mm-959

Subject: PROOF that numbers are countable
This list is equivalent to 0.3..
0.3
0.33
0.333
0.3333
0.33333
...
It converges to 1/3. 0.3.. literally means 0.3 repeating 3.
But this list doesn¹t contain 1/3.
0.3
0.2
0.33
0.2
0.333
0.2
0.3333
0.2
It doesn¹t converge.
The computuble numbers contains all expansions of the
anti_diagonal
number.
Say the computables is like so :
0 . 3 5 6 7 4 ..
0 . 2 6 5 6 7 ..
0 . 1 2 3 4 5 ..
0 . 6 5 4 3 2 ..
0 . 3 6 4 6 4 ..
..
The Diag is 0 . 3 6 3 3 4 ..
Let Anti_Diag = 0. 4 7 4 4 5 ..
The computables contains the following numbers
0 . 4
0 . 4 7
0 . 4 7 4
0 . 4 7 4 4
..
Call this list ADL (anti-diagonal-list).
Unfortunately the computables looks more like this :
0 . 4
0 . 2
0 . 4 7
0 . 5
0 . 4 7 4
0 . 4
0 . 4 7 4 4
0 . 8
But, THERE EXISTS A COMPUTABLE PROCESS that
from the list of computable numbers can form ADL.
ADL is equivalent to the diagonal number,
THEREFORE the computable numbers CONTAIN ALL NUMBERS.
Herc
QED + 1/2
--
oo
____|mn
/ /_/ / _
/ K-9/ /_/ - www.YeOldeCoffeeShoppe.com -
/____/_____
--------------
===
Subject: Re: PROOF that numbers are countable
: The computuble numbers contains all expansions of the
anti_diagonal
: number.
You have NOT DEFINED expansion, DUMBASS.
How you could have an expansion of anything
(like the decimal representation of a real)
THAT WAS *ALREADY* INFINITE to begin with is
more than usually mysterious.
: Say the computables is like so :
:
: 0 . 3 5 6 7 4 ..
: 0 . 2 6 5 6 7 ..
: 0 . 1 2 3 4 5 ..
: 0 . 6 5 4 3 2 ..
: 0 . 3 6 4 6 4 ..
: ..
:
: The Diag is 0 . 3 6 3 3 4 ..
:
: Let Anti_Diag = 0. 4 7 4 4 5 ..
:
: The computables contains the following numbers
:
: 0 . 4
: 0 . 4 7
: 0 . 4 7 4
: 0 . 4 7 4 4
: ..
:
: Call this list ADL (anti-diagonal-list).
:
: Unfortunately the computables looks more like this :
:
: 0 . 4
: 0 . 2
: 0 . 4 7
: 0 . 5
: 0 . 4 7 4
: 0 . 4
: 0 . 4 7 4 4
: 0 . 8
:
: But, THERE EXISTS A COMPUTABLE PROCESS that
: from the list of computable numbers can form ADL.
Right.
BUT THAT IS A CONTRADICTION.
This means that the list of computable numbers
is itself NOT effectively computable. The class of
computable numbers IS NOT recursively enumerable.
: ADL is equivalent to the diagonal number,
Right.
: THEREFORE the computable numbers CONTAIN ALL NUMBERS.
That simply does not follow.
===
Subject: Re: PROOF that numbers are countable
Discussion, linux)
> THEREFORE the computable numbers CONTAIN ALL NUMBERS.
Fascinating.
I can¹t wait to see the algorithm that deÞnes
Sum 2^{-K(i)}
where K(i) = 0 iff the i¹th TM halts on input i and 1
otherwise.
(This is more or less Chaitin¹s constant Omega.)
--
Jesse F. Hughes
What I represent is the unknowable future--the power of
change. In
that sense I¹m a force of Nature, a force of the Universe, a
living
emodiment of change itself. --James Harris and his sense of
humility
===
Subject: Re: PROOF that numbers are countable
> THEREFORE the computable numbers CONTAIN ALL NUMBERS.
> Fascinating.
> I can¹t wait to see the algorithm that deÞnes
> Sum 2^{-K(i)}
> where K(i) = 0 iff the i¹th TM halts on input i and 1
otherwise.
> (This is more or less Chaitin¹s constant Omega.)
Funny you should say that, because if you line up all the
stars from our
sun
to further and further stars in our possibly inÞnite
universe, and give a
1
if the sun explodes (halts), and a 0 if it doesn¹t, and make
an equivalent
number
from that sequence, you get the same useless defn of another
inÞnite
oracle.
Keep on waiting, those digits will appear real soon.
Herc
===
Subject: Re: PROOF that numbers are countable
<40696067$0$20345$afc38c87@news.optusnet.com.au>
Discussion, linux)
>> THEREFORE the computable numbers CONTAIN ALL NUMBERS.
>> Fascinating.
>> I can¹t wait to see the algorithm that deÞnes
>> Sum 2^{-K(i)}
>> where K(i) = 0 iff the i¹th TM halts on input i and 1
otherwise.
>> (This is more or less Chaitin¹s constant Omega.)
> Funny you should say that, because if you line up all the
stars from our
sun
> to further and further stars in our possibly inÞnite
universe, and give a
1
> if the sun explodes (halts), and a 0 if it doesn¹t, and
make an equivalent
number
> from that sequence, you get the same useless defn of
another inÞnite
oracle.
> Keep on waiting, those digits will appear real soon.
Are you suggesting that my[1] number is not a number?
Your number may be a bit more dubious, depending on whether
it is a
fact *now* that a particular sun will (or will not) explode.
But it
is surely less dubious whether a particular deterministic
automaton
will or will not halt. Your analogy just doesn¹t do it for
me, sorry.
Footnotes:
[1] Well, Chaitin¹s or whoever¹s number.
--
Jesse F. Hughes
Contrariwise, continued Tweedledee, if it was so, it might
be, and
if it were so, it would be; but as it isn¹t, it ain¹t. That¹s
logic!
-- Lewis Carroll
===
Subject: Re: PROOF that numbers are countable
In sci.logic, |-|erc
<contactvia@wwwadamskingdom.com>
<4068d879$0$16966$afc38c87@news.optusnet.com.au>:
> This list is equivalent to 0.3..
> 0.3
> 0.33
> 0.333
> 0.3333
> 0.33333
> ...
> It converges to 1/3. 0.3.. literally means 0.3 repeating 3.
> But this list doesn¹t contain 1/3.
> 0.3
> 0.2
> 0.33
> 0.2
> 0.333
> 0.2
> 0.3333
> 0.2
> It doesn¹t converge.
> The computuble numbers contains all expansions of the
anti_diagonal
> number.
The above is clear as mud. Are you trying to deÞne Cauchy
sequences
in some obscure fashion?
> Say the computables is like so :
> 0 . 3 5 6 7 4 ..
> 0 . 2 6 5 6 7 ..
> 0 . 1 2 3 4 5 ..
> 0 . 6 5 4 3 2 ..
> 0 . 3 6 4 6 4 ..
> ..
> The Diag is 0 . 3 6 3 3 4 ..
> Let Anti_Diag = 0. 4 7 4 4 5 ..
> The computables contains the following numbers
> 0 . 4
> 0 . 4 7
> 0 . 4 7 4
> 0 . 4 7 4 4
> ..
> Call this list ADL (anti-diagonal-list).
> Unfortunately the computables looks more like this :
> 0 . 4
> 0 . 2
> 0 . 4 7
> 0 . 5
> 0 . 4 7 4
> 0 . 4
> 0 . 4 7 4 4
> 0 . 8
> But, THERE EXISTS A COMPUTABLE PROCESS that
> from the list of computable numbers can form ADL.
> ADL is equivalent to the diagonal number,
> THEREFORE the computable numbers CONTAIN ALL NUMBERS.
> Herc
> QED + 1/2
So you¹re suggesting that a list of computable numbers
will contain all Þnite preÞxes of a computable number,
therefore the diagonal number (which is computable)
is in the list?
Nice try, but the counterexample is your very own:
0.3
0.33
0.333
...
which doesn¹t contain the number 1/3, although it gets
arbitrarily close. Now, one can try to get cute and try
to deÞne 1/3 in some fashion to be able to compare it
to a decimal expansion. The simplest for this case is
arguably to deÞne Q(1/3, n) and R(1/3, n):
Q(1/3, 0) = 0
R(1/3, 0) = 1
Q(1/3, n+1) = þoor(10 * R(1/3, n) / 3)
R(1/3, n+1) = 10 * R(1/3, n) - 3 * Q(1/3, n+1)
which, as it turns out, is a form of long division,
which is more conventionally expressed:
0.333333333333333333333...
3|1.000000000000000000000...
- 9
.10
-.09
. 10
-. 09
. 10
-. 09
.
.
.
[.sigsnip]
--
#191, ewill3@earthlink.net
It¹s still legal to go .sigless.
===
Subject: Re: PROOF that numbers are countable
> This list is equivalent to 0.3..
0.3
> 0.33
> 0.333
> 0.3333
> 0.33333
> ...

> It converges to 1/3. 0.3.. literally means 0.3 repeating 3.

> But this list doesn¹t contain 1/3.
0.3
> 0.2
> 0.33
> 0.2
> 0.333
> 0.2
> 0.3333
> 0.2
It doesn¹t converge.

> The computuble numbers contains all expansions of the
anti_diagonal
> number.
> The above is clear as mud. Are you trying to deÞne Cauchy
sequences
> in some obscure fashion?
I¹ve given you 5 different versions of the inductive proof of
this already.
Even Barb gets it :
> ALL Þnite
> sequences of digits are computable (e.g. in order of length
and then
> lexicographically).
Say the computables is like so :
0 . 3 5 6 7 4 ..
> 0 . 2 6 5 6 7 ..
> 0 . 1 2 3 4 5 ..
> 0 . 6 5 4 3 2 ..
> 0 . 3 6 4 6 4 ..
> ..
The Diag is 0 . 3 6 3 3 4 ..
Let Anti_Diag = 0. 4 7 4 4 5 ..
The computables contains the following numbers
0 . 4
> 0 . 4 7
> 0 . 4 7 4
> 0 . 4 7 4 4
> ..
Call this list ADL (anti-diagonal-list).
Unfortunately the computables looks more like this :
0 . 4
> 0 . 2
> 0 . 4 7
> 0 . 5
> 0 . 4 7 4
> 0 . 4
> 0 . 4 7 4 4
> 0 . 8
But, THERE EXISTS A COMPUTABLE PROCESS that
> from the list of computable numbers can form ADL.
ADL is equivalent to the diagonal number,
> THEREFORE the computable numbers CONTAIN ALL NUMBERS.
Herc
> QED + 1/2
> So you¹re suggesting that a list of computable numbers
> will contain all Þnite preÞxes of a computable number,
> therefore the diagonal number (which is computable)
> is in the list?
> Nice try, but the counterexample is your very own:
> 0.3
> 0.33
> 0.333
> ...
Please read proofs more carefully in future.
> It converges to 1/3
Herc
===
Subject: Re: PROOF that numbers are countable
Don¹t swear on my proof, this is living history.
The 0.3333 analogy was made on 3 30 04
I.e the date is 3300 The fact I proved it on 3300 with the
example 3300
conÞrms it.
The proof is valid.
0.333..
is exactly equivalent to
0.3
0.33
0.333
0.3333
0.33333
..
There is a function that can map between the 2. .. means
recurring.
The computable numbers contains all Þnite sequences of the
anti_diagonal
digit.
If you are having trouble with my 5,000 words conÞrming this
then prove
that it doesn¹t. Give me the value for n where the 1st n
digits of
anti_diag
is not covered in the list of computable numbers.
You are all having difÞculty equating ALL FINITE with
INFINITE.
n = 1
get 1st n digits of anti_diag
n = n + 1
repeat
The proof is there before you, you only have to rethink the
myriad of
techniques
based on Cantor. Go to the 1st point where you concluded
there must be
uncountable inÞnity of real numbers, and *think* is there is
another
possibility.
Why does a sequence of symbols to denote a position cause this
extraordinary
claim about multiple inÞnities?
Herc
===
Subject: Re: PROOF that numbers are countable
<SNIP Please read proofs more carefully in future.
> > It converges to 1/3
But never actually reaches it. The item in the nth position
in the sequence
0.3
0.33
0.333
.
.
.
differs from 1/3 by at least 3/10^n.
And no, there is no inÞnitieth position in a list, by
deÞnition of
what
a list is (and what a Natural number is).
If you are saying that you can construct a sequence where the
limit
approaches a real number, congratulations, that¹s one way of
deÞning real
numbers (as limits of rational sequences).
However, this isn¹t what the Cantor construction is about;
its not about
the
limits of sequences, its about the numbers in the sequence
itself. You
can¹t
change a proof, and say see, my modiÞed version no longer
works, so the
original is crap.
===
Subject: Re: PROOF that numbers are countable
> This list is equivalent to 0.3..
0.3
> 0.33
> 0.333
> 0.3333
> 0.33333
> ...

> It converges to 1/3. 0.3.. literally means 0.3 repeating 3.

> But this list doesn¹t contain 1/3.
0.3
> 0.2
> 0.33
> 0.2
> 0.333
> 0.2
> 0.3333
> 0.2
It doesn¹t converge.

The Þrst list didn¹t contain 1/3, either.
> The computuble numbers contains all expansions of the
anti_diagonal
> number.
Say the computables is like so :
0 . 3 5 6 7 4 ..
> 0 . 2 6 5 6 7 ..
> 0 . 1 2 3 4 5 ..
> 0 . 6 5 4 3 2 ..
> 0 . 3 6 4 6 4 ..
> ..
The Diag is 0 . 3 6 3 3 4 ..
Let Anti_Diag = 0. 4 7 4 4 5 ..
The computables contains the following numbers
0 . 4
> 0 . 4 7
> 0 . 4 7 4
> 0 . 4 7 4 4
> ..
Call this list ADL (anti-diagonal-list).
>
This list doesn¹t contain the anti-diagonal number.
> Unfortunately the computables looks more like this :
0 . 4
> 0 . 2
> 0 . 4 7
> 0 . 5
> 0 . 4 7 4
> 0 . 4
> 0 . 4 7 4 4
> 0 . 8
But, THERE EXISTS A COMPUTABLE PROCESS that
> from the list of computable numbers can form ADL.
>
But the list of computable numbers isn¹t computable, so the
anti-diagonal number isn¹t computable.
> ADL is equivalent to the diagonal number,
> THEREFORE the computable numbers CONTAIN ALL NUMBERS.
>
No, the anti-diagonal number is not computable.
> Herc
> QED + 1/2
===
Subject: Re: Non-euclidean Geometry
> OK, so I¹m not even a undergrad yet, but I have what I
think should be
> a very simple question: If a non-euclidean geometry in N
dimensions is
> simply a geometry that occurs on the N-dimensional surface
of a N+1
> dimensional object, why is it such a big deal to say that
Euclid was
> wrong if it is simply a question of what the surface that
you are
> working with is?
Your premiss is already wrong. In the general case you need
greater
than N+1 dimensions to isometrically embed an N manifold,
usually much
greater (John Nash: The imbedding problem for Riemannian
manifolds,
Annals of Mathematics, 63 (1965), pp 20-63.) (Yes, A
Beautiful Mind
Nash).
For example, the Flat Torus. Zero Gaussian curvature
everywhere:
-Bruce
===
Subject: Re: Maple notation what is this?
>If anyone knows Maple would you tell me what this output
means?
>> f:= x^( j+r-1 ) * exp^(-t*x);
>f := x^(j+r-1)*exp^(-t*x)
>> int( f, x=0..inÞnity);
It means somebody made a mistake, using exp as a variable
instead
of a function. It should have been exp(-t*x), not exp^(-t*x).
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Re: Diagonal proof is nonsense
>This is the argument to why mathematicians have failed for a
century.
>some rationals aren¹t on simple lists : : some reals aren¹t
on rationals

>list : : some irrats aren¹t on computables list
>Its a long deductive sequence to go from former to latter
which you have
not
>done.
>0.3.. is a simple algorithm, otherwise you could not specify
it enough to
show
>its not on that list. 0.3 repeat 3.
Correct. Indeed, the set of all the rationals is recursively
enumerable
(which is what your on a list seems to mean). Exercise: prove
that fact
(or
look it up in any text on computability theory).
>How can you show 0.48498744343984744376487450985947..
>is not on the list of computables? YOU CAN¹T. Why? because
>every possible sequence occurs on the list of computables.
No... it... DOESN¹T. That is the whole point, that there are
some inÞnite

sequences of digits which are NOT computable at all (e.g.
Chaitin¹s omega,
the
halting number you¹ve referred to).
Furthermore, as I and others have unsuccessfully tried to
show you, even the

countable set of all the computable sequences is not itself
computable (on
a
list in your terms).
>You cannot specify a missing sequence ERGO there is no
missing number.
The WHOLE POINT is that there are sequences which can NOT be
Þnitely
speciÞed AT ALL (e.g. by any TM). Therefore OF COURSE you
can¹t be SHOWN
one, since anything you can be shown must be Þnitely speciÞed
(unless you

happen to be an inÞnite being, but that¹s another and more
twisted
story...)
This really isn¹t hard to understand. Please at least TRY.
--
---------------------------
| BBB b Barbara at LivingHistory stop co stop uk
| B B aa rrr b |
| BBB a a r bbb |
| B B a a r b b |
| BBB aa a r bbb |
-----------------------------
===
Subject: Re: Diagonal proof is nonsense
Barb Knox <see@sig.belowThis is the argument to why
mathematicians have failed for a century.
some rationals aren¹t on simple lists : : some reals aren¹t on
rationals
>list : : some irrats aren¹t on computables list
Its a long deductive sequence to go from former to latter
which you have
not
>done.
>0.3.. is a simple algorithm, otherwise you could not specify
it enough to
show
>its not on that list. 0.3 repeat 3.
> Correct. Indeed, the set of all the rationals is
recursively enumerable
> (which is what your on a list seems to mean). Exercise:
prove that
fact (or
> look it up in any text on computability theory).
>How can you show 0.48498744343984744376487450985947..
>is not on the list of computables? YOU CAN¹T. Why? because
>every possible sequence occurs on the list of computables.
> No... it... DOESN¹T. That is the whole point, that there
are some
inÞnite
> sequences of digits which are NOT computable at all (e.g.
Chaitin¹s omega,
the
> halting number you¹ve referred to).
> Furthermore, as I and others have unsuccessfully tried to
show you, even
the
> countable set of all the computable sequences is not itself
computable
(on a
> list in your terms).
>You cannot specify a missing sequence ERGO there is no
missing number.
> The WHOLE POINT is that there are sequences which can NOT
be Þnitely
> speciÞed AT ALL (e.g. by any TM). Therefore OF COURSE you
can¹t be
SHOWN
> one, since anything you can be shown must be Þnitely
speciÞed (unless
you
> happen to be an inÞnite being, but that¹s another and more
twisted
story...)
> This really isn¹t hard to understand. Please at least TRY.
long week but she¹s civil at last. I comprehend what you¹re
tying to say
but its bull.
the numbers are missing, you¹d need inÞnite paper to write one
down, they go
in between the numbers we know of --- its pure tooth fairy.
I¹m not asking for the inÞnite expansion, I¹m asking for a
sample of it.
If every sample
of something of any size is a computable sequence, what does
that tell you
about
the sequence you are sampling?
This is God¹s magic number!!!
What is it?
Though shalt not know the mind of God!
Oh go on!
Ok, it goes 0.3479584743785745848494877484849858487
Wait a minute, thats just computable sequence # 52!
Stop interrupting, I haven¹t Þnished yet
398474374387494959584748598595874
Hold on, that¹s computable sequence #114534!
Are you going to keep interrupting?
How long will this take until you come up with a sequence I
can¹t compute?
Herc
===
Subject: Re: Diagonal proof is nonsense
>Barb Knox <see@sig.below>This is the argument to why
mathematicians have failed for a century.
>>some rationals aren¹t on simple lists : : some reals aren¹t
on
rationals
>>list : : some irrats aren¹t on computables list
>>Its a long deductive sequence to go from former to latter
which you have

>>not
>>done.
>>0.3.. is a simple algorithm, otherwise you could not
specify it enough
to
>>show
>>its not on that list. 0.3 repeat 3.
>> Correct. Indeed, the set of all the rationals is
recursively enumerable
>> (which is what your on a list seems to mean). Exercise:
prove that
fact
>> (or
>> look it up in any text on computability theory).
>>How can you show 0.48498744343984744376487450985947..
>>is not on the list of computables? YOU CAN¹T. Why? because
>>every possible sequence occurs on the list of computables.
>> No... it... DOESN¹T. That is the whole point, that there
are some
inÞnite
>> sequences of digits which are NOT computable at all (e.g.
Chaitin¹s
omega,
>> the
>> halting number you¹ve referred to).
>> Furthermore, as I and others have unsuccessfully tried to
show you, even
the
>> countable set of all the computable sequences is not
itself computable
(on
>> a
>> list in your terms).
>>You cannot specify a missing sequence ERGO there is no
missing number.
>> The WHOLE POINT is that there are sequences which can NOT
be Þnitely
>> speciÞed AT ALL (e.g. by any TM). Therefore OF COURSE you
can¹t be
SHOWN
>> one, since anything you can be shown must be Þnitely
speciÞed (unless
you
>> happen to be an inÞnite being, but that¹s another and more
twisted
>> story...)
>> This really isn¹t hard to understand. Please at least TRY.
>long week but she¹s civil at last. I comprehend what you¹re
tying to say
but
>its bull.
>the numbers are missing, you¹d need inÞnite paper to write
one
down,
>they go
>in between the numbers we know of --- its pure tooth fairy.
>I¹m not asking for the inÞnite expansion, I¹m asking for a
sample of it.
If
>every sample
>of something of any size is a computable sequence, what does
that tell you

>about
>the sequence you are sampling?
Listen carefully: it tells you exactly NOTHING. That is
because ALL Þnite

sequences of digits are computable (e.g. in order of length
and then
lexicographically). I presume you agree with that. When you
say any
size
you really are meaning any FINITE size. If so, then showing
how to compute
a
set of Þnite-length preÞxes of a sequence DOES NOT show you
how to compute

the entire sequence.
For example, for any given n I can compute Chaitin¹s Omega to
n digits
(since
I can compute EVERY n-digit binary number), but no-one can
compute the
entire
INFINITE sequence. By their nature, ALL non-computable things
are
inÞnite,
since we can easily exhaustively enumerate all the Þnite ones.
>How long will this take until you come up with a sequence I
can¹t compute?
You clearly didn¹t really understand the 2nd-to-last
paragraph of my
previous
post. For me to come up with a sequence means that I can
compute it,
right?
If I can compute it then so can you. NO uncomputable thing is
come-upable

with as a Þnite speciÞcation -- THAT¹S precisely what it
means for it to
be
uncomputable in the Þrst place!
You seem to admit the (uncomputable) existence of SOME
uncomputable things
(e.g. Chaitin¹s Omega, the Halts(tm) function). If so, than
what¹s the big

deal about accepting the proofs that there are other
uncomputable
sequences?
--
---------------------------
| BBB b Barbara at LivingHistory stop co stop uk
| B B aa rrr b |
| BBB a a r bbb |
| B B a a r b b |
| BBB aa a r bbb |
-----------------------------
===
Subject: Re: Diagonal proof is nonsense
>>This is the argument to why mathematicians have failed for
a century.
>>some rationals aren¹t on simple lists : : some reals aren¹t
on
rationals
>>list : : some irrats aren¹t on computables list
>>Its a long deductive sequence to go from former to latter
which you
have
>>not
>>done.
>>0.3.. is a simple algorithm, otherwise you could not
specify it enough
to
>>show
>>its not on that list. 0.3 repeat 3.
>> Correct. Indeed, the set of all the rationals is
recursively
enumerable
>> (which is what your on a list seems to mean). Exercise:
prove that
fact
>> (or
>> look it up in any text on computability theory).
>>How can you show 0.48498744343984744376487450985947..
>>is not on the list of computables? YOU CAN¹T. Why? because
>>every possible sequence occurs on the list of computables.
>> No... it... DOESN¹T. That is the whole point, that there
are some
inÞnite
>> sequences of digits which are NOT computable at all (e.g.
Chaitin¹s
omega,
>> the
>> halting number you¹ve referred to).
>> Furthermore, as I and others have unsuccessfully tried to
show you,
even the
>> countable set of all the computable sequences is not
itself computable
(on
>> a
>> list in your terms).
>>You cannot specify a missing sequence ERGO there is no
missing
number.
>> The WHOLE POINT is that there are sequences which can NOT
be Þnitely
>> speciÞed AT ALL (e.g. by any TM). Therefore OF COURSE you
can¹t be
SHOWN
>> one, since anything you can be shown must be Þnitely
speciÞed (unless
you
>> happen to be an inÞnite being, but that¹s another and more
twisted
>> story...)
>> This really isn¹t hard to understand. Please at least TRY.
>
>long week but she¹s civil at last. I comprehend what you¹re
tying to say
but
>its bull.
>the numbers are missing, you¹d need inÞnite paper to write
one
down,
>they go
>in between the numbers we know of --- its pure tooth fairy.
I¹m not asking for the inÞnite expansion, I¹m asking for a
sample of it.
If
>every sample
>of something of any size is a computable sequence, what does
that tell
you
>about
>the sequence you are sampling?
> Listen carefully: it tells you exactly NOTHING. That is
because ALL
Þnite
> sequences of digits are computable (e.g. in order of length
and then
> lexicographically). I presume you agree with that. When you
say any
size
> you really are meaning any FINITE size. If so, then showing
how to
compute a
> set of Þnite-length preÞxes of a sequence DOES NOT show you
how to
compute
> the entire sequence.
Sure it does.
n = 1
compute the sequence for n digits.
n = n + 1
repeat
> For example, for any given n I can compute Chaitin¹s Omega
to n digits
(since
> I can compute EVERY n-digit binary number), but no-one can
compute the
entire
> INFINITE sequence. By their nature, ALL non-computable
things are
inÞnite,
> since we can easily exhaustively enumerate all the Þnite
ones.
busy beaver outputs just an integer.
>How long will this take until you come up with a sequence I
can¹t
compute?
> You clearly didn¹t really understand the 2nd-to-last
paragraph of my
previous
> post. For me to come up with a sequence means that I can
compute it,
right?
> If I can compute it then so can you. NO uncomputable thing
is
come-upable
> with as a Þnite speciÞcation -- THAT¹S precisely what it
means for it
to be
> uncomputable in the Þrst place!
halt is Þnitely speciÞed
> You seem to admit the (uncomputable) existence of SOME
uncomputable
things
> (e.g. Chaitin¹s Omega, the Halts(tm) function). If so, than
what¹s the
big
> deal about accepting the proofs that there are other
uncomputable
sequences?
Because unknowns aren¹t numbers. Omega is merely a glimpse
into the
future,
its not physically a number at all. Remember the inÞnite non
computable
number
I constructed. If the nearest star will explode thats 1, if
not thats 0,
then add
digits for each star going further out. (assume an inÞnite
sized
universe). That¹s
almost identical deÞnition of the halting number, will the
processes
terminate?
The future¹s not ours to see. Its really just babble thrown
in because
Cantor¹s
proof was accepted, so everything that came after was
tailored to Þt
theory.
Herc
Accept no deÞnitions
===
Subject: Maple help?
Since I don¹t have access to Maple right now and won¹t until
tomorrow
night,
would someone please please calculate for me the integral of
f(x) = x^(j+r-1) * e^(-tx)
from x = 0 to inÞnity
So Integral of f(x) dx
from x = 0 to inÞnity?
I think it¹s pretty obvious that integration by parts would
be used here,
but I don¹t trust my bookkeeping here and need precision
(which is why I
resort to Maple)
(j is an integer greater than or equal to zero, and r,t are
both real
numbers greater than or equal to zero)
Moshe
===
Subject: Re: Maple help?
>Since I don¹t have access to Maple right now and won¹t until
tomorrow
night,
>would someone please please calculate for me the integral of
>f(x) = x^(j+r-1) * e^(-tx)
>from x = 0 to inÞnity
>(j is an integer greater than or equal to zero, and r,t are
both real
>numbers greater than or equal to zero)
t^(-j-r) Gamma(j+r)
assuming t > 0. Just use the substitution u = t x and the
deÞnition of
the Gamma function.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Help with Butterworth Filter
Hi All,
I¹ve been plowing through pages of theory in dsp books trying
to work
out how to create a butterworth Þlter. They all seem to use
different symbols and many different forms of the same
equation. Lots
of theory but very little of practical examples.
Basically I have a series of 1500 points taken at even
intervals.
Let¹s call it f(t). Eg
t f(t)
1 3412
2 3432
3 3300
4 3433
......... etc
Basically I¹m looking to numerically produce g(t) which is
f(t) with
all frequencies higher than 0.01 (seconds/data point) Þltered
out.
Keeping it simple - lets say I want a second order (ie 2 pole)
Butterworth Filter this gives (from a table in a book)
a0= 8.663387E-04
a1= 1.732678E-03 b1= 1.919129E+00
a2= 8.663387E-04 b2= -9.225943E-01
Question is - what is the formula that I plug these Þve
constants
into to give me g(t), ie the lowpass Þlter output at a point
t???
===
Subject: Re: Help with Butterworth Filter
Hii Stewart,
look at
http://www.mathworks.com/access/helpdesk/help/toolbox/signal/
butter.html
for general information about implementing a butterworth
Þlter.
Then check
http://www.mathworks.com/access/helpdesk/help/toolbox/signal/
dÞlt.df1.html
and you¹ll see a graphical representation which explains the
use of
the b[] and a[] coefÞcients.
Be aware: some books name the coefÞcients and or their signs
differently.
I guess that the source of your coefÞcients uses a and b the
other
way round.
Hope this gives some idea how to go on.
Bernhard
===
Subject: Re: Help with Butterworth Filter
> Hi All,
I¹ve been plowing through pages of theory in dsp books trying
to work
> out how to create a butterworth Þlter. They all seem to use
> different symbols and many different forms of the same
equation. Lots
> of theory but very little of practical examples.
Basically I have a series of 1500 points taken at even
intervals.
> Let¹s call it f(t). Eg
t f(t)
> 1 3412
> 2 3432
> 3 3300
> 4 3433
> ......... etc
Basically I¹m looking to numerically produce g(t) which is
f(t) with
> all frequencies higher than 0.01 (seconds/data point)
Þltered out.
Keeping it simple - lets say I want a second order (ie 2 pole)
> Butterworth Filter this gives (from a table in a book)
a0= 8.663387E-04
> a1= 1.732678E-03 b1= 1.919129E+00
> a2= 8.663387E-04 b2= -9.225943E-01
Question is - what is the formula that I plug these Þve
constants
> into to give me g(t), ie the lowpass Þlter output at a
point t???
First observation: that doesn¹t look right. The transfer
function
should be normalized (see OP) so that the highest-order term
in the
denominator is one:
b_2 z^2 + b_1 z
H(z) = -------------------
z^2 + a_1 z + a_0
Then g(t) = a_1 * g(t-1) + a_0 * g(t-2) + b_2 * f(t) + b_1 *
f(t-1).
Note that this is recursive, and that g(t) never goes to zero
once f(t)
has been wiggled. That makes it an inÞnite impulse response
Þlter.
Second observation: Why in the heck are you messing with IIR
Þlters if
you have a DSP? Unless you¹re designing a closed-loop control
system
where delay is critical you can get much better attenuation
performance
(and better phase matching) by using a FIR (Þnite impulse
response
Þlter). And if you _are_ designing a closed-loop control
system you
want to avoid the Butterworth and it¹s ilk like the plague.
Not only
that, but the whole motivation for a DSP having a blazing fast
multiply-accumulate instruction is so you can implement FIR
Þlters
efÞciently.
--
Tim Wescott
Wescott Design Services
http://www.wescottdesign.com
===
Subject: Re: Help with Butterworth Filter
...
> a0= 8.663387E-04
> a1= 1.732678E-03 b1= 1.919129E+00
> a2= 8.663387E-04 b2= -9.225943E-01
Question is - what is the formula that I plug these Þve
constants
> into to give me g(t), ie the lowpass Þlter output at a
point t???
First observation: that doesn¹t look right. The transfer
function
> should be normalized (see OP) so that the highest-order
term in the
> denominator is one:
b_2 z^2 + b_1 z
> H(z) = -------------------
> z^2 + a_1 z + a_0
Signal processing guys prefer the technically realizable form:
H(z) = (a_0 + a_1 z^{-1} + a_2 z^{-2} )/(1 - (b_1 z^{-1} + b_2
z^{-2}))
and this looks perfectly allright to me. This system has a
frequency
response of a bilinear transformed lowpass Þlter with cutoff
frequency Fc = 0.01 ( = fc / fs). It is realized by this
recursion
equation:
g(t) = a_0 f(t) + a_1 f(t-T) + a_2 f(t-2T) + b_1 g(t-T) + b_2
g(t-2T)
where T is the sampling period.
> Second observation: Why in the heck are you messing with
IIR Þlters if
> you have a DSP?
There are many good reasons for IIR Þlters. If done right they
perform as expected (btw, the DSP MAC instruction is just as
applicable to IIR as to FIR Þlters).
Andor
===
Subject: Re: Help with Butterworth Filter

> ...
>
>a0= 8.663387E-04
>a1= 1.732678E-03 b1= 1.919129E+00
>a2= 8.663387E-04 b2= -9.225943E-01
>Question is - what is the formula that I plug these Þve
constants
>into to give me g(t), ie the lowpass Þlter output at a point
t???
>>First observation: that doesn¹t look right. The transfer
function
>>should be normalized (see OP) so that the highest-order
term in the
>>denominator is one:
>> b_2 z^2 + b_1 z
>>H(z) = -------------------
>> z^2 + a_1 z + a_0

> Signal processing guys prefer the technically realizable
form:
H(z) = (a_0 + a_1 z^{-1} + a_2 z^{-2} )/(1 - (b_1 z^{-1} + b_2
> z^{-2}))
and this looks perfectly allright to me. This system has a
frequency
> response of a bilinear transformed lowpass Þlter with cutoff
> frequency Fc = 0.01 ( = fc / fs). It is realized by this
recursion
> equation:
g(t) = a_0 f(t) + a_1 f(t-T) + a_2 f(t-2T) + b_1 g(t-T) + b_2
g(t-2T)
where T is the sampling period.

>>Second observation: Why in the heck are you messing with
IIR Þlters if
>>you have a DSP?

> There are many good reasons for IIR Þlters. If done right
they
> perform as expected (btw, the DSP MAC instruction is just as
> applicable to IIR as to FIR Þlters).
Andor
Boy I¹m getting a lot of grief for that comment -- which I
suppose I
should, since I probably implement more IIR Þlters than FIR
ones
myself. And yes, the DSP MAC instruction is just as
applicable, but if
you¹re only multiplying a few things together you spend as
many clock
ticks in extra setup for the 1-cycle MAC is you save by
having it happen
in one cycle. If _all_ you¹re doing is one or two IIR Þlters
you may
as well do it on a fast RISC processor as a DSP -- you only
see value
from the one-cycle MAC if you can express a bunch of IIR
Þlters as a
matrix multiply or if you¹re doing FIR Þlters.
--
Tim Wescott
Wescott Design Services
http://www.wescottdesign.com
===
Subject: Re: Help with Butterworth Filter
> Hi All,
I¹ve been plowing through pages of theory in dsp books trying
to work
> out how to create a butterworth Þlter. They all seem to use
> different symbols and many different forms of the same
equation. Lots
> of theory but very little of practical examples.
Basically I have a series of 1500 points taken at even
intervals.
> Let¹s call it f(t). Eg
t f(t)
> 1 3412
> 2 3432
> 3 3300
> 4 3433
> ......... etc
Basically I¹m looking to numerically produce g(t) which is
f(t) with
> all frequencies higher than 0.01 (seconds/data point)
Þltered out.
Keeping it simple - lets say I want a second order (ie 2 pole)
> Butterworth Filter this gives (from a table in a book)
a0= 8.663387E-04
> a1= 1.732678E-03 b1= 1.919129E+00
> a2= 8.663387E-04 b2= -9.225943E-01
Question is - what is the formula that I plug these Þve
constants
> into to give me g(t), ie the lowpass Þlter output at a
point t???
> First observation: that doesn¹t look right. The transfer
function
> should be normalized (see OP) so that the highest-order
term in the
> denominator is one:
> b_2 z^2 + b_1 z
> H(z) = -------------------
> z^2 + a_1 z + a_0
> Then g(t) = a_1 * g(t-1) + a_0 * g(t-2) + b_2 * f(t) + b_1
* f(t-1).
> Note that this is recursive, and that g(t) never goes to
zero once f(t)
> has been wiggled. That makes it an inÞnite impulse response
Þlter.
http://ccrma-www.stanford.edu/~jos/Þlters/BiQuad_Section.html
has some
more
info. Note, the author uses beta and alpha instead of a and b.
Here¹s some C code if that is of use to you:
http://www.dspguru.com/sw/lib/biquad.c
> Second observation: Why in the heck are you messing with
IIR Þlters if
> you have a DSP? Unless you¹re designing a closed-loop
control system
> where delay is critical you can get much better attenuation
performance
> (and better phase matching) by using a FIR (Þnite impulse
response
> Þlter). And if you _are_ designing a closed-loop control
system you
> want to avoid the Butterworth and it¹s ilk like the plague.
Not only
> that, but the whole motivation for a DSP having a blazing
fast
> multiply-accumulate instruction is so you can implement FIR
Þlters
> efÞciently.
There are plenty of applications where using IIR Þlters makes
sense on a
DSP.
IIR Þlters can be much more efÞcient than FIRs, especially
when the
cut-off
frequency is low compared to the sample rate. They also have
a simplicity
and
ease of parameterization that FIRs don¹t if you need to
calculate Þlter
coefÞcients on the þy.
===
Subject: Re: Help with Butterworth Filter
>
>Hi All,
>I¹ve been plowing through pages of theory in dsp books
trying to work
>out how to create a butterworth Þlter. They all seem to use
>different symbols and many different forms of the same
equation. Lots
>of theory but very little of practical examples.
>Basically I have a series of 1500 points taken at even
intervals.
>Let¹s call it f(t). Eg
>t f(t)
>1 3412
>2 3432
>3 3300
>4 3433
>......... etc
>Basically I¹m looking to numerically produce g(t) which is
f(t) with
>all frequencies higher than 0.01 (seconds/data point)
Þltered out.
>Keeping it simple - lets say I want a second order (ie 2
pole)
>Butterworth Filter this gives (from a table in a book)
>a0= 8.663387E-04
>a1= 1.732678E-03 b1= 1.919129E+00
>a2= 8.663387E-04 b2= -9.225943E-01
>Question is - what is the formula that I plug these Þve
constants
>into to give me g(t), ie the lowpass Þlter output at a point
t???
>>First observation: that doesn¹t look right. The transfer
function
>>should be normalized (see OP) so that the highest-order
term in the
>>denominator is one:
>> b_2 z^2 + b_1 z
>>H(z) = -------------------
>> z^2 + a_1 z + a_0
>>Then g(t) = a_1 * g(t-1) + a_0 * g(t-2) + b_2 * f(t) + b_1
* f(t-1).
>>Note that this is recursive, and that g(t) never goes to
zero once f(t)
>>has been wiggled. That makes it an inÞnite impulse response
Þlter.

>
http://ccrma-www.stanford.edu/~jos/Þlters/BiQuad_Section.html
has some
more
> info. Note, the author uses beta and alpha instead of a and
b.
> Here¹s some C code if that is of use to you:
> http://www.dspguru.com/sw/lib/biquad.c

>>Second observation: Why in the heck are you messing with
IIR Þlters if
>>you have a DSP? Unless you¹re designing a closed-loop
control system
>>where delay is critical you can get much better attenuation
performance
>>(and better phase matching) by using a FIR (Þnite impulse
response
>>Þlter). And if you _are_ designing a closed-loop control
system you
>>want to avoid the Butterworth and it¹s ilk like the plague.
Not only
>>that, but the whole motivation for a DSP having a blazing
fast
>>multiply-accumulate instruction is so you can implement FIR
Þlters
>>efÞciently.

> There are plenty of applications where using IIR Þlters
makes sense on a
DSP.
> IIR Þlters can be much more efÞcient than FIRs, especially
when the
cut-off
> frequency is low compared to the sample rate. They also
have a simplicity
and
> ease of parameterization that FIRs don¹t if you need to
calculate Þlter
> coefÞcients on the þy.

True, but this guy is obviously a beginner, and his comments
make it
sound like he¹s looking for a brick-wall Þlter. Both of these
indicate
that his life would be simpliÞed with a FIR.
--
Tim Wescott
Wescott Design Services
http://www.wescottdesign.com
===
Subject: Re: Help with Butterworth Filter
The constants seem scaled already, should be normalized to 1.
(These are locations of zeros and poles in the complex plane
for the
Þlter).
> Hi All,
> I¹ve been plowing through pages of theory in dsp books
trying to work
> out how to create a butterworth Þlter. They all seem to use
> different symbols and many different forms of the same
equation. Lots
> of theory but very little of practical examples.
> Basically I have a series of 1500 points taken at even
intervals.
> Let¹s call it f(t). Eg
> t f(t)
> 1 3412
> 2 3432
> 3 3300
> 4 3433
> ......... etc
> Basically I¹m looking to numerically produce g(t) which is
f(t) with
> all frequencies higher than 0.01 (seconds/data point)
Þltered out.
> Keeping it simple - lets say I want a second order (ie 2
pole)
> Butterworth Filter this gives (from a table in a book)
> a0= 8.663387E-04
> a1= 1.732678E-03 b1= 1.919129E+00
> a2= 8.663387E-04 b2= -9.225943E-01
> Question is - what is the formula that I plug these Þve
constants
> into to give me g(t), ie the lowpass Þlter output at a
point t???
===
Subject: John Baez¹s This Week in Mathematics - 200
algebraic theory: http://math.ucr.edu/home/baez/week200.html
ŒBut Lawvere¹s really big idea was that there¹s a certain
category with
Þnite products whose only goal in life is to contain a group
object. To
build this category, Þrst we put in an object
G
Since our category has Þnite products this automatically
means it gets
objects 1, G, G x G, G x G x G, and so on. Next, we put in a
binary
operation called multiplication, namely a morphism
m: G x G -> G
We also put in a unary operation called inverse:
inv: G -> G
and a nullary operation called the unit:
i: 1 -> G
And then we say a bunch of diagrams commute, which express
all the axioms
for a group listed above.¹
In this section, is Prof. Baez talking about the classifying
category
for the algebraic theory of groups?
===
Subject: Re: Rotations via Dissections Without Rotations
>> If you take a square, and then cut it as you wish into a
Þnite number
>> of pieces, then rearrange the pieces *without rotating any
of them*, I
>> wonder if you can form another square that is at
45-degrees to the
>> original.
>[ As a possible hint, my method is a nine-piece dissection.
Other
>students at the same training camp came up with different
solutions,
>at least one of which involved fewer pieces. ]
On a related but not quite identical note, Laczkovich in 1990
proved that
any two polygons of equal area are equidecomposable without
rotation. In
the same paper he gives the stunning result that a circle and
a square are
also likewise equidecomposable! I presume that the pieces are
nothing even
remotely resembling simply-connected polygons, as the
construction makes
extensive use of AC, IIRC.
-- Erick
===
Subject: Re: Rotations via Dissections Without Rotations
) On a related but not quite identical note, Laczkovich in
1990 proved that
) any two polygons of equal area are equidecomposable without
rotation.
If I recall correctly, once the AoC gets involved, they need
not be of
equal area. Or did that only work with spheres ?
SaSW, Willem
--
Disclaimer: I am in no way responsible for any of the
statements
made in the above text. For all I know I might be
drugged or something..
No I¹m not paranoid. You all think I¹m paranoid, don¹t you !
#EOT
===
Subject: Re: Rotations via Dissections Without Rotations
> ) On a related but not quite identical note, Laczkovich in
1990 proved
that
> ) any two polygons of equal area are equidecomposable
without rotation.
If I recall correctly, once the AoC gets involved, they need
not be of
> equal area. Or did that only work with spheres ?
IIRC, it only works when the dimension is greater than 2.
Jose Carlos Santos
===
Subject: Re: Rotations via Dissections Without Rotations
>> If I recall correctly, once the AoC gets involved, they
need not be of
>> equal area. Or did that only work with spheres ?
IIRC, it only works when the dimension is greater than 2.
Krzysztof Ciesielski (The Mathematical Intelligencer 11,
No.2, 54-58
(1989)).
Jose Carlos Santos
===
Subject: Re: Rotations via Dissections Without Rotations
> On a related but not quite identical note, Laczkovich in
1990 proved
> that any two polygons of equal area are equidecomposable
without
> rotation. In the same paper he gives the stunning result
that a
> circle and a square are also likewise equidecomposable! I
presume
> that the pieces are nothing even remotely resembling
simply-connected
> polygons, as the construction makes extensive use of AC,
IIRC.
The pieces are certainly non-measurable. He raises the
question in
that paper of when two polygons are equidecomposable if one is
restricted to measurable pieces.
Certainly a handwaving argument can convince one that, if we
restrict
to Þnite numbers of pieces, translations, and scissors
congruence
(i.e., each piece looks like something we could cut out --
topologically
speaking, a disk) that certain dissections are impossible;
for instance,
no triangle is equicomposable with a square under these
conditions.
Indeed, I think that the condition for equidecomposability
ends up
being something like this:
Associate to every direction (i.e., unit vector in the plane)
a
formal value +-theta (0 <= theta < pi). Travel around the
polygon,
calulating the formal sum
Sigma(length(edge)*direction(edge)). Then
two polygons are equidecomposable (under the above
restrictions) iff
the two formal sums are the same.
[ Examples: the formal sum of any parallelogram (and hence
square)
is 0, since the opposite edges are parallel and of equal
length, but
traversed in opposing directions. So the formal sum looks like
f(parallelogram) = e1*[theta1] + e2*[theta2] - e1*[theta1] -
e2*[theta2]
= (e1 - e1)*[theta1] + (e2 - e2)*[theta2]
= 0
The triangle between (0,0), (0,1) and (1,0), however, has the
formal
sum:
f(triangle) = 1*[0] + Sqrt(2)*[3*pi/4] - 1*[pi/2]
!= 0
]
Geoff.
-------------------------------------------------------------
---------------
-
Geoff Bailey (Fred the Wonder Worm) | Programmer by trade --
ftww@maths.usyd.edu.au | Gameplayer by vocation.
-------------------------------------------------------------
---------------
-
===
Subject: Re: Rotations via Dissections Without Rotations
Can you tell me if there are books or links about dissection
puzzles or
geometry of dissections?
Giorgio
===
Subject: Re: Rotations via Dissections Without Rotations
> Can you tell me if there are books or links about
dissection puzzles or
> geometry of dissections?
Do a Google search with dissection puzzle.
Jose Carlos Santos
Distribution: world
===
Subject: Re: Variables inside and outside the exponent?
> So your f(x) is something like x + 3. If your g(x) were
similarly
simple
> -- say, something like x + n -- then the Lambert W function
could be used
> to solve the equation symbolically.
I haven¹t been taught about the Lambert W function, and
though I¹ve
Þddled with it, not much luck. The exact form I need to solve
is the
following (in terms of v)
v - k1 = k1 e ^ (k2 * D(v) / v)
where D(v) is an anti-derivative of v. I need to solve for v
in terms
of its anti-derivative.
How to apply the W function to that, I¹m not sure.
Starling
===
Subject: Lehigh University Geometry and Topology Conference
Lehigh University Geometry and Topology Conference
The conference will start at 11:00 am on Thursday, June 10.
The Þrst
talk will be at 11:00 am Thursday (a change from previous
years), and the
last talk will end before 5:00 Saturday, June 12.
Principal Speakers:

Colin Adams, Williams College

Jesper Grodal, Univ. of Chicago

Peter Li, Univ. of California, Irvine

Yair Minsky, Yale Univ.

Shing-Tung Yau, Harvard Univ.

Wolfgang Ziller, Univ. of Pennsylvania
In addition, there will be parallel sessions of 40-minute
contributed
talks, divided roughly into Differential and Complex
Geometry, Algebraic
Topology, and Geometric Topology.
Breakfast will be provided Friday and Saturday mornings, and
lunch will be
provided Thursday, Friday and Saturday noons. Dinner will be
the only
meal not provided gratis. On Thursday, expeditions to nearby
restaurants
will be arranged, followed by a party. On Friday there will
be a banquet
at a cost of $30.
More information will be available at the website
http://www.lehigh.edu/~math/geotop.html as it becomes
available. Please
check there from time to time.
The conference is supported by Lehigh University, and the
NSF. Limited
travel travel is available for recent PhD¹s, current graduate
students
and members of underrepresented groups. Participants
interested in being
considered for this support should complete the request for
travel support
--
David L. Johnson
__o | There is always an easy solution to every human problem
- neat,
_`(,_ | plausible, and wrong. --H.L. Mencken
(_)/ (_) |

===
Subject: BRILLLLIANT ARTICLE BY DR THOMAS FRIEDMAN OF NYTIMES
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BRILLLLIANT ARTICLE BY DR THOMAS FRIEDMAN OF NYTIMES
He is twice pullitzer winner
OP-ED COLUMNIST
> Awaking to a Dream
> By THOMAS L. FRIEDMAN
Columnist Page: Thomas L. Friedman
>
> I have a confession to make: I am the foreign affairs
columnist for The
> New York Times and I didn¹t listen to one second of the
9/11 hearings
> and I didn¹t read one story in the paper about them. Not
one second. Not
> one story.
Lord knows, it¹s not out of indifference to 9/11. It¹s
because I made up
> my mind about that event a long time ago: It was not a
failure of
> intelligence, it was a failure of imagination. We could
have had perfect
> intelligence on all the key pieces of 9/11, but the fact is
we lacked ?
> for the very best of reasons ? people with evil enough
imaginations to
> put those pieces together and realize that 19 young men
were going to
> hijack four airplanes for suicide attacks against our
national symbols
> and kill as many innocent civilians as they could, for no
stated reason
> at all.
Imagination is on my mind a lot these days, because it seems
to me that
> the only people with imagination in the world right now are
the bad
> guys. As my friend, the Middle East analyst Stephen P.
Cohen, says,
> That is the characteristic of our time ? all the
imagination is in the
> hands of the evildoers.
I am so hungry for a positive surprise. I am so hungry to
hear a
> politician, a statesman, a business leader surprise me in a
good way. It
> has been so long. It¹s been over 10 years since Yitzhak
Rabin thrust out
> his hand to Yasir Arafat on the White House lawn. Yes, yes,
I know,
> Arafat turned out to be a fraud. But for a brief, shining
moment, an old
> warrior, Mr. Rabin, stepped out of himself, his past, and
all his scar
> tissue, and imagined something different. It¹s been a long
time.
I have this routine. I get up every morning around 6 a.m.,
Þre up my
> computer, call up AOL¹s news page and then hold my breath
to see what
> outrage has happened in the world overnight. A massive
bombing in Iraq
> or Madrid? More murderous violence in Israel? A hotel going
up in þames
> in Bali or a synagogue in Istanbul? More U.S. soldiers
killed in Iraq?
I so hunger to wake up and be surprised with some really good
news ? by
> someone who totally steps out of himself or herself,
imagines something
> different and thrusts out a hand.
I want to wake up and read that President Bush has decided to
offer a
> real alternative to the stalled Kyoto Protocol to reduce
global
> warming. I want to wake up and read that 10,000 Palestinian
mothers
> marched on Hamas headquarters to demand that their sons and
daughters
> never again be recruited for suicide bombings. I want to
wake up and
> read that Crown Prince Abdullah of Saudi Arabia invited
Ariel Sharon to
> his home in Riyadh to personally hand him the Abdullah
peace plan and
> Mr. Sharon responded by freezing Israeli settlements as a
good-will
> gesture.
I want to wake up and read that General Motors has decided it
will no
> longer make gas-guzzling Hummers and President Bush has
decided to
> replace his limousine with an armor-plated Toyota Prius, a
hybrid car
> that gets over 40 miles to the gallon.
I want to wake up and read that Dick Cheney has apologized to
the
> U.N. and all our allies for being wrong about W.M.D. in
Iraq, but then
> appealed to our allies to join with the U.S. in an even
more important
> project ? helping Iraqis build some kind of democratic
framework. I want
> to wake up and read that Tom DeLay called for a tax hike on
the rich in
> order to save Social Security and Medicare for the next
generation and
> to Þnance all our underfunded education programs.
I want to wake up and read that Justice Antonin Scalia has
recused
> himself from ruling on the case involving Mr. Cheney¹s
energy task force
> when it comes before the Supreme Court ? not because Mr.
Scalia did
> anything illegal in duck hunting with the V.P., but because
our Supreme
> Court is so sacred, so vital to what makes our society
special ? its
> rule of law ? that he wouldn¹t want to do anything that
might have even
> a whiff of impropriety.
I want to wake up and read that Mr. Bush has announced a
Manhattan
> Project to develop renewable energies that will end
America¹s addiction
> to crude oil by 2010. I want to wake up and read that Mel
Gibson just
> announced that his next Þlm will be called Moses and all
the proÞts
> will be donated to the Holocaust Museum.
Most of all, I want to wake up and read that John Kerry just
asked John
> McCain to be his vice president, because if Mr. Kerry wins
he intends
> not to waste his four years avoiding America¹s hardest
problems ? health
> care, deÞcits, energy, education ? but to tackle them, and
that can
> only be done with a bipartisan spirit and bipartisan team.

===
Subject: Re: BRILLLLIANT ARTICLE BY DR THOMAS FRIEDMAN OF
NYTIMES
> BRILLLLIANT ARTICLE BY DR THOMAS FRIEDMAN OF NYTIMES
He is twice pullitzer winner

>>OP-ED COLUMNIST
>>Awaking to a Dream
>>By THOMAS L. FRIEDMAN
>>Columnist Page: Thomas L. Friedman
>>
>>I have a confession to make: I am the foreign affairs
columnist for The
>>New York Times and I didn¹t listen to one second of the
9/11 hearings
>>and I didn¹t read one story in the paper about them. Not
one second. Not
>>one story.
>>Lord knows, it¹s not out of indifference to 9/11. It¹s
because I made up
>>my mind about that event a long time ago: It was not a
failure of
>>intelligence, it was a failure of imagination. We could
have had perfect
>>intelligence on all the key pieces of 9/11, but the fact is
we lacked ?
>>for the very best of reasons ? people with evil enough
imaginations to
>>put those pieces together and realize that 19 young men
were going to
>>hijack four airplanes for suicide attacks against our
national symbols
>>and kill as many innocent civilians as they could, for no
stated reason
>>at all.
>>Imagination is on my mind a lot these days, because it
seems to me that
>>the only people with imagination in the world right now are
the bad
>>guys. As my friend, the Middle East analyst Stephen P.
Cohen, says,
>>That is the characteristic of our time ? all the
imagination is in the
>>hands of the evildoers.
>>I am so hungry for a positive surprise. I am so hungry to
hear a
>>politician, a statesman, a business leader surprise me in a
good way. It
>>has been so long. It¹s been over 10 years since Yitzhak
Rabin thrust out
>>his hand to Yasir Arafat on the White House lawn. Yes, yes,
I know,
>>Arafat turned out to be a fraud. But for a brief, shining
moment, an old
>>warrior, Mr. Rabin, stepped out of himself, his past, and
all his scar
>>tissue, and imagined something different. It¹s been a long
time.
>>I have this routine. I get up every morning around 6 a.m.,
Þre up my
>>computer, call up AOL¹s news page and then hold my breath
to see what
>>outrage has happened in the world overnight. A massive
bombing in Iraq
>>or Madrid? More murderous violence in Israel? A hotel going
up in þames
>>in Bali or a synagogue in Istanbul? More U.S. soldiers
killed in Iraq?
>>I so hunger to wake up and be surprised with some really
good news ? by
>>someone who totally steps out of himself or herself,
imagines something
>>different and thrusts out a hand.
>>I want to wake up and read that President Bush has decided
to offer a
>>real alternative to the stalled Kyoto Protocol to reduce
global
>>warming. I want to wake up and read that 10,000 Palestinian
mothers
>>marched on Hamas headquarters to demand that their sons and
daughters
>>never again be recruited for suicide bombings. I want to
wake up and
>>read that Crown Prince Abdullah of Saudi Arabia invited
Ariel Sharon to
>>his home in Riyadh to personally hand him the Abdullah
peace plan and
>>Mr. Sharon responded by freezing Israeli settlements as a
good-will
>>gesture.
>>I want to wake up and read that General Motors has decided
it will no
>>longer make gas-guzzling Hummers and President Bush has
decided to
>>replace his limousine with an armor-plated Toyota Prius, a
hybrid car
>>that gets over 40 miles to the gallon.
>>I want to wake up and read that Dick Cheney has apologized
to the
>>U.N. and all our allies for being wrong about W.M.D. in
Iraq, but then
>>appealed to our allies to join with the U.S. in an even
more important
>>project ? helping Iraqis build some kind of democratic
framework. I want
>>to wake up and read that Tom DeLay called for a tax hike on
the rich in
>>order to save Social Security and Medicare for the next
generation and
>>to Þnance all our underfunded education programs.
>>I want to wake up and read that Justice Antonin Scalia has
recused
>>himself from ruling on the case involving Mr. Cheney¹s
energy task force
>>when it comes before the Supreme Court ? not because Mr.
Scalia did
>>anything illegal in duck hunting with the V.P., but because
our Supreme
>>Court is so sacred, so vital to what makes our society
special ? its
>>rule of law ? that he wouldn¹t want to do anything that
might have even
>>a whiff of impropriety.
>>I want to wake up and read that Mr. Bush has announced a
Manhattan
>>Project to develop renewable energies that will end
America¹s addiction
>>to crude oil by 2010. I want to wake up and read that Mel
Gibson just
>>announced that his next Þlm will be called Moses and all
the proÞts
>>will be donated to the Holocaust Museum.
>>Most of all, I want to wake up and read that John Kerry
just asked John
>>McCain to be his vice president, because if Mr. Kerry wins
he intends
>>not to waste his four years avoiding America¹s hardest
problems ? health
>>care, deÞcits, energy, education ? but to tackle them, and
that can
>>only be done with a bipartisan spirit and bipartisan team.

>
What butt munch would cross post this crap in a language
group. Like I
care about your political views.
===
Subject: Re: BRILLLLIANT ARTICLE BY DR THOMAS FRIEDMAN OF
NYTIMES
> BRILLLLIANT ARTICLE BY DR THOMAS FRIEDMAN OF NYTIMES
>
Friedman¹s a fool. Knows a lot yet consistently draws the
wrong
conclusions.
===
Subject: Re: BRILLLLIANT ARTICLE BY DR THOMAS FRIEDMAN OF
NYTIMES
charset=Windows-1252
BRILLLLIANT ARTICLE BY DR THOMAS FRIEDMAN OF NYTIMES
> OP-ED COLUMNIST
> Awaking to a Dream
> By THOMAS L. FRIEDMAN
> I want to wake up and read that Mr. Bush has announced a
Manhattan
> Project to develop renewable energies that will end
America¹s addiction
> to crude oil by 2010. I want to wake up and read that Mel
Gibson just
> announced that his next Þlm will be called Moses and all the
proÞts
> will be donated to the Holocaust Museum.
>
..........ahahahaha.....ahahaha......what we have here is a
very
enterprising advocate of combining environmentalism with the
holocaust industry, both being exclusive machinations to make
$$$$ but really not much else.......
I have seen¹em come and I have seen¹em go, and now Admirer
and Friedman, 2 Yiddles, are trying to get into the
$$$action........
Hey guys, the good money times in these schemes have gone.
ahahahahaha.....ahahahanson
===
Subject: Are you crazy? useless question.
um......
i saw the book The man who loved only numbers - Paul Erdos.
In the contents of a book ,
we mathematicians are all a little bit crazy
um.....are you really crazy??
Can i also crazy if i have studied hard?
um..........
===
Subject: Re: Are you crazy? useless question.
>i saw the book The man who loved only numbers - Paul Erdos.
Now read the movie!
>In the contents of a book ,
>we mathematicians are all a little bit crazy
>um.....are you really crazy??
No we¹re not.
Yes we are.
dave
dave
===
Subject: Re: Are you crazy? useless question.
Ernst Bloch...
After he killed a few of his relatives with the sabre that
was part of
his army uniform, he was comitted to a Paris hospital for the
--
G. A. Edgar
http://www.math.ohio-state.edu/~edgar/
===
Subject: Re: Are you crazy? useless question.
hot-girl
> i saw the book The man who loved only numbers - Paul Erdos.
> In the contents of a book ,
> we mathematicians are all a little bit crazy
> um.....are you really crazy??
Yes, enough, but I don¹t class myself as a mathematician.
> Can i also crazy if i have studied hard?
It helps to be Hungarian... But really, in my experience
of crazy people, which is very extensive, mathematicians
do not stand out. Maybe the subject itself has a saning
effect upon them. Erdos was original, but not at all in the
real weirdo class of Brouwer, Bieberbach, Kaczynski,
Fourier, and a few others, most of whom, notice, had
political notions on the brain, in addition to mathematics.
It seems to be a dangerous mix.
===
Subject: Re: Are you crazy? useless question.
> Erdos was original, but not at all in the
> real weirdo class of Brouwer, Bieberbach, Kaczynski,
> Fourier, and a few others, most of whom, notice, had
> political notions on the brain, in addition to mathematics.
> It seems to be a dangerous mix.
Bieberbach was a Nazi, but that was not particularly unusual
for a
German of his time.
Brouwer¹s _mathematical_ opinions (intuitionism) were
certainly
strange, but did he qualify as a weirdo as a person? I simply
don¹t
know anything about his life, except that it was a VERY long
one.
Kaczynski - wasn¹t that the Unabomber? But then, he wasn¹t in
any
sense an outstanding mathematician, like the other ones you
mention.
As for Fourier, I suspect you may be confusing the
mathematician with
his namesake the utopian socialist. The former was anything
but a
weirdo, and made a very competent pr.8efet (governor) of the
French
d.8epartement of Is.8fre under Napoleon. You might replace
him on your
list with Grothendieck - an eminent French mathematician who
at some
point gave up on science and acedemia and withdrew to a
shepherd¹s hut
in the Pyrenees, or so it was rumored in the late sixties...
Angelos TSIRIMOKOS, Brussels
===
Subject: Re: A Klein Bottle has no inside or outside.
> Please correct me if I am in error, and where.
I will attempt to correct you with gentleness, though it is
hard to do so
with one who shows both considerable expertise and also great
humility.
> Coming to Cross-cap KB :
> The KB (IMO) is made thus: Take a þexible tube , introduce a
> cut/incision on the side of tube big enough to push back
one end of
> tube into the sectioned area, bring together tube ends and
glue/weld/fuse
Yes, that is a standard physical model.
Alas! The lovely thingy you end up with is NOT a Klein
bottle, it is
a physical object which attempts to model a Klein bottle, and
worse
still does so with more inaccuracy than it needs to, in order
to make
an amusing toy that will actually hold liquids.
But it contains various falsities, both physical and
conceptual.
A major conceptual falsity with it is, that it models only an
immersion
of a KB into 3-space, NOT a true KB itself. A true KB cannot
be modelled
in physical 3 space, though it could be in 4-space, if we
lived in such
a universe, which we don¹t, (alas!)
A physical falsity is that, given this, and thus the need for
such a
surface
to have self-intersection, (which is what makes it an
immersion rather than
an embedding), it would STILL be more accurate to have the
cervix
closed
over so that liquid would pour in no further than that, and
leave the
further
depths of the (physical) inside totally dry. They do not do
this though,
just
because it is more fun to have it the way they do, even
though it is a lie!
A mathematical KB is a collection of points which are locally
topologically
2-Euclidean, and globally as suggested by the model. But it
would be
better to think of it merely as a þat rectangle where two
opposite sides
were pointwise identiÞed directly, and the other two opposite
sides
identiÞed reþectively. This identiÞcation is rather like that
of many computer war/search games where the cyber-universe
goes round and round forever.
One advantgae of this approach is that it reminds us that the
KB (& the
torus)
are actually FLAT - that their total (Gaussian) curvature is
actually zero.
I hope that is some use to you.
I don¹t fully understand your remarks concerning the Þgure-8
model,
but once again remember (i) it is only a physical model, (ii)
it is a model
of an immersion not an embedding, so has the unavoidable
self-intersection.
I¹m sure it is this latter matter that is putting you off,
imaginatively.
An immersion *can* have a true inside and outside - think of
a sphere
elongated into a tube, then bent around so that one (thinner)
end
goes through and out the other side of the other (fatter) end.
This is a self-intersecting immersion of the sphere into
3-space
which nonetheless still maintains an inside volume. Note that
if one
is to measure the actual physical volume properly, one would
have to
count the overlapping volume twice, to make it all come out
right!
However, the KB is not of this type. However it is immersed,
it will have
no deÞnable inside. Here is a cross-section of your Þgure-8...
_..-----.._ _..-----.._
.-¹ a b`-. .-¹ `-. ...this is a cross-section;
/ c `. .¹ it continues on through
| d >< | the back of your screen,
.¹e `. / round like an inner tube,
`-._ _.-¹ f `-._ _.-¹ back through the plane of
``-----¹¹ g `------¹¹ the screen but lower down,
and Þnally back to go into the screen from the air, at the
same place
here.
However, this Þgure-8 (model of a) KB has NO inside. This is
indicated by
the letters I have drawn in alphabetical order, showing how
they pass
through
the line of self-intersection (though NOT through their own
local
wall!),
and emerge from the apparent inside to the apparent outside.
?Es claro? (I expect the mailer has munged the characters!)
-------------------------------------------------------------
---------------
--
Bill Taylor W.Taylor@math.canterbury.ac.nz
-------------------------------------------------------------
---------------
--
Semantic continuity:- Garbage in, garbage out.
-------------------------------------------------------------
---------------
--
===
Subject: Re: A Klein Bottle has no inside or outside.
>A true KB cannot be modelled
>in physical 3 space,
if we insist that physical 3 space be locally Euclidian
(as is quite usually required) AND have certain particular
global restrictions--speciÞcally, the vanishing of its
Þrst Stiefel-Whitney class--on its topology (and there is
at this date no consensus on what such global restrictions
either ought to be, or in fact *are*).
For instance, if physical 3 space happened to be the
manifold RP3 (de Sitter¹s sphere modded out by the antipodal
map), then a true KB would be modelled by the boundary
of a tubular neighborhood of a projective line RP1 in RP3.
Lee Rudolph
===
Subject: Re: A Klein Bottle has no inside or outside.
Excellent ASCII portrayal indeed !
> Let me put it as it appeared in both cases when I Þrst saw
it.This is
> still my impression ...
Coming to the Figure of Eight/InÞnity symbol cross section KB:
> It seems to have two annular connected compartments forming
a
> continuous inside volume, but it is a single compartment
connected on
> the side along a Moebius twisted line. By a 0, 2 Pi
continuous v
> parameter variation, one can see and appreciate that two
revolutions
> make one closed volume that appears as toroidal continuous
volume with
> a clearly demarcated closed inside space. Still it does not
convince
> (at least to me right now) about no inside/outside
situation.
Well, literally speaking there *is* an enclosed volume
in the Þgure, as you say. But the surface that encloses
it has only one side; if you tried to paint just the outside
of one tube and kept painting along that surface through the
intersection line, you would Þnd yourself painting the inside
of the other tube -- and since they¹re really the same tube,
you end up painting everything.
The enclosed volume is an artifact of the immersion and
not intrinsic to the topology. You could do an analogous
thing,
say, with a cylinder in R^3 by making a dent in it and pushing
the dent all the way through the opposite side; there would be
a volume enclosed between the outside surface of the dent, and
the outside of the opposite wall of the cylinder, i.e. it
would
be *outside* the cylinder yet enclosed.
However,instead of continuity for v at 0, 2 Pi allowing a
> v-discontinuity
> rt=ParametricPlot3D[{x,y,z},{u,0,Pi},{v,.3,
6},PlotPoints->{25,13}];
> lt=ParametricPlot3D[{x,y,z},{u,0,-
Pi},{v,.3,6},PlotPoints->{25,13}];
> the S-shaped sections are now opened out and it is quite
convincing
> that the entire surface is a topological single sheet
without inside
> and outside. What I did not, or still perhaps unable to
properly
> express (in view of the subject being new)is that
parametrization
> should include/specify to cover every possibility of
determining of
> volume openness.
I see how it could be an interesting question, what immersions
in 3D do or do not contain enclosed volumes when physically
modeled. I¹m not the person to speculate on such things,
perhaps others in the group are.
Coming to Cross-cap KB :
As Bill Taylor pointed out earlier, the term crosscap refers
to something else. Best to avoid that term here.
> The KB (IMO) is made thus: Take a þexible tube , introduce a
> cut/incision on the side of tube big enough to push back
one end of
> tube into the sectioned area, bring together tube ends and
> glue/weld/fuse them to form neck/zero gauss curvature þat
cervix.
> Acme glass KBs are topologically representative as no
open/close space
> surfaces. A þy, after entering the cervix and going through
the bent
> tube gets inside the þask. If there is still a gap between
the tube
> and þask, the þy can escape through it , and it is here OK
to say
> Cross-cap KB no inside or outside. Else ( i.e., if þask is
fused to
> the tube as seen in the Acme KB photographs), the hapless
þy is
> trapped in the bottle,
Well actually, no, in this case. The þy could exit
by retracing his steps and going out the way he came
in. This immersion of the KB does not have an enclosed
volume. Btw I think you got this wrong in a previous
post, too.
and it is not OK to say Acme KB at least, has
> no inside/outside.
Saying there is no inside/outside means that you observe
the painting convention that I used above -- you follow
along the surface even if it intersects itself. If I had
done that with my cylinder example, it would still be
clear that only one side is the outside, and that is the
side that would get painted. But in this immersion of the
Klein bottle, as in the other one, the entire surface would
get painted; i.e. it is all outside (and all inside).
In other words, think of no inside/outside in this context
as a statement about the *surface*, not about any enclosed
volume that might be there by accident due to some
intersection
of the surface with itself.
===
Subject: Re: A Klein Bottle has no inside or outside.
[regarding the openness of the bent-tube Klein bottle]

> Well actually, no, in this case. The þy could exit
> by retracing his steps and going out the way he came
> in. This immersion of the KB does not have an enclosed
> volume. Btw I think you got this wrong in a previous
> post, too.
On second thought, what I said here isn¹t quite
kosher mathematically. While it is true that the
fabricators of the Acme KB punch a hole in the outer
surface so the bent tube can pass through, this hole
really is *not* there in the parameterization. So,
probably there should be an enclosed volume in the
model, strictly speaking. Of course in that case our
hypothetical þy would not be able to walk into the
enclosed volume.
===
Subject: Lebesgue dominated convergence and measures
Let f_n be a sequence of integrable functions and suppose
that {f_n}
converges to a function f. (m=measure) Show that:
If lim int( | f_n - f | dm) =0 then int ( | f| dm) = lim int
( |f_n| dm)
We notice that:
| |f_n| - |f| | <=|f_n - f|
So we can apply lebesgue dominated convergence theorem:
int( f dm) = lim int ( |f_n| - |f| dm) <= int( |f_n -f | dm)
(the
inequality
follows from the monoticity of the integral)
By asssumption: int ( | f_n - f| dm) =0
so lim int ( |f_n| - |f| dm ) =0
I don¹t know if the following step is valid (i.e linearity of
limits):
lim int( |f_n) dm) - lim int (|f| dm )=0
but lim int ( | f| dm ) = int ( |f| dm) (because |f| doesn¹t
depends on n)
so int(| f| dm)= lim int( | f_n| dm)
===
Subject: Re: Lebesgue dominated convergence and measures
>Let f_n be a sequence of integrable functions and suppose
that {f_n}
>converges to a function f. (m=measure) Show that:
>If lim int( | f_n - f | dm) =0 then int ( | f| dm) = lim int
( |f_n| dm)
>We notice that:
>| |f_n| - |f| | <=|f_n - f|
>So we can apply lebesgue dominated convergence theorem:
Let f_n(x) = 1 / x^[2-(n-1)/n]. Then:
f_n(x) is integrable for all n, but lim f_n(x) = 1/x is not
integrable. It follows that f - f_n is not integrable for any
n. How
do you use the dominated convergence theorem here?
--
I¹m not interested in mathematics that might have anything
to do with reality. -- Russell Easterly, in sci.math
===
Subject: graph theory: duality between indepdent set and
graph matching?
I try to establish duality between independent set and graph
matching.
Conceptually they are true duals.
If my problem is independent set problem but I want to use
graph matching
to
solve it( I already have graph matching program), how do I do
the
transformation of the graph?
My confusion comes from that if you map all edges to in graph
G to vertices
in dual graph G¹, and map all shared vertices in G to edges
in G¹, the dual
graph G¹ loses edges.
The vertices with odd-degree in G cannot be mapped to be
edges in graph G¹.
How do I solve this mapping problem?
Joenyim
===
Subject: Re: assignment problem with grouping
Your requirement (distributed algo with private cost
information) would
probably be satisÞed with some auction algorithm, where
disclosure of cost
beforehand is not required. Have you seen Bertsekas¹ auction
algorithms for
solving gen. assignment problem?
-Samik
> Looking for algorithms or whatnot to solve a many-to-one
assignment
> problem. Given m persons and n objects, where each person i
> associates with each object j a cost a_ij; Þnd the best
object x in
> terms of overall cost (i.e.
n
> x = arg min sum a_ij
> j i
I know this is trivial if you have all the cost data; I¹m
actually
> looking for a distributed algorithm (w/ each person being a
processor)
> that can solve this and where each person/processor only
has access to
> it¹s own cost data.
The more important problem would be to subdivide a population
of r*q
> persons so that r of the n objects each have q persons
assigned to
> them (and the rest zero). Again, looking for a distributed
algorithm
> as described above.
--
Samik Raychaudhuri
University of Wisconsin, Madison
http://samik.freeshell.org/
===
Subject: Re: assignment problem with grouping
That¹s what I started looking at, but I don¹t see how you can
ensure
the grouping constraint is met with a continuous network þow
model.
A couple of different ways to do it with a discrete network...
> Your requirement (distributed algo with private cost
information) would
probably be satisÞed with some auction algorithm, where
disclosure of cost
beforehand is not required. Have you seen Bertsekas¹ auction
algorithms for
solving gen. assignment problem?
> -Samik
===
Subject: Re: Jarry relation
> In R. I. G. Hughes¹ book The Structure and Interpretation
of Quantum
> Mechanics pages 106-107 he says that two observables are
mutually
> transformable if (a) they are representable in a Hilbert
space H by
> operators A and B, and (b) there exists a unitary operator
U on H such
> that A = U B U^ -1. This relation, which it is tempting to
call the
> Jarry-relation, is reþexive, symmetric, and transitive. If
A = U B U^
> -1, then B = U^ -1 A U.
> What I¹d like to know is why he is tempted to call this the
Jarry
> relation. Who is/was Jarry?
Robin Chapman
===
Subject: Re: Jarry relation
X-URL:
http://mygate.mailgate.org/mynews/sci/sci.math/
5a19e50d7bb90e6494265280f095a8
20.35661%40mygate.mailgate.org
> What I¹d like to know is why he is tempted to call this the
Jarry
> relation. Who is/was Jarry?

What possible connection could there be between a French
dramatist and
unitary operators in quantum mechanics?
David.
--
===
Subject: Re: Jarry relation
<5a19e50d7bb90e6494265280f095a820.35661@mygate.mailgate.org>,
David

> > What I¹d like to know is why he is tempted to call this
the Jarry
> > relation. Who is/was Jarry?

> What possible connection could there be between a French
dramatist and
> unitary operators in quantum mechanics?
> David.
Very little, except for the symbolic similarities:
> that A = U B U^ -1
===
Subject: Re: Is this SET of computable numbers computable?
In sci.logic, |-|erc
<contactvia@wwwadamskingdom.com>
<4068d42e$0$25657$afc38c87@news.optusnet.com.au>:
>> So : does the set CONTAIN all computable numbers and is it
amenible to
diagonalisation?
>> Herc
>> The list may not contain every computable number.
>> There might be a computable number such that for each j,
the Þrst j
>> digits are eventually computed, but only after the Þrst j
digits of a
>> different computable number have been computed, so that it
is never
>> the Þrst computable number to have j digits computed and
never gets
>> on the list.
>> In fact, the diagonal number for the list will deÞnitely
be such a
>> computable number.
> No, any computable number will eventually halt on enough
input values
> (j) (digits) and be counted.
> n i 1 2 3 4 5 6 7 ...
> 1 1 2 4 7
> 2 3 5 8
> 3 6 9
> 4 10
> ...
Do you understand the difference between:
for any i, there exists a j such that F(i,j) = G(j)
and
there exists a j such that for all i, F(i,j) = G(j)
?
They¹re quite different.
[rest snipped]
--
#191, ewill3@earthlink.net
It¹s still legal to go .sigless.
===
Subject: Re: Is this SET of computable numbers computable?
>> So : does the set CONTAIN all computable numbers and is it
amenible
to diagonalisation?
>> Herc
>> The list may not contain every computable number.
>> There might be a computable number such that for each j,
the Þrst j
>> digits are eventually computed, but only after the Þrst j
digits of a
>> different computable number have been computed, so that it
is never
>> the Þrst computable number to have j digits computed and
never gets
>> on the list.
>> In fact, the diagonal number for the list will deÞnitely
be such a
>> computable number.
No, any computable number will eventually halt on enough
input values
> (j) (digits) and be counted.

> n i 1 2 3 4 5 6 7 ...
> 1 1 2 4 7
> 2 3 5 8
> 3 6 9
> 4 10
> ...
> Do you understand the difference between:
> for any i, there exists a j such that F(i,j) = G(j)
This is completely unrelated to the post you replied to.
G shares a common digit with all numbers.
> and
> there exists a j such that for all i, F(i,j) = G(j)
G shares the same common digit with all numbers
> ?
> They¹re quite different.
Herc
===
Subject: Re: Is this SET of computable numbers computable?
> > So : does the set CONTAIN all computable numbers and is
it amenible to
diagonalisation?
Herc
The list may not contain every computable number.
There might be a computable number such that for each j, the
Þrst j
> digits are eventually computed, but only after the Þrst j
digits of a
> different computable number have been computed, so that it
is never
> the Þrst computable number to have j digits computed and
never gets
> on the list.
In fact, the diagonal number for the list will deÞnitely be
such a
> computable number.
No, any computable number will eventually halt on enough
input values (j)
(digits)
> and be counted.

Not necessarily. You could have a computable number, such
that, before
its Þrst digit has been computed, the program has already
found the
1st number to put in the list, and before the Þrst 2 digits
have been
computed, the program has already found the 2nd number to put
in the
list, and before the Þrst 3 digits have been computed, the
program
has already found the 3rd number to put in the list, and so
on. In
fact, the diagonal number will deÞnitely be like this.
<snip>
===
Subject: Re: Is this SET of computable numbers computable?
> > So : does the set CONTAIN all computable numbers and is
it amenible
to diagonalisation?
Herc
The list may not contain every computable number.
There might be a computable number such that for each j, the
Þrst j
> > digits are eventually computed, but only after the Þrst j
digits of
a
> > different computable number have been computed, so that
it is never
> > the Þrst computable number to have j digits computed and
never gets
> > on the list.
In fact, the diagonal number for the list will deÞnitely be
such a
> > computable number.
No, any computable number will eventually halt on enough
input values
(j) (digits)
> and be counted.

> Not necessarily. You could have a computable number, such
that, before
> its Þrst digit has been computed, the program has already
found the
> 1st number to put in the list, and before the Þrst 2 digits
have been
> computed, the program has already found the 2nd number to
put in the
> list, and before the Þrst 3 digits have been computed, the
program
> has already found the 3rd number to put in the list, and so
on. In
> fact, the diagonal number will deÞnitely be like this.
I addressed this issue in the post you replied to:
your case arises in this example when digits 1247, 358 all
halt before
69. This would give ECN :
j
1 1247 (the values computed from these parameter pairs)
2 358
3 xxx
so j is too large to accept 69 even if both digits have been
found,
(halted).
Only a 3 digit ECN will be accepted from here on.
..
Every paramater pair is given unlimited processing. The
scheduling is such
that more and more processing goes towards the existing
numbers as more
numbers are slowly introduced.
Herc
===
Subject: Re: Is this SET of computable numbers computable?
> > > So : does the set CONTAIN all computable numbers and is
it
amenible to diagonalisation?
> > Herc
The list may not contain every computable number.
There might be a computable number such that for each j, the
Þrst
j
> > digits are eventually computed, but only after the Þrst j
digits of
a
> > different computable number have been computed, so that
it is never
> > the Þrst computable number to have j digits computed and
never
gets
> > on the list.
In fact, the diagonal number for the list will deÞnitely be
such a
> > computable number.
No, any computable number will eventually halt on enough
input values
(j) (digits)
> > and be counted.

Not necessarily. You could have a computable number, such
that, before
> its Þrst digit has been computed, the program has already
found the
> 1st number to put in the list, and before the Þrst 2 digits
have been
> computed, the program has already found the 2nd number to
put in the
> list, and before the Þrst 3 digits have been computed, the
program
> has already found the 3rd number to put in the list, and so
on. In
> fact, the diagonal number will deÞnitely be like this.
>
I addressed this issue in the post you replied to:

> your case arises in this example when digits 1247, 358 all
halt before
> 69. This would give ECN :
j
> 1 1247 (the values computed from these parameter pairs)
> 2 358
> 3 xxx
so j is too large to accept 69 even if both digits have been
found,
(halted).
> Only a 3 digit ECN will be accepted from here on.
> ..
>
That¹s right. And if the program Þnds its 3rd computable
number
before the 3rd Turing machine computes its Þrst 3 digits,
then the
3rd Turing machine won¹t be 3rd in the list. And if the
program Þnds
its 4th computable number before the 3rd Turing machine
computes its
Þrst 4 digits, then the 3rd Turing machine won¹t be 4th in
the list.
And so on. It is possible for it to be nowhere in the list.
In fact,
this deÞnitely happens with the diagonal number.
Every paramater pair is given unlimited processing. The
scheduling is
such
> that more and more processing goes towards the existing
numbers as more
> numbers are slowly introduced.
Herc
===
Subject: Re: Is this SET of computable numbers computable?
> > > So : does the set CONTAIN all computable numbers and is
it
amenible to diagonalisation?
> > Herc
> > The list may not contain every computable number.
> > There might be a computable number such that for each j,
the Þrst
j
> > > digits are eventually computed, but only after the Þrst
j digits
of a
> > > different computable number have been computed, so that
it is
never
> > > the Þrst computable number to have j digits computed
and never
gets
> > > on the list.
> > In fact, the diagonal number for the list will deÞnitely
be such
a
> > > computable number.
No, any computable number will eventually halt on enough input
values (j) (digits)
> > and be counted.

Not necessarily. You could have a computable number, such
that,
before
> > its Þrst digit has been computed, the program has already
found the
> > 1st number to put in the list, and before the Þrst 2
digits have
been
> > computed, the program has already found the 2nd number to
put in the
> > list, and before the Þrst 3 digits have been computed,
the program
> > has already found the 3rd number to put in the list, and
so on. In
> > fact, the diagonal number will deÞnitely be like this.
>
> I addressed this issue in the post you replied to:

> your case arises in this example when digits 1247, 358 all
halt before
> 69. This would give ECN :
j
> 1 1247 (the values computed from these parameter pairs)
> 2 358
> 3 xxx
so j is too large to accept 69 even if both digits have been
found,
(halted).
> Only a 3 digit ECN will be accepted from here on.
> ..
That¹s right. And if the program Þnds its 3rd computable
number
> before the 3rd Turing machine computes its Þrst 3 digits,
then the
> 3rd Turing machine won¹t be 3rd in the list. And if the
program Þnds
> its 4th computable number before the 3rd Turing machine
computes its
> Þrst 4 digits, then the 3rd Turing machine won¹t be 4th in
the list.
> And so on. It is possible for it to be nowhere in the list.
In fact,
> this deÞnitely happens with the diagonal number.
go back a few posts where I refuted this. The amount of
processing per
digit goes up polynomially to the increase in amount of new
numbers.
i.e. the area of the triangle is proportional to the
processing time, new
numbers
are proportional to the side of this triangle.
There could be odd functions that have exponential complexity
with
its parameter. Technically you¹re right, practically all
permutations are
covered with polynomial algorithms. ECN may only cover
polynomials.
Extending to higher complexity algorithms would be possible
but tedious.
New numbers would have to be fed in at a logarithmic rate.
Every paramater pair is given unlimited processing. The
scheduling is
such
> that more and more processing goes towards the existing
numbers as more
> numbers are slowly introduced.
Herc
Herc
===
Subject: Matrix equation solvable?
Is this matrix equation solvable? The problem is of course
that it isn¹t
commutative.
R = S^-1 * A * S
Are there numerical methods?
===
Subject: Re: Matrix equation solvable?
>Is this matrix equation solvable? The problem is of course
that it isn¹t
>commutative.
>R = S^-1 * A * S
>Are there numerical methods?
(Solvable for S that is, as Dan added later)
This equation says R and A are similar. Two matrices (over
the complex
numbers, or in general over a Þeld where the characteristic
polynomials
factor completely) are similar if and only if their Jordan
canonical forms
are the same. So a method, in principle, is
1) Find the Jordan canonical forms of R and A, which means
Þnding
J_i and P_i such that J_1 = P_1^(-1) R P_1 and J_2 = P_2^(-1)
A P_2
where J_1 and J_2 are in Jordan canonical form.
2) If J_1 and J_2 are not the same, R and A are not similar.
If they
are, then you can take S = P_2 P_1^(-1).
Finding the Jordan canonical form is not a numerically stable
operation in general. But if the matrices R and A both have
complete sets of eigenvectors (which is generically the
case), then
all you have to do is
1) Find the eigenvalues lambda_i of R with corresponding
eigenvectors u_i,
and the eigenvalues mu_i of A with corresponding eigenvectors
v_i.
2) If the lambda_i and mu_i are different, R and A are not
similar.
If they are the same (perhaps after some permutation), then S
is the
matrix such that S u_i = v_i.
If U and V are the matrices with columns u_i and v_i
respectively, this
says S U = V, i.e. S = V U^(-1).
Dan also added:
| In this
| particular case I¹m most interested in I know that S is
unitary, that is
| S*S^H=I.
The usual case where you know S is unitary is when the
matrices R and A
are normal. In this case, R and A will certainly have
complete sets of
eigenvectors, and in fact the eigenvectors can be taken to
form
orthonormal bases.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Re: Matrix equation solvable?
> Is this matrix equation solvable? The problem is of course
that it isn¹t
> commutative.
R = S^-1 * A * S
If S and A are given, R is solvable. :-)
If S and R are given, A is solvable too. :-|
If R and A are given, S will not be solvable in general. :-(
For instance, since the determinant of S cancels that of
S^-1, the
determinants of A and R will have to be equal for a solution
to exist at
all. Not quite sure how to proceed though. (I don¹t suppose
that you
know beforehand that S is orthonormal? That would have
helped.)
--
M.vr.gr.
Dave
(d-dot-langers-at-wxs-dot-nl)
===
Subject: Re: Matrix equation solvable?
> If R and A are given, S will not be solvable in general. :-(
> For instance, since the determinant of S cancels that of
S^-1, the
> determinants of A and R will have to be equal for a
solution to exist at
> all. Not quite sure how to proceed though. (I don¹t suppose
that you
> know beforehand that S is orthonormal? That would have
helped.)
That¹s the case: I want to solve for S. I forgot to mention
that :)
A necessary and sufÞcient condition would ne nice of course.
In this
particular case I¹m most interested in I know that S is
unitary, that is
S*S^H=I.
===
Subject: Re: another boring critisism of Cantor¹s Theorem
> Here is another question: I look on the web and Þnd a ton
of papers
> trying to refute Cantor¹s Theorem. (Just Google Cantor
diagonal
> wrong and see what you get.) I have never seen one of them
that
> convinced me. Why is there so much opposition to this
theorem when
> most other famous theorems are pretty much universally
believed?
There¹s a slight confusion here. One thing is Cantor¹s
theorem stating
that the reals are uncountable and another thing is Cantor¹s
diagonal
argument, which was actually Cantor¹s second proof of his
theorem; see
http://en.wikipedia.org/wiki/Cantor%27s_Þrst_uncountability_
proof
(American Mathematical Monthly 101, No.9, 819-832 (1994)).
What I fail to understand is this: why is it that those who
attack
Cantor¹s diagonal argument do not attack also is other proof
of his
theorem?
Jose Carlos Santos
===
Subject: Re: another boring critisism of Cantor¹s Theorem
> Here is another question: I look on the web and Þnd a ton
of papers
> trying to refute Cantor¹s Theorem. (Just Google Cantor
diagonal
> wrong and see what you get.) I have never seen one of them
that
> convinced me. Why is there so much opposition to this
theorem when
> most other famous theorems are pretty much universally
believed?
>There¹s a slight confusion here. One thing is Cantor¹s
theorem stating
>that the reals are uncountable and another thing is Cantor¹s
diagonal
>argument, which was actually Cantor¹s second proof of his
theorem; see
>http://en.wikipedia.org/wiki/Cantor%27s_Þrst_uncountability_
proof
>(American Mathematical Monthly 101, No.9, 819-832 (1994)).
>What I fail to understand is this: why is it that those who
attack
>Cantor¹s diagonal argument do not attack also is other proof
of his
>theorem?
I would suggest that either they do not know about it, or
they do not
understand it. The one person whom I have seen arguing about
it on
sci.math did not understand it.
David
But I¹m always true to you, darlin¹, in my fashion,
Yes, I¹m always true to you, darlin¹, in my way.
-- Lois Lane
-----
===
Subject: Re: another boring critisism of Cantor¹s Theorem
> Here is another question: I look on the web and Þnd a ton
of papers
> trying to refute Cantor¹s Theorem. (Just Google Cantor
diagonal
> wrong and see what you get.) I have never seen one of them
that
> convinced me. Why is there so much opposition to this
theorem when
> most other famous theorems are pretty much universally
believed?
Fundamentally it is the incorrigible stupidity of humans.
But I have recently noted a verbal tic afþicting many of
the anti-Cantor cretins: they often speak of the Cantor
number.
A mathematician seeing that would say what Cantor number? what
the **** is that guy talking about? But it is a revealing
gaffe.
It shows the idiot is thinking that Cantor¹s diagonal argument
the Cantor number) that is not on any list of numbers, and
that¹s
absurd. Of course it is but that isn¹t what the diagonal
argument shows.
The crackpot fails to realize the distinction between the
statement
for each list there is a number not appearing in it and
there is a number which does not appear in any list. In other
words,
the moron is confused about order of quantiÞcation, and so is
attack a straw man diagonal argument (the second in quotes)
much to the amusement of passing mathematicians.
Robin Chapman
===
Subject: Re: Well-founded and Induction principle
>>Let¹s have (X,<) an ordered set that is supposed to be
well-founded
(every
>>non empty subset has a minimal element).
>>I can prove that (X,<) satisÞes the Induction Principle
stated as
follow:
>>For any B subset of X, then if Vx(Vy( y in x --> y in B)
--> x in B) then
>>X=B
>I assume y in x was supposed to be y<x...
>>I am not able to follow the reciproqual: if (X,<) satisÞes
the Induction
>>Principle, then (X,<) is well founded.
>>Any help much appeciated!
>Let A be a nonempty subset of X; we want to show that A has
a least
>element.
We want to show that A has a minimal element.
>Let B = X - A. Since A is nonempty, B is not equal to
>X. Applying the contrapositive of the Induction Principle you
>enunciated we have that there exists an x in X such that,
for all y,
>y<x -> y in B, but x not in B.
>Let x_0 be such an element. Since x_0 is not in B, it is in
A. I claim
>it is a minimal element of A: for is y in X with y<x_0, then
by choice
>of x_0 we must have y in B, and therefore y not in A.
At this point, you have proven that A has a minimal element.
You need
no more.
> That is: if y in
>A, then y>= x_0. This proves that x_0 is a minimal element
of A.
Here, you have added the assumption that (X,<) is a totally
ordered,
or linearly ordered set. As far as I can see, the assumption
is that
(X,<) is a partially ordered set. This means that A can have
elements
not comparable to x_0, so that it is not necessarily true
that y >= x_0
for all y in A.
David
But I¹m always true to you, darlin¹, in my fashion,
Yes, I¹m always true to you, darlin¹, in my way.
-- Lois Lane
-----
===
Subject: Re: Well-founded and Induction principle
Adjunct Assistant Professor at the University of Montana.
>Let¹s have (X,<) an ordered set that is supposed to be
well-founded
(every
>non empty subset has a minimal element).
>I can prove that (X,<) satisÞes the Induction Principle
stated as
follow:
>For any B subset of X, then if Vx(Vy( y in x --> y in B) -->
x in B)
then
>X=B
>>I assume y in x was supposed to be y<x...
>I am not able to follow the reciproqual: if (X,<) satisÞes
the
Induction
>Principle, then (X,<) is well founded.
>Any help much appeciated!
>>Let A be a nonempty subset of X; we want to show that A has
a least
>>element.
>We want to show that A has a minimal element.
Oops. Yeah, depends on your conventions... (Some people use
ordered/partially ordered rather than totally
ordered/ordered...)
>>Let B = X - A. Since A is nonempty, B is not equal to
>>X. Applying the contrapositive of the Induction Principle
you
>>enunciated we have that there exists an x in X such that,
for all y,
>>y<x -> y in B, but x not in B.
>>Let x_0 be such an element. Since x_0 is not in B, it is in
A. I claim
>>it is a minimal element of A: for is y in X with y<x_0,
then by choice
>>of x_0 we must have y in B, and therefore y not in A.
>At this point, you have proven that A has a minimal element.
You need
>no more.
>> That is: if y in
>>A, then y>= x_0. This proves that x_0 is a minimal element
of A.
>Here, you have added the assumption that (X,<) is a totally
ordered,
>or linearly ordered set. As far as I can see, the assumption
is that
>(X,<) is a partially ordered set. This means that A can have
elements
>not comparable to x_0, so that it is not necessarily true
that y >= x_0
>for all y in A.
An easy Þx is to say That is: if y in A, then y is not
smaller than
x_0 rather than what I did say.
--
=============================================================
=========
It¹s not denial. I¹m just very selective about
what I accept as reality.
--- Calvin (Calvin and Hobbes)
=============================================================
=========
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Constant Gauss Curvature Moebius Band
Is there a parameterization (closed form or differential) for
a
non-orientable Moebius Band with constant negative Gauss
Curvature
in R3 ?
One would expect the band with cuspidally sharp edges.To make
a model
perhaps it needs one full twist before bending to re-join
ends of an
extruded U- section.
===
Subject: Re: What is the number derivative ???
> Some things one obtains from this are:
> an algebraic group over F_1 should be the corresponding
Weyl group,
> the K-groups of F_1 should be the stable homotopy groups of
spheres,
> sheaves over Spec(F_1) should be Þnite sets,
> algebras over F_1 should be monoids,
> etc.
>
So we make n-stacks over Spec(F_1) out to be n-categories,
and n-gerbes over Spec(F_1) out to be n-groupoids?
David CorÞeld
===
Subject: Re: Is this proof of Hadwiger¹s Conjecture right?
> Any other question?
It¹s unreadable. The author needs to get a English Speaker to
> proofread. I can¹t make heads or tails of many of his
sentences.
Nathan

The proof of Hadwiger¹s Conjecture is very simple. In this
paper, I
just want to say, HC is a geometrical problem, it moughtn¹t
be proved
in graph theory. I describe the proof of 4CT here:

1. The relation between K^n and n-colorable graphs couldn¹t be
represented in graph theory in a nutshell. However, it did
here.
Taking n pieces of plasticene colored in different colors,
and making
them adjacent each other, we obtained a K^n prototype. Cut
it, and we
can Þnd that every n-colorable planar map is its section.

2. NOTICE: It is the plane that has the discrete property.
There
haven¹t Þve countries adjacent each other. Could we construct
a
special solid space that has the same discrete property? Yes,
I called
it C^4R^3. The plane is C^4R^2. Of course, the plane is the
section
subset of C^4R^3. It is very important. A little imaginative
ability
is need here.

3. If a planar map is 5-colorable, well, a solid K^5
prototype is
need. However, the plane is the section subset of C^4R^3, and
a solid
K^5 couldn¹t be embedded into C^4R^3. So there hasn¹t a
5-colorable
planar map.

In short:
1. A planar map is in the plane.
2. The plane subset {the section set of C^4R^3}. (Lost in
graph
theory)
3. A 5-colorable planar map is in the section set of a solid
K^5.
(Lost in graph theory)
4. The solid K^5 is out of C^4R^3.
So a 5-colorable planar map is out of the plane.

I am not good at English. If there have any wrong sentences
in the
===
Subject: Re: Dirk, making mistakes again.
> Here is a link to one of Dirk van de moortels Immortal
Fumbles pages
> where he attempts to ridicule one of my posts.
>
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/
DisCont.html
> He highlights the words space and discrete, because he
believes
> spaces cannot be discrete.
> Discrete spaces in fact do exist, as any competent
mathematican can
> tell you.
> Here are some links:
> http://planetmath.org/encyclopedia/Discrete.html
> http://en.wikipedia.org/wiki/Discrete_space
> After I revealed Dirks malicious ignorance, mistake making
Dirk
> responded by changing his story and stating that discrete
spaces have
> no place in quantum theory. Here is a quote:
> HAHAHAHAHA, I¹m probably the last one in the
> world who needs links to show me that mathematicians
> use discrete spaces :-)
> But last time I looked quantum theory didn¹t use a
> discrete space for its setting. When and where was
> the last time you looked?
> Here is a link to that post:
24zj4.1264%40news.cpqcorp.net
> But once again, Dirk made a mistake.
> Discrete spaces have long been investigated in Quantum
Theory, as any
> competent Physicist can tell you.
> Here are some links:
> http://www.wolframscience.com/reference/notes/1027c
> http://adela.karlin.mff.cuni.cz/~motl/Gibbs/discrete.htm
> http://www.weburbia.demon.co.uk/press/html/g09.htm
> JS
We are used to Dinky¹s arrogance and lack of knowledge. I
think he has a
Œfumble¹ of mine listed where I stated that 1 has two square
roots, 1
and -1, and I¹m still amazed to Þnd it doesn¹t.
Androcles.
===
Subject: Re: Dirk, making mistakes again.
message
> We are used to Dinky¹s arrogance and lack of knowledge. I
think he has a
> Œfumble¹ of mine listed where I stated that 1 has two
square roots, 1
> and -1, and I¹m still amazed to Þnd it doesn¹t.
> Androcles.
Are you saying that you are amazed to Þnd that 1 doesn¹t
have two square roots?
Where did you Þnd that?
Franz
===
Subject: Re: Dirk, making mistakes again.
> message
> We are used to Dinky¹s arrogance and lack of knowledge. I
> think he has a
> Œfumble¹ of mine listed where I stated that 1 has two
> square roots, 1
> and -1, and I¹m still amazed to Þnd it doesn¹t.
> Androcles.
> Are you saying that you are amazed to Þnd that 1 doesn¹t
> have two square roots?
> Where did you Þnd that?
hm, forgot to mention this one:
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/
AndersenLogic.html
;-)
Dirk Vdm
===
Subject: Re: Dirk, making mistakes again.
Franz Heymann <notfranz.heymann@btopenworld.com> escribi.97
en el
mensaje
> message
> We are used to Dinky¹s arrogance and lack of knowledge. I
> think he has a
> Œfumble¹ of mine listed where I stated that 1 has two
> square roots, 1
> and -1, and I¹m still amazed to Þnd it doesn¹t.
> Androcles.
> Are you saying that you are amazed to Þnd that 1 doesn¹t
> have two square roots?
> Where did you Þnd that?
To be precise, the square roots of 1 in C are exp(i* n * pi)
where n>=0 and integer. :-)
> Franz
===
Subject: Re: Dirk, making mistakes again.
> We are used to Dinky¹s arrogance and lack of knowledge. I
think he has a
> Œfumble¹ of mine listed where I stated that 1 has two
square roots, 1
> and -1, and I¹m still amazed to Þnd it doesn¹t.
> Androcles.
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/
SqrtAnswers.html
Dirk Vdm
===
Subject: Re: Dirk, making mistakes again.
Dirk Van de moortel
message
message
We are used to Dinky¹s arrogance and lack of knowledge.
I think he has a
> Œfumble¹ of mine listed where I stated that 1 has two
square roots, 1
> and -1, and I¹m still amazed to Þnd it doesn¹t.
> Androcles.
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/
SqrtAnswers.html
extreme stupidity.
And he still does not see it, which compounds his failure
even more.
Franz
===
Subject: Re: Dirk, making mistakes again.
We are used to Dinky¹s arrogance and lack of knowledge. I
think he has
a
> Œfumble¹ of mine listed where I stated that 1 has two
square roots, 1
> and -1, and I¹m still amazed to Þnd it doesn¹t.
> Androcles.
>
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/
SqrtAnswers.html
Oh, my!!!
This is more serious that I had thought. There is evidence
that Androcles
didn¹t
even pass High School maths, and probably Elementary School
maths either.
> Dirk Vdm
===
Subject: Re: Dirk, making mistakes again.
Androcles <jp006f9750@nospamblueyonder.co.uk> escribi.97 en
el mensaje
Here is a link to one of Dirk van de moortels Immortal Fumbles
pages
> where he attempts to ridicule one of my posts.
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/
DisCont.html
He highlights the words space and discrete, because he
believes
> spaces cannot be discrete.
Discrete spaces in fact do exist, as any competent
mathematican can
> tell you.
Here are some links:
> http://planetmath.org/encyclopedia/Discrete.html
> http://en.wikipedia.org/wiki/Discrete_space

> After I revealed Dirks malicious ignorance, mistake making
Dirk
> responded by changing his story and stating that discrete
spaces have
> no place in quantum theory. Here is a quote:
HAHAHAHAHA, I¹m probably the last one in the
> world who needs links to show me that mathematicians
> use discrete spaces :-)
But last time I looked quantum theory didn¹t use a
> discrete space for its setting. When and where was
> the last time you looked?
Here is a link to that post:
24zj4.1264%40news.cpqcorp.net

> But once again, Dirk made a mistake.
Discrete spaces have long been investigated in Quantum
Theory, as any
> competent Physicist can tell you.
Here are some links:
> http://www.wolframscience.com/reference/notes/1027c
> http://adela.karlin.mff.cuni.cz/~motl/Gibbs/discrete.htm
> http://www.weburbia.demon.co.uk/press/html/g09.htm

> JS
> We are used to Dinky¹s arrogance and lack of knowledge. I
think he has a
> Œfumble¹ of mine listed where I stated that 1 has two
square roots, 1
> and -1, and I¹m still amazed to Þnd it doesn¹t.
> Androcles.
Knowing Dirk, maybe he expects the solution written in the
form exp(i*pi*n)
with
n >= 0 and integer. But he will not tell and will have laugh
about it. :-)
===
Subject: Re: A Paradox of The Central Limit Theorem
[snips]
> Incidentally you have repestedly failed to answer my
previous question.
Just curious, was that a Freudian typo or was it deliberate?
I¹ve never encountered it before. I am making note of this
happy coinage, for the beneÞt of future OED researchers.
===
Subject: algebra problem...
Þnite Þeld Z_2 cannot be made into an ordered Þeld.
------------------------
um.....i am anxious about reason
thank you.
===
Subject: Re: algebra problem...
> Þnite Þeld Z_2 cannot be made into an ordered Þeld.
------------------------
um.....i am anxious about reason
thank you.
Okay, that notation means the Þeld consisting of just 1 and
0, right?
That¹s what it means here, anyway, but I know that,
especially in
undergrad texts, they can go way off the wall with notation.
In order for it to be an ordered Þeld, it must satisfy this
axiom
(this isn¹t sufÞcient, of course, but it must satisfy this
axiom):
If a < b, then a + c < b + c, for all c.
Try the ordering 0 < 1, and let c = 1. Then we¹d need 0 + 1 <
1 + 1,
or 1 < 0, which contradicts the assumption. If we try the
other
ordering (1 < 0), we get the same problem. So there¹s no way
to
satisfy the axiom, so you can¹t make F_2 an ordered Þeld.
===
Subject: Re: algebra problem...
> Þnite Þeld Z_2 cannot be made into an ordered Þeld.
------------------------
um.....i am anxious about reason
thank you.
Squares must be positive, and positive plus positive must be
positive.
Do you see a problem here with any Þnite Þeld - and many more
besides?
Achava
===
Subject: Re: algebra problem...
* hot-girl
> Þnite Þeld Z_2 cannot be made into an ordered Þeld.
------------------------
um.....i am anxious about reason
I¹m not. Start with the deÞnition of an ordered Þeld. Do you
know it?
--
Jon Haugsand
Dept. of Informatics, Univ. of Oslo, Norway,
mailto:jonhaug@iÞ.uio.no
http://www.iÞ.uio.no/~jonhaug/, Phone: +47 22 85 24 92
===
Subject: Re: algebra problem...
> * hot-girl
> Þnite Þeld Z_2 cannot be made into an ordered Þeld.
> Start with the deÞnition of an ordered Þeld. Do you know it?
It could be given the trivial order. Otherwise if order isn¹t
trivial,
there¹s some distince a,b in Z_2, with a < b. Guess what a,b
have to be.
Notice 1 = -1, 0 = -0 in Z_2 and prove that in a (partially)
ordered group
a <= b iff -b <= -a
Conclude Z_2 = Z_1, oh my, oh my.
Excercise: show the group Z_n can¹t be order, even partially,
except with the trivial order, equality.
===
Subject: Re: algebra problem...
>> * hot-girl
>> Þnite Þeld Z_2 cannot be made into an ordered Þeld.
>> Start with the deÞnition of an ordered Þeld. Do you know
it?
>It could be given the trivial order.
Uh, right. What _is_ the deÞnition of ordered Þeld, exactly?
>Otherwise if order isn¹t trivial,
>there¹s some distince a,b in Z_2, with a < b. Guess what a,b
have to be.
>Notice 1 = -1, 0 = -0 in Z_2 and prove that in a (partially)
ordered group
> a <= b iff -b <= -a
>Conclude Z_2 = Z_1, oh my, oh my.
>Excercise: show the group Z_n can¹t be order, even partially,
>except with the trivial order, equality.
************************
David C. Ullrich
===
Subject: Re: algebra problem...
> * hot-girl
>> Þnite Þeld Z_2 cannot be made into an ordered Þeld.
>> Start with the deÞnition of an ordered Þeld. Do you know
it?
>It could be given the trivial order.
> Uh, right. What _is_ the deÞnition of ordered Þeld, exactly?
What? You¹ve the notion it has to be totally ordered?
The only order, totaled or not for a cyclic group is the
trivial order.
a <=_Zn b iff a = b.
>Otherwise if order isn¹t trivial,
>there¹s some distince a,b in Z_2, with a < b. Guess what a,b
have to
be.
>Notice 1 = -1, 0 = -0 in Z_2 and prove that in a (partially)
ordered
group
> a <= b iff -b <= -a
>Conclude Z_2 = Z_1, oh my, oh my.
Excercise: show the group Z_n can¹t be order, even partially,
>except with the trivial order, equality.
===
Subject: Re: algebra problem...
> * hot-girl
> Þnite Þeld Z_2 cannot be made into an ordered Þeld.
>> Start with the deÞnition of an ordered Þeld. Do you know
it?
>>It could be given the trivial order.
>> Uh, right. What _is_ the deÞnition of ordered Þeld,
exactly?
>What? You¹ve the notion it has to be totally ordered?
No, What? You¹ve the notion it has to be totally ordered?
is not the deÞnition of ordered Þeld. You can try again
if you want.
>The only order, totaled or not for a cyclic group is the
trivial order.
> a <=_Zn b iff a = b.
>>Otherwise if order isn¹t trivial,
>>there¹s some distince a,b in Z_2, with a < b. Guess what
a,b have to
be.
>>Notice 1 = -1, 0 = -0 in Z_2 and prove that in a
(partially) ordered
group
>> a <= b iff -b <= -a
>>Conclude Z_2 = Z_1, oh my, oh my.
>>Excercise: show the group Z_n can¹t be order, even
partially,
>>except with the trivial order, equality.
************************
David C. Ullrich
===
Subject: Re: algebra problem...
> * hot-girl
> Þnite Þeld Z_2 cannot be made into an ordered Þeld.
> Start with the deÞnition of an ordered Þeld. Do you know it?
>>It could be given the trivial order.
>> Uh, right. What _is_ the deÞnition of ordered Þeld,
exactly?
> What? You¹ve the notion it has to be totally ordered?
It¹s not a notion. Look at
<http://en.wikipedia.org/wiki/Ordered_Þeld>. That is the
standard
deÞnition.
> The only order, totaled or not for a cyclic group is the
trivial order.
> a <=_Zn b iff a = b.
We are talking about ordered Þelds, not groups. The order on
an ordered
Þeld is required to satisfy the trichotomy law, which
excludes the
trivial order.
>>Otherwise if order isn¹t trivial,
>>there¹s some distince a,b in Z_2, with a < b. Guess what
a,b have to
be.
>>Notice 1 = -1, 0 = -0 in Z_2 and prove that in a
(partially) ordered
group
>> a <= b iff -b <= -a
>>Conclude Z_2 = Z_1, oh my, oh my.
>>Excercise: show the group Z_n can¹t be order, even
partially,
>>except with the trivial order, equality.
So extrapolite your Þndings to Þnite Þelds.
--
Dave Seaman
Judge Yohn¹s mistakes revealed in Mumia Abu-Jamal ruling.
<http://www.commoncouragepress.com/index.cfm?action=book&
bookid=228>
===
Subject: Re: algebra problem...
>
> * hot-girl
> Þnite Þeld Z_2 cannot be made into an ordered Þeld.
>> Start with the deÞnition of an ordered Þeld. Do you know
it?
>>It could be given the trivial order.
>> Uh, right. What _is_ the deÞnition of ordered Þeld,
exactly?
> What? You¹ve the notion it has to be totally ordered?
This seems to be the established meaning for ordered Þeld.
I guess, the concept dates from a time when people used
(a) ordered and partially orderd
instead of
(b) totally ordered and ordered.
Marc
===
Subject: Re: algebra problem...
>>
> * hot-girl
>> Þnite Þeld Z_2 cannot be made into an ordered Þeld.
>> Start with the deÞnition of an ordered Þeld. Do you know
it?
>It could be given the trivial order.
> Uh, right. What _is_ the deÞnition of ordered Þeld, exactly?
>> What? You¹ve the notion it has to be totally ordered?
> This seems to be the established meaning for ordered Þeld.
> I guess, the concept dates from a time when people used
> (a) ordered and partially orderd
> instead of
> (b) totally ordered and ordered.
I think that¹s the wrong way to look at it. The term ordered
Þeld has
to be understood as an inseparable unit.
You can¹t understand what an ordered Þeld is by merely
looking up the
terms ordered and Þeld and then combining the deÞnitions. It¹s
like trying to understand what an open mapping is by just
looking up
the deÞnitions for open and mapping.
--
Dave Seaman
Judge Yohn¹s mistakes revealed in Mumia Abu-Jamal ruling.
<http://www.commoncouragepress.com/index.cfm?action=book&
bookid=228>
===
Subject: Re: algebra problem...
>Þnite Þeld Z_2 cannot be made into an ordered Þeld.
>------------------------
>um.....i am anxious about reason
Can you prove that 0 < 1 < 1 + 1 in any ordered Þeld?
>thank you.
************************
David C. Ullrich
===
Subject: Re: Why do most quant analysist jobs require a PhD ?
>> If you new how to detect short squeezes, even just a
single one,
>> you¹ll be sitting around a pool bar in some Carib isle now
and you
>> wouldn¹t even mention it to anybody :)
>Trading is an art, not a gamble.
I spell it gamble.
> .. Nobody in his right mind would put
>tons of money on volatile trades. If you put 1 million on
the buy side
>of a short squeeze, you will break the pattern and it won¹t
work.
>Smart traders will put no more than $10k on a short squeeze.
Also
>obvious short squeezes (the ones you know for a fact that
you are
>going to make money on) are not very frequent. Only a few
dozens
>stocks are good candidates, and the squeeze usually happens
less than
>once every 45 days.
>
>>
>> By the way, you have no way of knowing whether some buying
activity is
>> a short sqeeze or just a normal reversal but only on
hindsight.
>A 25% drop over 48 hours on no news and moderate volume when
similar
>stocks experience no movements is the starting point of a
short
>squeeze. You can safely buy when the drop has reached 25%
and sell
>within 48 hours.
>
>>
>> Every strategy to beat the roulette has a probability of
total ruin.
>> It will get you sooner or later. Since you have made it
clear you have
>> a Ph.D from a prestigious body shop can you calculate how
proÞtable a
>> strategy must be to have a probability of total ruin less
than 10^-20?
>> (By the way even 10-^20 is not low enough but I settle for
that)
>I am an applied statistician and a real trader. You are a
theoretical
>statistician and obviously not a trader (not an experienced
one at
>least).
You are a trader and not an investor. You are also not one of
those whose business receives the commodity (such as oil,
grain, meat, etc.). There is a reason your kind of trading
uses the term hedge.
Subtract a hundred and four for e-mail.
===
Subject: Re: Why do most quant analysist jobs require a PhD ?
>> If you new how to detect short squeezes, even just a
single one,
>> you¹ll be sitting around a pool bar in some Carib isle now
and you
>> wouldn¹t even mention it to anybody :)
Trading is an art, not a gamble.
I spell it gamble.
Trading *receipts* has never been anything but a gamble.
Which is still the leading reason that the people
who build Cruise Missiles, also build Cray computers,
Intel, Oregon Tree Farms, IRS Forms, Xerox Machines,
South African gold mines, Nuclear Reactors, AT&T,
International Harvester, G.E., G.M., MIT, and
Bill Gates gambles.
===
Subject: Re: Why do most quant analysist jobs require a PhD ?

>One of my favorites is detecting short squeezes on certain
types of
> stocks. You can make 25% return in 5 days :) Another one is
robust
cross-correlations with time-lag, involving two similar
stocks and one
index.

If you new how to detect short squeezes, even just a single
one,
> you¹ll be sitting around a pool bar in some Carib isle now
and you
> wouldn¹t even mention it to anybody :)
By the way, you have no way of knowing whether some buying
activity is
> a short sqeeze or just a normal reversal but only on
hindsight.
Every strategy to beat the roulette has a probability of
total ruin.
In _Fractals. Chaos, and Power Laws_ (page 151 in my edition)
Manfred
Schroeder claims to have developed a system to beat roulette,
using a
small analog computer in the 1960¹s. He claims to have not
used it for
more than a proof-of-concept demonstation for himself and a
cohort,
thus leaving room for your eventual total ruin scenario to
remain
unchallenged. FWIW.
> It will get you sooner or later. Since you have made it
clear you have
> a Ph.D from a prestigious body shop can you calculate how
proÞtable a
> strategy must be to have a probability of total ruin less
than 10^-20?
> (By the way even 10-^20 is not low enough but I settle for
that)
Russell
===
Subject: . it never will be and it never was.
--
oo
____|mn
/ /_/ / _
/ K-9/ /_/ - www.YeOldeCoffeeShoppe.com -
/____/_____
--------------
===
Subject: . So far anti_diag is no match for the list of
computables
--
oo
____|mn
/ /_/ / _
/ K-9/ /_/ - www.YeOldeCoffeeShoppe.com -
/____/_____
--------------
===
Subject: . the 1st couple digits of anti_diag.
--
oo
____|mn
/ /_/ / _
/ K-9/ /_/ - www.YeOldeCoffeeShoppe.com -
/____/_____
--------------
===
Subject: . the computables can easily COVER
--
oo
____|mn
/ /_/ / _
/ K-9/ /_/ - www.YeOldeCoffeeShoppe.com -
/____/_____
--------------
===
Subject: . the 1st digit of anti_diag, and from those
--
oo
____|mn
/ /_/ / _
/ K-9/ /_/ - www.YeOldeCoffeeShoppe.com -
/____/_____
--------------
===
Subject: . contains 100s of numbers, over 10 of which matched
--
oo
____|mn
/ /_/ / _
/ K-9/ /_/ - www.YeOldeCoffeeShoppe.com -
/____/_____
--------------
===
Subject: . this digit too, but it doesn¹t matter, computables
easily
--
oo
____|mn
/ /_/ / _
/ K-9/ /_/ - www.YeOldeCoffeeShoppe.com -
/____/_____
--------------
===
Subject: . digit on the second number, anti_diag has to
counter
--
oo
____|mn
/ /_/ / _
/ K-9/ /_/ - www.YeOldeCoffeeShoppe.com -
/____/_____
--------------
===
Subject: . that are computable covering it. Go to the second
--
oo
____|mn
/ /_/ / _
/ K-9/ /_/ - www.YeOldeCoffeeShoppe.com -
/____/_____
--------------
===
Subject: . because there¹s atleast 10 other numbers
--
oo
____|mn
/ /_/ / _
/ K-9/ /_/ - www.YeOldeCoffeeShoppe.com -
/____/_____
--------------
===
Subject: . It doesn¹t matter what digit anti_diag is,
--
oo
____|mn
/ /_/ / _
/ K-9/ /_/ - www.YeOldeCoffeeShoppe.com -
/____/_____
--------------
===
Subject: . anti_diag has to provide a counter digit.
--
oo
____|mn
/ /_/ / _
/ K-9/ /_/ - www.YeOldeCoffeeShoppe.com -
/____/_____
--------------
===
Subject: . the 1st digit of the 1st computable number,
--
oo
____|mn
/ /_/ / _
/ K-9/ /_/ - www.YeOldeCoffeeShoppe.com -
/____/_____
--------------
===
Subject: . Think of the problem this way. When you lay down
--
oo
____|mn
/ /_/ / _
/ K-9/ /_/ - www.YeOldeCoffeeShoppe.com -
/____/_____
--------------
===
Subject: prime numbers form :pa*pb+pc
is there a proof for the next proposition:
A=pa*pb+pc is prime number if pa/=pb/=pc (diferent from /=)
pa,pb,pc prime numbers
===
Subject: Re: prime numbers form :pa*pb+pc
> is there a proof for the next proposition:
A=pa*pb+pc is prime number if pa/=pb/=pc (diferent from /=)
pa,pb,pc prime numbers
Proposition false.
If pa and pb and pc are all odd then A is even
===
Subject: Re: prime numbers form :pa*pb+pc
X-Mimeole: Produced By Microsoft MimeOLE V6.00.2800.1106
X-Msmail-Priority: Normal
> is there a proof for the next proposition:
> A=pa*pb+pc is prime number if pa/=pb/=pc (diferent from /=)
> pa,pb,pc prime numbers
If 2 is not in {pa,pb,pc} then pa*pb is odd, and pc is odd,
so pa*pb + pc
is
even. There aren¹t too many even primes.
--
+---- Kevin C. Saff ----+ F-15 | |Eagle
| Engineer, Boeing, StL | _____|_^_|_____
| Tracking/Fleet Support| * + [_(x)_] + *
===
Subject: Re: prime numbers form :pa*pb+pc
>is there a proof for the next proposition:
>A=pa*pb+pc is prime number if pa/=pb/=pc (diferent from /=)
>pa,pb,pc prime numbers
So for example 2*5+11 is prime. Proving this is going to be
challenging...
************************
David C. Ullrich
===
Subject: Re: prime numbers form :pa*pb+pc

>>is there a proof for the next proposition:
>>A=pa*pb+pc is prime number if pa/=pb/=pc (diferent from /=)
>>pa,pb,pc prime numbers

> So for example 2*5+11 is prime. Proving this is going to be
> challenging...
This could be a nice homework example - or a research topic:
For n in N let M_n = {<p,q,r> : p,q,r prime and p<q and
råÂ=p and råÂ=q}
M_n is the disjoint union of examples E_n and couter-examples
C_n.
Now, what is the asymptotic behaviour of #E_n/#C_n ?
The answer would allow qualifying statements like the
conjecture is
nearly always false which is what I believe.
Helmut Richter
===
Subject: Re: prime numbers form :pa*pb+pc
<oe1j609u2vmfr4rmmlsis8jbuolslg6lu8@4ax.com>is there a proof
for the next proposition:
>>A=pa*pb+pc is prime number if pa/=pb/=pc (diferent from /=)
>>pa,pb,pc prime numbers
>So for example 2*5+11 is prime. Proving this is going to be
>challenging...
In general, A will be even. It¹s only odd if 2 is in
{pa,pb,pc}.
--Keith Lewis klewis {at} mitre.org
The above may not (yet) represent the opinions of my employer.
===
Subject: Buying petrol at the lowest price.
Hi everyone,
Here¹s a problem I encounter every day: Buying petrol at the
lowest price.
The petrol prices are very irregular at the moment due to an
ongoing price
war involving all the major suppliers. Prices may change by
up to 20 %
within a single day. I have too often experienced the
frustation of Þlling
the car up with expensive petrol, and shortly thereafter pass
another
supplier where the price is much lower.
I would like to model - and solve - the decision making
problem: When I
approach a petrol station, should I buy petrol, and if so,
how much?
My current strategy is to buy a small amount when the price
is high, and
Þll up the car when the price is low. However, if the car is
running low
on
petrol, I¹m forced to buy, irrespective of the price.
So the decision seems to depend on two observables: The
current price at
the
petrol station, and the current quantity of petrol in the car.
I¹m going to make some explicit assumptions: When commuting,
I pass one
petrol station once each way. Each time I pass the petrol
station, the
current price is a stochastic variable, independent of all
previous values.
In particular, the price in the afternoon is independent of
the price in
the
morning. This agrees well with my experience.
I have a fair distance to commute, so I need to refuel at
least once a
week;
the car carries petrol enough for ten trips altogether (Þve
each way). If
we need to, we can make the problem discrete by requiring
that the amount
of
petrol bought is an integral multiple of one tenth of the
cars capacity.
The probability distribution of the price is deÞnitely not
symmetric. The
most probable price is also the largest price. The price
seems always to
lie
in the range [3/4, 1], where the value 1 corresponds to the
maximum price.
This could be modeled with the probability function
32*(x-3/4), in the
speciÞed range.
So to summarize, I¹m looking for the optimal decision
strategy to use, each
time I approach the petrol supplier. Each time I must decide
how much
petrol
to buy. The object is to minimize the total amount spent on
petrol over a
long period of time.
Any helpful comments, or solutions?
-Michael.
===
Subject: Re: Buying petrol at the lowest price.
I tried simplifying the problem by considering just two
states: Petrol
level
in car is above/below half full. This appears to look like a
Markov problem
with the transition probabilities:
P(A->A) = alpha(P)
P(A->B) = 1 - alpha(P)
P(B->B) = beta(P)
P(B->A) = 1 - beta(P)
Here P is the current price, a stochastic variable. The two
functions
alpha(P) and beta(P) are to be determined.
With each transition the associated cost is:
C(A->A) = P/2
C(A->B) = 0
C(B->B) = P/2
C(B->A) = P
I¹m stuck because the transition probabilities are not
constant, but
instead
stochastic variables.
The expected amount of money to pay each time depends on the
current state
of the car: If in state A the expected cost is P(A->A) *
C(A->A) = alpha(P)
* P/2. If in state B the expected cost is P(B->A) * C(B->A) +
P(B->B) *
C(B->B) = P * (1 - beta(P)/2).
But what are the probabilities of being in either state?
I¹m hoping to get the expected cost in terms of the two
unknown functions
alpha(P) and beta(P). Then perhaps some variational method
should give the
optimal expressions.
But is this problem really well-deÞned? Can I hope to
determine an
explicit
optimal solution? Or will the optimum just be given in terms
of mean and
(co-) variance of alpha(P) and beta(P)?
I¹m stuck here on this one. Any helpful suggestions are most
welcome.
-Michael.
===
Subject: Re: The First Proof of Fermat¹s Last Theorem to Exist
> 44[ 6.It was Proven, by a Pthagorean, that, for
Now that it has been changed, it is no longer the Þrst
proof...
Wiles¹ proof is. And since Wiles¹ is actually a proof, ken¹s
is of no further interest.
===
Subject: Re: The First Proof of Fermat¹s Last Theorem to Exist
> 44[ 6.It was Proven, by a Pthagorean, that, for
> Now that it has been changed, it is no longer
> the Þrst proof...
That¹d be the case =iff= we were doing ŒSemantics¹,
and not Maths.
The Error was one word, which, if anyone looks,
they will see was not actually an Error in Maths,
but a but an Error in Œlanguage¹-interface dynamics
-- because, in lines 11-19 of the Proof, the =Maths=
is clearly-stated to be any Right-Triangle =except=
in some cases where n = 2.
I do =not= Œexcuse¹ myself, but are we doing Maths
or ŒSemantics¹? [I do understand that Semantics is
QuantiÞable.]
And, if we¹re doing the latter, where does that
leave Maths?
What it comes down to is that I¹ve not had the
BeneÞt of Collegial interaction [no one to read
my Proofs before they are Published, no one to
do Œlanguage¹-interfacing with, and, so, Practice
it], nor the BeneÞt of Formal Training in Maths
Proofs.
Still, I Love Maths, and have done a =lot= of
SigniÞcant work in Maths.
What your position comes down to is that you¹d
Equate my having to work without the BeneÞts
of Collegial interaction and Formal Training with
Maths.
But such does not Compute, does it?
Nope.
Your position would reduce all of the Worth that¹s
in the Proof of FLT that I¹ve Published to a word
Carlessly-used, Cherish that, and throw-out the
Maths.
But such doesn¹t Compute, does it?
Nope.
Your position has everything to do with absence-
of-understanding with respect to how and why
nervous systems process information [which is
one of the primary foci of the Maths I do], and
=nothing= to do with Maths.
So, where does that leave your position, with
respect to Maths?
It¹s VeriÞably not in-Maths, is it?
Nope.
> Wiles¹ proof is. And since Wiles¹ is actually
> a proof, ken¹s is of no further interest.
Nope.
I Published this Proof of FLT, using different
words, in CompuServe¹s Sci/Math Forum,
before Wiles Published his Proof that invokes
Eliptical Functions.
The Proof of FLT that I¹ve Published not only
was the First Proof of FLT, it¹s also the First
Proof that higher Maths [including the Eliptical
Functions invoked by Wiles] and simple Geom-
etry are, at least partially, Unitary, and I Know,
with Certainty, that this partiality is signiÞcant-
ly-non-partial
It¹s actually =this= Proof that I¹m Œexcited¹ about.
And that¹s only the beginning of the Proof I¹ve
Published, and that you would Œthrow out¹.
[My Conjecture is that all Maths is Unitary -
that, rather than being a Œshow-stopper¹,
Incommensurability is as a ŒKey¹ that Œun-
locks¹ Maths - and that Unity can be ac-
cessed via the Methods that are discussed
in the Proof of FLT that I¹ve Published. I
understand that this seems Œwildly¹-too-much,
but it¹s =Exactly= what is done, albeit, in a
partial case, in the Proof of FLT that I¹ve
Published.]
The Thing is to =Advance= Maths.
Beyond stating Truth, as I¹m Obliged to do,
I¹m =not= interested in Œpunishing¹ folks with
respect to what¹s past [it does no Good to
do so, and can shunt things off in directions
that can do Considerable Harm, and, not in
the least, to the coming-forward of Newness].
I¹m, here in sci.math to =Advance= Maths,
and, using Maths, to =Advance= Science.
You¹d Œthrow-out¹ all of that Œbecause¹ I¹ve
never had the BeneÞts of Collegial interac-
tion or Formal Training?
Such does not Compute.
k. p. collins
===
Subject: IS THERE A PROOF FOR THE FOLLOWING CONJECTURE
:A=2*q+z (PRIME)
IF....
IF q,z are prime numbers diferent each other than every
number of the form
A=2*q+z is prime for every q,z
for example
A=2*3+5=11
A=2*5+3=13
A=2*5+7=17
A=2*7+5=19
.....
===
Subject: Re: IS THERE A PROOF FOR THE FOLLOWING CONJECTURE
:A=2*q+z (PRIME)
IF....
>IF q,z are prime numbers diferent each other than every
number of the form
>A=2*q+z is prime for every q,z
>for example
>A=2*3+5=11
>A=2*5+3=13
>A=2*5+7=17
>A=2*7+5=19
>.....
2*3+2 = 8
2*13+7 = 33
I can see you have really put a lot of thought into your
conjecture.
--Lynn
===
Subject: Re: IS THERE A PROOF FOR THE FOLLOWING CONJECTURE
:A=2*q+z (PRIME)

IF....
IF q,z are prime numbers diferent each other than every
number of the
form
A=2*q+z is prime for every q,z
for example
A=2*3+5=11
> A=2*5+3=13
> A=2*5+7=17
> A=2*7+5=19
> .....
Did you forget that 2 is prime?
Anyhow, you don¹t have to go much further in your sequence to
Þnd a counterexample even for odd primes:
A=2*11+3=25
===
Subject: Lenstra¹s ECM
Could somebody explain to me how Lenstra¹s ECM of
factorisation works
or point me towards a place where I could Þnd out more about
it.
DMcG.
===
Subject: Re: Lenstra¹s ECM
> Could somebody explain to me how Lenstra¹s ECM of
factorisation works
> or point me towards a place where I could Þnd out more
about it.
Here¹s my brief attempt.
Suppose we are trying to factorise n;
and suppose
E(Q): y^2 = x^3 + bx + c (b,c in Z)
is an elliptic curve,
and P = (x,y) is a point on E with x,y in Z.
(In practice choose x, y and b, and then compute c.)
Now suppose the prime p | n,
and suppose the group on the elliptic curve
E(F_p): y^2 = x^3 + bx + c mod p
over the Þnite Þeld F_p has order m that is b-smooth,
ie if p^e | m then p^e <= b.
Let k = lcm{1,2,...,b}.
Then kP = 0 = [0,1,0] on E(F_p).
In other words, if we compute kP as a point of E(Q)
then its x-coordinate, x_k say, is divisible by p.
Hence gcd(x_k,n) > 1, giving a factor of n.
In practice, compute kP mod n.
(If you like, compute kP assuming n is a prime.)
If the computation breaks down at some point,
this is good, because it (almost certainly)
gives you a factor of n.
For example, you might need to work out a^{-1} mod n.
This will fail if gcd(a,n) > 1,
but in that case gcd(a,n) is a factor of n.
The idea is to try this with various elliptic curves,
and perhaps increasing b.
Hopefully one curve will have have b-smooth order mod p
for some p dividing n,
and then gcd(x_k,n) will give a factor of n.
--
Timothy Murphy
e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie
tel: +353-86-2336090, +353-1-2842366
s-mail: School of Mathematics, Trinity College, Dublin 2,
Ireland
===
Subject: Re: Wanting proof of an interesting fact.
induction step.
T_(n+1) = Sum 0<=i<=n d_(n-i) T_i
= Sum 0<=i<=n-1 e_i T_i
where e_i = d_(n-i) - d_0 d_(n-i-1)
This does not reduce to the n-1 cases. It is because T_i,
0<=i<=n-1, are
all functions of d_i¹s. So d_i¹s serve both as the
coefÞcients in the
summation and also argument to each T_i¹s.
i.e. minimizing
T_(n+1) = Sum 0<=i<=n-1 e_i T_i
is different from minimizing
T_(n) = Sum 0<=i<=n-2 d_i T_i
>
>>I have done some programming which supports the following
conjecture,
>>but I can¹t prove it. I hope some of you may prove it (or
disprove it,
>>which I don¹t want to see).
>>Suppose
>>d_0 >= d_1 >= d_2 >= ... >= d_n = 0
>>d_0 + d_1 + ... + d_{n-1} = C ,for some positive constant C.
>>Set S_{-1}=1 and recursively deÞne
>>S_i = d_0 * S_{i-1} + d_1 * S_{i-2} + ... + d_i * S_{-1}
>>So the Þrst few S¹s are
>>S_{-1} = 1
>>S_0 = d_0
>>S_1 = d_1 + d_0^2
>>S_2 = d_2 + 2d_0*d_1 + d_0^3
>>Our conjecture is:
>>S_n - S_{n-1} is minimized by setting
d_0=d_1=...=d_{n-1}=C/n
>>Best Wishes,
>>Daniel Fan

> I have a very *rough* solution. The basic outline should be
okay, you
may
> need to repair some messes and potholes. It¹s probably more
complicated
> than it needs to be.
I think this will be a bit simpler if you rephrase the problem
equivalently:
****Problem:
> Suppose:
> d_0 >= d_1 >= ... >= d_n >= 0
> Sum 0<=i<=n (d_i) = C_n >= C_(n-1)
DeÞne S to be:
> S_0 = 1
> S_(n+1) = Sum 0<=i<=n d_i S_(n-i) = Sum 0<=i<=n d_(n-i) S_i
the recurrence in the two forms, because sometimes it will be
easier to
> understand the subscripts one way, sometimes the other.
****Proof:
> Now if T_(n+1) = S_(n+1) - S_(n), then for n > 1:
1) T_(n+1) = Sum 0<=i<=n (d_i S_(n-i)) - Sum 0<=i<=n-1 (d_i
S_(n-i-1))
> = d_n S_0 + Sum 0<=i<=n-1 d_i [S_(n-i) - S_(n-i-1)]
> = d_n + Sum 0<=i<=n-1 d_i T_(n-i)
For convenience, we can deÞne T_0 = 1, and this becomes:
T_(n+1) = Sum 0<=i<=n d_i T_(n-i) = Sum 0<=i<=n d_(n-i) T_n
The same recurrence as for S_(n+1)! However, T_n is not the
same as S_n,
> since we have not yet deÞned the seed T_1, which is a
special case not
> covered by the recurrence. Plugging in gives us T_1 = S_1 -
S_0 = d_0 -
1.
Now it looks like we¹re headed towards a proof by strong
induction.
> Clearly the statement is true for T_1 = S_1 - S_0, since
there is only
one
> d_i here.
****Induction step:
> Assume the statement holds for all T_i from T_1 to T_n.
Then we seek to
> minimize
> T_(n+1) = Sum 0<=i<=n d_(n-i) T_i
> = Sum 0<=i<=n-1 d_(n-i) T_i + d_0 T_n
> = Sum 0<=i<=n-1 d_(n-i) T_i + d_0 Sum 0<=i<=n-1 d_(n-1-i)
T_i
> = Sum 0<=i<=n-1 (d_(n-i) - d_0 d_(n-1-i) T_i
Let e_i = d_(n-i) - d_0 d_(n-i-1)
> = Sum 0<=i<=n-1 e_i T_i
If the e_i¹s were independent, we¹d know the way to minimize
the above
> expression over them, since the problem reduces to the n-1
case.
Let¹s try setting them equal, and see if we can make it work.
> 0 = e_i - e_j = d_(n-i) - d_0 d_(n-i-1) - d_(n-j) + d_0
d_(n-j-1)
> = d_(n-i) - d_(n-j) + d_0 (d_(n-j-1) - d_n-i-1)
This describes a system of n equations, but it¹s apparent
setting all d¹s
> equal to each other is sufÞcient for the e¹s to be equal,
too!
So, setting the d¹s equal minimizes T_(n+1).
That concludes the induction step.
****Closed form solutions:
> We can easily Þnd closed form solutions for these optimal S
and T¹s.
An alternate formulation from (1) gives:
> T_(n+1) = Sum 0<=i<=n (d_i S_(n-i)) - Sum 1<=i<=n (d_(i-1)
S_(n-i))
> = d_0 S_n + Sum 1<=i<=n (d_i - d_(i-1)) S_(n-i)
> = Sum 0<=i<=n D_i S_(n-i)
> where D_0 = d_0; D_i = d_i - d_(i-1).
For an optimal set of d¹s:
> D_i = 0 (i!=0)
> T_(n+1) = d_0 S_n
> And so
> S_(n+1) = d_0 S_n + S_n = (1 + d_0) S_n
> Since S_1 = d_0, we get:
> S_(n+1) = (1 + d_0)^n d_0
Since d_0 is so important, let¹s just call it d.
> S_(n+1) = d (1 + d)^n
> T_(n+1) = d (1 + d)^n - d (1 + d)^(n-1)
> = d (1 + d)(1 + d)^(n-1) - d (1 + d)^(n-1)
> = d^2 (1 + d)^(n-1)

> Hope you can Þnd something useful in this mess.
--
> +---- Kevin C. Saff ----+ F-15 | |Eagle
> | Engineer, Boeing, StL | _____|_^_|_____
> | Tracking/Fleet Support| * + [_(x)_] + *

>
===
Subject: Re: Wanting proof of an interesting fact.
Your suggestion of changing the subscript does help.
But I Þnd a possible error in the induction step.
T_(n+1) = Sum 0<=i<=n d_(n-i) T_i
= Sum 0<=i<=n-1 e_i T_i
where e_i = d_(n-i) - d_0 d_(n-i-1)
This does not reduce to the n-1 case, because T_i, 0<=i<=n-1,
are all
function of d_i¹s, so when you minimize T_n with respect to
d_i¹s, you
are both changing the d_i¹s in the summation as well the
d_i¹s within
every previous T_i¹s.
>
>>I have done some programming which supports the following
conjecture,
>>but I can¹t prove it. I hope some of you may prove it (or
disprove it,
>>which I don¹t want to see).
>>Suppose
>>d_0 >= d_1 >= d_2 >= ... >= d_n = 0
>>d_0 + d_1 + ... + d_{n-1} = C ,for some positive constant C.
>>Set S_{-1}=1 and recursively deÞne
>>S_i = d_0 * S_{i-1} + d_1 * S_{i-2} + ... + d_i * S_{-1}
>>So the Þrst few S¹s are
>>S_{-1} = 1
>>S_0 = d_0
>>S_1 = d_1 + d_0^2
>>S_2 = d_2 + 2d_0*d_1 + d_0^3
>>Our conjecture is:
>>S_n - S_{n-1} is minimized by setting
d_0=d_1=...=d_{n-1}=C/n
>>Best Wishes,
>>Daniel Fan

> I have a very *rough* solution. The basic outline should be
okay, you
may
> need to repair some messes and potholes. It¹s probably more
complicated
> than it needs to be.
I think this will be a bit simpler if you rephrase the problem
equivalently:
****Problem:
> Suppose:
> d_0 >= d_1 >= ... >= d_n >= 0
> Sum 0<=i<=n (d_i) = C_n >= C_(n-1)
DeÞne S to be:
> S_0 = 1
> S_(n+1) = Sum 0<=i<=n d_i S_(n-i) = Sum 0<=i<=n d_(n-i) S_i
the recurrence in the two forms, because sometimes it will be
easier to
> understand the subscripts one way, sometimes the other.
****Proof:
> Now if T_(n+1) = S_(n+1) - S_(n), then for n > 1:
1) T_(n+1) = Sum 0<=i<=n (d_i S_(n-i)) - Sum 0<=i<=n-1 (d_i
S_(n-i-1))
> = d_n S_0 + Sum 0<=i<=n-1 d_i [S_(n-i) - S_(n-i-1)]
> = d_n + Sum 0<=i<=n-1 d_i T_(n-i)
For convenience, we can deÞne T_0 = 1, and this becomes:
T_(n+1) = Sum 0<=i<=n d_i T_(n-i) = Sum 0<=i<=n d_(n-i) T_n
The same recurrence as for S_(n+1)! However, T_n is not the
same as S_n,
> since we have not yet deÞned the seed T_1, which is a
special case not
> covered by the recurrence. Plugging in gives us T_1 = S_1 -
S_0 = d_0 -
1.
Now it looks like we¹re headed towards a proof by strong
induction.
> Clearly the statement is true for T_1 = S_1 - S_0, since
there is only
one
> d_i here.
****Induction step:
> Assume the statement holds for all T_i from T_1 to T_n.
Then we seek to
> minimize
> T_(n+1) = Sum 0<=i<=n d_(n-i) T_i
> = Sum 0<=i<=n-1 d_(n-i) T_i + d_0 T_n
> = Sum 0<=i<=n-1 d_(n-i) T_i + d_0 Sum 0<=i<=n-1 d_(n-1-i)
T_i
> = Sum 0<=i<=n-1 (d_(n-i) - d_0 d_(n-1-i) T_i
Let e_i = d_(n-i) - d_0 d_(n-i-1)
> = Sum 0<=i<=n-1 e_i T_i
If the e_i¹s were independent, we¹d know the way to minimize
the above
> expression over them, since the problem reduces to the n-1
case.
Let¹s try setting them equal, and see if we can make it work.
> 0 = e_i - e_j = d_(n-i) - d_0 d_(n-i-1) - d_(n-j) + d_0
d_(n-j-1)
> = d_(n-i) - d_(n-j) + d_0 (d_(n-j-1) - d_n-i-1)
This describes a system of n equations, but it¹s apparent
setting all d¹s
> equal to each other is sufÞcient for the e¹s to be equal,
too!
So, setting the d¹s equal minimizes T_(n+1).
That concludes the induction step.
****Closed form solutions:
> We can easily Þnd closed form solutions for these optimal S
and T¹s.
An alternate formulation from (1) gives:
> T_(n+1) = Sum 0<=i<=n (d_i S_(n-i)) - Sum 1<=i<=n (d_(i-1)
S_(n-i))
> = d_0 S_n + Sum 1<=i<=n (d_i - d_(i-1)) S_(n-i)
> = Sum 0<=i<=n D_i S_(n-i)
> where D_0 = d_0; D_i = d_i - d_(i-1).
For an optimal set of d¹s:
> D_i = 0 (i!=0)
> T_(n+1) = d_0 S_n
> And so
> S_(n+1) = d_0 S_n + S_n = (1 + d_0) S_n
> Since S_1 = d_0, we get:
> S_(n+1) = (1 + d_0)^n d_0
Since d_0 is so important, let¹s just call it d.
> S_(n+1) = d (1 + d)^n
> T_(n+1) = d (1 + d)^n - d (1 + d)^(n-1)
> = d (1 + d)(1 + d)^(n-1) - d (1 + d)^(n-1)
> = d^2 (1 + d)^(n-1)

> Hope you can Þnd something useful in this mess.
--
> +---- Kevin C. Saff ----+ F-15 | |Eagle
> | Engineer, Boeing, StL | _____|_^_|_____
> | Tracking/Fleet Support| * + [_(x)_] + *

>
===
Subject: hyperplane intersection with ellipsoid
I have an ellipsoid centered at the origin, (x^T)Dx <= 1
where D is a diagonal matrix of size n x n. I have a
hyperplane x.w = 0 passing thru the origin.
[ Assume that all diagonal entries of D are +ve ].
How do I compute the intersection of the plane with the
ellipsoid
and get the one lower dimensional ellipsoid? What is the most
numerically stable way of doing this?
--Elijah
===
Subject: Between T0 and T1
If S is a T0 space, is the set
U_y = { x | y in (cl x)x }
open?
When S is T1, then U_y = nulset is open.
Conversely, thesis implies space is T0. Proof:
if x not in U_y:
x = y or y not in cl x
x = y or some open U nhood y with x not in U
if x in open U_y:
some open U nhood x with x in U subset U_y
y not in U_y
y not in U
----
===
Subject: Re: Can this be solved for x?
> .
Given an equation in the form: A^x + B^x = C
Can this solved for x?
Yes
> But if you want real solutions A,B and C must be greater
than 0,
> because we shall take logarithms of A, B and C.
> As Log(A) appears in the denomiantor, A must greater than 1.
> My preferred method of solutions is by iteration.
> Take A^x as common factor:
A^x.(1 + (B/A)^x) = C ; if A > B take R = B/A otherwise R =A/B
>
> A^x = C / (1 + R^x) --> x Log(A) = Log(C) - Log(1 + R^x)
x = Log(C) / Log(A) - Log(1 + R^x) / Log(A)

> Call X0 = Log(C) / Log(A)
Beginning with x = X0
If A > B iterate:
>
> x = X0 - Log(1 + R^x)/ Log(A)
Otherwise X0 = Log(C) / Log(B) and iterate:
>
> x = X0 - Log(1 + R^x)/Log(B)
>
The conditions for the convergence of that iteration are:
A > 0 ; B > 0 ; C > 0

If C = 1 then A <> 1 and B <> 1
If A > B then A <> 1
If B > A then B <> 1 L.Rodriguez
===
Subject: When 100% isn¹t 100%
I read in a book it¹s possible to prove that the 100% of the
zeros to
the Zeta function lie on the 1/2 + ib line. But 100% != all
--> there
can still be inÞnitely many exceptions..
I accept the result on face value at the moment but i¹d like
to see a
proof that saying 100% of all integers have a property !=
all. It
doesn¹t matter if it isn¹t to do with the Zeta function.. any
proof
demonstrating this concept would be nice.
Another question: Does this problem with 100%!=all only occur
in
inÞnite sets?
Simon.
===
Subject: Re: When 100% isn¹t 100%
> I read in a book it¹s possible to prove that the 100% of
the zeros to
> the Zeta function lie on the 1/2 + ib line. But 100% != all
--> there
> can still be inÞnitely many exceptions..
I accept the result on face value at the moment but i¹d like
to see a
> proof that saying 100% of all integers have a property !=
all. It
> doesn¹t matter if it isn¹t to do with the Zeta function..
any proof
> demonstrating this concept would be nice.
Another question: Does this problem with 100%!=all only occur
in
> inÞnite sets?
Simon.
What it means is this...
Let A be a subset of N = {1,2,3,...}
for each n, let a_n = |A intersect {1,2,...,n}| be the
number of elements of A from 1 to n.
We say Þguratively that A is k percent of N if the
limit of a_n/n is k/100.
For example, let A be all natural numbers except the squares.
Then a_n/n approaches 1, even though a_n < n for every n.
a_10000/10000 = 9900/10000 = 0.99, so 99 percent of the
numbers
from 1 to 10000 are non-squares. As n gets bigger, this
fraction increases, with limit 1.
Yet there are still inÞnitely many squares...
zero percent of the integers are squares...
--
G. A. Edgar
http://www.math.ohio-state.edu/~edgar/
===
Subject: Distribution 2D and Correlation
Hello everybody,
Does anybody knows how to express the correlation between two
random
variables:
Cxy = E[XY] (where E[.] is the expectation)
from the joint distribution:
Fxy(xy) = P(X < x, Y < y)
Sparkes
===
Subject: who is captain in here?
where is sci.math server?
who is a administrator?
I wonder about how to operate sci.math.
here is miraculous.
===
Subject: Re: who is captain in here?
> where is sci.math server?
There are many , in many countries; they are news-servers, not
just math-servers.
> who is a administrator?
Your admin. is at Hanaro. He/she administrates all Hanaro,
but not US or Japan company or university news-servers.

> I wonder about how to operate sci.math.
The servers talk to each other, spreading news over
a large part of the world.
> here is miraculous.
I read that the following book has had good reviews:
Where Wizards Stay Up Late: The Origins of the Internet, by
Katie Hafner and Matthew Lyon
David Bernier
from the headers of your message:
===
Subject: who is captain in here?
administrator¹s e-mail)
[etc.]
===
Subject: Re: who is captain in here?
> where is sci.math server?
Babylon.
who is a administrator?
The Beast.
I wonder about how to operate sci.math.
and power was given him over all kindreds, and tongues, and
nations.
here is miraculous.
Here is wisdom. Let him that hath understanding count the
number
of the beast: for it is the number of a man; and his number is
Six hundred threescore and six.
Using the simple rule a=1, b=2, c=3, etc. and ignoring
punctuation,
s c i m a t h = 19 3 9 13 1 20 8
Now group and sum the numbers
primes: 19 + 3 + 13 = 35
composites: 9 + 20 + 8 = 37
units: 1 = 1
Note that the primes plus units equals 36 and that the
composites minus the units also equals 36.
Coincidence?
Of course not! The sum of integers from 1 to 36 equals 666.
And he causeth all, both small and great, rich and poor,
free and bond, to receive a mark in their right hand,
or in their foreheads. And that no man might buy or sell,
save he that had the mark, or the name of the beast,
or the number of his name.
As you can see, if you want math help, you must accept
the Mark of the Beast.
===
Subject: Re: who is captain in here?
We are an anarcho-syndicalist collective....
===
Subject: Re: who is captain in here?
> We are an anarcho-syndicalist collective....
===
Subject: Re: who is captain in here?
>where is sci.math server?
Nowhere, yet everywhere.
>who is a administrator?
No one, yet everyone.
>I wonder about how to operate sci.math.
No way, yet it still operates.
>here is miraculous.
Any miracles are due to people, not computers.
--
I¹m not interested in mathematics that might have anything
to do with reality. -- Russell Easterly, in sci.math
===
Subject: Re: um......a little problem..0^0=1
>hello.......
>i want a problem that use 0^0 = 1
>i don¹t ask to reason of 0^0 = 1
>if there is deÞned 0^0 =1 , what¹s advantage?
>advice...please...thank you.
You say you don¹t want the reason why 0^0 = 1, but then you
ask what is
the advantage of deÞning 0^0 = 1. These seem somewhat
contradictory.
The advantage of deÞning 0^0 = 1 is that it is the most
useful in the
most cases. Strictly speaking, 0^0 is not deÞned since the
limit as
x and y approach 0 of x^y depends on how x and y approach 0.
However,
in most formulas, for example the binomial theorem, the
exponent, y, is
either a constant or restricted to the integers. In either
case, x^y
is 1 for any applicable values of x and y near 0. In most
cases where
the exponent is real and variable, the expression is often in
the form
of an exponential, e^f(x,y).
So usually when you encounter 0^0, the most useful value is
1. It is
such a common convention, that if an author means anything
else, it
is usually stated explicitly. However, if the author is a
student in
a calculus class, the intended value of 0^0 should be stated
explicitly
in any case, accompanied by a proof if necessary.
Rob Johnson <rob@trash.whim.org>
take out the trash before replying
===
Subject: Re: um......a little problem..0^0=1
> The advantage of deÞning 0^0 = 1 is that it is the most
useful in the
> most cases. Strictly speaking, 0^0 is not deÞned since the
limit as
> x and y approach 0 of x^y depends on how x and y approach 0.
You are assuming that 0^0 cannot be deÞned in any other way
except by
invoking limits. If that¹s the case, then how do you deÞne
x^y for
positive x and y without invoking limits and hence being
circular?
Strictly speaking, 0^0 = 1 because the cardinality of the
set of mappings from the empty set to itself is 1.
--
Dave Seaman
Judge Yohn¹s mistakes revealed in Mumia Abu-Jamal ruling.
<http://www.commoncouragepress.com/index.cfm?action=book&
bookid=228>
===
Subject: probability question
Believe it or not, this is a real-world question even though
the
numbers are a bit wonky.
I¹m drawing from a deck of 39 cards, 16 of which are good
cards, 23
of which are bad cards.
If I have a hand of 4 cards, I will draw 4 cards, then
discard as many
of those cards as I wish and draw the same number of cards to
reÞll
my hand.
What are the odds that I¹ll have at least 2 good cards in my
hand
after discarding any bad cards and reÞlling my hand?
I¹d love to hear (and understand) a method of Þnding the
answer to
the more general deck of X cards, Y of which are good, Z of
which
are bad with a hand of M cards, what are the odds that N of
those
cards are good.
Obviously, I would only keep good cards that I draw in my
Þrst hand
and discard the bad cards.
Right now, I can Þgure the odds of having at least 1 good
card by
calculating the odds of having no good cards at all, but I
can¹t
Þgure out how to calculate having 2 or more good cards in my
ending
hand after the discard.
Michael Wilson
===
Subject: Re: probability question
> Believe it or not, this is a real-world question even
though the
> numbers are a bit wonky.
> I¹m drawing from a deck of 39 cards, 16 of which are good
cards, 23
> of which are bad cards.
> If I have a hand of 4 cards, I will draw 4 cards, then
discard as many
> of those cards as I wish and draw the same number of cards
to reÞll
> my hand.
> What are the odds that I¹ll have at least 2 good cards in
my hand
> after discarding any bad cards and reÞlling my hand?
> I¹d love to hear (and understand) a method of Þnding the
answer to
> the more general deck of X cards, Y of which are good, Z of
which
> are bad with a hand of M cards, what are the odds that N of
those
> cards are good.
> Obviously, I would only keep good cards that I draw in my
Þrst hand
> and discard the bad cards.
> Right now, I can Þgure the odds of having at least 1 good
card by
> calculating the odds of having no good cards at all, but I
can¹t
> Þgure out how to calculate having 2 or more good cards in
my ending
> hand after the discard.
> Michael Wilson
So if you calculate the probability of exactly 1 good card,
then you can
subtract the result from the probability of at least 1 good
card to get the
answer you¹re looking for. You need to consider the
probability of getting
one good card on the Þrst draw, and no further good cards on
the second
draw; and the probability of getting no good cards initially,
and one good
card on the second draw.
Duncan
===
Subject: Re: Totally ordered sets
>
>> True or false? Given any two totally ordered sets, one can
be
>> embedded in the other.

> Of course, all I needed to do was post to think of an
answer to my own
> question. The answer is false: Neither of omega-1 and the
rationals
> can be embedded in the other.
For a more trivial example, consider the positive integers
and the negative integers.
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
Lacan, Jacques, 79, 91-92; mistakes his penis for a square
root, 88-9
Francis Wheen, _How Mumbo-Jumbo Conquered the World_
===
Subject: Re: Totally ordered sets
> True or false? Given any two totally ordered sets, one can
be embedded
> in the other.
Is there a class of prototypical totally ordered sets such
that any
> totally ordered set is isomorphic to a unique set in this
class? I am
> thinking of something similar to the way that ordinals are
prototypical
> well ordered sets.
See other posts for the answer to this speciÞc question.
However, changing the question gives an interesting answer:
Any countable total order can be embedded (order preserving)
in the
rationals.
The proof is very nice, and can be extended to many other
situations
(random graph embedds all countable graphs (as graphs) etc.).
You get various other properties aswell
Google Fraisse Limit for more information.
(It was the work of Roland Fraiss¹e, though I don¹t know the
reference, you can Þnd it in Hodges Model Theory, and
probably in
Theory of relations. Revised edition. With an appendix by
Norbert
Sauer. Studies in Logic and the Foundations of Mathematics,
145.
North-Holland Publishing Co., Amsterdam, 2000. ii+451 pp.
ISBN:
0-444-50542-3 which is a translation from the French I think).
To get more off topic, Phillip Hall [J. London Math. Soc. 34
(1959),
305--319; Zbl 88, 23] shows there is exactly one universal
locally
Þnite group. This paper interested me a great deal, as the
technique
was in some ways similar to the model theoretic technique of
Fraisse
but with a non elementary class of structures.
===
Subject: Re: Totally ordered sets
>True or false? Given any two totally ordered sets, one can
be embedded
>in the other.
False. The real numbers with their natural ordering, and any
well-ordered
set of cardinality aleph(c), where aleph denotes Hartog¹s
aleph and c is
the cardinality of R (the power of the continuum), serve as a
counterexample.
David
But I¹m always true to you, darlin¹, in my fashion,
Yes, I¹m always true to you, darlin¹, in my way.
-- Lois Lane
-----
===
Subject: Re: Totally ordered sets
Stephen J. Herschkorn
> Is there a class of prototypical totally ordered sets such
that any
> totally ordered set is isomorphic to a unique set in this
class? I am
> thinking of something similar to the way that ordinals are
prototypical
> well ordered sets.
I don¹t have a classiÞcation, but something can be said. Given
a set X with a total order T (identiÞed with a subset of XxX),
there exists a pair of orders on X, say R and S, such that
1) R is well-founded
2) the opposite of S (call it S~) is well-founded
3) T is the least upper bound of R and S~.
Here the l.u.b. means the weakest order which is stronger
than both of the given orders.
There are about 8 equivalent deÞnitions of a well-founded
order, but one of them is: an order which can be extended
to a well-order.
LH
===
Subject: Re: Totally ordered sets
>
>>
>> True or false? Given any two totally ordered sets, one can
be
>> embedded in the other.
>>
> Of course, all I needed to do was post to think of an
answer to my
> own question. The answer is false: Neither of omega-1 and
the
> rationals can be embedded in the other.
>> and... er... what¹s omega-1 ?
> The least uncountable ordinal, also called aleph-1.
Oh. And the ordering on omega_1, is that then the same
ordering
as that on the reals?
Then just by cardinality I see that the rationals cannot be
embedded into omega_1, but then wouldn¹t it seem like the
rationals
could be embedded in omega_1 (trivially)?
--
Mitch Harris
(remove q to reply)
===
Subject: Re: Totally ordered sets
>>
>
> True or false? Given any two totally ordered sets, one can
be
> embedded in the other.
>
>> Of course, all I needed to do was post to think of an
answer to my
>> own question. The answer is false: Neither of omega-1 and
the
>> rationals can be embedded in the other.
> and... er... what¹s omega-1 ?
>> The least uncountable ordinal, also called aleph-1.
>Oh. And the ordering on omega_1, is that then the same
ordering
>as that on the reals?
Not the natural ordering on the reals. The natural ordering
on the
reals is dense. An ordering of type omega-1 is not dense.
>Then just by cardinality I see that the rationals cannot be
>embedded into omega_1, but then wouldn¹t it seem like the
rationals
>could be embedded in omega_1 (trivially)?
You can¹t use cardinality to prove that the rationals cannot
be embedded
into omega-1. The rationals have the same cardinality as
omega-0, so
that there are deÞnitely 1-1 mappings from the rationals to
omega-1.
There is no order-preserving 1-1 mapping since there exist
inÞnite
descending chains of rational numbers, but there is no
inÞnite descending
chain in omega-1.
David
But I¹m always true to you, darlin¹, in my fashion,
Yes, I¹m always true to you, darlin¹, in my way.
-- Lois Lane
-----
===
Subject: Re: Totally ordered sets
>
>Oh. And the ordering on omega_1, is that then the same
ordering
>as that on the reals?
Not the natural ordering on the reals. The natural ordering
on the
> reals is dense. An ordering of type omega-1 is not dense.
Also, it goes almost but (IMHO) not quite without saying
that this ordering exists only under CH. Indeed its
existence *is* the CH.
===
Subject: Re: Totally ordered sets
>>
>>Oh. And the ordering on omega_1, is that then the same
ordering
>>as that on the reals?
>>
>> Not the natural ordering on the reals. The natural
ordering on the
>> reals is dense. An ordering of type omega-1 is not dense.
>Also, it goes almost but (IMHO) not quite without saying
>that this ordering exists only under CH. Indeed its
>existence *is* the CH.
That depends on what you are saying. If you are disussing
whether
the reals can be given an ordering of type omega-1, then yes,
such
an ordering only exists under CH, and its existence is
equivalent
to CH. If, on the other hand, you are just discussing
orderings of
type omega-1, then these exist without assuming AC or CH.
David
But I¹m always true to you, darlin¹, in my fashion,
Yes, I¹m always true to you, darlin¹, in my way.
-- Lois Lane
-----
===
Subject: Re: Totally ordered sets
> Also, it goes almost but (IMHO) not quite without saying
> that this ordering exists
on the reals
only under CH. Indeed its
> existence *is* the CH.
===
Subject: Re: Question about associativity of cartesian product
> What I mean by canonical is that it commutes with the
projections.
> In a cartesian product there are projections
p_A: A*B -> A deÞned by (a,b) |-> a
> p_B:A*B -> B deÞned by (a,b) |-> b
Likewise for cartesian products (A*B)*C, A*(B*C) there are
projections
> into each of its factors. The canonical isomorphism
i:(A*B)*C -> A*(B*C)
commutes with the projections. In details: we have the
equality of
> functions
p_A = p_A compose i
where p_A are the relevant projections of the cartesian
products, and
> the similar equations for p_B and p_C. In fact i is the
*unique* map
> (A*B)*C -> A*(B*C) that commutes with the projections -
that is what I
> mean by being uniquely deÞned. It is a theorem by S.
Maclane that
> given two ways to parenthesize an n-fold cartesian product
(for
> example, for n=3, there are only 2 ways: (A*B)*C and
A*(B*C) but for
> n=4 there are already 5 ways) there is a *unique*
isomorphism between
> them commuting with the projections. This justiÞes the
usual practice
> of treating the cartesian product of sets as if it were a
strictly
> associative operation.
By similar argument wouldn¹t there exist *unique* map A*B ->
B*A that
commutes with the projections? Is Catesian Product
commutative?
===
Subject: Re: Question about associativity of cartesian product
>> What I mean by canonical is that it commutes with the
projections.
>> In a cartesian product there are projections
>>
>> p_A: A*B -> A deÞned by (a,b) |-> a
>> p_B:A*B -> B deÞned by (a,b) |-> b
>>
>> Likewise for cartesian products (A*B)*C, A*(B*C) there are
projections
>> into each of its factors. The canonical isomorphism
>>
>> i:(A*B)*C -> A*(B*C)
>>
>> commutes with the projections. In details: we have the
equality of
>> functions
>>
>> p_A = p_A compose i
>>
>> where p_A are the relevant projections of the cartesian
products, and
>> the similar equations for p_B and p_C. In fact i is the
*unique* map
>> (A*B)*C -> A*(B*C) that commutes with the projections -
that is what I
>> mean by being uniquely deÞned. It is a theorem by S.
Maclane that
>> given two ways to parenthesize an n-fold cartesian product
(for
>> example, for n=3, there are only 2 ways: (A*B)*C and
A*(B*C) but for
>> n=4 there are already 5 ways) there is a *unique*
isomorphism between
>> them commuting with the projections. This justiÞes the
usual practice
>> of treating the cartesian product of sets as if it were a
strictly
>> associative operation.
>By similar argument wouldn¹t there exist *unique* map A*B ->
B*A that
>commutes with the projections? Is Catesian Product
commutative?
Yes.
G. Rodrigues
===
Subject: Re: How many integers are there?
Originator: richard@cogsci.ed.ac.uk (Richard Tobin)
Blimey guv, there must be dozens of Œem.
-- Richard
===
Subject: Re: How many integers are there?
How many integers are there? All of them.
David Ames
===
Subject: Re: How many integers are there?
> > Is 1 the last integer?
> > Is 2 the last integer?
> > Is 3 the last integer?
none of them. There are inÞnite integers!!!

There is a computable number that agrees with the diagonal
number to
> the Þrst digit.
> There is a computable number that agrees with the diagonal
number to
> the Þrst 2 digits.
> There is a computable number that agrees with the diagonal
number to
> the Þrst 3 digits.
And so on.
But there is no computable number that agrees with the
diagonal number
> to inÞnitely many digits.
non sequitur
>
It doesn¹t follow from what I said before, but it is true and
easy to
prove.
> clarify why there are inÞnitely many integers and not
inÞnitely
computable diag. number digits.
The diagonal number is not computable. There is no computable
function
G(j) such that G(j) is the j-th digit of the diagonal number.
> I.e. how do you reason by extension there are an inÞnite
number of
integers?
>
I don¹t know what you¹re talking about. Of course there are
inÞnitely
many integers but this is irrelevant.
> InÞnite means no limit correct?
PROPOSITION :
> There is no limit to the number of digits that computable
numbers agree
with the diagonal number.
T/F ?
>
What I said before.
There is a computable number that agrees with the diagonal
number to
the Þrst digit.
There is a computable number that agrees with the diagonal
number to
the Þrst 2 digits.
There is a computable number that agrees with the diagonal
number to
the Þrst 3 digits.
And so on.
But there is no computable number that agrees with the
diagonal number
to inÞnitely many digits.
> Herc
===
Subject: Re: How many integers are there?
I don¹t actually agree with you (or what you said).
You give me a list of the computable numbers, and a simple
algorithm
produces the diagonal number.
There is no problem with computing the diagonal number per
se; the problem
is that the list of computable numbers is itself not
computable - you can¹t
give me the list.
> > Is 1 the last integer?
> > Is 2 the last integer?
> > Is 3 the last integer?
none of them. There are inÞnite integers!!!

> There is a computable number that agrees with the diagonal
number to
> > the Þrst digit.
> > There is a computable number that agrees with the
diagonal number to
> > the Þrst 2 digits.
> > There is a computable number that agrees with the
diagonal number to
> > the Þrst 3 digits.
And so on.
But there is no computable number that agrees with the
diagonal
number
> > to inÞnitely many digits.
non sequitur
It doesn¹t follow from what I said before, but it is true and
easy to
> prove.
> clarify why there are inÞnitely many integers and not
inÞnitely
computable diag. number digits.
> The diagonal number is not computable. There is no
computable function
> G(j) such that G(j) is the j-th digit of the diagonal
number.
> I.e. how do you reason by extension there are an inÞnite
number of
integers?
I don¹t know what you¹re talking about. Of course there are
inÞnitely
> many integers but this is irrelevant.
> InÞnite means no limit correct?
PROPOSITION :
> There is no limit to the number of digits that computable
numbers agree
with the diagonal number.
T/F ?
What I said before.
> There is a computable number that agrees with the diagonal
number to
> the Þrst digit.
> There is a computable number that agrees with the diagonal
number to
> the Þrst 2 digits.
> There is a computable number that agrees with the diagonal
number to
> the Þrst 3 digits.
> And so on.
> But there is no computable number that agrees with the
diagonal number
> to inÞnitely many digits.
> Herc
===
Subject: Re: How many integers are there?
> I don¹t actually agree with you (or what you said).
You give me a list of the computable numbers, and a simple
algorithm
> produces the diagonal number.
There is no problem with computing the diagonal number per
se; the
problem
> is that the list of computable numbers is itself not
computable - you
can¹t
> give me the list.
>
How does that contradict what I said?
> > Is 1 the last integer?
> > > Is 2 the last integer?
> > > Is 3 the last integer?
> > none of them. There are inÞnite integers!!!

There is a computable number that agrees with the diagonal
number
to
> > the Þrst digit.
> > There is a computable number that agrees with the
diagonal number
to
> > the Þrst 2 digits.
> > There is a computable number that agrees with the
diagonal number
to
> > the Þrst 3 digits.
And so on.
But there is no computable number that agrees with the
diagonal
number
> > to inÞnitely many digits.
non sequitur
>
> It doesn¹t follow from what I said before, but it is true
and easy to
> prove.
> clarify why there are inÞnitely many integers and not
inÞnitely
> computable diag. number digits.
The diagonal number is not computable. There is no computable
function
> G(j) such that G(j) is the j-th digit of the diagonal
number.
> I.e. how do you reason by extension there are an inÞnite
number of
> integers?
>
> I don¹t know what you¹re talking about. Of course there are
inÞnitely
> many integers but this is irrelevant.
> InÞnite means no limit correct?
PROPOSITION :
> > There is no limit to the number of digits that computable
numbers
agree
> with the diagonal number.
T/F ?
>
> What I said before.
There is a computable number that agrees with the diagonal
number to
> the Þrst digit.
> There is a computable number that agrees with the diagonal
number to
> the Þrst 2 digits.
> There is a computable number that agrees with the diagonal
number to
> the Þrst 3 digits.
> And so on.
But there is no computable number that agrees with the
diagonal number
> to inÞnitely many digits.
> Herc
===
Subject: Re: How many integers are there?
--------------
> I don¹t actually agree with you (or what you said).
> You give me a list of the computable numbers, and a simple
algorithm
> produces the diagonal number.
> There is no problem with computing the diagonal number per
se; the
problem
> is that the list of computable numbers is itself not
computable - you
can¹t
> give me the list.
why not?
I can give you the indexed set of computable numbers with
some other data
Þlling in
some of the spaces? will this do?
You¹re in the wrong thread.
Re: Is this SET of computable numbers computable?
Herc
===
Subject: Re: How many integers are there?
> PROPOSITION :
> There is no limit to the number of digits that computable
numbers agree
with the diagonal
number.
T/F ?
How many sequential digits of the diag are in the computable
numbers list?
How many Natural numbers are there?
same answer.
1
22
333
4444
55555
..
= there are an inÞnite number of Naturals
0.1
0.12
0.123
0.1234
0.12345
..
= there are inÞnite number of sequential digits of the
diagonal in the
computables.
Herc
===
Subject: Re: How many integers are there?
> > PROPOSITION :
> > There is no limit to the number of digits that computable
numbers
agree with the diagonal
> number.
T/F ?

> How many sequential digits of the diag are in the
computable numbers
list?
How many Natural numbers are there?
same answer.
1
> 22
> 333
> 4444
> 55555
> ..
= there are an inÞnite number of Naturals
0.1
> 0.12
> 0.123
> 0.1234
> 0.12345
> ..
= there are inÞnite number of sequential digits of the
diagonal in the
computables.

> Herc
For every j, there is a number on the list that agrees with
the
diagonal number to j digits.
But there is no number on the list that agrees with the
diagonal
number to inÞnitely many digits.
The diagonal number is not on the list.
===
Subject: Re: How many integers are there?
> > PROPOSITION :
> > There is no limit to the number of digits that computable
numbers
agree with the diagonal
> number.
T/F ?

> How many sequential digits of the diag are in the
computable numbers
list?
How many Natural numbers are there?
same answer.
1
> 22
> 333
> 4444
> 55555
> ..
= there are an inÞnite number of Naturals
0.1
> 0.12
> 0.123
> 0.1234
> 0.12345
> ..
= there are inÞnite number of sequential digits of the
diagonal in the
computables.

> Herc
> For every j, there is a number on the list that agrees with
the
> diagonal number to j digits.
> But there is no number on the list that agrees with the
diagonal
> number to inÞnitely many digits.
There are numbers on the list that agree with the diagonal
number
to unlimited amount of digits.
> The diagonal number is not on the list.
Its repeating representation is.
Herc
===
Subject: Complexity of refuting Turing machine impostors
Originator: tchow@lagrange.mit.edu.mit.edu (Timothy Chow)
Pick your favorite EXPTIME-complete language L, and deÞne a
function n as
follows: Given any Turing machine M and any polynomial p, let
n(M,p) be
the length of the shortest x that refutes M, i.e., the
shortest x on which
M fails to halt with the correct answer (x in L or x not in L)
after
p(|x|) steps. Such an x must exist, by the time hierarchy
theorem, and
can be effectively computed by exhaustive search.
Is there anything known, or even conjectured, about the
complexity or the
growth rate of n(M,p)? Naively, I would think that there
might exist
machines M that are very difÞcult to distinguish from
machines that
correctly decide L, since without the time bound p it is
undecidable
whether a given machine correctly decides L.
Of course, I picked PTIME and EXPTIME just for illustration;
I am also
interested in the analogous question for any other pair of
complexity
classes (uniform or not) that are provably or conjecturally
separated.
--
Tim Chow tchow-at-alum-dot-mit-dot-edu
The range of our projectiles---even ... the
artillery---however great, will
never exceed four of those miles of which as many thousand
separate us from
the center of the earth. ---Galileo, Dialogues Concerning Two
New Sciences
X-mailer: xrn 9.02
===
Subject: Re: Non-Abelian Groups [Was: Proving associativity]
Mail-To-News-Contact: postmaster@nym.alias.net
>> I started with aba=b, and looked for a contradiction.
Unfortunately, I
>> can¹t seem to do much with it....
> There¹s no contradiction. Did you read my earlier suggestion
>right through?
Sorry, but I didn¹t do that.
I read your initial (and quite helpful) post far enough to
get an idea
of how to tackle this proof. Then, I quite deliberately
closed my eyes
to the rest.
Even though I¹m doing this study on my own, I feel compelled
to try to
do as much of the work as possible. This is based on my
experience that
not doing the work leads directly and inevitably to not
really learning
the material.
So, I took the initial hints, did what I could, and struggled
with the
rest for several more days. When I did Þnally throw my hands
up in
frustration, I didn¹t think to review the skipped parts of
previous
posts. Rather, I just came right back here and asked
questions that had
already been answered.
I realize now that this was a mistake; one which I will
attempt to
avoid in the future.
--
Michael F. Stemper
#include <Standard_Disclaimer>
Outside of a dog, a book is man¹s best friend.
Inside of a dog, it¹s too dark to read.
===
Subject: summing fractions
prove: 1/1 + 1/2 + 1/3 + ... + 1/n<80, if you throw away the
fractions
with a 9 as a digit in the fraction¹s denominator Þrst.
Can this be put generally [i.e. for other individual digited
fractions
being thrown out?]
===
Subject: Re: summing fractions
Quddus Ali <quddusaliquddus@hotmail.com> escribi.97:
> prove: 1/1 + 1/2 + 1/3 + ... + 1/n<80, if you throw away
the fractions
> with a 9 as a digit in the fraction¹s denominator Þrst.
> Can this be put generally [i.e. for other individual
digited fractions
> being thrown out?]
Hint: How many integers n without nines there are in [10^k,
10^(k+1) - 1]?
For that n, set a bound B(k) for A(k) = Sum(1/n, n, 10^k,
10^(k+1) - 1),
and
then sum B(k), for k = 0 to inf.
--
Ignacio Larrosa Ca.96estro
A Coru.96a (Espa.96a)
ilarrosaQUITARMAYUSCULAS@mundo-r.com
===
Subject: (k+a/b)^n is not an integer?
For a,b,k,n positive integers such that a<b show that
(k+a/b)^n is not an
integer.
My suggestion for proof:
1 0<a<b => 0<a/b<1
2 k^n<(k+a/b)^n<(k+1)^n => k<k+a/b<k+1 (I take n-root of
inequalities)
3 k+a/b is not an integer
4 (k+a/b)^n is not an integer
Now I can¹t argument from 3 to 4. Could someone help me to
complete this
proof?
===
Subject: Re: (k+a/b)^n is not an integer?
Sikari <sikari.h@spray.se> escribi.97:
> For a,b,k,n positive integers such that a<b show that
(k+a/b)^n is
> not an integer. My suggestion for proof:
> 1 0<a<b => 0<a/b<1
> 2 k^n<(k+a/b)^n<(k+1)^n => k<k+a/b<k+1 (I take n-root of
inequalities)
> 3 k+a/b is not an integer
> 4 (k+a/b)^n is not an integer
> Now I can¹t argument from 3 to 4. Could someone help me to
complete
> this proof?
Obviously, you can go from 1 to 3 withou pass throw 2 ...
For the step 3 to 4, write k + a/b as a reduced fraction.
What about factor
descomposition of numerator an denominator of that fraction?
--
Ignacio Larrosa Ca.96estro
A Coru.96a (Espa.96a)
ilarrosaQUITARMAYUSCULAS@mundo-r.com
===
Subject: Lebesgue convergence
Sorry Tony, I forgot to state f must be integrable too.
Let f_n be a sequence of integrable functions and suppose
that {f_n}
converges to an *integrable* function f. (m=measure) Show
that:
If lim int( | f_n - f | dm) =0 then int ( | f| dm) = lim int
( |f_n| dm)
Is the following proof correct?
We notice that:
| |f_n| - |f| | <=|f_n - f|
So we can apply lebesgue dominated convergence theorem:
int( f dm) = lim int ( |f_n| - |f| dm) <= int( |f_n -f | dm)
(the
inequality
follows from the monoticity of the integral)
By asssumption: int ( | f_n - f| dm) =0
so lim int ( |f_n| - |f| dm ) =0
I don¹t know if the following step is valid (i.e linearity of
limits):
lim int( |f_n) dm) - lim int (|f| dm )=0
but lim int ( | f| dm ) = int ( |f| dm) (because |f| doesn¹t
depends on n)
so int(| f| dm)= lim int( | f_n| dm)
===
Subject: Re: Lebesgue convergence
> Let f_n be a sequence of integrable functions and suppose
that {f_n}
> converges to an *integrable* function f. (m=measure) Show
that:
> If lim int( | f_n - f | dm) =0 then int ( | f| dm) = lim
int ( |f_n| dm)
> Is the following proof correct?
You should note one thing: depending on what is your
deÞnition of the
Lebesgue (Borel) integral the above problem (like many others
of the same
kind) sums up to the following proposition, which shows that
the deÞnition
of the integral (same thing than below, but we suppose f_n
are simple
integrable functions instead) cannot be enriched by assuming
that the
functions f_n are integrable functions (in other words, you
won¹t get
more
integrable functions using the proposition below as a
deÞnition)) (Th
2.8.2, Foundations of Modern Analysis, Avner Friedman):
If a sequence of integrable functions (f_n) converges almost
everywhere
(or
in measure) to a function f, and if (f_n) is Cauchy in the
mean, then f is
integrable and lim int(f_n) = int(f).
===
Subject: Re: surface Þtting polynomial function
It¹s unusual today to use polynomials to Þt surfaces.
When you Þt a polynomial to a set of points, you can
make it go through the points, but it¹s likely to
be all over the place elsewhere.
Look into splines. Much better behaved.
John Nagle
Animats
===
Subject: tensor product question
I am trying to prove :
for a PID, R, if x,y are in R, and z = gcd(x,y).
then map R/(z) -> (R/(x),R/(y)) deÞned by 1 -> (1,1) in tensor
product induces an isomorphism. Please comment :
Proof : since the map is R-balanced, for r1,r2 in R; a1,a2 in
R/(z);
r1.a1+r2.a2 -> r1(a1 mod x,1) + r2(1,a2 mod y) ((can also use
some
other bracketed expressions for tensor)
this proves the R-module homomorphism part.
then since the kernel is 0, this is an isomorphism.
I wonder if some dimensionality argument can also be used
that is more
general, or uses some invariant factor or primary divisor
arguments.
X-mailer: xrn 9.02
===
Subject: Re: Prime numbers
Mail-To-News-Contact: postmaster@nym.alias.net
>Conjecture: Every number is the sum of a prime number plus a
non prime
>number.
>Intuitively, this conjecture is true. But proving it may or
may not
>be difÞcult.
Not too difÞcult, even for somebody that bombed out of a math
program
after three semesters:
For any given number, n, let p be the smallest prime greater
than n.
Let q = n-p. Since p>n, q<0 and hence non-prime.
n = p + q
> That is not the point!
What is the point of your post, then?
>My problem is - So what if the conjecture its true or false?
If you don¹t care about the truth or falsity of your
conjectures, why do
you bother to make them? (Serious question)
>If the conjecture has already been tested and discarded, so
much the
>better.
Well, now here, you appear to be saying that you do care: you
say that
it¹s better (by some measurement) if your conjecture is false.
--
Michael F. Stemper
#include <Standard_Disclaimer>
This message contains at least 95% recycled bytes.
Distribution: world
===
Subject: Re: Prime numbers
>> Conjecture: Every number is the sum of a prime number plus
a non prime
>> number.
>> Assuming that number means positive integer, 2 is a
>> counter-example.
> So is 5.
> Conjecture: Every integer > 5 is a
> sum of a positive prime plus a positive non prime.
Nope. 10 is a counter example.
1 + 9
2 + 8
3 + 7 (both prime)
4 + 6
5 + 5 (both prime)
What the real problem here is, is we¹re trying to deÞne a
trick,
without knowing about the underlying meaning of the
mathematics behind
it.
I don¹t know much about theory, but here¹s the Þrst few of
the series
of numbers that can¹t be summed with a prime and a non-prime:
0 1 2 3 4 5 6 8 10
...well I¹ll be. Looks like 10 is the last one. At least
nothing
between 10 and 10000 can¹t be summed with a prime and a
non-prime.
Any Fermats out there want to prove this?
Starling
===
Subject: Re: Prime numbers
>> Conjecture: Every integer > 5 is a
>> sum of a positive prime plus a positive non prime.
>Nope. 10 is a counter example.
>2 + 8
2 is prime, 8 is composite.
>I don¹t know much about theory, but here¹s the Þrst few of
the series
>of numbers that can¹t be summed with a prime and a non-prime:
>0 1 2 3 4 5 6 8 10
It helps if you know that 2 is a prime.
6 = 4 + 2
8 = 6 + 2
10 = 8 + 2
--
I¹m not interested in mathematics that might have anything
to do with reality. -- Russell Easterly, in sci.math
===
Subject: Maths self-study
Hello all,
This is my Þrst posting here, so I hope I am not stepping on
any toes
here. I don¹t know whether this is the correct place to ask
this
question, but I would like know what you guys think are the
best maths
books for self-study? I am trying to teach myself mathematical
analysis, modern and linear algebra. I have a background in
applied
physics, but the course did not give me a sufÞcient
mathematical
foundation for graduate work. I think a lot of people here are
graduate students, or even PhDs and reseachers, and your
experiences
===
Subject: Re: Maths self-study
Hello Zhong Yuan,
Here is one online book on abstract algebra from a more modern
viewpoint: http://www.math.miami.edu/~ec/book/
Are you interested in studying category theory? It is a great
uniÞer
of diverse areas e.g. topology, algebra, etc PLUS it is a
candidate
as a foundation for mathemtaics. If so, I can look for some
tutorials
for you.
> Hello all,
> This is my Þrst posting here, so I hope I am not stepping
on any toes
> here. I don¹t know whether this is the correct place to ask
this
> question, but I would like know what you guys think are the
best maths
> books for self-study? I am trying to teach myself
mathematical
> analysis, modern and linear algebra. I have a background in
applied
> physics, but the course did not give me a sufÞcient
mathematical
> foundation for graduate work. I think a lot of people here
are
> graduate students, or even PhDs and reseachers, and your
experiences
===
Subject: Re: Uri Geller & Jack Sarfatti, Zero Point Energy in
London
>> Normally Geller is doing the hanging on.
>>I would agree. Any enterprise into which Geller is welcomed
is
>instantly suspect. Although, he did teach me how to bend
spoons
>and stuff, though it seems childish, now.
>>John Vreeland
>> So how does he do it John?
>> Please tell.
>> It¹s simple slight-of-hand and misdirection. Any good
magician can do
>> it. It really helps to get to the cutlery before the event
and þex
>> the spoons to fatigue and weaken the metal.
>> Most people don¹t know how easy it is to bend spoons and
keys, because
>> they¹ve never tried it - usually, you avoid bending your
keys and
>> cutlery! Bent keys don¹t work well, and it¹s hard to eat
with a bent
>> spoon.
>> CM
>You¹ll notice also that spoons are easier to bend by hand
than knives and
>forks, with no tines or sharp edges to worry about. This
should not be an
>issue for a real spoonbender.
Ahm, so how does he do it john?
===
Subject: Re: Uri Geller & Jack Sarfatti, Zero Point Energy in
London
> Normally Geller is doing the hanging on.
>>I would agree. Any enterprise into which Geller is welcomed
is
>instantly suspect. Although, he did teach me how to bend
spoons
>and stuff, though it seems childish, now.
>>John Vreeland
>> So how does he do it John?
>> Please tell.
>> It¹s simple slight-of-hand and misdirection. Any good
magician can do
>> it. It really helps to get to the cutlery before the event
and
þex
>> the spoons to fatigue and weaken the metal.
>> Most people don¹t know how easy it is to bend spoons and
keys, because
>> they¹ve never tried it - usually, you avoid bending your
keys and
>> cutlery! Bent keys don¹t work well, and it¹s hard to eat
with a bent
>> spoon.
>> CM
>
You¹ll notice also that spoons are easier to bend by hand
than knives
and
>forks, with no tines or sharp edges to worry about. This
should not be
an
>issue for a real spoonbender.

> Ahm, so how does he do it john?
Wrong John... but I agree with the other one.
===
Subject: Re: Legal fallacies ...
> It depends on the size of the sample space. If all 500,000
people are
> suspects, then the 1-in-500 argument applies. But if the
list of
suspects
> is really only 100, then the probability is much higher.
Model the
number
> of suspects with the rare blood type as a Poisson variable
(why would
this
> be a goo approximation?), remove the probbility for a zero
outcome, and
> normalize the remaining probability. Chances are, so to
speak, that
there
> is only one suspect with the blood type, and we¹ve got our
man.
Not necessarily. It depends on how the suspects were picked
in the Þrst
place. If they found, through traditional detective work, 100
people with
motive, opportunity, who were seen in the area of the crime,
etc., and then
they Þnd of those exactly 1 matches the blood evidence, then
you are
right.
If, however, they happened to have a database of blood types
of 100 people
that they have no reason to believe are connected to the
crime, and decide
to check the blood type from the crime scene against those,
and get exactly
1 match, then that¹s not very convincing at all.
--
--Tim Smith
===
Subject: I don¹t understand this....A User¹s Guide for
Measure Theoretic
Probability
I encountered the example which it is hard to follow, can
someone explain
this?
Example. Suppose mu is a Þnite measure and f is a
non-negative measureble
function.
Then int f d(mu) < infty if and only if Sum(n=1...infty)
mu{f>=n}
and it goes;
The assertion is just a pointwise inequality in disguise. By
considering
separately values for which k =< f(x) =< k+1 for
k=0,1,2...you can
verify the pointwise inequality between functions
Sum(n=1...infty){f>=n} =< f =< 1 + Sum(n=1...infty){f>=n}
I can¹t understand how this equation comes from.
I can¹t keep up with bare expression you can verify...
I would be appreciated if someone explains this to me.
===
Subject: Re: I don¹t understand this....A User¹s Guide for
Measure Theoretic
Probability
>I encountered the example which it is hard to follow, can
someone explain
this?
>Example. Suppose mu is a Þnite measure and f is a
non-negative measureble
>function.
>Then int f d(mu) < infty if and only if Sum(n=1...infty)
mu{f>=n}
Presumably that was supposed to be
int f d(mu) < infty if and only if
Sum(n=1...infty) mu{f>=n} < infty.
>and it goes;
>The assertion is just a pointwise inequality in disguise. By
considering
>separately values for which k =< f(x) =< k+1 for
k=0,1,2...you can
>verify the pointwise inequality between functions
>Sum(n=1...infty){f>=n} =< f =< 1 + Sum(n=1...infty){f>=n}
>I can¹t understand how this equation comes from.
>I can¹t keep up with bare expression you can verify...
>I would be appreciated if someone explains this to me.
The sets k =< f(x) < k+1 are disjoint and their union is
the whole space, right?
Suppose that k =< f(x) < k+1. What is Sum(n=1...infty){f>=n} ?
(If k =< f(x) < k+1 then how many n satisfy f(x) >= n?)
************************
David C. Ullrich
===
Subject: Re: is this a valid proof?
>
>PROVE there is no computable function that enumerates all
>the computable real numbers between 0 and 1 (as decimal
fractions).
>>assume there is a
>TM F such that F(i,j) always halts with the j¹th decimal
>digit of the i¹th computable real number
>>I.e., PROVE Ai Ej F(i,j) /= G(j) for some G
>let G(j) = 5 if F(j,j) = 4, and = 4 otherwise
>Therefore Ai, Ej F(i,j) /= G(j)
>>Why wouldn¹t it be?
>because Ei, F(i, 1) = G(1)
>Ei, F(i,2) = G(2) & F(i, 1) = G(1)
>Ei, F(i, 3) = G(3) & F(i,2) = G(2) & F(i, 1) = G(1)
>FORALL J
>Ei, F(i, j) = G(j) & F(i, j-1) = G(j-1) ... & F(i, 1) = G(1)
>i.e. No matter how many digits you specify, that number is
contained in
F.
> -- ------ --- ---- ----- --- ----- ---- ----- --
------- -- -
>>That¹s no reason why the proof is invalid.
>This means G is in F. Its just more grounded because it
analyses the
>numbers in a countable way, it doesn¹t rely on Ej with j
ranging to oo.
>Because of the cyclic defn of G speciÞcally constructed to
rebuke
computable
>numbers, you cannot specify a J, which means you cannot
specify a number
>not on the list.
>>Yes, you can, for each i you can specify a j such that
F(i,j)!=G(j),
>>which means that the number whose j-th digit is G(j) is a
number not
>>on the list.

> That doesn¹t make sense.
Tell me the number on the list and I¹ll show you a different
number!
You¹re close: Tell me ALL the numbers on the list, and I¹ll
show you a
number not on the list.
Or: Give me a way to compute any number on the list and I¹ll
produce
one your method doesn¹t produce.
All the diag does is this little trick, it HAPPENS to always
show the
same
> number but its still just playing tell me 1st and I¹ll rig
my answer.
Basically. For any list you make, I¹ll make something you
missed.

> For each J, *any number* up to J digits you can specify,
that number
> is on the list. Examine this statement, it literally means
on the list
does it not?
Except it¹s not true for all functions. Given the Þrst J
digits of J
numbers, there is SOMETHING with J digits not on your list.
You are not
comparing all possible J digit strings, just some of them.
To PROVE the number is not on the list, you must not be able
to show that
> from that number it can be found. The operative part is
*any number*.
You are given a particular list. Therefor, I can construct a
particular
omission. Given a different list, I will construct a
different omission.
How many digits of the diagonal must you show before it is
shown to be a
unique
> number?
not 1, not 2,
> You HAVE to show inÞnite digits!
All I need is a way to make it different from each number on
the list.
You are confusing an inÞnite digit number, with an impossible
process.
What is the 1st digit position on the Diag to not match a
number on
computables list?
> Its not 1 0.0 0.1 0.2 0.3... are all present
> Its not 2 0.00 0.01 0.02 ... 0.99 are all present
> Its not 3 a few thousand computable numbers will cover
every digit
possibility of
> the 1st 3 digits
> Its not 4, 5, 6
>
Here is where you make your mistake. You are comparing the
Þrst digit
of diag to ten numbers. This is not what is done in the
construction.
Diag is constructed by comparing the Þrst digit to *one*
number. The
second digit to a different number, etc.
> There is no incompatible digit for n, n e N.
Your statements above having nothing to do with the argument
being
presented.
Therefore the diag failed to specify a number not on the list.
Perhaps you are a constructivist trying to tear down what you
view as
non-constructive mathematics? Are you working from different
axioms
than the rest of us? Or do you simply not understand how the
argument
works (as indicated by your misrepresentation of the
diagonalization
method)?
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: is this a valid proof?
>
>PROVE there is no computable function that enumerates all
>the computable real numbers between 0 and 1 (as decimal
fractions).
>>assume there is a
>TM F such that F(i,j) always halts with the j¹th decimal
>digit of the i¹th computable real number
>>I.e., PROVE Ai Ej F(i,j) /= G(j) for some G
>let G(j) = 5 if F(j,j) = 4, and = 4 otherwise
>Therefore Ai, Ej F(i,j) /= G(j)
>>Why wouldn¹t it be?
>>because Ei, F(i, 1) = G(1)
>>Ei, F(i,2) = G(2) & F(i, 1) = G(1)
>>Ei, F(i, 3) = G(3) & F(i,2) = G(2) & F(i, 1) = G(1)
>>FORALL J
>>Ei, F(i, j) = G(j) & F(i, j-1) = G(j-1) ... & F(i, 1) = G(1)
>>i.e. No matter how many digits you specify, that number is
contained in
F.
> -- ------ --- ---- ----- --- ----- ---- ----- --
------- -- -
>That¹s no reason why the proof is invalid.
>It¹s his probabilistic argument. It¹s a strange argument,
somewhat
>akin to proving that one can¹t throw an inÞnite number of
heads
>in a row with a fair coin.
>I don¹t think it¹s valid but can¹t prove it either way.
>>What can¹t you prove either way? We have a valid proof that
the number
>>whose j-th digit is G(j) is not on the list.

I.e., PROVE Ai Ej F(i,j) /= G(j) for some G
But this condition holds for this also :
F
> 0.99999999
> 0.99999999
> 0.99999999
> 0.99999999
> 0.99999999
G = 1.00000000
It proves Ai Ej F(i,j) /= G(j) for some G
but its obviously wrong because that G is on the list.
Herc
Which is why G(j) =4 or 5 in the original construction.
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: is this a valid proof?
>
>PROVE there is no computable function that enumerates all
>the computable real numbers between 0 and 1 (as decimal
fractions).
>assume there is a
>TM F such that F(i,j) always halts with the j¹th decimal
>digit of the i¹th computable real number
>I.e., PROVE Ai Ej F(i,j) /= G(j) for some G
>let G(j) = 5 if F(j,j) = 4, and = 4 otherwise
>Therefore Ai, Ej F(i,j) /= G(j)
>>Why wouldn¹t it be?

> because Ei, F(i, 1) = G(1)
This need not be true. It is possible that Ai F(i,1)=3. Then
G(1)=4.
You cannot assume anything about what F produces. F may be
the function
F(i,j)= (j mod 9) +1 if j<i, 0 if j>=i.
Ei, F(i,2) = G(2) & F(i, 1) = G(1)
Ei, F(i, 3) = G(3) & F(i,2) = G(2) & F(i, 1) = G(1)
FORALL J
Ei, F(i, j) = G(j) & F(i, j-1) = G(j-1) ... & F(i, 1) = G(1)
i.e. No matter how many digits you specify, that number is
contained in
F.
> -- ------ --- ---- ----- --- ----- ---- ----- --
------- -- -
>
Only for some F, not for ALL F.
This means G is in F. Its just more grounded because it
analyses the
> numbers in a countable way, it doesn¹t rely on Ej with j
ranging to oo.
Because of the cyclic defn of G speciÞcally constructed to
rebuke
computable
> numbers, you cannot specify a J, which means you cannot
specify a number
> not on the list.
Herc
This argument is invalid because you are imposing properties
on F that
were not given.
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: is this a valid proof?
>PROVE there is no computable function that enumerates all
>the computable real numbers between 0 and 1 (as decimal
fractions).
>>assume there is a
>TM F such that F(i,j) always halts with the j¹th decimal
>digit of the i¹th computable real number
>>I.e., PROVE Ai Ej F(i,j) /= G(j) for some G
>let G(j) = 5 if F(j,j) = 4, and = 4 otherwise
>Therefore Ai, Ej F(i,j) /= G(j)
>>Why wouldn¹t it be?

> because Ei, F(i, 1) = G(1)
> This need not be true. It is possible that Ai F(i,1)=3.
Then G(1)=4.
> You cannot assume anything about what F produces. F may be
the function
> F(i,j)= (j mod 9) +1 if j<i, 0 if j>=i.
Ei, F(i,2) = G(2) & F(i, 1) = G(1)
Ei, F(i, 3) = G(3) & F(i,2) = G(2) & F(i, 1) = G(1)
FORALL J
Ei, F(i, j) = G(j) & F(i, j-1) = G(j-1) ... & F(i, 1) = G(1)
i.e. No matter how many digits you specify, that number is
contained in
F.
> -- ------ --- ---- ----- --- ----- ---- ----- --
------- -- -
Only for some F, not for ALL F.
This means G is in F. Its just more grounded because it
analyses the
> numbers in a countable way, it doesn¹t rely on Ej with j
ranging to oo.
Because of the cyclic defn of G speciÞcally constructed to
rebuke
computable
> numbers, you cannot specify a J, which means you cannot
specify a
number
> not on the list.
Herc
> This argument is invalid because you are imposing
properties on F that
> were not given.
F is all computable numbers, that¹s always been my argument.
UTM(1)
UTM(2)
UTM(3)
...
there¹s my list, tell me a sequence it doesn¹t cover fool.
Herc
===
Subject: Re: is this a valid proof?
> PROVE there is no computable function that enumerates all
the computable
real
> numbers between 0 and 1 (as decimal fractions).
assume there is a
> TM F such that F(i,j) always halts with the j¹th decimal
digit of the
i¹th
> computable real number
I.e., PROVE Ai Ej F(i,j) /= G(j) for some G
> let G(j) = 5 if F(j,j) = 4, and = 4 otherwise
> Therefore Ai, Ej F(i,j) /= G(j)
>
F is computable (by assumption) and gives an enumeration of
some real
numbers between 0 and 1. G is deÞned by F, so it is also
computable
and is deÞned in such a way that it enumerates a number
between 0 and
1. G enumerates a number that F does not, so F does not
enumerate all
computable real numbers between 0 and 1.
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: is this a valid proof?
In sci.logic, Rupert
<rupertmccallum@yahoo.com> In sci.logic, Rupert
>> <rupertmccallum@yahoo.com>> > PROVE there is no computable
function that enumerates all
> > the computable real numbers between 0 and 1 (as decimal
fractions).
> >> > assume there is a
> > TM F such that F(i,j) always halts with the j¹th decimal
> > digit of the i¹th computable real number
> >> > I.e., PROVE Ai Ej F(i,j) /= G(j) for some G
> > let G(j) = 5 if F(j,j) = 4, and = 4 otherwise
> > Therefore Ai, Ej F(i,j) /= G(j)
>> Why wouldn¹t it be?
>
> because Ei, F(i, 1) = G(1)
>
> Ei, F(i,2) = G(2) & F(i, 1) = G(1)
>
> Ei, F(i, 3) = G(3) & F(i,2) = G(2) & F(i, 1) = G(1)
>
> FORALL J
>
> Ei, F(i, j) = G(j) & F(i, j-1) = G(j-1) ... & F(i, 1) = G(1)
>
> i.e. No matter how many digits you specify, that number is
contained
in F.
> -- ------ --- ---- ----- --- ----- ---- ----- --
------- -- -
>
>> That¹s no reason why the proof is invalid.
>>
>> It¹s his probabilistic argument. It¹s a strange argument,
somewhat
>> akin to proving that one can¹t throw an inÞnite number of
heads
>> in a row with a fair coin.
>>
>> I don¹t think it¹s valid but can¹t prove it either way.
>What can¹t you prove either way? We have a valid proof that
the number
> whose j-th digit is G(j) is not on the list.
Because it is a probabilistic argument and works for all
integers j.
It does *not* work for a constructible inÞnite b, given a
constructible inÞnite list c of constructible numbers.
However,
that¹s a different problem, and b can be approximated to any
length desired -- and that particular variant of b is
somewhere
on his list. (Presumably.)
At least, such is my understanding of his exposition of the
problem.
It¹s quite ridiculous, in some respects -- akin to a proof
that
each and every Þnite decimal expansion is a rational number
(which,
as it turns out, it is -- but those aren¹t the interesting
ones).
In short, it¹s dead wrong but how to show it in a fashion that
makes sense to those of us (who are most of us, methinks )
who can¹t write down an inÞnite decimal sequence? :-)
I could attempt more rigor, something along the lines
of the following.
Assume a function F, domain N x N, range
{0,1,2,3,4,5,6,7,8,9}.
Construct a function G, domain N, range {0,1,2,3,4,5,6,7,8,9},
such that, if F(i,i) = Œ4¹ then G(i) = Œ5¹ else G(i) = Œ4¹.
Does there exist an i in N, such that, for *every* j in N,
G(j) = F(i,j) ?
By construction, no; take j = i and kaboom.
Is it the case that, for every j in N, there exists an i such
that G(j) = F(i,j)?
Depends, but probably yes, if F() is sufÞciently interesting.
A trivial example would be F(i,j) = the j¹th digit of the
fraction 123456789 / 10^(i + 8). Since F(i,i) is always 1,
G(i) = Œ4¹, G represents 4/9, and F(i,i+3) = G(i) for every
positive integer i.
(The description of sufÞciently interesting I¹ll have
to leave to the reader. It is a sufÞcient but not
necessary condition that, while enumerating F(i,j) over
j, that the digits 0 through 9 appear at least once.)
This is not a useful result, obviously, except to show that
generalization is non-commutative. :-)
If F represents a list of all computable numbers (where
i is a program description on, say, a modiÞed Turing
machine which can only move forward, never backward, on
its primary output tape (but it also has a scratch tape)
and must write a digit before moving) and j is the j¹th
cell on the tape), then clearly G is a computable function
(using a universal machine) if the machines represented by
F all (eventually) halt when computing any desired digit --
and we have a contradiction.
If F represents a list of all real numbers, then clearly
G is a real number not equal to any of them -- and we have
another contradiction.
That¹s probably as good a proof as any, given my sluggardiness
(it¹s 11:30 PM over here and I should be sleeping).
>
> This means G is in F. Its just more grounded because it
analyses the
> numbers in a countable way, it doesn¹t rely on Ej with j
ranging to
oo.
>
> Because of the cyclic defn of G speciÞcally constructed to
> rebuke computable numbers, you cannot specify a J, which
means
> you cannot specify a number not on the list.
>
>> Yes, you can, for each i you can specify a j such that
F(i,j)!=G(j),
>> which means that the number whose j-th digit is G(j) is a
number not
>> on the list.
>>
>> For j=1, about 1/10 of F(i,j) are such that F(i,1) = G(1);
>> for j=2, about 1/100 of F(i,j) are such that F(i,1) = G(1)
>> and F(i,2) = G(2);
>> for j=3, about 1/1000 of F(i,j) are such that F(i,1) = G(1)
>> and F(i,2) = G(2) and F(i,3) = G(3);
>> etc.
>>
>> This clearly only works for Þnite j.
>>
> Herc
--
#191, ewill3@earthlink.net
It¹s still legal to go .sigless.
===
Subject: Re: is this a valid proof?
> > PROVE there is no computable function that enumerates all
> > the computable real numbers between 0 and 1 (as decimal
fractions).
> >> > assume there is a
> > TM F such that F(i,j) always halts with the j¹th decimal
> > digit of the i¹th computable real number
> >> > I.e., PROVE Ai Ej F(i,j) /= G(j) for some G
> > let G(j) = 5 if F(j,j) = 4, and = 4 otherwise
> > Therefore Ai, Ej F(i,j) /= G(j)
>> Why wouldn¹t it be?
>> because Ei, F(i, 1) = G(1)
>> Ei, F(i,2) = G(2) & F(i, 1) = G(1)
>> Ei, F(i, 3) = G(3) & F(i,2) = G(2) & F(i, 1) = G(1)
>> FORALL J
>> Ei, F(i, j) = G(j) & F(i, j-1) = G(j-1) ... & F(i, 1) =
G(1)
>> i.e. No matter how many digits you specify, that number is
contained
in F.
> -- ------ --- ---- ----- --- ----- ---- -----
-- ------- -- -
> That¹s no reason why the proof is invalid.
>> It¹s his probabilistic argument. It¹s a strange argument,
somewhat
>> akin to proving that one can¹t throw an inÞnite number of
heads
>> in a row with a fair coin.
>> I don¹t think it¹s valid but can¹t prove it either way.
>
> What can¹t you prove either way? We have a valid proof that
the number
> whose j-th digit is G(j) is not on the list.
> Because it is a probabilistic argument and works for all
integers j.
> It does *not* work for a constructible inÞnite b, given a
> constructible inÞnite list c of constructible numbers.
However,
> that¹s a different problem, and b can be approximated to any
> length desired -- and that particular variant of b is
somewhere
> on his list. (Presumably.)
> At least, such is my understanding of his exposition of the
problem.
> It¹s quite ridiculous, in some respects -- akin to a proof
that
> each and every Þnite decimal expansion is a rational number
(which,
> as it turns out, it is -- but those aren¹t the interesting
ones).
> In short, it¹s dead wrong but how to show it in a fashion
that
> makes sense to those of us (who are most of us, methinks )
> who can¹t write down an inÞnite decimal sequence? :-)
> I could attempt more rigor, something along the lines
> of the following.
> Assume a function F, domain N x N, range
{0,1,2,3,4,5,6,7,8,9}.
> Construct a function G, domain N, range
{0,1,2,3,4,5,6,7,8,9},
> such that, if F(i,i) = Œ4¹ then G(i) = Œ5¹ else G(i) = Œ4¹.
> Does there exist an i in N, such that, for *every* j in N,
> G(j) = F(i,j) ?
> By construction, no; take j = i and kaboom.
> Is it the case that, for every j in N, there exists an i
such
> that G(j) = F(i,j)?
> Depends, but probably yes, if F() is sufÞciently
interesting.
> A trivial example would be F(i,j) = the j¹th digit of the
> fraction 123456789 / 10^(i + 8). Since F(i,i) is always 1,
> G(i) = Œ4¹, G represents 4/9, and F(i,i+3) = G(i) for every
> positive integer i.
> (The description of sufÞciently interesting I¹ll have
> to leave to the reader. It is a sufÞcient but not
> necessary condition that, while enumerating F(i,j) over
> j, that the digits 0 through 9 appear at least once.)
> This is not a useful result, obviously, except to show that
> generalization is non-commutative. :-)
> If F represents a list of all computable numbers (where
> i is a program description on, say, a modiÞed Turing
> machine which can only move forward, never backward, on
> its primary output tape (but it also has a scratch tape)
> and must write a digit before moving) and j is the j¹th
> cell on the tape), then clearly G is a computable function
> (using a universal machine) if the machines represented by
> F all (eventually) halt when computing any desired digit --
> and we have a contradiction.
> If F represents a list of all real numbers, then clearly
> G is a real number not equal to any of them -- and we have
> another contradiction.
> That¹s probably as good a proof as any, given my
sluggardiness
> (it¹s 11:30 PM over here and I should be sleeping).
> This means G is in F. Its just more grounded because it
analyses
the
> numbers in a countable way, it doesn¹t rely on Ej with j
ranging to
oo.
>> Because of the cyclic defn of G speciÞcally constructed to
> rebuke computable numbers, you cannot specify a J, which
means
> you cannot specify a number not on the list.
> Yes, you can, for each i you can specify a j such that
F(i,j)!=G(j),
>> which means that the number whose j-th digit is G(j) is a
number not
>> on the list.
>> For j=1, about 1/10 of F(i,j) are such that F(i,1) = G(1);
>> for j=2, about 1/100 of F(i,j) are such that F(i,1) = G(1)
>> and F(i,2) = G(2);
>> for j=3, about 1/1000 of F(i,j) are such that F(i,1) = G(1)
>> and F(i,2) = G(2) and F(i,3) = G(3);
>> etc.
>> This clearly only works for Þnite j.
>>
Good reply. I thought of the non reversible tape design too.
I think of the problem this way. When you lay down the 1st
digit of
the 1st computable number, anti_diag has to provide a counter
digit.
It doesn¹t matter what digit anti_diag is, because there¹s
atleast 10 other
numbers that are computable covering it. Go to the second
digit on the
second number, anti_diag has to counter this digit too, but
it doesn¹t
matter,
computables easily contains 100s of numbers, over 10 of which
matched
the 1st digit of anti_diag, and from those the computables
can easily COVER
the 1st couple digits of anti_diag.
So far anti_diag is no match for the list of computable
numbers, and it
never will be
and it never was.
Herc
===
Subject: Re: is this a valid proof?
> Good reply. I thought of the non reversible tape design too.
I think of the problem this way. When you lay down the 1st
digit of
> the 1st computable number, anti_diag has to provide a
counter digit.
It doesn¹t matter what digit anti_diag is, because there¹s
atleast 10
other
> numbers that are computable covering it. Go to the second
digit on the
> second number, anti_diag has to counter this digit too, but
it doesn¹t
matter,
> computables easily contains 100s of numbers, over 10 of
which matched
> the 1st digit of anti_diag, and from those the computables
can easily
COVER
> the 1st couple digits of anti_diag.
So far anti_diag is no match for the list of computable
numbers, and it
never will be
> and it never was.
>
Let¹s see if I can rephrase this correctly:
You have the computable numbers, and you are assuming they
are ALL on a
list. We now enumerate anti-diag in order, noting at each
stage of
enumeration that there is a computable number that it could
crank out to
be. Therefore, it¹s on the list.
Is this a correct summary of your argument?
Here¹s my take on why the above is wrong:
Let¹s let A be the anti-diag and A(j) be the j¹th digit of
anti-diag.
L(i) is the i¹th number on the list, L(i,j) is the j¹th digit
of the
i¹th number. Your assumption is that A is on the list. Fine,
where?
Before we start enumerating it, it *could* be the Þrst
number. Ok,
let¹s check. A(1) /= L(1,1), so A wasn¹t Þrst. Ah ha! A could
be
some other number L(n1). Fine, let¹s check that: A(n1) /=
L(n1,n1).
Woops! That wasn¹t it. Well, if we enumerate A out to A(n1)
we have
something that matches some other number¹s Þrst n1 digits,
say L(n2).
Let¹s check it. A(n2) /= L(n2,n2). Repeat ad nauseum.
Every time you *think* you¹ve found where A is on the list,
you Þnd
that the n¹th digit doesn¹t match. So where is it on the
list? You
said it was there, but we can¹t seem to Þnd it. By induction,
in fact,
you can easily show that for arbitrary N, A is not one of the
Þrst N
numbers on the list, so A is *NOT* on the list.
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: is this a valid proof?
oo
____|mn
/ /_/ / _
/ K-9/ /_/ - www.YeOldeCoffeeShoppe.com -
/____/_____
--------------
> Good reply. I thought of the non reversible tape design too.
I think of the problem this way. When you lay down the 1st
digit of
> the 1st computable number, anti_diag has to provide a
counter digit.
It doesn¹t matter what digit anti_diag is, because there¹s
atleast 10
other
> numbers that are computable covering it. Go to the second
digit on the
> second number, anti_diag has to counter this digit too, but
it doesn¹t
matter,
> computables easily contains 100s of numbers, over 10 of
which matched
> the 1st digit of anti_diag, and from those the computables
can easily
COVER
> the 1st couple digits of anti_diag.
So far anti_diag is no match for the list of computable
numbers, and it
never will be
> and it never was.
Let¹s see if I can rephrase this correctly:
> You have the computable numbers, and you are assuming they
are ALL on a
> list. We now enumerate anti-diag in order, noting at each
stage of
> enumeration that there is a computable number that it could
crank out to
> be. Therefore, it¹s on the list.
> Is this a correct summary of your argument?
> Here¹s my take on why the above is wrong:
> Let¹s let A be the anti-diag and A(j) be the j¹th digit of
anti-diag.
> L(i) is the i¹th number on the list, L(i,j) is the j¹th
digit of the
> i¹th number. Your assumption is that A is on the list.
Fine, where?
> Before we start enumerating it, it *could* be the Þrst
number. Ok,
> let¹s check. A(1) /= L(1,1), so A wasn¹t Þrst. Ah ha! A
could be
> some other number L(n1). Fine, let¹s check that: A(n1) /=
L(n1,n1).
> Woops! That wasn¹t it. Well, if we enumerate A out to A(n1)
we have
> something that matches some other number¹s Þrst n1 digits,
say L(n2).
> Let¹s check it. A(n2) /= L(n2,n2). Repeat ad nauseum.
> Every time you *think* you¹ve found where A is on the list,
you Þnd
> that the n¹th digit doesn¹t match. So where is it on the
list? You
> said it was there, but we can¹t seem to Þnd it. By
induction, in fact,
> you can easily show that for arbitrary N, A is not one of
the Þrst N
> numbers on the list, so A is *NOT* on the list.
They completely contradict each other. If you can¹t specify a
Þnite
contradiction
to the list you can¹t then go ahead and specify an inÞnite
one.
Your argument is bogus. You want me to completely demonstrate
the list,
and you don¹t even have to show your number at all. You CAN¹T
show
any number. Show me the 1st million digits of anti_diag,
that¹s on the
list.
By induction you can show inÞnite number of digits of
anti_diag and they
are all covered too. The anti_diag IS present to UNLIMITED
number of
digits, don¹t blame me for not giving an index to it when you
can¹t even
tell me what number you have.
Herc
===
Subject: Re: is this a valid proof?
> >> > PROVE there is no computable function that enumerates
all
> >> > the computable real numbers between 0 and 1 (as decimal
fractions).
> >> > assume there is a
> >> > TM F such that F(i,j) always halts with the j¹th
decimal
> >> > digit of the i¹th computable real number
> >> > I.e., PROVE Ai Ej F(i,j) /= G(j) for some G
> >> > let G(j) = 5 if F(j,j) = 4, and = 4 otherwise
> >> > Therefore Ai, Ej F(i,j) /= G(j)
> > >> Why wouldn¹t it be?
> > >> because Ei, F(i, 1) = G(1)
> > >> Ei, F(i,2) = G(2) & F(i, 1) = G(1)
> > >> Ei, F(i, 3) = G(3) & F(i,2) = G(2) & F(i, 1) = G(1)
> > >> FORALL J
> > >> Ei, F(i, j) = G(j) & F(i, j-1) = G(j-1) ... & F(i, 1)
= G(1)
> > >> i.e. No matter how many digits you specify, that
number is
contained in F.
> >> -- ------ --- ---- ----- --- ----- ---- -----
-- ------- -- -
> >>
That¹s no reason why the proof is invalid.
It¹s his probabilistic argument. It¹s a strange argument,
somewhat
> > akin to proving that one can¹t throw an inÞnite number of
heads
> > in a row with a fair coin.
I don¹t think it¹s valid but can¹t prove it either way.
>
> What can¹t you prove either way? We have a valid proof that
the number
> whose j-th digit is G(j) is not on the list.
>

> I.e., PROVE Ai Ej F(i,j) /= G(j) for some G
But this condition holds for this also :
F
> 0.99999999
> 0.99999999
> 0.99999999
> 0.99999999
> 0.99999999
G = 1.00000000
It proves Ai Ej F(i,j) /= G(j) for some G
but its obviously wrong because that G is on the list.
>
Here we have a situation where Ai Ej F(i,j) != G(j), but the
number
whose j-th digit is G(j) is the same as one of the numbers on
the
list, even though it has a different decimal expansion.
You need to take into account the fact that a number with its
decimal
expansion ending in inÞnitely many 9s can be equal to number
with its
decimal expansion ending in inÞnitely many 0s. You did this by
saying, let G(j)=4 if F(j,j) does not equal 4, 5 otherwise.
This
guarantees that not only will the number whose j-th digit is
G(j) have
a different decimal expansion to every number on the list,
but it will
actually be a different number.
<snip>
===
Subject: Re: is this a valid proof?
> > > PROVE there is no computable function that enumerates
all
> > > the computable real numbers between 0 and 1 (as decimal
fractions).
> > assume there is a
> > > TM F such that F(i,j) always halts with the j¹th decimal
> > > digit of the i¹th computable real number
> > I.e., PROVE Ai Ej F(i,j) /= G(j) for some G
> > > let G(j) = 5 if F(j,j) = 4, and = 4 otherwise
> > > Therefore Ai, Ej F(i,j) /= G(j)
Why wouldn¹t it be?
because Ei, F(i, 1) = G(1)
Ei, F(i,2) = G(2) & F(i, 1) = G(1)
Ei, F(i, 3) = G(3) & F(i,2) = G(2) & F(i, 1) = G(1)
FORALL J
Ei, F(i, j) = G(j) & F(i, j-1) = G(j-1) ... & F(i, 1) = G(1)
i.e. No matter how many digits you specify, that number is
contained
in F.
> > -- ------ --- ---- ----- --- ----- ---- ----- --
------- -- -
>
> That¹s no reason why the proof is invalid.
>
This means G is in F. Its just more grounded because it
analyses the
> > numbers in a countable way, it doesn¹t rely on Ej with j
ranging to
oo.
Because of the cyclic defn of G speciÞcally constructed to
rebuke
computable
> > numbers, you cannot specify a J, which means you cannot
specify a
number
> > not on the list.
>
> Yes, you can, for each i you can specify a j such that
F(i,j)!=G(j),
> which means that the number whose j-th digit is G(j) is a
number not
> on the list.
>
That doesn¹t make sense.
>
It makes perfect sense.
> Tell me the number on the list and I¹ll show you a
different number!
>
No, give me a computable list of computable numbers and I¹ll
give you
a computable number not on the list.
> All the diag does is this little trick, it HAPPENS to
always show the
same
> number but its still just playing tell me 1st and I¹ll rig
my answer.

Given any computable list of computable numbers, we can
construct the
diagonal number, which will be a computable number not on the
list.
That shows that no computable list can contain all the
computable
numbers.
What¹s rigged about that?
> For each J, *any number* up to J digits you can specify,
that number
> is on the list. Examine this statement, it literally means
on the list
does it not?
>
For each j, given j digits, there will be a number which
starts with
those j digits. But that¹s irrelevant. There is a computable
number
not on the list.
> To PROVE the number is not on the list, you must not be
able to show that
> from that number it can be found. The operative part is
*any number*.
>
To prove the number is not on the list, you must show that
for each
number on the list there¹s a digit where it doesn¹t agree
with that
number. That¹s easy to do.
> How many digits of the diagonal must you show before it is
shown to be a
unique
> number?
>
This question doesn¹t mean very much.
To specify the diagonal number fully, you need inÞnitely many
digits.
> not 1, not 2,
> You HAVE to show inÞnite digits!
You are confusing an inÞnite digit number, with an impossible
process.
What is the 1st digit position on the Diag to not match a
number on
computables list?
For any number on the list, there¹s a digit of the diagonal
which
doesn¹t match the corresponding digit of the number.
> Its not 1 0.0 0.1 0.2 0.3... are all present
> Its not 2 0.00 0.01 0.02 ... 0.99 are all present
> Its not 3 a few thousand computable numbers will cover
every digit
possibility of
> the 1st 3 digits
> Its not 4, 5, 6
There is no incompatible digit for n, n e N.
Therefore the diag failed to specify a number not on the list.
>
For any Þnite segment of the diagonal, there¹s a number on
the list
which agrees with the diagonal on that segment.
But there¹s no number on the list that agrees with the
diagonal to
inÞnitely many digits.
> Herc
===
Subject: Re: is this a valid proof?
> > > PROVE there is no computable function that enumerates
all
> > > the computable real numbers between 0 and 1 (as decimal
fractions).
> > assume there is a
> > > TM F such that F(i,j) always halts with the j¹th decimal
> > > digit of the i¹th computable real number
> > I.e., PROVE Ai Ej F(i,j) /= G(j) for some G
> > > let G(j) = 5 if F(j,j) = 4, and = 4 otherwise
> > > Therefore Ai, Ej F(i,j) /= G(j)
> > Why wouldn¹t it be?
because Ei, F(i, 1) = G(1)
Ei, F(i,2) = G(2) & F(i, 1) = G(1)
Ei, F(i, 3) = G(3) & F(i,2) = G(2) & F(i, 1) = G(1)
FORALL J
Ei, F(i, j) = G(j) & F(i, j-1) = G(j-1) ... & F(i, 1) = G(1)
i.e. No matter how many digits you specify, that number is
contained
in F.
> > -- ------ --- ---- ----- --- ----- ---- -----
-- ------- -- -

> > That¹s no reason why the proof is invalid.

> > This means G is in F. Its just more grounded because it
analyses
the
> > numbers in a countable way, it doesn¹t rely on Ej with j
ranging to
oo.
Because of the cyclic defn of G speciÞcally constructed to
rebuke
computable
> > numbers, you cannot specify a J, which means you cannot
specify a
number
> > not on the list.

> > Yes, you can, for each i you can specify a j such that
F(i,j)!=G(j),
> > which means that the number whose j-th digit is G(j) is a
number not
> > on the list.
>
> That doesn¹t make sense.
It makes perfect sense.
> Tell me the number on the list and I¹ll show you a different
number!
No, give me a computable list of computable numbers and I¹ll
give you
> a computable number not on the list.
no you can¹t, give == Þnite
> All the diag does is this little trick, it HAPPENS to
always show the
same
> number but its still just playing tell me 1st and I¹ll rig
my answer.

> Given any computable list of computable numbers, we can
construct the
> diagonal number, which will be a computable number not on
the list.
> That shows that no computable list can contain all the
computable
> numbers.
> What¹s rigged about that?
you cannot demonstrate that its not on the list
> For each J, *any number* up to J digits you can specify,
that number
> is on the list. Examine this statement, it literally means
on the list
does it not?
For each j, given j digits, there will be a number which
starts with
> those j digits. But that¹s irrelevant. There is a
computable number
> not on the list.
its not irrelevant, it contradicts your next sentence
> To PROVE the number is not on the list, you must not be
able to show
that
> from that number it can be found. The operative part is
*any number*.
To prove the number is not on the list, you must show that
for each
> number on the list there¹s a digit where it doesn¹t agree
with that
> number. That¹s easy to do.
no its not. only Þnite samples of the list of computables can
ever be
physically
processed, this is pure conceptual argument.
the list of computables can be generated to any length, but
it wont ever
physically
be complete. ERGO given any number it can be found on the
list of
computables.
> How many digits of the diagonal must you show before it is
shown to be a
unique
> number?
This question doesn¹t mean very much.
> To specify the diagonal number fully, you need inÞnitely
many digits.
> not 1, not 2,
> You HAVE to show inÞnite digits!
You are confusing an inÞnite digit number, with an impossible
process.
What is the 1st digit position on the Diag to not match a
number on
computables list?
> For any number on the list, there¹s a digit of the diagonal
which
> doesn¹t match the corresponding digit of the number.
> Its not 1 0.0 0.1 0.2 0.3... are all present
> Its not 2 0.00 0.01 0.02 ... 0.99 are all present
> Its not 3 a few thousand computable numbers will cover
every digit
possibility of
> the 1st 3 digits
> Its not 4, 5, 6
There is no incompatible digit for n, n e N.
Therefore the diag failed to specify a number not on the list.
For any Þnite segment of the diagonal, there¹s a number on
the list
> which agrees with the diagonal on that segment.
> But there¹s no number on the list that agrees with the
diagonal to
> inÞnitely many digits.
no single number perhaps if your ideal conceptualisation is
allowed.
But the diagonal is on the list to 1 place, 2 places, 3
places....
This does not stop, the number of places goes to inÞnity.
Herc
===
Subject: Re: is this a valid proof?
> > > > PROVE there is no computable function that enumerates
all
> > > > the computable real numbers between 0 and 1 (as
decimal
fractions).
> > > > assume there is a
> > > > TM F such that F(i,j) always halts with the j¹th
decimal
> > > > digit of the i¹th computable real number
> > > > I.e., PROVE Ai Ej F(i,j) /= G(j) for some G
> > > > let G(j) = 5 if F(j,j) = 4, and = 4 otherwise
> > > > Therefore Ai, Ej F(i,j) /= G(j)
> > Why wouldn¹t it be?
> > because Ei, F(i, 1) = G(1)
> > Ei, F(i,2) = G(2) & F(i, 1) = G(1)
> > Ei, F(i, 3) = G(3) & F(i,2) = G(2) & F(i, 1) = G(1)
> > FORALL J
> > Ei, F(i, j) = G(j) & F(i, j-1) = G(j-1) ... & F(i, 1) =
G(1)
> > i.e. No matter how many digits you specify, that number is
contained in F.
> > > -- ------ --- ---- ----- --- ----- ---- -----
-- ------- -- -
>
> > That¹s no reason why the proof is invalid.
> > This means G is in F. Its just more grounded because it
analyses
the
> > > numbers in a countable way, it doesn¹t rely on Ej with
j ranging
to oo.
> > Because of the cyclic defn of G speciÞcally constructed
to rebuke
computable
> > > numbers, you cannot specify a J, which means you cannot
specify a
number
> > > not on the list.
>
> > Yes, you can, for each i you can specify a j such that
F(i,j)!=G(j),
> > which means that the number whose j-th digit is G(j) is a
number
not
> > on the list.

> > That doesn¹t make sense.
>
> It makes perfect sense.
> Tell me the number on the list and I¹ll show you a different
number!
>
> No, give me a computable list of computable numbers and
I¹ll give you
> a computable number not on the list.
no you can¹t, give == Þnite
>
Suppose I accept this for the sake of argument. So what?
Computable
numbers can be Þnitely given.
If the list can be given, the diagonal number can be given.
>
> All the diag does is this little trick, it HAPPENS to
always show the
same
> > number but its still just playing tell me 1st and I¹ll
rig my answer.

Given any computable list of computable numbers, we can
construct the
> diagonal number, which will be a computable number not on
the list.
That shows that no computable list can contain all the
computable
> numbers.
What¹s rigged about that?
you cannot demonstrate that its not on the list

Yes I can. Its Þrst digit is not the same as the Þrst digit
of the
Þrst number. So it¹s not the Þrst number. Its 2nd digit is
not the
same as the 2nd digit of the 2nd number. So it¹s not the 2nd
number.
And so on.
> For each J, *any number* up to J digits you can specify,
that number
> > is on the list. Examine this statement, it literally
means on the
list does it not?
>
> For each j, given j digits, there will be a number which
starts with
> those j digits. But that¹s irrelevant. There is a
computable number
> not on the list.
its not irrelevant, it contradicts your next sentence

No, it doesn¹t.
>
> To PROVE the number is not on the list, you must not be
able to show
that
> > from that number it can be found. The operative part is
*any
number*.
>
> To prove the number is not on the list, you must show that
for each
> number on the list there¹s a digit where it doesn¹t agree
with that
> number. That¹s easy to do.
no its not.
Yes it is, I did it above.
> only Þnite samples of the list of computables can ever be
physically
> processed, this is pure conceptual argument.
the list of computables can be generated to any length, but
it wont ever
physically
> be complete. ERGO given any number it can be found on the
list of
computables.
That¹s a ridiculous argument.

> How many digits of the diagonal must you show before it is
shown to be
a unique
> > number?
>
> This question doesn¹t mean very much.
To specify the diagonal number fully, you need inÞnitely many
digits.
> not 1, not 2,
> > You HAVE to show inÞnite digits!
You are confusing an inÞnite digit number, with an impossible
process.
What is the 1st digit position on the Diag to not match a
number on
computables list?
For any number on the list, there¹s a digit of the diagonal
which
> doesn¹t match the corresponding digit of the number.
> Its not 1 0.0 0.1 0.2 0.3... are all present
> > Its not 2 0.00 0.01 0.02 ... 0.99 are all present
> > Its not 3 a few thousand computable numbers will cover
every digit
possibility of
> > the 1st 3 digits
> > Its not 4, 5, 6
There is no incompatible digit for n, n e N.
Therefore the diag failed to specify a number not on the list.
>
> For any Þnite segment of the diagonal, there¹s a number on
the list
> which agrees with the diagonal on that segment.
But there¹s no number on the list that agrees with the
diagonal to
> inÞnitely many digits.

> no single number perhaps if your ideal conceptualisation is
allowed.
> But the diagonal is on the list to 1 place, 2 places, 3
places....
This does not stop, the number of places goes to inÞnity.
Herc
But the diagonal number still isn¹t on the list.
===
Subject: Re: is this a valid proof?
>
>How many digits of the diagonal must you show before it is
shown to be a
unique
>number?
>>This question doesn¹t mean very much.
>>To specify the diagonal number fully, you need inÞnitely
many digits.
>not 1, not 2,
>You HAVE to show inÞnite digits!
>You are confusing an inÞnite digit number, with an
impossible process.
>What is the 1st digit position on the Diag to not match a
number on
computables list?
>>For any number on the list, there¹s a digit of the diagonal
which
>>doesn¹t match the corresponding digit of the number.
>Its not 1 0.0 0.1 0.2 0.3... are all present
>Its not 2 0.00 0.01 0.02 ... 0.99 are all present
>Its not 3 a few thousand computable numbers will cover every
digit
possibility of
> the 1st 3 digits
>Its not 4, 5, 6
>There is no incompatible digit for n, n e N.
>Therefore the diag failed to specify a number not on the
list.
>>For any Þnite segment of the diagonal, there¹s a number on
the list
>>which agrees with the diagonal on that segment.
>>But there¹s no number on the list that agrees with the
diagonal to
>>inÞnitely many digits.
no single number perhaps if your ideal conceptualisation is
allowed.
> But the diagonal is on the list to 1 place, 2 places, 3
places....
This does not stop, the number of places goes to inÞnity.
Herc
Suppose you give me the following list:
.30000000000000...
.33000000000000...
.33300000000000...
.
.
.
I will construct: .4444444444444444...
The FIRST decimal place is different from all your numbers,
therefor it
is not on your list. Your objection is incorrect.
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: is this a valid proof?
In sci.math, |-|erc
<contactvia@wwwadamskingdom.com>
<40690bb9$0$28242$afc38c87@news.optusnet.com.au>:
>> > > PROVE there is no computable function that enumerates
all
>> > > the computable real numbers between 0 and 1 (as decimal
fractions).
>> > > > > assume there is a
>> > > TM F such that F(i,j) always halts with the j¹th
decimal
>> > > digit of the i¹th computable real number
>> > > > > I.e., PROVE Ai Ej F(i,j) /= G(j) for some G
>> > > let G(j) = 5 if F(j,j) = 4, and = 4 otherwise
>> > > Therefore Ai, Ej F(i,j) /= G(j)
>> > > > > Why wouldn¹t it be?
>> > > because Ei, F(i, 1) = G(1)
>> > > Ei, F(i,2) = G(2) & F(i, 1) = G(1)
>> > >> > Ei, F(i, 3) = G(3) & F(i,2) = G(2) & F(i, 1) = G(1)
>> > > FORALL J
>> > > Ei, F(i, j) = G(j) & F(i, j-1) = G(j-1) ... & F(i, 1)
= G(1)
>> > > i.e. No matter how many digits you specify, that
number is
contained in F.
>> > -- ------ --- ---- ----- --- ----- ---- -----
-- ------- -- -
>> > > > That¹s no reason why the proof is invalid.
>> > > > This means G is in F. Its just more grounded because
it
>> > analyses the numbers in a countable way, it doesn¹t rely
>> > on Ej with j ranging to oo.
>> > > Because of the cyclic defn of G speciÞcally
constructed to
>> > rebuke computable numbers, you cannot specify a J, which
>> > means you cannot specify a number not on the list.
>> > > > Yes, you can, for each i you can specify a j such
that F(i,j)!=G(j),
>> > which means that the number whose j-th digit is G(j) is
a number not
>> > on the list.
>> > That doesn¹t make sense.
>> It makes perfect sense.
>> Tell me the number on the list and I¹ll show you a
different
number!
>> No, give me a computable list of computable numbers and
I¹ll give you
>> a computable number not on the list.
> no you can¹t, give == Þnite
If you give him a method by which he can compute to any
desired
precision a list of computable numbers, he¹ll give you another
number that cannot be represented by that particular method.
>> All the diag does is this little trick, it HAPPENS to
always show the
same
>> number but its still just playing tell me 1st and I¹ll rig
my answer.
>> Given any computable list of computable numbers, we can
construct the
>> diagonal number, which will be a computable number not on
the list.
>> That shows that no computable list can contain all the
computable
>> numbers.
>> What¹s rigged about that?
> you cannot demonstrate that its not on the list
DeÞne two functions F(i,i) and G(i), given
F: dom NxN, range {0,1,2,3,4,5,6,7,8,9}
G: dom N, range {0,1,2,3,4,5,6,7,8,9}
and G(i) = Œ5¹ if F(i,i)=¹4¹, G(i) = Œ4¹ otherwise.
G is clearly computable if F is. Where is G on F¹s list
(as indexed by its Þrst argument)?
Is G = F(i,*)? No, because G(i) != F(i,i). Therefore there is
no such i and F() cannot describe all computable numbers.
A modiÞcation of this proof should show that no Þnite set of
F works, either (construct H(i,j) such that H(i,j) = F_k(m,j)
for some k and m, dependent on i).
Dunno about inÞnite sets, at this point.
[rest snipped]
--
#191, ewill3@earthlink.net
It¹s still legal to go .sigless.
===
Subject: Re: difÞcult taylor series
>I have the function
>f(h,t) =
> (2*h+3) ( (2+sqrt(1+2*h))^(-t/h) - (2-sqrt(1+2*h))^(-t/h) )
>log
-------------------------------------------------------------
------------
> 2 sqrt(1+2*h)
>where t>=0 and h>=0 are real. How can I Þnd the Taylor series
>of f in h at h=0 for Þxed Þnite t? The packages I have tried
failed.
There is a good reason why. One actually can get the
Taylor series, but it will not represent the function; the
Þrst term in the long parenthesis, while not identically
0, will have all its Taylor coefÞcients 0, because
3^(-t/h) and all its derivatives approach 0 as h -> 0+ for
t>0.
The other terms can be handled well at h=0, but the package
might not handle (2-sqrt(1+2*h))^(-t/h) without help. The
expression 2-sqrt(1+2*h) and its logarithm can be expanded
in a Taylor series about h=0, convergent for |h| <= .5, with
leading term of the logarithm being -h. This makes it
possible for the power to be expanded itself in a Taylor
series.
--
This address is for information only. I do not claim that
these views
are those of the Statistics Department or of Purdue
University.
Herman Rubin, Department of Statistics, Purdue University
hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
===
Subject: Re: difÞcult taylor series
>I have the function
>f(h,t) =
>
> (2*h+3) ( (2+sqrt(1+2*h))^(-t/h) - (2-sqrt(1+2*h))^(-t/h) )
>log
-------------------------------------------------------------
------------
> 2 sqrt(1+2*h)
>
>where t>=0 and h>=0 are real. How can I Þnd the Taylor series
>of f in h at h=0 for Þxed Þnite t? The packages I have tried
failed.
Why do you think there is such a Taylor series? To me it
looks rather
> singular at h=0.
Robert Israel israel@math.ubc.ca
> Department of Mathematics http://www.math.ubc.ca/~israel
> University of British Columbia
> Vancouver, BC, Canada V6T 1Z2
This came from an investigation into modiÞed differential
equations
(MoDE) of multivalue methods for ODE systems. MoDEs begin life
as retarded difference-differential (RDD) equations, then
become
inÞnite-order ODEs, Þnally Þnite-order ODEs, which can be
occasionally integrated. Got a Taylor series for the OP from
a different method (recursive descent) , and was trying to
conÞrm
the result by going the other way. But it seem that ordinary
CAS run
into problems handling these singularities.
Here is a example of a similar but much easier case, starting
from
the RDD form in u=u(t) for scalar Forward Euler:
RDD MoDE: u(t+h) - u(t) = h L u(t) L = complex constant,
h=step
descent > du/dt = (1 - (1/2) h L + (1/3) h^2L^2 - ... ) L u =
(log
(1+ hL)/h ) u
integ> u(t) = u0 (1+hL) ^(t/h) -> u0 e ^(Lt) as h ->0
where u0=u(0). Reverse: start with u = u0 (1+hL) ^(t/h) and
get the
two MoDE forms above. This one is easily done by any CAS, or
by hand.
In fact it was done by Euler, so CAS are catching up with the
XVIII
century.
===
Subject: Re: difÞcult taylor series
|>>I have the function
|>>f(h,t) =

|>> (2*h+3) ( (2+sqrt(1+2*h))^(-t/h) - (2-sqrt(1+2*h))^(-t/h)
)
|>>log
-------------------------------------------------------------
------------
|>> 2 sqrt(1+2*h)

|>>where t>=0 and h>=0 are real. How can I Þnd the Taylor
series
|>>of f in h at h=0 for Þxed Þnite t? The packages I have
tried
failed.

|>> Why do you think there is such a Taylor series? To me it
looks rather
|>> singular at h=0.

|>This came from an investigation into modiÞed differential
equations
|>(MoDE) of multivalue methods for ODE systems. MoDEs begin
life
|>as retarded difference-differential (RDD) equations, then
become
|>inÞnite-order ODEs, Þnally Þnite-order ODEs, which can be
|>occasionally integrated. Got a Taylor series for the OP from
|>a different method (recursive descent) , and was trying to
conÞrm
|>the result by going the other way. But it seem that
ordinary CAS run
|>into problems handling these singularities.
It¹s not just that the CAS have problems, it¹s that there¹s a
mathematical
problem: the function is not analytic at h=0. Note that a
Taylor series
expansion about h=0 is not just for h > 0 but for all h in a
neighbourhood
of 0. And your function will be rather bad for small negative
values of
h. However, you can get an asymptotic series which will be
good for
positive h.
Something like this (in Maple):
> q1:= -t/h*ln(2 + sqrt(1+2*h)); # so (2+sqrt(1+2*h))^(-t/h)
= exp(q1)
> series(q1,h);
t ln(3) 1 2 19 2 97 3 2059 4 / 5
- ------- - - t + - t h - -- t h + --- t h - ---- t h + Oh /
h 3 9 81 324 4860

> q2:= -t/h*ln(2 - sqrt(1+2*h)); # so (2-sqrt(1+2*h))^(-t/h)
= exp(q2)
> series(q2,h);
1 2 1 3 9 4 / 5
t + - t h - - t h + -- t h + Oh /
3 4 20
For h -> 0+ with t > 0, the -t ln(3)/h term makes exp(q1) =
O(h^n) for
all h. So for the asymptotic series we can ignore q1.
> f:= ln(-(2*h+3)*exp(q2)/(2*sqrt(1+2*h)));
> map(simplify,series(%,h));
1 /7 1 2 / 100 1 3
(ln(3) - ln(2) + ln(-exp(t))) - - h + |- + - t| h + |- --- -
- t| h
3 9 3 / 81 4 /
/9 158 4 / 5 3856 5 / 6
+ |-- t + ---| h + |- - t - ----| h + Oh /
20 81 / 8 1215/
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Re: difÞcult taylor series
> I have the function
> f(h,t) =
(2*h+3) ( (2+sqrt(1+2*h))^(-t/h) - (2-sqrt(1+2*h))^(-t/h) )
> log
-------------------------------------------------------------
------------
> 2 sqrt(1+2*h)
where t>=0 and h>=0 are real. How can I Þnd the Taylor series
> of f in h at h=0 for Þxed Þnite t? The packages I have
tried failed.
It does not have a Taylor series in h at h=0 for Þxed Þnite
t. I
assume that you tried this using Mathematica and Maple? In
that case, as
Mathematica will have told you, it has an essential
singularity there.
This is due the term
(Sqrt[2 h + 1] + 2)^(-(t/h))
as you will see if you enter
(Sqrt[2 h + 1] + 2)^(-(t/h)) + O[h]

into Mathematica (adding an order term, O[h], coerces series
expansion).
The Log of this expression _does_ have a series expansion
which involves
1/h, leading to the essential singularity:
PowerExpand[Log[(Sqrt[2 h + 1] + 2)^(-(t/h))]] + O[h]
You get essentially the same behaviour for Exp[-1/x]
Exp[-1/x] + O[x]
As pointed out by another respondent you are probably
interested in
(2 - Sqrt[1 + 2 h])^(-t/h) - (2 + Sqrt[1 + 2 h])^(-t/h)
for otherwise you will be taking the log of a negative number.
As an aside, the limit as h -> 0 does exist (which may be the
reason you
are expecting a Taylor series). In fact
(2 - Sqrt[1 + 2 h])^(-t/h) - (2 + Sqrt[1 + 2 h])^(-t/h) ->
Exp[t]
as h -> 0 (for t > 0).
Paul
--
Paul Abbott Phone: +61 8 9380 2734
School of Physics, M013 Fax: +61 8 9380 1014
The University of Western Australia (CRICOS Provider No
00126G)

35 Stirling Highway
Crawley WA 6009 mailto:paul@physics.uwa.edu.au
AUSTRALIA http://physics.uwa.edu.au/~paul
===
Subject: Re: difÞcult taylor series
> I have the function
> f(h,t) =
(2*h+3) ( (2+sqrt(1+2*h))^(-t/h) - (2-sqrt(1+2*h))^(-t/h) )
> log
-------------------------------------------------------------
------------
> 2 sqrt(1+2*h)
where t>=0 and h>=0 are real.
If t = 1 = h, you¹re taking the log of a negative number?
===
Subject: Re: Zero Point Energy Debate
> We missed you in London. Make sure you and Hal come next
time. :-)
Nick Cook can moderate a TV discussion.
I will deÞnitely include what you say below in Super Cosmos
lest there
> be no confusion.
What I meant of course was in a humorous vein since you are
the younger
> man and work in the same building with Hal on a daily
basis. You know
> like Batman and Robin.
See photos at http://qedcorp.com/London/

> Jack:
>
> I am not Hal Puthoff¹s sidekick. We share a common
employer, and a
> common attitude to much of our physics; Hal and I spend
much time in
> creative & vigorous exchanges. But I have complete autonomy
to pursue
> ideas I believe relevant to the broad goals of the
Institute. If you
> look at what I have published, I think you will Þnd it hard
to justify
> the sidekick characterization. Nor do I think Hal would
agree with you.
> In fact the opposite is closer to the truth: The
publications [1-5]
> could be regarded as attempts to steer a reluctant SED
community onto a
> new - and I believe more promising - course.
Nobody has ever said than anybody doing QM research is
a sidekick of anybody else. The claim is still that
*ALL* QM researchers are *sidelobes* of Van Neumann.
Einstone is irrelevent to the equation,
do to well-known, historically-documented, dice
logic, and geometry problems.
Feynmann is irrelevent to the equation due to
well-known robotics incompetance, and for
having the bizzarre belief that gaskets have
some non-zero coorelation with brakets and electrons.
Carl Sagan is irrelevant to the equation
due to well-known failure to understand not only
Hal, but also pi.
===
Subject: Re: Zero Point Energy Debate
> Why do you believe anyone on Earth cares about your personal
> correspondence?
xxein: What are you responding to?
===
Subject: Re: Zero Point Energy Debate
>
>>Why do you believe anyone on Earth cares about your personal
>>correspondence?

> xxein: What are you responding to?
Hey, if you had to care what someone said in order to
ridicule them,
99% of the NG trafÞc would dry up.
-E
===
Subject: Very simple question.
60 is 40% of what number? And how do I go about answering it.
===
Subject: Re: Very simple question.
>60 is 40% of what number?
11, exactly.
>And how do I go about answering it.
You do what I did: lie with conviction.
The reason to do that is that it will force you to think
about how you
would CHECK such an answer. When you do that you¹ll discover
how to
FIND the right answer. That¹s a very general tool for solving
math
problems!
In this example, you¹re probably ready to tell me that 11
can¹t
be the right answer. Why? Because you know what 40% of 11 IS,
and it ain¹t 60. In fact, it¹d be 0.40 times 11 (which is
4.4).
So whatever the right answer _really is_, it¹d be a number N
for which exactly the same checking process will give you 60,
that is, it¹d make 0.40 times N = 60. So now you have the
equation that needs to be solved:
0.40 x N = 60
so
N = 60 / 0.40 = 150.
Sure enough,
0.40 x 150 = 60
so this time we know we have the right answer.
Remember that principle: lie Þrst, then watch yourself Þgure
out
why the Þrst guess is wrong. It¹ll help you Þgure out what¹s
right!
dave
===
Subject: Re: Very simple question.

>>60 is 40% of what number?
11, exactly.

>>And how do I go about answering it.
You do what I did: lie with conviction.
The reason to do that is that it will force you to think
about how you
> would CHECK such an answer. When you do that you¹ll
discover how to
> FIND the right answer. That¹s a very general tool for
solving math
problems!
In this example, you¹re probably ready to tell me that 11
can¹t
> be the right answer. Why? Because you know what 40% of 11
IS,
> and it ain¹t 60. In fact, it¹d be 0.40 times 11 (which is
4.4).
> So whatever the right answer _really is_, it¹d be a number N
> for which exactly the same checking process will give you
60,
> that is, it¹d make 0.40 times N = 60. So now you have the
> equation that needs to be solved:
> 0.40 x N = 60
> so
> N = 60 / 0.40 = 150.
> Sure enough,
> 0.40 x 150 = 60
> so this time we know we have the right answer.
Remember that principle: lie Þrst, then watch yourself Þgure
out
> why the Þrst guess is wrong. It¹ll help you Þgure out
what¹s right!
>
That¹s very nice, Dave. I¹ll have to remember it for use in
my classes.
Made my evening,
Rick
===
Subject: Re: Very simple question.
Originator: richard@cogsci.ed.ac.uk (Richard Tobin)
>60 is 40% of what number? And how do I go about answering it.
40% is just another way of writing 0.40. So you want to solve
for X in
60 = 0.40 X
-- Richard
===
Subject: Re: Very simple question.
> 60 is 40% of what number? And how do I go about answering
it.
1) Always replace the percent symbol
%
with
* 1/100 (times 1/100).
Example:
40% = 40 * 1/100 = 40/100 = 0.4
2) Replace
of what
with
* x (times x)
So you get:
60 = 40 * 1/100 * x
You can Þnd x as follows:
Multiply both sides with 100:
60 * 100 = 40 * x
divide both sides by 40:
60 * 100 / 40 = x
and Þnd
x = 150.
Dirk Vdm
===
Subject: Re: Very simple question.
Adjunct Assistant Professor at the University of Montana.
>60 is 40% of what number? And how do I go about answering it.
Use the so-called Rule of Three.
x is to 100
as
60 is to 40
That means that x/100 = 60/40, or that 60*100/40 = x.
--
=============================================================
=========
It¹s not denial. I¹m just very selective about
what I accept as reality.
--- Calvin (Calvin and Hobbes)
=============================================================
=========
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Re: Very simple question.
* Arturo Magidin
>60 is 40% of what number? And how do I go about answering it.
Use the so-called Rule of Three.
x is to 100
> as
60 is to 40
That means that x/100 = 60/40, or that 60*100/40 = x.
Alternatively, if 60 is 40%, how much is 20%? Clearly 30. How
much
is 10%... Go down to 1%. Finally, Þnd 100%
--
Jon Haugsand
Dept. of Informatics, Univ. of Oslo, Norway,
mailto:jonhaug@iÞ.uio.no
http://www.iÞ.uio.no/~jonhaug/, Phone: +47 22 85 24 92
===
Subject: Re: Help needed with nonlinear Þt (continued)
Why write your own program? Try TableCurve 2D from
http://www.systat.com/products/TableCurve2D/
-Joel
> many ways to Þt a curve to data, so after doing some
research I believe
> that the best way for me to Þt my data is by using the
quadratic. I
would
> like to see how this Þts the datapoints, and if that
doesn¹t work well,
> then I will need to go for the cubic.
Now the only problem I have is how to symbolically solve ax^2
+ bx + c for
a
> b and c (given some data set of x and y values) so that I
can implement
this
> into a function in c++. I have the same problem solving for
ax^3 + bx^2
+cx
> + d.
I apologize for my algebra and calculus are horrible after so
long out of
> school. Once I can symbolically solve these two guys for
their
> coefÞcients, then I can implement the code.
Just in case anyone is interested, the data set is from
maldi-tof mass
spec,
> and is after I have processed the data for peaks. I need to
be able to
> estimate what peak intensity -should- be in regions where
no peaks have
been
> found based on the overall curve Þtted to the existing data.
If anyone again, can help out directly or point me to a good
site that
can
> walk me through this...
Bryan
P.S. Checked out the numerical recipes in C, helpful but
assumes I know
how
> to do some things that I do not, like it leaves out the
matrix
> calculations...
===
Subject: Re: n-points
>> Given n-points in space. Find the smallest possible circle
to enclose
>> all these points and locate the centre of this circle and
give the
>> radius of this circle.
JR proposed placing the center of the circle to be
constructed at the
center of mass of the n points. That¹s obviously not the
smallest
circle in general: consider (n-1) points near the origin and
one
point near (0,1); the C-o-M circle will have a radius near 1,
whereas
the minimal circle will have radius near 1/2.
The correct circle will typically be a circle drawn through
three of
the n points; the precise coordinates of the other n-3 points
won¹t matter, so you can¹t expect an algebraic formula
involving all
n points symmetrically to tell you where the center ought to
be.
(Occasionally the correct circle will pass through only two
of the
points -- typically when one pair of points is much farther
apart than
any other pair.)
If Find the smallest circle means Give an algorithm to Þnd...,
then
clearly a systematic enumeration of all triples of points
will eventually
stumble on the smallest circle, but there are better ways.
Look for
minimum bounding circle algorithms (or change circle to sphere
for
higher-dimensional analogues).
dave
===
Subject: Solving equations degree 2 and 4 mod 2^n
Can someone give me an algorithm about solving equations of
type x^4+
x+ a=0 mod 2^n, where a is known in F_2^n and also for
equation of the
form x^2+x+b=0 mod 2^n where b is also known.
A
===
Subject: Re: Solving equations degree 2 and 4 mod 2^n
> Can someone give me an algorithm about solving equations of
type x^4+
> x+ a=0 mod 2^n, where a is known in F_2^n and also for
equation of the
> form x^2+x+b=0 mod 2^n where b is also known.
Do you realise that F_{2^n} is not the same as Z mod 2^n ?
--
Timothy Murphy
e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie
tel: +353-86-2336090, +353-1-2842366
s-mail: School of Mathematics, Trinity College, Dublin 2,
Ireland
===
Subject: Re: What is the effect of nesting quantiÞers?
>> What is the difference between the axiom of extensionality
(#1) and
>> expression #2.
>>
>> #1 AaAb[Ax(x in a <->x in b) -> a = b)]
>> #2 AaAbAx((x in a <-> x in b) -> a = b)
>>
>> How are these two things read differently in natural
language?
>Let¹s just say there is a reason natural language is often
avoided when
>dealing with quantiÞers.
Right, unless you happen to think that the way mathematicians
talk is
natural (putting all the bound variables at the beginning of a
sentence). In natural language, the universal quantiÞers tend
to
migrate to the end of the sentence, giving us such helpful
sentences as:
There¹s an even number bigger than every odd number.
dave
===
Subject: Re: How many reduced basis in a cubic subset of Z^3?
I realize there was a little mistake in my previous post:
|> I need to count the number of all reduced 3D basis with
integral
coordinates
|> whose vectors are contained in a given cube of edge n
centered on
origin.
--------------------^
I had a cube of edge 2n in mind, not n.
Any hint, pointer, solution is welcome.
--
Here¹s where you can reach me (Þxed-width font required).
. ,-. .
. . ,-. ,-. ,-. ,-. |- /,-. ,-. ,-. ,-| ,-. ,-. ,-.
| | `-. |-¹ | | |-¹ | |,-|| | | -- |-¹ -- | | | | | | |
`-^ `-¹ `-¹ Œ Œ `-¹ `¹ `-^¹ `-| `-¹ `-^ :; `-¹ Œ `-|
`-- | ,|
` `¹
===
Subject: Recommendations wanted: Complex analysis textbook
for undergraduate
course
Would anyone like to make a recommendation for a textbook for
an
undergraduate Þrst course in complex analysis? I¹m teaching
it in the
fall. We¹ve been using Churchill & Brown and I¹d like to at
least look
at what else might be available before using the same old
thing.
Students will have had calc 1-2-3 (single and multivariable)
plus
linear algebra (we use David Lay¹s book). The thrust of the
course is
purely mathematical, keeping in mind that the students may
not have had
any proof based course.
-Lotofun
===
Subject: Re: Recommendations wanted: Complex analysis
textbook for
undergraduate course
@yahoo.com:

> Students will have had calc 1-2-3 (single and
multivariable) plus
> linear algebra (we use David Lay¹s book). The thrust of the
course is
> purely mathematical, keeping in mind that the students may
not have had
> any proof based course.
Even though you said purely mathematical, I assume that
doesn¹t
preclude all applications, but rather, that this is not a
course
that¹s servicing engineers. I have used Saff and Snider and
really liked it for my math majors. Each chapter concluded
with
some sort of application or other optional section (Chapter 1
has
some linkage problems) which do help the students get a
handle on
things. Otherwise it hits all the high points and follows the
usual order.
I don¹t recall any serious problems or any complaints from the
students with the exposition. I think it¹s a fairly clean
and straightforward text.
Bart
===
Subject: Solution to second order linear ODE with non
constant coef?
Any ideas of how to solve analyticaly this ODE:
1/2 x^2 f(x)¹¹ + ( x - 1) f(x)¹ - f(x) = 0
It¹s similar to Euler-Cauchy but I got this agly -1 in the
Þrst derivative
that make it harder.
Boundary condition: f(0) = 1
also lim_x->00 f(x) = 0
I think I can apply the method of variation of parameters as
soon as I know
two linearly independent solutions. I just got one f1(x) =
c(x-1) where
c=constant.
Any ideas, sugestions?
--pan
===
Subject: Re: Solution to second order linear ODE with non
constant coef?
>Any ideas of how to solve analyticaly this ODE:
>1/2 x^2 f(x)¹¹ + ( x - 1) f(x)¹ - f(x) = 0
>It¹s similar to Euler-Cauchy but I got this agly -1 in the
Þrst
derivative
>that make it harder.
>Boundary condition: f(0) = 1
>also lim_x->00 f(x) = 0
>I think I can apply the method of variation of parameters as
soon as I
know
>two linearly independent solutions. I just got one f1(x) =
c(x-1) where
>c=constant.
Hint: knowing f1(x) is a great place to start. Now try f2(x)
= u(x) f1(x)
and see what that gets you (this is called reduction of
order: you
shoud get an ODE for u¹¹ and u¹ with no term in u, and if you
set
v = u¹ you¹ll have a Þrst order equation in v which you
should be able
to solve).
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Equivalent class question
Say S is a set and P(S) its power set. u is a non-negative,
sub-additive function on P(S). + means symmetric difference
of sets
and e means belongs to.
DeÞne [A] = {X e P(S): u(A+X)=0 and A e P(S)}
DeÞne Q = {[A]: A e P(S)}
If X and Y e Q, does it follow that if u(X+Y)=0 then X=Y? Why?
I¹m studying Rudin¹s chapter on measure, where he asserts
that by
deÞning Q this way, it is a metric space and this is not
clearly
explained. All help appreciated.
===
Subject: Re: Equivalent class question
>Say S is a set and P(S) its power set. u is a non-negative,
>sub-additive function on P(S). + means symmetric difference
of sets
>and e means belongs to.
>DeÞne [A] = {X e P(S): u(A+X)=0 and A e P(S)}
>DeÞne Q = {[A]: A e P(S)}
>If X and Y e Q, does it follow that if u(X+Y)=0 then X=Y?
Why?
This is a bit of an abuse of notation. What you should really
have is
that if [X] and [Y] e Q, and u(X+Y)=0, then [X]=[Y].
>I¹m studying Rudin¹s chapter on measure, where he asserts
that by
>deÞning Q this way, it is a metric space and this is not
clearly
>explained. All help appreciated.
You are deÞning an equivalence relation on P(S): A~B if and
only if
u(A+B)=0. Your deÞnition [A] corresponds to the equivalence
class of
A.
To see this, note that A is in [A] (I assume u(empty)=0); is
A~X then
X~A; and if A~B and B~C, then u(A+B) = u(B+C) = 0; and
u(A+C) = u(A+(B+B)+C) = u((A+B)+(B+C)) <= u(A+B) + u(B+C) =
0+0 = 0.
so A~C.
So Q is the quotient set of P(S) under this equivalence
class. That
means that Y in [X] if and only if X~Y if and only if Y~X if
and only
if X in [Y] if and only if [X]=[Y].
So, let [X] and [Y] be in Q, and assume that u(X+Y)=0. By
deÞnition
of ~, this means that Y is in [X], which means that [Y]=[X].
--
=============================================================
=========
It¹s not denial. I¹m just very selective about
what I accept as reality.
--- Calvin (Calvin and Hobbes)
=============================================================
=========
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: how do i solve...
How could I solve X^2 = 2^X ?
I have tried using calculus and algebra, but I am either
doing it
help!
scott
===
Subject: known inequality?
I have found a short, elementary proof that:
(n^n)/e^n < n! < (n+1)^(n+1)/e^n
for all natural numbers n.
I believe the lower bound on n! is well-known. However,
I haven¹t explicitly seen the upper bound before.
I know that Stirling¹s formula can give you something like
n! < n^(n+1)/e^n for n large enough,
but this both:
(1) is not true for _all_ n (speciÞcally it fails for n<7),
and
(2) is much harder to prove (i.e. proving Stirling¹s formula
is
not super-easy).
Besides Stirling, is there any other known bound on n! which
supersedes this one?
-Tyler