| Topic: |
Science > Physics |
| User: |
"Tomoko Kanazawa" |
| Date: |
07 Jun 2007 12:14:37 AM |
| Object: |
Question about the field of complex numbers C |
Seems like the Field of complex numbers can be considered the "union" of two
fields.
for any z in C, z = Re(z) + Im(z), seems like Re(z) is a field, after all
it is just R, but Im(z) is also a field in and of itself.
Does that sound bogus ?
The reason I'm interested is because of the silly explorations I've been
conducting into the complex-like randoms expressed as ( a + ~b).
If a is real but ~b is an uncertainty, then ~b is not a number but this does
not mean that it cannot behave like a field.
Instead of Re(z) and Im(z), you can consider NonRan(z) and Ran(z), where
NonRan(z) is the nonrandom part of z and Ran(z) is the random part of z.
So, it would be interesting if NonRan(z) acts like a field, and Ran(z) acts
like a field, and very interesting if the collection of all z = (NonRan(z) +
Ran(z)) also acts like a field.
Some deeper results for C might carry over to "the randoms".
How stupid does that sound ?
.
|
|
| User: "Arturo Magidin" |
|
| Title: Re: Question about the field of complex numbers C |
07 Jun 2007 09:54:51 AM |
|
|
In article <xL6dnXnla9E2CPrbnZ2dnUVZ_hqdnZ2d@comcast.com>,
Tomoko Kanazawa <T.Kanazawa@eodomo.gsfc.nasa.gov> wrote:
Seems like the Field of complex numbers can be considered the "union" of two
fields.
Then appearances can be deceiving.
for any z in C, z = Re(z) + Im(z),
Hmmm.. Are you sure? I would have said
z = Re(z) + i*Im(z).
Both Re and im, as functions defined on the complex plane, have real
range.
seems like Re(z) is a field, after all
"Re(z)" is a single real number, namely the real part of the complex
number z. What you mean, perhaps, is that the ->range<- of the
function Re is a field, namely, the real numbers.
it is just R, but Im(z) is also a field in and of itself.
With the same interpretation of what you said and correcting the fact
that the values of Im(z) are reals, it is the ->same<- field, namely
the reals.
Perhaps you are thinking of the imaginary axis? It's not a field,
because it is not closed under multiplication. The complex square root
of negative one, i, is in the imaginary axis, but the product with
itself is not.
Does that sound bogus ?
It actually sounds very confused. You don't even seem to have the
meaning of "Im(z)" correct (at least, up to the usual conventions).
The reason I'm interested is because of the silly explorations I've been
conducting into the complex-like randoms expressed as ( a + ~b).
If a is real but ~b is an uncertainty, then ~b is not a number but this does
not mean that it cannot behave like a field.
Is the expression unique?
If not, you could be in trouble.
--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
======================================================================
Arturo Magidin
magidin-at-member-ams-org
.
|
|
|
| User: "Tomoko Kanazawa" |
|
| Title: Re: Question about the field of complex numbers C |
09 Jun 2007 10:52:26 PM |
|
|
"Arturo Magidin" <magidin@math.berkeley.edu> wrote in message
news:f4967r$1utu$1@agate.berkeley.edu...
In article <xL6dnXnla9E2CPrbnZ2dnUVZ_hqdnZ2d@comcast.com>,
Tomoko Kanazawa <T.Kanazawa@eodomo.gsfc.nasa.gov> wrote:
Seems like the Field of complex numbers can be considered the "union" of
two
fields.
Then appearances can be deceiving.
for any z in C, z = Re(z) + Im(z),
Hmmm.. Are you sure? I would have said
z = Re(z) + i*Im(z).
Both Re and im, as functions defined on the complex plane, have real
range.
seems like Re(z) is a field, after all
"Re(z)" is a single real number, namely the real part of the complex
number z. What you mean, perhaps, is that the ->range<- of the
function Re is a field, namely, the real numbers.
it is just R, but Im(z) is also a field in and of itself.
With the same interpretation of what you said and correcting the fact
that the values of Im(z) are reals, it is the ->same<- field, namely
the reals.
Perhaps you are thinking of the imaginary axis? It's not a field,
because it is not closed under multiplication. The complex square root
of negative one, i, is in the imaginary axis, but the product with
itself is not.
Does that sound bogus ?
It actually sounds very confused. You don't even seem to have the
meaning of "Im(z)" correct (at least, up to the usual conventions).
I stand corrected. I was thinking that you have i*Im(z), and -i*Im(z), so
you have an additive inverse, i*0 is your additive identity, but you dont
have closure under multiplication, I humbly apologize for being an idiot and
PWI, posting while intoxicated.
The reason I'm interested is because of the silly explorations I've been
conducting into the complex-like randoms expressed as ( a + ~b).
If a is real but ~b is an uncertainty, then ~b is not a number but this
does
not mean that it cannot behave like a field.
Is the expression unique?
If not, you could be in trouble.
I should not even be reaching for things until I've got a very solid grasp
on just what the heck (~b) is, and how (~) behaves.
If (~b) is an uncertainty then I really dont think that you can call (~b) a
number, unless it is considered a number in the same sense that quaternions
are numbers. Can (~) be considered as something similar to i ?
Seems reasonable that b must be chosen from either a continuous or a
discrete domain.
You must have that either
b = (~)[ b1, b2 ] ,
b chosen at random from the interval [ b1, b2 ],
or
b = (~){ b1, b2, b3,...b_n },
b chosen at random from the discrete set { b1, b2, b3,...b_n }.
I think that it actually makes sense to let this be arbitrary. That you have
this distinction, but you let it remain intact, and you just allow this
wierd duality.
What I need to do is show whether you have a field with either or both of
the above cases, one case at a time.
But I cant do that until I figure out :
what (~)[ b1, b2 ] * (~)[ b1, b2 ] is equal to,
what (~)[ b1, b2 ] + (~)[ b1, b2 ] is equal to,
what (~){ b1, b2, b3,...b_n } * (~){ b1, b2, b3,...b_n } is equal to,
what (~){ b1, b2, b3,...b_n } + (~){ b1, b2, b3,...b_n } is equal to,
whether (~){ b1, b2, b3,...b_n } + (~)[ b1, b2 ] makes any sense,
whether (~){ b1, b2, b3,...b_n } * (~)[ b1, b2 ] makes any sense,
what is (~) + (~),
what is (~) * (~),
does (~) have an inverse,
does (~)[ b1, b2 ] have an additive or multiplicative inverse,
does (~){ b1, b2, b3,...b_n } have an additive or multiplicative inverse,
Does it make more sense to invent an inverse operation based on some
philosophical justifications, simply because you would like to shoehorn this
"number" into satisfying nice properties,
or is that approach so artificial that it undermines the objective which is
ultimately an attempt to model reality ,...........
or, is it possible that the notion of field in this case is just slightly
different (dont laugh), because uniqueness is somehow related to scale which
we havent talked about much but ultimately we _are_ working on a kind of
scalar calculus and so things are quite different .......?
And yes I'm very glad that I have no career at this point becauuse they
would have fired me a long time ago !!! : )
.
|
|
|
| User: "Tomoko Kanazawa" |
|
| Title: Re: Question about the field of complex numbers C |
10 Jun 2007 10:36:18 PM |
|
|
Seems reasonable that b must be chosen from either a continuous or a
discrete domain.
You must have that either
b = (~)[ b1, b2 ] ,
b chosen at random from the interval [ b1, b2 ],
or
b = (~){ b1, b2, b3,...b_n },
b chosen at random from the discrete set { b1, b2, b3,...b_n }.
I think that it actually makes sense to let this be arbitrary. That you
have
this distinction, but you let it remain intact, and you just allow this
wierd duality.
What I need to do is show whether you have a field with either or both of
the above cases, one case at a time.
But I cant do that until I figure out :
what (~)[ b1, b2 ] * (~)[ b1, b2 ] is equal to,
what (~)[ b1, b2 ] + (~)[ b1, b2 ] is equal to,
what (~){ b1, b2, b3,...b_n } * (~){ b1, b2, b3,...b_n } is equal to,
what (~){ b1, b2, b3,...b_n } + (~){ b1, b2, b3,...b_n } is equal to,
whether (~){ b1, b2, b3,...b_n } + (~)[ b1, b2 ] makes any sense,
whether (~){ b1, b2, b3,...b_n } * (~)[ b1, b2 ] makes any sense,
I have a geometric justification for the above 6 cases, and I will posit
that there may very well be a "mixed type" of number which I will try to
explain using a geometric approach.
Let |~~~~~~| represent length which is everywhere probabilistic.
Let |---------| represent well behaved length, some bounded subset of R1.
Let |###| represent an uncertainty.
Now, I have been arguing that these different kinds of length can be
equivalent, specifically that
(a)*(~b) = (a + ~b)
|~~~~~~~~~~~~~~| = |-----------| + |#######|
That is probably a relatively easy result to think about. Lets add one
additional concept, that you can have a "mixed type" which starts to
resemble something like a polynomial representation.
For example
[(a)*(~b)]^2 + (a+~b)^2 +
(a+~b)
|~~~~~|*|~~~~~| + ( |-----| + |##| )*( |-----| + |##| ) + ( |-----| + |##| )
Yes, I do realize that there is a danger that two different polynomial
expressions might represent the same thing, I dont think that it would be a
deal breaker if it did. This approach actually suggests some possible
experiments for guys like JSH who should be spending their time in a laser
lab somewhere concocting setups with beamsplitters to better understand the
wave-particle duality instead of trying to solve FLT and stumbling around
claiming that the universe "makes no sense".
what is (~) + (~),
I still dont understand that, or if there is such a thing as -(~), I'm
clueless on that,
what is (~) * (~),
I still dont understand that, or if there is such a thing as (~)^ -1, I'm
clueless on that as well,
does (~) have an inverse,
does (~)[ b1, b2 ] have an additive or multiplicative inverse,
does (~){ b1, b2, b3,...b_n } have an additive or multiplicative inverse,
Does it make more sense to invent an inverse operation based on some
philosophical justifications, simply because you would like to shoehorn
this
"number" into satisfying nice properties,
or is that approach so artificial that it undermines the objective which
is
ultimately an attempt to model reality ,...........
or, is it possible that the notion of field in this case is just slightly
different (dont laugh), because uniqueness is somehow related to scale
which
we havent talked about much but ultimately we _are_ working on a kind of
scalar calculus and so things are quite different .......?
What I am trying to say is that the amount of order or disorder is related
to scale. On the quantum scale you have a very large (~b) and (a) is near
zero. On ordinary scales you have (~b) near zero, and (a) dominates. The
aamount of order and disorder is variable, according to scale somehow.
.
|
|
|
| User: "Tomoko Kanazawa" |
|
| Title: Re: Question about the field of complex numbers C |
10 Jun 2007 11:46:38 PM |
|
|
Seems reasonable that b must be chosen from either a continuous or a
discrete domain.
You must have that either
b = (~)[ b1, b2 ] ,
b chosen at random from the interval [ b1, b2 ],
or
b = (~){ b1, b2, b3,...b_n },
b chosen at random from the discrete set { b1, b2, b3,...b_n }.
I think that it actually makes sense to let this be arbitrary. That you
have
this distinction, but you let it remain intact, and you just allow this
wierd duality.
What I need to do is show whether you have a field with either or both
of
the above cases, one case at a time.
But I cant do that until I figure out :
what (~)[ b1, b2 ] * (~)[ b1, b2 ] is equal to,
what (~)[ b1, b2 ] + (~)[ b1, b2 ] is equal to,
what (~){ b1, b2, b3,...b_n } * (~){ b1, b2, b3,...b_n } is equal to,
what (~){ b1, b2, b3,...b_n } + (~){ b1, b2, b3,...b_n } is equal to,
whether (~){ b1, b2, b3,...b_n } + (~)[ b1, b2 ] makes any sense,
whether (~){ b1, b2, b3,...b_n } * (~)[ b1, b2 ] makes any sense,
I have a geometric justification for the above 6 cases, and I will posit
that there may very well be a "mixed type" of number which I will try to
explain using a geometric approach.
Let |~~~~~~| represent length which is everywhere probabilistic.
Let |---------| represent well behaved length, some bounded subset of R1.
Let |###| represent an uncertainty.
Now, I have been arguing that these different kinds of length can be
equivalent, specifically that
(a)*(~b) = (a + ~b)
|~~~~~~~~~~~~~~| = |-----------| + |#######|
That is probably a relatively easy result to think about. Lets add one
additional concept, that you can have a "mixed type" which starts to
resemble something like a polynomial representation.
For example
[(a)*(~b)]^2 + (a+~b)^2 +
(a+~b)
|~~~~~|*|~~~~~| + ( |-----| + |##| )*( |-----| + |##| ) + ( |-----| +
|##| )
Yes, I do realize that there is a danger that two different polynomial
expressions might represent the same thing, I dont think that it would be
a
deal breaker if it did. This approach actually suggests some possible
experiments for guys like JSH who should be spending their time in a
laser
lab somewhere concocting setups with beamsplitters to better understand
the
wave-particle duality instead of trying to solve FLT and stumbling around
claiming that the universe "makes no sense".
what is (~) + (~),
I still dont understand that, or if there is such a thing as -(~), I'm
clueless on that,
what is (~) * (~),
I still dont understand that, or if there is such a thing as (~)^ -1, I'm
clueless on that as well,
does (~) have an inverse,
does (~)[ b1, b2 ] have an additive or multiplicative inverse,
does (~){ b1, b2, b3,...b_n } have an additive or multiplicative
inverse,
Does it make more sense to invent an inverse operation based on some
philosophical justifications, simply because you would like to shoehorn
this
"number" into satisfying nice properties,
or is that approach so artificial that it undermines the objective which
is
ultimately an attempt to model reality ,...........
or, is it possible that the notion of field in this case is just
slightly
different (dont laugh), because uniqueness is somehow related to scale
which
we havent talked about much but ultimately we _are_ working on a kind of
scalar calculus and so things are quite different .......?
What I am trying to say is that the amount of order or disorder is related
to scale. On the quantum scale you have a very large (~b) and (a) is near
zero. On ordinary scales you have (~b) near zero, and (a) dominates. The
aamount of order and disorder is variable, according to scale somehow.
One last thing, this last passage may lead to some type of notion of
"inverse". I wont say anything else about that because I am just babbling
like an idiot, but the notion of inverse may be fundamentally linked to the
notion of scale in this scheme.
.
|
|
|
|
|
|
|
| User: "=?ISO-8859-1?Q?Jos=E9_Carlos_Santos?=" |
|
| Title: Re: Question about the field of complex numbers C |
07 Jun 2007 03:08:29 AM |
|
|
On 6/7/2007 6:14 AM, Tomoko Kanazawa wrote:
Seems like the Field of complex numbers can be considered the "union" of two
fields.
for any z in C, z = Re(z) + Im(z), seems like Re(z) is a field, after all
it is just R, but Im(z) is also a field in and of itself.
If _z_ in C, then what you have is z = Re(z) + Im(z)i, instead of
Re(z) + Im(z).
Besides, if _z_ in C, then Re(z) is *not* a field; it is a real number.
It is the *set* of all numbers of the form Re(z) that is a field. The
set of all numbers f the form Im(z) is also a field, but it is the same
field.
Does that sound bogus ?
Yes.
Best regards,
Jose Carlos Santos
.
|
|
|
|
| User: "Rotwang" |
|
| Title: Re: Question about the field of complex numbers C |
09 Jun 2007 12:04:45 AM |
|
|
On 7 Jun, 06:14, "Tomoko Kanazawa" <T.Kanaz...@eodomo.gsfc.nasa.gov>
wrote:
Seems like the Field of complex numbers can be considered the "union" of two
fields.
[...]
There are a number of nice replies in this thread raising issues with
your first post, but I don't think that anyone has been pedantic
enough to point out this particular problem: when you write "union" I
think you mean "(set theoretic) product". The union of {Re(z) | z in
C} and {i*Im(z) | z in C} consists only of numbers that are real or
pure imaginary.
.
|
|
|
| User: "Tomoko Kanazawa" |
|
| Title: Re: Question about the field of complex numbers C |
09 Jun 2007 06:26:33 PM |
|
|
Seems like the Field of complex numbers can be considered the "union" of
two
fields.
[...]
There are a number of nice replies in this thread raising issues with
your first post, but I don't think that anyone has been pedantic
enough to point out this particular problem: when you write "union" I
think you mean "(set theoretic) product". The union of {Re(z) | z in
C} and {i*Im(z) | z in C} consists only of numbers that are real or
pure imaginary.
Quite likely the case.
I have to make a confession. I have always believed that math is based on
fundamentally simplistic things, and I've always believed that there are
probably many things which are so simple that they become incomprehensible
and so they remain unknown.
The fact that I'm not a professional mathematician has allowed me to pursue
simplicity by intentionally trying to make myself as simplistic as possible,
and so I have been intentionally trying to become dumb, and I've been doing
this for many years. That may sound stupid, but just imagine the leap which
was made when man first learned how to add abstractly, or when man first
began the practice of counting.
I am very interested in things like that. And so I have no shame whatsoever
when I say that I have been trying to become as dumb as possible, it is
true. But the tradeoff is that I wont have the expertese of others who's
paths wont allow them to wander off and be dumb as I have done.
You're absolutely correct, in my sloppiness I called it a union, it is
indeed a set theoretic product.
I guess the question that remains is that if uncertainty is treated as an
object, then it might behave like a number in some ways. The uncertainty has
all kinds of properties, it is either continuous or discretely chosen, it is
bounded or unbounded, there is an expected value, all kinds of things. Do
you really need random variables ? Is it possible to treat an uncertainty
for what it is ?
.
|
|
|
|
|
| User: "smn" |
|
| Title: Re: Question about the field of complex numbers C |
07 Jun 2007 01:26:18 AM |
|
|
On Jun 6, 10:14 pm, "Tomoko Kanazawa"
<T.Kanaz...@eodomo.gsfc.nasa.gov> wrote:
Seems like the Field of complex numbers can be considered the "union" of two
fields.
for any z in C, z = Re(z) + Im(z), seems like Re(z) is a field, after all
it is just R, but Im(z) is also a field in and of itself.
No z = Re(z) + i Im(z) and i is nor a real number since ii=-1 smn
Does that sound bogus ?
The reason I'm interested is because of the silly explorations I've been
conducting into the complex-like randoms expressed as ( a + ~b).
If a is real but ~b is an uncertainty, then ~b is not a number but this does
not mean that it cannot behave like a field.
Instead of Re(z) and Im(z), you can consider NonRan(z) and Ran(z), where
NonRan(z) is the nonrandom part of z and Ran(z) is the random part of z.
So, it would be interesting if NonRan(z) acts like a field, and Ran(z) acts
like a field, and very interesting if the collection of all z = (NonRan(z) +
Ran(z)) also acts like a field.
Some deeper results for C might carry over to "the randoms".
How stupid does that sound ?
.
|
|
|
| User: "smn" |
|
| Title: Re: Question about the field of complex numbers C |
07 Jun 2007 02:16:59 AM |
|
|
On Jun 6, 11:26 pm, smn <smnewber...@comcast.net> wrote:
On Jun 6, 10:14 pm, "Tomoko Kanazawa"
<T.Kanaz...@eodomo.gsfc.nasa.gov> wrote:
Seems like the Field of complex numbers can be considered the "union" of two
fields.
for any z in C, z = Re(z) + Im(z), seems like Re(z) is a field, after all
it is just R, but Im(z) is also a field in and of itself. No z = Re(z) + i Im(z) and i is nor a real number since ii=-1 smn
Does that sound bogus ?
The reason I'm interested is because of the silly explorations I've been
conducting into the complex-like randoms expressed as ( a + ~b).
If a is real but ~b is an uncertainty, then ~b is not a number but this does
not mean that it cannot behave like a field.
Instead of Re(z) and Im(z), you can consider NonRan(z) and Ran(z), where
NonRan(z) is the nonrandom part of z and Ran(z) is the random part of z.
So, it would be interesting if NonRan(z) acts like a field, and Ran(z) acts
like a field, and very interesting if the collection of all z = (NonRan(z) +
Ran(z)) also acts like a field.
Some deeper results for C might carry over to "the randoms".
How stupid does that sound ?- Hide quoted text -
- Show quoted text -
.
|
|
|
|
|
| User: "" |
|
| Title: Re: Question about the field of complex numbers C |
08 Jun 2007 05:22:00 PM |
|
|
On Jun 7, 12:14 am, "Tomoko Kanazawa"
<T.Kanaz...@eodomo.gsfc.nasa.gov> wrote:
Seems like the Field of complex numbers can be considered the "union" of two
fields.
The complex numbers are the field comprising the base 3 "decimal"
numerals formed from the following digits
L = -i-1,
upside-down L = i-1
backwards L = 1-i
backwards upside-down L = 1+i
O = 0
C = -1
backwards C = 1
U = -i
upside-down U = i
As this shows, they can be conceived as an essential unity in and of
themselves, rather than as an amalgamation of ordered pairs, as they
usually are.
They can also be defined by base 3i numerals...
The Untold Story of Nineary Numerals
http://federation.g3z.com/Mathematics/Nineary.htm
.
|
|
|
|
| User: "Timothy Golden BandTechnology.com" |
|
| Title: Re: Question about the field of complex numbers C |
07 Jun 2007 08:29:17 AM |
|
|
On Jun 7, 1:14 am, "Tomoko Kanazawa" <T.Kanaz...@eodomo.gsfc.nasa.gov>
wrote:
Seems like the Field of complex numbers can be considered the "union" of two
fields.
for any z in C, z = Re(z) + Im(z), seems like Re(z) is a field, after all
it is just R, but Im(z) is also a field in and of itself.
Does that sound bogus ?
The reason I'm interested is because of the silly explorations I've been
conducting into the complex-like randoms expressed as ( a + ~b).
If a is real but ~b is an uncertainty, then ~b is not a number but this does
not mean that it cannot behave like a field.
Instead of Re(z) and Im(z), you can consider NonRan(z) and Ran(z), where
NonRan(z) is the nonrandom part of z and Ran(z) is the random part of z.
So, it would be interesting if NonRan(z) acts like a field, and Ran(z) acts
like a field, and very interesting if the collection of all z = (NonRan(z) +
Ran(z)) also acts like a field.
Some deeper results for C might carry over to "the randoms".
How stupid does that sound ?
Hi Tomoko.
I still think that your fundamental construction
x + ~ y
puts the result in the same domain as a and b are each in.
But that being said, you can also instantiate them into as many
dimensions as you like.
So, instead of x being in the reals why not just allow both x and y to
have two degrees of freedom?
Now you have a center point x and a radius of convergence y. The two
can be interactive no different than standard analysis does. The
multidimensional
~ y
has a nice hyperspherical quality about it. So you have statistical
geometry. Trying to grant these superposed qualities uniqueness while
superposing them forces one to this. It is also somewhat relativistic
to treat this as a basis so I think your idea is coherent but
dimension is a seperate concept. Still, your construction has a
consequence on dimension in that the kernel must now be spherical
rather than cubical.
This opens up a question: What does the random density of a circle
even look like projected to one dimension? Is it Gaussian? Certainly
it peaks at the center x but it looks like it contains a sinusoidal
component. Are these relations trivial?
-Tim
.
|
|
|
| User: "Tomoko Kanazawa" |
|
| Title: Re: Question about the field of complex numbers C |
08 Jun 2007 07:46:40 AM |
|
|
"Timothy Golden BandTechnology.com" <tttpppggg@yahoo.com> wrote in message
news:1181222957.248202.324000@q69g2000hsb.googlegroups.com...
On Jun 7, 1:14 am, "Tomoko Kanazawa" <T.Kanaz...@eodomo.gsfc.nasa.gov>
wrote:
Seems like the Field of complex numbers can be considered the "union" of
two
fields.
for any z in C, z = Re(z) + Im(z), seems like Re(z) is a field, after
all
it is just R, but Im(z) is also a field in and of itself.
Does that sound bogus ?
The reason I'm interested is because of the silly explorations I've been
conducting into the complex-like randoms expressed as ( a + ~b).
If a is real but ~b is an uncertainty, then ~b is not a number but this
does
not mean that it cannot behave like a field.
Instead of Re(z) and Im(z), you can consider NonRan(z) and Ran(z), where
NonRan(z) is the nonrandom part of z and Ran(z) is the random part of z.
So, it would be interesting if NonRan(z) acts like a field, and Ran(z)
acts
like a field, and very interesting if the collection of all z =
(NonRan(z) +
Ran(z)) also acts like a field.
Some deeper results for C might carry over to "the randoms".
How stupid does that sound ?
Hi Tomoko.
Well, there are some things which just cannot be added together so nicely.
If you add together things such as a + ib, the result (a+ib) behaves like
a number very nicely, but you cannot really get rid of the addition symbol
in this number unless you express it in polar form.
I dont really want to confuse the issue, there's a very straightforward and
easily understood geometric justification for using numbers in the form (a+
~b). I'll just post it one more time, I'm sure everyoen has aalready seen
this :
Case #1
(a)*(~b)
| ~ ~ ~ ~ ~~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ |
Length is everywhere fuzzy. The existence of any given point is
probabilistic.
Case #2
(a + ~b)
|---|---|---|---|---|---|---|---|---| + [~ ~ ~)
Length is everywhere well behaved except at the endpoint, which is
probabilistic.
So Tim, if you have a geometric justification for expanding degrees of
freedom etc please post. But this approach is extremely simple. It says that
if you have a given length "a" and there is an uncertainty associated with
the magnitude of that length "~b", then you have two ways to model this,
case#1 and case #2, and that they are equivalent.
Many dualities follows directly from this, continuous-discrete duality, and
I believe wave-particle duality as well.
The approach is justified by allowing existential indeterminacy, otherwise
it would be impossible.
.
|
|
|
| User: "Timothy Golden BandTechnology.com" |
|
| Title: Re: Question about the field of complex numbers C |
08 Jun 2007 09:41:39 AM |
|
|
On Jun 8, 8:46 am, "Tomoko Kanazawa" <T.Kanaz...@eodomo.gsfc.nasa.gov>
wrote:
"Timothy Golden BandTechnology.com" <tttppp...@yahoo.com> wrote in messagenews:1181222957.248202.324000@q69g2000hsb.googlegroups.com...
On Jun 7, 1:14 am, "Tomoko Kanazawa" <T.Kanaz...@eodomo.gsfc.nasa.gov>
wrote:
Seems like the Field of complex numbers can be considered the "union" of
two
fields.
for any z in C, z = Re(z) + Im(z), seems like Re(z) is a field, after
all
it is just R, but Im(z) is also a field in and of itself.
Does that sound bogus ?
The reason I'm interested is because of the silly explorations I've been
conducting into the complex-like randoms expressed as ( a + ~b).
If a is real but ~b is an uncertainty, then ~b is not a number but this
does
not mean that it cannot behave like a field.
Instead of Re(z) and Im(z), you can consider NonRan(z) and Ran(z), where
NonRan(z) is the nonrandom part of z and Ran(z) is the random part of z.
So, it would be interesting if NonRan(z) acts like a field, and Ran(z)
acts
like a field, and very interesting if the collection of all z =
(NonRan(z) +
Ran(z)) also acts like a field.
Some deeper results for C might carry over to "the randoms".
How stupid does that sound ?
Hi Tomoko.
Well, there are some things which just cannot be added together so nicely.
If you add together things such as a + ib, the result (a+ib) behaves like
a number very nicely, but you cannot really get rid of the addition symbol
in this number unless you express it in polar form.
I dont really want to confuse the issue, there's a very straightforward and
easily understood geometric justification for using numbers in the form (a+
~b). I'll just post it one more time, I'm sure everyoen has aalready seen
this :
Case #1
(a)*(~b)
| ~ ~ ~ ~ ~~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ |
Length is everywhere fuzzy. The existence of any given point is
probabilistic.
Case #2
(a + ~b)
|---|---|---|---|---|---|---|---|---| + [~ ~ ~)
Length is everywhere well behaved except at the endpoint, which is
probabilistic.
So Tim, if you have a geometric justification for expanding degrees of
freedom etc please post. But this approach is extremely simple. It says that
if you have a given length "a" and there is an uncertainty associated with
the magnitude of that length "~b", then you have two ways to model this,
case#1 and case #2, and that they are equivalent.
Many dualities follows directly from this, continuous-discrete duality, and
I believe wave-particle duality as well.
The approach is justified by allowing existential indeterminacy, otherwise
it would be impossible.
I was just hoping you would consider your construction in n
dimensions. I do see your energy going toward two dimensions and was
trying to shake it up a little. I don't quite see your first
representation with the whole line being fuzzy and at that level I do
see that when b varies you do have a more dynamic fuzziness though the
resolution is always back on the line. If the line is then permitted a
thickness you could have an envelope. Is that more like where you are
trying to go? You could consider a function b(a) taking 'a' alone to
be a straight and narrow line. In some ways this is a wavelet style
since b should not only effect this graphic at a single point a but
across a swath of a. That is challenging to categorize. It almost
looks like a shape would have to be chosen to superpose and hence the
wavelet style. Is this more like the struggle toward 2D you make?
Sorry if this sounds like more nonsense. I would like a continuous-
discrete duality but I don't see the discrete part yet in this
construction.
-Tim
.
|
|
|
|
|
|
| User: "" |
|
| Title: Re: Question about the field of complex numbers C |
07 Jun 2007 08:05:33 AM |
|
|
On Jun 7, 1:14 am, "Tomoko Kanazawa" <T.Kanaz...@eodomo.gsfc.nasa.gov>
wrote:
Seems like the Field of complex numbers can be considered the "union" of two
fields.
for any z in C, z = Re(z) + Im(z), seems like Re(z) is a field, after all
it is just R, but Im(z) is also a field in and of itself.
Does that sound bogus ?
The reason I'm interested is because of the silly explorations I've been
conducting into the complex-like randoms expressed as ( a + ~b).
If a is real but ~b is an uncertainty, then ~b is not a number but this does
not mean that it cannot behave like a field.
Instead of Re(z) and Im(z), you can consider NonRan(z) and Ran(z), where
NonRan(z) is the nonrandom part of z and Ran(z) is the random part of z.
So, it would be interesting if NonRan(z) acts like a field, and Ran(z) acts
like a field, and very interesting if the collection of all z = (NonRan(z) +
Ran(z)) also acts like a field.
Some deeper results for C might carry over to "the randoms".
How stupid does that sound ?
Some people did some research some time ago called probabilistic
analysis or something. The idea was to use probability density
functions as objects and combine them to get other pdfs. So
"addition" would be by convolution. You don't get a field this way
but you do get a semigroup.
.
|
|
|
|
| User: "" |
|
| Title: Re: Question about the field of complex numbers C |
07 Jun 2007 05:30:38 AM |
|
|
On Jun 7, 1:14 am, "Tomoko Kanazawa" <T.Kanaz...@eodomo.gsfc.nasa.gov>
wrote:
Seems like the Field of complex numbers can be considered the "union" of two
fields.
for any z in C, z = Re(z) + Im(z), seems like Re(z) is a field, after all
it is just R, but Im(z) is also a field in and of itself.
Does that sound bogus ?
The reason I'm interested is because of the silly explorations I've been
conducting into the complex-like randoms expressed as ( a + ~b).
If a is real but ~b is an uncertainty, then ~b is not a number but this does
not mean that it cannot behave like a field.
Instead of Re(z) and Im(z), you can consider NonRan(z) and Ran(z), where
NonRan(z) is the nonrandom part of z and Ran(z) is the random part of z.
So, it would be interesting if NonRan(z) acts like a field, and Ran(z) acts
like a field, and very interesting if the collection of all z = (NonRan(z) +
Ran(z)) also acts like a field.
Some deeper results for C might carry over to "the randoms".
How stupid does that sound ?
Well, since you asked, it sounds pretty stupid. Here is just a little
information on fields and random numbers.
A field F has the following properties:
1) If a and b are in F then c = a+b = b+a is in F
2) 0 (the additive identity) is in F
3) If a is in F then -a (additive inverse) is in F
4) If a and b are in F then c = a*b = b*a is in F
5) 1 (multiplicative identity) is in F
6) If a is in F and is not 0 then a^(-1) (multiplicative inverse) is
in F
Now, if z is a member of the field of complex numbers one can write:
z = Re(z) + i*Im(z), where i*i = -1
Re(z) and Im(z) are both members of the field of real numbers.
An "uncertainty", as you call it, can be a real. Or it can be an
imaginary, or any other object that you chose. If you want to
consider quantities of the form a + b, where a is certain and b is
uncertain, then you must require that a and b be members of the same
field, otherwise the operation of addition would not make sense.
Random _numbers_ can be members of a field. The fact that they are
random does not affect whether they are members of a field or not.
"Random" simply means that the number is generated by a process in
which one cannot know what the value is until the process produces
that number.
The quantity c = a + b where a is certain and b is uncertain is also
uncertain since you cannot know the value of c until the random
process produces b.
As for NonRand(c) and Rand(c), you could apply the expectation
operator to define NonRand() and Rand(), to wit:
NonRand(c) = E(c) = the expected value, or mean value, of c.
Rand(c) = c - E(c) = a zero-mean random quantity.
- MO
.
|
|
|
| User: "Tomoko Kanazawa" |
|
| Title: Re: Question about the field of complex numbers C |
07 Jun 2007 07:50:02 AM |
|
|
<beeworks@hotmail.com> wrote in message
news:1181212238.383044.321270@q69g2000hsb.googlegroups.com...
On Jun 7, 1:14 am, "Tomoko Kanazawa" <T.Kanaz...@eodomo.gsfc.nasa.gov>
wrote:
Seems like the Field of complex numbers can be considered the "union" of
two
fields.
for any z in C, z = Re(z) + Im(z), seems like Re(z) is a field, after
all
it is just R, but Im(z) is also a field in and of itself.
Does that sound bogus ?
The reason I'm interested is because of the silly explorations I've been
conducting into the complex-like randoms expressed as ( a + ~b).
If a is real but ~b is an uncertainty, then ~b is not a number but this
does
not mean that it cannot behave like a field.
Instead of Re(z) and Im(z), you can consider NonRan(z) and Ran(z), where
NonRan(z) is the nonrandom part of z and Ran(z) is the random part of z.
So, it would be interesting if NonRan(z) acts like a field, and Ran(z)
acts
like a field, and very interesting if the collection of all z =
(NonRan(z) +
Ran(z)) also acts like a field.
Some deeper results for C might carry over to "the randoms".
How stupid does that sound ?
Well, since you asked, it sounds pretty stupid. Here is just a little
information on fields and random numbers.
A field F has the following properties:
1) If a and b are in F then c = a+b = b+a is in F
2) 0 (the additive identity) is in F
3) If a is in F then -a (additive inverse) is in F
4) If a and b are in F then c = a*b = b*a is in F
5) 1 (multiplicative identity) is in F
6) If a is in F and is not 0 then a^(-1) (multiplicative inverse) is
in F
Now, if z is a member of the field of complex numbers one can write:
z = Re(z) + i*Im(z), where i*i = -1
Re(z) and Im(z) are both members of the field of real numbers.
An "uncertainty", as you call it, can be a real. Or it can be an
imaginary, or any other object that you chose. If you want to
consider quantities of the form a + b, where a is certain and b is
uncertain, then you must require that a and b be members of the same
field, otherwise the operation of addition would not make sense.
Random _numbers_ can be members of a field. The fact that they are
random does not affect whether they are members of a field or not.
"Random" simply means that the number is generated by a process in
which one cannot know what the value is until the process produces
that number.
The quantity c = a + b where a is certain and b is uncertain is also
uncertain since you cannot know the value of c until the random
process produces b.
As for NonRand(c) and Rand(c), you could apply the expectation
operator to define NonRand() and Rand(), to wit:
NonRand(c) = E(c) = the expected value, or mean value, of c.
Rand(c) = c - E(c) = a zero-mean random quantity.
- MO
Thanks for that. I'm trying to do something which may or may not make sense.
Trying to treat an uncertainty as a mathematical object, distinct from the
notion of random number. It seems that an uncertainty cannot be a number,
but may behave like one in ways.
Also, thanks Jose, you're tight. I dont think that i*Im(z) is a field
because i^(2n) is not imaginary, and multiplicative inverse would probably
be 1 which is not imaginary,......
.......you're right, it's bogus.
Unless you allow i*i*i*i to be the multiplicative inverse, and ignore the
fact that i*i*i*i = 1, which you cant, unless you drop some properties like
associativity or something, but then you dont have a field for that reason.
I'm convinced, it's a bogus idea, in this case.
In fact, my whole approach to randomness might very well be bogus, but I
dont know why it would.
But I'm still curious, lets say that you have 2 Reals. Upper case Reals and
Lower case Reals. So, for every upper case number, you have a lower case
number of equal magnitude. Then, the Uppers and the Lowers are both fields,
and is their union also a field ?
But maybe not because you might be sacrificing uniqueness of identity
element or something like that.
Ahh well, it is bogus bogus bogus, life is so unfair - : )
.
|
|
|
|
|
|
Related Articles |
|
|
