mm-105
==



> Could a proof of the Goldbach Conjecture be worth of a Fields prize?
DonOt you know the regulations for your own prize?
>> Fids Medals are restricted to those under age 40.
Is this in the rules or is it mery a habit that the prize
givers have
>got into?
>Rules. HereOs what it states at
>http://www.emis.math.ca/EMIS/mirror/IMU/medals/
> At the 1924 International Congress of Mathematicians in
Toronto, a
>> resolution was adopted that at each ICM, two gold medals
should be
>> awarded to recognize outstanding mathematical achievement.
Professor
>> J. C. Fids, a Canadian mathematician who was Secretary of
the 1924
>> Congress, later donated funds establishing the medals
which were
>> named in his honor. Consistent with FidsOs wish that the
awards
>> recognize both existing work and the promise of future
achievement,
>> it was agreed to restrict the medals to mathematicians not
over
>> forty at the year of the Congress. In 1966 it was agreed
that, in
>> light of the great expansion of mathematical research, up
to four
>> medals could be awarded at each Congress.
This is all very interesting - but does it answer the
original question?
Wl, I was only answering the most recent question. Nesting
and all,
you see. (-:
>I would say that a proof of GoldbachOs conjecture by
somebody less
>than 40 years old might or might not result in a Fids Medal
being
>awarded. It would be more liky to do so if the proof
involved some
>fundamental new ideas which opened up major new directions
>for future research.
Quite so. I seem to remember a similar point being made about
FermatOs
Last Theorem, that if someone could find
FermatOs remarkable
proof,
and it turned on some kind of trick or curious observation,
that it
would be a disappointment to most mathematicians. That one
reason
people were interested in a possible proof of FLT was that the
consensus was that in order to prove it, it would require
major new
insights and major new work (as indeed it did), and that the
excitement was on those insights rather than the actual
result.
Looking at the list of Fids winners in
http://www.accessscience.com/Awards/Fids/MATH/main.html
I see that out of 44 winners, 3 are cited for specific
problems as in
particular probably signifying major theoretical advances
along the
way to a particular theorem (Richard Borcherds in 1998 for
his work
in the fids of algebra and geometry, especially his proof of
the
OMoonshine conjectureO; Efim Zmanov
in 1994 for work as an
algebraist and in group theory; in particular, for solving the
restricted Burnside problem, one of the fundamental questions
in group
theory; and a Freedman in 1986 for work in topological
analysis, in particular the proof of the Poincar conjecture
for
four-dimensional topological manifolds). In addition, 5 are
cited
specifically for a particular result (Jesse Douglas in 1936
for
solving the plateau problem; K.F. Roth in 1958 for solving the
Thue-Sieg problem concerning the approximations of algebraic
numbers
by rational numbers; Alan Baker in 1970 for generalizing the
Gfond-Schneider theorem, generating previously unidentified
transcendental numbers, and showing how the theory could be
used in
the fid of diophantine equations; Gerd Faltings in 1986 for
his proof of the Mordl conjecture using arithmetic algebraic
geometry; and Laurent Lafforgue in 2002 for proving the global
Langlands correspondence for function fids, a major advance
toward
the realization of the OLanglands Program[.]O).
Which does suggest that a proof of the Goldbach Conjecture is
not per
se a guarantee of a Fids Medal.
==
==
TOP POSTED since quote is long
With a little ingenuity, the state space can be reduced to
size 2 x 3^5
= 486: Let the state be the numbers of values already seen in
each set
{2, 12}, {3, 11}, {4, 10}, {5, 9}, {6, 8}, and {7}. That is,
the state
is a sextuple: each of the first five components
can be 0, 1,
or 2;
the last component can be either 0 or 1. v(2,2,2,2,2,1) = 0;
we want
v(0,0,0,0,0,0).
> We would expect to throw 1 six-sided die 6(1 + 1/2 + 1/3 +
1/4 + 1/5 +
>> 1/6) = 14.7 times before throwing all the numbers 1 to 6
at least once.
> How many times would we expect to roll 2 six-sided dice
before we have
>> thrown all the numbers from 2 to 12 at least once?
> This is combinatorally a much harder question, since the
distribution
> is no longer uniform. However, the space is not so big here
that one
> cannot program it.
Let p_k = min(k-1, 13-k) /36, the probability of rolling k.
Let v(S)
> = the expected remaining number of rolls until you roll
numbers,
> where you have already rolled the numbers in S, a subset of
{2, 3,
> 4,..., 12}.
v(S) = 1 + sum(k in S, p_k) v(S) + sum(k not in S, p_k v(S U
{k}))
= [1 + sum(k not in S, p_k v(S U {k}))] / sum(k not in S, p_k)
v({2, 3, 4,..., 12}) = 0
We want v({}).
Note that there are 2^11 = 2048 states, but it can be done.
If I
> had access to the computer (which is down) with Mathematica
right
> now, I could write up a little program fast.
==
Stephen J. Herschkorn herschko@rutcor.rutgers.edu
> TOP POSTED since quote is long
With a little ingenuity, the state space can be reduced to
size 2 x
> 3^5 = 486: Let the state be the numbers of values already
seen in
> each set {2, 12}, {3, 11}, {4, 10}, {5, 9}, {6, 8}, and
{7}. That
> is, the state is a sextuple: each of the first
five
components can
> be 0, 1, or 2; the last component can be either 0 or 1.
> v(2,2,2,2,2,1) = 0; we want v(0,0,0,0,0,0).
> We would expect to throw 1 six-sided die 6(1 + 1/2 + 1/3 +
1/4 + 1/5 +
> 1/6) = 14.7 times before throwing all the numbers 1 to 6 at
least once.
How many times would we expect to roll 2 six-sided dice
before we have
> thrown all the numbers from 2 to 12 at least once?
> This is combinatorally a much harder question, since the
distribution
>> is no longer uniform. However, the space is not so big
here that one
>> cannot program it.
> Let p_k = min(k-1, 13-k) /36, the probability of rolling k.
Let v(S)
>> = the expected remaining number of rolls until you roll
numbers,
>> where you have already rolled the numbers in S, a subset
of {2, 3,
>> 4,..., 12}.
> v(S) = 1 + sum(k in S, p_k) v(S) + sum(k not in S, p_k v(S
U {k}))
> = [1 + sum(k not in S, p_k v(S U {k}))] / sum(k not in S,
p_k)
> v({2, 3, 4,..., 12}) = 0
> We want v({}).
>
Now that I have access to Mathematica again, here is the more
efficient
solution:
We devop a recursive formulation by conditioning on each
roll. Our
state space is {0,1,2}^5 x {0,1}, as noted above.
We will need to update the state after a roll. Let d(i,j) be
the
Kronecker dta (1 if i= j, 0 otherwise).
u(f,k)_j = min(f_j + d(j,k), 2) for 1 <= j <= 5
u(f,k)_6 = min(f_6 + d(6,k), 1)
Here, k = 1 represents a two or a twve rolled; k = 2
represents a
three or an even rolled; etc.
v(2,2,2,2,2,1) = 0
v(f) = 1 + 1/36 sum(k=1..6, k f_k) v(f) + 1/36 sum(k=1..5,
(2-f_k) k
v(u(f,k))) + 1/6 (1-f_6) v(u(f,6))
= [36 + sum(k=1..5, (2-f_k) k v(u(f,k))) + 6(1-f_6)
v(u(f,6))] / [36 -
sum(k=1..6, k f_k)]
v(0,0,0,0,0,0) is our answer.
Lickety split, Mathematica gave me 61 +
2733466759/12574325400 =
61.2174 (approx.) This was lower than I expected: a lower
bound is 54,
the expected number of rolls until both a two and a twve
appear. (The
expected number of rolls until 2, 3, 11, and 12 is 59.4;
until 2, 3, 4,
10, 11, and 12, the expectation is 60.6455.)
For those interested, here is the Mathematica code:
u[f_,k_]:= ReplacePart[f, f[[k]]+1, k]
v[{2,2,2,2,2,1}]= 0
v[f_]:= (v[f]= 0) /; f[[6]]>1 || Max[f]>2
v[f_]:= v[f]= (36 + Sum[(2-f[[k]]) k v[u[f,k]], {k,5}] +
6(1-f[[6]])
v[u[f,6]]) /
(36 - Sum[k f[[k]], {k,6}])
v[{0,0,0,0,0,0}]
It is very easy to program recursive formulae in Mathematica.
==
Stephen J. Herschkorn herschko@rutcor.rutgers.edu
A little hp here please? What am I missing in this
discussion? Why
all this talk about programming? ArenOt these math problems?
ItOs
been a long time since I took probability and combinatorics,
but
except for using a calculator for factorials, it was strictly
a pencil
and paper enterprise.
> TOP POSTED since quote is long
With a little ingenuity, the state space can be reduced to
size 2 x 3^5
> = 486: Let the state be the numbers of values already seen
in each set
> {2, 12}, {3, 11}, {4, 10}, {5, 9}, {6, 8}, and {7}. That
is, the state
> is a sextuple: each of the first five components
can be 0, 1,
or 2;
> the last component can be either 0 or 1. v(2,2,2,2,2,1) =
0; we want
> v(0,0,0,0,0,0).
>> We would expect to throw 1 six-sided die 6(1 + 1/2 + 1/3
+ 1/4 + 1/5 +
>> 1/6) = 14.7 times before throwing all the numbers 1 to 6
at least
once.
>> How many times would we expect to roll 2 six-sided dice
before we have
>> thrown all the numbers from 2 to 12 at least once?
> This is combinatorally a much harder question, since the
distribution
> is no longer uniform. However, the space is not so big here
that one
> cannot program it.
> Let p_k = min(k-1, 13-k) /36, the probability of rolling
k. Let v(S)
> = the expected remaining number of rolls until you roll
numbers,
> where you have already rolled the numbers in S, a subset of
{2, 3,
> 4,..., 12}.
> v(S) = 1 + sum(k in S, p_k) v(S) + sum(k not in S, p_k
v(S U {k}))
> = [1 + sum(k not in S, p_k v(S U {k}))] / sum(k not in S,
p_k)
> v({2, 3, 4,..., 12}) = 0
> We want v({}).
> Note that there are 2^11 = 2048 states, but it can be
done. If I
> had access to the computer (which is down) with Mathematica
right
> now, I could write up a little program fast.
> A little hp here please? What am I missing in this
discussion? Why
> all this talk about programming? ArenOt these math
problems? ItOs
> been a long time since I took probability and
combinatorics, but
> except for using a calculator for factorials, it was
strictly a pencil
> and paper enterprise.
It *has* been a long time - I donOt think anybody actually
does the math by
hand anymore, except to design the algorithm in the first
place and/or
double-check the computerOs answers.
- Sir Bob.
>A little hp here please? What am I missing in this
discussion? Why
>all this talk about programming? ArenOt these math
problems?
ItOs
>been a long time since I took probability and combinatorics,
but
>except for using a calculator for factorials, it was
strictly a pencil
>and paper enterprise.
>
Because the OP posed a problem the method of whose solution
one can
readily pose, but the actual computations to be carried out
are
extensive - one would never want to do them by hand. What do
you have
against the use of computers for carrying out calculations?
As I say
bow:
>This is combinatorally a much harder question, since the
distribution
>is no longer uniform. However, the space is not so big here
that one
>cannot program it.
>TOP POSTED since quote is long
>With a little ingenuity, the state space can be reduced to
size 2 x 3^5
>>= 486: Let the state be the numbers of values already seen
in each set
>>{2, 12}, {3, 11}, {4, 10}, {5, 9}, {6, 8}, and {7}. That
is, the state
>>is a sextuple: each of the first five components
can be 0, 1,
or 2;
>>the last component can be either 0 or 1. v(2,2,2,2,2,1) =
0; we want
>>v(0,0,0,0,0,0).
We would expect to throw 1 six-sided die 6(1 + 1/2 + 1/3 +
1/4 + 1/5 +
>>1/6) = 14.7 times before throwing all the numbers 1 to 6 at
least once.
>How many times would we expect to roll 2 six-sided dice
before we have
>>thrown all the numbers from 2 to 12 at least once?
 This is combinatorally a much harder question, since the
distribution
>is no longer uniform. However, the space is not so big here
that one
>cannot program it.
Let p_k = min(k-1, 13-k) /36, the probability of rolling k.
Let v(S)
>= the expected remaining number of rolls until you roll
numbers,
>where you have already rolled the numbers in S, a subset of
{2, 3,
>4,..., 12}.
v(S) = 1 + sum(k in S, p_k) v(S) + sum(k not in S, p_k v(S U
{k}))
= [1 + sum(k not in S, p_k v(S U {k}))] / sum(k not in S, p_k)
v({2, 3, 4,..., 12}) = 0
We want v({}).
Note that there are 2^11 = 2048 states, but it can be done.
If I
>had access to the computer (which is down) with Mathematica
right
>now, I could write up a little program fast.

==
Stephen J. Herschkorn herschko@rutcor.rutgers.edu
nicolas bigeard
Nowadays,some are seeking and seeking for the highest
possible prime.
I think you mean the highest _known_ prime.
> So I guess that this means lotOs of work to know how to
discard some of
the big
> numbers...
Those problems are quiet a thrill and they even could have
some nice
> applications in cryptography. So, I just wonder, is there
any wl known
> theorem about the density of easily decomposable numbers.
By this I
think
> to numbers that we could easly discard, in some prime
search...
Dunno about density, but some big number are easily factored,
at least
partly. If b^n-1 is a prime (and b,n > 0) then n must be
prime. To see why,
just consider what the number looks like in base b notation.
We get some
restrictions on b as wl. I guess this is why the
big-prime-hunters, like
the big-game-hunters of old, are interested particularly in
Mersenne
numbers
2^n-1.
LH
> I yet again forgot the command to block messages by a
certain person, as
> rary do people annoy me. This guy obviously understands a
few
interesting
> features of quantum mechanics and string theory, as wl as
concepts of a
> sinularity-based collective conciousness, but this juvenile
continued
> obsessing on trivial details is closer to psychosis than
enlightenment,
in
> my opinion.
there is only trivial. 1=0. your ego destroys insight.
0=1
1=8
8=0
0=totality.
i annoy you because you speak from a viewpoint of
individualism. if you
had payed attention you would know truth. it is right in
front of you; a
part of everything - all energy, all words, all that exists.
Exercise 13, chapter 6:
Let L^infty = L^infty(m), where m is Lebesgue measure on I =
[0,1].
Show that there is a bounded linear functional lambda!=0 on
L^infty
that is 0 on C(I), and that therefore there is no g in L^1(m)
that
satisfies lambda f=int_I fg dm for every f in L^infty. Thus
(L^infty)* != L^1.
nojb.
Ojeda
>Exercise 13, chapter 6, Real and Complex Analysis (Rudin):
>Prove that L^inf([0,1],m) is not isometric to L^1([0,1],m)
by showing
>that there exists a bounded lineal nonzero functional T on
the first
>space such that Tf=0 for all f continuous on [0,1].
No, thatOs not what the exercise says. I canOt
tl you what it
> _does_ say, because the book is at the office and
IOm at home
> (and itOs raining), but the exercise doesnOt
say that, I
guarantee
> you.
Any hp appreciated!
What does the exercise actually ask?
==
>Exercise 13, chapter 6:
Let L^infty = L^infty(m), where m is Lebesgue measure on I =
[0,1].
>Show that there is a bounded linear functional lambda!=0 on
L^infty
>that is 0 on C(I), and that therefore there is no g in
L^1(m) that
>satisfies lambda f=int_I fg dm for every f in L^infty. Thus
>(L^infty)* != L^1.
ThatOs more or less what I thought the exercise must say.
Note that there were _two_ problems with the way you stated
the exercise, one just a typo and one more serious: The typo
was you said (L^infty) where you meant to say (L^infty)* .
The more serious problem is this: The (informally stated)
conclusion
(L^infty)* != L^1 does _not_ say that (L^infty)* is not
isometric
to L^1. It means that the _canonical_ mapping from L^1 into
(L^infty)* does not map onto (L^infty)*, but it does not say
that there could not be some other isometric isomorphism.
Honest, there actually is a difference: There exists a Banach
space X such that X _is_ isometric to X**, although X is
_not_ re?xive (which is to say that the canonical map
from X into X** is not surjective). IOm pretty sure the
example is known as the James tree space, if you
want to try to find it somewhere. The exercise in Rudin
shows that L^1 is not re?xive, but the James tree space
shows that this is not the e as showing that L^1
is not isometric to (L^1)**, which is what you said it
said.
(Of course itOs not hard to show that (L^infty)* and L^1
are not isometric, but it has very little to do with what this
exercise asks you to do.)
>nojb.
Ojeda
>>
Exercise 13, chapter 6, Real and Complex Analysis (Rudin):
Prove that L^inf([0,1],m) is not isometric to L^1([0,1],m) by
showing
>that there exists a bounded lineal nonzero functional T on
the first
>space such that Tf=0 for all f continuous on [0,1].
No, thatOs not what the exercise says. I canOt
tl you what it
>> _does_ say, because the book is at the office and
IOm at
home
>> (and itOs raining), but the exercise
doesnOt say that, I
guarantee
>> you.
>Any hp appreciated!
What does the exercise actually ask?
>nojb.
>> ************************
David C. Ullrich
************************
David C. Ullrich
Honest, there actually is a difference: There exists a Banach
> space X such that X _is_ isometric to X**, although X is
> _not_ re?xive (which is to say that the canonical map
> from X into X** is not surjective). IOm pretty sure the
> example is known as the James tree space, if you
> want to try to find it somewhere.
Better would be The James Space without the trees.
This space X has the property that the canonical image of X
in X**
has codimension 1, but X is isometric to X**.
The James Tree Space repeats the James space along every
branch of a binary tree. It produces a Banach space Y
where Y is separable but Y** is non-separable. Just as in
the James space you get one extra dimension in X**, in Y you
get
one extra dimension for every branch of the tree.
> The exercise in Rudin
> shows that L^1 is not re?xive, but the James tree space
> shows that this is not the e as showing that L^1
> is not isometric to (L^1)**, which is what you said it
> said.
(Of course itOs not hard to show that (L^infty)* and L^1
> are not isometric,
It is hard in the sense that no ement of (L^infty)* not
in L^1 can actually be written down... They all live way out
in Axiom of Choice Land.
> but it has very little to do with what this
> exercise asks you to do.)
>
Perhaps, perhaps not. Depending on how you do this exercise
to produce
one functional, it may in fact produce uncountably many of
them all having distance 1 from each other.
==
http://www.math.ohio-state.edu/~edgar/
Honest, there actually is a difference: There exists a Banach
>> space X such that X _is_ isometric to X**, although X is
>> _not_ re?xive (which is to say that the canonical map
>> from X into X** is not surjective). IOm pretty sure the
>> example is known as the James tree space, if you
>> want to try to find it somewhere.
Better would be The James Space without the trees.
>This space X has the property that the canonical image of X
in X**
>has codimension 1, but X is isometric to X**.
>The James Tree Space repeats the James space along every
>branch of a binary tree. It produces a Banach space Y
>where Y is separable but Y** is non-separable. Just as in
>the James space you get one extra dimension in X**, in Y you
get
>one extra dimension for every branch of the tree.
Ok, .
>> The exercise in Rudin
>> shows that L^1 is not re?xive, but the James tree space
>> shows that this is not the e as showing that L^1
>> is not isometric to (L^1)**, which is what you said it
>> said.
(Of course itOs not hard to show that (L^infty)* and L^1
>> are not isometric,
It is hard in the sense that no ement of (L^infty)* not
>in L^1 can actually be written down... They all live way out
>in Axiom of Choice Land.
Wl roght - of course in the present context weOre assuming
the Hahn-Banach theorem, and as you point out bow
itOs not hard given that.
>> but it has very little to do with what this
>> exercise asks you to do.)
>>
Perhaps, perhaps not. Depending on how you do this exercise
to produce
>one functional, it may in fact produce uncountably many of
>them all having distance 1 from each other.
Wl, yes, I should have stopped with this is not what
the exercise asks you to do. For example, for any real
x there exists lambda_x in (L^infinity)* such that
lambda_x(f) = g(x+) - g(x-)
for all f such that f = g ae for some function g such that
the two one-sided limits g(x+) and g(x-) both exist. Now
letting f be the characteristic function of the interval
(x, infinity) shows that ||lambda_x - lambda_y|| >= 1 for
x <> y, hence (L^infinity)* is not separable.
************************
David C. Ullrich
nojb.
>
<arbitrary snip
>> The exercise in Rudin
>> shows that L^1 is not re?xive, but the James tree space
>> shows that this is not the e as showing that L^1
>> is not isometric to (L^1)**, which is what you said it
>> said.
>> (Of course itOs not hard to show that (L^infty)* and L^1
>> are not isometric,
>It is hard in the sense that no ement of (L^infty)* not
>in L^1 can actually be written down... They all live way out
>in Axiom of Choice Land.
Wl roght - of course in the present context weOre assuming
> the Hahn-Banach theorem, and as you point out bow
> itOs not hard given that.
> but it has very little to do with what this
>> exercise asks you to do.)
>Perhaps, perhaps not. Depending on how you do this exercise
to produce
>one functional, it may in fact produce uncountably many of
>them all having distance 1 from each other.
Wl, yes, I should have stopped with this is not what
> the exercise asks you to do. For example, for any real
> x there exists lambda_x in (L^infinity)* such that
lambda_x(f) = g(x+) - g(x-)
for all f such that f = g ae for some function g such that
> the two one-sided limits g(x+) and g(x-) both exist. Now
> letting f be the characteristic function of the interval
> (x, infinity) shows that ||lambda_x - lambda_y|| >= 1 for
> x <> y, hence (L^infinity)* is not separable.
> ************************
David C. Ullrich
> I yet again forgot the command to block messages by a
certain person, as
> rary do people annoy me. This guy obviously understands a
few
interesting
> features of quantum mechanics and string theory, as wl as
concepts of a
> sinularity-based collective conciousness, but this juvenile
continued
> obsessing on trivial details is closer to psychosis than
enlightenment,
in
> my opinion.
eusian wants to be famous one day. Satan uses his ego to
suppress
totality. he knows that i speak truth; that is why he is
angry at
these trivial details . .
>>BTW: How could a set of axioms of ZFC not be finite (since
there are
>>only finite many ZFC axioms in the first place)?
And what if
we use
>>all of the ZFC axioms for the subset? (unconfuse me)
>In spite of TorkOs response, the correct answer is that
there are
>infinity many axioms (the axioms of separation and
collections are
>both actually axiom schemas: there are infinity many of each,
as each is
>dependent on a set theoretic formula, and there are infinity
many such
>formulae).
The axiom of replacement is another such axiom shema, i.e.
the axiom of
replacement is also infinity many axioms.
David McAnally
Despite anything you may have heard to the contrary,
the rain in Spain stays almost invariably in the hills.
I am betting that this is easily proved somehow,...maybe ...
(I wonder if I posted it already a long while back...)
Let each S(0,m) = a(m), which is any rational you like, as
long as
a(m) is defined for every positive integer m, and each
{m!*a(m)} is an
integer.
Let S(n,m) = sum{j=1 to m} S(n-1,j)
for every positive integer n.
then (perhaps), if q and r are fixed integers where 0<= q < r:
m! *S(q,m)
is always congruent to
m! *S(r,m) (mod {r-q})
I may very wl have made a mistake, since I was caress;
but this result seems as if it *must* be a trivial
consequence of the
fact that:
S(n,m) =
sum{k=1 to m} binomial(m+n-k-1,n-1) * a(k).
But I mysf did not use this particular sum to get the result.
,
> I am betting that this is easily proved somehow,...maybe ...
(I wonder if I posted it already a long while back...)
Let each S(0,m) = a(m), which is any rational you like, as
long as
> a(m) is defined for every positive integer m, and each
{m!*a(m)} is an
> integer.
> Let S(n,m) = sum{j=1 to m} S(n-1,j)
for every positive integer n.
> then (perhaps), if q and r are fixed integers where 0<= q <
r:
m! *S(q,m)
is always congruent to
m! *S(r,m) (mod {r-q})
I may very wl have made a mistake, since I was caress;
> but this result seems as if it *must* be a trivial
consequence of the
> fact that:
S(n,m) =
 sum{k=1 to m} binomial(m+n-k-1,n-1) * a(k).
But I mysf did not use this particular sum to get the result.
,
>
Here is a proof using immediaty previous sum.
S(n,m) =
sum{k=1 to m} sum{j=0 to m-k} n^j *S(m-k,j) a(k) /(m-k)!,
where S() is an unsigned Stirling number of the 1st kind.
(All that is important here is that S() is an integer.)
So,
m!(S(r,m) - S(q,m)) =
sum sum S(m-k,j) *a(k) *k! *binomial(m,k) *(r^j -q^j).
So, if {a(k) *k!} = integer, each term of double sum is
integer multiplied
by
(r^j - q^j), and (r^j - q^j) is divisible by (r-q).
QED
,
Leroy
IOm taking an ectronics course, and there is a spot about
Instantaneous
Values of a Sine
Wave.
IOm stuck on a trig problem, I donOt remember
anything about
trig from
school.
a formula to solve for instantaneous value of current is:
i = Ip sin 0 (with the 0 having a dash through the center)
example:
What is the intantaneous value of a sinusoidal current at
25degrees if
Ip=2.2A (I is
current, p is peak, Ip is peak current measured in amps)
i = Ip sin 0
i = 2.2A sin 25degrees
i = 0.93A
Ummm, that is the example out of the text book, but I am
lost?!?! heheh,
need some hp.
I do have a scientific calculator with a SIN key, but I
canOt
figure it
out.
Any hp would be greatly appreciated.
Davis.
IOm taking an ectronics course, and there is a spot about
Instantaneous
Values of a
Sine
> Wave.
IOm stuck on a trig problem, I donOt remember
anything about
trig from
school.
> a formula to solve for instantaneous value of current is:
i = Ip sin 0 (with the 0 having a dash through the center)
example:
What is the intantaneous value of a sinusoidal current at
25degrees if
Ip=2.2A (I is
> current, p is peak, Ip is peak current measured in amps)
i = Ip sin 0
i = 2.2A sin 25degrees
i = 0.93A
Ummm, that is the example out of the text book, but I am
lost?!?! heheh,
need some hp.
> I do have a scientific calculator with a SIN key, but I
canOt figure it
out.
Any hp would be greatly appreciated.
Davis.
OK!!!!!!!!!!!!! I think I actually figured it out, after about
an hour of
trying and
trying...
each degree of a cycle = .833ms
So, 25degrees x .833ms = 20.825
i = Ip sin 0
i = 2.2A sin 25 degrees (which means 2.2/20.825 and then
press COS on my
calculator)
got me .93A?!?!?
I think IOve got it!
Imagine what life would be like without the internet,
ummmmmm........
> OK!!!!!!!!!!!!! I think I actually figured it out, after
about an hour of
trying and
> trying...
each degree of a cycle = .833ms
So, 25degrees x .833ms = 20.825
i = Ip sin 0
i = 2.2A sin 25 degrees (which means 2.2/20.825 and then
press COS on my
calculator)
got me .93A?!?!?
>
OK, that is all wrong.
Enter 25 press SIN, then multiply x 2.2A . Got it!! hahah, I
fe better
now.
> An obvious counter-example is v(t) = v_0 sin(wt). In the
> fourth quadrant, the vocity is obviously increasing, since
> itOs going from more negative to less negative values
because
> the acceration is positive. Since v is negative and a is
> positive, v.a is obviously negative.
[PREVIOUS POST HAD TYPO] vocity _magnitude_ is decreasing, so
it is
decerating, so your counter-example is false.
John Schoenfd:
>> An obvious counter-example is v(t) = v_0 sin(wt). In the
>> fourth quadrant, the vocity is obviously increasing, since
>> itOs going from more negative to less negative values
because
>> the acceration is positive. Since v is negative and a is
>> positive, v.a is obviously negative.
[PREVIOUS POST HAD TYPO] vocity _magnitude_ is decreasing,
so it is
>decerating, so your counter-example is false.
Magnitude is not sufficiemt to define a vector.
Vector
functions have a
direction, too. Why do you think I told you to take the
derivative of v^2
and avoid the silliness of your assertion about vector
functions? Just
> John Schoenfd:
>> An obvious counter-example is v(t) = v_0 sin(wt). In the
>> fourth quadrant, the vocity is obviously increasing, since
>> itOs going from more negative to less negative values
because
>> the acceration is positive. Since v is negative and a is
>> positive, v.a is obviously negative.
>[PREVIOUS POST HAD TYPO] vocity _magnitude_ is decreasing,
so it is
>decerating, so your counter-example is false.
Magnitude is not sufficiemt to define a vector.
Vector
functions have a
> direction, too. Why do you think I told you to take the
derivative of v^2
> and avoid the silliness of your assertion about vector
functions?
Just
I didnOt realize I was dealing with either,
1. A kid
2. A spastic
3. A crackpot
JS
>Here is the proof that v.a < 0 = deceration.
 An obvious counter-example is v(t) = v_0 sin(wt). In the
> fourth quadrant, the vocity is obviously increasing, since
> itOs going from more negative to less negative values
because
> the acceration is positive. Since v is negative and a is
> positive, v.a is obviously negative.
I am shocked to see you make this blunder, Bilge.
If the vocity is going from more negative numbers to less
negative
numbers then the vocity is decreasing as it is approaching 0.
Your counter-example is blindingly false.
>let v = vocity vector
>let a = acceration vector
>We know that decleration occurs when d|v|/dt is a negative
vector.
>If vector a0 is the projection of a onto v, then a0 =
d|v|/dt.
>a0 = [(a.v) / |v|] <v
> YouOve attempted to obtain a simple result by starting
with
a
> lot of nonsense about vector-valued functions and obtained
one
> which is incorrect (or at best, stated awkwardly and in
> con?ct with your first assertion about v.a).
If you bieve that I have posted nonsense, do your credibility
a
favour and show precisy where the mathematical error lies in
the
proof given bow.
> You canOt discard the sign if you are talk about
vector-valued
> functions.
You need to point out exactly where the sign was being
discarded, not
just claim it is being discarded.
> Your absolute value signs are an attempt to use v^2 without
> acknowledging that you are really talking about v.a
representing
> the slope of the scalar, v^2.
The absolute signs are derived from the fact that
(v.a) = |v| |a| cos(x).
Or perhaps you have a problem with vector projection. Do you
know what
that is, Bilge? (because it sure doesnOt look like based on
what you
are saying).
========================
PROOF:
let v(t) = vocity vector function
let a(t) = acceration vector function
Theorem:
|v(t)| is decreasing if (v(t).a(t)) < 0.
We know that |v(t)| is decreasing if d|v|/dt < 0.
let a0(t) = projection of a(t) onto v(t) = (v(t).a(t))/|v(t)|
<v(t)>
So,
d|v|/dt = |a0(t)|
= (v(t) . a(t)) / |v(t)|
Since |v(t)| > 0, the only way d|v|/dt < 0 is iff v(t) . a(t)
< 0.
So there we have it, proof.
================================
We can further determine that for v(t) . a(t) < 0, the angle
between v(t) and a(t) must be interior to the interval [pi/2,
3pi/2]
v(t) . a(t) = |v(t)| |a(t)| cos(x)
Since |v(t)| > 0 and |a(t)| > 0, the only way v(t) . a(t) < 0
is iff
cos(x) < 0, thus x must be interior to the interval [pi/2,
3pi/2].
Point out the mathematical error(s), Bilge.
JS
John Schoenfd:
>>Here is the proof that v.a < 0 = deceration.
>> An obvious counter-example is v(t) = v_0 sin(wt). In the
>> fourth quadrant, the vocity is obviously increasing, since
>> itOs going from more negative to less negative values
because
>> the acceration is positive. Since v is negative and a is
>> positive, v.a is obviously negative.
I am shocked to see you make this blunder, Bilge.
If the vocity is going from more negative numbers to less
negative
>numbers then the vocity is decreasing as it is approaching 0.
Look, youOre the one who chose to say vector function rather
than
use v^2. A vocity which goes from a larger negative value to
a smaller
one is an increasing vocity. Just draw a line for such a
vocity and
youOll see it has a positive slope.
>Your counter-example is blindingly false.
All you have to do key a few values in from your calculator
and
it will obvious that what I said is correct. If I have write
down
the values of the sine, cosine for you, IOm not going to
very
polite about it.
>let v = vocity vector
>>let a = acceration vector
>>We know that decleration occurs when d|v|/dt is a negative
vector.
>>If vector a0 is the projection of a onto v, then a0 =
d|v|/dt.
>>a0 = [(a.v) / |v|] <v> YouOve attempted to obtain a simple result by starting
with a
>> lot of nonsense about vector-valued functions and obtained
one
>> which is incorrect (or at best, stated awkwardly and in
>> con?ct with your first assertion about v.a).
If you bieve that I have posted nonsense, do your
credibility a
>favour and show precisy where the mathematical error lies in
the
>proof given bow.
I already gave a counter example. That alone is sufficient to
falsify a proof. If your proof was correct, there would be no
counter example.
>> You canOt discard the sign if you are talk about
vector-valued
>> functions.
You need to point out exactly where the sign was being
discarded, not
>just claim it is being discarded.
Look idiot. I did exactly that in the _VERY NEXT PARAGRAPH_.
READ
WHATOS WRITTEN BEFORE MINDLESSLY POUNDING THE KEYBOARD AND
ASKING
ME TO REPEAT IT.
> Your absolute value signs are an attempt to use v^2 without
>> acknowledging that you are really talking about v.a
representing
>> the slope of the scalar, v^2.
The absolute signs are derived from the fact that
>(v.a) = |v| |a| cos(x).
v(t) = v_0 + at v_0 = -5m/sec, a = 1m/sec^2
at time t = 1, v = -5 m/sec + (1m/sec^2) 1 sec = -4m/sec
dv/dt = a
which IOve already defined to be positive. Since
v and a are
in opposite
directions, cos(theta) = cos(pi) = -1, v.a = -1 |v||a|, which
is
negative and the vocity is increasing.
>Or perhaps you have a problem with vector projection. Do you
know what
>that is, Bilge? (because it sure doesnOt look like based on
what you
>are saying).
YouOve apparently been studying pmbOs replies
since youOre
willing
to go to any length to deny whatOs staring you in face.
> John Schoenfd:
>>Here is the proof that v.a < 0 = deceration.
>> An obvious counter-example is v(t) = v_0 sin(wt). In the
>> fourth quadrant, the vocity is obviously increasing, since
>> itOs going from more negative to less negative values
because
>> the acceration is positive. Since v is negative and a is
>> positive, v.a is obviously negative.
>I am shocked to see you make this blunder, Bilge.
>If the vocity is going from more negative numbers to less
negative
>numbers then the vocity is decreasing as it is approaching 0.
 Look, youOre the one who chose to say vector function
rather than
> use v^2. A vocity which goes from a larger negative value
to a smaller
> one is an increasing vocity. Just draw a line for such a
vocity and
> youOll see it has a positive slope.
I find it unbievable that you persist with your ridiculous
position
that
v(t)=sin(t) is somehow a counter-example to v.a<0 being
deceration.
Listen idiot, something decerates when d|v|/dt < 0
d|v|/dt = (v.a)/|v|
So clearly, IF d|v|/dt < 0 THEN (v.a) < 0.
Your counter-example is perfectly explained with my
definition.
v = sin(x)
Lets do some calculations for when x is in the fourth
quadrant.
v(270) = sin(270) = -1
v(271) = sin(271) = -0.99
v(300) = sin(300) = -0.88
Now i ask of you to CONSIDER SLOWLY:
Are we speeding up or slowing down?
Clearly, it is slowing down. So as expected, d|v|/dt < 0
because (v.a)
< 0, so the conclusion is that the motion is DECERATING IN THE
FOURTH QUADRANT.
>Your counter-example is blindingly false.
 All you have to do key a few values in from your calculator
and
> it will obvious that what I said is correct. If I have
write down
> the values of the sine, cosine for you, IOm not going to
very
> polite about it.
Do you see why your position is not only absurd, but worthy
of it
being immortalized on those crank web sites?
I used to think you knew what you were talking about, but if
you make
these kinds of errors on such a basic topic, i suspect you
make all
sorts of errors on all other topics.
>let v = vocity vector
>let a = acceration vector
>We know that decleration occurs when d|v|/dt is a
negative vector.
>If vector a0 is the projection of a onto v, then a0 =
d|v|/dt.
>a0 = [(a.v) / |v|] <v >> YouOve attempted to obtain a simple result by starting
with a
>> lot of nonsense about vector-valued functions and obtained
one
>> which is incorrect (or at best, stated awkwardly and in
>> con?ct with your first assertion about v.a).
>If you bieve that I have posted nonsense, do your
credibility a
>favour and show precisy where the mathematical error lies in
the
>proof given bow.
 I already gave a counter example. That alone is sufficient
to
> falsify a proof. If your proof was correct, there would be
no
> counter example.
Your counter-example supports my argument, because the SPEED
IS
DECREASING AND VOCITY DECERATING.
Lets go through the proof again:
THEOREM: v(t).a(t) < 0 denotes deceration.
PROOF:
Deceration is when d|v|/dt < 0
d|v|/dt = (v(t).a(t)) / |v(t)|
So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0
Do you understand this blindingly simple logic yet?
>> Your absolute value signs are an attempt to use v^2 without
>> acknowledging that you are really talking about v.a
representing
>> the slope of the scalar, v^2.
>The absolute signs are derived from the fact that
>(v.a) = |v| |a| cos(x).
 v(t) = v_0 + at v_0 = -5m/sec, a = 1m/sec^2
at time t = 1, v = -5 m/sec + (1m/sec^2) 1 sec = -4m/sec
dv/dt = a
 which IOve already defined to be positive.
Since v and a are
in opposite
> directions, cos(theta) = cos(pi) = -1, v.a = -1 |v||a|,
which is
> negative and the vocity is increasing.
What about the magnitude of the vocity as wl, genious?
Of course it is not, because it is decerating and therefore
your
counter-example is totally and utter agreement with my
theorem.
An increase in vocity does NOT imply an increase in vocity
magnitude.
>Or perhaps you have a problem with vector projection. Do you
know what
>that is, Bilge? (because it sure doesnOt look like based on
what you
>are saying).
 YouOve apparently been studying pmbOs
replies since youOre
willing
> to go to any length to deny whatOs staring you in face.
Point out the error, imbecile.
THEOREM: v(t).a(t) < 0 denotes deceration.
PROOF:
Deceration is when d|v|/dt < 0
d|v|/dt = (v(t).a(t)) / |v(t)|
So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0
Do you understand this blindingly simple logic yet?
Did you get that, crackpot? Let me post it again just incase
you
forget to quote it and forget to disprove it.
THEOREM: v(t).a(t) < 0 denotes deceration.
PROOF:
Deceration is when d|v|/dt < 0
d|v|/dt = (v(t).a(t)) / |v(t)|
So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0
Do you understand this blindingly simple logic yet?
THEOREM: v(t).a(t) < 0 denotes deceration.
PROOF:
Deceration is when d|v|/dt < 0
d|v|/dt = (v(t).a(t)) / |v(t)|
So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0
Do you understand this blindingly simple logic yet?
THEOREM: v(t).a(t) < 0 denotes deceration.
PROOF:
Deceration is when d|v|/dt < 0
d|v|/dt = (v(t).a(t)) / |v(t)|
So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0
Do you understand this blindingly simple logic yet?
THEOREM: v(t).a(t) < 0 denotes deceration.
PROOF:
Deceration is when d|v|/dt < 0
d|v|/dt = (v(t).a(t)) / |v(t)|
So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0
Do you understand this blindingly simple logic yet?
THEOREM: v(t).a(t) < 0 denotes deceration.
PROOF:
Deceration is when d|v|/dt < 0
d|v|/dt = (v(t).a(t)) / |v(t)|
So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0
Do you understand this blindingly simple logic yet?
THEOREM: v(t).a(t) < 0 denotes deceration.
PROOF:
Deceration is when d|v|/dt < 0
d|v|/dt = (v(t).a(t)) / |v(t)|
So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0
Do you understand this blindingly simple logic yet?
THEOREM: v(t).a(t) < 0 denotes deceration.
PROOF:
Deceration is when d|v|/dt < 0
d|v|/dt = (v(t).a(t)) / |v(t)|
So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0
Do you understand this blindingly simple logic yet?
THEOREM: v(t).a(t) < 0 denotes deceration.
PROOF:
Deceration is when d|v|/dt < 0
d|v|/dt = (v(t).a(t)) / |v(t)|
So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0
Do you understand this blindingly simple logic yet?
Where is the error?
JS
> YouOve apparently been studying pmbOs
replies since youOre
willing
> to go to any length to deny whatOs staring you in face.
IOm sorry to inform you bilge but that is a characteristic
that you
have. It is not one that I possess. In fact youOve
demonstrated a lack
of reading comprehension. However this is probably a direct
result of
your arrogance in that you simply refuse to pay close enough
attention
to what youOre reading before you quickly jump to your usual
?me-mode. You prove that in almost every single post that
IOve ever
seen you write in our discussions.
Previous post had a typo:
d|v|/dt = (v.a)/|v|, in scalar form
d|v|/dt <v> = (v.a)/|v| <v>, in vector form
> Here is the proof that v.a < 0 = deceration.
> let v = vocity vector
> let a = acceration vector
> We know that decleration occurs when d|v|/dt is a
negative vector.
Not true. d|v|/dt is a *scalar*. So unless you have qualms, I
will
assume
> the above sentence says is negative.
IOve just shown you that if you project the acceration
vector
a onto
the vocity vector v, the resultant vector is d|v|/dt and it is
ear with v. So d|v|/dt is certainly a vector and is defined
as:
d|v|/dt = (v.a) / |a| <v > If vector a0 is the projection of
a onto v, then a0 = d|v|/dt.
Not true. *All* the change in v is not a *limited* change in
v. You have
> constrained a to be ear with v, for all further
consideration. It
> amounts to v x a = 0, as I had done. It is no more general.
David, pay attention closy.
I have not constrained acceration to be ear with vocity, I
have PROJECTED the acceration onto the vocity, and the
resultant
vector is the revant component of a that is ear with v.
> Your proof is DOA.
Do your credibility a favour and point out the mathematical
error. All
you have done is claimed
1. you have constrained a to be ear with v, which is false.
> d|v|/dt < 0 is still deceration, is egant, and sufficient.
And
you
> canOt patent it.
d|v|/dt = (v.a) / |a| <v>
Transform to 1-dimension where <v> is positive axis.
d|v|/dt = (v.a) / |a|
Any competent mathematician can see the only way d|v|/dt can
be
negative is when the angle between v and a is interior to the
interval
[pi/2, 3pi/2].
>
JS
>Here is the proof that v.a < 0 = deceration.
> let v = vocity vector
> let a = acceration vector
> We know that decleration occurs when d|v|/dt is a
negative vector.
> Not true. d|v|/dt is a *scalar*. So unless you have
qualms, I will
assume
> the above sentence says is negative.
IOve just shown you that if you project the acceration
vector
a onto
> the vocity vector v, the resultant vector is d|v|/dt and it
is
> ear with v. So d|v|/dt is certainly a vector and is defined
as:
d|v|/dt = (v.a) / |a| <v>
Accepted with your posted correction.
>If vector a0 is the projection of a onto v, then a0 =
d|v|/dt.
> Not true. *All* the change in v is not a *limited* change
in v. You
have
> constrained a to be ear with v, for all further
consideration. It
> amounts to v x a = 0, as I had done. It is no more general.
David, pay attention closy.
I have not constrained acceration to be ear with vocity, I
> have PROJECTED the acceration onto the vocity,
... or vice versa...
> and the resultant
> vector is the revant component of a that is ear with v.
> Your proof is DOA.
Do your credibility a favour and point out the mathematical
error. All
> you have done is claimed
> 1. you have constrained a to be ear with v, which is false.
I am not up to this challenge. I will stand down and study.
Perhaps next
time I will be a better antagonist.
> what are the sums of the measures of the angles in a convex
29-gon?
(29*180)-360
Let G(0,m) =
(1 +1/2 +1/3 +...+1/m) /(1+m)
for every positive integer m.
Let G(n,m) = sum{j=1 to m} G(n-1,j)
for every positive integer n.
And let
H(0,m) = 1/m
for every positive integer m,
and H(n,m) = sum{j=1 to m} H(n-1,m)
for every positive integer n.
(So, G(0,m) = H(1,m)/(1+m).
And H(1,m) = the m_th harmonic number.)
Then, for q and r = any fixed nonnegative integers:
sum{k=1 to m} H(r,k) *H(k+q+r,m+1-k) *(-1)^(1+k)
= 2 *G(q,m).
(I apologize if this result has already been posted by me
years ago.)
(It IS true, is it not??)
==

Consider, for m and n = positive integers:
h(n,m) =
sum{j=0 to ?or((m-1)/n)} 1/(m -j*n),
which is
1/m for m <= n, h(n,m) = h(n,m-n) + 1/m;
and
h(n,m) =
1/m + 1/(m-n) + 1/(m-2n) + 1/(m-3n) +....,
the sum continuing as long as the denominator is a positive
integer.
(So, h(1,m) = H(m), the standard harmonic number.)
One result:
sum{k=1 to m} h(n,k) =
(-m + sum{j=1 to n} h(n,m+j) *(m+j)) /n - 1.
Another result, a congruence involving h(2,m) (and h(1,m)),
is at:
threadm=b4be2fdf
For one thing, can the congruence in the linked-to post be
generalized
for other nOs?
What other interesting things can be said about these
h(n,m)Os?
(I already know that h(n,mn) = H(m)/n.)
==
suggested to consider, for positive integers m and n,
h(n,m) = sum{j=0..?or((m-1)/n)} 1/(m-j*n),
and noted in particular the rations h(1,m)=H(m),
h(n,n*m)=H(m)/n,
where
H(m) = sum{j=1..m} 1/j
are the standard harmonic numbers.
I would like to offer the following intriguing sums,
sum{n=1..m} mu(n) * h(n,m+1-n) = 1/m,
sum{n=1..m} phi(n) * h(n,m+1-n) = (m+1)*H(m)-m,
sum{n=1..m} n^k * h(n,m+1-n) = sum{n=1..m} sigma[k](n)/(m+1-n)
where mu(n) is the Moebius function, phi(n) is the Euler
totient
function, and sigma[k](n) is the sum of k-th powers of the
divisors of
n, for k>=0. These formulas can be derived from the generating
function
sum{m=1..oo} h(n,m) z^m = -ln(1-z)/(1-z^n)
by using the Lambert series for mu(n) etc, and comparing
coefficients
of powers of z. The third sum can also be directly verified by
substituting sigma[k](n)=sum{d|n}d^k on the right-hand side
and
exchanging the order of summations.
http://mathworld.wolfram.com/MoebiusFunction.html
http://mathworld.wolfram.com/TotientFunction.html
http://mathworld.wolfram.com/DivisorFunction.html
http://mathworld.wolfram.com/LambertSeries.html
Icke
Paola Zizzi, in several visionary papers that by a miracle
escaped
GinspargOs censor, proposes that Hawkings
Mind of GOD is conscious,
but alas only for a brief moment. This is an idea not
unfamiliar to my
own thinking. However, I suspect GOD(D) (I.J. Good) is
conscious
continually and the large-scale structures (voids, walls,
filaments) of
weaved dark energy/dark matter exotic zero point energy vacua
we see are
like clular structures (Solaris, VALIS, Star Maker, Black
Cloud et-al
in Sci Fi) However we have some significant differences of
principle
around the issue of signal locality or no cloning a qubit.
Excerpts from:
EMERGENT CONSCIOUSNESS: FROM THE EARLY UNIVERSE
TO OUR MIND
P. A. Zizzi
Dipartimento di Astronomia dlO Universit.88 di Padova
Vicolo dlO Osservatorio, 5
35122 Padova, Italy
zizzi@giunone.pd.astro.it
==
In a previous paper (gr-qc/9907063) we described the early
inflationary
universe in terms
of quantum information. In this paper, we analyze those
results in more
detail, and we
stress the fact that, during in?tion, the universe can be
described as
a superposed state of
quantum registers. The sf-reduction of the superposed quantum
state is
consistent with
the Penroses Objective Reduction (OR) mod.
The quantum gravity threshold is reached at the end of
in?tion, and
corresponds to a
superposed state of 10^9 quantum registers. This is also the
number of
superposed
tubulins-qubits in our brain, which undergo the
Penrose-Hameroffs
Orchestrated
Objective Reduction, (Orch OR), leading to a conscious event.
Then, an
analogy naturally arises between the very early quantum
computing universe,
and our mind.
2
1. INTRODUCTION
What is consciousness? Everybody knows about his own
consciousness, but
it is
impossible to communicate our subjective knowledge of it to
others.
Moreover, a
complete scientific definition of consciousness is
still
missing.
However, quite
recently, it has been realized that the study of
consciousness should
not be restricted to
the fids of cognitive science, philosophy and biology, but
enlarged to
physics, more
precisy, to quantum physics.
The most popular (and conventional) description of
consciousness is
based on the
classical computing activities in the brains neural
networks,
corrated with mental
states. In that picture, mind and brain are identified, and
are compared
to a classical
computer. That approach (see for example [1]) is called in
various ways:
physicalism,
reductionism, materialism, functionalism, computationalism.
However, although the brain can actually support classical
computation,
there is an
ement of consciousness which is non-computable (in the
classical
sense), as it was
shown by Penrose [2]. Moreover, the seminal paper by Stapp
[3] clarified
why
classical mechanics cannot accommodate consciousness, but
quantum
mechanics can.
Finally, reductionism cannot explain the hard problem of
consciousness, which deals
with our inner life, as it was illustrated by Chalmers [4].
A quite different line of though about consciousness is the
one which
comprises
pamnpsychism, pan-experientalism, idealism, and
funda-mentalism.
Pan-experientalism states that consciousness (or better
proto-consciousness) is
intrinsically unfolded in the universe, and that our mind can
grasp
those proto-conscious
experiences. This line of though goes back to Democritus,
Spinoza [5],
Leibniz [6],
until Whitehead [7] who re-interpreted the Leibnizs
monads as
occasions of
experience. Shimony [8] compared Whiteheads occasions of
experience to
quantum
jumps. More recently, Penrose interpreted the occasions of
experience as
the quantum state reductions occurring at the Planck scale,
where spin
networks [9] encode protoconsciousness.
This is a pan-experiential approach to consciousness which is
consistent
with quantum gravity, and is called Objective Reduction (OR)
[2]. A
further
devopment is the Penrose-Hameroff Orchestrated Objective
Reduction
(Orch OR)
[10] which deals with the sf-collapse of superposed tubulins
in the
brain. Superposed
tubulins are qubits, and perform quantum computation, until
they reach
the quantum
gravity threshold, then they collapse to classical bits,
giving rise to
a conscious event.
Finally, Chalmers [4] claimed that physical systems which
share the e
organization
will lead to the e kind of conscious experience (Principle of
Invariance). As physical systems, which have the e
organization (no
matter what
they are made of ) encompass the e information, it follows,
from the
above
principle, that information is the source of consciousness.
The present paper is on the e line with Chalmers
conclusions. This
is of course valid also for an immaterial system, like the
vacuum-dominated early in?tionary universe which, as it was
shown in
[11], is a superposed quantum state of qubits. At this point,
a
conjecture arises very naturally: the early universe had a
conscious
experience at the end of in?tion, when the superposed
quantum state of
10^9 = n quantum gravity registers underwent Objective
Reduction. The
striking point is that this value of n equals the number of
superposed
tubulins-qubits in our brain, which undergo Orchestrated
Objective
Reduction, leading to a conscious event. Then, we make the
conjecture
that the early universe and our mind share the e organization,
encompass the e quantum information, and undergo similar
conscious
experiences. In other words, consciousness might have a
cosmic origin,
with roots in the pre-consciousness ingrained directly from
the Planck
time.
The paper is organized as follows. In Sect.2, we describe the
early
in?tionary
universe as a superposed state of quantum gravity registers.
In Sect.3,
we show that a quantum gravity register follows some
cybernetic
together with the no-cloning theorem, and Chalmers Principle
of
Computational Modes which begets consciousness. In Sect.4, we
claim
that the superposed state of quantum gravity registers
undergoes
sf-reduction as in the Penroses OR mod at the end of
in?tion, and
show that this fact is responsible for the actual entropy of
our universe.
In Sect.5, we shortly review the Orch OR mod, pointing out
that the
number of
tubulins involved in the Orch OR, equals the number of
quantum gravity
registers,
involved in the OR, and we make the conjecture that the
universe might
have achieved
consciousness at the end of in?tion.
In Sect.6, we interpret the Boolean observer as a necessary
product of
the postin?tionary
universe. Moreover, we show that the functor Past can be
defined only
once a particular quantum gravity register is sected. Sect.7
is
devoted to some concluding remarks.
2. QUANTUM GRAVITY REGISTERS
Go to original paper e.g.
http://www.consciousness.arizona.edu/hameroff/New/Big_Wow/Big
_Wow.htm
http://arxiv.org/abs/gr-qc/0007006
http://arxiv.org/abs/gr-qc/9907063
3. QUANTUM GRAVITY COMPUTATION AND CYBERNETIC
As we pointed out in Sect.2, the quantum gravity registers
are not
proper quantum
computers. Basically, they resemble quantum clular automata,
as time
is discrete.
Quantum gravity registers do perform quantum computation, but
in a
rather particular
way, that we shall call quantum gravity computation (QGC). The
peculiarity of a
quantum system which performs QGC, is that it shares some
features of
sf-organizing
systems. We recall that sf-organization is a process of
evolution
taking place basically inside the system, with minimal or
even null
effect of the environment. In fact, the dynamical behaviour
of a quantum
gravity register follows some cybernetic
principles:
i) Autocatalytic growth
At each computational time step, the presence of a Planckian
black hole
(which acts as
a creation operator), makes the quantum gravity register grow
autocatalytically.
As N qubits represent here a de Sitter horizon with an area
of N pixs,
the
autocatalytic growth, in this case, is exponential expansion,
i.e.,
in?tion.
ii) Autopoiesis (or sf-production)
The quantum gravity register reproduces itsf. The components
of the
quantum gravity
register generate recursivy the e network of processes
(applications of the
Hadamard gate to the vacuum state) which produced them.
In this case recursion is defining the program in such a way
that it may
call itsf:
.
This is on the e line of thought of KauffmannOs Fourth Law [
15]:
The
hypothesis that the universe as a whole might be a
sf-constructing
coevolving
community of autonomous agents that maximizes the sustainable
growth.
For Kauffmann, the autonomous agents are knotted structures
created of
spin networks
which act on one another and become collectivy autocatalytic.
The
picture given in
this paper and the KauffmannO s picture, are equivalent,
because spin
networks pierce
the de Sitter horizonsO surfaces [12].
iii) Sf-similarity
This mod of the early in?tionary universe is based on the
holographic principle
[16], more in particular, on the quantum holographic
principle [11]
[17]. But each part
of a hologram carries information about the whole hologram.
So, there is
a physical
correspondence between the parts and the whole.
iv) Sf-reproduction
Can a quantum gravity register, as a unit, produce another
unit with a
similar
organization? This possibility, which could be taken into
account
because the quantum
gravity register is an autopoietic system, (and only
autopoietic systems
can sf-reproduce), is in fact forbidden by the no-cloning
theorem [18]
(quantum information
cannot be copied).
Sarfatti Commentary: Only if you assume signal
locality, i.e.
sub-quantal heat death (Antony Valentini
http://www.edge.org/3rd_culture/bios/valentini.html
http://www.edge.org/discourse/information.html
http://www.fourmilab.ch/rpkp/valentini.html
http://arxiv.org/abs/quant-ph/0112151
PZ: However, there is a way out. When the sected quantum
gravity
register collapses to
classical bits, it is not just an ordinary quantum register
which
collapses, but an
autopoietic one. The outcomes (classical bits) carry along the
autopoiesis. The
resulting classical automaton is then autopoietic and, in
principle, can
sf-reproduce.
Moreover, as it was pointed out in the Introduction, the
Chalmers
Principle of
organization, the e amount of information, and the e conscious
experience of
the original one.
and from the
no-cloning theorem, we get The principle of alternating
computational
modes: A unit
produced by an autopoietic classical computing system built
up from the
outcomes of a
decohered quantum autopoietic system, shares the e
organization, the
e
amount of information, and the e conscious experience of the
producing unit.
Moreover, in order to share the e conscious experience of the
decohered quantum
system, the produced unit must alternate quantum and classical
computational modes
at least once.
JS: This principle is subject to doubt because it posits the
signal
locality of micro-quantum theory. The latter may be to
quantum gravity
as special rativity is to general rativity. On the other
hand, Zizzi
may prove correct. It is too soon to decide.
PZ: The above arguments are summarized in the following
scheme:
Autopoietic quantum register
no-cloning theorem
no sf-reproduction
decoherence
autopoietic classical clular automaton
sfreproduction
produced unit with the e organization
principle of organizational invariance
the produced unit shares the e information content, and the e
conscious experience
the produced unit gets both quantum and classical
computational modes,
the former from the autopoietic quantum register, the latter
from the
autopoietic classical clular automata
the modes alternate to each other.
Then, we are lead to make the conjecture that the final
outcome of a
quantum gravity
register might be a brain. In fact, tubulins in the brain
alternate
classical and quantum
computational modes [10 ].
A rated paper on the issue of a cybernetic approach to
consciousness
can be found in
[19].
4. OBJECTIVE REDUCTION AND DECOHERENCE
The superposed state of quantum gravity registers represents
the early
in?tionary
universe which is a closed system. Obviously then, the
superposed
quantum state
cannot undergo environmental decoherence.
However, we know that at the end of the in?tionary epoch,
the universe
reheated by
getting energy from the vacuum, and started to be
radiation-dominated
becoming a
Friedmann universe.
This phase transition should correspond to decoherence of the
superposed
quantum
state. The only possible reduction mod in this case is
sf-reduction,
as in the OR
mod [2] which invokes quantum gravity.
4.1 The discrete energy spectrum and the quantum gravity
threshold
Go to original paper http://arxiv.org/abs/gr-qc/0007006
4.2 Entanglement with the environment
The superposed state of 10^18 qubits will collapse to
classical bits by
getting entangled
with the emergent environment (radiation).
Sarfatti Commentary: I independently got 10^18 qubit
consciousness
threshold and it is published in my two books from 2002
Destiny Matrix
and Space-Time and Beyond II. I am not sure
if I still bieve my
original cosmology argument for it. However, Hameroff told me
that there
are 10^18 hydrophobic caged single ectron qubits in the human
brain. I
am unclear how Paola jumped from 10^9 to 10^18? I suppose she
means a
macro-coherence effect ~ N^2 where N = 10^9 that fits my own
ideas. On
the other hand the generalized phase
rigidity of a giant ODLRO cosmic
vacuum wave will be immune to decoherence from the CMB so I
do not
really buy, as yet, what Paola proposes here in detail. She
may,
however, be correctly applying Penroses ideas?
PZ: This entanglement process with the environment can be
interpreted
as the action of a XOR (or controlled NOT) gate, as it was
illustrated
in [11], which gives the output of
the quantum computation in terms of classical bits: the
source of
classical information
in the post-in?tionary universe.
4.3 Holography and Clular Automata
Clular automata (CA) were originally conceived by von Neumann
[20], to
provide a
mathematical framework for the study of complex systems.
go to original
The rules force patterns to emerge (sf-organization). By
taking into
account the classical holographic principle, we are lead to
bieve
that at the end of in?tion, the universe starts to behave as
a
classical CA which sf- organizes and evolves complexity and
structure.
We call it Classical Holographic Clular Automata (CHCA).
It should be noted that the CHCA is made out of the bits
which are the
outcomes of the
collapse of the qubits of the quantum gravity register which
is an
autopoietic quantum
system. Then, the CHCA is an autopoietic classical system.
There are two important consequences.
i) The CHCA, being autopoietic, undergoes autocatalytic
growth, and the
classical
universe is still expanding. However, as classical
computation is slower
than quantum
computation, the expansion is not anymore exponential
(post-in?tionary
universe).
ii) The CHCA, beeing a classical autopoietic system, can
sf-reproduce.
According to the Principle of alternating computational modes
discussed in Sect.3,
the produced units will be able to perform both quantum and
classical
computation.
We conclude by saying that in our mod, the post-in?tionary,
classically holographic
universe, follows the laws of classical complex adaptive
systems
(systems at the edge
of chaos).
5. CONSCIOUSNESS AND TUBULINS/QUBITS
So far, consciousness was studied in the context of
neuroscience, and
was described as
an emergent feature of classical computing in the brainOs
neural networks.
But neuroscience fails to explain some features of
consciousness as, for
example,
subjective experience (ChalmersO hard problem [4]).
A new, different approach to the study of consciousness is
due to
Penrose and
Hameroff [10] and it is based on quantum effects occurring in
tubulins.
In a brainO s neuron there is the cytosketon, which is made
of protein
networks. The
most important components of the cytosketon are microtubules.
Microtubules are
hollow cylindrical polymers of proteins called
tubulins.Tubulins are
dipoles and they
can be in (at least) two different states (or conformations).
Tubulins have been studied in classical computing. In fact
simulations
suggest that
tubulins behave as a classical CA.
But tubulins can also be in a superposition of the two (or
more)
conformation states.
Sarfatti Commentary: This last assertion is in doubt because
of Max
Tegmarks work.
http://arxiv.org/abs/quant-ph/9907009
http://www.hep.upenn.edu/~max/
Author:
Max Tegmark
Abstract:
Based on a calculation of neural decoherence rates, we argue
that that
the degrees of freedom of the human brain that rate to
cognitive
processes should be thought of as a classical rather than
quantum
system, ie, that there is nothing fundamentally wrong with
the current
classical approach to neural network simulations. We find that
the
decoherence timescales ~10^{-13}-10^{-20} seconds are
typically much
shorter than the revant dynamical timescales (~0.001-0.1
seconds),
both for regular neuron firing and for kink-like polarization
excitations in microtubules. This conclusion disagrees with
suggestions
by Penrose and others that the brain acts as a quantum
computer. Our
result is also difficult to reconcile with the Stapp mod where
thought
processes correspond to top-lev multi-neuron quantum events.
To place this problem in context, we also discuss a
decomposition of the
global density matrix into three subsystems as in the figure
on the
right, where the three interaction Hamiltonians cause
qualitativy
different effects.
Rated papers:
Reference info: quant-ph/9907009
Rated papers:
This site also contains the latest versions of some closy
rated
papers of mine:
Tegmark 1993 describes how decoherence looks like, fes like
and
smls like wavefunction collapse, thereby iminating one of the
main
motivations for the Copenhagen interpretation.
Tegmark 1998 describes the so-called many-worlds
interpretation of
quantum mechanics.
Tegmark 1997 proposes a theory with an even larger ensemble,
making
even the many worlds of quantum mechanics seem plain and
close to home
in comparison!
Tegmark & Yeh 1994 and Tegmark & Shapiro 1994 are marginally
rated,
giving explicit examples of the effects interaction with the
environment.
Tegmark 1996 is also marginally rated, discussing how
decoherence
can be measured in practice.
PZ: In this case they are qubits, and they behave as a
biological
quantum clular automata . Indeed, tubulins can perform both
classical
and quantum computing.
In a classical computing mode, patterns of tubulins move,
evolve,
interact and lead to
new patterns. Quantum coherence emerges from resonance in
classical
patterns.
Sarfatti Commentary: I would like to see the math of that
interesting
remark.
PZ: When the quantum gravity threshold is reached,
sf-collapse occurs
and the
eigenstate evolves as a classical CA. In the orchestrated
objective
reduction (Orch OR) mod by Penrose and Hameroff [10], the
number of
tubulins/cl involved in the threshold is 10^9 = n , with a
coherence
time T=500 msec (1/2 sec). As tubulins are qubits, we can
indulge in
speculating about the brain-universe, with 10^9 = n quantum
gravity
registers, and a coherence time 10^-34 sec = T , which might
have a
conscious experience. Then, if the in?tionary universe, which
performed quantum computation, was able to achieve
consciousness, so
will do any quantum computer?
For the moment, the only possible answer is no, for three
reasons:
1. Because quantum computers are very difficult to build in
practice, as
the technology
is not yet so advanced to maintain coherence for a
sufficiently long time.
2. Because quantum computers (at least the first generations)
will not
have enough
mass.
3. For a quantum system to be able to get a conscious
experience, it is
a necessary but
non sufficient condition that it performs quantum computation.
The extra
requirement
is that the quantum computing system should be
quantum-autopoietic.
However, we cannot really foresee anything: In the long run
quantum
computers might
have conscious experiences.
6. CONSCIOUSNESS ARISES IN THE BITS ERA
There have been some attempts in appending consciousness to
the universe
as a
whole [21]. In our mod, during the qubits era there are no
events in
the usual sense, (occasions of experience, in the
philosophical language
of Whitehead [7]) although, there can be events in a
non-Boolean sense
(some work is in progress on that [22]).
So, if we, Boolean-minded beings, conceive consciousness in
terms of
occasions of
experience (events in the Boolean sense), we can argue that
in the
qubits era there was
no consciousness at all in the universe (perhaps, there was
just
sub-consciousness).
Sarfatti Commentary: Why Boolean-minded. Our
motor activity is
Boolean, our nerve impulses are Boolean, but our dreams are
certainly
non-Boolean-minded IMHO.
PZ: Consciousness appeared in the classical bits era: it
was the
projection in the past of future observers who had to be
programmed by
the sf-organizing CA, in order to observe the emergent
events.
Sarfatti: See Wheer on dayed choice of
the observer-participator.
PZ: 6.1 The Boolean observer
To observe the events in the post-in?tionary universe, the
observers
should be
Boolean. This means that the qubits-tubulins of the
observers brain
should collapse
to classical bits at a rather high frequency.
Of course, the observers being Boolean, they will not be able
to grasp
the unfolding
quantum computing structure of spacetime at the fundamental
lev (the
Planck scale)
by the use of tools like causal sets [23] and the functor
Past [24],
which become
usess at that scale, where locality and causality are missing
(some
work is in
progress on this issue [25]).
What the Boolean observers can do, is just to recognize the
large scale
structure of the
universe, and, by the use of the functor Past, go backward in
time.
But, anyway, the trav will stop at the end of in?tion: the
big
bang will never be
reached because then the multiverse started, for which the
functor
Past loses any
meaning.
The problem is that a Boolean observer is endowed with the
concept of
time, which is
a mere artifact of his own perception, and moreover, he tends
to extend
this concept to
regions of reality where it is meaningless.
6.2 The functor Past and the quantum registers
The functor Past [24] is the functor from a causal set C [23]
to the
category of sets,
go to original
and the properties of re?xivity, antisymmetry and
transitivity are
satisfied. .
In this context, the Functor Past can be defined, but there is
no
quantum computation.
In fact, as in this case the logic gates are connecting the
quantum
gravity registers one
to another, the quantum gravity registers cannot process
information
individually.
This picture is strictly rated to the idea of an internal
observer
[24], which is not an
adequate assumption as far as the very early universe is
considered.
ii) The Fock space interpretation
go to original
In this atemporal picture, the early in?tionary universe is
interpreted as an ensemble of
quantum registers in parall, and quantum computation is runned
overall.
Sarfatti: runned?
PZ: and the causal past is not unique [26].
6.3 Objective reduction, decoherence and the emergence of Past
go to original ..
At this point, the functor Past can be defined. Thus, although
the
quantum past is not
unique, in the world W the past can be re-constructed
univocally.
6.4 The analogy
In?tion (the qubits era) is for the universe what
pre-consciousness
(superposed
tubulins) is for our mind. The end of in?tion (beginning of
the bits
era) is for the universe what consciousness (Orch OR of
superposed
states of tubulins) is for our mind.
The analogy goes like that:
For tubulins in the brain:
CLASSICAL CA
EMERGENCE OF QUANTUM COHERENCE (PRE- CONSCIOUSNESS) QUANTUM CA
SF-COLLAPSE BY ORCHESTRATED OBJECTIVE REDUCTION
CONSCIOUS EXPERIENCE
CLASSICAL CA.
For qubits in the early universe:
CLASSICAL BIT (THE VACUUM)
QUBITBEGINNING OF INFLATION (THE UNIVERSE IS A SUPERPOSED
STATE OF
QUANTUM REGISTERS)
SF-REDUCTION BY OBJECTIVE REDUCTION (END OF INFLATION)
CONSCIOUS
EXPERIENCE
COLLAPSE OF QUBITS TO BITS (THE XOR GATE)
CLASSICAL CA.
Of course, the analogy between our mind and the universe is
very
speculative at this
stage, but the emergent picture is quite exciting: it seems
that our
brain owes its
structure and organization to the very early universe. This
is in
agreement with the Penrose-Hameroffs bief that
consciousness is a
fundamental property of reality, and has its roots in the
spacetime
structure at the Planck scale.
Then, although we can be just classical as observers, we can
be also
quantum as
thinkers (for example, we can conceive quantum computation).
This fact
must be the
result of some kind of imprinting we received from the
quantum computing
early
universe.
If we had not both quantum and classical computational modes
available
in our brain,
in other words, if we were always conscious and Boolean, we
would not be
able to
think quantum.
Sarfatti: I think thinking is non-Boolean [CapitalEth] at
least creative
thinking.
PZ: 7. CONCLUSIONS
In this paper, we described the early in?tionary universe as
an
ensemble of quantum
gravity registers in parall. At the end of in?tion, the
superposed
state sf-reduces by reaching the quantum gravity threshold as
in the
Penroses Objective Reduction mod. This sf-reduction can
be
interpreted as a primordial conscious experience. Actually,
the number
of quantum gravity registers involved in the OR equals the
number of
superposed
tubulins in our brain, which are involved in the Orch OR,
leading to a
conscious experience. Further, the qubits of the sected
quantum
gravity register get entangled
with the emergent environment and collapse to classical bits.
This
environmental
collapse is the source of classical information and Boolean
logic in the
actual universe.
Thus, we make the conjecture that the post-in?tionary
universe starts
to organize
itsf, very liky as a classical Clular Automata, and
necessarily
produces sf-similar
computing systems (our minds). In this way, the actual
universe and its
products use the e (Boolean) logic so that the past can be
recorded,
and information can be stored.
It should be noted that, in this mod, the quantum gravity
registers in
parall are
parall universes. This interpretation is very much on the e
line
with Deutsch idea
rating quantum computers to parall universes (the multiverse)
[27].
However, at the end of in?tion, only one universe is sected,
the one
which is
endowed with that particular amount of entropy which makes it
our world.
ACKNOWLEDGMENTS
I am grateful to R. E. Zimmermann for hpful discussions.
I thank the Department of Astronomy, University of Padova,
Italy, for
hospitality.
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of the
Mind-
Brain, Cambridge, MA, MIT Press (1986).
P.S. Churchland, Brainshy: On non-neural theories of the
mind. In:
Toward a
Science of Consciousness II-The Second Tucson Discussions and
Debates,
Eds. S.
Hameroff, A. Kaszniak, A. Scott, MIT Press (1998).
[2] R. Penrose, The Emperors New Mind , Oxford Press,
Oxford, UK (1989).
R. Penrose, Shadows of the Mind, Oxford Press, Oxford, U.K.
(1994).
[3] H. P. Stapp, Why Classical Mechanics Cannot Naturally
Accommodate
Consciousness but Quantum Mechanics Can, Psyche 2(5), May
1995.
H. P. Stapp, Mind, Matter, and Quantum Mechanics, Springer-
Verlag,
Berlin (1993).
[4] D. Chalmers, Facing Upto the Problem of Consciousness,
Journal of
Consciousness
Studies (1995), and in: Toward a Science of Consciousness-The
First
Tucson Discussions
and debates, Eds. S. Hameroff, A. Kaszniak, A. Scott, MIT
Press,
Cambridge, MA, pp 5-
28, also available online, at: http://www.Starlab.org/
D. Chalmers, The Conscious Mind-In search of a fundamental
theory,
Oxford University
Press, New York (1996).
[5] B. Spinoza, (1677), Ethica in Opera quotque reperta sunt.
3rd
edition, Eds J. van
Vloten and J. P. N. Land (Netherlands: Den Haag).
[6] G.W. Leibniz, (1768), Opera Omnia, 6 volumes, Louis
Dutens, Ed. Geneva.
[7] A. N. Whitehead, Process and Reality, Macmillan,
N.Y.(1929).
[8] A. Shimony, Search for a Naturalistic World View-Volume
II. Natural
Science and
Metaphysics. Cambridge University Press, Cambridge, U.K.
(1993).
[9] C. Rovli and L. Smolin, Spin networks in quantum gravity,
Phys.
Rev. D52
(1995) 5743.
[10] S. Hameroff and R. Penrose, Orchestrated reduction of
quantum
coherence in brain
microtubules: A mod for consciousness. In: Toward a Science of
Consciousness-The
First Tucson Discussions and Debates, Eds. S. Hameroff, A.
Kaszniak, and
A. Scott. MIT
Press, Cambridge, MA (1996).
S. Hameroff and R. Penrose, Conscious events as orchestrated
spacetime
sections,
Journal of Consciousness Studies 3(1) (1996) pp 36-53.
S. Hameroff, Funda-mental geometry: The Penrose-Hameroff Orch
OR mod of
consciousness. In: Geometry and the foundations of Science:
Contributions from an
Oxford Conference honouring Roger Penrose. Oxford Press
(1997).
[11] P. A. Zizzi, Holography, Quantum Geometry, and Quantum
Information
Theory,
gr-qc/9907063; Entropy (2) (2000) pp 39-69.
[12] P. A. Zizzi, Quantum Foam and de Sitter-like Universe,
hep-th/9808180; IJTP
Vol. 38, N 9 (1999) pp 2333-2348.
[13] H. Everett III, Rative State Formulation of Quantum
Mechanics,
Rev. of Modern
Phys. Vol. 29 (1957) pp 454-462.
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R. E. Zimmermann, Loops and Knots as Topoi of Substance.
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Revisited, grqc/
0004077.
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large?,
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Math. Phys.
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spacetime,
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9910146.
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802-819.
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I recently came across a book which recasts music theory
using categories
as a foundation==different topologies for rhythm, harmony,
mody and so
forth are given. ItOs called The Topos of Music by Guerino
Mazzola, and
finds
it quite
easily). ItOs all very interesting though much of it is way
over my head.
The author devops some computer-implemented analytic methods
that
represent data structures of music as objects. It struck me
at times that
the language used to discuss manipulating these data
structures was very
similar to the language of category theory itsf. (Not that
IOm an expert
in either of those fids.) Does anyone se think that categories
and
object-oriented programming are a natural pair? Is the charge
ever
levled that some people only use category theory because it
fits the
data structures sected? (Not that this is a bad reason; just
that if
this is how it is, then for me at least it removes some of
the deepness
about the applicability of categories in this case.)
Curious to see your comments, about this question and about
the work
mentioned, if anyone has had a look at it. ==Henri.
>I recently came across a book which recasts music theory
using categories
>as a foundation==different topologies for rhythm, harmony,
mody and so
>forth are given. ItOs called The Topos of Music by Guerino
Mazzola,
and
finds
it quite
>easily). ItOs all very interesting though much of it is way
over my head.
The author devops some computer-implemented analytic methods
that
>represent data structures of music as objects. It struck me
at times that
>the language used to discuss manipulating these data
structures was very
>similar to the language of category theory itsf. (Not that
IOm an expert
>in either of those fids.) Does anyone se think that
categories and
>object-oriented programming are a natural pair? Is the
charge ever
>levled that some people only use category theory because it
fits the
>data structures sected? (Not that this is a bad reason; just
that if
>this is how it is, then for me at least it removes some of
the deepness
>about the applicability of categories in this case.)
Curious to see your comments, about this question and about
the work
>mentioned, if anyone has had a look at it. ==Henri.
>
I shared this with a good friend who is wl-versed in both
music and
> I suspect those people think too much. IOve not yet seen
an
answer to
> the question why does a certain chord progression make me
cry?.
==
Stephen J. Herschkorn herschko@rutcor.rutgers.edu
> I recently came across a book which recasts music theory
using categories
> as a foundation==different topologies for rhythm, harmony,
mody and so
> forth are given. ItOs called The Topos of Music by Guerino
Mazzola,
and
finds
it quite
> easily).
I searched with:
The Topos of Music Guerino Mazzola +.pdf
and only got links to the for-sale book.
l8r, N. toff
> I recently came across a book which recasts music theory
using categories
> as a foundation==different topologies for rhythm, harmony,
mody and so
> forth are given. ItOs called The Topos of Music by Guerino
Mazzola,
and
finds
it quite
> easily). ItOs all very interesting though much of it is
way
over my head.
The author devops some computer-implemented analytic methods
that
> represent data structures of music as objects. It struck me
at times that
> the language used to discuss manipulating these data
structures was very
> similar to the language of category theory itsf. (Not that
IOm an
expert
> in either of those fids.) Does anyone se think that
categories and
> object-oriented programming are a natural pair? Is the
charge ever
> levled that some people only use category theory because it
fits the
> data structures sected? (Not that this is a bad reason;
just that if
> this is how it is, then for me at least it removes some of
the deepness
> about the applicability of categories in this case.)
>
HavenOt seen the work you reference. But IOve
always thought
the
category/oop connection is very superficial. Both involve
thingies with
arrows (morphisms/messages) between them, but I think thatOs
really
about as far as it goes.
> HavenOt seen the work you reference. But
IOve always
thought the
> category/oop connection is very superficial. Both involve
thingies with
> arrows (morphisms/messages) between them, but I think
thatOs really
> about as far as it goes.
ItOs not really a necessary part of category theory or OOP,
or
necessarily absent from other foundations or programming
paradigms, but:
my impression is that people who use category theory (as I do
not) and
oop (as I sometimes do) are more willing to think in terms of
systems
defined by the behavior of the objects, and less about the
detailed
implementations of those objects == for imperative
programmers, in
contrast, the detailed layout of bits in a struct can be
important, and
for set-theoretic foundationalists, it can be important
exactly how one
chooses to represent an ordered pair as a certain kind of set.
==
David Eppstein http://www.ics.uci.edu/~eppstein/
Univ. of California, Irvine, School of Information & Computer
Science
> I recently came across a book which recasts music theory
using categories
> as a foundation==different topologies for rhythm, harmony,
mody and so
> forth are given. ItOs called The Topos of Music by Guerino
Mazzola,
and
finds
it quite
> easily).
This sounds very interesting, but == sorry == I have tried a
Google search
and have only been able to find advertisements for the
hardcover book to be
purchased. Could you please supply a URL for a downloadable
PDF of the
Amittai Aviram
>Only an absolute moron would design a language
>(e.g. Python) where white space is significant.
>> Wl of course your post was not meant to be taken
>> seriously (or rather it seems you may have a serious
>> point, but these comments are not meant to be
>> taken literally).
>I didnOt have a serious point, I just threw out
>a bunch of languages and insulted each one. In
>some cases I used the canonical insult for the languague.
>For Python the canonical insult is the use of significant
>white space.
>I am a Python fan. I donOt particularly like the
significant
>white space, but it doesnOt bother me either.
It only bothers people who havenOt tried it.
The biggest problem is that the significant white space issue
>distracts from other much more important issues.
Does it? I used to read comp.lang.python, and there was
> plenty of discussion of actual issues, with no distraction
> that I can recall.
>
As you point out, once youOve tried it you no longer
consider
it an issue. So on comp.lang.python where the vast majority
of people have tried python, it is not considered an issue.
However, in comparative language discussions the issue
frequently arises.
- William Hughes
>>Only an absolute moron would design a language
>(e.g. Python) where white space is significant.
Wl of course your post was not meant to be taken
>> seriously (or rather it seems you may have a serious
>> point, but these comments are not meant to be
>> taken literally).
I didnOt have a serious point, I just threw out
>a bunch of languages and insulted each one. In
>some cases I used the canonical insult for the languague.
>For Python the canonical insult is the use of significant
>white space.
I am a Python fan. I donOt particularly like the
significant
>white space, but it doesnOt bother me either.
It only bothers people who havenOt tried it.
>The biggest problem is that the significant white space issue
>distracts from other much more important issues.
Does it? I used to read comp.lang.python, and there was
>> plenty of discussion of actual issues, with no distraction
>> that I can recall.
>>
As you point out, once youOve tried it you no longer
consider
>it an issue. So on comp.lang.python where the vast majority
>of people have tried python, it is not considered an issue.
>However, in comparative language discussions the issue
>frequently arises.
Oh. Yes, in comparative language discussions it may wl
distract from more serious issues.
> - William Hughes
************************
David C. Ullrich
>Hey, I have an idea. LetOs combine Python and C. Use Python
>for the complicated high lev stuff where speed of devopment
is
important
>but execution speed isnOt, and C for the simpler low lev
stuff where
>execution speed is important. Wow, what a concept, combine an
interpreted
>language with a compiled language. IOm brilliant,
>IOm going to be famous, IOm going to be rich
....
>What do you mean someone se thought of this first?
Heh. Yes, for the benefit of those who are not in on the joke,
one
> of the things that people who like Python like about it is
that itOs
> specifically designed to make it easy to extend Python with
> modules compiled in C (or whatever), and also to embed
Python
> in C (or whatever) programs, allowing exactly what you
suggest,
> using Python for the complicated high-lev stuff and C (or
> whatever) for the inner loop.
>
And to be fair, the idea of using different languages for
different
levs of the e problems predates Python.
- William Hughes
>Oh wl, send back the Rolls.
>> (For example, trying to write a record-breaking
PrimePi(n) calculator
>> in Python would probably be stupid (at least doing it in
straight
>> Python would be, having no experience with numpy I
couldnOt
>> say how wl suited it would be for something like that.
>Numpy will allow you to do a lot of matrix stuff quickly
without
>having to go to C (or equivalent). However, if you have
something
>specialized that needs to be done quickly, Python/numpy is
not
>the way to go.
> - William Hughes
> ************************
David C. Ullrich
...
> I didnOt have a serious point, I just threw out
> a bunch of languages and insulted each one. In
> some cases I used the canonical insult for the languague.
> For Python the canonical insult is the use of significant
> white space.
 I am a Python fan. I donOt particularly like the
significant
> white space, but it doesnOt bother me either.
LetOs have a bit of background here. (Yes, it all happened
at
CWI.
Why, nobody knows. Perhaps because we are allowed to do what
we
want, as long as we do what must be done too?)
Back in the 80Os Lambert Meertens (who had much to do with
Algol 68)
started to design his own language: ABC. The intent was to
make
programming simple for the user. It worked so wl that when it
became improbable to support ABC on our systems, most of the
secretarial staff revolted. This experience did lead Guido
van Rossum
to design Python, which was loosy based on ABC.
Significant white space is especially eschewed by users of
other
languages. But they always complain when you produce a
program in
their language of choice without proper indentation.
Indentation
is visually important, so why not make it syntactically
important?
A further question is: what is large scale. For some time
already
we have running a Python program that multiplexes TCP/IP
streams to
four RS-232 ports and the reverse. It runs on a 286, and
performs
wl. (That it does not run now is due to a broken motor unit
which
it attempts to control.)
==
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam,
nederland,
+31205924131
home: bovenover 215, 1025 jn amsterdam, nederland;
http://www.cwi.nl/~dik/
>...
> I didnOt have a serious point, I just threw out
> a bunch of languages and insulted each one. In
> some cases I used the canonical insult for the languague.
> For Python the canonical insult is the use of significant
> white space.
 I am a Python fan. I donOt particularly like the
significant
> white space, but it doesnOt bother me either.
LetOs have a bit of background here. (Yes, it all happened
at
CWI.
>Why, nobody knows. Perhaps because we are allowed to do what
we
>want, as long as we do what must be done too?)
Back in the 80Os Lambert Meertens (who had much to do with
Algol 68)
>started to design his own language: ABC. The intent was to
make
>programming simple for the user. It worked so wl that when it
>became improbable to support ABC on our systems, most of the
>secretarial staff revolted. This experience did lead Guido
van Rossum
>to design Python, which was loosy based on ABC.
Significant white space is especially eschewed by users of
other
>languages. But they always complain when you produce a
program in
>their language of choice without proper indentation.
Indentation
>is visually important, so why not make it syntactically
important?
Precisy. Since weOre going to be indenting code anyway to
make
it readable, why _also_ require the {}?
(Not that the whitespace thing is the main issue in this
thread...)
>A further question is: what is large scale. For some time
already
>we have running a Python program that multiplexes TCP/IP
streams to
>four RS-232 ports and the reverse. It runs on a 286, and
performs
>wl.
I suspect that if one is convinced that Python is no good for
large
systems then any system that it works wl for is going to turn
out
to be small, by definition. (I _am_ curious _why_
itOs said
that
Python is no good for large systems...)
>(That it does not run now is due to a broken motor unit which
>it attempts to control.)
Now _this_ strikes me as a serious ?w in Python. Any language
that would attempt to control a broken motor unit should have
its head examined.
************************
David C. Ullrich
> Hey, I have an idea. LetOs combine Python and C. Use
Python
> for the complicated high lev stuff where speed of devopment
is
important
> but execution speed isnOt, and C for the simpler low lev
stuff where
> execution speed is important. Wow, what a concept, combine
an
interpreted
> language with a compiled language. IOm brilliant,
> IOm going to be famous, IOm going to be rich
....
> What do you mean someone se thought of this first?
> Oh wl, send back the Rolls.
Forget the Rolls. Send it to Guido van Rossum. I have used
quite a bit
of python; I have seen the devopment (look where I come from
and where
Guido did come from). It appears to be quite simple (although
I have
never done it) to interface Python with a library of C
routines. In
fact, most of the interfaces Python has with the standard C
library
and other libraries are based on it. They are only
preprogrammed for your
pleasure. There is simply a wl-defined Python -> C interface.
==
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam,
nederland,
+31205924131
home: bovenover 215, 1025 jn amsterdam, nederland;
http://www.cwi.nl/~dik/
> Hey, I have an idea. LetOs combine Python and C. Use
Python
> for the complicated high lev stuff where speed of devopment
is
important
> but execution speed isnOt, and C for the simpler low lev
stuff where
> execution speed is important. Wow, what a concept, combine
an
interpreted
> language with a compiled language. IOm brilliant,
> IOm going to be famous, IOm going to be rich
....
> What do you mean someone se thought of this first?
> Oh wl, send back the Rolls.
Forget the Rolls. Send it to Guido van Rossum. I have used
quite a bit
>of python; I have seen the devopment (look where I come from
and where
>Guido did come from). It appears to be quite simple
(although I have
>never done it) to interface Python with a library of C
routines. In
>fact, most of the interfaces Python has with the standard C
library
>and other libraries are based on it. They are only
preprogrammed for your
>pleasure. There is simply a wl-defined Python -> C interface.
For the record itOs not just C. Most of my Python thinglets
happen in
an environment where Python has been embedded in Dphi. (So
the Dphi actually gets used for the user interface and also
sometimes for low-lev calculations, and the Python does the
stuff in between.)
************************
David C. Ullrich
Is it true that sums with uncountably many nonzero terms
diverge always?
How
do you prove this?
> Is it true that sums with uncountably many nonzero terms
diverge always?
How
> do you prove this?
If for each natural n there are only countably many nonzero
terms
greater than 1/n, then the union over N of {nonzero terms
greater than
1/n} is countable. But all the terms are in that union.
Therefore here must be uncountably many nonzero terms greater
than 1/n
for some value of n, hence the sum is infinite.
 Is it true that sums with uncountably many nonzero terms
diverge always?
How
> do you prove this?
 If for each natural n there are only countably many nonzero
terms
> greater than 1/n, then the union over N of {nonzero terms
greater than
> 1/n} is countable. But all the terms are in that union.
Therefore here must be uncountably many nonzero terms greater
than 1/n
> for some value of n, hence the sum is infinite.
This would necessarily be true if all the terms were
positive, or if you
are considering only absolute convergence, but what of
conditional
convergence?
Suppose a wl ordering of the terms so that positive and
negative terms
keep cancling out?
Is it true that sums with uncountably many nonzero terms
diverge
always? How
> do you prove this?
If for each natural n there are only countably many nonzero
terms
>> greater than 1/n, then the union over N of {nonzero terms
greater than
>> 1/n} is countable. But all the terms are in that union.
Therefore here must be uncountably many nonzero terms greater
than 1/n
>> for some value of n, hence the sum is infinite.
This would necessarily be true if all the terms were
positive, or if you
>are considering only absolute convergence, but what of
conditional
>convergence?
Suppose a wl ordering of the terms so that positive and
negative terms
>keep cancling out?
That does give a more general notion of sum than the
definition
Jode gave (the definition he gave is equivalent to the
definition I
gave in another post). Of course itOs not the sum of a set
of
numbers,
itOs the sum of a sequence of numbers. To specify things a
little
more precisy:
Suppose weOre given a real number x_a for every ordinal a <
b.
Define S_0 = 0, if a = c + 1 define S_a = S_c +
x_a, and if
a is a limit ordinal define S_a to be the limit of S_c as c
approaches a from bow (for a sum to be convergent
we require that this limit always exist.) Then say the sum
is S_b. (Or S_b if b is a limit ordinal and the obvious
thing if b is a successor.)
Now a sum certainly can converge in this sense without
being absoluty convergent, hence without converging
in the sense of the other definition. IOm pretty
sure that
if a sum converges in this sense it can nonethess
have at most countably many nonzero terms.
Ah, hereOs a proof. You prove the following by induction
on c:
Fact: For every c <= b there exist at most countably
many a < c with x_a <> 0.
Pf: If c is a successor ordinal then the induction step
is clear; a countable set plus one ement is countable.
Suppose that c is a limit oridinal. Now, the e
proof as in calc 1 shows that x_a -> 0 as a -> c
from bow. So for each natural number n there
exists a_n < c such that |x_d| < 1/n whenever
a_n < d < c. But (by induction!) there are only
countably many non-zero terms _before_ a_n;
hence there are only countably many terms
with absolute value >= 1/n before c, and hence
only countably many non-zero terms before c.
QED.
Now let c = b, QED.
************************
David C. Ullrich
Suppose a wl ordering of the terms so that positive and
negative terms
> keep cancling out?
Wl-ordering? Aha! That was never mentioned before in this
thread...
Use the parital sums, and then use this: If a net in R
indexed by the
first uncountable ordinal converges, then it is eventually
constant.
To prove this use the fact that R is first-countable.
==
http://www.math.ohio-state.edu/~edgar/
Suppose a wl ordering of the terms so that positive and
negative terms
>> keep cancling out?
Wl-ordering? Aha! That was never mentioned before in this
thread...
Use the parital sums, and then use this: If a net in R
indexed by the
>first uncountable ordinal converges, then it is eventually
constant.
>To prove this use the fact that R is first-countable.
That was the first thing I thought of saying -
itOs not quite
enough.
This shows that there are only countably many non-zero terms
before the first uncountable ordinal, but it _can_ happen that
more non-zero terms appear later. (One can jack up the
argument to give a proof, as I did in my reply to Virgil.)
************************
David C. Ullrich
> Suppose a wl ordering of the terms so that positive and
negative terms
> keep cancling out?
>Wl-ordering? Aha! That was never mentioned before in this
thread...
>Use the parital sums, and then use this: If a net in R
indexed by the
>>first uncountable ordinal converges, then it is eventually
constant.
>>To prove this use the fact that R is first-countable.
That was the first thing I thought of saying -
itOs not quite
enough.
>This shows that there are only countably many non-zero terms
>before the first uncountable ordinal, but it _can_ happen
that
>more non-zero terms appear later. (One can jack up the
>argument to give a proof, as I did in my reply to Virgil.)
(Perhaps I should add that one _does_ need to actually give
a new proof, one canOt just say that the e proof appplies.
Because itOs not true that a convergent net indexed by _any_
uncountable ordinal is eventually constant.)
>************************
David C. Ullrich
************************
David C. Ullrich
>Is it true that sums with uncountably many nonzero terms
diverge always?
How
>do you prove this?
>If for each natural n there are only countably many nonzero
terms
>>greater than 1/n, then the union over N of {nonzero terms
greater than
>>1/n} is countable. But all the terms are in that union.
>Therefore here must be uncountably many nonzero terms
greater than 1/n
>>for some value of n, hence the sum is infinite.
> This would necessarily be true if all the terms were
positive, or if you
> are considering only absolute convergence, but what of
conditional
> convergence?
Suppose a wl ordering of the terms so that positive and
negative terms
> keep cancling out?
I suppose that the concept of sum which is being used here is
this:
given a set S of complex numbers, we say that its sum is a
complex
number s if, for every r > 0, there is some finite subset I of
S such
that, for every finite subset J of S which contains I we have
| s - (sum of all ements of J) | < r.
This definition is equivalent to the following one: a set S of
complex
numbers has a sum if, for every r > 0, there is some finite
subset I of
S such that, for every finite subset J of S which does not
intersect I,
we have
| sum of all ements of J | < r.
So, if n is a natural number, there is a subset I(n) of S
such that, for
every finite subset J of S which does not intersect I(n), we
have
| sum of all ements of J | < 1/n.
If x does not bong to the reunion of all I(n), it follows
from the
definition of I(n) that |x| < 1/n for every natural number n;
therefore
x = 0. So, outside the reunion of all I(n), which is a
countable set,
every ement of a convergent sum must be equal to zero.
When I tried to prove it, I needed to rearrange the terms in
the sum to
show
that it diverged, but, if the sum diverges, then you canOt
always rearrange
the terms (or so I think...)
> Is it true that sums with uncountably many nonzero terms
diverge always?
How
> do you prove this?
>When I tried to prove it, I needed to rearrange the terms in
the sum to
show
>that it diverged, but, if the sum diverges, then you canOt
always
rearrange
>the terms (or so I think...)
YouOve put your finger on a reason why your
original question
is
not wl-defined: You need to first
define what you _mean_ by
the sum of an uncountable number of numbers.
For that matter, you need to first define what you
mean by the
sum of a _set_ of numbers in general; the sum of an ordinary
infinite series is not the sum of a countable set of numbers,
itOs the sum of a _sequence_ of numbers (and
thereOs a
difference, because rearranging the terms can change the
answer).
If S is any set of _non-negative_ reals then thereOs really
only
one reasonable definition of the sum of the ements of S:
the sum is the sup of the sums of finite subsets of S. With
this definition you can easily prove the result you ask about,
using the arguments I suspect youOve come up with already.
What about the sum of a set of reals, not necessarily
non-negative? The only reasonable definition that comes
to mind is essentially the Lebesgue integral with respect
to counting measure - with that definition if a sum
converges then the sum of the absolute values also
converges, reducing the question to the previous paragraph.
(If you had another definition in mind you should say
what it is...)
>> Is it true that sums with uncountably many nonzero terms
diverge always?
>How
>> do you prove this?
>
************************
David C. Ullrich
>When I tried to prove it, I needed to rearrange the terms in
the sum to
show
>that it diverged, but, if the sum diverges, then you canOt
always
rearrange
>the terms (or so I think...)
YouOve put your finger on a reason why your
original question
is
> not wl-defined: You need to first
define what you _mean_ by
> the sum of an uncountable number of numbers.
For that matter, you need to first define what you
mean by the
> sum of a _set_ of numbers in general;
Let K be a set of reals, then the Osum of the
setO is:
Sum[x in K] x
> the sum of an ordinary
> infinite series is not the sum of a countable set of
numbers,
> itOs the sum of a _sequence_ of numbers (and
thereOs a
> difference, because rearranging the terms can change the
> answer).
Why would the order matter given that addition is commutative
in the reals?
l8r, N. toff
> When I tried to prove it, I needed to rearrange the terms
in the sum to
>show
>that it diverged, but, if the sum diverges, then you canOt
always
>rearrange
>the terms (or so I think...)
> YouOve put your finger on a reason why your
original
question is
>> not wl-defined: You need to first
define what you _mean_ by
>> the sum of an uncountable number of numbers.
> For that matter, you need to first define what
you mean by the
>> sum of a _set_ of numbers in general;
Let K be a set of reals, then the Osum of the
setO is:
>Sum[x in K] x
Uh, thatOs not a definition - what does Sum[x in
K] x mean?
>> the sum of an ordinary
>> infinite series is not the sum of a countable set of
numbers,
>> itOs the sum of a _sequence_ of numbers (and
thereOs a
>> difference, because rearranging the terms can change the
>> answer).
Why would the order matter given that addition is commutative
in the
reals?
The order matters even for ordinary infinite sums (ie sums of
sequences of reals). Addition is commutative, but the sum
of an infinite series is not an actual sum, itOs
the limit of
some
sums.
In fact if sum(x_j) is any conditionally convergent series
of reals (converges but not absoluty) then itOs not hard
to show that for any S there is a rearrangement of the
sequence (x_j) with sum S.
>l8r, N. toff
************************
David C. Ullrich
>When I tried to prove it, I needed to rearrange the terms in
the sum
to
>show
>that it diverged, but, if the sum diverges, then you canOt
always
>rearrange
>the terms (or so I think...)
>> YouOve put your finger on a reason why your
original
question is
>> not wl-defined: You need to first
define what you _mean_ by
>> the sum of an uncountable number of numbers.
>> For that matter, you need to first define what
you mean by
the
>> sum of a _set_ of numbers in general;
>Let K be a set of reals, then the Osum of the
setO is:
>Sum[x in K] x
Uh, thatOs not a definition - what does Sum[x in
K] x mean?
>
Ok. Using limits I can see why youOd need an order. But I
have recently
encountered similar things like, for example,:
Let {A_j | j in J} be an uncountable collection of members of
a topology
T1,
for some index set J. Prove that Union[j in J] A_j in T1.
Does Union[j in J] A_j, on its own, constitute a definition
assuming a
definition for finite index sets has been properly
defined? Or
does the
switch to an infinite index set require some other machinery
(ie: a limit
needed to be taken for summation of numbers in an infinite set
which is not
required when summing numbers in finite sets)?
>Why would the order matter given that addition is
commutative in the
reals?
The order matters even for ordinary infinite sums (ie sums of
> sequences of reals). Addition is commutative, but the sum
> of an infinite series is not an actual sum,
itOs the limit
of some
> sums.
In fact if sum(x_j) is any conditionally convergent series
> of reals (converges but not absoluty) then itOs not hard
> to show that for any S there is a rearrangement of the
> sequence (x_j) with sum S.
>
l8r, N. toff
>When I tried to prove it, I needed to rearrange the terms in
the sum
>to
>show
>that it diverged, but, if the sum diverges, then you canOt
always
>rearrange
>the terms (or so I think...)
> YouOve put your finger on a reason why your
original
question is
>> not wl-defined: You need to first
define what you _mean_ by
>> the sum of an uncountable number of numbers.
> For that matter, you need to first define what
you mean by the
>> sum of a _set_ of numbers in general;
Let K be a set of reals, then the Osum of the
setO is:
>Sum[x in K] x
> Uh, thatOs not a definition - what does Sum[x
in K] x mean?
Ok. Using limits I can see why youOd need an order. But I
have recently
>encountered similar things like, for example,:
Let {A_j | j in J} be an uncountable collection of members of
a topology
T1,
>for some index set J. Prove that Union[j in J] A_j in T1.
Does Union[j in J] A_j, on its own, constitute a definition
assuming a
>definition for finite index sets has been
properly defined? Or
does the
>switch to an infinite index set require some other machinery
(ie: a limit
>needed to be taken for summation of numbers in an infinite
set which is
not
>required when summing numbers in finite sets)?
Unions and sums are very different. Infinite sums are not
really
sums, theyOre limits of finite sums. But
infinite unions can
easily
be defined without reference to finite unions:
If S is a collection of sets then the Union{x: x in S}
(written by
set theorists as just Union S is the set of all things which
are ements of at least one ement of S. ThatOs the
definition,
it works equally wl whether S contains finity many sets
or infinity many, and it has nothing to do with how the ements
of S are ordered.
(Usually people first define A union B to be the
set of all
things
which are in A or in B - if thatOs where you start then you
could
get the idea that the union of infinity many sets needs to be
defined in terms of finite unions somehow. But in
actual official
set theory you start with just Union S, meaning the union
of all the ements of S as defined above, and then
A union B is defined as Union {A, B}.)
>Why would the order matter given that addition is
commutative in the
>reals?
> The order matters even for ordinary infinite sums (ie sums
of
>> sequences of reals). Addition is commutative, but the sum
>> of an infinite series is not an actual sum,
itOs the limit
of some
>> sums.
> In fact if sum(x_j) is any conditionally convergent series
>> of reals (converges but not absoluty) then itOs not hard
>> to show that for any S there is a rearrangement of the
>> sequence (x_j) with sum S.
l8r, N. toff
************************
David C. Ullrich
> Why would the order matter given that addition is
commutative in the
reals?
look up conditional convergence in your calculus textbook
> Why would the order matter given that addition is
commutative in the
reals?
look up conditional convergence in your calculus textbook
l8r, N. toff
> On a supercalifragilisticexpialidocious day, after dancing
> about singing Bibbety bobbety boo!, ClueStickMan
> ishkabibbled:
> [ snipped...not the point ]
>-
>The Queen of DXers, as wl as
>Queen of the Commonwealth of Virginia, as wl as
>The Ruler of A.D.P., as wl as
>Saint Debbe, as wl as
>Our Lady of the Black Hole Exploratory Input Services
> as OhFishAlly Appointed by the Psychedic Pope, a/k/a
> Saint Isidore of Seville
>An Ointed Minister of the Universal Life Church
>Reverant of the Church of the SubGenius, UnOrthodox
>Superior Mutha Superior of the Little Sistahs of the
> Politically Incorrect
>Worshipper of Eris, Goddess of Discord
I WONOT grow up!! == Peter Pan
Just when it begins to seem that maybe lame has been
complety defined, you come along and extend it.
> A TRANSFINITE METHOD TO COUNT THE POWER SET OF N
> Proposition:
> The power set of N is countable.
Proof:
Being P(N) the set of the subsets of N (power set of N), we
will
> count its ements in an orderly way. First we will count the
subset
> with 0 ements, and then the subsets with one ement
(singletons),
> then the subsets with two ements, and so on.
> 1) 1 <-> {}
> 2) In order to count the subsets with 1 ement {n}, we will
use the
> naturals that are multiple of 2. We have 2 <-> {1}, 4 <->
{2}, 6 <-{3} ? 2k <-> {k}
> 3) In order to count the subsets with 2 ements {n1, n2} we
will use
> the naturals multiple of 3 that they are not divisible by
2. We have
> 3 <-> {1, 2}, 9 <-> {1, 3}, 15 <-> {1, 4}, ?
> 4) In order to count the subsets with 3 ements {n1, n2, n3}
we will
> use the naturals multiple of 5 that they are not divisible
by 2 and
> 3. We have 5 <-> {1, 2, 3}, 25 <-> {1, 2, 4}, 35 <-> {1, 2,
5}, ?
> 4) In order to count the subsets with 4 ements {n1, n2, n3,
n4} we
> will use the naturals multiple of 7, that they are not
divisible by
> 2, 3 and 5, and so on.
Conclusion:
> As the naturals are infinite and the prime numbers too, the
bijection
> N <-> P(N) is possible, so P(N) is countable.
Nicolas de la Foz
ThatOs pretty good.
Now what Cantor says is that if x is an ement of f(x), then
it isnOt
an ement of S, which is a subset of X, and if x isnOt an
ement of
f(x), then it is an ement of S.
So, for example as 1 is not an ement of {}, Cantor sects an
ement of the powerset, S, and one of its ements is 1, because
your
function f has 1->{} and the set of all ements of the empty
set is
the empty set. So S={1, ...}, more ements than one havenOt
been
shown.
S canOt be a singleton in your mapping, because the
singletons are
mapped to mutiples of two. Many of those multiples of two,
even
numbers, x=2, 4, 6, ..., wonOt be ements of f(x), or se some
wouldnOt be mapped. It is certain that you can map the even
numbers
to the singletons containing ements of N. What happens though
is
that any process would have at least one ement, and infinity
many
ements, where x is not an ement of f(x), and is thus adding
to the
proposed value of S. To minimize the ements added to S, each
even
number x is mapped to {x}, f(x)={x}, and S is {1, 3, 5, ...},
the set
of odd numbers.
In mapping the unordered pairs to odd multiples of three,
x=3, 9, 15,
..., then it is not as clear as the case of the even numbers
that any
more ements would be contributed to S, and so on and so forth
for
the following pOth primes to map to p-sets, or rather
kOth
primes to
map to k-subsets of N.
It seems clear that the minimal set S, a subset of X or in
this case
N, is the set of odd integers. Then, to which prime is mapped
this
infinite subset of N? There are infinity many
prime numbers, but
each natural integer is finite, and there are
infinity many odd
integers.
Now what you might consider is mapping N to Z and Z to P(N),
establishing a bridge, as it were. Here it seems you are
using N as
{1, 2, 3, ...} instead of {0, 1, 2, 3...}. LetOs say you
consider
mapping N union {-1} to P(N). What you can then do is have
each kOth
prime map to a k-subset, which is fair as N maps to NxNxN...
finity
many times, in a way thus that S=N, not the odds, as above,
or evens
or primes or other infinite subset of the naturals, the whole
thing.
Then, set f(-1)=N, or whatever other infinite subset of N it
may have
been. The mapping between N and N U {-1} is f(x)=x-1. ItOs
only one
ement difference. Then youOve described bijections between N
U {any
ement not in N} and P(N) and between N and N U {any ement not
in
N}.
ThatOs pretty good.
WeOve talked about this somewhat before, couched in terms of
immediate and deferred.
Grime raises the important point: to what values then are
mapped all
the other infinite subsets of N? As we discussed above each
prime is
finite. (This is in lieu of consideration of a prime, and/or
composite, at infinity.) What I have in mind is a restriction
of the
powerset definition, a construct, thus that it is, for an
(countably)
infinite set, the set of all finite subsets, and
(unioned with)
any
finite set of the infinite subsets of the
infinite set. If this
subset of the powerset has only finity many
infinite ements, it
is
countable, otherwise, it is uncountable.
For example, the set of all finite subsets of an (countably)
infinite
set is countable, the set of all infinite subsets of an
infinite set,
uncountable. A subset of all infinite subsets of an
infinite
set is
countable or uncountable, depending on which they are.
Chapman then chimes in with that cranks can map to the
cofinite sets
as wl of infinite sets. One method for this would be in
mapping the
evens to some finite set and the odds to their binary
complement as
infinite binary sequences representing subsets of the
infinite
set.
What are left to map in that sense are non-cofinite (as a
subset of an
infinite set, the complement within the superset is not
finite)
infinite subsets of the infinite set. In the past
I suggested
something along the lines of a tier progression hierarchy.
40hot.rr.com
IOm reminded of statements that uncountable sets are
countable unions
of countable sets, as was mentioned sewhere.
Virgil: ...for the Snark would be a Boojum, you see.
It also leads into considerations like if there are infinity
many
integers, are there infinite integers, surreal
numbersO
generative
day omega, p-adics, hyperintegers, etcetera.
For proponents of the utility of countability and cardinals,
what you
need to show Missourrians (from the Show-Me state) its
validity is
solve a story problem with it.
Why is the powerset mapping the object of so much scrutiny? I
think
much of it has to do with concern about the contents of the
number
system: What are the real numbers?, How many of the real
numbers
are integers?, etcetera. Others associate breakthrough
arguments
about it with recognition. Me, I think infinite sets are
equivalent.
ItOs convenient for some extensions of analysis to the
infinite.
Also, I think that besides cardinality there are other,
useful,
measures of infinite quantities. Generally,
thatOs accepted,
for
example, the number-theoretic asymptotic density of the even
integers
within the integers is 1/2, corresponding to a natural
language
statement: half of the integers are even. We can say that
there are
more rationals than integers, for various reasons. IOm
reminded of
KatzO paper.
We can examine the infinite binary sequences, with a
beginning,
representing numbers between zero and one inclusive, [0,1),
depending
upon various interpretations, and examine their
characteristics and
propensities, even determining asymptotic (read: analytical
infinitesimal) expressions of them.
Good luck.
Ross
Since Nicolas puts so much store in his proof that it is
trivial to
show
that there is no bijection between N and R, using decimal
expansions of
real numbers in [0,1), and since others have pointed out that
he left out
all the infinite subsets of N in his bijection between N and
P(N) (and
I
also pointed out that he has also left out many finite subsets
of N), I
thought that I would concentrate this particular response on
his PROOF 1,
and demonstrate exactly why it does not hold up to scrutiny.
>PROOF 1
> A METHOD TO PROVE THAT NATURALS CANNOT UNDERTAKE THE
COUNTING OF THE
>REALS DIRECTLY FROM ITS DECIMAL EXPANSION
>It is possible to represent any real number using a given
positional
>system of numeration. For ease we will use the decimal
notation.
>Proposition:
>The transfinite construction N <-> R is not possible
>Proof:
>Every real number within the interval (0, 1) has as first
decimal
As can be seen from bow, 0 should be an ement of the interval
since
it is being listed bow. The correct interval is [0,1).
>digit one of the ten digits of the decimal system of
numeration, i.
>e. it must be of the form 0.0, 0.1, 0.2, ..., 0.9. As you
can see for
>this first digit there are 10 ^ 1 possibilities. For the
second digit
>we have 0.00, 0.01, 0.02,..., 0.99. Consequently, there are
10 ^2
>possible combinations for the two first digits, 10 ^3 for the
first
>three digits, and so on. When the number of digits is
infinite we get
>R.
When the number of digits is infinite, we get the ements of R
with a
nonterminating decimal expansion.
>Now, when we have only a digit of the decimal expansion of
the reals,
>we make the one-to-one correspondence 0 <-> 0.0, 1 <-> 0.1,
2 <-0.2, ..., 9 <-> 0.9. Next, with two digits we begin the
1-1
>correspondence 0 <-> 0.00, 1 <-> 0.01,..., 10 <-> 0.10, 11
<-> 0.11, ...,
>99 <-> 0.99. With three digits, we continue doing the e, and
so on.
>And now is when the impossible transfinite construction
appears.
>While we keep doing the 1-1 correspondence within the finite
decimal
>expansion, everything will go fine.
What Nicolas has written above will NOT go fine. He will not
end up with
a bijection between N and the terminating decimals in the
interval [0,1).
I urge him to think carefully about it, and ask himsf what
natural
number corresponds to 0.1 in his intended limiting bijection
between N and
the terminating decimals in the interval [0,1), or what
terminating
decimal corresponds to 1 in his intended limiting bijection.
The answer
is that neither of these questions has a legitimate answer
since there is
no limiting bijection. But the error here is easily fixed.
Just take
0 <-> 0, 1 <-> 0.1, ..., 9 <-> 0.9, 10 <-> 0.01, 11 <-> 0.11,
12 <-> 0.21,
...., 254987631 <-> 0.136789452, .... Note that I am
REVERSING the order
of the digits, and placing them after the decimal place. THIS
will
determine a bijection such as Nicolas requires. NicolasO way
certainly
wonOt.
And note that I can immediaty tl what natural number
corresponds to a
given terminating decimal, and vice versa.
>But, what natural number would
>correspond to the decimal expansion of the number e?
NO natural number would correspond to e. e is between 2 and
3, and so e
is not an ement of [0,1). It follows that e is not in the
range. A
better question would have been: What natural number would
correspond to
the decimal expansion of the number e-2? The reason why this
would have
been a better question is that e-2 DOES fall in the interval
[0,1).
>By induction it
>is obvious that it should be 71828182...,
With NicolasO ill-defined, and nonexistent,
bijection. With my
corrected
bijection, the answer would be ...828182817, i.e. not a
natural number.
>i. e., a number with infinite
>nonzero digits, which doesnOt bong to N (Point 1).
This is true, but does not lead where Nicolas thinks it does.
>Conclusion:
>The transfinite construction N <->R is not possible, i. e.
the naturals
>cannot undertake the counting of the reals directly from its
decimal
>expansions.
Here, Nicolas would use the fact that there exists a
bijection between N
and the terminating decimals (specifically, one
specific
bijection
supplied by ME, by the way, since Nicolas supplied no such
bijection at
all), to assert that there is no bijection between N and R.
NicolasO
argument is based on the statement that there cannot be a
bijection
between N and a subset of [0,1) which properly contains as a
subset the
set of all terminating decimals in [0,1). In other words, if
A is a
subset of [0,1), and A contains all the terminating decimals
in [0,1) as
ements, and A contains at least one nonterminating decimal as
an
ement, then Nicolas would have us bieve that the existence of
my
bijection above forbids the existence of a bijection between
N and A. So
let us take a look at a set B whose ements are the
terminating decimals
in [0,1), and also e-2. Can I find a bijection between N and
B? Since
Nicolas expect us to buy his argument, then he would
presumably answer
No. But the correct answer is Yes. Take the correspondence 0
<->
e-2,
1 <-> 0, 2 <-> 0.1, ..., 9 <-> 0.8, 10 <-> 0.9, 11 <-> 0.01,
12 <-> 0.11,
13 <-> 0.21, ..., 1259000 <-> 0.9998521, ..., 12356987 <->
0.68965321,
.... Note the rule: 0 corresponds to e-2, and for natural
numbers greater
than zero, subtract 1, reverse the order of the digits, and
place after
the decimal point.
The point that I am making is that just because I can come up
with a
bijection between N and the terminating decimals in [0,1),
that does not
of itsf forbid any bijections between N and [0,1), since we
have not
iminated the possibility that I can manipulate the bijection
so that it
is a bijection between N and [0,1). Of course, I cannot so
manipulate the
bijection in this manner, but more sophisticated tools are
required to
prove it than Nicolas have used in his proof.
In short, Nicolas ried on the unstated assumption that N is
not
bijective with a proper subset of itsf, an assumption which
is known to
be false.
PROOF 1 is worthless.
David McAnally
Despite anything you may have heard to the contrary,
the rain in Spain stays almost invariably in the hills.
I have the follwing problem.
A rocket of rest length 600m move directly away from earth
with
constant vocity. A radar signal sent from earth is rected off
the
rear and front of the rocket, with a 17.4 microsecond day. The
first pulse is received back 200 sec after emission. Calculate
the
distance of the rocket from the earth when the signal reaches
it.
Calculate the veocity of the rocket rative to earth.
For calculating the distance it was easy. Since it takes the
first
pulse 200 sec to be received back. It only takes 100sec for
the pulse
to get to the rocket. So the distance of the rocket from the
earth
when the signal reaches it is just: d = c*t = 3.0x10^8 m/s *
100 s
= 3.0x10^10m
I am having trouble with the second question...calculate the
vocity
of the rocket rative to the earth. I am trying to use the
length
contraction formula to calculate the vocity (i.e. Dta t =
sqrt(1-v^2)* (dta tO). We know the length of the rocket is
600m in
the earth frame of reference. But I am not sure how to
calculate the
length of the rocket in the frame of reference of the rocket.
Can
anyone hp?? How do I use he 17.4 microsecond day. Am I
approaching the problem wrong?
>A rocket of rest length 600m move directly away from earth
with
>constant vocity. A radar signal sent from earth is rected
off the
>rear and front of the rocket, with a 17.4 microsecond day.
The
>first pulse is received back 200 sec after emission.
Calculate the
>distance of the rocket from the earth when the signal
reaches it.
>Calculate the veocity of the rocket rative to earth.
For calculating the distance it was easy. Since it takes the
first
>pulse 200 sec to be received back. It only takes 100sec for
the pulse
>to get to the rocket. So the distance of the rocket from the
earth
>when the signal reaches it is just: d = c*t = 3.0x10^8 m/s *
100 s
>= 3.0x10^10m
I am having trouble with the second question...calculate the
vocity
>of the rocket rative to the earth. I am trying to use the
length
>contraction formula to calculate the vocity (i.e. Dta t =
>sqrt(1-v^2)* (dta tO). We know the length of the rocket is
600m in
>the earth frame of reference. But I am not sure how to
calculate the
>length of the rocket in the frame of reference of the
rocket. Can
>anyone hp?? How do I use he 17.4 microsecond day. Am I
>approaching the problem wrong?
The length of the rocket in its frame of reference is 600m.
The length
in our frame of reference is 600m sqrt(1-(v/c)^2). The time
it takes
the light to catch the front of the ship, 8.7 microseconds, is
600m sqrt(1-(v/c)^2)/(c-v) so
1+v/c
8.7e-6 c = 600 sqrt( ====- )
1-v/c
1+v/c
4.35 = sqrt( ====- )
1-v/c
v/c = .899610993
Since the day is given to 2 places, we should only use 2
places of
this result, v/c = .90. Therefore, I would say that the
rocket is
moving away from us at about 270000 km/sec.
Rob Johnson <rob@trash.whim.org>
take out the trash before replying
>A rocket of rest length 600m move directly away from earth
with
>constant vocity. A radar signal sent from earth is rected
off the
>rear and front of the rocket, with a 17.4 microsecond day.
The
>first pulse is received back 200 sec after emission.
Calculate the
>distance of the rocket from the earth when the signal
reaches it.
>Calculate the veocity of the rocket rative to earth.
>For calculating the distance it was easy. Since it takes
the first
>pulse 200 sec to be received back. It only takes 100sec for
the pulse
>to get to the rocket. So the distance of the rocket from the
earth
>when the signal reaches it is just: d = c*t = 3.0x10^8 m/s *
100 s
>= 3.0x10^10m
>I am having trouble with the second question...calculate
the vocity
>of the rocket rative to the earth. I am trying to use the
length
>contraction formula to calculate the vocity (i.e. Dta t =
>sqrt(1-v^2)* (dta tO). We know the length of the rocket is
600m in
>the earth frame of reference. But I am not sure how to
calculate the
>length of the rocket in the frame of reference of the
rocket. Can
>anyone hp?? How do I use he 17.4 microsecond day. Am I
>approaching the problem wrong?
The length of the rocket in its frame of reference is 600m.
The length
> in our frame of reference is 600m sqrt(1-(v/c)^2). The time
it takes
> the light to catch the front of the ship, 8.7 microseconds,
is
> 600m sqrt(1-(v/c)^2)/(c-v) so
1+v/c
> 8.7e-6 c = 600 sqrt( ====- )
> 1-v/c
1+v/c
> 4.35 = sqrt( ====- )
> 1-v/c
v/c = .899610993
Since the day is given to 2 places, we should only use 2
places of
> this result, v/c = .90. Therefore, I would say that the
rocket is
> moving away from us at about 270000 km/sec.
Rob Johnson <rob@trash.whim.org take out the trash before
replying
How did you find out that it takes 8.7 microseconds for the
light to
reach the front of the rocket? Did you just divide 17.4
microseconds
by 2? If so why? Is it because it takes 200 sec for the pulse
to
return, so it takes 100 sec for the pulse to reach the rocket
=> half
the day should be in the 100 sec?
>A rocket of rest length 600m move directly away from earth
with
>constant vocity. A radar signal sent from earth is rected
off the
>rear and front of the rocket, with a 17.4 microsecond day.
The
>first pulse is received back 200 sec after emission.
Calculate the
>distance of the rocket from the earth when the signal
reaches it.
>Calculate the veocity of the rocket rative to earth.
For calculating the distance it was easy. Since it takes the
first
>pulse 200 sec to be received back. It only takes 100sec for
the pulse
>to get to the rocket. So the distance of the rocket from the
earth
>when the signal reaches it is just: d = c*t = 3.0x10^8 m/s *
100 s
>= 3.0x10^10m
I am having trouble with the second question...calculate the
vocity
>of the rocket rative to the earth. I am trying to use the
length
>contraction formula to calculate the vocity (i.e. Dta t =
>sqrt(1-v^2)* (dta tO). We know the length of the rocket is
600m in
>the earth frame of reference. But I am not sure how to
calculate the
>length of the rocket in the frame of reference of the
rocket. Can
>anyone hp?? How do I use he 17.4 microsecond day. Am I
>approaching the problem wrong?
The length of the rocket in its frame of reference is 600m.
The length
>> in our frame of reference is 600m sqrt(1-(v/c)^2). The
time it takes
>> the light to catch the front of the ship, 8.7
microseconds, is
>> 600m sqrt(1-(v/c)^2)/(c-v) so
1+v/c
>> 8.7e-6 c = 600 sqrt( ====- )
>> 1-v/c
1+v/c
>> 4.35 = sqrt( ====- )
>> 1-v/c
v/c = .899610993
Since the day is given to 2 places, we should only use 2
places of
>> this result, v/c = .90. Therefore, I would say that the
rocket is
>> moving away from us at about 270000 km/sec.
How did you find out that it takes 8.7 microseconds for the
light to
>reach the front of the rocket? Did you just divide 17.4
microseconds
>by 2? If so why? Is it because it takes 200 sec for the
pulse to
>return, so it takes 100 sec for the pulse to reach the
rocket => half
>the day should be in the 100 sec?
It takes the light the e time to go out to a point in space
as it
takes to come back from that point in space. Thus, the time
to the
close end of the ship is 100 sec (200/2 sec) and the time to
the far
end of the ship is 100.0000087 sec (200.0000174/2 sec).
Therefore, the
time it took to trav the length of the ship, at a rative
vocity
of c-v in our frame of reference, was 8.7 microseconds. The
length of
the ship in our frame of reference is 600m sqrt(1-(v/c)^2).
Equating
the rative distances travled, we get
8.7e-6 (c-v) = 600 sqrt(1-(v/c)^2)
which leads to the solution above.
Rob Johnson <rob@trash.whim.org>
take out the trash before replying
> Again, for the sake of argument, letOs assume that a set
>closed under the successor function must contain some
>ement that is not a natural number.
>PA and ZF do not rule out this possibility.
> Let @ and # represent two unnatural numbers.
>@ and # do not have to have successors because
>they are not natural numbers.
> Define two sets:
>x = (0,1,2,...,@)
>y = (0,1,2,...,#)
>Both x and y are closed using my definition.
> Inductivy closed sets must contain the successor of every
memnber,
and
> that successor must be different from the member itsf.
> What are the successors of @ and #?
> So that your definition is ?wed.
The induction axiom only says that an inductive set contains
> every natural number. It says nothing about members
> that are not natural numbers. @ and # donOt have to have
> successors.
There are lots of sets that contain the natural numbers. That
is not the
issue. We are looking for sets that satisfy the Peano
postulates. They
are rarer, and contain exactly one member without a
predecessor in the
e set, and contain no members without successors in the e set.
Let z be the intersection of x and y.
>z = (0,1,2,...)
> z is not closed under the definition I give above.
>It is impossible to prove z contains every natural number.
> As your sets are not inductive, they are also not revant.
Yes they are. Sets x and y contain every natural number.
> (Set z can not be shown to have this property.)
What is your definiton of inductive then? The common
definition
requires an inductive set to contain the successors of every
member of
the set, Your sets do not, so are not inductive in the common
meaning of
that word.
I am not trying to prove there are unnatural numbers.
> I am just trying to show why the proofs given that N
> only contains natural numbers are not sufficient.
Only when you change critical definitions! If you are allowd
to change
definitions any wqay you want, you can prove anything.
> If every inductive set also includes a member that is not a
natural
number
> then the intersection could contain a member that is not a
natural
number.
That is a misunderstanding of the definition. The
definition
says that
if x is in every inductive set, then this very fact is what
makes x a
natural number.
>> We can exclude any non-natural number, and all its
successors other than
>> zero and still have the inductive property.
That only works if the number being excluded is not itsf a
successor.
> But, you would lose the ability to show this set is closed
under the
> successor
> function. Without this you can not prove the set is
inductive.
Any member that is not a successor can be excluded, along
with all its
successors. The natural numbers have the property that 0 is
the only
non-successor.
>> So how that intersection contain anything except natural
numbers? THAT
>> is the set we call N, and its members are the objects we
call natural
>> numbers, and it canOt contain anything se.
> I havenOt seen a proof of this that doesnOt
assume what it
is trying to
> prove.
We do not intend to prove that the natural numbers are the
intersection
of all inductive sets. We mery assert that the intersection
exists.
Whatever that set is, its members are defined to be the
natural numbers.
>> Inductivy closed sets must contain the successor of every
memnber, and
>> that successor must be different from the member itsf.
>> What are the successors of @ and #?
>> So that your definition is ?wed.
> The induction axiom only says that an inductive set contains
> every natural number.
HeOs talking about the definition of an
inductive set, as
mentioned at
<http://mathworld.wolfram.com/InductiveSet.html>. You are
talking about
the fifth Peano axiom (the induction axiom), which is not the
e thing.
The definition we are talking about here is the one that says
N, the set
of natural numbers, is identical to the smallest inductive
set. That
means that if x bongs to every inductive set, then it follows
that x is
a natural number.
>It says nothing about members
> that are not natural numbers. @ and # donOt have to have
> successors.
They must have successors in order for the set to be
inductive.
> Let z be the intersection of x and y.
> z = (0,1,2,...)
> z is not closed under the definition I give above.
> It is impossible to prove z contains every natural number.
>> As your sets are not inductive, they are also not revant.
> Yes they are. Sets x and y contain every natural number.
> (Set z can not be shown to have this property.)
No, they are not. The definition mentions the intersection of
all
inductive sets. Sets that are not inductive do not get to
participate in
the intersection. Your sets are not inductive.
> I am not trying to prove there are unnatural numbers.
> I am just trying to show why the proofs given that N
> only contains natural numbers are not sufficient.
But you proved instead that you have misunderstood the
definition.
==
Heıs talking about the definition of an
inductive set, as mentioned at
> <http://mathworld.wolfram.com/InductiveSet.html>. You are
talking about
> the fifth Peano axiom (the induction axiom), which is not
the e thing.
The definition we are talking about here is the one that says
N, the set
> of natural numbers, is identical to the smallest inductive
set. That
> means that if x bongs to every inductive set, then it
follows that x is
> a natural number.
This is one reason I dislike the term inductive set. It
suggests
that the set in question satisfies some induction scheme, but
it isnOt
so (not necessarily).
Also, too many authors have used variations of the term. I
*think*
you are using RoitmanOs definition (as given on
the mathworld
page).
Another area of confusion is over the successor operation. I
think
that you have a fixed successor in mind (the usual y u {y}
successor),
while Virgil prefers to discuss sets with arbitrary
distinguished
ement and arbitrary successor. This makes some of the
statements
the two of you make evidently con?ct (no example at hand,
but IOve
noticed a time or two in the last couple days).
Wait: I think this is an example of some confusion. You say
that any
inductive set which satisfies the induction axiom of Peano is
necessarily N. I assume that this must be so only because we
have
fixed the successor operation to be y u {y} and so
non-standard mods
are ruled out by foundation (what if weOre in
anti-wl-founded
set
theory, I wonder?). Is this right?
Personally, my bias lies with Virgil on this. The natural
numbers
really have nothing to do with particular mods in N, so we
shouldnOt
fix our notion of successor to particular operators on Set.
But, I
canOt say anything nearly so clever as, oh, Benacerraf (What
Numbers
Could Not Be), so I wonOt bother to say anything more.
Structuralism rulez, dood.
==
After years of arguing I realize that your intlects are too
limited
to fully grasp my work. [...]Still, no matter how child-like
your minds
are, [...] since you have language, [...] thereOs a chance
that IOll
be able to find something that your minds can handle. ==
>> HeOs talking about the definition of an
inductive set, as
mentioned at
>> <http://mathworld.wolfram.com/InductiveSet.html>. You are
talking about
>> the fifth Peano axiom (the induction axiom), which is not
the e
thing.
>> The definition we are talking about here is the one that
says N, the set
>> of natural numbers, is identical to the smallest inductive
set. That
>> means that if x bongs to every inductive set, then it
follows that x
is
>> a natural number.
> This is one reason I dislike the term inductive set. It
suggests
> that the set in question satisfies some induction scheme,
but it isnOt
> so (not necessarily).
> Also, too many authors have used variations of the term. I
*think*
> you are using RoitmanOs definition (as given
on the
mathworld page).
> Another area of confusion is over the successor operation.
I think
> that you have a fixed successor in mind (the usual y u {y}
successor),
Not always; itOs just a convenient example.
> while Virgil prefers to discuss sets with arbitrary
distinguished
> ement and arbitrary successor. This makes some of the
statements
> the two of you make evidently con?ct (no example at hand,
but IOve
> noticed a time or two in the last couple days).
> Wait: I think this is an example of some confusion. You say
that any
> inductive set which satisfies the induction axiom of Peano
is
> necessarily N.
I would say any inductive set which satisfies the induction
axiom is an
acceptable mod of N. That doesnOt mean all such sets are
identical,
even if we confine our attention to a specific mod
of ZF.
>I assume that this must be so only because we have
> fixed the successor operation to be y u {y} and so
non-standard mods
> are ruled out by foundation (what if weOre in
anti-wl-founded set
> theory, I wonder?). Is this right?
I donOt object to nonstandard mods, but sury we need to
concentrate
on just one mod at a time. Within any given mod, I think my
statement is correct. That doesnOt mean that the mod of PA
that you
get in RobinsonOs NSA is identical to the one we are used
to,
but any
other mod of PA that is internal to RobinsonOs theory will
look
essentially like the one you get in that theory by using the
s(y) = y U
{y} successor operation. That is, there is an internal
isomorphism
between the two.
> Personally, my bias lies with Virgil on this. The natural
numbers
> really have nothing to do with particular mods in N, so we
shouldnOt
> fix our notion of successor to particular operators on Set.
But, I
> canOt say anything nearly so clever as, oh, Benacerraf
(What Numbers
> Could Not Be), so I wonOt bother to say anything more.
> Structuralism rulez, dood.
I think the only apparent con?ct here has to do with what I
mean when I
say that the set N *is* the integers. By that I mean, up to
isomorphism,
rative to the chosen mod.
ItOs like the situation we have with
real-numbers-as-Dedekind-cuts vs.
real-numbers-as-equivalence-classes-of-Cauchy-sequences. We
are
justified in saying that either of those constructions gives
us R.
==
==
>> Wait: I think this is an example of some confusion. You
say that any
>> inductive set which satisfies the induction axiom of Peano
is
>> necessarily N.
I would say any inductive set which satisfies the induction
axiom is an
> acceptable mod of N. That doesnOt mean all such sets are
identical,
> even if we confine our attention to a specific
mod of ZF.
I think the confusion has been mine, due to my laziness. I
never read
the Peano axioms as given on mathworld, but finally Russl
quoted
them. I see that the induction axiom there is for all sets,
not for
all formulas. Therefore, PA really does characterize N up to
isomorphism, right?
>> Structuralism rulez, dood.
I think the only apparent con?ct here has to do with what I
mean when I
> say that the set N *is* the integers. By that I mean, up to
isomorphism,
> rative to the chosen mod.
to isomorphism.
> ItOs like the situation we have with
real-numbers-as-Dedekind-cuts vs.
> real-numbers-as-equivalence-classes-of-Cauchy-sequences. We
are
> justified in saying that either of those constructions gives
us R.
==
I think the burden is on those people who think he didnOt
have
weapons of mass destruction to tl the world where they are.
== White House spokesman Ari Fleischer
|I think the confusion has been mine, due to my laziness. I
never read
|the Peano axioms as given on mathworld, but finally Russl
quoted
|them. I see that the induction axiom there is for all sets,
not for
|all formulas. Therefore, PA really does characterize N up to
|isomorphism, right?
No. Be aware that PA, although it stands for Peano arithmetic,
does not consist of the axioms originally given by Peano.
Peano
induction axiom refers either to all properties or all
subsets of the
domain.
PA has induction only for first-order formulas, and
doesnOt
characterize
N up to isomorphism. One main reason I suppose for considering
PA is that PA is a formal system, and PeanoOs axioms
donOt
constitute a formal system. A formal system is an
axiomatization
in which all the consequences can be computably enumerated.
> |I think the confusion has been mine, due to my laziness. I
never read
> |the Peano axioms as given on mathworld, but finally Russl
quoted
> |them. I see that the induction axiom there is for all
sets, not for
> |all formulas. Therefore, PA really does characterize N up
to
> |isomorphism, right?
No. Be aware that PA, although it stands for Peano arithmetic,
> does not consist of the axioms originally given by Peano.
Peano
> induction axiom refers either to all properties or all
subsets of the
> domain.
PA has induction only for first-order formulas, and
doesnOt
characterize
> N up to isomorphism. One main reason I suppose for
considering
> PA is that PA is a formal system, and PeanoOs axioms
donOt
> constitute a formal system. A formal system is an
axiomatization
> in which all the consequences can be computably enumerated.
Alrighty, then. Now *thatOs* how I always used the term PA,
too ==
first-order induction. But then Russl quoted the
PeanoOs Axioms
page from Mathworld, which presumably gives the original
axioms. So,
evidently, one must be careful: PA != the theory of PeanoOs
axioms.
In fact, PA != Peano Arithmetic, according to the entry at
Mathworld,
since Peano arithmetic is defined there as the theory of
PeanoOs
axioms.
==
Analysis/ editors have evaluated the paper, they accepted it
for
publication and they have the copyright of its contents - and
thus
they are responsible for its correctness,O she [said].
> Wait: I think this is an example of some confusion. You say
that any
> inductive set which satisfies the induction axiom of Peano
is
> necessarily N.
>> I would say any inductive set which satisfies the induction
axiom is an
>> acceptable mod of N. That doesnOt mean all such sets are
identical,
>> even if we confine our attention to a specific
mod of ZF.
> I think the confusion has been mine, due to my laziness. I
never read
> the Peano axioms as given on mathworld, but finally Russl
quoted
> them. I see that the induction axiom there is for all sets,
not for
> all formulas. Therefore, PA really does characterize N up to
> isomorphism, right?
I think for an isomorphism to exist, the two objects being
compared have
to be in the e mod. It doesnOt make sense to ask whether the
usual
N is isomorphic to RobinsonOs *N, since N is not an internal
set and
therefore there are not even any mappings between the two
according to
RobinsonOs internal set theory. If there are no mappings,
there canOt be
any isomorphisms.
Therefore, when I say the Peano axioms characterize N up to
isomorphism, I mean, in all contexts in which the concept of
an
isomorphism even makes sense. I may be missing something, but
thatOs
what I meant by the statement.
==
Seaman
Judge YohnOs mistakes revealed in Mumia Abu-Jamal ruling.
<http://www.commoncouragepress.com/index.cfm?action=book&
bookid=228 HeOs talking about the definition of an
inductive set, as
mentioned at
> <http://mathworld.wolfram.com/InductiveSet.html>. You are
talking
about
> the fifth Peano axiom (the induction axiom), which is not
the e
thing.
> The definition we are talking about here is the one that
says N, the
set
> of natural numbers, is identical to the smallest inductive
set. That
> means that if x bongs to every inductive set, then it
follows that x
is
> a natural number.
This is one reason I dislike the term inductive set. It
suggests
> that the set in question satisfies some induction scheme,
but it isnOt
> so (not necessarily).
Also, too many authors have used variations of the term. I
*think*
> you are using RoitmanOs definition (as given
on the
mathworld page).
Another area of confusion is over the successor operation. I
think
> that you have a fixed successor in mind (the usual y u {y}
successor),
> while Virgil prefers to discuss sets with arbitrary
distinguished
> ement and arbitrary successor. This makes some of the
statements
> the two of you make evidently con?ct (no example at hand,
but IOve
> noticed a time or two in the last couple days).
Wait: I think this is an example of some confusion. You say
that any
> inductive set which satisfies the induction axiom of Peano
is
> necessarily N. I assume that this must be so only because
we have
> fixed the successor operation to be y u {y} and so
non-standard mods
> are ruled out by foundation (what if weOre in
anti-wl-founded set
> theory, I wonder?). Is this right?
Personally, my bias lies with Virgil on this. The natural
numbers
> really have nothing to do with particular mods in N, so we
shouldnOt
> fix our notion of successor to particular operators on Set.
But, I
> canOt say anything nearly so clever as, oh, Benacerraf
(What Numbers
> Could Not Be), so I wonOt bother to say anything more.
Structuralism rulez, dood.
These are PeanoOs Axioms as given at Mathworld:
http://mathworld.wolfram.com/PeanosAxioms.html
1. Zero is a number.
2. If a is a number, the successor of a is a number.
3. zero is not the successor of a number.
4. Two numbers of which the successors are equal are themsves
equal.
5. (induction axiom.) If a set S of numbers contains zero and
also the
successor of every number in S, then every number is in S.
Stupid questions about PA.
1) What prevents a natnum from being its own successor?
2) Why canOt a natnum have more than one successor?
3) What prevents circular successors (a successor of b, b
successor of a)?
4) What prevents a natnum from not having a successor (like
0)?
5) Why do we need axiom 5?
I assume these questions are answered by the definition of
successor.
(except the last question.)
This is the definition at Mathworld:
http://mathworld.wolfram.com/Successor.html
It doesnOt answer any of these questions.
One would think that something as important as the definition
of successor would be part of the axioms.
Russl
- 2 many 2 count
> Heıs talking about the definition of an
inductive set, as
mentioned
at
><http://mathworld.wolfram.com/InductiveSet.html>. You are
talking
about
>the fifth Peano axiom (the induction axiom), which is not the
e
> thing.
> The definition we are talking about here is the one that
says N, the
set
>of natural numbers, is identical to the smallest inductive
set. That
>means that if x bongs to every inductive set, then it
follows that
x
> is
>a natural number.
> This is one reason I dislike the term inductive set. It
suggests
> that the set in question satisfies some induction scheme,
but it isnOt
> so (not necessarily).
> Also, too many authors have used variations of the term.
I *think*
> you are using RoitmanOs definition (as given
on the
mathworld page).
> Another area of confusion is over the successor
operation. I think
> that you have a fixed successor in mind (the usual y u {y}
successor),
> while Virgil prefers to discuss sets with arbitrary
distinguished
> ement and arbitrary successor. This makes some of the
statements
> the two of you make evidently con?ct (no example at hand,
but IOve
> noticed a time or two in the last couple days).
> Wait: I think this is an example of some confusion. You
say that any
> inductive set which satisfies the induction axiom of Peano
is
> necessarily N. I assume that this must be so only because
we have
> fixed the successor operation to be y u {y} and so
non-standard mods
> are ruled out by foundation (what if weOre in
anti-wl-founded set
> theory, I wonder?). Is this right?
> Personally, my bias lies with Virgil on this. The natural
numbers
> really have nothing to do with particular mods in N, so we
shouldnOt
> fix our notion of successor to particular operators on Set.
But, I
> canOt say anything nearly so clever as, oh, Benacerraf
(What Numbers
> Could Not Be), so I wonOt bother to say anything more.
> Structuralism rulez, dood.
> These are PeanoOs Axioms as given at Mathworld:
> http://mathworld.wolfram.com/PeanosAxioms.html
> 1. Zero is a number.
> 2. If a is a number, the successor of a is a number.
> 3. zero is not the successor of a number.
> 4. Two numbers of which the successors are equal are
themsves equal.
> 5. (induction axiom.) If a set S of numbers contains zero
and also the
> successor of every number in S, then every number is in S.
Stupid questions about PA.
1) What prevents a natnum from being its own successor?
One can show that if any natnum is its own successor then
also 0 must be
its own successor, which is prohibited by rule (3).
> 2) Why canOt a natnum have more than one successor?
The rules say the successor, not a successor.
> 3) What prevents circular successors (a successor of b, b
successor of
a)?
By rule (3), such a circle cannot contain 0, and by rule (5)
we can get
all natnumOs without ever including any members of such a
circle.
> 4) What prevents a natnum from not having a successor (like
0)?
What do you mean by saying that 0 has no sucessor? Rule (2)
guarantees
that every natnum has a successor.
> 5) Why do we need axiom 5?
(a) To answer some of your questions.
(b) To deal with the totality of the set of natnums.
(c) To allow inductive proofs.
I assume these questions are answered by the definition of
successor.
> (except the last question.)
> This is the definition at Mathworld:
> http://mathworld.wolfram.com/Successor.html
It doesnOt answer any of these questions.
> One would think that something as important as the
definition
> of successor would be part of the axioms.
What you would think is no measure of what the vast majority
of us would
think. There are lots of wildly different mods which satisfy
the Peano
axioms with quite different interpretations of successor so
it is only
the properties required by the axioms themsves for successors
that are
important.
For example:
Mod 1: the 0 ement is the empty set, { }, and the successor
of x is
the set whose only member is x, {x}.
Mod 2: the 0 ement is again the empty set, { }, but now the
successor of x is the set { {x},x}.
Mod 3: the zero ement is the real number pi, and the
successor of x
is x^2, i.e., the square of x.
All of these mods satisfy all the axioms quite nicy.
the real number 3 is the zero ement and successor of x is x/2
> These are PeanoOs Axioms as given at Mathworld:
> http://mathworld.wolfram.com/PeanosAxioms.html
> 1. Zero is a number.
> 2. If a is a number, the successor of a is a number.
> 3. zero is not the successor of a number.
> 4. Two numbers of which the successors are equal are
themsves equal.
> 5. (induction axiom.) If a set S of numbers contains zero
and also the
> successor of every number in S, then every number is in S.
Jesse Hughes has answered your questions, but I wanted to
comment on the
following:
> I assume these questions are answered by the definition of
successor.
> (except the last question.)
> This is the definition at Mathworld:
> http://mathworld.wolfram.com/Successor.html
> It doesnOt answer any of these questions.
Actually, it does, if you take that approach. Jesse (and
Virgil) take
the approach of defining the naturals to be a Peano set, where
the exact
meaning of 0 and successor are unspecified. I (and mathworld)
take
the approach that 0 = {} and s(n) = n U {n}, from which it
can be shown
that these definitions result in a Peano set.
> One would think that something as important as the
definition
> of successor would be part of the axioms.
Definitions and axioms are different things.
Definitions are
just a way
of assigning names to concepts; they do not allow us to
deduce new
things that were not already implied by the axioms.
But letOs take, for example, your question about why we
canOt
have
circularity: s(a) = b and s(b) = a. Jesse proved that this
canOt
happen by using the Peano axioms, but itOs also possible to
prove that
circularity is incompatible with the definition of successor
given at
mathworld (assuming regularity).
If s(a) = b, then we have a U {a} = b. Since we are also
assuming s(b) =
a, we must also have b U {b} = a. But then we have a =
s(s(a)) = s(b) =
s(a U {a}) = (a U {a}) U {a U {a}} = a U {a} U {{a}}. In
particular,
this means a is a member of itsf, which violates regularity.
==
> These are PeanoOs Axioms as given at Mathworld:
> http://mathworld.wolfram.com/PeanosAxioms.html
> 1. Zero is a number.
> 2. If a is a number, the successor of a is a number.
> 3. zero is not the successor of a number.
> 4. Two numbers of which the successors are equal are
themsves equal.
> 5. (induction axiom.) If a set S of numbers contains zero
and also the
> successor of every number in S, then every number is in S.
Stupid questions about PA.
1) What prevents a natnum from being its own successor?
Proof by induction. Let S be the set of all natural numbers n
such
that s(n) != n. Then 0 is in S. Suppose that n is in S, and
we must
show that s(n) is in S, i.e., that s(s(n)) != s(n). But, by
(4), is
s(s(n)) = s(n), then s(n) = n, contradicting our supposition
that n is
in S. Hence, S is inductive, i.e., S = N.
> 2) Why canOt a natnum have more than one successor?
Successor is a function by definition.
> 3) What prevents circular successors (a successor of b, b
successor
> of a)?
A proof by induction similar to (1).
> 4) What prevents a natnum from not having a successor (like
0)?
Successor is a function by definition.
> 5) Why do we need axiom 5?
To give proofs like that in (1). To ensure that every natural
number
is given by a finite iteration of successor applied to 0.
> It doesnOt answer any of these questions.
> One would think that something as important as the
definition
> of successor would be part of the axioms.
No, itOs entiry inconsequential. A mod for PA is any set X
together with a distinguished ement 0 in X and a *function*
s:X -> X
satisfying (1) - (5).
Question for people that know such things: is (5) the standard
induction axiom for PA? I expected induction over formulas
rather
than arbitrary sets. (5) is what I thought one may call
second-order
induction.
==
Analysis/ editors have evaluated the paper, they accepted it
for
publication and they have the copyright of its contents - and
thus
they are responsible for its correctness,O she [said].
> These are PeanoOs Axioms as given at Mathworld:
> http://mathworld.wolfram.com/PeanosAxioms.html
> 1. Zero is a number.
> 2. If a is a number, the successor of a is a number.
> 3. zero is not the successor of a number.
> 4. Two numbers of which the successors are equal are
themsves equal.
> 5. (induction axiom.) If a set S of numbers contains zero
and also the
> successor of every number in S, then every number is in S.
> Stupid questions about PA.
> 1) What prevents a natnum from being its own successor?
Proof by induction. Let S be the set of all natural numbers n
such
> that s(n) != n. Then 0 is in S. Suppose that n is in S, and
we must
> show that s(n) is in S, i.e., that s(s(n)) != s(n). But, by
(4), is
> s(s(n)) = s(n), then s(n) = n, contradicting our
supposition that n is
> in S. Hence, S is inductive, i.e., S = N.
At first I thought (4) prevented this, but it
doesnOt.
If there exists an n that is its own successor then (4)
proves that every natnum is equal to zero.
This may be boring, but I donOt see an axiom
that prevents it. Using induction we prove N=(0).
> 2) Why canOt a natnum have more than one successor?
Successor is a function by definition.
OK
> 3) What prevents circular successors (a successor of b, b
successor
> of a)?
A proof by induction similar to (1).
Again, I think all induction would prove is that every natural
number is equal to zero.
> 4) What prevents a natnum from not having a successor (like
0)?
Successor is a function by definition.
I meant to say what prevents a natnum from not being
a successor, like zero. (3) says zero is not the successor
of any natural number, but it doesnOt say that zero is
the ONLY such natural number.
> 5) Why do we need axiom 5?
To give proofs like that in (1). To ensure that every natural
number
> is given by a finite iteration of successor applied to 0.
So, all of this falls apart without induction.
> It doesnOt answer any of these questions.
> One would think that something as important as the
definition
> of successor would be part of the axioms.
No, itOs entiry inconsequential. A mod for PA is any set X
> together with a distinguished ement 0 in X and a *function*
s:X -> X
> satisfying (1) - (5).
Any function satisfying (1)-(5) is a successor function?
Sounds like a challenge.
Russl
- 2 many 2 count
> These are PeanoOs Axioms as given at Mathworld:
> http://mathworld.wolfram.com/PeanosAxioms.html
> 1. Zero is a number.
> 2. If a is a number, the successor of a is a number.
> 3. zero is not the successor of a number.
> 4. Two numbers of which the successors are equal are
themsves equal.
> 5. (induction axiom.) If a set S of numbers contains zero
and also the
> successor of every number in S, then every number is in S.
> Stupid questions about PA.
> 1) What prevents a natnum from being its own successor?
Proof by induction. Let S be the set of all natural numbers n
such
> that s(n) != n. Then 0 is in S. Suppose that n is in S, and
we must
> show that s(n) is in S, i.e., that s(s(n)) != s(n). But, by
(4), is
> s(s(n)) = s(n), then s(n) = n, contradicting our
supposition that n is
> in S. Hence, S is inductive, i.e., S = N.
> 2) Why canOt a natnum have more than one successor?
Successor is a function by definition.
> 3) What prevents circular successors (a successor of b, b
successor
> of a)?
A proof by induction similar to (1).
> 4) What prevents a natnum from not having a successor (like
0)?
Successor is a function by definition.
> 5) Why do we need axiom 5?
To give proofs like that in (1). To ensure that every natural
number
> is given by a finite iteration of successor applied to 0.
> It doesnOt answer any of these questions.
> One would think that something as important as the
definition
> of successor would be part of the axioms.
No, itOs entiry inconsequential. A mod for PA is any set X
> together with a distinguished ement 0 in X and a *function*
s:X -> X
> satisfying (1) - (5).
Question for people that know such things: is (5) the standard
> induction axiom for PA? I expected induction over formulas
rather
> than arbitrary sets. (5) is what I thought one may call
second-order
> induction.
==
> Analysis/ editors have evaluated the paper, they accepted
it for
> publication and they have the copyright of its contents -
and thus
> they are responsible for its correctness,O she [said].
> Question for people that know such things: is (5) the
standard
> induction axiom for PA? I expected induction over formulas
rather
> than arbitrary sets. (5) is what I thought one may call
second-order
> induction.
Peano of course stated it for sets. (And started with 1 not
0.)
If you want a first-order axiom system you have to restrict to
formulas. Probably only logicians are concerned about that.
>> Inductivy closed sets must contain the successor of every
memnber,
and
>> that successor must be different from the member itsf.
>> What are the successors of @ and #?
>> So that your definition is ?wed.
> The induction axiom only says that an inductive set contains
> every natural number.
HeOs talking about the definition of an
inductive set, as
mentioned at
> <http://mathworld.wolfram.com/InductiveSet.html>. You are
talking about
> the fifth Peano axiom (the induction axiom), which is not
the e thing.
OK. This is the definition given at Mathworld:
an inductive set is a nonempty partially ordered set in which
every ement
has a successor
> The definition we are talking about here is the one that
says N, the set
> of natural numbers, is identical to the smallest inductive
set. That
> means that if x bongs to every inductive set, then it
follows that x is
> a natural number.
This clears up my confusion. If you are defining every member
of N to
be a natural number then you ARE assuming N only contains
natural
numbers. Why did you bother to give a proof of this
definition?
>It says nothing about members
> that are not natural numbers. @ and # donOt have to have
> successors.
They must have successors in order for the set to be
inductive.
Easy enough. Let $ be the successor of @ and @ be the
successor of $. Similarly, % is the successor of # and #
is the successor of %.
Now:
x = [0,1,2,...,@,$]
y = [0,1,2,...,#,%]
> Let z be the intersection of x and y.
> z = (0,1,2,...)
> z is not closed under the definition I give above.
> It is impossible to prove z contains every natural number.
>> As your sets are not inductive, they are also not revant.
> Yes they are. Sets x and y contain every natural number.
> (Set z can not be shown to have this property.)
No, they are not. The definition mentions the intersection of
all
> inductive sets. Sets that are not inductive do not get to
participate in
> the intersection. Your sets are not inductive.
> I am not trying to prove there are unnatural numbers.
> I am just trying to show why the proofs given that N
> only contains natural numbers are not sufficient.
But you proved instead that you have misunderstood the
definition.
We were using different definitions.
I was using PAOs fifth axiom and
ZFOs AoI.
The definition you are using also assumes
every member of N is a natural number.
Would you consider this set to be inductive?
[0,1,2,...,w,w+1,w+2,...]
Russl
- 2 many 2 count
>> Inductivy closed sets must contain the successor of
every memnber,
> and
>that successor must be different from the member itsf.
>> What are the successors of @ and #?
>> So that your definition is ?wed.
> The induction axiom only says that an inductive set
contains
>every natural number.
> Heıs talking about the definition of an
inductive set, as
mentioned at
> <http://mathworld.wolfram.com/InductiveSet.html>. You are
talking
about
> the fifth Peano axiom (the induction axiom), which is not
the e
thing.
OK. This is the definition given at Mathworld:
an inductive set is a nonempty partially ordered set in which
every
ement
> has a successor
 The definition we are talking about here is the one that
says N, the
set
> of natural numbers, is identical to the smallest inductive
set. That
> means that if x bongs to every inductive set, then it
follows that x
is
> a natural number.
This clears up my confusion. If you are defining every member
of N to
> be a natural number then you ARE assuming N only contains
natural
> numbers. Why did you bother to give a proof of this
definition?
It says nothing about members
>that are not natural numbers. @ and # donOt have to have
>successors.
> They must have successors in order for the set to be
inductive.
Easy enough. Let $ be the successor of @ and @ be the
> successor of $. Similarly, % is the successor of # and #
> is the successor of %.
> Now:
> x = [0,1,2,...,@,$]
> y = [0,1,2,...,#,%]
>
Now consider the intersection of x and y.
>Let z be the intersection of x and y.
>z = (0,1,2,...)
> z is not closed under the definition I give above.
>It is impossible to prove z contains every natural number.
>> As your sets are not inductive, they are also not revant.
Yes they are. Sets x and y contain every natural number.
>(Set z can not be shown to have this property.)
> No, they are not. The definition mentions the intersection
of all
> inductive sets. Sets that are not inductive do not get to
participate
in
> the intersection. Your sets are not inductive.
> I am not trying to prove there are unnatural numbers.
>I am just trying to show why the proofs given that N
>only contains natural numbers are not sufficient.
> But you proved instead that you have misunderstood the
definition.
We were using different definitions.
> I was using PAOs fifth axiom and
ZFOs AoI.
> The definition you are using also assumes
> every member of N is a natural number.
He is using the fact that there is a minimal inductive set
containing 0.
Would you consider this set to be inductive?
> [0,1,2,...,w,w+1,w+2,...]
Yes, if you mean the set {0,1,2,...,w,w+1,w+2,...} with the
usual
successor operation, but it is by no means minimal in that
respect.
The set of all integers, Z, is also inductive. Now consider
the
intersection of your set and Z.
If N is the zero-origin set of naturals, then every inductive
set that
contains 0 contians N as a subset, and N is the intersection
of all
inductive sets containing 0.
Since for anyobject that is not a natural number, there is an
inductive
set containing 0 which does not have that object as a member,
the
intersection, N, also does not have it as a member.
>> The definition we are talking about here is the one that
says N, the set
>> of natural numbers, is identical to the smallest inductive
set. That
>> means that if x bongs to every inductive set, then it
follows that x
is
>> a natural number.
> This clears up my confusion. If you are defining every
member of N to
> be a natural number then you ARE assuming N only contains
natural
> numbers. Why did you bother to give a proof of this
definition?
You donOt prove definitions. You may be confused
by the fact
that I
suggested an alternate approach. Here are three ways to go
about
definining the natural numbers. What is deduced in one
approach may turn
out to be part of the definition in a different approach.
1. N = smallest inductive set (the definition we have been
using).
a. Define inductive set.
b. Show that the intersection of all inductive sets is a
nonempty
set.
c. Define this set to be N.
d. Show that N is a Peano set.
2. N = set of all finite ordinals (the alternate approach I
suggested).
a. Define ordinal.
b. Define natural number as an ordinal that remains wl
ordered when you reverse the order.
c. Define N to be the set of all natural numbers as
defined in
step b.
d. Show that N is a Peano set.
3. N = w = the first limit ordinal (yet another approach).
a. Show that, by the axiom of infinity, there exists a limit
ordinal. Let w be the smallest such.
b. Define N = w.
c. Show that N is a Peano set.
Notice that each method has a different starting point and
begins with
different definitions, all derived from ZF. In each case, we
define a
certain set N (the exact method differs), and in each case we
wind up
proving that N is a Peano set. ThatOs what matters.
> We were using different definitions.
> I was using PAOs fifth axiom and
ZFOs AoI.
PeanoOs fifth axiom is not a part of ZF. In
order to show that
PAOs
fifth axiom applies, you first have to say what N
is in terms
of ZF and
then show that your N satisfies the Peano axioms. I suggested
three
different ways of doing that above.
> The definition you are using also assumes
> every member of N is a natural number.
ThatOs a strange way of putting it. A definition
doesnOt assume
anything. When we define a triangle as a three-sided polygon,
we are not
assuming that triangles are polygons with three sides; we are
mery
stating that thatOs what the word triangle means.
> Would you consider this set to be inductive?
> [0,1,2,...,w,w+1,w+2,...]
Yes. ThatOs the e set that I called w+w in a previous post.
ItOs the
next limit ordinal after w.
==
> an inductive set is a nonempty partially ordered set in
which every
ement
> has a successor
>
[...]
Easy enough. Let $ be the successor of @ and @ be the
> successor of $. Similarly, % is the successor of # and #
> is the successor of %.
> Now:
> x = [0,1,2,...,@,$]
> y = [0,1,2,...,#,%]
The ration < defined by your successor is not a partial order.
You
have $ < @ < $ and hence $ < $.
==
So, at this time, IOd like to assure you that I am not
interested in
making sure mathematicians worldwide get fired.
IOve rethought
my
desire to go to Congress and try to get funding for
mathematicians
cut. == is a reasonable man. Whew!
> Again, for the sake of argument, letOs assume that a set
> closed under the successor function must contain some
> ement that is not a natural number.
No, letOs not.
> PA and ZF do not rule out this possibility.
Yes, they do.
> Let @ and # represent two unnatural numbers.
> @ and # do not have to have successors because
> they are not natural numbers.
> Define two sets:
> x = (0,1,2,...,@)
> y = (0,1,2,...,#)
> Both x and y are closed using my definition.
They are not inductive sets.
> Let z be the intersection of x and y.
> z = (0,1,2,...)
> z is not closed under the definition I give above.
Is the intersection of two green sets also green? Who cares?
We were
discussing inductive sets.
> It is impossible to prove z contains every natural number.
N is defined to be the intersection of all inductive sets. We
know this
collection is nonempty because of the axiom of infinity. The
intersection of any nonempty collection of sets is a set.
Each member of this intersection is, by virtue of its
membership in the
intersection, a natural number. ThatOs how the natural
numbers are
defined.
I did point out that there is more than one way to define the
naturals,
and that the various definitions are equivalent. But in the
context of
the specific definition that we started with, your
question
makes no
sense.
==
> The set of all linear combinations of a set of vectors
forms a vector
> space called the _span_ of the set.
> The dimension of this space is the largest number of
linearly
> independent vectors in the original set, which may be less
than the
> total number of vectors in the set.
> In your example, the 4(?) vectors only span a 3
dimensional space.
> And any 3 independent vectors in that 3-space form a
_basis_ for that
> space.
So in my case this could be L((1,0,0),(0,1,0),(0,0,1))
(without the
fourth
> coordinates, because they are usess in |R^3.) Right?
Wrong.
Because that underlying space is R^4, all the vectors have to
be vectors
in R^4.
And, in general, a three dimensional subspace of R^4 need not
have any
of {(1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1)} in a basis.
Note, by analogy, that a plane through origin (2-space) in
R^3 need not
contain any of the coordinate axes.
The best bet is to simply take an independent subset of the
given 4
vectors. In your case, IIRC, I think that any 3 of the 4 form
a linearly
independent set, so any any 3 of the 4 given vectors should
do as a
basis.
> Karl
> [P.S. (?) ShouldnOt it be 5, or did I misunderstand
something?
 By the way, I know this is another theme, but I didnOt
want
to start an
> extra
> thread for this: Do you, or does anybody know a good
internet-site about
> the basics of descriptive matrices and basis-transformation
matrices? IOm
> looking
> for a site with lots of examples but didnOt
find anything so
far.
> ]
>
<bubqk5$pm3$03$1@news.t-online.com If itOs correct, how can I build a 3D-space from 3 vectors
with 4
> dimensions? Or did I actually found a basis for a 4D-plane
and not
> a space?
>
The easiest example of a 3d space built with 3 vectors of a
4d space is
the space spanned by
(1,0,0,0), (0,1,0,0), (0,0,1,0)
In three space the xy plane is spanned by
(1,0,0) and (0,1,0)
> A second more of thought. Pick any non-abian group of finite
order n.
> Then (ab)^n = 1 = a^n b^n and (ab)^(n+1) = ab =
a^(n+1)b^(n+1).
Or, if you donOt want the initial exponent to be a multiple
of the
order, you could use the quaternion group (8 ements) with
exponents of
4 and 5. (I wonder whether it can be proven that exponents
like that
are the only ones that work in a non-abian group.)
==
http://members.iquest.net/~panoptes/
> A second more of thought. Pick any non-abian group of finite
order n.
> Then (ab)^n = 1 = a^n b^n and (ab)^(n+1) = ab =
a^(n+1)b^(n+1).
Hmmm...yeah, that seems to do it. I always miss solutions
like this==they
seem somehow unfair, and I overlook them. :-)
This reminds me of a homework problem I failed to get back in
my college
days. We were asked to show that some property that youOd
expect to imply
convergence of a sequence didnOt. I was unable to
find a
counterexample.
The answer? The sequence {1,1/2,1/3,...,1/n,...}, on the
interval (0,1].
It doesnOt converge because the number it would converge to,
0, is outside
the interval. Bah...seems like cheating to me! :-)
==
>Just a question: CouldnOt finding GCD of pairs
of the
>numbers and repeating the process be a better procedure?
It could be, depending on how you divided the numbers up
into pairs.
If
>you
>get it wrong, you take just as long as you would with the
naive
method,
>plus you use additional storage space for the intermediate
GCDs.
Any suggestions for how to divide up the numbers?
I am not yet sure of that. You have much less number of
>calls of GCD. As to intermediate results, they could be
>stored at where the pairs originally were.
 If you have n numbers, then you need exactly n-1 calls of
GCD, no
matter
> how you arrange the grouping. Some of them may be carried
out in
> parall, but it doesnOt decrease the number of calls that
need to be
> made.
round of GCD computation of the pairs to enter the
> GCD computation of the next round will become however
> less and less in magnitude. This could be an advantage
> in the general case, I would think.
M. K. Shen
Particularly if early on you hit on two which are rativy
prime.

>Just a question: CouldnOt finding GCD of pairs
of the numbers
>and repeating the process be a better procedure?
It could be, depending on how you divided the numbers up into
>pairs. If you get it wrong, you take just as long as you
would
>with the naive method, plus you use additional storage space
>for the intermediate GCDs.
Any suggestions for how to divide up the numbers?
I am not yet sure of that. You have much less number of
calls of
>GCD. As to intermediate results, they could be stored at
where
>the pairs originally were.
 If you have n numbers, then you need exactly n-1 calls of
GCD, no
> matter how you arrange the grouping. Some of them may be
carried
> out in parall, but it doesnOt decrease the number of calls
that
> need to be made.
computation of the pairs to enter the GCD computation of the
next
> round will become however less and less in magnitude. This
could be
> an advantage in the general case, I would think.
When doing GCD calculations in sequence, the number in your
accumulator will become small very fast, and thus only the
first few
calculations will tend to be time consuming with typically
distributed data. With typically distributed data, youOll
hit
1
pretty soon, anyway, and need not even look at the rest.
==
David Kastrup, Kriemhildstr. 15, 44793 Bochum
> In sci.math, Aunty Allyson
> Uncle Al seems to picture himsf as *the* omnipotent
> sf-declared garbage collector on this forum.
Omnipotent, definity not. Omniscient, maybe. Admittedly,
> I was under the impression his regular haunts were more
> along the lines of sci.physics, where he routiny
> excoriates cranks == but he can be quite pleasant with
> intligent questions. Of course they do have to be
> intligent questions == something along the lines of
> how does a rocket work? would not quite qualify as the
> answers are readily available on Google for those willing
> to search.
>
This is from a recent letter by closet pedophile Uncle Al:
<quote>
Look on the bright side: Niggers canOt be trained, spics
have
dignity, and the kikes are dying in Florida. Your basic wog
still
occupies the e vital niche it did under the British Raj -
loyal
trainable dog.
</quote>
By the way, wog is pedophile-AlOs term for Indians.
l8r, N. toff
Is the general family of ovals r1/r2^n =Constant studied when
r1 and
r2 are bipolar coordinates? n=1 are Apollonius circles, n=-1
are
Cassinian ovals and n=0 are Circles.
>Definition 2.2: The definition of
JO(k, p) doesnOt make sense.
JO(k,p)
>also depends on the numbers a and b. And you cannot define
that the
>number of ements of JO(k, p) is >= 2; this will only be
true
for
>certain choices of k, p, a and b.
More clearly, yes, there do exist, according to the
intersection
> notation in Definition 2.2, coverings with one ement,
indeed.
> But we just donOt use them for the sake of the proof for
the problem.
> It turns out that if we do NOT use it, THEN the argument
bow Definition
2.2
> works.
Thus, if we are given a=11 and b = 13, and J(0,11), then we
have,
> according to the right hand side of Definition 2.2,
JO(0,11). But we
> just do not use it. Use it or not has nothing to do with
> contradictions.
Say we are going to define something. Will we use the closure
axiom? Y
> or N. Will we use the associativity axiom? Y or N. Will we
use the
> the existence-of-identity-ement axiom? Y or N. Will we use
the
> existence-of-inverse-ement-for-every-ement axiom? Y or N.
> Just as history (or my heroes or our heroes or masters or
whatever)
> created a group with Y, Y, Y, Y, we just choose not to use
certain
things.
I think you are missing the point.
We are not discussing whether errors in your paper can be
fixed or not,
we are discussing whether there are errors in your paper.
When studying
a proof for the Goldbach conjecture everyone will stop at the
first
error. Until you fix this problem, your proof is worthless.
I showed you two errors; that was very generous of me because
I should
have stopped at the first one. Fix the problems, then try
again.
MATRIX THEORY!
http://slashdot.org/~MatrixTheorist/journal/58765
Been there - done that!
The Psychedick Pope
Saint Isidore of Laytonville
^.85^ Patron Saint of the Internet ^.85^
^.85^
http://apple2.org.za/gswv/me
AOXOMOXOA and ENESSA QUA ONNICA
IMO the most evil number is the 7.
I always have many 7s when I cannot solve something.
I think the 7 is an evil number.
> IMO the most evil number is the 7.
ThatOs obvious[1]:
**** THE PROOF THAT Seven IS EVIL ****
S E V E N
19 5 22 5 14 - as numbers
1 5 4 5 5 - digits added
_/ _/ _/ _/ _/
1 5 4 5 5 - digits added
Thus, Seven is 15455.
Turn the number backwards, divide by 11 - the symbol of
judgment and
disorder. The number is now 5041.
Turn the number backwards, and add 1904 - the year
Oppenheimer, the man
who
created the atomic bomb, was born. The number is now 3309.
Turn the number backwards, divide by 3 - the symbol of
fulfillment. The
number is now 3011.
Subtract 2591 from the number - this is the year killer fog
haunted
London,
written backwards. It gives 420.
Turn the number backwards, and add 1865 - the year Lincoln
was shot.
The
number is now 1889.
This clearly proves how evil the subject is. QED.
[1] http://lcamtuf.coredump.cx/evilfinder/ef.shtml
==
==Tim Smith
> IMO the most evil number is the 7.
ThatOs obvious[1]:
**** THE PROOF THAT Seven IS EVIL ****
S E V E N
> 19 5 22 5 14 - as numbers
> 1 5 4 5 5 - digits added
> _/ _/ _/ _/ _/
> 1 5 4 5 5 - digits added
Thus, Seven is 15455.
Turn the number backwards, divide by 11 - the symbol of
judgment and
> disorder. The number is now 5041.
Now 5041 = 7! + 1. This angst provoking discrepancy bodes ill
and allows
me
to shorten TimOs proof. QED
JB
> IMO the most evil number is the 7.
ThatOs obvious[1]:
**** THE PROOF THAT Seven IS EVIL ****
S E V E N
> 19 5 22 5 14 - as numbers
> 1 5 4 5 5 - digits added
> _/ _/ _/ _/ _/
> 1 5 4 5 5 - digits added
Thus, Seven is 15455.
Turn the number backwards, divide by 11 - the symbol of
judgment and
> disorder. The number is now 5041.
Turn the number backwards, and add 1904 - the year
Oppenheimer, the
man who
> created the atomic bomb, was born. The number is now 3309.
Turn the number backwards, divide by 3 - the symbol of
fulfillment.
The
> number is now 3011.
Subtract 2591 from the number - this is the year killer fog
haunted
London,
> written backwards. It gives 420.
Turn the number backwards, and add 1865 - the year Lincoln
was shot.
The
> number is now 1889.
> This clearly proves how evil the subject is. QED.
[1] http://lcamtuf.coredump.cx/evilfinder/ef.shtml
==
> ==Tim Smith
>
I would not have bieved it had I not seen this
incontrovertible proof
with
my own eyes. And here I was thinking 7 was lucky! What a fool
IOve been!
When I think of all the wasted years....
l8r, N. toff
Can someone please explain to me what
generators are? For example if the
question involved the sum of generators in p - where p is
prime,
what would that mean? How would i find the generators of
numbers such
as 5, 7, 11 and 13? :shock:
==-= 19 East/West-Coast Specialized Servers - Total Privacy
via Encryption
==
What class is this for? There are, to my knowledge, no such
things as
generators in p if p is a number. What I think youOre
talking
about
are generators of Z_p, where p is prime. Z_p is the cyclic
group of
order p == that is, the set of numbers (symbols, ements,
whatever
you want to call them) { 1, 2, 3, ... p } with an operaton
(IOll call
it # to avoid confusing you early on, although in normal math
weOd
just use a + sign). Here are the rules for #:
If a + b <= p, then a # b = a + b. Otherwise, a + b is more
than p,
but not more than twice p (since a <= p and b <= p), so p < a
+ b <=
2p. Since we want a number in Z_p = { 1, 2, 3, ... p }, that
is, a
number 0 < n <= p, we can subtract p from each side to get 0
< a + b -
p <= p, so a # b = a + b - p. (Sometimes, itOs more popular
to write
0 instead of p. It doesnOt really make any difference, since
adding p
to a number and then subtracting it (because youOve gone
over
p) is
the e as just adding 0 to the number)
You can think of this like how a clock works == if itOs 3
hours past
5, then itOs 5 + 3= 8 oOclock, but if
itOs 3 hours past even,
then
itOs 11 + 3 - 12 = 2 oOclock.
So, consider, say, Z_10. This is the set { 1, 2, 3, ... 10 }.
Consider what happens if we take, say, 1 and keep #Oing it
to
itsf:
1 = 1
1 # 1 = 2
1 # 1 # 1 = 1 # 2 = 3
1 # 3 = 4
etc...
1 # 9 = 10
If we do it again, we get back to 1. So we go in a loop
around Z_10,
covering every single ement. Now consider if we pick 3:
3 = 3
3 # 3 = 6
3 # 3 # 3 = 3 # 6 = 9
3 # 9 = 12 - 10 = 2
3 # 2 = 5
3 # 5 = 8
3 # 8 = 11 - 10 = 1
3 # 1 = 4
3 # 4 = 7
3 # 7 = 10
And if we did it again, weOd get back to 3. Again, we have
gone in a
loop around Z_10, covering every single ement. Now, one last
time,
try 4.
4 = 4
4 # 4 = 8
4 # 8 = 12 - 10 = 2
4 # 2 = 6
4 # 6 = 10
And if we did it once more, weOd have 4 again. But look! We
went in
a complete circle, but it only covered 2, 4, 6, 8, and 10 ==
not 1, 3,
5, 7, or 9.
Numbers like 1 and 3 that are capable of generating the
entire set
by just adding to themsves over and over again are called
generators, while numbers like 4, which do NOT, are not
generators.
There is a rule that shows which numbers are generators and
which are
not, which you should figure out on your own. As a hint, if p
is
prime, EVERY number other than p is a generator of Z_p.
Make sure to write back and tl me the context. Since I donOt
think
thereOs such a thing as say, a generator of 5 or something
like
that, IOm pretty sure whatever you were reading has a
mistake
in it,
but what it meant is a little hard to tl.
> Can someone please explain to me what
> generators are? For example if the
> question involved the sum of generators in p - where p is
prime,
> what would that mean? How would i find the generators of
numbers such
> as 5, 7, 11 and 13? :shock:
==-= 19 East/West-Coast Specialized Servers - Total Privacy
via Encryption
===-
>True, my apologies. If you are ever forced to use MS Windows
(yuk) and
^^^
I agree, but not fanatically, as you can see by my headers.
>TeX (also yuk) IOd recommend it. While there are
alternatives to MS
^^^
I do *not* agree, period!
>Windows, there donOt seem to be any serious ones to TeX==or
are there?
Not really, depending on oneOs aim (IOm told
that *roffs are
far from
being dead and buried, as some naivy think!). But who cares?
also
yuk is not a thing of mine...
==
> Comments should say _why_ something is being done.
Oh? My comments always say what _really_ should have
happened. :)
- Tore Aursand on comp.lang.perl.misc
>However quite about the only thing I know about Farey
sequences is
>that if aO/bO is the next Farey fraction,
then baO-abO=1, so
(1) b|abO+1
and then aO=(abO+1)/b. Now, if
bOO satisfies (1) in place of
bO and
>aOO is defined accordingly,
then
(2) aO/bO <
aOO/bOO <=>
bOO(abO+1) <
bO(abOO+1) <=> bO
> bOO.
OTOH, since bO is in {1..b-1} and (a,b)=1... (2) is more or
less
*complete rubbish*! You just have to solve (1) for bO, thus
bO = -a^{phi(b)-1}%b,
where m%n is the remainder of the division of m by n and phi
is
EulerOs totient function.
For example if a/b = 300/5069 then
5069=37*137 => phi(b)-1 = 36*136-1 = 4895,
a^4895 = 4579 (mod b) => bO = b-4579=490,
aO = (300*490+1)/5069 = 29.
Hence aO/bO = 29/490.
HTH,
==
> Comments should say _why_ something is being done.
Oh? My comments always say what _really_ should have
happened. :)
- Tore Aursand on comp.lang.perl.misc
I want to fit some pure complex data x (only imaginary part)
to some other
complex data, y, by a polynomial with linear least squares.
y=p0 T0(x) +p1 T1(x)+...+pn Tn(x)
where Ti(x) represents the iOth Chebyshev polynomial.
(i thought this would provide me with better numerical
conditioning,
since i get Vandermonde system to solve).
I need real coefficients in my polynomial, so i assume p0,
..., pn and the
coefficients of Ti(x) have to be real,
while x is complex. I calculate coefficients with (Ax=b)
With left matrix A :
Re(T0(x)) Re(T1(x)) .... Re(Tn(x))
Im(T0(x)) Re(T1(x)) .... Re(Tn(x))
... for all x
unknown coefficients are X=[p0 p1 ... pn]O
And right columnvector vector b :
Re(y)
Imag(y)
... for all y
Every equation is split in real and imaginary part to make
coefficients
real.
And thereOs my problem. This orthogonal technique works wl
if
x is real
and scaled in [0,1], since Ti(x) doesnOt return very high
values (only
between
0 and 1). But if x is complex, this property is not longer
valid !
e.g. T2(x)=2x^2-1 if x=1*i then T2(x)=-3
In fact, this method gets every sooner illconditioned than
when i donOt
use Chebyshev polynomials.
Does anyone know what iOm doing wrong, or does this
technique
only work
for real data??
Please hp me.
Alfred.
>>Awwww... ainOt that cute? The two little cranks are
playing
together!
> When I was working at MIT, a professor of my acquaintance
(IOll say
> The idea was to be kind, and yet at the e time, not waste
his own
> time which was better spent on other things. He thought
that just
> ignoring people or dismissing them would be unkind. And
since he
> worked at the AI lab and was high profile in various ways,
he gets
> look at it, see what the topic was, and pull the file of
other letters
> about the e topic. Then he would reply with a form letter
saying
> something like: IOm sorry, but I donOt have
the expertise
to address
> your letter. The following people may be better suited to
discuss
> your points. Then he would add the names and addresses of a
few of
> Now this struck me as wonderfully brilliant. Simultaneously
being
> kind, moreover, giving the nut someone to talk to who will
listen and
> respond in the e vein, satisfying both, keeping his own
time free
> for other things==amazing.
usenet.
Sury there are general programs for the purpose of
shortening strings with some given simple axioms
(with the usual problem that some strings must be
heavily lengthened before they can be shortened)?
==
Hauke Reddmann <:-EX8 fc3a501@uni-hamburg.de
als man ankam wollte man werden, die geschichte schreiben,
die doofen sollen sterben, der plan als man damals nach
hamburg kam
(Kettcar)
What you want is called a theorem prover. There are several
available, but mainly as parts of larger programs that use
them to
advice;
I donOt use them mysf very often.
On the other hand, if you want to prove that you CANOT
reduce
one
string to another, youOll probably need a program based that
searches
for invariants; IOm not sure if these are around, but if the
axioms
are simple enough, itOd be doable, if long and ugly.
> Sury there are general programs for the purpose of
> shortening strings with some given simple axioms
> (with the usual problem that some strings must be
> heavily lengthened before they can be shortened)?
>>I am using hyperlatex (and latex2html in addition) to
create web pages.
>>But hyperlatex seems to age without evolving. Do you know a
software
>>which could replace hyperlatex ?
> Not that thig ng is really *wrong*, but my general answer
to posts
> like this one is to check comp.text.tex, especially since
your
> particular question is not math-rated, unlike the one of
the poster
> who asked, a few days ago, for a particular symbol...
However IOve heard *very good* cmts of TeX4ht.
These are PeanoOs Axioms as given at Mathworld:
>http://mathworld.wolfram.com/PeanosAxioms.html
N.B. successor is a FUNCTION from N to N, where throughout, I
am using
ONO to denote the set of natural numbers.
>1. Zero is a number.
>2. If a is a number, the successor of a is a number.
>3. zero is not the successor of a number.
>4. Two numbers of which the successors are equal are
themsves equal.
>5. (induction axiom.) If a set S of numbers contains zero
and also the
>successor of every number in S, then every number is in S.
>Stupid questions about PA.
Denote the successor of the natural number n by nO.
>1) What prevents a natnum from being its own successor?
Let S = {n in N : nO != n}, where x != y means that x is not
equal to
y,
so S is the set of all natural numbers not equal to their
successors.
Since 0O != 0 (by axiom 3), then 0 is an ement of S. If n is
an ement
of S, then nO != n, so nOO !=
nO (by axiom 4: since nO and n
are not
equal, then their successors nOO and
nO are not equal), and
so nO is an
ement of S. Since 0 is an ement of S, and nO is an ement of
S
for
all ements n of S, then S = N (by the induction axiom). It
follows that
if n is a natural number, then n is an ement of S, so n is
not equal to
its successor (i.e. n is not its own successor).
>2) Why canOt a natnum have more than one successor?
Because successor is a function from N to N. The answer
follows from
the definition of function.
>3) What prevents circular successors (a successor of b, b
successor of a)?
Let S = N-{a,b} = {n in N : n != a and n != b}, so that S is
the set of
all natural numbers not equal to a or b. Since bO = a, then
0
!= a (by
axiom 3). Since aO = b, then 0 != b (by axiom 3). So 0 is an
ement
of S. Suppose n is an ement of S, then n != a and n != b.
Since
n != b, then nO != bO = a (by axiom 4), and
since n != a, then
nO != aO = b (by axiom 4). It follows that
nO != a and nO !=
b, and so
nO is an ement of S. Since 0 is an ement of S, and
nO is an
ement
of S for all ements n of S, then S = N. It follows that no
ement of
N is equal to a or b, contradicting the assumption of their
existence.
It follows that such natural numbers a and b cannot be found.
In a similar fashion, longer circular sequences of successors
are also
forbidden.
>4) What prevents a natnum from not having a successor (like
0)?
Because successor is a function from N to N. The answer
follows from
the definition of function.
>5) Why do we need axiom 5?
In order to restrict the set to N. Without Axiom 5, there are
other sets
which satisfy the first 4 axioms, e.g. the set of ordinals
less than
w2 = w+w, where w (usually written as omega) is the ordinal
denoting the
ordering of the natural numbers. Axiom 5 iminates all
possibilities
except those isomorphic to N.
David McAnally
Despite anything you may have heard to the contrary,
the rain in Spain stays almost invariably in the hills.
> In sci.logic, Garry Denke
>> Still made more sense than GarryOs. :-)
 Your logic is 0/0 = 0, and 0/0 = 1, which makes no sense.
lim(x->0) x/x = 0/0 = 1.
> lim(x->0) 2*x/x = 0/0 = 2.
> lim(x->0) x/(2*x) = 0/0 = 1/2 (or {1}over{2} for TeX-users,
perhaps).
> lim(x->0) x^2/x = 0/0 = 0.
> lim(x->0) x/x^2 = 0/0 = oo (or infty for TeX-users).
You were saying?
now youOre getting it, got any more, or should i say, the
rest?
>> To be sure, one has to be careful.
>> lim (x->+oo) x = +oo.
>> lim (x->+oo) 1/x = 0.
>> lim (x->+oo) x * 1/x = 1 but
>> lim (x->+oo) x * lim(x->+oo) 1/x = +oo * 0
>> which equals 0 if one follows Rudin, and AFAIK is undefined
>> if one follows others.
>> Fortunaty, if lim(x->+oo) a(x) = A, and lim(x->+oo) b(x) =
B,
>> where A and B are finite, one can prove lim(x->+oo)
a(x)b(x) = AB.
 My logic is 0/0 = 0/0, which makes sense.
 Go figure?
Yeah, go figure. Is 0.000...1 [*] equal to
infinity? Zero?
> Is 1.000...1 equal to infinity? One? Zero? Do such
> questions even make sense?
the solution to every algebraic equation makes total sense.
> [*] the lipses presumably refer to an infinite sequence of
digits,
> in this case. Obviously for a finite sequence the answer
> is clearly no in all cases.
good luck with it, 0/0, go eagles!
garry denke, geologist
denoco inc. of texas
In sci.logic, Garry Denke
<garrydenke@catholic.org> In sci.logic, Garry Denke
>> Still made more sense than GarryOs. :-)
Your logic is 0/0 = 0, and 0/0 = 1, which makes no sense.
lim(x->0) x/x = 0/0 = 1.
>> lim(x->0) 2*x/x = 0/0 = 2.
>> lim(x->0) x/(2*x) = 0/0 = 1/2 (or {1}over{2} for
TeX-users, perhaps).
>> lim(x->0) x^2/x = 0/0 = 0.
>> lim(x->0) x/x^2 = 0/0 = oo (or infty for TeX-users).
You were saying?
now youOre getting it, got any more, or should i say, the
rest?
Oh, I see. You donOt bieve in infinities. You
canOt represent
lim (x->0) x/x at all. You donOt have a clue as to what to
do
with lim (x->1) (x^2 - x)/(x - 1). Perish the thought that
one can do anything at all with lim(n->+oo) (1 - 10^(-n)).
And 1/3 != 0.333... .
> To be sure, one has to be careful.
lim (x->+oo) x = +oo.
>> lim (x->+oo) 1/x = 0.
>> lim (x->+oo) x * 1/x = 1 but
>> lim (x->+oo) x * lim(x->+oo) 1/x = +oo * 0
which equals 0 if one follows Rudin, and AFAIK is undefined
>> if one follows others.
Fortunaty, if lim(x->+oo) a(x) = A, and lim(x->+oo) b(x) = B,
>> where A and B are finite, one can prove lim(x->+oo)
a(x)b(x) = AB.
 My logic is 0/0 = 0/0, which makes sense.
Go figure?
Yeah, go figure. Is 0.000...1 [*] equal to
infinity? Zero?
>> Is 1.000...1 equal to infinity? One? Zero? Do such
>> questions even make sense?
the solution to every algebraic equation makes total sense.
Sure. Just let x = 0/0. Simple, albeit not horribly useful,
perhaps.
> [*] the lipses presumably refer to an infinite sequence of
digits,
>> in this case. Obviously for a finite sequence the answer
>> is clearly no in all cases.
good luck with it, 0/0, go eagles!
garry denke, geologist
> denoco inc. of texas
==
#191,
ItOs still legal to go .sigless.
Apologies if this question has been answered before...
I am trying to improve my perception of imaginary numbers. I
already have
done an extensive search in the web but I keep finding
information about
how
to manipulate them (add, multiply etc), about their
representation on a
plane with two axes and other things like that, that I am
already familiar
with.
Can anybody point me to any resource describing their TRUE
physical
meaning,
how the need for their introduction into mathematics rose and
generally the
whole concept behind them?
I just canOt bieve that mathematicians came up with them
just
because
they
couldnOt solve x^2+1=0...
In sci.math, George Savvas
<ch6l00hju24vg9ic5g2cncb729tdiphbb7@4ax.com>:
> Apologies if this question has been answered before...
I am trying to improve my perception of imaginary numbers. I
already have
> done an extensive search in the web but I keep finding
information about
how
> to manipulate them (add, multiply etc), about their
representation on a
> plane with two axes and other things like that, that I am
already
familiar
> with.
What is the meaning of number?
I canOt hold a number. I canOt go around
chasing them with
butter?
nets. All numbers are pure imaginary, and products of
intligent
imagination (IOd say human but SETI is still looking :-) ).
If I have 3 rocks, I can toss them 1, 2, 3. But that doesnOt
mean
I have a 3. It just means I have triple the number of rocks
that
one would have were one holding a single rock.
If I have 1/2 a cube of butter I could put two of these
half-cubes
together and create a whole cube. But I canOt hold a 1/2.
If I hold 1.37 cups of milk, I have a certain quantity of
milk,
namy, 1.37 times the amount of 1 cup of milk. But I canOt
hold 1.37.
If I expand pi to a million digits (easy enough to do if one
has
the computer resources), and print them out, I have a printout
with a million digits of pi. I can hold that (itOs ink on
paper,
after all), but I canOt hold pi.
Can anybody point me to any resource describing their TRUE
physical
meaning,
> how the need for their introduction into mathematics rose
and generally
the
> whole concept behind them?
I just canOt bieve that mathematicians came up with them
just
because
they
> couldnOt solve x^2+1=0...
>
IOm not entiry sure, but presumably one had to start with
the
notion that
i^2 + 1 = 0
and then work in enough machinery so that one can prove
things like
e^(i*pi) = -1
(a beautiful formula in its strangeness, rating e, a
transcendental defined from factorials and/or integration of
1/x,
pi, a transcendental defined from Euclidean geometry,
and i, the aforementioned imaginary)
its cousin,
e^(i * theta) = cos theta + i * sin theta,
and the three cube roots of 1, which are
1, -1/2 + i * sqrt(3)/2, and -1/2 - i * sqrt(3)/2.
==
#191,
ItOs still legal to go .sigless.
>Apologies if this question has been answered before...
I am trying to improve my perception of imaginary numbers. ..
I have always ft that the way complex numbers are introduced
in
ementary algebra courses is a travesty. My introduction to
complex
numbers in my HS algebra course in the 50Os was along the
lines:
<story>
x^2 + 1 = 0 has no solution. Because if it did we would have
a perfect
square = -1, and we all know x^2 >= 0 for all ***real***
numbers. The
word real, of course, is loaded with the implication that
these
numbers are, wl, real. Nothing solves x^2 + 1 = 0. If we did
have
such a figment of our imagination, call it i, where i^2 = -1,
wl
that is certainly an imaginary thing. Just remember if you
ever see
one of those imaginary i things in a calculation though, that
i^2 =
-1, and carry on. Lets call it like it is, i is an imaginary
number!
Jeez, if we get these imaginary things mixed up with the real
things,
we get mysterious things like a + bi. Very confusing. Lets
call them
**complex** because they are so mysterious. Just keep your
feet on the
ground and remember what you have learned though and you will
be
alright. Just donOt forget that i^2 = -1 and keep the real
and
imaginary parts separate.
</story>
Think how much nicer it would be if students at that lev, and
who
are familiar with the xy cartesian coordinate system, were
asked to
notice that the x axis has an implicit algebraic addition,
subtraction, multiplication and division inherited from the
real
numbers. So we can perform a natural arithmetic on pair
numbers of
the form (a, 0). So call the points in the plane (a,b) some
new term
like pair numbers. See if you can figure out some useful way
to
extend the arithmetic operations to your new pair numbers so
you can
do arithmetic in 2D. The students will guess the appropriate
addition
formula and you will have to lead them through multiplication
and
division, but ANY WAY YOU DO IT WILL BE BETTER THAN WHAT IS
NOW BEING
DONE, and when you are finished there wonOt be
anything
imaginary.
Your students will have discovered pair number arithmetic and
can then
discover that there is in fact a pair number satisfying
(a,b)^2 =
(-1,0).
I was always going to get around and give a talk along these
lines to
the local high school algebra classes titled ThereOs nothing
imaginary about complex numbers, but I never got around to it.
==Lynn
> Apologies if this question has been answered before...
I am trying to improve my perception of imaginary numbers. I
already have
> done an extensive search in the web but I keep finding
information about
how
> to manipulate them (add, multiply etc), about their
representation on a
> plane with two axes and other things like that, that I am
already
familiar
> with.
Can anybody point me to any resource describing their TRUE
physical
meaning,
> how the need for their introduction into mathematics rose
and generally
the
> whole concept behind them?
I just canOt bieve that mathematicians came up with them
just
because
they
> couldnOt solve x^2+1=0...
>
Actually, the cubic equation was more instrumental for
promoting the
acceptance of the imaginary numbers if I recall correctly.
Imaginary numbers are used for many purposes besides
guaranteeing that
every polynomial over the reals has a root over the complex
numbers.
Another use is when Einstein introduced the unit imaginary in
the
interval (x,y,z,ict), which, when squared produces
(x,y,z,ict)^2 = x^2 + y^2 + z^2 - c^2t^2
EinsteinOs trick of introducing the unit imnaginary here is
not
favored much these days.
I bieve it was Gauss who mapped vectors in the plane into
points in
the complex plane to establish a means of multiplication of
vectors.
David Hestenes interprets imaginary numbers as rotation
operators of
90 degrees acting on vectors in the plane. The unit imaginary
i
applied twice maps a vector into its negative because i^2 =
-1. And
that means a total rotation of 180 degrees. Then he
generalized
line at
http://ModingNTS.la.asu.edu
So, what is the TRUE physical meaning of imaginaries? I
donOt
think
that the question is clear enough to give a proper answer.
The closest
to a physical meaning seems to be the rotation-operator
meaning.
Perhaps itOs clearer to think of this as a physical use
rather than
a physical meaning.
You might also consider unipotent numbers as wl. A unipotent
number
u is a noncomplex number which upon squaring is unity, or u^2
= 1, but
u is not plus or minus unity.
Patrick
> I bieve it was Gauss who mapped vectors in the plane into
points in
> the complex plane to establish a means of multiplication of
vectors.
This is usually associated with Jean Argand (1768-1832). See:
[
http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/
Argand.html
]
==
http://www.math.ohio-state.edu/~edgar/
> Apologies if this question has been answered before...
I am trying to improve my perception of imaginary numbers. I
already have
> done an extensive search in the web but I keep finding
information about
how
> to manipulate them (add, multiply etc), about their
representation on a
> plane with two axes and other things like that, that I am
already
familiar
> with.
Can anybody point me to any resource describing their TRUE
physical
meaning,
In the words of Mr. Natural It donOt mean 
Gib
<ch6l00hju24vg9ic5g2cncb729tdiphbb7@4ax.com> thusly:
> Apologies if this question has been answered before...
I am trying to improve my perception of imaginary numbers. I
already have
> done an extensive search in the web but I keep finding
information about
> how to manipulate them (add, multiply etc), about their
representation on
> a plane with two axes and other things like that, that I am
already
> familiar with.
Can anybody point me to any resource describing their TRUE
physical
> meaning, how the need for their introduction into
mathematics rose and
> generally the whole concept behind them?
I just canOt bieve that mathematicians came up with them
just
because
> they couldnOt solve x^2+1=0...
>
There are many applications where complex numbers can
simplify the analysis.
The classic case is AC circuit analysis, where the equations
include many
sines and cosines making them really long. A totally
equivalent
mathematical treatment is to use the much simpler equations
of DC circuits
but use complex numbers for the components of the circuits,
and the
voltages and currents supposed to be present therein.
For example, an inductance is handled as a pury imaginary
resistance
(though the more general word impedance is usually used). By
OhmOs Law,
this would imply a pure imaginary current for a real voltage.
In physical
terms this means that the current is 90 degrees ahead of the
voltage.
The trigonometric approach to the e problem would have the
voltage givend
by an expression containing a sine, and the current would
then be given by
an expression containing a cosine. Again, this would give a
current 90
degrees ahead of the voltage.
==
Townsend
I put it down there, and when I went back to it, there it was
GONE!
Interchange the alphabetic ements to reply
><ch6l00hju24vg9ic5g2cncb729tdiphbb7@4ax.com> thusly:
> Apologies if this question has been answered before...
I am trying to improve my perception of imaginary numbers. I
already
have
>> done an extensive search in the web but I keep finding
information about
>> how to manipulate them (add, multiply etc), about their
representation
on
>> a plane with two axes and other things like that, that I
am already
>> familiar with.
Can anybody point me to any resource describing their TRUE
physical
>> meaning, how the need for their introduction into
mathematics rose and
>> generally the whole concept behind them?
I just canOt bieve that mathematicians came up with them
just
because
>> they couldnOt solve x^2+1=0...
>There are many applications where complex numbers can
simplify the
analysis.
>The classic case is AC circuit analysis, where the equations
include many
>sines and cosines making them really long. A totally
equivalent
>mathematical treatment is to use the much simpler equations
of DC circuits
>but use complex numbers for the components of the circuits,
and the
>voltages and currents supposed to be present therein.
For example, an inductance is handled as a pury imaginary
resistance
>(though the more general word impedance is usually used). By
OhmOs Law,
>this would imply a pure imaginary current for a real
voltage. In physical
>terms this means that the current is 90 degrees ahead of the
voltage.
The trigonometric approach to the e problem would have the
voltage
givend
>by an expression containing a sine, and the current would
then be given by
>an expression containing a cosine. Again, this would give a
current 90
>degrees ahead of the voltage.
Another example from physics is the formulation of the
space-time
vector in the form (x,y,z,ict) so that the metric comes out as
x2+y2+z2-c2t2. I think one problem is that words used in
mathematics
or in physics should not be taken literally or from a naive
point of
view. In rativity it is not true that space is real and time
somehow imaginary. Similarly all types of numbers, real or
complex, are on equal footing as mathematical objects. There
are
certainly many mathematical ideas that are far more weird if
taken
literally than are complex numbers.
WasnOt there a thread recenty in this group about the origin
of the
term imaginary?
Can anybody point me to any resource describing their TRUE
physical
> meaning,
So physical meaning is TRUE and go stuff mathametical meaning
:-(
==
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
Needless to say, I had the last laugh.
Alan Partridge, _Bouncing Back_ (14 times)
You might enjoy the book by Barry Mazur titled Imagining
Numbers
(particularly
the square root of minus fifteen), published by Farrar Straus
Giroux, New
John Robertson
> Apologies if this question has been answered before...
> I am trying to improve my perception of imaginary numbers.
I already have
> done an extensive search in the web but I keep finding
information about
how
> to manipulate them (add, multiply etc), about their
representation on a
> plane with two axes and other things like that, that I am
already
familiar
> with.
> Can anybody point me to any resource describing their TRUE
physical
meaning,
> how the need for their introduction into mathematics rose
and generally
the
> whole concept behind them?
> I just canOt bieve that mathematicians came up with them
just because
they
> couldnOt solve x^2+1=0...
One of the early reasons for the acceptance of complex
numbers was that
they make it possible to express the roots of a cubic
equation using
CardanOs formula:
<http://mathworld.wolfram.com/CubicEquation.html>.
Even in the case where all three roots are real, the algebraic
expressions for these roots may turn out to contain complex
values in
which the imaginary parts happen to canc, but there is no
algebraic way
to simplify the expressions to use real values only.
==
Seaman
Judge YohnOs mistakes revealed in Mumia Abu-Jamal ruling.
<http://www.commoncouragepress.com/index.cfm?action=book&
bookid=228> Apologies if this question has been answered before...
> I am trying to improve my perception of imaginary numbers.
I already
have
>> done an extensive search in the web but I keep finding
information about
how
>> to manipulate them (add, multiply etc), about their
representation on a
>> plane with two axes and other things like that, that I am
already
familiar
>> with.
> Can anybody point me to any resource describing their TRUE
physical
meaning,
>> how the need for their introduction into mathematics rose
and generally
the
>> whole concept behind them?
> I just canOt bieve that mathematicians came up with them
just because
they
>> couldnOt solve x^2+1=0...
>One of the early reasons for the acceptance of complex
numbers was that
>they make it possible to express the roots of a cubic
equation using
>CardanOs formula:
<http://mathworld.wolfram.com/CubicEquation.html>.
>Even in the case where all three roots are real, the
algebraic
>expressions for these roots may turn out to contain complex
values in
>which the imaginary parts happen to canc, but there is no
algebraic way
>to simplify the expressions to use real values only.
As a consequence of the identity e^(ix)= cos(x)+isin(x)
complex number
expressions are useful in describing oscillatory behavior
(harmonic
motion) of all kinds in mechanics and ectrical engineering.
Such
expressions are generally useful in solving key differential
equations
in all parts of physics. The theory of functions of a complex
variable contributes many useful methods, such as conformal
mapping,
to the solution of problems in heat ?w, ectrostatics,
ectrodynamics, and so on. Also quantum mechanics is
formulated from
the beginning in terms of complex valued functions.
>I am trying to improve my perception of imaginary numbers.
[...]
>Can anybody point me to any resource describing their TRUE
physical
meaning,
They have no physical meaning. Many people argue the e holds
for
real numbers.
>how the need for their introduction into mathematics rose
and generally
the
>whole concept behind them?
Complex numbers form an algebraically closed fid, which makes
most
operations with them very nice. Physicists like nice
mathematical
operations, so they adopted them. Need follows the innovation.
> Physicists like nice mathematical
> operations, so they adopted them.
Wllllll......
Complex numbers make the description of waves and vibrations,
including
ectrical phenomena, a lot easier. That sounds like a better
reason to
adopt them, and it wouldnOt surprise me if physicists had
actually
invented something like complex numbers for this purpose.
V.
==
homepage: cs utk edu tilde lastname
> Apologies if this question has been answered before...
I am trying to improve my perception of imaginary numbers. I
already have
> done an extensive search in the web but I keep finding
information about
how
> to manipulate them (add, multiply etc), about their
representation on a
> plane with two axes and other things like that, that I am
already
familiar
> with.
Can anybody point me to any resource describing their TRUE
physical
meaning,
> how the need for their introduction into mathematics rose
and generally
the
> whole concept behind them?
I just canOt bieve that mathematicians came up with them
just
because
they
> couldnOt solve x^2+1=0...
Why not? They *are* a weird bunch, you know.
>In sci.logic, Garry Denke
> I have one number, zero, that I like better than others.
> 0Os the solution to every algebraic equation.
> ThatOs why 0Os my favorite number.
> WhatOs your favorite?
> And why?
> x^2 = 1 has a solution x = 0?
> the knowledge of nothing is everything, so
> nothing squared plus one equals nothing.
> here, i will write the algebraic equation out so i can
solve it for
you.
> x^2 + 1 = 0
> solution for the equation is 0 the number
> (the number to the right of equals sign).
> challenge: post any algebraic equation in the form of it
equaling
> 0 the number, and i will solve your algebraic equation.
> Somehow, I doubt it.
> here are some more algebraic equations whose solution is
0 the number
> (the number to the right of the equals sign):
Actually, 1 is the solution to every equation:
radical equation... sqrt(x - 10) - 3 = 1
> quadratic equation... x^2 - 5x + 4 = 1
> exponential equation... e^2x - 3e^x + 3 = 1
> logarithmic equation... 6log(x^2 + 1) - x + 1 = 1
> trigonometric equation... 3tan^2 x = 1
no , wonOt work, you must have 0/0 on one side
garry denke, geologist
denoco inc. of texas
>I have one number, zero, that I like better than others.
> 0Os the solution to every algebraic equation.
ThatOs why
0Os my
> favorite
> number.
> WhatOs your favorite?
> My favorite is 17.
>And why?
> ItOs the first random number.
> Frank
> I was going to say 137, since it proves the existence of
God and gives
> meaning to the universe
I thought that number was 42? Damn! My entire system of biefs
is
> shattered! Oh wl. All praise the mighty 137!!
l8r, N. toff
I like 137 as wl,
since it is
3*4*5*(1 +1/2 +1/3 +1/4 +1/5).
But even better is the rated 1764,
which is
1*2*3*4*5*6*(1 +1/2 +1/3 +1/4 +1/5 +1/6),
and, amazingly,
1764 also = ....
42 *42 (!)...
Speaking of 42, I heard years ago that scientists found that
42 REALLY
did have some astrophysical
significance, rated to universal-expansion or something.
I do not know if the 42 constant was exact or if the result
was
refuted or just a joke.
What was up with this?...
,
Leroy
> How can you show that exp takes B_eps = {X in M_n(C) such
that ||X|| <
eps}
> injectivy into M_n(C), where eps < log(2)?
M_n(C) denotes the n x n matrices with complex entries, and
exp denotes
> exponential, eps denotes epsilon.
Is it like this? :
Of course we need eps < log(2) because this is where log is
analytic.
Now,
> to show injectivity show that if e^A = I, then A = 0.
As the other poster said, you need to show that if e^A=e^B
then A=B.
Wl, take log of both sides and done.
>
The trick here is to show that if ||A||<eps, then log(e^A)=A.
You can do
this
using the Riesz-Dunford functional calculus. But I have a
feing that you
have
not yet been taught this, and that you should show this
explicitly. I think
that you are supposed to use the assumption to show that
||e^A-I||<1, and
then
use the power series for log(1+x) to show (by subsituting one
power series
into
another) that log(I+(e^A-I))=A.
>How can you show that exp takes B_eps = {X in M_n(C) such
that ||X|| <
eps}
>injectivy into M_n(C), where eps < log(2)?
>M_n(C) denotes the n x n matrices with complex entries, and
exp denotes
>exponential, eps denotes epsilon.
>Is it like this? :
>Of course we need eps < log(2) because this is where log is
analytic.
It is? Actually the result is true for eps < log(Pi).
> Now,
>to show injectivity show that if e^A = I, then A = 0.
How would this show exp(A) <> exp(B) if A <> B in B_eps?
Note, for
example, that exp(A) = exp(B) does not imply exp(A-B) = I if
A and B
donOt commute.
>Wl, take log of both sides and done.
Which log?
U = C (-infinity, 0]. Show that exp(tA) = (exp(A))^t
for real t (use the principal branch of z -> z^t,
which is analytic in U).
Robert Isra isra@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~isra
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
> While playing with the Fibonacci sequence in a spreadsheet,
the
> following interesting pattern appeared.
 When the Fibonacci number is divided by its position in the
list, the
> result is a whole number when the position corresponds to a
left
> truncatable prime*. I ran out of accuracy to test this past
the 67th
> number and I know little about left truncatable primes.
Does anyone
> know if this property is anything worth digging into? A
> on fibonacci and primes numbers returns a few things but
nothing of
> great depth.
 If anyone has more info about Fibonacci and primes I would
appreciate
> anything you care to share.
 The numbers from the spreadsheet are bow. Primes(*) Column
> A=Fibonacci Column B=Position Column C=A/B Also notice the
primes
> appear the intervals of 6,4,6,14,6,4,6,14... wonder if that
would
> continue? Do I have too much time on my hands?
 Many ,
 Bob Carlson
 Fibonacci
> Sequence Position
> A B A/B
> 1 0
> 1 1 1.000000
> 2 2 1.000000
> 3 3* 1.000000 ====-
> 5 4 1.250000
> 8 5 1.600000
> 13 6 2.166667
> 21 7* 3.000000======-
> 34 8 4.250000
> 55 9 6.111111
> 89 10 8.900000
> 144 11 13.090909
> 233 12 19.416667
> 377 13* 29.000000======
> 610 14 43.571429
> 987 15 65.800000
> 1597 16 99.812500
> 2584 17* 152.000000====-
> 4181 18 232.277778
> 6765 19 356.052632
> 10946 20 547.300000
> 17711 21 843.380952
> 28657 22 1302.590909
> 46368 23* 2016.000000====-
> 75025 24 3126.041667
> 121393 25 4855.720000
> 196418 26 7554.538462
> 317811 27 11770.777778
> 514229 28 18365.321429
> 832040 29 28691.034483
> 1346269 30 44875.633333
> 2178309 31 70268.032258
> 3524578 32 110143.062500
> 5702887 33 172814.757576
> 9227465 34 271396.029412
> 14930352 35 426581.485714
> 24157817 36 671050.472222
> 39088169 37* 1056437.000000====-
> 63245986 38 1664368.052632
> 102334155 39 2623952.692308
> 165580141 40 4139503.525000
> 267914296 41 6534495.024390
> 433494437 42 10321296.119048
> 701408733 43* 16311831.000000====-
> 1134903170 44 25793253.863636
> 1836311903 45 40806931.177778
> 2971215073 46 64591632.021739
> 4807526976 47* 102287808.000000====
> 7778742049 48 162057126.020833
> 12586269025 49 256862633.163265
> 20365011074 50 407300221.480000
> 32951280099 51 646103531.352941
> 53316291173 52 1025313291.788460
> 86267571272 53* 1627690024.000000====
> 139583862445 54 2584886341.574070
> 225851433717 55 4106389703.945450
> 365435296162 56 6525630288.607140
> 591286729879 57 10373451401.386000
> 956722026041 58 16495207345.534500
> 1548008755920 59 26237436541.016900
> 2504730781961 60 41745513032.683300
> 4052739537881 61 66438353080.016400
> 6557470319842 62 105765650320.032000
> 10610209857723 63 168416029487.667000
> 17167680177565 64 268245002774.453000
> 27777890035288 65 427352154389.046000
> 44945570212853 66 680993488073.530000
> 72723460248141 67* 1085424779823.000000 ====
> 117669030460994 68 1730426918544.030000
Bob,
Using your e list above, where A is the Fibonacci sequence
> in the first column and B is the second column of positions
then
> where if any B is (ODD) then B = prime if A == 0 or 1 (mod
B).
Correction!
Except of course, prime (5) which I overlooked.
> At least up to B = 67. ;-)
I donOt know if this fails for any higher (ODD) values of B.
Dan
for any sequence {An},
let sequence {Bn} : any n in N , Bn = A1+A2+...+An
show that if lim An = A => lim Bn = A (n->infinite : all
application)
====================================
itOs my thinking process. please confirm and
check up.
if itOs error, tl me right process....
let An = Cn + A
A1+A2+...+An = (C1+A)+(C2+A)+...+(Cn+A)
(A1+A2+...+An)/n = {(C1+A)+(C2+A)+...+(Cn+A)}/n =
{(C1+C2+...+Cn)/n} +
A
thus
if lim (C1+C2+...+Cn)/n = 0 => lim (A1+A2+...+An)/n = A
thus
we must show that if lim Cn = 0 => lim (C1+C2+...+Cn)/n = 0
{C1 + C2 +...+ Cn}/n = {(C1 + C2 +...+ Cp)/n} + {(Cp+1 + Cp+2
+...+
Cn)/n}
{C1 + C2 +...+ Cn}/n <=
(&#12643;C1 + C2 +...+ Cp&#12643;/n) +
{&#12643;Cp+1&#12643;+&#12643;Cp+2&#12643;+...+&#12643;Cn&#
12643;}/n
for any e>0 , (because limCn = 0)
choose p such that n>p => Cn < e/2
{&#12643;Cp+1&#12643;+&#12643;Cp+2&#12643;+...+&#12643;Cn&#
12643;}/n <
{e/2 + e/2 + ... + e/2}/n < e/2========(1)
choose N such that
n>N>p => &#12643;C1 + C2 + ... + Cp&#12643;/n < e/2====-(2)
(Archimedes theorem)
thus n>N =>
{C1 + C2 +... + Cn}/n < e/2 + e/2 = e by (1),(2)
thank you very much
> for any sequence {An},
let sequence {Bn} : any n in N , Bn = A1+A2+...+An
show that if lim An = A => lim Bn = A (n->infinite : all
application)
This is false.
Let. for example An = 1, then lim An = 1 also.
But then Bn = n, and lim Bn does not exist as n -> oo.
> for any sequence {An},
let sequence {Bn} : any n in N , Bn = A1+A2+...+An
show that if lim An = A => lim Bn = A (n->infinite : all
application)
This is false.
Let. for example An = 1, then lim An = 1 also.
But then Bn = n, and lim Bn does not exist as n -> oo.
meant was Bn = (A1+A2+...+An)/n. Fixing this typo, the rest
of the
Rob Johnson <rob@trash.whim.org>
take out the trash before replying
>for any sequence {An},
let sequence {Bn} : any n in N , Bn = A1+A2+...+An
show that if lim An = A => lim Bn = A (n->infinite : all
application)
====================================
itOs my thinking process. please confirm and
check up.
if itOs error, tl me right process....
>let An = Cn + A
A1+A2+...+An = (C1+A)+(C2+A)+...+(Cn+A)
(A1+A2+...+An)/n = {(C1+A)+(C2+A)+...+(Cn+A)}/n =
{(C1+C2+...+Cn)/n} +
>A
thus
>if lim (C1+C2+...+Cn)/n = 0 => lim (A1+A2+...+An)/n = A
thus
>we must show that if lim Cn = 0 => lim (C1+C2+...+Cn)/n = 0
{C1 + C2 +...+ Cn}/n = {(C1 + C2 +...+ Cp)/n} + {(Cp+1 + Cp+2
+...+
>Cn)/n}
{C1 + C2 +...+ Cn}/n <=
This should be: |{C1 + C2 +...+ Cn}/n| <=
>(|C1 + C2 +...+ Cp|/n) +
>{|Cp+1|+|Cp+2|+...+|Cn|}/n
for any e>0 , (because limCn = 0)
choose p such that n>p => Cn < e/2
This should be: n>p => |Cn| < e/2
>{|Cp+1|+|Cp+2|+...+|Cn|}/n <
>{e/2 + e/2 + ... + e/2}/n < e/2========(1)
choose N such that
>n>N>p => |C1 + C2 + ... + Cp|/n < e/2====-(2)
>(Archimedes theorem)
thus n>N = {C1 + C2 +... + Cn}/n < e/2 + e/2 = e by (1),(2)
Looking up &#12643; in the Unicode tables, it is character
3163 (hex) in
Hangul Compatibility Jamo. It appears there as a vertical
bar, so I
have translated it as O|O above and it seems to
be what was
intended.
Other than a couple of missing absolute values, which are
pointed out
above, your proof looks good.
Rob Johnson <rob@trash.whim.org>
take out the trash before replying
In sci.math, a N. toff
<mtoff@sympatico.caREMOVETHIS>
<7RhOb.2879$XZ.421988@news20.blglobal.com>:
> In sci.math, Garry Denke
>> It is the Denke head that is ?t, here!
Here are some more algebraic equations whose solution, in
accordance
> with the rules of the mathematicians here, is zero (0) the
number:
[snip for brevity]
>> had no zero; neither did the Greeks == although the ancient
>> Babylonians, from whence we derive our admittedly rather
>> weird hour-minute-second time and degree-minute-second
circle
>> subdivisions, had a double-wedge zero symbol apparently.
> The modern positional notation we got from the Indians.
 Are you having a problem with the equals sign or 0 the
number to the
right
> of the equals sign?
Just kidding. But seriuously Ghost, youOre wasting your
time.
He is
> obviously a lost cause.
>
Ah, wl...I suspect Garry Denke will never understand me. :-)
Hopefully some of the lurkers do. :-)
> l8r, N. toff
>
==
#191,
ItOs still legal to go .sigless.
> In sci.math, Garry Denke
>> It is the Denke head that is ?t, here!
 Here are some more algebraic equations whose solution, in
accordance
> with the rules of the mathematicians here, is zero (0) the
number:
 quadratic equation;
 x^2 + bx + c = 0
You *are* familiar with completing the square and/or the
> Quadratic Formula, arenOt you?
yep, but I would rather get ready for the game.
> Given the equation A * x^2 + B * x + C = 0,
divide both sides by 0
> the solution is not 0, but
A * x^2 + B * x + C is not equal to 0?
divide both sides by 0, short way, (A * x^2 + B * x + C)/0 =
0/0
and now your long way(s)
> -B sqrt(B^2 - 4*A*C)
> x = ======================
> 2A
Most people know this solution. The method of deriving it
isnOt
> difficult, either:
[1] A * x^2 + B * x + C = 0
> [2] x^2 + B/A * x + C/A = 0
> [3] x^2 + B/A * x = - C/A
> [4] x^2 + B/A * x + B^2/(4*A^2) = -C/A + B^2/(4*A^2) = (B^2
-
4*A*C)/(4*A^2)
> [5] (x + B/(2*A))^2 = (B^2 - 4*A*C)/(4*A^2)
> [6] x + B/(2*A) = sqrt((B^2 - 4*A*C)/(4*A^2)) =
sqrt(B^2 - 4*A*C) / (2*A)
The usual term is completing the square, which is an apt
description
> for step [4].
Of course Mathworld has an entry:
http://mathworld.wolfram.com/QuadraticEquation.html
which basically reprises and expands what IOve said here,
> then goes into some interesting sidights; for example,
> one can also represent the solution as
2C
> x = ======================
> -B sqrt(B^2 - 4*A*C)
which would not have occurred to me personally.
much too long, and too time consuming, for me
> cubic equation;
 x^3 + bx^2 + cx + d = 0
http://mathworld.wolfram.com/CubicEquation.html
just divide both sides by 0
> quartic equation;
 x^4 + bx^3 + cx^2 + dx + e = 0
http://mathworld.wolfram.com/QuarticEquation.html
again, divide both sides by 0
> quintic equation;
 x^5 + bx^4 + cx^3 + dx^2 + ex + f = 0
http://mathworld.wolfram.com/QuinticEquation.html
and again, divide both sides by 0
> Dare I mention I took some Galois theory in college? :-P
thatOs great, really, youOre smart
> (Admittedly, IOve forgotten most of it... :-) )
probably a good thing, unless youOre hooked on long versions
> Algebra is easy because 0 is an now a number.
0Os been a number ever since someone discovered it way back.
> That someone had to discover it is in itsf interesting,
> however; a fairly large number of ancient cultures simply
> had no representation therefor. The Romans in particular
> had no zero; neither did the Greeks == although the ancient
> Babylonians, from whence we derive our admittedly rather
> weird hour-minute-second time and degree-minute-second
circle
> subdivisions, had a double-wedge zero symbol apparently.
diophantus and abu jafar muhammad ibn musa al-khwarizmi
perhaps knew
the shortcut, though they failed to publish it, or refused to
(you can
bet there will be a big argument over that here in the ng)
> The modern positional notation we got from the Indians.
a spur under the greeksO and babyloniansO
saddle
garry denke, geologist
denoco inc. of texas
In sci.math, Garry Denke
<garrydenke@catholic.org> In sci.math, Garry Denke
>> It is the Denke head that is ?t, here!
Here are some more algebraic equations whose solution, in
accordance
> with the rules of the mathematicians here, is zero (0) the
number:
quadratic equation;
x^2 + bx + c = 0
You *are* familiar with completing the square and/or the
>> Quadratic Formula, arenOt you?
yep, but I would rather get ready for the game.
Happy watching. :-) IOll admit my 9ers arenOt
in there this
year. :-/
> Given the equation A * x^2 + B * x + C = 0,
divide both sides by 0
> the solution is not 0, but
A * x^2 + B * x + C is not equal to 0?
It *is* equal to zero because it was *set* equal to zero as
a problem constraint.
One can also deduce such things as the value of
A * x^2 + B * x
and
x^2 + B/A * x
(the former is -C, the latter -C/A.)
divide both sides by 0, short way, (A * x^2 + B * x + C)/0 =
0/0
and now your long way(s)
> -B sqrt(B^2 - 4*A*C)
>> x = ======================
>> 2A
Do tl how one could get from
(A * x^2 + B*x +C)/0 = 0/0
to
x = (-B sqrt(B^2 - 4*A*C)) / (2*A)
Or perhaps youOd mery want to *multiply* by 0, then *divide*
again,
yiding the following rather uninteresting expression:
A * 0/0 * x^2 + B*0/0 * x + C*0/0 = 0*0/0
or
0/0*x^2 + 0/0 * x + 0/0 = 0/0
or
0/0 = 0/0
[rest snipped]
==
#191,
ItOs still legal to go .sigless.
> Are you having a problem with the equals sign or 0 the
number to the
right
> of the equals sign?
Just kidding. But seriously Ghost, youOre wasting your time.
He is
> obviously a lost cause.
http://sports.espn.go.com/n?playoffs03/series?series=carphi&
partnersite=es
pn
0/0, Go Eagles!
Garry Denke, Geologist
Denoco Inc. of Texas
> In sci.math, Garry Denke
> Consider the equation: x - 3 = 0
> The solution for x - 3 is the number 0
>> No, itOs x = 3.
 Are you having a problem with the equals sign
> or 0 the number to the right of the equals sign?
The equation
x - 3 = 0
cannot be solved by declaring that the solution is Ox - 3 =
0O
> or just O0O. ThatOs just
stupid.
So you are having a problem with the = 0. The answer is not
stupid,
the answer is 0. If you want to solve for x, fe free to do
so, but
the fact of the matter is that the answer is 0. ItOs not my
fault, it
just is. ThatOs the problem with mathematicians (and
computer
programmers), they do not stand behind their statements. They
hmmm,
and haaa, and rant away...
> Of course for such a simple equation itOs fairly obvious
anyway
> that the solution is x = 3; that sort of thing is probably
> learned in the third grade == although IOm not entiry sure
> when simple algebra is first taught anymore. (*I* learned it
> in third grade, but thatOs me...and IOm not
entiry sure if
> the taught material in the third grade is where I learned
it,
> as I have an old algebra book which now I canOt
find.)
Yep. The old personal attack manuever. Second grade?
> For example:
 Consider the equation x^3 - 3x - 4 = 0
> The solution for x^3 - 3x - 4 is the number 0
>> IOd have to work it out, but x = 0 wonOt
work here (0^3 -
3*0 - 4 =
-4).
 Are you having a problem with the equals sign
> or 0 the number to the right of the equals sign?
Are *you* having a problem with the notion of solving
> an arbitrary equation f(x) = 0, f(x) = K, or f(x) = g(x),
> for the variable x?
No, thatOs boring to me. No, my study is 0, and 0/0.
> For example:
 Consider the equation x - log x = 0
> The solution for x - log x is the number 0
>> That has no solution at all in the reals. IOm not sure
regarding
>> the complex plane minus the origin.
 Are you having a problem with the equals sign
> or 0 the number to the right of the equals sign?
Polly wanna cracker?
Yea. The old personal attack manuever. First grade?
> For example:
 Consider the equation y - 4 x + 1 = 0
> The solution for y - 4 x + 1 is the number 0
>> That solution is an infinite line (x,y) which passes
through
>> the points (0, -1), (1/4, 0) and (1, 3), among uncountably
>> infinite others. Or, if you prefer, the slope is 4 and the
>> y-intercept is -1.
 Are you having a problem with the equals sign
> or 0 the number to the right of the equals sign?
Polly wanna cracker?
Yes. The old personal attack manuever. Playground?
> Consider the equation x^2 + 1 = 0
> The solution for x^2 + 1 is the number 0
Polly wanna cracker?
Oh no, not again.
> ObSheesh: Sheesh.
Pre-school?
> Write again if you need hp.
I got a degree in this stuff, Rocks-In-Head-Boy. :-P
> I would hope to have at least half a clue.
Good for you, really.
Yep, 0 is the big rock of 0123456789, 8 a couple of smaller
rocks, and
6Os and 9Os swinging rocks. But whenever
youOre finished
playing with
your xOs, yOs, and zOs, write
again, and I will hp you with
the
solution to every algebraic equation, namy, = 0, and 0/0.
Garry Denke, Geologist
Denoco Inc. of Texas
In sci.math, Garry Denke
<garrydenke@catholic.org> In sci.math, Garry Denke
>Consider the equation: x - 3 = 0
>The solution for x - 3 is the number 0
No, itOs x = 3.
Are you having a problem with the equals sign
> or 0 the number to the right of the equals sign?
The equation
x - 3 = 0
cannot be solved by declaring that the solution is Ox - 3 =
0O
>> or just O0O. ThatOs just
stupid.
So you are having a problem with the = 0. The answer is not
stupid,
> the answer is 0. If you want to solve for x, fe free to do
so, but
> the fact of the matter is that the answer is 0.
You are confusing solving the equation
x - 3 = 0
with evaluating the equation
x - 3
given the initial constraints, one of which simply states that
x - 3 = 0
therefore
x - 3 = 0
A tautology, and not allthat useful. Of course the answerOs
0; itOs
part of the problem!
> ItOs not my fault, it
> just is. ThatOs the problem with mathematicians (and
computer
> programmers), they do not stand behind their statements.
They hmmm,
> and haaa, and rant away...
Depends on what you want to do: solve for x or evaluate the
formula equated for 0. The latter is trivial.
> Of course for such a simple equation itOs fairly obvious
anyway
>> that the solution is x = 3; that sort of thing is probably
>> learned in the third grade == although IOm not entiry
sure
>> when simple algebra is first taught anymore. (*I* learned
it
>> in third grade, but thatOs me...and IOm not
entiry sure if
>> the taught material in the third grade is where I learned
it,
>> as I have an old algebra book which now I canOt
find.)
Yep. The old personal attack manuever. Second grade?
Probably. Of course I still donOt understand your
methodology
for
solving equations.
Given that
x^2 + 4*x + 4 = 0
I can state that
x^2 + 4*x + 4 = 0
and even
x^2 + 4*x + 3 = -1
but what is x? Wl, it turns out x = -2 in this case, as can
readily be demonstrated by either guessing:
(-2)^2 + 4*(-2) + 4 = 4 - 8 + 4 = 0
or by deploying the formula:
x = (-4 sqrt(4^2 - 4 * 1 * 4) ) / 2 = (-4
sqrt(0))/2 = -2 [twice]
> For example:
>Consider the equation x^3 - 3x - 4 = 0
>The solution for x^3 - 3x - 4 is the number 0
IOd have to work it out, but x = 0 wonOt work
here (0^3 - 3*0
- 4 =
-4).
Are you having a problem with the equals sign
> or 0 the number to the right of the equals sign?
Are *you* having a problem with the notion of solving
>> an arbitrary equation f(x) = 0, f(x) = K, or f(x) = g(x),
>> for the variable x?
No, thatOs boring to me. No, my study is 0, and 0/0.
0 is 0. It is neither negative nor positive.
It is the arithmetic identity.
0/0 is undefined, without more info; it can be any number.
x = 0/0 solves any equation.
For example, x^2 + 4*x + 4 = (0/0)^2 + 4*(0/0) + 4
= 0/0 + 0/0 + 4 = 0/0 = 0
since 0/0 equals any number. (Recall that a + b/c = (ac+b)/c
so 4 + (0/0) = (4*0 + 0)/0 = 0/0. Also, a = ac/c so a =
(a*0)/0 = 0/0.)
Is this useful? What do you think?
[rest snipped]
==
#191,
ItOs still legal to go .sigless.
The inverse function of f(x)= tanh(x) is the rection about y=x
graphically where tanh(x)=
sinh(x)/cosh(x)=(e^x-e^-x)/(e^x+e^-x). I
have a problem which asks to justify the existance of f^-1(x)
using
derivatives.
fO(x)= 1/cosh^2(x)and I assume f^-1O(x)=
cosh^2(x). I think
IOm
missing something that makes the connection, possibly the
expression
for f^-1(x) but it could be something simpler. Anyone?
Phil Holman
> The inverse function of f(x)= tanh(x) is the rection about
y=x
> graphically where tanh(x)=
sinh(x)/cosh(x)=(e^x-e^-x)/(e^x+e^-x). I
> have a problem which asks to justify the existance of
f^-1(x) using
> derivatives.
> fO(x)= 1/cosh^2(x)and I assume f^-1O(x)=
cosh^2(x). I think
IOm
> missing something that makes the connection, possibly the
expression
> for f^-1(x) but it could be something simpler. Anyone?
Phil Holman
The hyperbolic tangent function has an explicit inverse:
y = tanh(x) if and only if x = ln((1+y)/(1-y))/2, for -1 < y
< 1.
This can be derived as follows:
y = Tanh(x) = (e^x - e^-x)/(e^x + e^-x) =
(e^(2*x)-1)/(e^(2*x)+1)
Let u = e^(2*x), then y = (u-1)/(u+1), and u = (1 + y)/(1 - y)
e^(2*x) = (1 + y)/(1 - y)
2*x = ln((1 + y)/(1 - y))
x = ln((1 + y)/(1 - y))/2
And since -1 < tanh(x) < 1, we must have -1 < y < 1.
Then one can find dx/dy = 1/(1 - y^2) directly and proceed
from there.
> The inverse function of f(x)= tanh(x) is the rection
> about y=x graphically where tanh(x)
> = sinh(x)/cosh(x)=(e^x-e^-x)/(e^x+e^-x).
> I have a problem which asks to justify
> the existance of f^-1(x) using derivatives.
> fO(x)= 1/cosh^2(x)and I assume f^-1O(x)=
cosh^2(x).
> I think IOm missing something that makes the connection,
> possibly the expression for f^-1(x) but it could be
> something simpler. Anyone?
Here is an idea:
For all x, cosh(x) > 0.
So, (cosh(x))^2 > 0 also.
Since its derivative is positive,
f^-1 is increasing.
>The inverse function of f(x)= tanh(x) is the rection about
y=x
>graphically where tanh(x)=
sinh(x)/cosh(x)=(e^x-e^-x)/(e^x+e^-x). I
>have a problem which asks to justify the existance of
f^-1(x) using
>derivatives.
>fO(x)= 1/cosh^2(x)and I assume f^-1O(x)=
cosh^2(x). I think
IOm
>missing something that makes the connection, possibly the
expression
>for f^-1(x) but it could be something simpler. Anyone?
Rob Johnson <rob@trash.whim.org>
take out the trash before replying
Holman)
>The inverse function of f(x)= tanh(x) is the rection about
y=x
>graphically where tanh(x)=
sinh(x)/cosh(x)=(e^x-e^-x)/(e^x+e^-x). I
>have a problem which asks to justify the existance of
f^-1(x) using
>derivatives.
Weird problem. How does one take a derivative of something
without
knowing whether it even exists or not?
>fO(x)= 1/cosh^2(x)and I assume f^-1O(x)=
cosh^2(x).
Is that really what the derivative rule for inverse functions
says?
Holman)
>The inverse function of f(x)= tanh(x) is the rection about
y=x
>graphically where tanh(x)=
sinh(x)/cosh(x)=(e^x-e^-x)/(e^x+e^-x). I
>have a problem which asks to justify the existance of
f^-1(x) using
>derivatives.
Weird problem. How does one take a derivative of something
without
> knowing whether it even exists or not?
No kidding, the fact that I can visualize it as a ?ction on
y=x
obviously insnOt enough.
>fO(x)= 1/cosh^2(x)and I assume f^-1O(x)=
cosh^2(x).
Is that really what the derivative rule for inverse functions
says?
IOll check but a re?ction of the function about y=x
will
have an
inverse slope. 1/2 maps to 2 etc.
Others allude to the fact that if the inverse function is
increasing on
all x then it therefore exists. Unlike say a re?ction of
sin(x) which
would have multiple values of y for a single x and is not
allowed.
Phil Holman
> Others allude to the fact that if the inverse function
> is increasing on all x then it therefore exists.
The theorem he needs to cite is:
If f is an increasing function, then f^-1 exists.
> Others allude to the fact that if the inverse function
> is increasing on all x then it therefore exists.
The theorem he needs to cite is:
> If f is an increasing function, then f^-1 exists.
>
He is me and for the simple citation.
Phil Holman
hi guys,
2^(x+1)+2^(3-x)=17
logging gives
(x+1)lg2+(3-x)lg2=lg17
expanding gives
lg2x+lg2+3lg2-lg2x=lg17
but lg2x-lg2x=0.....goinn mad...graphed it so i know x=3 but
canOt get
there
numerically.....please hp i know iOm missing something
simple
shaun
> hi guys,
2^(x+1)+2^(3-x)=17
logging gives
(x+1)lg2+(3-x)lg2=lg17
expanding gives
lg2x+lg2+3lg2-lg2x=lg17
but lg2x-lg2x=0.....goinn mad...graphed it so i know x=3 but
canOt get
there
> numerically.....please hp i know iOm missing something
simple
shaun
You are missing that lg(a+b) is not equal to lg(a) + lg(b).
The rule is that lg(a*b) = lg(a) + lg(b), which does not hp
you.
What you can do is write 2^(x+1) as 2*2^x = 2*y, where y = 2^x
and write 2^(3-x) as 8/2^x = 8/y, and procede from there to
solve for y,
and then solve y = 2^x for x.
>hi guys,
2^(x+1)+2^(3-x)=17
logging gives
(x+1)lg2+(3-x)lg2=lg17
expanding gives
lg2x+lg2+3lg2-lg2x=lg17
but lg2x-lg2x=0.....goinn mad...graphed it so i know x=3 but
canOt get
there
>numerically.....please hp i know iOm missing something
simple
log(a+b) is not in general log(a) + log(b).
Let y = 2^x, then your equation becomes
2y + 8/y = 17
Multiplying by y and collecting terms, we get
2
2y - 17y + 8 = 0
Using the quadratic formula, we get
17 +/- sqrt(289-64)
y = ==================-
4
17 +/- 15
= ========-
4
So 2^x is either 1/2 or 8, that means that x is either -1 or
3.
Rob Johnson <rob@trash.whim.org>
take out the trash before replying
> hi guys,
2^(x+1)+2^(3-x)=17
logging gives
(x+1)lg2+(3-x)lg2=lg17
>
You canOt do this. You have to do lg(2^(x+1)+2^(3-x)) =/=
(x+1)lg2+(3-x)lg2
ie:
log 100 + log 1000 = 5
log 1100 ~= 3
l8r, N. toff
|It is my impression that magma is a bit more
|powerful, but I think that one is not for free, unless you
have
|very good connections with the University of Australia.
Wl, I work for an international company based in Australia.
Er, and I have a T-shirt from the Australian National
University.
Probably not good enough to get a copy for free, eh? :-)
Given the eigenvalues, can we know the rank of a matrix?
And another question, given the eigenvalues, we can determine
the
eigenvalues of B times B, but can we also determine the
eigenvalues of
BOB? (Transpose of B times B)?
>Given the eigenvalues, can we know the rank of a matrix?
I assume itOs a square matrix...
Given the eigenvalues _with their algebraic multiplicities_,
we can
(of course we only need the multiplicity of 0).
>And another question, given the eigenvalues, we can
determine the
>eigenvalues of B times B, but can we also determine the
eigenvalues of
>BOB? (Transpose of B times B)?
No. For example, the eigenvalues of
[ 0 a ]
B = [ b 0 ]
depend only on ab, but the eigenvalues of
[ a^2 0 ]
BOB = [ 0 b^2 ]
donOt.
Robert Isra isra@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~isra
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
> Given the eigenvalues, can we know the rank of a matrix?
No. The matrices
0 0
0 0
and
0 1
0 0
have the e eigenvalues, but different ranks.
Let P(x) ba a polynomial with real coefficients. Is there
a simple way of finding a number R > 0 such that every real
root of P(x) is in [-R,R]? IOm thinking in terms of R being
obtained as a function of the coefficients of P(x).
Let P(x) ba a polynomial with real coefficients. Is there
> a simple way of finding a number R > 0 such that every real
> root of P(x) is in [-R,R]? IOm thinking in terms of R
being
> obtained as a function of the coefficients of P(x).
Yes, of course.
If P(x) = x^n + a_1 x^{n-1} + .... + a_n,
take R = max(1, |a_1| + |a_2| + ... + |a_n|).
==
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
Needless to say, I had the last laugh.
Alan Partridge, _Bouncing Back_ (14 times)
Can anyone name these four formulas for operators of
exponents, they
cover the basic rule for addition, subtraction, division and
multiplication of exponents
using simple addition, subtraction, division and
multiplication.
The following formula is a division formula for exponents.
a^m = a^y(^(m/y))
The following formula is a subtraction formula used to find
exponent
factors for any exponent, the number of exponent factor sets
within
this equation is equal to m/2
a^m = a^(m-n) x a^n
The following formula is an addition formula used to find
equal
exponents
a^m = a(m+n) / a^n
The following formula is a multiplication formula.
a^m = a(m*n) / a^((m*n) -n)
http://www.sollog.com/exponentformulas.pdf
Sollog Immanu Adonai-Adoni
> Can anyone name these four formulas for operators of
exponents, they
> cover the basic rule for addition, subtraction, division and
> multiplication of exponents
>
A bunch of blah. Stick with
a^(m+n) = a^m * a^n
(a^n)^m = a^(nm)
a^(-m) = 1/a^m
and the stuff bow will come quickly if perhaps itOs needed.
> The following formula is a division formula for exponents.
a^m = a^y(^(m/y))
>
Correction: a^m = (a^y)^(m/y)
Toss it out, itOs redundant to
(a^n)^m = a^(nm)
> The following formula is a subtraction formula used to find
exponent
> factors for any exponent, the number of exponent factor
sets within
> this equation is equal to m/2
a^m = a^(m-n) x a^n
> (a^n)^m = a^(nm)
DonOt use x, it could be confused with the variable x.
Toss it out, itOs redundant to
a^(m+n) = a^m * a^n
> The following formula is an addition formula used to find
equal
> exponents
a^m = a(m+n) / a^n
>
Huh? ItOs in error.
Set a = 2, n = 0, m = 3.
> The following formula is a multiplication formula.
a^m = a(m*n) / a^((m*n) -n)
>
Error. Set n = 0, a = 1 = m.
Does a(m*n) = a*(m*n) or something se?
How about
a^m = a^(m+n) / a^n ?
and
a^m = a^(mn) / a^(mn - n) ?
But thatOs wrong, itOs
a^m = a^(mn) / a^(mn - m)
HereOs a generalization
a^p = a^(mn) / a^(mn - p)
Both are of minimal to negligible value and superceeded by
a^p = a^m / a^(m-p)
Exercise: quickly proof the generalization from the three
equations I
gave at start.
I consider that referenced web page to be of dubious value.
Frankly, thereOs much better stuff to be found on the web.
Look around
http://mathforum.org
http://www.mathpages.com
and forget the page you dredged up.
Show in general that the inclusions
.... C^{n}(U) subset of C^{n-1}(U) subset of ... subset of
C^{2}(U) subset
of C^{1}(U)
are proper. (Clearly if I deted are proper this would hold)
But otherwise? Can I look for a function that is in C^{1}(U)
and not in
C^{2}(U) and then somehow extend it in general to C^{n}(U)?
C^n denotes functions whose partial derivatives up to order n
are all
continuous and exist.
Moshe
> Show in general that the inclusions
.... C^{n}(U) subset of C^{n-1}(U) subset of ... subset of
C^{2}(U)
subset
> of C^{1}(U)
are proper. (Clearly if I deted are proper this would hold)
But otherwise? Can I look for a function that is in C^{1}(U)
and not in
> C^{2}(U) and then somehow extend it in general to C^{n}(U)?
> C^n denotes functions whose partial derivatives up to order
n are all
> continuous and exist.
Moshe
IOll suppose youOre talking about real-valued
functions,
defined on an
open subset of R (or R^k). If it were complex functions,
defined on an
open subset of C, then differentiability is the e as
analyticity,
which in turn implies differentiability of infinite order.
In the real context, take a function with a simple
discontinuity, say
{ 0 for x <= 0,
H_(-1)(x) = {
{ 1 for x > 0.
Next, let
H_0(x) = integral( H_(-1)(t), t=0..x )
Then H_0 is continuous, but not continuously differentiable
at 0.
In other words, H_0 is in C^0 but not in C^1.
For each k>0, define
H_k(x) = integral( H_(k-1) (t) , t=0..x )
By the Fundamental Theorem of Calculus, (H_k)O = H_(k-1),
which is (by induction) in C^(k-1) but not C^k. Thus,
we find H_k is in C^k but not C^(k+1).
This is all wrt a single variable. ItOs not a big deal to
piece together a function for which this behavior exists
for the partials wrt each of the variables, if only by
adding together:
F(x1,x2,...,xn) = H_k1(x1) + H_k2(x2) + H_k3(x3) + ... +
H_kn(xn)
IOm sure other folks can concoct fancier examples.
Dale.
>Show in general that the inclusions
.... C^{n}(U) subset of C^{n-1}(U) subset of ... subset of
C^{2}(U) subset
>of C^{1}(U)
are proper. (Clearly if I deted are proper this would hold)
But otherwise? Can I look for a function that is in C^{1}(U)
and not in
>C^{2}(U) and then somehow extend it in general to C^{n}(U)?
>C^n denotes functions whose partial derivatives up to order
n are all
>continuous and exist.
>
Let a be any value in the first projection of U. Let f_0(x) =
|x-a|, f_(n+1) (x) = integral(a..x, f_n). Let g_n(x) =
f_n(x1). Then
g_n is in C^n, not in C^(n+1)
==
Stephen J. Herschkorn herschko@rutcor.rutgers.edu
[.snip.]
>> Proofs that your definition of coprime and
VirgilOs
definition
>>are equivalent in the algebraic integers have been known
since the
>>time of Dedekind.
> You have never shown the slightest awareness of the
Euclidean
>>algorithm, one of the absoluty basic tools in algebraic
number
>>theory and the real reason that the definition that Virgil
uses
>>for coprime is the standard accepted definition.
No, the Euclidean algorithm is not really at issue here.
Yes I know that. The point was, Harris has never shown any
awareness of it or of the fact that in the integers, the E.
algorithm
is used to show that c = GCD(a,b) can be written as a*s + b*t
= c.
He has seen this often enough in many posts but has never
shown any sign of appreciating it, and my guess is it just
went over his head every time. As you may recall he never
step may have been that it had nothing to do with coprimeness
as he understands it, so Virgil must have been assuming what
he wanted to prove.
>For the
>Euclidean algorithm to apply, you must have a division
algorithm,
>which means the domain is an Euclidean domain; but Euclidean
implies
>PID, so the ring of all algebraic integers is certainly not
>Euclidean. In fact, few number fids are Euclidean, even fewer
than
>are PIDs.
(In fact, thatOs one reason why finding
gcdOs is hard in the
ring of
>all algebraic integers: if we had the Euclidean algorithm,
it would
>not be so hard...)
>
Yes, I complety agree on that. In fact it seems like
incredible luck that gcdOs can be found at all in the
algebraic
integers. I wonder if something even deeper, not yet
discovered,
might be another explanation.
==
>==
>ItOs not denial. IOm just very sective about
> what I accept as reality.
> ==- Calvin (Calvin and Hobbes)
>==
Arturo Magidin
>magidin@math.berkey.edu
... stuff deted ...
>>For the
>>Euclidean algorithm to apply, you must have a division
algorithm,
>>which means the domain is an Euclidean domain; but
Euclidean implies
>>PID, so the ring of all algebraic integers is certainly not
>>Euclidean. In fact, few number fids are Euclidean, even
fewer than
>>are PIDs.
>(In fact, thatOs one reason why finding
gcdOs is hard in the
ring of
>>all algebraic integers: if we had the Euclidean algorithm,
it would
>>not be so hard...)
>
Yes, I complety agree on that. In fact it seems like
> incredible luck that gcdOs can be found at all in the
algebraic
> integers. I wonder if something even deeper, not yet
discovered,
> might be another explanation.
>
Most liky the Object Ring. (GROAN)
>==
>>==
>>ItOs not denial. IOm just very sective
about
>>what I accept as reality.
>> ==- Calvin (Calvin and Hobbes)
>>==
>Arturo Magidin
>>magidin@math.berkey.edu
>
Dale
> [.snip.]
> Proofs that your definition of coprime and
VirgilOs definition
>are equivalent in the algebraic integers have been known
since the
>time of Dedekind.
You have never shown the slightest awareness of the Euclidean
>algorithm, one of the absoluty basic tools in algebraic
number
>theory and the real reason that the definition that Virgil
uses
>for coprime is the standard accepted definition.
>No, the Euclidean algorithm is not really at issue here.
> Yes I know that. The point was, Harris has never shown any
>awareness of it or of the fact that in the integers, the E.
algorithm
>is used to show that c = GCD(a,b) can be written as a*s +
b*t = c.
>He has seen this often enough in many posts but has never
>shown any sign of appreciating it, and my guess is it just
>went over his head every time.
My apologies; it is indeed clear from your syntax that you
did not
mean to imply that the Euclidean algorithm was rated to the
fact
that the Algebraic integers are a Bezout domain... I jus read
it a bit
too quickly.
==
==
==
<40080921.2060303@hamilton.edu>
<87smifkkwf.fsf@phiwumbda.org> Of course, James will never look up the reference unless
itOs a web
>> page, but his argument will be that much more blatantly
silly once a
>> clear reference is given.
>>
I donOt know that itOs on a web page. Lots of
important things
> are not.
Wl, you know that and I know that, but I bet we have a common
Usenet
acquaintance who seems not to know that (just as he seems to
bieve
that, whenever a new theorem or, say, way to count primes is
discovered, textbook presses everywhere swing into gear to
replace the
outdated texts currently used).
==
Jesse Hughes
Casting [Demi] Moore as a woman who has come to the New World
so that
she can Oworship without fear or persecutionO
in
_The_Scarlet_Letter_
is like casting Bruce Willis as Young Rene Descartes. -Joe
Queenan
<40080921.2060303@hamilton.edu>
<87smifkkwf.fsf@phiwumbda.org No one needs further evidence. This is past mere
mathematical
> error. Deep inside, I think you know it. This is
dishonesty, even
> unto yoursf. This is denial of reality. This is sickness.
You sound like youOve been reading Kierkegaard or something.
> I havenOt. You disagree with the statements?
No, no, itOs just that Kierkegaard has a famous book titled
The
Sickness Unto Death. ItOs been a while since I read it, but
I
think
it was about prostate cancer.
Anyway, your word choices (and contents) reminded me of that
book.
==
Jesse F. Hughes
What I represent is the unknowable future==the power of
change. In
that sense IOm a force of Nature, a force of the Universe, a
living
emodiment of change itsf. == and his sense of humility
[.snip.]
LetOs consider a similar statement to Lemma
1O
>in a general ring R.
> Lemma A:
> Let f,a and b be ements of R and let f divide ab in R.
> Then there exist s and t in R, such
> that s divides a in R, t divides b in R and f=st.
[.snip.]
>P.S. Is there a known characterization of rings R for which
Lemma A
holds?
They are the pre-Schreier rings. If, in addition, R is
integrally
> closed in its fid of fractions, then you have a Schreier
ring.
> There is a fair amount of work on them, as was pointed out
to me by
> Bill Dubuque; see his post with references:
40nestle.ai.mit.edu
P.M. CohnOs paper is particularly interesting (it is
mentioned in
> BillOs post, but I thought I would repeat it here):
P.M. Cohn: Bezout rings and their subrings. Proc. Cambridge
> Philos. Soc. 65 (1968), pp. 251-264. MR 36#5117.
>
Next question. I made that (implied) claim that
(i} a,b coprime and a | bc => a|c
is obvious, however this may have been rash. The result is
certainly obvious for the integers
(no matter which definition of coprime we use), and if we use
the
standard definition of coprime it is true for a large number
of rings (commutative with identity is sufficient).
However, consider the rated claim.
(ii) a,b share only unit factors and a | bc => a|c
Now in any commutative ring with identity in which
(iii) a,b share only unit factors => a,b coprime
(ii) must be true (the algebraic integers being an obvious
example).
Does this hold more generally, i.e. does (ii) imply (iii)?
-William Hughes
[.snip.]
>LetOs consider a similar statement to Lemma
1O
>in a general ring R.
> Lemma A:
Let f,a and b be ements of R and let f divide ab in R.
> Then there exist s and t in R, such
> that s divides a in R, t divides b in R and f=st.
[.snip.]
>P.S. Is there a known characterization of rings R for which
Lemma A
holds?
They are the pre-Schreier rings. If, in addition, R is
integrally
>> closed in its fid of fractions, then you have a Schreier
ring.
>> There is a fair amount of work on them, as was pointed out
to me by
>> Bill Dubuque; see his post with references:
40nestle.ai.mit.edu
P.M. CohnOs paper is particularly interesting (it is
mentioned in
>> BillOs post, but I thought I would repeat it here):
P.M. Cohn: Bezout rings and their subrings. Proc. Cambridge
>> Philos. Soc. 65 (1968), pp. 251-264. MR 36#5117.
>
Next question. I made that (implied) claim that
(i} a,b coprime and a | bc => a|c
is obvious, however this may have been rash. The result is
>certainly obvious for the integers
>(no matter which definition of coprime we use), and if we use
the
>standard definition of coprime it is true for a large number
>of rings (commutative with identity is sufficient).
However, consider the rated claim.
(ii) a,b share only unit factors and a | bc => a|c
Now in any commutative ring with identity in which
(iii) a,b share only unit factors => a,b coprime
(ii) must be true (the algebraic integers being an obvious
example).
Does this hold more generally, i.e. does (ii) imply (iii)?
No. Any UFD which is not a PID will probably give a
counterexample,
but certainly Z[x] will do.
Let a and b be two ements of Z[x] which have only unit
factors, and
assume that a|bc. We want to show that a|c (to establish that
(ii)
holds in Z[x]). Factor a and b into irreducibles
a = u*p_1*...*p_r
b = v*q_1*...*q_s
u,v units, p_i,q_j irreducibles, each p_i is different from
each q_j;
r,s>=0. Then factor c into irreducibles
c = w*P_1*...P_t
w a unit, P_i irreducibles. Since a divides b*c =
(uv)*q_1*...*q_s*P_1*...*P_t, by unique factorization, for
each
i=1,...,r, p_i is equal to some P_j; wlog we may reorder the
P_j so
p_i=P_i for each i, so a|c. This shows that (ii) holds in
Z[x].
However, (iii) does not hold in Z[x]: 2 and x share only unit
factors,
but they are not coprime in the sense that (2,x) is not the
unit ideal.
==
==
==
> You have never shown the slightest awareness of the
Euclidean
>algorithm, one of the absoluty basic tools in algebraic
number
>theory and the real reason that the definition that Virgil
uses
>for coprime is the standard accepted definition.
 No, the Euclidean algorithm is not really at issue here.
For the
> Euclidean algorithm to apply, you must have a division
algorithm,
> which means the domain is an Euclidean domain; but
Euclidean implies
> PID, so the ring of all algebraic integers is certainly not
> Euclidean. In fact, few number fids are Euclidean, even
fewer than
> are PIDs.
==
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam,
nederland,
+31205924131
home: bovenover 215, 1025 jn amsterdam, nederland;
http://www.cwi.nl/~dik/
> You have never shown the slightest awareness of the
Euclidean
>algorithm, one of the absoluty basic tools in algebraic
number
>theory and the real reason that the definition that Virgil
uses
>for coprime is the standard accepted definition.
 No, the Euclidean algorithm is not really at issue here.
For the
> Euclidean algorithm to apply, you must have a division
algorithm,
> which means the domain is an Euclidean domain; but
Euclidean implies
> PID, so the ring of all algebraic integers is certainly not
> Euclidean. In fact, few number fids are Euclidean, even
fewer than
> are PIDs.
UFD. But when I compiled the list not all UFDOs with
positive
n were
>known; all Euclidean fids were known.
They are still not known, unless somethingOs happened in the
last few
months... The most recent progress I can find in MathSciNet is
a pair
the form
Q(sqrt(4m^2+1)), with 4m^2+1 squarefree, the class number is
1 (hence
a UFD and so a PID) if and only if m=1,2,3,5,7, or 13. In
YokoiOs
where he proves that a fid of the form Q(sqrt(m^2+4)) with
m^2+4
squarefree has class number 1 if and only if m=1,3,5,7,13, or
17.
==
==
==
> A few years ago I posted that the difference between
JamesO
IQ and mine
> is at least thirty. IOll raise that number to fourty.
Seems
he gets more
> and more stupid every year.
(A statement like that looks bad with a spling mistake in it
:) ItOs
not the size, itOs how you use it. In JSHOs
case, the
limitations are
probably more emotional/psychological than IQ.
Gib
>IOve concluded that the problem IOm facing is
that the
human
> brain
>isnOt built to handle Mathematics.
> Wl, I take it you are now claiming that you donOt have a
Ohuman
> brainO?
> Think about it some more.
> I think James means that he is the only one with a human
brain.
> Think about it some more.
>
ThereOs not a lot to think about here Jimbo.
> Your statement, put mathematically (I know something you
abhor) is:
If a thing is human, then that thing isnOt built to handle
Mathematics
> The contrapositive is:
> If the thing is built to handle Mathematics, then itOs not
human.
>
If itOs a horse, then it has four legs.
Therefore if it has four legs, then itOs a horse?
Keith K
> Now letOs consider an instantiation of this statement and
consider you.
> Two possibilities exist.
> Either you arenOt built to handle Mathematics Or you are
built to handle
> Mathematics
> If you are built to handle Mathematics, then youOre not
human.
> Which is obviously false (although there are times when
your submental
> arguments make us wonder)
> Therefore you arenOt built to handle Mathematics must be
true.
> (yes, and this would indeed give you the aforementioned
problem.)
> As always James, you have a firm grip on the obvious
(although like most
> things you, since you do them incorrectly, you think there
is some
> additional hidden meaning in it that nobody except you is
smart enough to
> understand.)
Jack
>IOve concluded that the problem IOm facing is
that the human
> brain
>isnOt built to handle Mathematics.
> Wl, I take it you are now claiming that you donOt have a
Ohuman
> brainO?
> Think about it some more.
>I think James means that he is the only one with a human
brain.
> Think about it some more.
> ThereOs not a lot to think about here Jimbo.
> Your statement, put mathematically (I know something you
abhor) is:
> If a thing is human, then that thing isnOt built to handle
Mathematics
> The contrapositive is:
> If the thing is built to handle Mathematics, then itOs not
human.
 If itOs a horse, then it has four legs.
> Therefore if it has four legs, then itOs a horse?
Keith K
Wl, I read that too quickly. DidnOt notice the negations.
Jack Zamat
had
it right. (similar to: If it is a 2-legged animal, then itOs
not a
horse
has the contrapositive:
If it is a horse, then it is not a 2-legged
animal
Keith K
> Now letOs consider an instantiation of this statement and
consider you.
> Two possibilities exist.
> Either you arenOt built to handle Mathematics Or you are
built to
handle
> Mathematics
> If you are built to handle Mathematics, then youOre not
human.
> Which is obviously false (although there are times when
your submental
> arguments make us wonder)
> Therefore you arenOt built to handle Mathematics must be
true.
> (yes, and this would indeed give you the aforementioned
problem.)
> As always James, you have a firm grip on the obvious
(although like
most
> things you, since you do them incorrectly, you think there
is some
> additional hidden meaning in it that nobody except you is
smart enough
to
> understand.)
> Jack

A mathematically-inclined glove-fetishist has a number of
pairs of
gloves of
> identical design, but of several (at least three) different
colors.
She has
> at least three pairs of each color. In the dark she can
distinguish
the
> handedness of a glove, but not its color. Unfortunaty, she
keeps
the
> gloves jumbled up in a drawer in an unlit clar.
>She knows that if she takes out 21 gloves, in the dark,
she can be
sure of
> getting at least one pair.
>What is the maximum number of pairs of gloves that she
might have?
> (Rot-13):
>ragl-svir.
>Fur unf fvk cnvef va rnpu bs guerr pbybhef, naq frira cnvef
va n
sbhegu
>pbybhe.
>Ol gnxvat >ragl yrsg-unaq tybirf fur pna or fher bs univat
ng yrnfg
bar
>bs rnpu pbybhe. Gur >ragl-svefg vf n evtug-unaq tybir, juvpu
vf
obhaq
>gb zngpu.
> You can do three better if you use one fewer color and
the e
strategy.
That is not the correct answer.
I think that *is* the correct answer: 28.
The answer to a generalized form of the puzzle seems to me to
be:
The maximum number of pairs of gloves that she might have, is
given by
the integral part of
(n-2)*(c/(c-1))
Where n is the number of gloves she takes out in the dark,
and c is the
minimum number of different colours. (The minimum number of
pairs of
each colour is immaterial.)
So for n=21 and c=4 the answer is 25 (which is where I went
wrong).
But for n=21 and c=3 the answer is 28.
==
Alec McKenzie
> A mathematically-inclined glove-fetishist has a number of
pairs
of
gloves of
>identical design, but of several (at least three) different
colors. She has
>at least three pairs of each color. In the dark she can
distinguish the
>handedness of a glove, but not its color. Unfortunaty, she
keeps
the
>gloves jumbled up in a drawer in an unlit clar.
> She knows that if she takes out 21 gloves, in the dark,
she can
be
sure of
>getting at least one pair.
> What is the maximum number of pairs of gloves that she
might
have?
>(Rot-13):
>>ragl-svir.
> Fur unf fvk cnvef va rnpu bs guerr pbybhef, naq frira cnvef
va n
sbhegu
> pbybhe.
> Ol gnxvat >ragl yrsg-unaq tybirf fur pna or fher bs univat
ng
yrnfg
bar
> bs rnpu pbybhe. Gur >ragl-svefg vf n evtug-unaq tybir,
juvpu vf
obhaq
> gb zngpu.
> You can do three better if you use one fewer color and
the e
strategy.
> That is not the correct answer.
I think that *is* the correct answer: 28.
The answer to a generalized form of the puzzle seems to me to
be:
The maximum number of pairs of gloves that she might have, is
given by
> the integral part of
(n-2)*(c/(c-1))
Where n is the number of gloves she takes out in the dark,
and c is the
> minimum number of different colours. (The minimum number of
pairs of
> each colour is immaterial.)
So for n=21 and c=4 the answer is 25 (which is where I went
wrong).
But for n=21 and c=3 the answer is 28.
I think that the answer is 29: 3 of one color, 13 of another,
and 13 of
another.
If she then takes out 4 left gloves and 17 right ones... she
must have a
pair.
==
Clive Tooth
http://www.clivetooth.dk
> I think that the answer is 29: 3 of one color, 13 of
another, and 13 of
> another.
If she then takes out 4 left gloves and 17 right ones... she
must have a
> pair.
Yes, of course, that works. Hence the restriction of at least
3 of each
colour, otherwise there could be 30 pairs, needing only 20
taken out.
But can it be proved that 29 is the maximum?
==
Alec McKenzie
> I think that the answer is 29: 3 of one color, 13 of
another, and 13 of
> another.
> If she then takes out 4 left gloves and 17 right ones...
she must have
a
> pair.
Yes, of course, that works. Hence the restriction of at least
3 of each
> colour, otherwise there could be 30 pairs, needing only 20
taken out.
But can it be proved that 29 is the maximum?
Yes. I have programmatically generated the 317 partitions of
30 into at
least three parts, each part being at least 3.
Programmatically solving the
problem for each partition shows that 21 is never the least
number of
gloves
to be picked. And thus, clearly, it cannot be the least
number of gloves
for
any number greater than 30.
I do not know of any more egant proof.
==
Clive Tooth
http://www.clivetooth.dk
> james Randi owes me $1,000,000
Not unless you prove scientifically verifiable
psychic
abilities. Obviously
you havenOt done that.
==
Peacenik
> Neat code. Since you have raised the stakes,
A tling phrase, indeed.
> I shall see your Numerical solution, and raise you an
> EXACT SYMBOLIC solution (given parameter values). See:
http://www.mathStatica.com/Sumof2Betas/ [...]
Wl, hereOs some Maxima (http://maxima.sourceforge.net) code
for this problem. We approach the problem by stating it as a
convolution, and then carrying out the integration.
(C1) f(t):=1/beta(a,b)*t^(a-1)*(1-t)^(b-1)$
(C2) i:f(t)*f(u-t)$
(C3) assume(u>0)$
(C4) I1:integrate(ev(i,a=1/2,b=1),t,0,u);
%PI
(D4) ==-
4
(C5) forget(u>0)$ assume(u>1,u<2)$
(C6)
(C7) I2:integrate(ev(i,a=1/2,b=1),t,u-1,1);
1
2 ATAN(==========-) - 2 ATAN(SQRT(u - 1))
SQRT(u - 1)
(D7) ========================================-
4
(C8) g(u):=if u<1 then ev(I1) se ev(I2);
(D8) g(u) := IF u < 1 THEN EV(I1) SE EV(I2)
(C9) plot2d(O(g(u)),[u,0,2])$
For these parameters (a=1/2, b=1) Maxima is happy. However
Maxima canOt solve it with a=3/4, b=1. Rats!
Now what would be -really- useful in this context would be to
have the Maxima simplifier know about how indicator functions
in
integrals work. Then the limits would be handled
automatically,
and repeated integrations (sums of three or more variables in
this context) would be much simpler.
Incidentally Maxima is derived from the old (circa 1980)
MIT/DOE Macsyma source code. Commercial Macsyma was a fork of
that e code. It appears that Wolfram was heavily inspired by
Macsyma; maybe he should have taken the opportunity to revise
the
programming language == itOs rather strange in
Macsyma/Maxima, and
for better or worse Mathematica shares a lot of that
strangeness.
Robert Dodier
==
If I have not seen as far as others, it is because
giants were standing on my shoulders. == Hal Abson
|>Concerning a numerical solution, Rose < @mathStatica.com>
|>Wl, hereOs some Maxima (http://maxima.sourceforge.net)
code
|>for this problem. We approach the problem by stating it as a
|>convolution, and then carrying out the integration.
|>(C1) f(t):=1/beta(a,b)*t^(a-1)*(1-t)^(b-1)$
|>(C2) i:f(t)*f(u-t)$
...
|>For these parameters (a=1/2, b=1) Maxima is happy. However
|>Maxima canOt solve it with a=3/4, b=1. Rats!
OK, hereOs some Maple.
> f:= (a,b,t) ->
1/Beta(a,b)*piecewise(t<0,0,t<1,t^(a-1)*(1-t)^(b-1),0);
> int(f(1/2,1,t)*f(1/2,1,u-t),t=0..1) assuming u>0,u<1;
Pi
====
4
> int(f(1/2,1,t)*f(1/2,1,u-t),t=0..1) assuming u>1,u<2;
u - 2
-1/2 arcsin(====-)
u
# However, Maple isnOt clever enough to do the integral
assuming u::real.
# For a=3/4,b=1, it wonOt do it assuming u>0,u<1 either,
but:
> v:= f(3/4,1,t)*f(3/4,1,u-t) assuming t>0,t<1,u>t,u-t<1;
9
v := ==================
1/4 1/4
16 t (u - t)
> F:= int(v,t);
3/4
t hypergeom([1/4, 3/4], [7/4], t/u)
F := 3/4 ======================================
1/4
u
> F1:= simplify(eval(F,t=u)-eval(F,t=0)) assuming u>0,u<1;
1/2 2
F1 := 9/8 ================
1/2
Pi
> F2:= simplify(eval(F,t=1)-eval(F,t=u-1)) assuming u>1,u<2;
/
F2 := 3/4 |hypergeom([1/4, 3/4], [7/4], 1/u)
(3/4) u - 1 / 1/4
- (u - 1) hypergeom([1/4, 3/4], [7/4], ====-)| / u
u / /
# Maple wonOt simplify this farther, although it can express
it in terms
# of the LegendreP function.
> plot(piecewise(u<1,F1,F2),u=0..2);
Robert Isra isra@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~isra
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
> Infprod (f(x)) = lim n=1 to 00 prod (over k) f(k/n)^(1/n)
> Infprod (f(x)) = e ^ (Integral log f(x))
Does anybody know of any applications for Inf calculus?
Thm: Suppose f(z) is analytic on the open unit disc in the
complex
plane and continuous on the closed unit disc. Let A = {|z| =
1: f(z) =
0}. If m is (Lebesgue) surface measure on the boundary of the
disc and
m(A) > 0, then f is the zero function.
Pf sketch: By applying Mobius transformations, it suffices to
show
that the above hypotheses imply that f(0) = 0. Define the
function
g(z) = Prod_{k=1}^n f(z*e^{2pi*i*k/n}).
By the maximum principle,
|g(0)| <= sup_{|z|=1} |g(z)|.
Or, in other words,
|f(0)| <= sup_{|z|=1} Prod_{k=1}^n |f(z*e^{2pi*i*k/n})|^(1/n).
Taking limits (you have to be careful here to make this
rigorous),
|f(0)| <= Infprod_0^1 |f(e^{2pi*i*x})|
= e^{Int log|f(e^{2pi*i*x})|}
= e^{-infty} = 0.
> This is a little novty isomorphism, one which is probably wl
known.
> But here goes anyway.
I was wondering about the limit version of the product
function, being
the
> equivalent of the integral function for sums. Lets call it
Infprod
instead
> of integral.
Integral (f(x)) = lim n=1 to 00 sigma (over k) f(k/n)/n
> Infprod (f(x)) = lim n=1 to 00 prod (over k) f(k/n)^(1/n)
What wonderful new properties would such a thing have?
Wl none, actually, because
Log (Inf(f(x)) = Log prod f(k/n)^(1/n)
> = sum log((k/n) ^ (1/n)
> = (1/n) sum (log(k/n))
> = Integral log(x)
> Therefore
Inf (f(x)) = e ^ (Integral log f(x))
How depressing! The whole theory of infinite products reduced
to the
branch
> of integral calculus that deals with functions of the form
log(f(x)).
But of course the reverse is equally true. We could have
devoped our
> product calculus (Inf) from the analogous definition of the
limit in
normal
> calculus. Then some dumbass posts a message in a newsgroup
which invents
> this new form of calculus based upon sums, and devops the
new
Integral
> operator as
Integral (f(x)) = Log (Inf(e^x))
So all of this new sum based integral is reduced to a single
branch of
Inf
> calculus that deals with functions of the form e^f(x).
So normal calculus is isomorphic to Inf calculus, as each is
isomorphic
(?)
> to a subset of the other.
Though it is interesting that although these two are
effectivy duals, I
> have never seen an example where infinite product integrals
appear. I can
> see one reason in analysis of physical situations, as
infinite products
do
> not preserve dimensions (metre, sec, kg) where sums
(integrals) do. But
of
> course, calculus isnOt just used to measure beam strength.
Does anybody know of any applications for Inf calculus?
Peter Webb
I do not bieve this is an extremy natural application of such
a
product-limit, but here is a recent post of mine:
http://mathforum.org/discuss/sci.math/a/t/557879
Discussed is
exp(integral ln(zeta(x)) dx)
and its Dirichlet-sum equivalent.
(I am surprised I was not ?med after posting that message,
for I
suggested the result MAYBE could be used hp to prove the
Riemann
hypothesis....{laughs!})
,
Leroy
< .b.andersen@hia.no Henry, here is the statement of mine you
responded to:
| The P-code (military) pseudo random pattern is transmitted
>| with 10.23Mbps, but this bit rate isnOt normally measured
>| from the ground.
>| The important point is that the 10.23MHz signal is
>| the frequency standard of the satlite clocks.
>| That is, the satlite clock counts 10 230 000 cycles for
>| each second it advances. The clock time is transmitted.
>| (Coded as a bit pattern modulated on two carriers.)
>| We know the frequency must be right since the satlite
>| clocks stay in synch with the GPS time, not because
>| the frequency is measured from the ground.
To this you have responded:
| It is all explained by the fact that lioght speed is source
dependent.
| Unless a GPS clock is directly overhead, it will have a
vocity
component wrt
>| the receiver.
>| To cut a long story short, the doppler method used by the
system confuses
this
>| effect with another that it tries to explain using
rativity.
| The error due to c+v may be rather small when the clocks
are near
vertical but
>| generally they are not. The transverse doppler correction
assumes light
speed
>| is c and not c+(3770.cos theta)
>Of course anybody understands that IF you from the ground
>measure the frequency of the carrier or anything modulated
onto it,
>it will be Doppler shifted, and the Doppler shift will
change all the time
>as the satlite moves rative to the receiver. Of course
anybody
>will understand that it therefore blazingly obvious is
practically
impossible
>to precisy measure the frequency emitted by the satlite with
a
receiver
>on the ground.
>THATOS WHY IT ISNOT DONE.
The trav time of the signal from the OC to the receiver is
c+v and not c.
Therefore the GC reading when the signal arrives is affected,
which causes
an
small error in the Otransverse doppler
correctionO
incorporated into the
positioning process. It turns out that, on average, this
error is of the
order
of 4x10^-10.
So what the heck are you babbling about, and what is the
revance to
>the posting of mine you responded to?
You are not seriously claiming that source dependency can
explain
>why an uncorrected clock in GPS orbit gains 38 us a day
every day
>compared to the GPS coordinated time, or that this have
anything
>whatsoever with Doppler shift to do, are you?
If you are, you must have lost your mind complety.
>Which, BTW, is a possibility I will not exclude.
ItOs up to you.
Do the calculation yoursf if you donOt bieve me.
>
Henri Wilson.
www.users.bigpond.com/hewn/index.htm
< .b.andersen@hia.no
< .b.andersen@hia.no
>> It refutes rativity.
>>Sorry, SR predicts the result precisy.
> Wishful thinking!
> You can twist anything if you have enough faith.
.. which you so vividly demonstrates. :-)
You have indeed to twist facts into fiction to make laser
gyros
>support your blind faith in source dependent speed of light.
It is actually quite simple.
>There are many types of what loosy can be called
>laser gyros or ring gyros, the two main types
>are the Sagnac ring gyro and the ring laser.
>Common to them all is that the light source
>is rotating with the ring (in ring lasers the sources
>are the atoms in the lasing gas), and that the speed
>of light is isotropic (c or c/n) in the non rotating
(inertial) frame.
>This is as predicted by SR and ether theories, but
>blatantly obvious as opposed to the source dependent
>light theory, where the speed of light should be isotropic
>in the rotating frame (the ring frame) where the source
>is stationary.
> Ah! , rotating frames are very dangerous things. They can
lead one up
a
>> blind alley.
>> Ring gyros are like 4-mirror sagnacOs but with an
infinite
number of
mirrors.
Did you have a point?
Thousands of ring lasers are in operation at any time, you
will
>find several of them in every commercial and military
aircraft.
>If the speed of light were dependent on the vocity of the
source,
>they wouldnOt work. They do.
They work perfectly wl under source depedency. See my reply
to George.
Why do you think Sagnac thought his experiment
>supported the ether theory?
The Sagnac experiment confirms SR.
> That isnOt the general view.
Of course it is.
>Claiming otherwise is plain stupidity.
>Or utter ignorance.
But we all know that anything said against rativity must be
stupid, donOt
we
? :-))))))
The Sagnac experiment confirms sonOs ether
theory
>The Sagnac experiment falsifies source dependent light
theory.
> Every experiment MUST support reality, ie, source
dependency.
Now, THATOS a beauty. :-)
Even SR is based on source dependency.
Do you not agree that light leaves its source at c?
And very illustrative of Henry WilsonOs way of reasoning:
> I, Henry Wilson, defines reality. My fantasy world IS
reality.
> All experiments MUST support my idea of reality.
> If they donOt, they are faked. Or
OmisinterpretedO. Or
?wed.
> Every experiment ever perfomed confirming SR falls in
> this category.
As a wise man stated recently (that Greek philosopher,
Androcles, IIRC) if
you
have enough faith in something, you will be able to see
evidence of it
everywhere.
The MMX confirms SR.
> Wl, indirectly, yes. SR ries on source dependency. It
clearly
requires that
>> light travs at c from any source. Any observer at rest
with that source
will
>> receive the e light at c.
>> If the light is returned to the source by a mirror at rest
wrt the source
that
>> the return trav time of that light will be constant,
irespective of the
speed
>> of the system.
It is an indisputable fact that the MMX confirms SR.
>ThatOs why it isnOt disputed.
The MMX proves source dependency, fair and square. No doubt
about it.
The MMX confirms source dependent light theory
> Absoluty correct. No doubt about that.
ThatOs why it isnOt disputed.
The MMX falsifies sonOs ether theory.
> Not exactly.
>> As I have previously explained, it is impossible to prove
the
non-existence of
>> anything.
>> Similarly, if something doesnOt exist, it is impossible
to
assign to it
>> properties that can be tested experimentally.
>> However, my alternative explanation refutes the
conventional argument for
the
>> null result. Mine takes into account the angular departure
of the
splitting
>> mirror from 45 degrees due to aetherian length contraction
and shows how
the
>> cross beam would actually be de?cted backwards even
though the return
trav
>> time of the beam ements remains constant.
Mindless babble.
It is an indisputable fact that the MMX falsifies sons ether
theory.
>ThatOs why it isnOt disputed.
The reason it isnOt disputed is that its NULL result was
easily manipulable
to
give the answer that the physics mafia wanted.
Null results prove very litle if anything.
They usually reveal a ?w in the experiment.
My demo shows that ?w. IF the son aether existed, then the 45
mirror
would not remain at 45 after its length was contracted. The
cross beam would
be
bent backwards as shown, thus exactly compensating for the
presumed
Odiagonal
effectO..
So which of the three mentioned theories is not
>falsified by one of these two experiments?
> OSource dependencyO of course.
Every experiment MUST support [my idea of] reality, ie, source
dependency.
>Thus even the experiments falsifying source dependency MUST
support it!
>So there!
There are NO experiments that falsify source dependency. De
Sitter was
wrong
and there are no others..
You are twisting great, Henry! :-)
But you can twist anything if you have enough faith.
>So you will keep bieving in your fantasy world, wonOt you?
> Wl I certainly havenOt twisted any of the above.
The answer to that is so obvious that I wonOt have to state
it. :-)
This was possibly the best of them all!
> Every experiment MUST support reality, ie, source
dependency.
>I think even you will have a problem with topping that one!
>Go for it!
Like I said, even SR agrees that light always leaves its
source at c.
Are you in sf-denial or just a hypocrite ?
>
Henri Wilson.
www.users.bigpond.com/hewn/index.htm
<george@briar.demon.co.uk
Correct, the only major difference is that in the gyro the
>speed would be c/n immediaty so your comments on extinction
>distance for binary star systems would not apply.
> They were not my comments.
Sorry, someone talked of a source-dependent mod and said the
>reason it didnOt show up in binary star tests was because
the
>speed adjusted to c over ~1 light second distance. I thought
it
>was you, sorry if it wasnOt.
The subject has been discussed here and sewhere but it isnOt
something I
take
seriously or know much about.. It has always been considered
a possible
explanation for the binary star behavior.
>No, the light approaches at c rative to the mirror so will
>either be re?cted at c or reset to c however you want to
>formulate the Osource dependencyO part. There
is no speed
>change rative to the mirrors or fibre (always c) or the
>lab (always c+v).
> Nah.
> ^ very small upward vocity
>> A
>> /============<c+v==S
>> |
>> | <==-c+v-dv
>> |
>> ============-c+v-2dv-> C
>> B
The vocity of the mirrors is parall to their surfaces
>(I canOt think of a way to add arrowheads but suppose the
>table is rotating anticlockwise in your diagram) and also
>applies to the source OSO and detector
OCO. If the horizontal
>component of v at S is vO then the light is moving at
c+vO
>to the left. The horizontal component at A is also vO so
>it approaches at c rative. The vertical component at A
>is also vO so the light moves down at c+vO
and so on:
v / A c+vO v
> / /==========<S
> |
> |
> | <==-c+vO
> |
> |
> ==========>C /
> v B c+vO / v
The other way round it is c-vO all the way.
You are ignoring the fact that the mirros rotate slightly
while the light is
in
transit. The vertical component of A is not quite vO.
>> The rays SA and BC are not quite parall and this causes
the fringe
>movement..
v / A SO v
> / /<==========S
> | ^
> | |
> | |
> | |
> v |
> ==========>/ /
> v B C / v
If the light is going the e way as the table, each angle
>has to be slightly less than 90 degrees so that the light
>gets to SO instead of S because S will have moved.
Now consider the light going the other way round the table:
v / A SO v
> / /==========>S
> ^ |
> | |
> | |
> | |
> | v
> ==========>/ /
> v B C / v
This time the angles each have to be slightly more than
>90 degrees for the light to reach SO.
Since one angle increases and the other decreases, put the
>two together to create fringes and the two beams meet at
>exactly 90 degrees regardless of the motion of the table.
>The only thing that affects the fringes is the time day
>and in Ritzian theory, there is none since the change of
>speed in the lab frame exactly matches the change of path
>length.
No! Every time there is a re?ction from a (moving) 45
mirror, the light
gets
an extra kick in the OforwardO direction or is
retarded
slightly in the
ObackwardO one.
In Ritzian theory, light gains a vocity when re?cted from a
moving
mirror.
If that is not wl known, it is now - because I said it.
This is obvious in the rotating frame where the speed is
>always c and the path length is always the circumference
>so the times taken are always equal.
Using the rotating frame is confusing. For instance, light
doesnOt move in
straight lines.
Light speed is NOT always c in this instance for the reasons
I have just
given.
Consider the replacing the usual sine wave of the light
>with a narrow pulse waveform, the effect of the phase
>change is then obvious.
> In the c+v mod, the two pulses would arrive at different
times and be
>> slightly displaced sideways. ThatOs what happens.
No, the pulses would arrive at the e time. Sideways
>displacement is parall to the wavefront so doesnOt move
>the fringes.
But the displacement is in opposite directions for the two
beams.
George
>
Henri Wilson.
www.users.bigpond.com/hewn/index.htm
Andersen
< .b.andersen@hia.no
>The only parameters used in the calculation is the
gravitational
>fid of the Earth, and the speed and altitude of the clocks.
> So what is the actual connection between a few maths
equations and the
physical
>> change in clock rates that is observed when they are in
free fall?
The connection is that the cause of the proper times of the
clocks
>is to be found in the gravitaional fid of the Earth and the
speed
>and altitude of the clocks, and that GR correctly accounts
for these.
> Does that mean that the e clocks would not work if they
were not in
the
>> EarthOs fid?
When cornered, talk nonsense, eh?
Are you jotting down your own thoughts, ?
> Or does it mean that one free fall is as good as any other,
as far as
clocks
>> are concerned?
Gravity probe A?
>HahA!
>So there!
It doesnOt work.
> Until you can actually specify how the physical change
occurs in these
orbiting
>> clocks, you are just making a complete fool of yoursf.
> It is quite obvious that the clock rates have physically
altered, as
seen in
>> the original Earth frame. Why? An equation? Fairies?
Kick and scream all you want, Henry.
Claiming that a correction calculated by GR is
> NOT a rativistic correction and that the cause
>of how the clocks behave is something not accounted
>for in the equations that predict the behaviour correctly
>to a precision better than 10^-12 is a stupidity beyond bief.
> If that is true, which I very much doubt because it was
never tested
properly,
>> it is pury coincidental. The prediction is wrong for other
orbits
anyway.
> For instance, it MUST be wrong for an orbit 1 centimetre
above the
EarthOs
>> surface since the gravitational potential is the e there
as o the
ground and
>> since we know that clock rates are not affected by
movement. (to
discourage
>> smartarse replies from the idiots here - you know who I
mean - assume the
earth
>> is perfectly round and has no atmosphere)
> So a clock circling the Earth at that height would not run
at a different
rate
>> from one at rest and so it woud remain always in synch
with the GC.
Beautiful, Henry. :-)
Oh thank you, . Yes, I thought it was quite clever mysf.
It complety annihilates your argument and brings down your
rigion.
Henry Wilson knows that the outcome of an unfeasible
experiment
>never done MUST support Henry WilsonOs idea of reality,
>and thus will falsify GR.
Do you not agre that the gravitational potential 1cm (better
still, 1 um)
ABOVE
the surface is near enough to that ON the surface, for the
purpose of this
experiment?
What about all the experiments that ARE done which falsifies
>Henry WilsonOs idea of how reality should have been?
>HahA!
>So there!
But keep bieving in fairies, if you whish.
>I will stick to GR until the day it is clear that the GPS
>never did work, but is a hoax made by the Great Conspiracy
>of GPS Users.
> , I am aware of you argument that Othe clock still runs at
the e
rate
>> but the length of a second varies with gravity and
speedO.
> Do you appreciate how pathetically weak that statement
sounds?
After all your vivid demonstrations I am getting to know
>how your mind works quite wl now.
>So I can imagine.
donOt avoid the question .
>
Henri Wilson.
www.users.bigpond.com/hewn/index.htm
At some point in my educational training, I thought I was
taught that
a metal can with height=radius (similar to a tuna can) was
special,
like the greatest volume for the least surface or something
counterintuitive like that. Alas, it really is the expected
tomato
sauce can (height=diameter) that has the greatest
volume/surface
ratio. So . . . is there anything special about the tuna can
or do I
just need to move on with my life?
>At some point in my educational training, I thought I was
taught that
>a metal can with height=radius (similar to a tuna can) was
special,
>like the greatest volume for the least surface or something
>counterintuitive like that. Alas, it really is the expected
tomato
>sauce can (height=diameter) that has the greatest
volume/surface
>ratio. So . . . is there anything special about the tuna can
or do I
>just need to move on with my life?
Consider the perimeter of the side view (2*(height+diameter)).
The tuna can has maximal volume for fixed perimeter.
I expect this has nothing whatever to do with the reason they
pack tuna in cans of this shape.
Robert Isra isra@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~isra
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
> My analysis continues to indicate that my research on
counting prime
> numbers *should* be more accessible than my other math
research which
> is more abstract, and clearly more difficult. Still I also
recognize
> that significant parts of my prime counting research are
beyond a lot
> of people simply because that research extends into partial
> differential equations.
To me the starting point is simple enough:
dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y, sqrt(y)) -
p(y-1,
> sqrt(y-1))],
S(x,1) = 0,
p(x, y) = ?or(x) - S(x, y) - 1,
and S(x,y) is the sum of dS from dS(x,2) to dS(x,y).
>
<deted
> For a while I pursued fast prime counting programs to see
if I
> couldnOt get progress by that route, and eventually I
really just got
> bored with figuring out fast algorithms, as
IOm more
interested in the
> math, but along the way I found some of the fastest
expressions
> possible for certain counts:
With even N,
N/2 - ?or((N-4)/6) - ?or((N-16)/10) + ?or((N-16)/30) -
> ?or((N-36)/14) + ?or((N-22)/42)
is basically the explicit result of summing an algorithmic
form of the
> dS(x,y) function and subtracting the resulting S(x,y) from
N, where
> evens are automatically handled, from 2 up to and including
10, which
> represents the primes 2, 3, 5 and 7.
That formula counts primes from N=36 up to N=174.
OOPS! It counts primes from N=24 to N=120.
> For instance, N=100, gives
50 - 16 - 8 + 2 - 4 + 1 = 25
as expected.
But beyond counting primes in a small range a slightly longer
formula
> works to give N minus the count of composites up to and
including N
> that have 2, 3, 5, or 7 as a factor out to positive infinity
for even
> N:
N/2 - ?or((N-4)/6) - ?or((N-16)/10) + ?or((N-16)/30) -
> ?or((N-36)/14) + ?or((N-22)/42) + ?or((N-106)/70) -
> ?or((N-106)/210) + 2
Modern mathematicians canOt find anything of
their own without
my
> research of similar length that works out to infinity.
Neat.
> Yup, despite the research done on prime numbers through the
entire
> math history of the world, mathematicians as a group can
only produce
> *longer* expressions if they use their own research on
counting prime
> numbers!
Want more on prime counting? Then see my blog archives:
> My analysis continues to indicate that my research on
counting prime
> numbers *should* be more accessible than my other math
research which
> is more abstract, and clearly more difficult. Still I also
recognize
> that significant parts of my prime counting research are
beyond a lot
> of people simply because that research extends into partial
> differential equations.
> To me the starting point is simple enough:
> dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y,
sqrt(y)) - p(y-1,
> sqrt(y-1))],
> S(x,1) = 0,
> p(x, y) = ?or(x) - S(x, y) - 1,
> and S(x,y) is the sum of dS from dS(x,2) to dS(x,y).
> <deted
> For a while I pursued fast prime counting programs to see
if I
> couldnOt get progress by that route, and eventually I
really just got
> bored with figuring out fast algorithms, as
IOm more
interested in the
> math, but along the way I found some of the fastest
expressions
> possible for certain counts:
> With even N,
> N/2 - ?or((N-4)/6) - ?or((N-16)/10) + ?or((N-16)/30) -
> ?or((N-36)/14) + ?or((N-22)/42)
> is basically the explicit result of summing an
algorithmic form of the
> dS(x,y) function and subtracting the resulting S(x,y) from
N, where
> evens are automatically handled, from 2 up to and including
10, which
> represents the primes 2, 3, 5 and 7.
> That formula counts primes from N=36 up to N=174.
OOPS! It counts primes from N=24 to N=120.
> For instance, N=100, gives
> 50 - 16 - 8 + 2 - 4 + 1 = 25
> as expected.
> But beyond counting primes in a small range a slightly
longer formula
> works to give N minus the count of composites up to and
including N
> that have 2, 3, 5, or 7 as a factor out to positive infinity
for even
> N:
> N/2 - ?or((N-4)/6) - ?or((N-16)/10) + ?or((N-16)/30) -
> ?or((N-36)/14) + ?or((N-22)/42) + ?or((N-106)/70) -
> ?or((N-106)/210) + 2
> Modern mathematicians canOt find anything of
their own
without my
> research of similar length that works out to infinity.
Neat.
> Yup, despite the research done on prime numbers through the
entire
> math history of the world, mathematicians as a group can
only produce
> *longer* expressions if they use their own research on
counting prime
> numbers!
Want more on prime counting? Then see my blog archives:
IsnOt replying to your own posts considered trolling?
David Moran
Situation:
Let y = a^n + b^n (1) where (a, b) = 1, b > a > 1 and odd n >
3.
Let a1, a2 are the two values of a and b1,b2 are the two
values of b
Here a2 > a1 and b2 > b1
y1 = a1^n + b1^n (2) and y2 = a2^n + b2^n (3)
Therefore, y2 > y1
Statement:
Wl, think about it... Pair the numbers up like this:
ODD - EVEN
1 ==== 2
3 ==== 4
5 ==== 6
7 ==== 8
.
etc...
.
999 == 1000
1001 is alone
wl, there are 1000 numbers between 1 and 1000, and obvious
half are
even and half are odd, so...
> how many even numbers are there between 1 and 1001
Many newsgroup readers show the threads which have had the
most recent
responses at the top of the list. This leads to a problem
when some
idiot comes on this board and says that .9999.. != 1, or that
heOs
squared the circle, or whatever, and like a million people
post
replies to messages like this.
My proposal is that, instead of replying immediaty, since
youOre
probably the only person on this thread who knows how to do
the
infinite geometric sum and show this guy that
heOs wrong, just
wait a
while and let the message die off the list, or at the most
send an
anybody who reads this thread, so why should the recent post
list be
?oded with them?
Dani McLaury
> Many newsgroup readers show the threads which have had the
most recent
> responses at the top of the list. This leads to a problem
when some
> idiot comes on this board and says that .9999.. != 1, or
that heOs
> squared the circle, or whatever, and like a million people
post
> replies to messages like this.
...
But thatOs the problem == such threads will get responses.
As
Iım sure the
late C. N. Parkinson would say, the length of a thread is
inversy
proportional to its value.
LH
>Many newsgroup readers show the threads which have had the
most recent
>responses at the top of the list. This leads to a problem
when some
>idiot comes on this board and says that .9999.. != 1, or
that heOs
>squared the circle, or whatever, and like a million people
post
>replies to messages like this.
My proposal is that, instead of replying immediaty, since
youOre
>probably the only person on this thread who knows how to do
the
>infinite geometric sum and show this guy that
heOs wrong,
just wait a
>while and let the message die off the list, or at the most
send an
>anybody who reads this thread, so why should the recent post
list be
>?oded with them?
OTOH, you could use a newsreader that doesnOt put the most
recent responses
at
the top.
==
Mensanator
Ace of Clubs
> When I cared about finding fast prime counting algorithms
versus
> concentrating on the core mathematics, I found formulas
like the
> following, which counts primes up to a given even N over a
certain
> range:
N/2 - ?or((N-4)/6) - ?or((N-16)/10) + ?or((N-16)/30) -
> ?or((N-36)/14) + ?or((N-22)/42) + ?or((N-106)/70) -
> ?or((N-106)/210) + 2.
And it counts primes for an even N from N=38 to N=120.
It actually works for NOs a bit smaller than N=38, though
N=120 is the
top limit as then you need 11.
My point at first was emphasizing *shortness* as it goes to
the speed
question which keeps popping up with my work.
However, IOve found that reasoning to be a leap few are
willing to
make, so IOll switch now to another subject which is the
built-in
compression.
One fascinating thing I can do, and I wonder if anyone knows
if others
can do it with explicit prime counting formulas is have it
*know* that
2 is prime, so that you can use even N, which shortens things
up.
Like consider
N/2 - ?or((N-4)/6) - ?or((N-16)/10) + ?or((N-16)/30)
which if you subtract it from N/2 - 1, you have
N/2 + ?or((N-4)/6) + ?or((N-16)/10) - ?or((N-16)/30) - 1
which gives you the count of composites with 2, 3 or 5 as a
factor for
even N from N=8 out to infinity.
For instance N=10 gives
5 + 1 - 1 = 5, and the 5 composites are 4,6,8,9, and 10.
Somehow it *knows* that 2 is even, so that you can get a
compressed
expression.
The technology required to produce such an expression, from
what IOve
seen is beyond mathematicians, which is strange to me, as
they have
figured out LOTS of ways to do fast counting, so maybe
IOm
wrong.
Iım curious to see if in this admittedly narrow area, any
mathematician can match or better what I have.
Just curious.
Want more prime counting? See my blog archives:
IOm a freshman in college this year, majoring in pure math
with an
emphasis on research, so I still have plenty of memories
about my
application process.
Most prestigious schools, i.e. the ones that cost $30,000 to
$40,000 a year, will cover all of your estimated financial
need,
based on how much money your family makes. While this is not
perfect
(due to some strange circumstances my family is considered by
the
government to be a LOT richer than we really are), it will
give you a
boost. Examples of these schools are ivies (Princeton,
Harvard,
Stanford, etc.) and big-name technical school (Caltech, MIT,
etc.),
and they have exclent math programs. On the other hand, they
are
rather hard to get into. To give you a clue about what kind of
qualifications you have to have, here are the
qualifications I
had
when applying to college, and my results:
* 2nd place, Oklahoma State Math Competition, Sr year
* Silver Medal, USA Math Talent Search, Jr and Sr years
* Oklahoma All-State Orchestra, Sr year
* 1560 SAT, 35 ACT, National Merit Scholar, 800 on SAT II
Math 2c
* Advanced coursework at a university (Early Entry program)
== 4
courses in * Abstract and Linear algebra, along with Calc 1 -
4.
* Princeton - Rejected
* Caltech - Accepted, attending
* MIT - Rejected
* Harvard - Wait list, then rejected
* U Chicago - Missed application deadline
* Stanford - Missed application deadline (yeah, IOm kinda
lazy)
* Yale - Rejected
These are in the order of my choices. If you donOt have any
clue
about how to order them, which I guess you donOt since
youOre
asking
about it, you should check out the US News rankings, which
give you a
good rule-of-thumb measurement. I should warn you that they
measure
all programs, not just math, so MIT and Caltech, which have
better
math programs, take huge hits for having virtually
nonexistant arts
and humanities programs (all though, as a classical musician,
I must
say that Caltech is still a great environment == we bring in
world-class performances all the time)
If you think you have the qualifications, you should go for
the
absolute top schools. Otherwise, still try for good
universities ==
Duke, Northwestern, Rice, etc., or for big-name public
universities
(UC Berkey, University of Virginia, etc.) Allow me to stress
this
== a community college will be an utter waste of your time
and will
provide you with NOTHING useful to a mathematician. At the
very
least, you can attend institutions like the University of
Oklahoma
(the only example IOm sure about, although I know there are
many
others like it), where if your test scores are high enough
you can get
a full ride. You will get a significant boost some places
because
youOre a girl; not very many girls do math and science
(Caltech, a
math-and-science only school, has about half as many girls as
boys.)
Another thing to consider is that you still have plenty of
time to
go out and seek qualifications. If you live near a university,
try to
competitions
if your school offers them, and DEFINITY participate in the
UTS
(see links bow). Get to know real mathematicians at
universities,
and look for opportunities to do math research. If youOre
not
good at
taking standardized tests, get better. Yeah, I hate them too,
but
many colleges filter out everyone with low test scores without
reading
the rest of their application. If youOre bad at writing
essays, get
better; essays are a major component of your application.
Grades
donOt matter as much as anything se, because some schools
give high
grades to everybody and they have no way of knowing, but
donOt blow
them off or anything. In other words, good grades wonOt hp
you,
but bad grades might hurt you. Also, pursue ALL of your
interests; it
hps to be wl rounded. This is not to say that you should go
out
and start trying to pick up random things you donOt enjoy
juts to
stuff them on your resume; itOll waste time you could be
learning
about math, or at least hanging out with your friends and
having fun.
A final thing to consider is quality-of-life. At Caltech, a
number of people love living in California == the weatherOs
nice,
weOre close to the beach, and you can take trips to LA all
the time.
On the other hand, the work is hard and thereOs a lot of it,
and
rationships (where I mean like boyfriend/girlfriend
rationships)
are kinda weird as there are twice as many guys as girls.
Remember,
this isnOt just a school, this is where you are living for
at
least 4
years of your life. Take advantage of programs that the
schools have
to ? you out and see their campus.
If you have any further questions, you can contact me; I
spend a
lot of time goofing off on my computer and IOd
be glad to do
something
at dan131m.
Useful Links:
USA Math Talent search, an individual math competition you
can do over
the internet. Definity get involved.
http://www.nsa.gov/programs/mepp/uts.html
US News and World Report College Rankings.
http://www.usnews.com/usnews/edu/college/rankings/ranknatudoc
_brief.php
This might be easy, but try to prove (which seems like it
might be fun
{in a math-proving way} to me):
For m = EVEN positive integer,
(m+1) always divides
m! * sum{k=1 to m} F(m+1-k) /k ,
where F(n) is the n_th Fibonacci number.
(F(1) = 1, F(2) = 1, F(m+2) = F(m+1) +F(m))
And in general, for more advanced players...:
Let H(0,m) = 1/m,
for all positive integers m.
Let H(n,m) = sum{k=1 to m} H(n-1,k) ,
for all positive integers n.
Then, for all EVEN positive integers m,
and for all nonnegative integers n,
(m+2n+1) always divides
m! (sum{k=1 to m} F(m+1-k) * H(n,k) ).
For example,
(m+3) always divides, for even m,
m! *sum{k=1 to m} F(m+1-k) H(k),
where H(k) = 1 +1/2 +1/3 +..+ 1/k,
the kth harmonic number.
,
Iıve been looking at this for a while. I canOt
seem to figure
it out.
I donOt think it should be that complicated :(
i have the following equation:
V31 = (V32 + V21) / (1 + V32*V21)
How do I verify that if -1 < V21, V32 < 1, then -1 < V31 <
1???
This is what I was thinking:
Wl we know the extreme values are
-2 < V32 + V21 < 2 and 0 < 1 + V32*V21 < 2