking about his
bookPendulum: Leon Foucault and the Triumph of Science.I'm
curious to know if anyone else watched it.Does anyone out
there know him?James HarrisMy math discoveries, found for
pro?http://mathforpro?.blogspot.com/
===
> Did anyone see
Amir Aczel on C-SPAN2? He was talking about his book>
Pendulum: Leon Foucault and the Triumph of Science.> I'm
curious to know if anyone else watched it.> Does anyone out
there know him?I attended a talk Amir Aczel gave at the B.U.
Bookstore last October. It was on his new book about
Foucault. It was a good talk and it sounds like a good book.
After the bookstore address, I happended to ride back on the
MBTA Greenline Train with him and we chatted some more.I am
waiting for the paperback edition.Bob Kolker
===
I have a
monic polynomial with integer coef?ients and I know thatits
root with greatest modulus is k and I also know that the rest
ofthe roots lie in the interval [-a,b], where k>a>0,
k>b>0.What can be said about the coef?ients of the
polynomial under thesecircumstances?Felix.as the ?st
combinatorist. See www.nytimes.com, National News, or the
urlJohn Robertson
===
[snip] Tom Van Flandern - Washington, DC
- see our web site on replacement> astronomy research at
http://metaresearch.orgTom,
seehttp://www.androc1es.pwp.blueyonder.co.uk/Fundamental_rv_
2.0.htmand its surrounding material.Androcles.
===
The
probability that a number n is divisible by a prime p is 1/p
for n> p. For n < p, the probability is zero. Is there a
function thatRussell
===
> The probability that a number n is
divisible by a prime p is 1/p for n> p. For n < p, the
probability is zero. Is there a function that> RussellNever
mind, I just realized how stupid the question was, head not
screwed onright... /R
===
> Students are supposed to do
homework assignments themselves.Polite question, harsh
answer. Fortunately many contributorsin this forum are not as
rude. Students asking about their homework are welcome,
IMO.
===
> Does anyone know a simple proof of the fact that
(at least over a> ?ld) the rank of the symmetric matrix A^T
A is the same as the rank> of A?> It is easy to see that
the null space of A^T A contains the null space> of A, so it
is enough to see the reverse inclusion (which I haven't> been
able to do).> If A^t A x = 0, then 0 = <A^t A x, x> =
<Ax,Ax> = ||Ax||^2, therefore> Ax = 0.> --Ron BruckCan this
be extended to A^T B A, where B is symmetric square and p.d.,
by taking the square root of B?
===
Actually, I guess that
2-by-2 example only works if p = 1 (mod 4),since to get a
nontrivial solution to b^2+d^2=0 (mod p), one needs -1to be a
quadratic residue mod p. But for any prime p, there are
3-by-3matrices A with rank(A) = 1 and rank(A^T A) = 0, since
the congruencea^2+b^2+c^2=0 (mod p) has non-zero solutions
for all primes p.This raises the (probably not too hard)
question:If p = 3 (mod 4) and if A is a 2-by-2 matrix, is it
true that A andA^T A have the same rank?Many other similar
questions come to mind. For example:For which primes p (if
any) do there exist 3-by-3 matrices A incharacteristic p with
the property that rank(A)=2 and rank(A^T A)=1?
===
>to be a
quadratic residue mod p. But for any prime p, there are
3-by-3>matrices A with rank(A) = 1 and rank(A^T A) = 0, since
the congruence>a^2+b^2+c^2=0 (mod p) has non-zero solutions
for all primes p.This raises the (probably not too hard)
question:If p = 3 (mod 4) and if A is a 2-by-2 matrix, is it
true that A and>A^T A have the same rank?Yes. It's pretty
easy to see because the only counterexample wouldhave to be
rank 1, which has the general form A = [a,b]^T [c,d], andthe
resulting a^2+b^2 scalar factors out since p is 3 (mod
4).>Many other similar questions come to mind. For
example:For which primes p (if any) do there exist 3-by-3
matrices A in>characteristic p with the property that
rank(A)=2 and rank(A^T A)=1?All primes p. For p=2, we use A =
[ [1 1 1] [1 1 0] [0 0 0] ]. Forany odd p, let (a,b,c) be an
all non-zero solution to a^2+b^2+c^2 = 0.Take A = [ [a b c]
[b -a 0] [0 0 0] ]. A^T A consists of all zeroesexcept for
one entry, a^2+b^2 = -c^2 which is necessarily non-zero. --
Erick
===
> Many people are fond of ?lds which contain a
square root of > -1, call it i. (You might also, of course,
call it 1 in case > the characteristic is 2.) Over such a
?ld, the matrix > A = [1, i; 0, 0] may make you unhappy.>
That answers the question then. It's always harder to prove
something > that isn't true.> (Actually in this case it is
A A^T that is zero, but [1, 0;i, 0] works > for the original
question.)> -ChadWhen over C you can take A* instead of
A^T to make it true. Basically,the adjoint operator w.r.t.
standard inner product should do the job.Felix.
===
Can
someone quote or point me to results about the inability to
de?ereasonable non-trivial measures on big spaces?I vaugely
recall a result due to Banach and Kuratowski in
whichreasonable was requiring that mu({x}) = 0, the sigma
algebra wasP(X) (where X was the space), and there were some
conditions on X, I'minterested in stuff like that.Also, given
a sequence of distribution functions F_n (that is, lim(t ->
-oo) F_n(t) = 0, lim (t -> oo) F_n(t) = 1, F_n are
monotoneincreasing and right continuous) am I guaranteed to
have a sequence of-independent- random variables X_n on some
probability space (Omega,F, P) such that F_(X_n)(x) = P(X_n
<= x) = F_n(x)? Pavel
===
[...]Also, given a sequence of
distribution functions F_n (that is, lim>(t -> -oo) F_n(t) =
0, lim (t -> oo) F_n(t) = 1, F_n are monotone>increasing and
right continuous) am I guaranteed to have a sequence
of>-independent- random variables X_n on some probability
space (Omega,>F, P) such that F_(X_n)(x) = P(X_n <= x) =
F_n(x)?Yes. You can take Omega to be a countable product of
real linesand P to be a product measure.>
Pavel************************David C. Ullrich
===
Happy
holidays to one and all. The only 2 days of the year
worthcelebrating are Plutonium days where 7 November 1990 was
the birth ofthe AtomTotality theory of 231Pu is God. For God
is science andScience is God and the best bible ever written
is the best physicstextbook. And 14 December 1940 was the
birth of thediscovery of the atoms of plutonium. Unlike
religions with theircelebrations and holidays each year based
upon some ?inexactdate such as the birth of a Roman
terrorist known as Jesus, the datesof 7 November and 14
December are precise and accurate dates. Jesuswas not even
born in the year 0 A.D. and the date of Christmas ispinned to
the Winter solstice and not some activity of Jesus.To
celebrate Plutonium day this year I am going to try to do
atradition for every year and get some photograph pictures of
my yearswork around the home of the trees and landscaping.
Analog pictures andtoday is a nice sunny day for doing
that.The word hallelujah was a nonsense word created by the
AtomTotalitysome 2 or 3 thousand years ago and was used in a
beautiful song of theHandel's Messiah. Saved for the purpose
that by 1990s this word wouldbe replaced by the word
plutonium for which god had saved since itssyllables
match.Another melody and song that needs to be re?ted with
plutonium isthe Lennonsong of Have a merry christmas and a
happy new year,,... war isover,....And another song that
needs a re?ting with plutonium is one sungby Pavorratiand
some other opera singer which seems dif?ult to sing and
havetried to locate the name of this song but unable to. I
have heard iton PBS sung by some soloist a week or two ago
and seems to be acrowning glory for a singer to be able to
sing it-- some soloist inNew Zealand of Q-2 sung it. So when
I ?d out the title of it and thelyrics will go and
replace.Sing Handel's Messiah of the Hallelujah chorus
withthese
lyrics:A-T-O-M-PLUTONIUM!A-T-O-M-PLUTONIUM!A-T-OMPLUTONIUM!
A-T-OMPLUTONIUM!A-T-O-M-P-L-UTONIUM!A-T-O-M-PLUTONIUM!
A-T-O-M-PLUTONIUM!A-T-OMPLUTONIUM!A-T-OMPLUTONIUM!
A-T-O-M-P-L-UTONIUM!DOT OF THE ELECTRON CLOUD, THE
5F6A-T-O-M-PLUTONIUM!A-T-O-M-PLUTONIUM!A-T-O-M-PLUTONIUM!
A-T-O-M-PLUTONIUM!DOT OF THE ELECTRON CLOUD, THE
5F6A-T-O-M-PLUTONIUM!A-T-O-M-PLUTONIUM!A-T-O-M-PLUTONIUM!
A-T-O-M-PLUTONIUM!DOT OF THE ELECTRON CLOUD, THE 5F6DOT OF
THE ELECTRON CLOUD, THE
5F6A-T-O-M-PLUTONIUM!A-T-O-M-PLUTONIUM!A-T-O-M-PLUTONIUM!
A-T-O-M-PLUTONIUM!DOT OF THE ELECTRON CLOUD, THE 5F6DOT OF
THE ELECTRON CLOUD, THE 5F6DOT OF THE ELECTRON CLOUD, THE
5F6A-T-O-M-PLUTONIUM!A-T-O-M-PLUTONIUM!A-T-O-M-PLUTONIUM!
A-T-O-M-PLUTONIUM!-ATOM HAS INFINITE
POTENTIALATOM HAS INFINITE POTENTIALATOM HAS INFINITE
POSSIBILITIES231Pu231PuAND IT SHALL NUCLEOSYNTHESIZE FOREVER
AND EVERAND IT SHALL NUCLEOSYNTHESIZE FOREVER AND EVERAND IT
SHALL NUCLEOSYNTHESIZE FOREVER AND EVERAND IT SHALL
NUCLEOSYNTHESIZE FOREVER AND EVERATOM OF ATOMSFOREVER, AND
EVERA-T-O-M-PLUTONIUM!A-T-O-M-PLUTONIUM!And Lord of
LightFOREVER, AND
EVERA-T-O-M-PLUTONIUM!A-T-O-M-PLUTONIUM!ATOM OF ATOMSFOREVER,
AND EVERA-T-O-M-PLUTONIUM!A-T-O-M-PLUTONIUM!And Lord of
LifeFOREVER, AND EVERA-T-O-M-PLUTONIUM!A-T-O-M-PLUTONIUM!ATOM
OF ATOMSFOREVER, AND
EVERA-T-O-M-PLUTONIUM!A-T-O-M-PLUTONIUM!And Lord of
LightFOREVER, AND
EVERA-T-O-M-PLUTONIUM!A-T-O-M-PLUTONIUM!ATOM OF ATOMSFOREVER,
AND EVERA-T-O-M-PLUTONIUM!A-T-O-M-PLUTONIUM!And Lord of
LifeFOREVER, AND EVERA-T-O-M-PLUTONIUM!A-T-O-M-PLUTONIUM!ATOM
OF ATOMS and Lord of LifeFOREVER, AND EVERAND ATOMS SHALL
NUCLEOSYNTHESIZE FOREVER AND EVERAND ATOMS SHALL
NUCLEOSYNTHESIZE FOREVER AND EVERAND ATOMS SHALL
NUCLEOSYNTHESIZE FOREVER AND EVERAND ATOMS SHALL
NUCLEOSYNTHESIZE FOREVER AND
EVERA-T-O-M-P-L-U-T-O-NIUM!
===
Excuse my poor english !!!Who
knows how to calculate int(exp(a*x+b/x),x=1..inf)=???a and b
are real or complex numbers.i have found several
results:int(exp(a*x+b/x),x=0..inf)=2*sqrt(b/a)*(BesselK(1,2*
sqrt(a*b))int(exp(a*x+b/x),x=1..inf)=Sum(b^n*E(n,-a)/n!,n=0..
inf) with E integral exponential.I don't know if theses
results can help to resolve the ?st expression.All ideas are
welcome.
===
>Who knows how to calculate
int(exp(a*x+b/x),x=1..inf)=???>a and b are real or complex
numbers.>i have found several
results:>int(exp(a*x+b/x),x=0..inf)=2*sqrt(b/a)*(BesselK(1,2*
sqrt(a*b))[ presumably for Re(a) < 0 and Re(b) < 0 ]OK, so if
you have that you just need to subtract the integral from 0 to
1. int(exp(a x+b/x),x=0..1) = sum_{n=0}^in?ity a^n/n! int(x^n
exp(b/x),x=0..1)Note that if C_n = int(x^n exp(b/x),x=0..1)
(and b < 0),C_0 = e^b + b Ei(1,-b) and (by integration by
parts)C_n = e^b/(1+n) + b/(1+n) C_{n-1}so thatC_n = e^b
sum_{j=0}^{n-1} (n-j)! b^j/(n+1)! + b^n/(n+1)! C_0 = e^b
sum_{j=0}^n (n-j)! b^j/(n+1)! + b^(n+1)/(n+1)! Ei(1,-b)
Robert Israel israel@math.ubc.caDepartment of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Robert Israel a .8ecrit :Who knows how to calculate
int(exp(a*x+b/x),x=1..inf)=???>a and b are real or complex
numbers.>i have found several
results:>int(exp(a*x+b/x),x=0..inf)=2*sqrt(b/a)*(BesselK(1,2
*sqrt(a*b))> [ presumably for Re(a) < 0 and Re(b) < 0
]yes> OK, so if you have that you just need to subtract the
integral from 0 to > 1. > int(exp(a x+b/x),x=0..1) > =
sum_{n=0}^in?ity a^n/n! int(x^n exp(b/x),x=0..1)> Note
that if C_n = int(x^n exp(b/x),x=0..1) (and b < 0),> C_0 =
e^b + b Ei(1,-b) and (by integration by parts)ok> C_n =
e^b/(1+n) + b/(1+n) C_{n-1}> so that> C_n = e^b
sum_{j=0}^{n-1} (n-j)! b^j/(n+1)! + b^n/(n+1)! C_0> = e^b
sum_{j=0}^n (n-j)! b^j/(n+1)! + b^(n+1)/(n+1)!
Ei(1,-b)I=sum_{n=0..inf} a^n*C_n/n!b*sum_{n=0..inf}
(a*b)^n/(n!*(n+1)!)=b*BesselI(1,2*sqrt(a*b))/sqrt(a*b)
okbutsum_{n=0..inf} sum_{j=0..n} (n-j)! * b^j*a^n/(n!*(n+1)!)
=??? i don't know. help !> Robert Israel
israel@math.ubc.ca> Department of Mathematics
http://www.math.ubc.ca/~israel > University of British
Columbia > Vancouver, BC, Canada V6T 1Z2>
===
> Excuse my
poor english !!!> Who knows how to calculate
int(exp(a*x+b/x),x=1..inf)=???> a and b are real or complex
numbers.> i have found several results:>
int(exp(a*x+b/x),x=0..inf)=2*sqrt(b/a)*(BesselK(1,2*sqrt(a*b)
)> int(exp(a*x+b/x),x=1..inf)=Sum(b^n*E(n,-a)/n!,n=0..inf)
with E integral> exponential.> I don't know if theses
results can help to resolve the ?st expression.> All ideas
are welcome.> What system you are using to get the ?st? Maple
gives-2/a*(-b)^(1/2)*(-a)^(1/2)*BesselK(1,2*(-b)^(1/2)*(-a)^(1
/2))which is ~ your answer. The second is Taylor series for
b?So what do you mean by ?resolve the ?st'? It is a kindof
?known' function (ie: it can be computed or what ever).Hm ..
what ?expression' would you like to have as answer?
===
Axel
Vogt a .8ecrit :>Excuse my poor english !!!>Who knows how
to calculate int(exp(a*x+b/x),x=1..inf)=???>a and b are real
or complex numbers.>i have found several
results:>int(exp(a*x+b/x),x=0..inf)=2*sqrt(b/a)*(BesselK(1,2
*sqrt(a*b))>int(exp(a*x+b/x),x=1..inf)=Sum(b^n*E(n,-a)/n!,n=
0..inf) with E integral>exponential.>I don't know if theses
results can help to resolve the ?st expression.>All ideas
are welcome.> What system you are using to get the ?st?
Maple gives>
-2/a*(-b)^(1/2)*(-a)^(1/2)*BesselK(1,2*(-b)^(1/2)*(-a)^(1/2))
it's result for int(exp(a*x+b/x),x=0..inf) but i have to
solve int(exp(a*x+b/x),x=0..inf).> which is ~ your answer.
The second is Taylor series for b?> So what do you mean by
?resolve the ?st'? It is a kind> of ?known' function 
(ie: it
can be computed or what ever).> Hm .. what ?expression' would
you like to have as answer?
===
> Excuse my poor english !!!
Who knows how to calculate int(exp(a*x+b/x),x=1..inf)=???> a
and b are real or complex numbers. i have found several
results:>
int(exp(a*x+b/x),x=0..inf)=2*sqrt(b/a)*(BesselK(1,2*sqrt(a*b)
) int(exp(a*x+b/x),x=1..inf)=Sum(b^n*E(n,-a)/n!,n=0..inf)
with E integral> exponential. I don't know if theses results
can help to resolve the ?st expression. All ideas are
welcome.Mathematica says that it doesn't converge.Lurch
===
>
Who knows how to calculate int(exp(a*x+b/x),x=1..inf)=???> a
and b are real or complex numbers.>Mathematica says that it
doesn't converge.Even for a=-1, b=0 in which case it's 
simply
1/e?
===
> Who knows how to calculate
int(exp(a*x+b/x),x=1..inf)=???> a and b are real or complex
numbers.>Mathematica says that it doesn't converge.??? That
sounded plausible to me. Computer algebra systems are, in my
ownexperience, terrible with this sort of thing. However,
here's what thecurrent version of Mathematica does.In[1]:=
Integrate[Exp[a*x + b/x], {x, 1, In?ity}]Out[1]=
Integrate[E^(b/x + a*x), {x, 1, In?ity}]Thus, at least no
unwarranted general claim of divergence is made!> Even for
a=-1, b=0 in which case it's simply 1/e?Ah, at least
Mathematica can handle that well.In[2]:= Assuming[a < 0 && b
== 0, Integrate[Exp[a*x + b/x], {x, 1, In?ity}]]Out[2]=
-(E^a/a)David
===
> [..., with reference to
Mathematica]In[2]:= Assuming[a < 0 && b == 0,
Integrate[Exp[a*x + b/x], {x, 1, In?ity}]]>Assuming?
That's a new one to me. It is not in my book (the second
edition, 1991), nor in the version (4.1 for Sun Solaris)
whereto I have access. What version do you use?-- Stephen J.
Herschkorn herschko@rutcor.rutgers.edu
===
[..., with
reference to Mathematica]In[2]:= Assuming[a < 0 && b == 0,
Integrate[Exp[a*x + b/x], {x, 1,>In?ity}]] Assuming? That's
a new one to me. It is not in my book (the second> edition,
1991), nor in the version (4.1 for Sun Solaris) whereto I>
have access. What version do you use?As I noted in the part
you snipped, I'm using the _current_ version, whichis version
5.0. Yes, Assuming is new to that version. But in version 4.1I
suspect that the following equivalent idea should work for
you:In[1]:=Integrate[Exp[a*x + b/x], {x, 1, In?ity},
Assumptions -> a < 0 && b == 0]Out[1]=-(E^a/a)David
===
Toni
Lassila a .8ecrit :>Who knows how to calculate
int(exp(a*x+b/x),x=1..inf)=???>a and b are real or complex
numbers.>Mathematica says that it doesn't converge.>
Even for a=-1, b=0 in which case it's simply 1/e?convergence
is not my problem. I choose a and b to converge.
===
Excuse
my poor english !!! Who knows how to calculate
int(exp(a*x+b/x),x=1..inf)=???> a and b are real or complex
numbers. i have found several results:>
int(exp(a*x+b/x),x=0..inf)=2*sqrt(b/a)*(BesselK(1,2*sqrt(a*b)
) int(exp(a*x+b/x),x=1..inf)=Sum(b^n*E(n,-a)/n!,n=0..inf)
with E integral> exponential. I don't know if theses results
can help to resolve the ?st expression. All ideas are
welcome. Mathematica says that it doesn't converge.For Re a <
0 it's easily seen to converge.But no, I don't know 
how to do
it... the b/x looks dif?ult. Sorry.- Arthur Lurch
===
By
computing z_1 + z_2 + z_3, or otherwise, determine the set
for which thestatement z_1^3 = z_2^3 = z_3^3 is false.a) z_1
= -sqrt3 + i, z_2 = sqrt3 + i, z_3 = -2ib) z_1 = 4, z_2 = -2
+sqrt12i, z_3 = -2 -sqrt12ic) z_1 = sqrt27 + 3i, z_2 =
-sqrt27 - 3i, z_3 = 6id) z_1 = -sqrt12i, z_2 = -3 +sqrt3i,
z_3 = 3 +sqrt3iWhat is the best method of solving this
problem? There must be a bettermethod than laboriously cubing
and comparing them all...?
===
Imagine that you have the polar
forms:z_1 = r_1 (cos @_1 + i sin @_1)and similarly for z_2
and z_3. What do you know about r_1, r_2, and r_3if z_1^3 =
z_2^3 = z_3^3? What do you know about the @
values?--OL
===
>By computing z_1 + z_2 + z_3, or otherwise,
determine the set for which the>statement z_1^3 = z_2^3 =
z_3^3 is false.a) z_1 = -sqrt3 + i, z_2 = sqrt3 + i, z_3 =
-2i>b) z_1 = 4, z_2 = -2 +sqrt12i, z_3 = -2 -sqrt12i>c) z_1 =
sqrt27 + 3i, z_2 = -sqrt27 - 3i, z_3 = 6i>d) z_1 = -sqrt12i,
z_2 = -3 +sqrt3i, z_3 = 3 +sqrt3iWhat is the best method of
solving this problem? There must be a better>method than
laboriously cubing and comparing them all...?Note that in
(a): z_1+z_2 = -z_3 in (b): z_2+z_3 = -z_1 in (c): z_1 = -z_2
in (d): z_2+z_3 = 2*sqrt(3)i = sqrt(12)i = -z_1So in (c) you
know that the statement z_1^3 = z_2^3 = z_3^3 is false,since
z_1 = -z_2 implies that z_1^3 = -z_2^3, and z_2^3 = -z_2^3
ifand only if z_2 = 0, which it is not in this case.Are you
sure you copied (c) correctly? Based on the other examples,
Iwould expect z_1 = sqrt(27) - 3i, so that z_1 + z_2 = -z_3.
So, in (a), (b), and (c), you have z_1+z_2+z_3 = 0. Cubing
this gives:0 = (z_1+z_2+z_3)^3 = (z_1+z_2)^3 +
3(z_1+z_2)^2*z_3 + 3(z_1+z_2)*z_3^2 + z_3^3 = z_1^3 +
3z_1^2*z_2 + 3z_1*z_2^2 + z_2^3 + 3z_1^2*z_3 + 6z_1z_2z_3 +
3z_1^2*z_3 + 3z_1*z_3^2 + 3z_2*z_3^2 + z_3^3 = z_1^3 + z_2^3
+ z_3^3 + 3z_1^2(z_2+z_3) + 3z_2^2(z_1+z_3) + 3z_3^2(z_1+z_2)
+ 6z_1*z_2*z_3 = z_1^3 + z_2^3 + z_3^3 + 3z_1^2(-z_1) +
3z_2^2(-z_2) + 3z_3^2(-z_3) + 6z_1*z_2*z_3 = z_1^3 + z_2^3 +
z_3^3 - 3z_1^3 - 3z_2^3 - 3z_3^3 + 6z_1*z_2*z_3 = -2(z_1^3 +
z_2^3 + z_3^3) + 6z_1*z_2*z3.Which means that3z_1*z_2*z_3 =
z_1^3 + z_2^3 + z_3^3.So if z_1^3 = z_2^3 = z_3^3, then
z_1*z_2*z_3 = z_1^3 = z_2^3 = z_3^3.Which means that
z_1*z_2*(-z_1-z_2) = z_1^3 -z_1^2*z_2 - z_1*z_2^2 = z_1^3
-z_1*z_2 - z_2^2 = z_1^2 z_1^2 + z_1*z_2 + z_2^2 = 0 (z_1 +
z_2)^2 - z_1*z_2 = 0 (z_1+z_2)^2 = z_1*z_2 (-z_3)^2 = z_1*z_2
z_3^2 = z_1*z_2And analogous for the other three.Conversely,
if z_3^2 = z_1*z_2, then z_1^3 = z_1*z_2*z_3, z_2^2 =
z_1*z_3, then z_2^3 = z_1*z_2*z_3 z_1^2 = z_2*z_3, then z_3^2
= z_1*z_2*z_3.and so they are all equal. Don't know if that is
simpler than cubing,though.--
===
===========================================
===
====It's not denial. I'm just very selective about 
what
I accept as reality. Calvin (Calvin and
Hobbes)
===
======================================
===
=========Arturo
Magidinmagidin@math.berkeley.edu
===
>By computing z_1 + z_2 +
z_3, or otherwise, determine the set for whichthe>statement
z_1^3 = z_2^3 = z_3^3 is false.a) z_1 = -sqrt3 + i, z_2 =
sqrt3 + i, z_3 = -2i>b) z_1 = 4, z_2 = -2 +sqrt12i, z_3 = -2
-sqrt12i>c) z_1 = sqrt27 + 3i, z_2 = -sqrt27 - 3i, z_3 =
6i>d) z_1 = -sqrt12i, z_2 = -3 +sqrt3i, z_3 = 3 +sqrt3iWhat
is the best method of solving this problem? There must be a
better>method than laboriously cubing and comparing them
all...? Note that in (a): z_1+z_2 = -z_3> in (b): z_2+z_3 =
-z_1> in (c): z_1 = -z_2> in (d): z_2+z_3 = 2*sqrt(3)i =
sqrt(12)i = -z_1 So in (c) you know that the statement z_1^3
= z_2^3 = z_3^3 is false,> since z_1 = -z_2 implies that
z_1^3 = -z_2^3, and z_2^3 = -z_2^3 if> and only if z_2 = 0,
which it is not in this case. Are you sure you copied (c)
correctly? Based on the other examples, I> would expect z_1 =
sqrt(27) - 3i, so that z_1 + z_2 = -z_3. So, in (a), (b), and
(c), you have z_1+z_2+z_3 = 0. Cubing this gives: 0 =
(z_1+z_2+z_3)^3> = (z_1+z_2)^3 + 3(z_1+z_2)^2*z_3 +
3(z_1+z_2)*z_3^2 + z_3^3> = z_1^3 + 3z_1^2*z_2 + 3z_1*z_2^2 +
z_2^3 + 3z_1^2*z_3 + 6z_1z_2z_3> + 3z_1^2*z_3 + 3z_1*z_3^2 +
3z_2*z_3^2 + z_3^3> = z_1^3 + z_2^3 + z_3^3 + 3z_1^2(z_2+z_3)
+ 3z_2^2(z_1+z_3)> + 3z_3^2(z_1+z_2) + 6z_1*z_2*z_3> = z_1^3 +
z_2^3 + z_3^3 + 3z_1^2(-z_1) + 3z_2^2(-z_2) + 3z_3^2(-z_3)> +
6z_1*z_2*z_3> = z_1^3 + z_2^3 + z_3^3 - 3z_1^3 - 3z_2^3 -
3z_3^3 + 6z_1*z_2*z_3> = -2(z_1^3 + z_2^3 + z_3^3) +
6z_1*z_2*z3. Which means that 3z_1*z_2*z_3 = z_1^3 + z_2^3 +
z_3^3. So if z_1^3 = z_2^3 = z_3^3, then z_1*z_2*z_3 = z_1^3
= z_2^3 = z_3^3. Which means that z_1*z_2*(-z_1-z_2) = z_1^3>
-z_1^2*z_2 - z_1*z_2^2 = z_1^3> -z_1*z_2 - z_2^2 = z_1^2>
z_1^2 + z_1*z_2 + z_2^2 = 0> (z_1 + z_2)^2 - z_1*z_2 = 0>
(z_1+z_2)^2 = z_1*z_2> (-z_3)^2 = z_1*z_2> z_3^2 = z_1*z_2
And analogous for the other three. Conversely, if z_3^2 =
z_1*z_2, then z_1^3 = z_1*z_2*z_3,> z_2^2 = z_1*z_3, then
z_2^3 = z_1*z_2*z_3> z_1^2 = z_2*z_3, then z_3^2 =
z_1*z_2*z_3. and so they are all equal. Don't know if that is
simpler than cubing,> though. -- >
===
===========================================
===
====> It's not denial. I'm just very selective about>
what I accept as reality.> Calvin (Calvin and Hobbes)>
===
===========================================
===
==== Arturo Magidin> magidin@math.berkeley.edu>Couldn't
you use De Moivre's Theorem?-- David MoranChief
MeteorologistOklahoma Storm Team
===
[.snip.]>By computing
z_1 + z_2 + z_3, or otherwise, determine the set for
which>the>statement z_1^3 = z_2^3 = z_3^3 is false.>a)
z_1 = -sqrt3 + i, z_2 = sqrt3 + i, z_3 = -2i>b) z_1 = 4, z_2
= -2 +sqrt12i, z_3 = -2 -sqrt12i>c) z_1 = sqrt27 + 3i, z_2 =
-sqrt27 - 3i, z_3 = 6i>d) z_1 = -sqrt12i, z_2 = -3 +sqrt3i,
z_3 = 3 +sqrt3i>Couldn't you use De Moivre's
Theorem?Certainly, though I though that would imply cubing
each. I'll skip(c), though I suspect it was a typo.In (a),
you have that arg(z_1) = pi-arg(z_2), and arg(z_3) = -pi/2.In
(b) you have that arg(z_1) = pi-arg(z_2), and arg(z_3) =
-pi/2In (d) you have arg(z_2) = pi-arg(z_3), and arg(z_1) =
-pi/2.In (a), |z_1| = |z_2| = |z_3| = 2In (b), |z_2| = |z_3|
= |z_1| = 4In (d), |z_1| = |z_2| = |z_3| = sqrt(12).In (a),
you would need to verify that arg(z_2) = pi/6; in (b)
thatarg(z_1)=pi/6, and in (d) that arg(z_3)=pi/6. If so, then
the cubesare equal; if not, then the cubes are not equal.--
===
===========================================
===
====It's not denial. I'm just very selective about 
what
I accept as reality. Calvin (Calvin and
Hobbes)
===
======================================
===
=========Arturo Magidinmagidin@math.berkeley.edu===>
I could be wrong, but I don't see much sign that he meant
varies within> a homotopy class when he said it depends on
the path. If he meant the> function element you arrive at
when you continue along a path, and we take> depends on the
path more literally, then that statement is correct, at>
least by itself.I did in fact mean that what you get depends
on the homotopy class ofthe path, not on the path within the
homotopy class.> I would say> that it's the fact that one is
taking the collection of all paths that makes> the
construction independent of a choice of any one given path.
It's like> making a manifold not depend on your choice of
coordinate system by de?ing> its atlas to be all possible
compatible coordinate charts. Before this sequence of posts I
was not aware that there was acomstruction for Riemann
surfaces which uses unrestricted analyticalcontinuation only,
bypassing cuts, sheets and boundaryidenti?ations. This may be
good in geometry but my feeling is that aRiemann surface so
de?ed is less likely be useful as a tool in thetheory of
functions of a complex variable. However this may be
adif?ult point to get across. Lee Rudolph has some
interesting andrelevant comments. Look in Google Groups for
Re: Riemann Surfacelrudolph.
===
> Before this sequence of
posts I was not aware that there was a>comstruction for
Riemann surfaces which uses unrestricted
analytical>continuation only, bypassing cuts, sheets and
boundary>identi?ations. This may be good in geometry but my
feeling is that a>Riemann surface so de?ed is less likely be
useful as a tool in the>theory of functions of a complex
variable. However this may be a>dif?ult point to get across.
Lee Rudolph has some interesting and>relevant comments. I do?
>Look in Google Groups for Re: Riemann Surface>lrudolph.I
guess you refer to
http://groups.google.com/groups?selm=asvepu%243gp%241%
40panix2.panix.com and, perhaps, speci?ally to In fact (at
least for Riemann surfaces spread over a domain in C or the
extended complex numbers, that is, for meromorphic functions
with *numerical*, rather than another-Riemann-surface,
values), there's sometimes a sort-of-canonical cut system
which can be generated in a sort-of-canonical way--roughly
speaking, sheets meet where real parts of branches coincide.
Notice the weasel words (sometimes, sort-of-canonical,
roughly)and please don't take too much comfort from my post.
In fact, the two applications of such a sort-of-canonical cut
system that I mentioned later in my post are (to me)
themselves very much geometry and not much at all concerned
with theory of functions of a complex variable (as I would
understand that phrase); I really don't thinkthey support
your feeling quoted above. (Since my own interest in Riemann
surfaces is primarily geometrical, indeed, topological, my
ignorance of interesting and/or signi?ant applications of
such a sort-of-canonical cut system to theory of functions
may onlymean that I haven't run into one, not that none
exists. But thereare plenty of people in sci.math, several
posting to this thread, whose *do* know theory of functions
of a complex variable insideout, and I suspect that they'd
know of such applications if therewere many of them.)Lee
Rudolph
===
> I've recently been looking at little and big
endian memory storage but have> encountered a little
confusion when it comes to transferring or copying data>
between architectures of different endianness and would be
grateful for some> clari?ation.I couldn't exactly follow
what you were trying to say. First, atthe conceptual level,
the issue is this. Suppose I tried to tellyou a number by
reading the digits one at a time. We would haveto agree
whether I began with the most signi?ant digit (readingthe
digits left-to-right) or wehther I began with the
leastsigni?ant digit (reading the digits from
right-to-left).So would I read the number 9307 as
nine-three-zero-sevenor seven-zero-three-nine? Clearly, we
need to agree on whatwe are doing or things will get very
confused. Note, butthe way, that there are some advantages to
reading right-to-left;for example, if you were ?ling in a
printed form where thedigits went into boxes, reading
right-to-left (little endian) wouldallow you to begin ?ling
in the 1's place, then the 10's place,etc, while 
reading
left-to-right (big endian), which may seemmore natural, would
require that you wait to hear the whole numberbefore you know
whether the ?st digit read goes in the boxfor 100's or
1000's etc.Ok, now to practical matters. Please note that
there reallyare no guarantees in this game, but this should
work--just keepan eye on things. Suppose one wanted to store
the 32-bit number0x110A0809 (in hexidecimal, base 16, right?)
in memory or ina ?e. On a big endian machine, if you store
this isa ?e, the ?st byte of the ?e will be 0x11 and
thesecond byte will be 0A, the third byte will be 08, andthe
fourth byte will be 09. On a little endian-machine,this order
will be reversed. Similarly, for memory.For example, in C
unsigned int val = 0x110A0809; unsigned int *p_val = &val;
unsigned char *p_char = (unsigned char*)(p_val);
printf(0x%02xn, *p_char);on a big-endian machine, this should
print 0x11, but on alittle endian machine this should give
0x%09. If youhave an unsigned int that was written to a ?e
on anmachine with one endian-ness and you have read that
backinto an unsigned int on a machine with a second
endian-ness,then the following function would allow you to
change theendian-ness:unsigned int
change_endian_uint(unsigned int n){ typedef union { unsigned
int value; char bytes[4]; } cheat_type; cheat_type input;
cheat_type output; input.value = n; output.bytes[3] =
input.bytes[0]; output.bytes[2] = input.bytes[1];
output.bytes[1] = input.bytes[2]; output.bytes[0] =
input.bytes[3]; return output.value;}For other situation you
may need to play other games with unions.Following up on a
question in your original posts, ifyou did mixed 16-bit and
32 bit objects, each object neesto be handled separately
(thus you might need another routinefor unsigned
shorts):unsigned short change_endian_ushort(unsigned short
n){ typedef union { unsigned short value; char bytes[2]; }
cheat_type; cheat_type input; cheat_type output; input.value
= n; output.bytes[1] = input.bytes[0]; output.bytes[0] =
input.bytes[1]; return output.value;}The above code fragments
assume, of course, that on your machinesint's are 32-bit
objects and short's are 16-bit objects.Several other notes
about things that can go wrong: The sizes of objects might
vary between machines. For exmaple, on some older machines,
int would be 16-bits, while on some high-end new machines,
int might be 64-bits. It is even possible that the order of
bits withing bytes can be reversed, but this is unlikely.
Machines have different alignment issues, which relate to
packing. For example, in C if you have struct a_struct{ int
a; char b; int c; }; then on most machines, sizeof(a_struct)
= 12, even though sizeof(int) + sizeof(char) + sizeof(int) =
9. This is done so that both int's can be placed in memory
locations whose addresses are divisble by 4 (a requirement
for reading 4-bytes at a time on some systems, and an
optimization on many others). The three bytes after the char
b contain unde?ed data.Two other useful facts. Intel is
probably the only architectureyou will meet that is
little-endian; most other machines (SPARC,and if memory
server Motorola and SGI) are big-endian.As a cultural note,
the term big-endian and little-endianpay homage to Gulliver's
travels, where kingdoms were at waras to whether to open eggs
at the little end or the big end.Best
wishes-Mike
===
-snip-===How to solve this equation (by
elementary methods!):n^p+m^p=(2n-m)^pn>0;m>0 et p>2
===
>
M?rio Amado Alves> I've learnt how to represent a map as the
set of pairs of connected> vertices. Accordingly, the object
{(A, B)} represents both (1) and> (2): (1) (2)> ___________
_____> | | | |> | A | | A |> | _____ | |_____|> | | | | | |>
| | B | | | B |> | |_____| | |_____|> |___________| But
clearly (1) and (2) differ with respect to the reachability
of> region B from the outside [(2) reachable, (1) not]. How
is this kind> of fact formally expressed usually? (I can
think of a couple of ways> but I'm sure this problem has been
dealt before so no need to reinvent> the wheel and/or use a
?strange' structure.) --Marius Amado Alves> (amateur
mathematician playing with map coloring)> Country A is simply
connected, as they say, in (2) but not in (1). If we>
represent countries by vertices rather than by faces, people
speak of a> bridge (resp. cutpoint) as an edge (resp. vertex)
which, if omitted, causes> the graph to become disconnected.
So, if there are other countries outside> A, A would be a
cutpoint....> Ok. But with this I still cannot represent
the fact explicitly.> Consider maps (3), (4) which are (1),
(2) resp. plus a region C> connected (only) to A. The
symbolic represention of *both* (3) and (4)> is {(A, B), (A,
C)}.> I'm playing with constructing (colored) maps by
adding one region at a> time (and exchanging colors). For
that I want to represent the> unreachability condition, for
example expressing the fact that I can> add pair (B, C) to
(4) but not to (3).> Maybe I should say a litle more. I
intuit that is possible to keep the> number of colors of the
periphery of any map, below 4, where periphery> is the set of
vertices available for connection to a newly added> vertex. I
intuit this is possible by exchanging colors in the newly>
created map. This is of course a quest for a different (?)
proof of> the four color theorem. Maybe I'm on a beaten
track. I don't know much> about four color theorem proving,
other then the ?fact' that existing> proofs are not elegant
enough for my taste.> The following website gives 43 proofs
of the Pythagorean
Theorem.http://www.cut-the-knot.org/pythagoras/
index.shtmlWhich proof(s) do you consider to be elegant?
===
>
Algebraic toplogy (AT herein) seems to be a good way of
formalizing> protocols in distributed systems (DS herein)
(such as decision problems like> ?consensus'). Do you know
about Petri nets, process algebras such asCCS, CSP and the
Pi-calculus, and more abstractcharacterizations such as
labelled transition systems (LTSs)?If not, those would be
important to know before plunging intotopology. There is even
a widely-accepted language (LOTOS)based on CCS and CSP that is
used for formally de?ingcommunication protocols. Moreover
PROMELA, the language forprotocol de?ition which is used by
the SPIN model checker, isalso based partly on CSP and is
also widely used. There may be some reason why these more
standard formalcharacterizations are not adequate for your
purposes, but in theabsence of other information, I would say
to try them ?st.--Jamie. (nel mezzo del cammin di nostra
vita)
===
> In comp.theory Michael N. Christoff
<mchristoff@sympatico.caremovethis Algebraic toplogy (AT
herein) seems to be a good way of formalizing> protocols in
distributed systems (DS herein) (such as decision
problemslike> ?consensus'). Do you know about Petri nets,
process algebras such as> CCS, CSP and the Pi-calculus, and
more abstract> characterizations such as labelled transition
systems (LTSs)?> If not, those would be important to know
before plunging into> topology. There is even a
widely-accepted language (LOTOS)> based on CCS and CSP that
is used for formally de?ing> communication protocols.
Moreover PROMELA, the language for> protocol de?ition which
is used by the SPIN model checker, is> also based partly on
CSP and is also widely used.>I actually have a book on CCS
(that I'll be taking another look at), but ifI'm not 
mistaken
its based on an interleaving model of concurrency. Also, Iam
most interested in architectures that do not use shared
memory in favourof distributed ?objects' with local hidden
state. One of the main reason Iwas interested in AT is that
it has given some very elegant proofs of somebasic
impossibility theorems (ie: impossibility of consensus with
one faultyprocess). Are there similar proofs based on CCS,
etc... ? Finally, on amore personal level, I have always been
interested in topology and grouptheory (did a lot of research
into the ?ite state automata / group theoryconnection). I
have also done a small amount of work with Petri nets.
Arethese process algebras a prerequisite to learning AT for
distributedcomputing?As far as AT goes, I'm going to start by
learning some basic point-settopology ?st. The only problem
is weeding out the material I need for DSfrom more ?pure
math' aspects of it.l8r, Mike N. Christoff
===
> Algebraic
toplogy (AT herein) seems to be a good way of formalizing>
protocols in distributed systems (DS herein) (such as
decision problemslike> ?consensus'). I am interested in
learning more, however AT is a huge?ld> and I am only
interested in learning the parts directly related to>
distributed computing. Can anyone suggest a book, that a)
assumes no> knowledge of algebraic toplogy b) assumes no more
than undergraduate level> math - ie: calculus, linear algebra,
basic geometry, ability to do proofs,> etc... c) focused on
showing how AT can be utilized to solve DS problemsand> does
not get into non-DS related aspects of AT (unless they are
required> background for understanding DS related AT topics).
I have found many introductions on the net, but they seem to
assume atleast> basic knowledge of topology, homotopy, and
other topics I am not familiar> with, so I think a full book
dedicated to the subject sounds more feasible> as a basis for
learning AT for DS. But any links or online books you may>
know of will be of great help as well.>A Tutorial on
Algebraic Topology and Distributed ComputingMaurice Herlihy,
1994http://citeseer.nj.nec.com/herlihy94tutorial.htmll8r,
Mike N. Christoff
===
Sub: Unsolvabve Geometrical Problems>
Squaring a given circle is unsolvable . There are volumes of
pages onthe subject which I have read - from Archimedes to
SrinivasaRamanujan.> Well! I have found out a solution. Could
not believe it ? My logic isvery simple.> Pi is an irrational
number. So is square root of 2. There is a simplegeometrical
construction to solve for square root of 2. Similarly I
havedeveloped a simple geometrical construction to solve for
pi.> My question is Where to send it? Is there any Maths
Forum where I canpresent my papers and answer questions of
the experts?Try sending it to James Harris to verify :
)--Paul V. S. TownsendInterchange the alphabetic elements to
reply
===
> Sub: Unsolvabve Geometrical Problems > Squaring a
given circle is unsolvable . There are volumes of pages on
the subject which I have read - from Archimedes to
SrinivasaRamanujan.> Well! I have found out a solution. Could
not believe it ? My logic is very simple.> Pi is an irrational
number. So is square root of 2. There is a simple geometrical
construction to solve for square root of 2. Similarly I have
developed a simple geometrical construction to solve for pi.>
My question is Where to send it? Is there any Maths Forum
where I can present my papers and answer questions of the
experts? > You're likely to run into the same sort of
small-minded attitude amongthe mathematicians here, that our
beloved JSH experienced.Why not follow his tack? Look into
this publication of the MegaFoundation:It worked for JSH.
Maybe it'll get your discoveries some clear air.BTW, the
other poster suggesting a web page containing your
argument,coupled by a discussion on sci.math dedicated to
exposing the ?the argument (sorry, if your construction
means to produce pi via aconstriction using compass and
[unmarked] straightedge only, then yourare mistaken, and
there is some error in the argument).Persevere, and be of
good faith (not to mention good humor), and nodoubt you'll
learn something.Dale.
===
> Sub: Unsolvabve Geometrical
Problems > Squaring a given circle is unsolvable . There are
volumes of pages on the subject which I have read - from
Archimedes to SrinivasaRamanujan.> Well! I have found out a
solution. Could not believe it ? My logic is very simple.> Pi
is an irrational number. So is square root of 2. There is a
simple geometrical construction to solve for square root of
2. Similarly I have developed a simple geometrical
construction to solve for pi.> My question is Where to send
it? Is there any Maths Forum where I can present my papers
and answer questions of the experts? > There's already a
geomtrical construciton of pi, take a compass andconstruct a
circle with a diameter of 1.
===
>Squaring a given circle is
unsolvable .Note that this unsolvability is unsolvability
using straightedge andcompass methods. If you permit certain
kinds of mechanicaldevices to be used, then it's possible to
construct pi and squarethe circle. The other standard
impossible constructions(trisecting an angle and duplicating
a cube) are also possibleif one is allowed the right kind of
tool in addition to straightedgeand compass.Keith
Ramsay
===
Squaring a given circle is unsolvable . Note that
this unsolvability is unsolvability using straightedge and>
compass methods. If you permit certain kinds of mechanical>
devices to be used, then it's possible to construct pi and
square> the circle. The other standard impossible
constructions> (trisecting an angle and duplicating a cube)
are also possible> if one is allowed the right kind of tool
in addition to straightedge> and compass. Keith Ramsay>Just
wondering, is it possible to create these tools using a
compass and astraightedge (and maybe a pair of scissors)?
/R
===
> Just wondering, is it possible to create these tools
using a compass and a> straightedge (and maybe a pair of
scissors)? /RWell, one angle trisection requires a
straightedge that you can mark; itis then slid along a circle
and line. (That not being one of theoperations described by
Euclid, it has no impact on the proof o?possibility.) I've
also seen a drawing of an angle trisection devicewith fairly
simple moving parts.-- Daniel W.
Johnsonpanoptes@iquest.nethttp://members.iquest.net/~panoptes
/039 53 36 N / 086 11 55 W
===
I've actually seen a trisection
device with no moving parts at all.It's basically a strip with
one end having square and circularattachments at the end. By
placing the object in the right positionrelative to a given
angle, the trisectors of the angle are identi?d.The
Archimedean trisection (using the marked straightedge)
algebraicllysolves any cubic equation with three real roots,
for the solution ofsuch an equation always reduces to an
angle trisection via Viete'sformulation. So the method can be
adapted to another old impossible(with unmarked straightedge
and compass) problem, the construction ofthe regular
heptagon.--OL
===
The problem is NOT that pi is irrational.
The problem instead is thatpi is not obtainable from integers
by any combination of arithmeticoperations plus solution of
quadratic equations, which are the onlyoperations that can be
implemented via a straightedge and compass. Thissituation
holds for any transcendental numbe, among others.You can't
square a circle. But you can get close to doing so since
theset of constructible numerical ratios (rational numbers
and someirrational ones) is dense.--OL
===
The ? that pi
is *not* an irrational number. It is transcendental.
===
>The
? that pi is *not* an irrational number. It is
transcendental.That should be pi is *not* just an irrational
number. It istranscendental. In other words, pi is
irrational, and it is a specialtype of irrational number (the
special type being a transcendental number).David
McAnally--
===
The ? that pi is *not* an
irrational number. It is transcendental.Transcendental
numbers ->are<- irrational. Not every irrational
istranscendental, but every transcedental is irrational.--
===
===========================================
===
====It's not denial. I'm just very selective about 
what
I accept as reality. Calvin (Calvin and
Hobbes)
===
======================================
===
=========Arturo Magidinmagidin@math.berkeley.edu=== (axiom-schema) is introduced...along with power
set...pairing is> redundant, they advocate keeping it
because it is elementary and> essential for the sequential
development and even for a scheme that> shuns replacement,
pairing makes sense. I think they are suggesting> that set
theory even without replacement is strong enough for almost>
all practical purposes. What do you think?>Oh, I see, I'm
used to the shunning replacement version, because I'm>used to
GBN set theory. So I think of replacement+comprehension
as>just GBN's comprehension, but of course, that isn't 
quite
right, and>this is an example why.>I believe that in GBN
set theory, pairing *is* required, is it not?> Yes, because
the NBG axioms do not have propositional> functions. Functions
are de?ed only by ordered pairs.Good, that's what I thought.
===
> Mathematicians dump on cranks with extreme vitriol.No.
Most mathematicians ignore them entirely.Thomas
===
>
[...]> But check this out:>
http://www.bearnol.pwp.blueyonder.co.uk/Math/> Wow,
James has been busy. I had no idea he'd, um, ?tackled' 
so
many problems!>For those who don't know Ramsden is an alias
of David Rusin, but I>don't see his point here in posting
that link.>Anyone have any idea what Rusin is up to here?Or
any idea how James got the idea that Ramsden and Rusin
were>the same person?Don't know, but it seems, from 
James's
post, that James Harris maybelieve that James Harris = James
Wanless (which is the person JohnRamsden was talking
about)...--
===
===========================================
===
====It's not denial. I'm just very selective about 
what
I accept as reality. Calvin (Calvin and
Hobbes)
===
======================================
===
=========Arturo Magidinmagidin@math.berkeley.edu
<3c65f87.0312131113.663a3a6c@posting.google.com>
<fmpmtvo37m5sgpjmtde23o9caooojpg5at@4ax.com>
<brg84v$1v52$1@agate.berkeley.edu>
===
> [...]> But
check this out:>
http://www.bearnol.pwp.blueyonder.co.uk/Math/> Wow,
James has been busy. I had no idea he'd, um, ?tackled' 
so
many problems!>For those who don't know Ramsden is an
alias of David Rusin, but I>don't see his point here in
posting that link.>Anyone have any idea what Rusin is up
to here?>Or any idea how James got the idea that Ramsden and
Rusin were>the same person? Don't know, but it seems, from
James's post, that James Harris may> believe that James
Harris = James Wanless (which is the person John> Ramsden was
talking about)...Yes, but also that John Ramsden = Larry
Hammick, since John didn'tpost any links at all, but Larry
did.How appropriate. The population of sci.math is plummeting
because ofa vengeful equivalence relation. The group sci.math
/ alias equivalence(the quotient of sci.math by the
equivalence relation generated by theis a sock puppet of
relation) won't have so many participants, Ifear.I wonder
which coset I'm in.-- Come on people!!! The US just blew up a
lot of people in Iraq, don'tyou realize that a person with my
exposure might just end up dead, bymysterious circumstances?
--James Harris, on the dangers of proving Fermat's last
theorem
===
> What do you mean by properly de?ed?> What
do you mean by predicative proof step?> And so on.I meant a
proof technique which could be modeled or otherwise aped(the
context does not require rigid precision) in a proof done
withina predicative proof system.Thomas
===
Herc, please
comment on the generalization of Cantor's diagonalargument.
Without doing that, everything you say could be trueand you
would still have multiple in?ity types.For all sets A, |A| <
|P(A)|.Let A be a set and P(A) be the powerset of A. A can be
in?ite,but it doesn't need to be.There are no bijections
between A and P(A) (see below). There isa bijection between A
and {{a}| a in A}, subset of P(A).By our de?ition of |A| <
|B|, it follows that |A| < |P(A)|.Let f: A -> P(A). f is not
a bijection, because there is at leastone element of P(A)
(one subset of A) to which f does not map.(See below.) Since
f is not speci?d beyond being a functionfrom A to P(A), this
is true of all such functions. Therefore,there are no
bijections between A and P(A) and |A| < |P(A)|.Consider the
set D_f, a subset of A, de?ed as D_f = { a in A| ~( a in
f(a) ) },which is equivalent to ( a in D_f ) iff ~( a in f(a)
).There is no d in A such that* f(d) = D_f.If there were, it
would follow that* ( d in D_f = f(d) ) iff ~( d in f(d) ).But
this would be a contradiction. Therefore, f is not a
bijection, no such function is a bijection, and |A| <
|P(A)|.See
alsohttp://mathworld.wolfram.com/CantorDiagonalMethod.htmlI'm
going to respond to your comments below, but I think it'sclose
to pointless, if your intention is to rule out morethan one
in?ity-type.> [...]> You say it's not more useful; I say
it is more useful. How> should we decide this point? I could
give you many examples> where the existence of the LUB is
important to the conclusion> drawn _in real analysis_. Can
you do the same with the> computability property?> do any
examples apply outside of number theory? what is> the
contribution to technology of uncountability? anything?up a
lot in quantum mechanics. I'm not sure it makes sense totalk
of that if the values that the norm takes, <psi|psi>,don't
form a complete ?ld. Completeness is the LUB guarantee.The
LUB guarantee leads to uncountability. You're sittingin front
of an application of quantum mechanics while youread this.
Good enough?On the other hand, while I believe that thinking
about computability and Turing Machines contributed to the
developmentof today's PCs and the Internet and all, there is
very littleuse for the distinction you are drawing,
betweencomputable-in-principle and
uncomputable-in-principle.The best ciphers we have today have
computable-in-principlealgorithms that will crack them, but,
until we havecomputable-in-practice algorithms, no one is
going to worry about it.[...]> However, several people have
pointed out that you do not need a> contradiction for this
result. In fact, Cantor did not use> contradiction. In the
latest version, the one you most recently> snipped, there is
no contradiction, just a demonstration that> every list
misses at least one real number. This is a > contradiction
only to the assumption that |N| = |R|.> misses an in?itely
long number, at a digit that it speci?s > itself must be
contradictory. the number is de?ed in terms of > itself.The
argument show that _at least_ one number is missed, yes. It
does this by showing it is false that this speci?d number is
in the list. If showing something is false is a contradiction,
then, OK, that's a contradiction. I don't see 
anything
paradoxical about that, though.Something else I've wanted to
mention: your argument sounds likethere is only this one
diagonal number to ? into the list, and then they'll all be
in there. In fact, almost all the reals do not ? in the
list. A rough way to see this is to look at all the diagonal
numbers, those numbers not equal to nth real at thenth
decimal. Map them all to the reals in base-9, skipping over
the nth digit of the nth real. 0 0.12234... (the diagonal
number) 0.17842... 1 0.27843... (one off-diagonal, base-10)
0.44454...2 0.34897... (maps to R, base-9) 0.34443...3
0.84747... 0 1 2 3 4 5 6 7 8 9 4 0.28682... | | | | / / / /
/... 0 1 2 3 4 5 6 7 8The number is _not_ de?ed in terms of
itself. It is de?ed in terms of the list the number is not
in. The number does not contradict _itself_; it contradicts
_being in the list_. If you assume that all reals are _in the
list_, then, yes, that would appear to be a paradox, a number
that is not equal to itself at one digit. That is the point
of the argument: that the number is _not_ in the list.> you
treat an in?itely long sequence like a simple object you >
can manipulate.If you don't like that, you'll hate how 
I
de?e the reals. Anyway,what's wrong with that? And how is
that different from what youdo with computable numbers?>
where is the contradictory bit? can you point it out? why >
cannot the number be computed to any precision speci?d? its
a > simple trick of enumeration, the number is part of the
list and > poorly de?ed at its own referential digit, it
doesn't lie > outside the list. why does logic con?e itself
to non self reference yet > mathematics yields it. there are
simple limitations to computer > programs when they self
reference, its a practical base of > limitations. natural
language and set theory both get around > self referential
paradoxes by specifying before hand their use in > the
domains. limitations to computers are very limited in scope,
> all other ?lds of science focus on them, they are
trivial.You have created the self-referential paradox by
assuming something false. It's you who says the number is
part of the list, that it contradicts itself, and so on. Your
reason for saying this is that _all the reals are on the
list_. But they aren't.[...]> The extra number is simply this
:> Given a number on a list, change it and put it back in the
> list without changing the list.[...] > Would it help to
tally the list of reals as a single number?> DIGIT 1 2 3 4
5 6> __________________________> UTM(1) 4 3 6 4 2 4> UTM(2) 7
4 3 4 3 2> UTM(3) 0 1 0 1 1 1> UTM(4) 1 2 2 2 2 2> UTM(5) 7 7
7 7 7 7> Working along the diagonal in this fashion :> 1
2 4 7> 3 5> 6> that would give the single real number :>
437640....This doesn't help at all. I still don't know 
what
you meant before.I don't know what I'm supposed to 
conclude
from this either.Your back-and-forth number isn't guaranteed
to either beon your list UTM(n) or to not be on your list.
You can only compare real numbers for equality by comparing
_correspondingdigits_, but your back-and-forth number mixes
up the order,leading to who knows what?[...]> Original
Format> alt.paranormal, alt.sci.physics I don't see anything
new here.> Its a diagonalisation proof of Godels theorom by
Roger Penrose, > Steven Hawkings assistant. Probably accepted
as a correct > proof by mainstream mathematicians to this
day, yet the > de?ition of the diagonal function is
obviously ?and > not well de?ed.> Read the
capitalisation part of the post and tell me if its a > valid
proof.I'm not sure the proof is valid, but for a different
reason thanyou use. I think step #2 jumps too quickly from
All mathematical truths are derivable to there is an algorith
to derive them. The step you object to I do not see a problem
with.Your own comment is just a restatement of what you said
above. a does not have a paradoxical bit because it is not in
the list. It is not in the list because we de?ed a so that we
could see it's not in the list. This is only a problem if we
insist, as you do, that the list is complete.: : > The
pattern is not legitimately created, it is obviously: : >
self referencing and a has a paradoxical bit when it
evaluates: : > its own number. Just because there's two steps
in seeing the: : > plausibility in a theorom, one of the steps
fails so the: : > theorem fails, not the whole encapsulation
of theoroms.Jim Burns
===
<^> <()> <^> -- Herc, please
comment on the generalization of Cantor's diagonal> argument.
Without doing that, everything you say could be true> and you
would still have multiple in?ity types. For all sets A, |A| <
|P(A)|. Let A be a set and P(A) be the powerset of A. A can be
in?ite,> but it doesn't need to be. There are no bijections
between A and P(A) (see below). There is> a bijection between
A and {{a}| a in A}, subset of P(A).> By our de?ition of |A|
< |B|, it follows that |A| < |P(A)|. Let f: A -> P(A). f is
not a bijection, because there is at least> one element of
P(A) (one subset of A) to which f does not map.> (See below.)
Since f is not speci?d beyond being a function> from A to
P(A), this is true of all such functions. Therefore,> there
are no bijections between A and P(A) and |A| < |P(A)|.
Consider the set D_f, a subset of A, de?ed as> D_f = { a in
A| ~( a in f(a) ) },> which is equivalent to> ( a in D_f )
iff ~( a in f(a) ).> There is no d in A such that> * f(d) =
D_f.> If there were, it would follow that> * ( d in D_f =
f(d) ) iff ~( d in f(d) ).> But this would be a
contradiction. Therefore, f is not a bijection,> no such
function is a bijection, and |A| < |P(A)|. See also>
http://mathworld.wolfram.com/CantorDiagonalMethod.html I'm
going to respond to your comments below, but I think it's>
close to pointless, if your intention is to rule out more>
than one in?ity-type. [...]> You say it's not more useful; I
say it is more useful. How> should we decide this point? I
could give you many examples> where the existence of the LUB
is important to the conclusion> drawn _in real analysis_. Can
you do the same with the> computability property? do any
examples apply outside of number theory? what is> the
contribution to technology of uncountability? anything? up a
lot in quantum mechanics. I'm not sure it makes sense to>
talk of that if the values that the norm takes, <psi|psi>,>
don't form a complete ?ld. Completeness is the LUB
guarantee.> The LUB guarantee leads to uncountability. You're
sitting> in front of an application of quantum mechanics while
you> read this. Good enough? On the other hand, while I
believe that thinking about> computability and Turing
Machines contributed to the development> of today's PCs and
the Internet and all, there is very little> use for the
distinction you are drawing, between> computable-in-principle
and uncomputable-in-principle.> The best ciphers we have today
have computable-in-principle> algorithms that will crack them,
but, until we have> computable-in-practice algorithms, no one
is going to> worry about it. [...]> However, several people
have pointed out that you do not need a> contradiction for
this result. In fact, Cantor did not use> contradiction. In
the latest version, the one you most recently> snipped, there
is no contradiction, just a demonstration that> every list
misses at least one real number. This is a> contradiction
only to the assumption that |N| = |R|. misses an in?itely
long number, at a digit that it speci?s> itself must be
contradictory. the number is de?ed in terms of> itself. The
argument show that _at least_ one number is missed, yes. It>
does this by showing it is false that this speci?d number is
in> the list. If showing something is false is a
contradiction, then,> OK, that's a contradiction. I 
don't see
anything paradoxical> about that, though. Something else I've
wanted to mention: your argument sounds like> there is only
this one diagonal number to ? into the list, and> then
they'll all be in there. In fact, almost all the reals do>
not ? in the list. A rough way to see this is to look at
all> the diagonal numbers, those numbers not equal to nth
real at the> nth decimal. Map them all to the reals in
base-9, skipping over> the nth digit of the nth real. 0
0.12234... (the diagonal number) 0.17842...> 1 0.27843...
(one off-diagonal, base-10) 0.44454...> 2 0.34897... (maps to
R, base-9) 0.34443...> 3 0.84747... 0 1 2 3 4 5 6 7 8 9> 4
0.28682... | | | | / / / / /> ... 0 1 2 3 4 5 6 7 8> The
number is _not_ de?ed in terms of itself. It is de?ed in>
terms of the list the number is not in. The number does not>
contradict _itself_; it contradicts _being in the list_. If
you> assume that all reals are _in the list_, then, yes, that
would> appear to be a paradox, a number that is not equal to
itself at> one digit. That is the point of the argument: that
the number is> _not_ in the list.As the number is de?ed it is
in the list.For a ?ite list you can say the number is not in
the list.For an in?ite list you cannot encapsulate the
entire number *********you can only provide an algorithm as
such : Get the 1st digit in the grid. do a transform go down
1 and right 1 repeatfor an in?ite list, the only possible
way to de?e the number isto *reference* the entire list. to
reference the entire list of programsmeans the number will
try to read a digit that it is supposed to writeCantors
method works on ?ite lists, simple 0 order objects,
constants.in?ite lists require an algorithm to encapsulate
the diagonal number.any algorithmic_number is automatically a
part of the list by theaxioms of the proof.I'm not giving you
an in?ite list to make a new number,I'm only giving you the
algorithm, UTM(Z). you treat an in?itely long sequence like
a simple object you> can manipulate. If you don't like that,
you'll hate how I de?e the reals. Anyway,> what's 
wrong with
that? And how is that different from what you> do with
computable numbers?Every step I take can be indexed by
computer. I'm not getting a paradoxfrom simply de?ed
?complete models' and inferring some beyond computerknowledge
technique only humans can comprehend.I am using ?ite methods
that can all be phsically achieved. I'm calling it?itism ~
unde?ed commentaries on in?ite lists are not accepted into
proofs. where is the contradictory bit? can you point it out?
why> cannot the number be computed to any precision speci?d?
its a> simple trick of enumeration, the number is part of the
list and> poorly de?ed at its own referential digit, it
doesn't lie> outside the list. why does logic con?e itself
to non self reference yet> mathematics yields it. there are
simple limitations to computer> programs when they self
reference, its a practical base of> limitations. natural
language and set theory both get around> self referential
paradoxes by specifying before hand their use in> the
domains. limitations to computers are very limited in scope,>
all other ?lds of science focus on them, they are trivial.
You have created the self-referential paradox by assuming>
something false. It's you who says the number is part of the
list,> that it contradicts itself, and so on. Your reason for
saying this> is that _all the reals are on the list_. But they
aren't.'but they aren't' is 
assumed here.you are assuming an
*algorithm* *off_the_list* to imply a new number. [...]> The
extra number is simply this :> Given a number on a list,
change it and put it back in the> list without changing the
list.> [...]> Would it help to tally the list of reals as a
single number? DIGIT 1 2 3 4 5 6> __________________________>
UTM(1) 4 3 6 4 2 4> UTM(2) 7 4 3 4 3 2> UTM(3) 0 1 0 1 1 1>
UTM(4) 1 2 2 2 2 2> UTM(5) 7 7 7 7 7 7 Working along the
diagonal in this fashion : 1 2 4 7> 3 5> 6 that would give
the single real number : 437640.... This doesn't help at all.
I still don't know what you meant before.> I don't 
know what
I'm supposed to conclude from this either.> Your
back-and-forth number isn't guaranteed to either be> on your
list UTM(n) or to not be on your list. You can only> compare
real numbers for equality by comparing _corresponding>
digits_, but your back-and-forth number mixes up the order,>
leading to who knows what?>I agree, this doesn't help at
all!> [...]> Original Format> alt.paranormal, alt.sci.physics
I don't see anything new here. Its a diagonalisation proof of
Godels theorom by Roger Penrose,> Steven Hawkings assistant.
Probably accepted as a correct> proof by mainstream
mathematicians to this day, yet the> de?ition of the
diagonal function is obviously ?and> not well de?ed.
Read the capitalisation part of the post and tell me if its
a> valid proof. I'm not sure the proof is valid, but for a
different reason than> you use. I think step #2 jumps too
quickly from All mathematical> truths are derivable to there
is an algorith to derive them.> The step you object to I do
not see a problem with. Your own comment is just a
restatement of what you said above.> a does not have a
paradoxical bit because it is not in the list.> It is not in
the list because we de?ed a so that we could see> it's not
in the list. This is only a problem if we insist, as you> do,
that the list is complete. : : > The pattern is not
legitimately created, it is obviously> : : > self referencing
and a has a paradoxical bit when it evaluates> : : > its own
number. Just because there's two steps in seeing the> : : >
plausibility in a theorom, one of the steps fails so the> : :
> theorem fails, not the whole encapsulation of theoroms. Jim
BurnsI'm not insisting the list is complete. I'm 
pointing out
that if youmake the axiom that the list is complete there is
no contradiction.In closed knowledge the list is complete.
There is nothing to de?e off the list.I don't follow the
power set bijection proof, I'm assuming its a simliar ?
Cantorsdiagonalisation.Herc
===
> ...> Maybe ... offhand ...
the computable numbers are indeed countable, and> we could
produce an unambiguous list giving the number of all _ ?ite algorithms_ which might produce one. But the proposed
algorithm of> the Cantor argument is not a ?ite algorithm,
since it requires us to> do an in?ite amount of work a
priori to make the list!> That just about is it. The
complete list requires us to do an in?ite> amount of work
just to determine the n-th element of the list when n is>
large enough. Suppose we de?e an algorithm as something that
has as> input an integer and as output (when it terminates) a
decimal digit. The> computable numbers are de?ed by all such
algorithms that terminate> for each ?ite input value. We can
enumerate the ?ite algorithms for> which it is obvious that
they de?e a computable number. But that is> not suf?ient,
we also have to enumerate the ?ite algorithms for which> it
is *not* obvious that they de?e a computable number.I'm
still not sure what we are demonstrating with the Cantor
argument!Accept your de?itions of algorithm and computable.
Assume for thesake of argument the problem algorithm de?es a
computable number. Then, since the problem algorithm is a
valid ?ite algorithm itshould get a number on the list, say
K. Call the j-th algorithm on thelist P(j) and its
corresponding computable number n(j). P(K) tells usto consult
the list, and augment the k-th decimal digit of n(k) by 1mod
10 (say), in order to get the k-th decimal digit of n(K).Ok
so far?So let's start computing. We realize we don't 
have to
compute the_entire_ representation of each computable number
in order to applyP(K), we only have to compute just enough
... so only a ?ite amountof work is required at each step
after all.First we take P(1), compute exactly one digit of
its correspondingnumber, n(1), and increment that digit by 1,
modulo 10. Next we takealgorithm P(2), compute the ?st two
digits of n(2) and increment thesecond digit by 1 mod 10,
etc. Eventually we get to K. Now P tellsus to consider
algorithm P itself, compute the ?st K digits of itsnumber,
then augment the K-th digit. We already have
digits1,2,...,K-1 , no problem; but we don't have digit K
yet, because wehaven't ?ished calculating it, so that spot
is blank! So we can'tcarry out the instructions. The
algorithm at this point simplybecomes nonsense, because it
also tells us that the K-th digit of n(K)is not equal to
itself. So the alleged algorithm P(K) may _look_ likean
algorithm, but in fact is nonsense.But maybe if we _don't_
insist on including it on the list in the?st place, we _can_
compute its number, and with a ?ite amount ofeffort at each
step! If the bad algoritm P _isn't_ on the list, thenwe can
get the k-th digit of its number by running the algorithm
inthe k-th slot for k places, no problem.So what have we
shown? If we assume a priori the valid algorithms
arecountable, including our problem child, the problem
algorithm becomesnonsense, hence we kick it off the list. But
once we don't try tocount it, it becomes sense again, hence a
valid algorithm which wasleft off the list!The family
resemblence to Russell's set paradox is striking, but 
I'mnot
sure what we are supposed to conclude: I'm leaning to
theconclusion that the computable numbers are not in fact
countable. Buthow can this be? All ?ite symbol strings (even
those drawn from acountably in?ite symbol set?) form a
countable class, and therecan't be more computable numbers
than there are ?ite algorithms, andthere can't be more ?ite
algorithms than there are ?ite symbolstrings ...> For
instance de?e the following (Collatz' hailstorm):> T(n):> c
:= 1;> while n != 1 do> c +:= 1;> if(n % 2 = 0) n /:= 2> else
n := n * 3 + 1> ? od;> return c % 10.> This is a ?ite
algorithm. Its output is always a decimal digit. Its> input
are the integers. Does it de?e a computable number? We do
not> know, because we do not know whether it terminates for
all n. Can we> determine in ?ite time whether it must be on
the list? Perhaps, our> mathematical knowledge is not good
enough to determine this. And there> are many more such ?ite
algorithms.I don't see that this observation solves our
dilemma. Non-terminating?ite algorithms may pad the list,
but if we can count _all_ ?itealgorithms, we can certainly
count the subset of ?ite algorithmswhich terminate for all
integer inputs.
===
... > For instance de?e the following
(Collatz' hailstorm): > T(n): > c := 1; > while n != 1 do > c
+:= 1; > if(n % 2 = 0) n /:= 2 > else n := n * 3 + 1 > ?> od;
> return c % 10. > This is a ?ite algorithm. Its output is
always a decimal digit. Its > input are the integers. Does it
de?e a computable number? We do not > know, because we do not
know whether it terminates for all n. Can we > determine in
?ite time whether it must be on the list? Perhaps, our >
mathematical knowledge is not good enough to determine this.
And there > are many more such ?ite algorithms. > I don't
see that this observation solves our dilemma. Non-terminating
> ?ite algorithms may pad the list, but if we can count _all_
?ite > algorithms, we can certainly count the subset of ?ite
algorithms > which terminate for all integer inputs.But is the
algorithm above non-terminating or not? I.e. should it beon
the list or not?-- dik t. winter, cwi, kruislaan 413, 1098 sj
amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn
amsterdam, nederland; http://www.cwi.nl/~dik/
===
>
_What_ is incorrect? > I think this has been clari?d. I
commented on your I think > you're thinking of. > Well,
*that's* confusing.Yup, isn't it? > Did you claim 
that
Richard was wrong when he > said that Edward Green was
thinking of, let's say, X? That is, Edward > was *not*
thinking of X? Or did you allege that Richard was wrong >
when he reported that he *thinks* Edward was thinking of X --
in fact, > Richard was *not* thinking that Edward was thinking
of X, but was > thinking something else, and he only
mistakenly reported that his > thoughts were about Edward and
his thinking of X?The ?st, I think. But I really do not know
what I am thinking of now. > I think I need a good
lie-down.Yes, a good idea.-- dik t. winter, cwi, kruislaan
413, 1098 sj amsterdam, nederland, +31205924131home:
bovenover 215, 1025 jn amsterdam, nederland;
http://www.cwi.nl/~dik/
===
Can a formula be found for the
numbers on the right given the value on the left?4 28 49 212
616 1218 420 1024 1225 427 628 1432 2436 3040 2044 2245 1048
3649 650 8Some simple observations: 8:4, 12:6, 20:10, 24:12,
28:14, 40:20, 44:22 follow 4:2 but 16:12, 32:24, 36:30 and
48:36 do not.18:4, 27:6, 45:10 follow 9:2 but 36:30 does
not16:12 and 32:24 and 48:36 seem to form a set. What about
64:48?9:2, 25:4 and 49:6 have value of sqrt(n)-1 since their
square roots are primes.Can anything de?ite be stated?
===
>
Can a formula be found for the numbers on the right given the
value on theleft? 4 2> 8 4> 9 2> 12 6> 16 12> 18 4> 20 10> 24
12> 25 4> 27 6> 28 14> 32 24> 36 30> 40 20> 44 22> 45 10> 48
36> 49 6> 50 8Yes, of course. You can ?d a polynomial of at
least degree 18 that wouldwork. However I do not think that
is what you want. You need to state thequestion more clearly.
Some simple observations: 8:4, 12:6, 20:10, 24:12, 28:14,
40:20, 44:22 follow 4:2 but 16:12, 32:24, 36:30 and 48:36 do
not. 18:4, 27:6, 45:10 follow 9:2 but 36:30 does not 16:12
and 32:24 and 48:36 seem to form a set. What about 64:48?
9:2, 25:4 and 49:6 have value of sqrt(n)-1 since their square
roots areprimes. Can anything de?ite be stated?>
===
>Can a
formula be found for the numbers on the right given the value
on the left?4 2>8 4>9 2>12 6>16 12>18 4>20 10>24 12>25 4>27
6>28 14>32 24>36 30>40 20>44 22>45 10>48 36>49 6>50 8Without
more knowledge, the answer is yes, but not in any
usefulway.For example, since you have given 19 points, one
may use Lagrangeinterpolation to ?d a polynomial f(x) of
degree 18 that will giveyou f(4)=2, f(8)=4, f(9)=2, etc. Have
the pairs(x_1,y_1) = (4,2)(x_2,y_2) = (8,4)(x_3,y_3) = (9,2) .
. .(x_19,y_19) = (50,8)and de?e f(x) as the sum of P_i(x),
i=1,...,19,whereP_i(x) = y_i* (product from k=1 to k=n,
skipping k=i)[(x-x_k)/(x_i-x_k)].But while that will give you
the answers you already have, it isunlikely to be what you
want. (A famous example from Mathematics made dif?ult, for
instance,points out that if you use that procedure to get a
polynomial thatwill correspond to the points1 22 43 84 16then
you will ?d that f(5) = 31).>Some simple observations: 8:4,
12:6, 20:10, 24:12, 28:14, 40:20, 44:22 follow 4:2How do they
follow, unless you are assuming some sort ofproportionality?
There is no reason to assume that if the k-thnumber is x_k,
then then k*l-th number will be x_k*l. So in what sensedo
they follow?> but 16:12, 32:24, 36:30 and 48:36 do not.18:4,
27:6, 45:10 follow 9:2 but 36:30 does not16:12 and 32:24 and
48:36 seem to form a set. What about 64:48?9:2, 25:4 and 49:6
have value of sqrt(n)-1 since their square roots are
primes.Can anything de?ite be stated?There are an in?ite
number of functions that will take the valuesyou specify at
the points you specify; they vary in terms of theirsimplicity
or obviousness. Without knowing more about the sequenceof
numbers you have or where they came from, what they are
supposed torepresent, etc, there is nothing de?ite that can
be stated. Just aswe cannot de?itely state that the number
that goes after 1, 2, 4,8, 16 is 32. (It is, if your sequence
is meant to represent subsequentpowers of 2 But if, say, the
n-th number gives you thenumber of areas into which a circle
gets dividied if you place n genericpoints in the
circumference and then draw all the lines that connectthem,
then the next number is 31).--
===
===========================================
===
====It's not denial. I'm just very selective about 
what
I accept as reality. Calvin (Calvin and
Hobbes)
===
======================================
===
=========Arturo Magidinmagidin@math.berkeley.edu===P
O S T>16 15 19 20 = 70 Oh dear. Shriner/Freemason Don Ocean
started calling me a pedophile<<<snipped a lot of psycho crap
from Shawn pedophile-boy Kabatoff>The following (courtesy of
Waxy.org) is sort of an unof?ial FAQexplaining the psychotic
nonsense posted to Usenet by Shawn DarylKabatoff AKA Dar, AKA
Probababbilities. And now AKA marcia andme.WARNING: Read below
before even thinking about responding to
thistwit.http://www.waxy.org/archive/2002/05/21/dar_
kaba.shtml#000643Usenet has the tendency to provide a public
forum for those who wouldnormally be scribbling in a closet.
For example, take Daryl ShawnKabatoff. For the last few
years, he's methodically gatheredstatistics from various
sources, ranging from local newspaperobituary pages to the
food court of the Saskatoon Midtown Plaza mall.With all the
raw data he's collected, he's attempting to prove 
dailythat
our full names are in mathematical harmony with our
birthdays.about, starting with calculations related to their
birthdate and fullnames, blending in whatever other personal
information about theirfamily members, spouses, birthplace,
and career he's been able tozealotry, and personal torment.
I've never seen anything like it.With all the prime numbers,
Fibonacci sequences and biblicalreferences, it's like reading
the notebooks of Maximillian Cohen andJohn Nash combined.
Unsurprisingly, several posts unfold to reveal ahistory of
painful mental illness. If you have some time, take a
look.I've detailed his posting history and a several sample
posts below. January 27, 1999 to July 5, 2000 as
Catsco@home.comDecember 9, 2000 to May 4, 2001 as
s.kabatoff@sk.sympatico.caOct 30, 2001 to Oct 31, 2001 as
kabatoff@the.link.caposts have been removed from Google
Groups archive)Selected Posts:Tessa Lynne SmithDastageer
Sakhizai and Helen SmithBrett David MakiAndrew Meredith
CottonAmanda Dawn NewtonMona Marie EtcheverryTony Peter
NusplLisa Charlene McMillanGrant Allyn Wood Comments scarier
still is that saskatoon is my hometown, though not my
currentresidence. and every single place he's mentioned in
his posts (mostnotably nervous harold's and the roastary)
were either places i'vebeen (as it's a small city of 
200K) or
hangouts, ie. the two placesout if they know of him, they (my
friends that is) being of thebroadway-centred slacker ilk.
myself, too, until i got out of there.eh, anyways. thought it
odd to see all this. midtown mall. i ate mymeals there, whilst
waiting several days in line for star wars episodeone, at the
theatre across the street.posted by andy raad on May 22, 2002
06:20 PMFascinating. It's like he's trying to take 
chaos and
bind it intowhatever rules he can ?d, religious, logical and
otherwise. Numbersand math have a reliable pattern, something
that can always be provento true or false. People and
religion do not. It reminds me of DarrenAronofsky's movie Pi.
It's the story of an paraniod genius who istrying to ?d a
pattern in Pi. A group that takes interest in hiswork is
convinced that the existence of Pi, a number whose
existencecan be proven but no quanti?d, is proof of the
existence of God.Kabatoff's hunt for patterns in something as
random as name selectionis a way to reconcile his deeply
logical thought process with hiscon?g religious
views.posted by matt on May 23, 2002 11:19 AMasking him if
he'd be willing to create a numerological analysis forme. I
also asked him if he had seen either Pi or A Beautiful Mind,
andwhat he thought of them. If he replies, I'll be sure to
post it.posted by Andy Baio on May 23, 2002 11:24 AMI baked
many pumpkin pies for Shawn (he likes pumpkin pies). I
rubbedpumpkin pie all over my breasts for him, and my breasts
turned orange.I am a pumpkin for Shawn.posted by Trisha
Blondie on July 24, 2002 10:41 PMUm, that's swell. So, 
you're
in love with him?posted by Andy Baio on July 25, 2002 07:10
AMShawn once went to a funeral for a Jehovah Witness that
shot himselfand the lemon tarts were very bad, they were not
only sour but wererubbery as well. Shawn said that the guy
was some kind of JehovahWitness prophet, he saw in advance
that the lemon tarts at his funeralwere to be very very bad,
and so he shot himself. Shawn said that henever ate pumpkin
pie at a funeral but would like to some day. Shawnlikes
pumpkin pie and so I have been practicing to make very
goodpumpkin pies.posted by Trisha Blondie on July 25, 2002
02:49 PMShawn said that the lemon tarts were sour, bitter and
rubbery.posted by Trisha Blondie on July 30, 2002 12:32 AMI
don't think this guy takes notes. I think he has Total
Recall, andit has driven him insane...posted by Todd Smith on
December 26, 2002 11:00 AMOh... I almost forgot... I didnt
spend thousands of dollars a daytormenting Daryl... We got a
deal on tormenting that ?cal year, itonly came to about
37cents a day....posted by Dr Claw on December 30, 2002 01:56
AMMr. Kabatoff attempts to portray himself as a victim, but in
fact heis a violent predatory pedophile who is well known to
his local lawenforcement. In his post to multiple newsgroups
with the subjectCollecting Mail For The Coming Anti-Christ,
he encourages mothers tosend him photos of their naked
daughters. Mr Kabatoff explains, IAnt-Christ) that were of
underage children unless the parent wassigning consent. He is
banned from virtually all the shopping mallsin his community
because he stalks young people and sexually harassesthem. He
has an extensive arrest record which includes
sexualmolestation charges. He's been hospitalized in mental
institutionsabout his contact with young girls in many posts.
Search newsgrouparchives for posts by him containing the word
nubile. As part of hisharrassment, he provides personal
details in a public forum, such asthe real names of real
children, in these and other posts. About onewanted her and
her sister
dead.http://groups.google.com/groups?q=Daryl+Shawn+Kabatoff+
dead+or+in+my+bed&hl=en&lr=&ie=UTF-8&selm=asqm35%24tjq5j%241%
40ID-136124.news.dfncis.de&rnuHe not only curses children and
prays for their death in his posts, healso enjoys attending
the funerals of young people: And so, sincenubile sweeties
are found in greatest abundance at the funerals ofhigh school
students, then it is the funerals of high school studentsthat
make the very very best funerals, especially if there is
food...I stuff my face (and my pockets) with all the good
food and look atall the pretty nubile sweeties and have the
time of my
life...http://groups.google.com/groups?q=Daryl+Shawn+Kabatoff
+nubile+sex&hl=en&lr=&ie=UTF-8&scoring=d&selm=LfXN8.63042%
24R53.25142039%40twister.socal.rr.com&rnum=1 Many of his
posts are sent to alt.teens.advice. However, he
liberallyspams, ?and crossposts his off-topic
threatening and offensivemissives to countless newsgroups.
Some people HAVE problems and somefolks ARE problems. Don't
dismiss Mr. Kabatoff as a harmless nut. Whenhe sends these
posts to any newgroup, please help by reporting him toI knew
of him when I was attending the University of
Saskatchewan.He'd hang out in the Arts computer lab and all
you'd see is screens ofnumbers racing by on his laptop. I
have an original copy of hisCollecting Mail for the Coming
Anti-Christ pamphlet, and have seenhim be hauled away by
campus security on more than one occasion. Myfriends and I
refer to him as Crazy Number Man.I've been posting to (and
about) Shawn for over two years with biggaps in between. He
has seen Pi and didn't like it and didn't think 
itresembled
him at all. (Wrong, it ?s him to a tee) He doesn't havetotal
recall and has stated that he travels with a lap top to
notateitems. Also, he uses cut n' paste a lot if you read all
the waythrough his ramblings. He is anti-social as shown by
his angrystatements towards those who, by his own admission,
have been kind(but not kind enough) to him. Still, he's
intelligent and seems to beable to take a joke on occassion.
That's where I came in. ALOHA if it comes from anyone not
already in my addressbook. I'll never even see it)
===
>
c6
===
===
I could
use some helpful hints on this problemLet G be a group which
operates transitively on a set S. Let H be thestabilizer of
s-sub 0 in S. Let g(s-sub 1, s-sub2) = (gs1, gs2) be
theaction of G on S x S. Show the bijection between double
cosets of H in G andG-orbits of S x S.I got this far with the
problem.I want to show O: G= union from i = 1 to n of H gi H
.88G-orbits. I know H =Stab(s0) = {g in S | gs0 = s0}, and
also that there exist g s/t gs1 = s2 forall s1, s2 in S. But
I don't understand what the orbits look like, ie I 
don't
understand how the elements of SxS fall into the G -orbits.
If I did, Ithink I would be able to ?d representatives from
each of the orbits tocorrespond to each HgiH, which would
prove the bijection.Steven
===
> I could use some helpful
hints on this problem> Let G be a group which operates
transitively on a set S. Let H be the> stabilizer of s-sub 0
in S. Let g(s-sub 1, s-sub2) = (gs1, gs2) be the> action of G
on S x S. Show the bijection between double cosets of H in G
and> G-orbits of S x S. I got this far with the problem. I
want to show O: G= union from i = 1 to n of H gi H G-orbits.
I know H => Stab(s0) = {g in S | gs0 = s0}, and also that
there exist g s/t gs1 = s2 for> all s1, s2 in S. But I don't
understand what the orbits look like, ie I don'> t understand
how the elements of SxS fall into the G -orbits. If I did, I>
think I would be able to ?d representatives from each of the
orbits to> correspond to each HgiH, which would prove the
bijection. StevenFirst, let's take a particular example.
Consider G= the group o?tegers under additions, which we
will call group-Z. ForS, let us take the set of integers,
set-Z. For the action,for m in group-Z and k in set-Z, let m
act on k to give theelement m+k in set-Z under normal
addition. Thus, group-Zacts on set-Z by translations (I know
it is cumbersome, but Iwant to emphasize the fact that we are
using Z is two distinct ways).Now SxS is pairs of elements
(k,l), where both k and l are inset-Z. For m in group-Z, m
acts on (k,l) producting (m+k, m+l).Ok, what are the orbits
in SxS?There is another hint several spacings down, but I am
addingspacing so that you might think about matters at this
pointbefore proceeding to the next hint.Ok, still with me.
Here is another hint. First, can we ?drepresentatives for
all orbits in SxS. Remember, we need toinvolve s 0. I had to
think about this for a bit, butthen I relazed that all
elements form (s 0, alpha s 0)are representatives of all
orbits, were alpha is an element of G.That is, any element of
SxS can be written as g (s 0, alpha s 0) = (g s 0, g alpha s
0).Ok, one more push in the right direction below, but take a
momentto think about this.Ok, so presumably we know that all
orbits can be represented byelements of the form (s 0, alpha
s 0), where s 0 is the elementselected at the beginning of
this discussing, and alpha ranges overall elements in G. The
question is, when do (s 0, alpha s 0) and (s 0, beta s
0)represent the same orbit. Well, for them to represent the
sameorbit, they must be in the same orbit (right? in the
discussionabove, the elements ?represented' the orbits by
being an elementof the orbit, so that all other elements of
the orbits could beobtained by applying the group action).
Now, for theseto be in the same orbit, this requires that
there exists g in Gs.t. g s 0 = s 0 and simultaneously g
alpha s 0 = beta s 0.Ok, can you do some algebra to learn
something about g?Best wishes, Mike
===
can someone help me
with these practice problems?1. Let f:X-->D for some
connected metric space X and some discretespace D. Show f is
continuous iff f is constant.2. Let f:X-->R for some metric
space X. If f is continuous, then |f|is continuous.On the
?st one I got one direction (namely <==) which was very
easy.I might just need a slight nudge in the right direction
on each, theseFSB
===
<usenetusr@nyc.rr.com> grava .88 la
saucisse et au marteau:> can someone help me with these
practice problems?> 1. Let f:X-->D for some connected
metric space X and some discrete> space D. Show f is
continuous iff f is constant.> 2. Let f:X-->R for some metric
space X. If f is continuous, then |f|> is continuous.For the
other direction, use the Intermediate value theorem (just
hopingthis is the way it is called in English)2. I'd say that
on a neighborhood of any point where the fonction isnonzero,
the function is of a constant sign. If there is a such
thatf(a) = 0, you can divide the neighbourhood in two
parts.-- Nicolas
===
[...]|> It's hard to do the kind of
reality-testing that Pat referred to when|> you don't have
people you let help you. ||I get help from mathematicians,
much of which I don't post about,|though I have posted about
some, but consistently when it comes time|to deliver, they
run away.Is any of them in a position to help you test the
basic issue, whetheryou have great mathematical discoveries,
or just some amateur-levelexperimentation? Not as far as I
can see. If they said the latter, wouldyou actually believe
them? I can't think of any way to test the qualityof your
work, that both you and most other people would be willing
toaccept the results of. Wouldn't it help a lot if there was
such a way?[...]|> They don't consider you to be a world
class mathematician. For a lot|> of mathematicians, the point
they start trying to ?ure out how to|> get you to leave them
alone is when they realize that you *think* of|> yourself as
a world class mathematician even though you're not.||Why
would they think that?||Basically I do a basic presentation,
often I do mention that I'm an|independent researcher.||What
makes you think that they would begin thinking I'm someone
who|thinks of himself as a world class mathematician?I'll get
to that shortly.|Usually I mention that I'm *not* a
mathematician, and am in need of|help or guidance.By
mathematician I didn't mean professional mathematician in
thesense of someone who gets paid to do mathematics.
Obviously you don'tthink you're getting paid to do 
your
stuff, yet. But you think whatyou're doing is worth getting
paid for.|What I ?d troubling is how *easily* you make
statements as if you|know!!!I likewise.You make
generalizations about professional mathematicians. I am
justdescribing what is common. Your idea is as far as I can
see based mainly onspeculations about what would cause people
to treat great work as if it werenothing extraordinary. Mine
is based on experience with how professionalsactually deal
with amateurs. For your version to be true, they would
havesecretly to realize you were onto something big. For mine
to be true, theyjust have to realize that it is much like
other things that they get on aroutine basis.|Please tell the
newsgroups if you've talked to Barry Mazur, Andrew|Granville
or any of the others to get your assessment.I haven't written
or talked to Mazur or Granville about you speci?ally.although
I don't remember getting into his reasons for not feeling
likeopening up further dialog with you.I am aware of
Granville's dealings with one fellow (of my
acquaintance,although not by choice) who repeatedly and
mistakenly thought he'd provencase 1 of Fermat's Last
Theorem. I was told he had published papers(although not
claiming to prove case 1). This seemed to be a little bit
inhis favor. The ?st time he thought he proved case 1, I
think there wasstill some hope among people acquainted with
him that he would turn outjust to have gotten overexcited
temporarily, but he kept slipping intothe same kind of error.
So I do know something about what Granville thinksof as
promising and not so promising.I would tend to expect Mazur
to be more tolerant than the averageprofessional for a little
eccentricity, and to be willing to take time forrandom
strangers. On the other hand he is adept at politely wrapping
thingsup so he can move along to the next thing he wants to
do. He does a varietyof interesting things. (Have a look at
his popular book about visualizingnumbers. I think it shows
more than minimal interest in the general public.)he had one
of the greatest mathematical discoveries of all time (and
didn't),but I think he would know enough not to make the
kinds of mistakes I'vemade in doing so, like arguing with you
on points where you evidently havemade up your mind
de?itely.?d this to be a common occurrence, from experience
with other examples ofamateurs sending in their work for
comment. Some professionals don't botherwith amateurs in any
way. A few will correspond with amateurs who they knowhave
the idea that they've squared the circle or something like
that. A lot,though, draw the line between the two, trying to
maintain friendly relationswith the general public in some
way, but avoiding arguing with the amateurswho are going to
think the professionals are prejudiced if they give
theirhonest opinion.Obviously in individual cases they might
correspond with some amateur fora bit and then stop quickly
for some other reason, but ?ding that theamateur thinks of
himself as the next Gauss (or as having discovered thenext
fundamental theorem of arithmetic, as it were), and just has
a littletrouble satisfying bureaucratic requirements like
writing up his proof topublication standards, or is *only*
having trouble because all the othermathematicians are
prejudiced against him, often ends the conversation
verysoon.You make it fairly obvious that you think of
yourself as havingworld-class mathematical abilities and
accomplishments. Here's justone example:|I'm afraid of 
having
made one of the greatest discoveries in the history
of|mathematics and having to live with the consequences.One
of the greatest discoveries in the history of
mathematics!We've seen this kind of posting from you many
times.What's more, we've seen this kind of pattern of 
writing
over and overfrom others. There's a pretty clean divide
between the amateurs who areamenable to feedback to a normal
degree, and are aware of being at anamateur level, and the
amateurs who think they are well above the peoplethey're
writing to in mathematical ability, and explain the feedback
theyget as the product of small, fearful, or envious minds.
Dealing with theformer can be a lot of fun, and dealing with
the latter tends mainly toconvince the amateur that one is
yet another person who fails to appreciatefailed attempt to
prove Fermat for even exponents only, this is cute. WhenU.S.,
he is unlikely to get a reply.Do you think you succeed in
passing yourself off as just a humble ordinaryamateur with an
interest in algebraic number theory? I don't doubt you cando
this for awhile. But I doubt you ever keep it up for long.
The drama ofthinking of yourself as a historic ?ure in the
history of mathematics isjust a *little* hard to conceal.
[...]|Also I've received replies from other notables and
famous people,|which are not relevant, so I don't discuss
them.Ok.|> But this is also not so unusual. Nearly any
educated person who bothers|> to write to people like this,
and is reasonably patient, sooner or later|> gets a response,
and not just a form letter. Not all famous people ever|>
answer, but enough do that just the fact you got to
correspond with them|> a little doesn't mean very much.||I
never said it did.Well, okay, but why did you write the
following, then?| I don't know how one sets about testing a
standpoint like his| empirically. So much of it hinges on his
evaluation of his own| work as being valid and valuable
mathematics. As far as I can| tell, the world continuing to
act just like he wasn't a world class| mathematician but just
a guy who thinks he is, is consistent| with his beliefs. He
just concludes his work is great but people| are choosing to
pretend it isn't.You replied:|Come on Keith Ramsay, 
I've sent
and received interesting replies from|Andrew Granville, Barry
Mazur, and others.[...]So what are you claiming is the
signi?ance of it? Come on what?What's the conclusion 
I'm
supposed to draw? You made it sound like youthought the fact
they corresponded a little with you served as some kindof
rebuttal to what I was saying.What they're doing is not what
they do with work of world class quality.What they're doing
is what they do either with ordinary interestedamateurs, or
with amateurs like circle squarers. There's nothing wrongwith
being an amateur, of course, but if you imagine that you've on
thelevel of John Nash, mathematically, when you're closer to
Russell Crowe[*],you're bound to be disappointed one way or
another.[* I have no idea how much math he knows, but I
assume it's somewhere inthe bottom 99.9% of the
population.][...]|You're leaving out the fact that I
questioned whether or not Iraq had|weapons of mass
destruction back in November of last year.Well, IMO that's a
point in your personal favor, no sarcasm intended,but that is
of course a discussion for another newsgroup.Keith
Ramsay
===
|> So you want the credibility of sources to be
*more* dependent|> on their credentials and occupation than
they are now?||Erik Max Francis is a computer
programmer.||Trying to defend him as if he were a
mathematician or otherwise an|expert over the ?lds he covers
is just another example of math|society's rather sad ability
to make up rules as it goes along.Not at all. I said what I
meant to say. I asked a question.I haven't seen an answer to
it.I don't know what Erik Max Francis has on his web site,
orwhether it's valid or not. I just think Tom Potter has
chosena peculiar way to criticize it. Questioning this
criticism is notspecial treatment I reserve for
mathematicians and experts.I would think you, of all people,
would have a problem with thenotion that in order to judge
whether he, you, me, Tom Potter,or anyone else is credible,
it's ever enough just to check theirjob titles and levels of
education.On the one hand, have many times have you
complained abouthow people will look at credentials and
positions, and conclude thatmathematicians who have them
should be believed just becauseof them? Should we say, Keith
Ramsay has a PhD in algebraicnumber theory, so whatever he
says about algebraic numbertheory must be correct? Well, I
would think it would lend*some* authority to it. But I think
people should remember thatthe word of authorities is not the
bottom line. Surely you agreewith that? You wouldn't want them
to start believing bulljust because someone with a PhD
said it, would you?On the other hand, should we say things
like, James Harrisdoes not have a PhD, and his BA is not in
mathematics, and hedoes [?l in the blank] for a living, so
we shouldn't listen to what hesays about algebraic numbers? I
don't think so. Don't you agreethat this is would be a 
bit
unfair? Maybe it's less likely that someonewith less
education knows what they're talking about, but 
oneshouldn't
assume that they don't.So I'll ask you too: do you 
really
want for the credibility of sourcesto depend more on
credentials and occupation than they do now?Keith
Ramsay
===
|> [...]|> |> What third rate California college?
Who rated it? What criteria?|> |>|> |Hey Wormley,|> |as you
use this programmer's web site as your primary rederence,|>
|it seems to me that you should know what college your
resident expert|> |attended.|> |> So you want the credibility
of sources to be *more* dependent|> on their credentials and
occupation than they are now?||It is interesting to see that
Keith Ramsay |agrees with Sam Wormly, |that if someone puts
up a web site that demeans folks |who have made a statement,
or statements, |that run counter to conventional wisdom,
|that such a person quali?s as an expert in the ?ld, |even
though they have not demonstrated their competence |in any
way.Don't put words in my mouth. I said what I meant to
say.You've been making heavy weathering of someone's 
levelof
education and occupation. Do you really think theseshould be
used *more heavily* to judge the credibility ofpeople's web
pages? I just think this is a somewhatstrange basis for
criticizing. If you have some better reasonto doubt whatever
it is he claims, then why treat where hewent to college as
such a big deal?[...]|I assert that having (Or being ) an
asshole |does not qualify one as an expert.||Apparently Keith
Ramsay and Sam Wormly feel otherwise.Again, don't put words in
my mouth.Keith Ramsay
===
|The key to being a crank is that you
can't know *nothing*...you have|to know *something*. For
example, you know that Mandarin literature|comes from China,
you might well know that the Mandarin's were the|bureaucratic
class, and you might know that Confucius was highly|revered
especially by the Mandarins.||I think that small bit of
knowledge (which just about exhausts mine, I|think) is plenty
to be a crank about Mandarin literature. ;)I can see it
now....Chinese communist Mandarins are subverting
freedom-lovingAmericans with their pagan/Maoist/Confucian
doctrines ofduty and respect for authority....A Google search
for communist mandarins turns up 22 hits,so there you go.Keith
Ramsay
===
> As a sidenote, I've not said that my proofs are
irrefutable proofs, as> I've said they are proofs. At other
times I've said that I have> proofs, and that since they are
proofs they are irrefutable.Ah, but they aren't proofs at
all.
===
> nice post James, good general FAQ or introductory
material about posting.> Herc> now will anyone look at MY
claim?Just look at the supporters you attract, James. Doesn't
that tell you something?> Webmasters help the TRUEman by
joining www.theBanner.net > Current:1 Goal:1000What can I
say?
===
WARNING ALERTAt New York's Kennedy airport today, an
individuallater discovered to be a public school teacher was
arrested tryingto board a ?while in possession of a
ruler, a protractor, asetsquare, a slide rule, and a
calculator.At a morning press conference, Attorney general
John Ashcroft said hebelieves the man is a member of the
notorious al-gebra movement.He is being charged by the FBI
with carrying weapons of mathinstruction. Al-gebra is a
fearsome cult, Ashcroft said.They desire average solutions by
means and extremes, andsometimes go off on tangents in a
search of absolutevalue. They use secret codenames likex and
y and refer tothemselves as unknowns, but we have determined
they belongto a common denominator of the axis of medieval
with coordinatesin every country.As the Greek philanderer
Isosceles used to say, there are 3sides to every triangle,
Ashcroft declared. When asked to comment onthe arrest,
President Bush said, If God had wanted us to have
betterweapons of math instruction, He would have given us
more ?gers andtoes. I am grati?d that our government has
given us a sinethat it is intent on protracting us from these
math-dogs who are willingto disintegrate us with calculus
disregard. Murky statisticians love toin?lane on every
sphere of in?, the President said,adding: Under the
circumferences, we must differentiate their root,make our
point, and draw the line. President Bush warned,These weapons
of math instruction have the potential to decimaleverything in
their math on a scale never before seen unless wefacts of
vertex. Attorney General Ashcroft said, As our GreatLeader
would say,read my ellipse. Here is one principle he
isuncertainty of: though they continue to multiply, their
days arenumbered as the hypotenuse tightens around their
necks.
===
>Would somebody please explain the
notation___>lim f(x)>x->clim f(x)>~~~>x->c> These are
presumably the lim sup and the lim inf.> The most intuitive
explanation for what the lim sup is is> probably this:> You
know that lim_{x->c} f(x) = L if and only if f(x_n) -> L> for
every sequence x_n such that x_n <> c but x_n -> c.> The lim
sup of f(x) as x -> c is the _largest_ lim f(x_n)> where x_n
-> c (and x_n <> c), while the lim inf is the> smallest such
lim f(x_n). (So lim_{x->c} exists if> and only if the lim sup
equals the lim inf; if and only> if there is only one such lim
f(x_n).)> Alternately, you can de?e> lim sup_{x->c} f(x)
= lim_{d ->0+} sup{f(t) : 0 < |t - c| < d}.
===
>Would
somebody please explain the notation>___>lim f(x)>x->clim
f(x)>~~~>x->c These are presumably the lim sup and the lim
inf. The most intuitive explanation for what the lim sup is
is> probably this: You know that lim_{x->c} f(x) = L if and
only if f(x_n) -> L> for every sequence x_n such that x_n <>
c but x_n -> c.> The lim sup of f(x) as x -> c is the
_largest_ lim f(x_n)> where x_n -> c (and x_n <> c), while
the lim inf is the> smallest such lim f(x_n). (So lim_{x->c}
exists if> and only if the lim sup equals the lim inf; if and
only> if there is only one such lim f(x_n).) Alternately, you
can de?e> lim sup_{x->c} f(x) = lim_{d ->0+} sup{f(t) : 0 <
|t - c| < d}. The reason these things are useful is that
_any_ function> has a lim sup and a lim inf at _every_ point
(if we allow> plus and minus in?ity as a value), while not
every function> has a limit at every point.>Seems same
discussion can be taylored for limsup's of
sequences.
===
Anyone read that book I mentioned in the
subject? I am reading it and havegotten up to about page 50.
It is my ?st time doing knot theory and itseems thus far
that the exercises in that book are nasty. For example inthe
?st chapter, called Introduction, there is Exercise 1.9
which asksto show that the knot in the ?ure is composite.
Wow, I don't think I caneven copy that scribbled knot
accurately let alone show such a proof. Howare people
handling this exercise. In fact I have been unable to do any
ofthe exercises so far in the book. Is this book known to
have unrealisticexercises considering the books title? Even
the second question in thebook asks to show that there are no
two-crossing nontrivial knots. Seemskind of advanced for a
second question in the book? Your assistance insetting me in
the right direction to solving some of these probs would
beappreciated. thanks.
===
I read a bit about this group and
wonder if there is a natural way tocount in it. In case you
dont know what it is:Let X be the set of all positive
integers and let S be the group ofall permutations of X; let
F be the subgroup of S consisting of allpermutations that
move only ?itely many elements of X; the In?iteAlternating
group is the subgroup of F generated by all 3-cyclesthe main
thing I'm wondering is if there's a natural and 
intuitiveway
to count in it, complete with a natural ?st element and
anatural next for each previous........ or if one is forced
to usecontrived methods, like the set of rationals.Please
forgive my ignorance, as I am far inferior to you
inmathematical knowledge. I assure you that I amyour humble
servantSniz Pilbora) How many digits are there in 1000!
(generalize, how many digits in n!)b) What is the digit in
the kth position of 1000!. for any k. (generalize,
samequestion but for n! instead of 1000!)thanks!
===
I'll try
to help you with your ?st question.> a) How many digits are
there in 1000! (generalize, how many digits in n!)In 1000!,
there are 2568 decimal digits. In general the number of
decimaldigits in n! is just ceiling( log_10(n!) ), where
log_10 denotes thedecimal (i.e., base 10) logarithm. Is that
OK, or do you want an expressionwhich avoids computing n! (or
a gamma function) perhaps? (If the latter, itmight be tricky
to get the expression exactly right.)David> b) What is the
digit in the kth position of 1000!. for any k.> (generalize,
same question but for n! instead of 1000!)
===
Given, where x
is in the ring of algebraic integers, I've shown
thefactorization(5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) =
49(300125 x^3 - 18375 x^2 - 360 x + 22)where b_3(x) = a_3(x) -
3 and the a's are roots ofa^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 -
147 x^2 + 3x)so when x=0, a_1(0) = a_2(0) = b_3(0) = 0.I'm
curious about the mental processes that allow *some* of you
toclaim that 49 divides off as a *variable* dependent on x,
so I'mgiving another opportunity for you to speak your
minds.To my knowledge, in the history of mathematics, no one
has everpresented such a proposition, so it is a unique one,
and I must saythat I'm intrigued.Speak your minds.James
Harris
===
bcc bcc PZ: If Yilmaz et al. are right that static
weak ?ld n-body Einstein GR solutions don't exist, then this
may turn out to be a real problem after all. JS: I think the
exponential metric in the isotropic radial coordinate has
serious problems independent of the precise metric model,
Yilmaz or Einstein because of topological reasons. PZ: So if
you are right you don't get full coordinate generality with
an exponential metric?JS: I do not understand coordinate
generality. I never saw that term before.This is in use by
the Wolfson group at Oxford. It's used as an alternative to
themore loaded general covariance which is traditionally
tangled up with generalrelativity.The erroneous con? of
formal covariance with physical relativity is almost
afounding principle of Einsteinan physics (later abandoned by
Einstein).There are several legitimate issues here. Why don't
you make a dictionary of these key terms and give
bestde?itions you can to avoid confusion. That book I
mentioned last time is very usefulin this regard BTW. More on
that anon. My debate with Hal on PV is really in relation to
the thepseudo-group of passive LOCAL coordinate
transformations at a ?ed point P and of world crystal.This
is not same as active diffeomorphisms P -> P' =/= P . The
active and passive transformations mustbe made mutually
consistent and this may solve the Kretchmann issue? Also
there is the issue of whetheror not the different points P
are distinguishable and what is an observable in GR? There is
the Einstein holeproblem. BTW Joy Christian is a male. Active
diffeomorphism invariants are NONLOCAL - a problem
ininterpretation. There is no consensus on these deep issues
and others among the Pundits. It's almost as bad asthe wars
over the interpretation of quantum theory.Do you mean general
coordinate transformations GCT? Be more precise if you
useplain English.The standard terminology is loaded. When I
say coordinate generality, I mean coordinategenerality: the
*desideratum* that laws should be formulated in such a manner
that their formdoes not depend on the particular choice of
coordinate system. Hence the use of coordinate-free devices
such as tensorsIt is not clear to me why a modern theory of
gravitation *must* be formulated in such a manner-- other
than as an expression of physical general relativity of
motion, which I contend doesnot exist.If GCT is what you mean
the answer is NO. What I mean is thatthe manifold looks
pathological and unphysical with at least a countable in?ity
of coordinatepatches outside the turning point r* = GM/c^2 for
curvature radial coordinate,which is analogous to event
horizon in Einstein's GR where there are onlyTWO patches
outside r* = 2GM/c^2 in that case (Einstein-Rosen Bridge,
i.e. non-traversable wormholein non-exotic vacuum case Ruv =
0 everywhere-when. It is clear to me that Hal is not really
thinkingabout the topology and differential geometry in his
naive engineering approach.OK, so you are saying that there
is an unavoidable pathological discontinuity in the
exponential PVsolution for a point mass?If by point mass you
mean taking the vacuum solution to the max yes. In the case
of GR I mean vacuum all the way i.e. solutionsof Ruv = 0 with
wormhole global topology of source Mass without mass (JA
Wheeler). And for PV what would correspond to that.I do not
think Dicke knew the differential geometry when he introduced
the exponential metric ~ 1961?I wasn't even aware that a
manifold was de?ed in PV. A physical metric, yes; but a
manifold?This is not a curved spacetime theory as far as I am
aware. The model is a polarizable vacuum withphysical rubber
rods and clocks.The problem is that Hal is completely obscure
to my mind on the fundamental world view of his model. He uses
metric notation after all? PZ: Although in the alternative
paradigm, general covariance looks more like a mathematical
fetish, since physical general relativity is absent.JS: There
is a lot about all this in Physics Meets Philosophy at the
Planck Scale Callender & Huggett Cambridge Press2001. I
suggest that we temporarily cease this line of inquiry until
we both digest what is in that book - some really GOOD
STUFF!OK, I've ordered it. But why stop the press?Because
those guys are pretty smart and have thought through a lot of
the issues you are interested in. So it's time to catch up.The
book was written prior to the realization of the new cosmology
of dark energy/matter - there is no mention of that.
However,it is good background stuff by hip philosophers and
some top physicists. Mathematician (in physicist's clothing)
John Baez also has a good n-categories and how they may make
an interesting formal connection between GR and QM. On the
other hand I see a lot of conceptual ?n the thinking of
the Pundits in both Q Gravity and M-Theory, one of which
being that they all assume Lp is a constant and not a
variable where perhapsLp*/Lp ~ e^(metric engineering control
parameter?)Another is that none of them seem to have read
P.W. Anderson's idea More is different and how it applies to
quantum measurement problem for example. Nevertheless, there
are many good relevant insights in the book.What's any of
that got to do with what I am talking about?I am talking
about internal tensions within orthodox GR.JS: We are also
talking about Hal's PV and also the book does deal also with
the internal tensions. Also my focusis how to combine quantum
theory and GR in order to solve the important real problems in
physics today:1. What is the Universe made of?2. What is the
physical nature of consciousness?3. How do we achieve the
kind of metric engineering we see in the UFO
observations?Studying the internal consistency of this or
that theory is secondary to these objectives.Such study may
well be necessary however. That seems to be so. I have
essentially hadmy eyes on this Golden Ring for 50 years and I
want to get some satisfaction!
:-)http://www.?dmidis.com/listen.go/589So does Hal and that
is why I am not letting him rest on the issues. PZ: Also, the
metric is not the ?ld; the tensor potential phi_uv represents
the physical ?ld and the gravitational-inertial metric is
derived from it. Non-linear coordinate transformations play a
fundamentally different role in this alternative model.I am
talking here about Yilmaz. JS: I mean there are an in?ity of
isotropic coordinate patches outside the turning point
boundary at GM/c^2 for a single curvature coordinate. In
Einstein's GR this ratio is only 2:1, i.e. 2 coordinate
patches outside the event horizon at 2GM/c^2 in the
Penrose-Kruskal diagram with 4 coordinate patches covering
the entire vacuum manifold.It's still interesting to me that
a coordinate discontinuity was originally mistaken for
aphysical event horizon.Even if you are right that the PV
solution is pathological, this does not necessarily applyto
Yilmaz's phi_uv. In Yilmaz's theory it is phi_uv that 
is
physically fundamental, whilethe exponential metric is
secondary and derivative.JS: Perhaps. Just what is the Yilmaz
theory in your understanding? I mean what is its world
view?What is the physical picture behind the obscure
formalism? PZ: Yet at the same time I think you get rid of
all the tricky properties of event horizons, since you get a
smooth solution for a point mass with no lightcone in?n
boundariesJS: There seems to be observational evidence of
event horizons? I am not up on the latest on this. But I sure
get the impression thatcompetent people like Martin Rees are
pretty con?ent on that score?JS: No because you still have
the turning point wheredr(isotropic)/dr(curvature) has a
critical point passing through zero and changing sign. This
acts spatially somewhat like an event horizon, i.e.dR >
in?ity at the turning point.dR = [1 -
GM/c^2r(isotropic)]^-1dr(curvature) TURNING POINTdT =
e^-GM/c^2r(isotropic) dt NO EVENT HORIZONThat is, Einstein's
event horizon is replaced by turning point in Hal's model.But
in any case there is no fundamental reason in PV for insisting
that every smoothcoordinate system is good.JS: This is not the
key point. I am talking about Hal's speci? SSS PV model.PZ:
Look, once general relativity is out of the picture, dogmatic
insistence on general covariancebegins to look like a
mathematical fetish. I see this as an example of
irrationality in contemporaryphysics.JS: The problem is
deeper than that. You cannot throw away differential
geometry. The problem is that PV's rules of the gameare
nebulous and shifting.PZ: Didn't Cartan produce a general
covariant version of Newtonian theory? Can't you do all
themetric tensor stuff within a purely Newtonian framework?
The metric tensor description isa mathematical truism. It
doesn't apply only to Einsteinian physics.JS: That's 
why I
brought up the distinction between the local pseudo-group of
coordinate transformations at a single P and theactive P ->
P' =/= P diffeomorphisms. That distinction may be important
in posing the relevant question here.There is the issue of
the relation of map to territory and even what is the
territory?PZ: If you go to the Newtonian limit in GR (if
there is such a limit), what do you get? You geta metric
tensor description of the Newtonian ?ld with non-vanishing
Riemann curvature(since there are still tidal forces).JS: The
Newtonian limit of GR has 2 aspects. Most important is c ->
in?ity so that we have Galilean relativity.Next is that mass
density is not so large that radii of curvature are too small
relative to scale of the measurement.Obviously the metric
interpretation will still hold although it becomes much less
rich.PZ: You could let c --> in?ity; then you get
instantaneous action at a distance.The fact that you re-write
the metric tensor for non-linear coordinate transformations
is, in andof itself, a mathematical truism. That you can
represent an inertial ?ld with such a transformedg_uv is in
itself trivial, and I think this can also be done in
Newtonian theory as well as ingeneralized SR.JS: The
arbitrary nonlinear coordinate transformation means a dust
cloud radars moving in arbitrary non-geodesic paths by ?ing
rockets and communicating with each other ontheir mutual
observations of the same non micro-quantum phenomena (so that
Heisenberg incompatibility does notget in the way).PZ: It's
just that this suggests a tensor theory of the permanent
gravitational ?ld *conceived as aphysical ?ld* -- which is
exactly what Yilmaz is proposing.JS: There is something very
ill-posed in what you are saying but I really can't at this
moment get a handle on it.It has an ineffable vagueness to my
mind. Again you do not seem to distinguish the two different
ideas of the non-tensorg-force at the connection level and
tidal curvature as certain tensor combinations of the partial
slopes of the g-force connection ?ld.PZ: This is also the
approach taken in bimetric theories, which distinguish
between the ?ckgroundgeometry (Minkowski metric) and the
*physical* metric ?ld.JS: Physically what is a bimetric
theory? How can you observationally tell if the world is mono
or bi?There is no dT = 0, but there is dR --> in?ity in Hal's
model at the turning point.And he has an in?ity of branches
not just two like Einstein has. It's a mess!I call it the
Medusa Manifold.PZ: If you're right I have to admit that this
has to be taken into account in evaluating PV.JS:r(curvature)
= K^1/2r(isotropic)This comes from angular part of PV
metricK(r(isotropic)^2[(dtheta)^2 +
sin^2(theta)(dphi)^2]inds^2 = K^-1(cdt^2) - K(dx^2 + dy^2 +
dz^2)IFdx^2 + dy^2 + dz^2 = dr(isotropic)^2 +
r(isotropic)^2[(dtheta)^2 + sin^2(theta)(dphi)^2]Here I am
using Ibison's argument as I recall it.This leads
toK^1/2r(isotropic) = r(curvature)more than a year ago.In the
PV model K = e^2GM/c^2r(isotropic)This is in effect an in?ite
order polynomial for the many roots r(isotropic) corresponding
to one value for r(curvature).On the other hand, I think you
can argue that you must really usedx^2 + dy^2 + dz^2 =
dr(curvature)^2 + r(curvature)^2[(dtheta)^2 +
sin^2(theta)(dphi)^2]because 4pir(curvature)^2 is area of
concentric sphere by de?ition.Then Hal has a weird looking
spatial metrice^2GM/c^2r(isotropic){dr(curvature)^2 +
r(curvature)^2[(dtheta)^2 + sin^2(theta)(dphi)^2]}and the
issue is, what is the functional dependence between
r(curvature) and r(isotropic).The point is that Hal's metric
form is ambiguous and ill-posed. It's like the tar patch in
Breire Rabbit.But I think my ?st argument was basically
Ibison's argument of over year or two ago? That is, in any
SSS metric theory look at the relation between isotropic and
curvature radial coordinates. You get a polynomial of degree
N with the curvature radial coordinate as a control
parameter. The N roots for isotropic r, at least when real,
each de?e a coordinate patch. In the exponential case that
Hal uses in his PV model, N --> in?ity. In Einstein's GR N =
2. PZ: OK, this is certainly a legitimate kind of criticism
and goes to the technical details, so it cannot be
ignored.JS: Hal has ignored it for more than a year. I
published all this inSpace-Time and Beyond II a year
ago.Hal?Hal Puthoff's reply to Zielinski's question
was:Jack's (mis)interpretation comes from trying to force ?
the PV non-curved-space results into standard GR curved-space
modeling, which takes coordinate-choice pathologies (e.g.,
event horizons) as physical. Ibison went to a great deal of
effort to educate Jack on this with a long attachment replete
with spacetime graphs, etc., but Jack just blew it off because
it didn't agree with his view as not even wrong.JS: Why not
educate Misner, Thorne and Wheeler, or Roger Penrose, Stephen
Hawking and Martin Rees on why event horizons are not
physical? Good luck. ;-) The point is that I know what GR
means. I have no idea what PV means from the kinds of vague
statements I see above? Hal seems to have a metric without
metric theory. He uses a metric formalism, but refuses to use
the differential geometry that it comes from. Again, I simply
cannot understand in a coherent way what Hal's1. Physical
world view is in PV?2. What mathematical rules he is using?I
mean if you do not inquire deeply into the meaning of Hal's
Tables I & II and things like the meaning of r and if you
ignore topology and basic metric geometry it looks like GR a
little in the large r limit. But that is simply not good
enough. It is not even theoretical physics as I understand
it. Even if you look at Hal's action formalism, at some point
one must ask what is r? I mean when GM/c^2r is no longer a
small dimensionless number.HP: So what else is new?! Jack
really can quote the dogma (with expertise, I might add, to
cut him some slack), but doesn't seem to understand where it
comes from, what it represents, how the underlying reality
can be modeled from many POVs (as you are showing
excellently), etc.Hal JS: I wish Hal would explain the basic
assumptions and physical picture of what he thinks he is
doing in PV in a way comparable to how Einstein explained
what he was doing. Also I wish he would show how his math
model leads to any interesting metric engineering
applications, which is really his main purpose.For the
record, did I correctly remember Ibison's formular(curvature)
= e^GM/c^2r(isotropic)r(isotropic) ?That is,ds^2 =
e^-2GM/c^2r(isotropic)(cdt)^2 -
e^2GM/c^2r(isotropic)dr(isotropic)^2 +
r^2(curvature)[(dtheta)^2 + sin^2(theta)(dphi)^2]Do you
accept that? If not, what is your formula in spherical
coordinate form? PZ: Why an exponential metric should create
such purely mathematical problems is, however, intuitively
not obvious (at least to me) -- especially when it seems to
solve so many others.JS: Why is not interesting really here.
Simple fact is that it's algebraically trivial - 11 grade
high school pre-calculus math. On Gravity Lensing The general
SSS metric, for simplicity, independent of the action and/or
local metrical ?ld equation, is in the curvature radial
coordinate ds^2 = gtt(r)(cdt)^2 - grr(r)dr^2 - r^2(dtheta^2 +
sin^2thetadphi^2) A light ray obeys, ds^2 = 0 PZ: OK. Null
geodesic. JS: Using Einstein's theory dT = gtt(r)^1/2dt dT is
what a clock at rest relative to the center of symmetry in the
LNIF measures between neighboring events P and P + dP. dR =
grr(r)^1/2dr dR is what a radar or a measuring rod measures.
where t, r are local coordinates in the rest LNIF that is a
point on a timelike non-geodesic. LNIF's can only exist if
there are non-gravity forces. PZ: This seems like a red
herring to me. Or are you just saying that you need to be
able to push a mass off a geodesic, and this push unavoidably
involves the mediation of EM forces?JS: That's what I am
saying. Why red herring? It seems a useful insight to methat
I have not seen made explicit in textbooks? In hindsight it
is obvious.PZ: It could be any force. It happens to be EM
forces, as the world is presently constituted.What is
important on the phenomenological level is that work has to
be done to pushmatter off its natural trajectory. It's just
that gravitational in?s alter what is natural.JS: No, I
mean I have a detailed model why it must be from QED (as the
dominant effect).I mean you need EM both to have light cones
in the ?st place in the off timelike geodesics. The latter
creates the instability in the Dirac virtual
electron-positron vacuum zero point ?tions to create
gravity and dark energy/matter exotic vacuum. PZ: The
abstract observer does not need to be physically pushed, so I
would think you can de?e an LNIF (<--> non-linear spacetime
coordinate system) in the abstract without reference to such
forces.JS: No you cannot. I mean Yes formally,PZ: Yes -- if
your formally is my abstract.JS: No physically.PZ: Ah hah! I
?d it interesting that you make this distinction -- which I
say is foreign toclassic Einsteinian physics.How do you
explain this distinction? What is your physical model?JS:
Gedankenexperiments. Einstein was a Master. How do you manage
to have an LNIF observer?I mean you can think of itbut not
make it.PZ: That is precisely the distinction between a
coordinate system and a physical frame ofreference in my
book. This seems like a very un-Einsteinian distinction.JS:
Not to me.PZ: It also parallels the difference between
general covariance and general relativity.JS: Again maybe
that has to do with distinction between nonlocal active
diffeomorphisms and pseudo-group of local passive coordinate
changes.The EEP has more to do with the latter than the
former! Hal's PV Tables vaguely have to do with the former -
has has P =/= P' where P' is very far away from P 
closer to
source M.I suppose, there would be nothing that could think
if there wereno e and no h and only G and c.Note, if Blackett
relation is truee = G*^1/2mThene/m = G(Newton)^1/2e^(metric
engineering parameter?) /2If e^2/hc invariant, then this
would control m = rest mass of the lepto-quarks.PZ: OK.JS:
BTW we WANT something like at any scale L at FRW cosmic time
timpose a Gaussian law of large numbers Eddington wavelet
?ter:2hG*(L)/c^3 = Lp^2[1 + (Lp*/Lp)^2 e^-(Lp* -
L)^2/Lp^2]whereLp*^2 = Lp^4/3(c/H(t))^2/3 i.e., world
hologram idea with Mach's Principle built in.H(t) =
R(t),t/R(t)in FRW metric of GRNear Big Bang c/H(t) = LpLp^2 =
hG(Newton)/c^3Note that as L/Lp > in?ity G*(L) >
G(Newton)Also as L/Lp > 1, G*(L) > G(Newton)but as L
> Lp*G*(L) >c^3Lp*^2/2h > G(Newton)and at present
epoch of UniverseLp* ~ 1 fermisincec/Ho ~ 10^28 cm PZ:
Although I suppose when you have physical measurement devices
that must physically accelerate through the vacuum, then EM
forces will as a matter of fact come into play. But this
points to what I regard as a fundamental conceptual problem
with the Einsteinian paradigm -- the demand for total
coordinate generality on the basis of the concept of
spacetime as a void (leading to an arbitrariness in the
choice of spacetime coordinates, and a conventionalist
approach to the de?ition of simultaneity, etc.), combined
with the evident physicality of accelerating measurement
devices through the vacuum.JS: There is much discussion of
these sorts of considerations in that book!JS: See the book
above. Lots of good discussions on these kinds of issues!PZ:
OK, but I think the arguments I have made independently are
quite cogent and self-suf?ient.JS: Looking at Hal's
incomprehensible, to my mind, remark above, cogency is, like
beauty and political truth, in the eye of the beholder. ;-)
PZ: As far as I can see this contradiction can *only* be
resolved by literally identifying gravitational and inertial
?lds -- and hence Einsteinian strict equivalence. Mach's
principle is supposed to refer the accelerated motion of a
mass to the average cosmic distribution of matter. But unless
you are prepared to accept action at a distance, this itself
implies the existence of a physical medium of propagation of
the inertial in?s (as later admitted by Einstein). Yet
once you admit such a medium, then the original basis for
Mach's approach is undermined. The snake eats its own tail.
And what is the basis for Mach's principle? A Machian
empiricism that refuses to attribute physical effects to
unobservables (i.e. the physical vacuum) since we cannot see
them, and insists that they be attributed to a relationship
to astronomically distant masses -- since we *can* see them.
A hopelessly naive, internally incoherent, and outdated
theory of science. Like the drunk who loses his keys at the
door but insists on looking for them down at the lamp post
where he can *see*. Absence of evidence is not quite the same
as evidence of absence. What this points to IMHO is a basic
problem with the entire concept of physical relativity of
motion, both special and general. That is why I believe
failure of the Einstein equivalence principle (always
conjectural in character) could ultimately bring down the
entire house of cards (i.e. relativity taken seriously). All
it would take is one reproducible experiment. And who can now
imagine that inertial and gravitational ?lds are physically
indistinguishable? The effects of accelerating a reference
frame propagate through transformed spacetime at the speed of
thought (Eddington) -- while time-varying gravitational
effects according to Einstein's theory propagate at a *?ite
speed*. Apples and oranges.JS: Julian Barbour has apparently
written books on this subject. Have you read them?PZ: No.JS:
I suggest you do. I should too but you are more interested in
this than me.I have not. Why don't you read them and
report?PZ: I am more interested in developing my own line of
deconstructive argument in reference toclassic primary
sources. What does Barbour have to add to what Eddington,
Laue, Synge,etc., etc. -- and even Einstein himself -- have
already written?JS: When you ?d out, let me know.PZ: Synge
-- who was a highly regarded relativist -- thought the
Einstein equivalence principlewas baloney and suggested that
it shouldn't even be taught. These sentiments were to
someextent echoed in Ohanian and Ruf?i (1995).Even in Landau
and Lifz's Classical Theory of Fields (1951) the
skepticism re: equivalenceis quite evident.JS: Quotes??PZ:
What we have here is a longstanding yet muf?volt against
Einsteinian dogma (to whichit appears even the born-again
realist Einstein did not himself subscribe after 1920). I
amsimply trying to push this repressed anti-thesis to its
full logical conclusions.As I said, even Ernst Mach himself
later disowned classic relativity, saying that it had
becometoo dogmatic. JS: Suppose the light ray also has a
component along a tangent to a longitudinal great circle at r
so that, dphi = 0 0 = gtt(r)(cdt)^2 - grr(r)dr^2 - r^2dtheta^2
c^2dT^2 = dR^2 + r^2dtheta^2 c^2 = (dR/dT)^2 +
r^2(dtheta/dT)^2 Therefore, the speed of light is always c.
However, we have a right triangle above; 1 = (vr/c)^2 + c^-2
r^2w^2 vr/c = cosChi w = physical angular speed of light
pulse = gtt^-1/2w' w' = dtheta/dt 1 = 
(grr/gtt)c^-2(dr/dt)^2
+ c^-2r^2gtt^-1(dtheta/dt)^2 = cos^2Chi + sin^2Chi tanChi =
sinChi/cosChi = grr^1/2(dr/dt)/(dtheta/dt) =
grr^1/2(dr/dtheta) Let's take Einstein's theory as a 
de?ite
example. gtt = 1 - 2GM/c^2r grr = (1 - 2GM/c^2r)^-1 outside
event horizon at r* where 1 - 2GM/c^2/r* = 0 only when r >
2GM/c^2 outside the throat of radius r* of this
non-traversable wormhole that pinches off crushing you if you
try to ?ough it. as an asymptotically ?herically
symmetric static wormhole Einstein-Bridge solution of Ruv = 0
i.e. Tuv(Non- Exotic Vacuum) = 0 The gravity lens effect has
one obvious measure tanChi = (1 - 2GM/c^2r)^-1/2(dr/dtheta)
Note that tanChi = in?ity at the event horizon, i.e. the
light ray is trapped in a circular orbit con?ed to the
surface of the event horizon. There is no radial component.
Now this is pretty. Much prettier than Hal's model IMHO.
Einstein is a genius in terms of pure aesthetics. PZ: Agreed.
I do not question the fact of Einstein's towering creative
genius. At the same time, it is well-established in the
theory of science that an idea that is heuristically useful
and even predictively powerful at a certain stage is not
necessarily relevant to the objective evaluation of a theory,
no matter how successful the program (e.g. Kekule's snake).
You have to separate the context of discovery (where creative
genius comes into play) from the context of evaluation (where
we must rely on logical reasoning and objective criteria).
JS: Beauty is Truth. PZ: That was the Florentine Platonist
Galileo's error.JS: Also Dirac's? Also Ed 
Witten's? :-)PZ:
Dirac is another good example. Of course, this kind of
approach can be successful for thelongest time before it
eventually ?es for heuristic bankruptcy.The problem with
this ?ation on mathematical elegance and aesthetics is that
it leads toa naive Platonism that does indeed confuse
physical truth with perceived mathematical beauty --leading
to persistent mirages like Einstein equivalence and
hypostatization of abstractchronogeometric models that I can
only view as fundamentally heuristic in character.JS:
Whitehead's fallacy of misplaced concreteness? Max Tegmark is
a good example of maximal rei?ation inout to be correct on
this in the long run. PZ: He was totally convinced that the
planets move in perfect concentric circles around the
absolute center of the world, and threw Kepler's work into
the trash bin. That was also Einstein's error. He really
believed (at least before 1920) that the inertial and
gravitational ?lds were one and the same, and that the
monolithic gravitational-inertial ?ld was exhaustively
reducible to the Riemannian metric g_uv on a curved 4-dim
spacetime manifold. Because it was all so beautiful. So much
for beauty is truth. Beauty is in the mind of the
beholder.JS: I still think there is equivalence between
inertial and gravity ?lds locally.The g-force is the
essentially the non-tensor connection for parallel
transport.PZ: What does this mean -- the g-force is the
non-tensor connection? Do you mean it canbe *described* in
terms of a non-tensor connection?JS: Yes this is the standard
formalism in
GR.http://mathworld.wolfram.com/Levi-CivitaConnection.htmlby
non-tensor connection I mean Christoffel symbols of the
second
kindhttp://mathworld.wolfram.com/
ChristoffelSymboloftheSecondKind.htmlWhat I am alluding to
and what you are looking for is equation (9) in the above
URLThe ?st term on RHS of eq 9 is the non tensor or
inhomogenous part of the metric connection ?ld.I suppose you
want to call that the curvilinear or inertial part?In the
timelike geodesic LIF the RHS of eq (9) is zero and that is
what you call cancellation?Start using this Wolfram on-line
resource to add mathematical meat to your informal
language.PZ: Isn't this is also true in Newtonian physics?
Just set up a space-time coordinate frameand write a metric
tensor expression for the invariant interval ds, i.e.ds^2 =
g_uv dx^u dx^vand see what happens under accelerative
coordinate transformations.Classical mechanics was fully
relativistic under velocity boosts. Even the material
aetherwas fully relativistic in this sense. In non-inertial
frames, we get *?titious forces*. Suchforces can be
described in terms of a metric-tensor transformation for
acceleratedframes (coordinate transformations that are
non-linear in the time coordinate). Straightinertial
trajectories are then described as geodesics that *look*
curved in those frames.You even get a connection ?ld (in ?ce
-time). The Newtonian ?titious force?ld can thus also
be described in terms of this
Riemann-Christoffel-Levi-Civitaconnection.So what? Inertial
trajectories are still *really* straight. They just *look*
curved in certainframes of reference -- from a certain
POV.But here it is the *POV* that is curved. In other words
-- formally, yes; physically, no.But the theory is still
beautiful. At least in the eyes of a Newtonian.JS:
d^2X^u/ds^2 = Connection^uwl(dX^w/ds)(dX^l/ds)Connection =/=
0 in LNIFPZ: OK.Connection = 0 in LIFPZ: OK.Same in covariant
formulation of Newtonian theory (with no gravity).JS:
Connection = 3rd rank Diff(4)Tensor part + Non-Tensor PartPZ:
Yes.JS: g-force = 0 in LIF where there is mutual cancellation
of the tensor with the non-tensorwhich de?es the free ?mel
ike geodesic extremum of the passive PZ: There is a
local cancellation of the metric gradients, yes. Just as
there is at a point when youadd a compensating sloped line to
a curve in an x-y graph. But a line is not a curve -- evenat
that point, where the curve has non-vanishing second
derivatives, while the line doesn't.pure geometry + Matter
Action + Exotic Vacuum Zero Point Stress-Energy Density
Action.PZ: I won't.geometry + Matter Action.JS: This is
characteristic of minimal coupling gauge theories BTW like in
EMP(kinetic)u = P(canonical)u - (e/c)AuAu is also a connection
in internal ?er space.Fuv(EM) is a curvature!PZ: Let me ask a
naive question. What, in your theory, results in inertial
effects? Do such effectsarise as the result of interaction
between matter and vacuum, or between matter and matter?JS:
Depends on what you mean by inertial effects? Please be more
speci?.Do you mean origin of inertia like trying to explain
m ~ 10^-27 grams forrest mass of the electron? I mean in the
sense of what Bernie Haisch wastrying to do?Or do you mean
gravity g-force is locally equivalent to an acceleration and
is likeany inertial force e.g. Coriolis and centrifugal? The
mass of the test out in the equation of motion of all of
them! That's what they have in common.In the former I mean
Wheeler's non-simply topological geometrodynamics of
wormholeMass without mass but with G* > G(Newton) on
micro-scale so that matter comesfrom attractive EXOTIC
VACUUM, i.e. zero point stress-energy density tensors.In the
latter I mean what Einstein meant as reconstructed in MTW.
Bottom line, gravity lensing does give information about r
the curvature radial coordinate where the concentric surface
area is 4pir^2 with an entropy of 4pir^2/Lp^2 c-bits. One
curvature radial coordinate has more than one isotropic
radial coordinate corresponding to it. The map is not 1-1.
PZ: Interesting. I'll have to give you a copy of Brian
Tupper's 1974 Nuovo Cimento paper which deals with a
parametrized class of geometrodynamic theories that are
consistent with the four classical tests of GR. Also, a 1999
paper by Alley, Yilmaz et al. that restates the entire n-body
argument in the context of more recent empirical data. ...
That is what I call semantic incoherence. JS: This is to be
expected from Noether's theorem! It is trivially so that the
bigger theory always violates Sacred Cows of the smaller
earlier theory e.g. Newton's Absolute Time as a slaughtered
sacred cow. But Einstein, The Great Rabbi, did it Kosher! PZ:
But then how do you explain the existence of these conserved
energy-momentum integrals in Newtonian theory -- arguably the
most successful physical theory of all time? JS: Trivial -
action at a distance in globally ?acetime. No problem
the translation group symmetry works. PZ: : Are you saying
that you don't need to recover the classical conservation
principles in the Newtonian limit? JS: No I said just the
opposite. I told you why it works. Curvature means violation
of the translational symmetry from which conservation of
momenergy comes.mainly an artifact of insisting on doing
*intrinsic* geometry on higher-dimensional curved
manifolds.JS: Right.PZ: OK. PZ: If you do intrinsic geometry
on the surface of an ordinary 3-sphere, you have exactly the
same problems. But as soon as you look at the same surface as
embedded in 3-space, all the Riemannian hocus pocus about
curvature connections and parallel transport becomes
irrelevant, since exactly the same vectors are then directly
comparable.JS: Huh? Only if the bigger space is globally ?ich
it is not in M-theory and evenhas fermion dimensions and
even may have non-commutative geometry.PZ: I am talking about
an ordinary sphere in ordinary space. If you insist on doing
purely intrinsicgeometry on the spherical surface, you are a
half-blind ?der who sees only the shadows(connection,
covariant derivatives, etc. etc), whereas viewed from
3-space, each vector hasperfectly well-de?ed components that
can be directly compared at every point on the surface.No need
for parallel transport and all that. All you have to do is
establish a Cartesian coordinatesystem in the 3-space, and
all the Riemannian apparatus becomes instantly redundant
--although of course it still applies to the dimensionally
challenged ?ders on the surface.JS: In GR 4D space-time
is curved and so are spacelike slices. The extrinsic
curvature of the spacelike slice isa kind of time push
forward observable.PZ: We can still imagine a 10-dim
spacetime in which this manifold can be embedded. Then we are
nolonger condemned to a life of purely intrinsic surface
geometry.JS: Yes but I doubt that the Calabi-Yau space of
string theory is such a ?clidean one with simply
connected global topology.11 Dim hyperspace of M-Theory is
not ?: Perhaps not. We can still *imagine* a ?he
r-dimensional spacetime in which the curved 4-dim
Riemannianmanifold of GR is embedded, and that is suf?ient
for my point.It puts things in *perspective*, so to speak.
The point being, let's not fetishize or ?ate upon
Riemanniangeometry.Golden Calf. PZ: The question of how to
de?e the energy-momentum of a physical gravitational ?ld is
not simply a question of Riemannian elegance.JS:The Question
is: What is The Question? (J.A. Wheeler).PZ:
Yes.Energy-momentum of gravity is a global ?tegral!PZ:
Of what density?!JS: I try to answer that below. But it is
half-baked at this point. I seem to be getting that you need
EXOTIC VACUUM to solve the problem locally. That is the
missing link -- maybe?Local stress-energy density Diff(4)
tensor for gravity is trivial it istuv(Gravity) = (String
Tension)GuvIn non-exotic vacuumtuv(Gravity) = 0.PZ: That is
exactly what is at issue here. The argument is that you do
not get asatisfactory Newtonian correspondence with t_uv(vac)
= 0.JS: Yeah, but maybe that is why we need exotic vacuum dark
energy/matter as a consistency requirement just like when
James Maxwell stuck in the displacement vacuum current to get
far ?ld EM radiation! This idea is in harmony with apparent
fact that Universe is 96% EXOTIC VACUUM. I do not think those
latest observations
http://space?ow.com/news/n0312/12darkenergy/ are being
properly interpreted BTW.In exotic dark energy/matter
vacuumtuv(Gravity) + tuv(Exotic Vacuum Zero Point
Stress-Energy Density) = 0PZ: OK, so you are saying that
there is indeed a t_uv(vac).JS: I have been saying this all
along for EXOTIC VACUUM. The effect is not there, I mean it's
ZERO in the ordinary NON-EXOTIC vacuum of Einstein's Ruv = 0.
Also my tuv(Exotic Vacuum) = (String Tension)/zpf guv(Curved
Space-Time)IS NOT YILMAZ's!My effect is a QUANTUM EFFECT it
is exactly INFINITE when quantum h -> 0 and G & c stay
?ite.Indeed, in large scale it gives an in?ite Einstein
cosmological constant when h -> 0.Vacuum Coherence -> 0 when
h -> 0 and Lp -> 0 when G, c ?ite, therefore /zpf -> -1/Lp^2
-> -in?ity (attractive dark matter limit catastrophe where
Universe cannot come into Being and Becoming at all!)That is
OK because there is no such thing as a consistent CLASSICAL
UNIVERSE. What we live in is a MACRO-QUANTUM UNIVERSE or
rather parallel brane worlds ?g in hyperspace on the IT
extra variable (rather than hidden variable) level. There is
also BIT.PZ: How does this t_uv(vac) self-gravitate in your
theory? Is it a tensor density?JS: Please Paul READ MY
EQUATIONS!tuv(Geometry) + Tuv(Matter) + tuv(Exotic Vacuum) =
0is a local tensor equation. All parts of it are tensors
under at least local passive general coordinate
transformations.tuv(Geometry) = (String Tension)[Ruv(Ricci) -
(1/2)Rguv]Tuv(Exotic Vacuum) = (String Tension)/zpfguv/zpf =
Lp^-2[Lp^3/2|Vacuum Coherence|^2 - 1]Do you mean the (-g)^1/2
Pauli tensor density formal thing?PZ: If it has a non-tensor
component, then where exactly does that come from,physically
speaking, in your theory?JS: You have again garbled the level
distinction between g-force connection ?lds and tidal
curvature tensor.My tuv(Exotic Vacuum) ~ guv(Curved
Space-Time)i.e. a second rank GR tensor. Your it ain't my it.
;-)What about the ?tegrals?Simple.tuv(Gravity) =
tuv(Gravity Near Field) + tuv(Gravity Far Field)Gravity
mom-energy ?tegral is only for gravity waves (LIGO &
LISA et-al)Pu (Far Field) = Spacelike 3D integral of
tvl(Gravity Far Field)(antisymmetric 3symbol uvl)dx^w/dx^l=
?d?tvl(far ?ld)?(u|vl)dx^v/dx^lWhere in some regions
contributing to the spacelike integraltuv(gravity far ?ld)
=/= 0tuv(gravity far-?ld) = - tuv(gravity near ?ld) -
tuv(Exotic Vacuum Zero Point Stress-Energy Density) =/= 0via
a kind of Gauss theorem relating surface to volume integrals
- there are complications doing that in curved space-time of
course.PZ: OK.JS: There is an interesting prediction from
this toy model I just now thought of for the ?st time.When
tuv(gravity near ?ld) > 0, one cannot see gravity waves
in the far ?ld locally unless there is anexotic-vacuum ?ld
tuv(Exotic Vacuum Zero Point Stress-Energy Density) locally
present at the surface for the ?tegral! This would seem
to predict a lack of gravity wave signal for Kip Thorne's
LISA and LIGO? Note, that is not same problem of 1913 type
pulsars, which is an indirect EM measurement of gravity waves
at that pulsar not a direct detection of gravity waves passing
through Earth. I am not sure if any of this makes sense yet.
It just popped up out of my sub-conscious mind, or maybe the
Jungian Collective Unconscious Mind of P.K. Dick's VALIS?
;-)PZ: OK, this is beginning to make sense.JS: Yeah, that's
AMAZING! Something like the gravity analog of Maxwell's
displacement current showing why there must be dark
energy/matter exotic vacuum regions - hence stargate time
travel with weightless warp drive not far behind and already
here in the UFO data? GR orthodoxy (many of whom are really
mathematicians) are confusing an artifact of insistence on
intrinsic geometry with a physical question of characterizing
the real permanent gravitational ?ld in terms of its
energy-momentum content and distribution. Of course if one
*de?es* the physical g-?ld in terms of the uni?d
gravitational- inertial Riemann-Christoffel-Levi-Civita
connection, the energy carried in this ?ld will be at least
locally frame-dependent. But from my POV this is an arti?e
that arises from a kind of *fetishization* of Riemannian
mathematics and general covariance. Think about it: take a
4-dim spacetime. Go to a 4-dim cylindrical coordinate system
for which the coordinates in a 2-dim spatial submanifold are
polar. You automatically get a non-vanishing connection that
is purely an artifact of the choice of coordinates. Of course
you have to re-write the metric components g_uv to take
account of this; and inside this polar subspace coordinate
system, straight lines look curved. But nowhere even in
orthodox GR is it suggested that this connection ?ld
constitutes any kind of *physical* ?ld. It is regarded as a
purely mathematical animal. Yet the whole apparatus of
connections, geodesics, etc. etc. applies just as it does to
the physically interpreted space-time coordinate connection
?ld. So the mere appearance of a connection ?ld in itself
need have no physical meaning at all -- even within orthodox
GR. All that glisters is not gold. JS: Of course one hopes
that the compensating gauge ?ld, in this case the
geometrodynamic guv ?ld restores the broken symmetry. It
does do that in the sense of locally conservation of
stress-energy density currents when all dynamical degrees of
freedom are included. PZ: OK. JS: Tuv(Non Exotic Vacuum) +
Tuv(Exotic Vacuum) + Tuv(Matter) = 0 The total covariant
4-divergence vanishes, but not the individual terms
separately in the practical metric engineering regime whose
technology we see in the saucers from the brane worlds next
door is my educated guess. Brian Greene thought he was joking
on NOVA, the joke may be on him - let us hope. :-) More anon.
JS: This is getting really interesting. Z.JS:
Yup.http://stardrive.org/cartoon/MagicBean.htmlhttp://
stardrive.org/cartoon/spectra.htmlhttp://stardrive.org/
cartoon/USSKron.htmlhttp://stardrive.org/cartoon/
bovines.htmlhttp://stardrive.org/cartoon/dan.htmlhttp://
stardrive.org/cartoon/coffee.htmlhttp://stardrive.org/cartoon
/Saturn.html
===
I have no idea how to even start this one.
Any hints will be helpful.Let a in S_n be a ?ed-point
involution or inversion where a^2 = e and a
===