Vectors summing to zero
TXB
These vectors are typically unit vectors of the dimensional space or the
difference of two unit vectors. I want to find all the three-vector
combinations that sum to zero. Currently I do this by selecting every
possible 3 vector combination to see if a non-trivial linear combination
exists that sums to zero. Specifically, I select a vector (v1) and perform
a least-squares solution to find a and b so that a*v2 + b*v3 = v1. If the
sum is zero (and a, b = +1 or -1), I record the grouping of (v1, v2, and
v3) and continue. This is working, however, I feel like there is a simpler
methodology that I am missing. Eventually I use this information to create
an adjacency matrix for these vectors and display it as a graph. I know
very little graph theory (I am an engineer, not a mathematician) so I am
not sure if I could use any graph theory to help.
Thanks.
txb
Gerry Myerson
TXB <txb...@gmail.com> wrote:
> I have a collection of vectors that span some n-dimensional space.
I take it this is a *finite* collection of vectors.
> These vectors are typically unit vectors of the dimensional space or the
> difference of two unit vectors. I want to find all the three-vector
> combinations that sum to zero. Currently I do this by selecting every
> possible 3 vector combination to see if a non-trivial linear combination
> exists that sums to zero. Specifically, I select a vector (v1) and perform
> a least-squares solution to find a and b so that a*v2 + b*v3 = v1.
I suspect it would take less work if instead you form a matrix
whose columns are the three vectors and then use row-reduction
(a.k.a. Gaussian elimination) to see whether you get a matrix of rank 2.
If you do, then there's a linear dependence relation for the three
vectors, and it's easy to work out what it is from the reduced matrix
and decide if it's three vectors adding to zero.
> If the
> sum is zero (and a, b = +1 or -1), I record the grouping of (v1, v2, and
> v3) and continue. This is working, however, I feel like there is a simpler
> methodology that I am missing.
Maybe this is better than both least squares and row reduction:
Form all the sums of 2 vectors and then look to see whether the list
of sums has any member in common with the list of negatives of
the original vectors. This may run faster if you sort the list of sums,
say, by value of 1st component, breaking any ties by sorting on
the value of the 2nd component, etc.
> Eventually I use this information to create
> an adjacency matrix for these vectors and display it as a graph.
I don't know what you mean by "an adjacency matrix for these vectors."
--
Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)
Tim Golden BandTech.com
The vectors will always be coplanar. This means that the cross product
of each combination will be colinear; the same value when normalized.
U( v1 x v2 ) = U( v1 x v3 ) .
Adjacency is good stuff. That is an interesting criterion.
- Tim
Gerry Myerson
<4945f2c3-43da-4fbf...@s9g2000yqd.googlegroups.com>,
And how, pray tell, do you propose to define/compute the cross product
in n-dimensional space, when n > 3?
Tim Golden BandTech.com
wrote:
> In article
> <4945f2c3-43da-4fbf-8748-2edc6c81c...@s9g2000yqd.googlegroups.com>,
> "Tim Golden BandTech.com" <tttppp...@yahoo.com> wrote:
>
>
>
> > On Aug 31, 1:25 pm, TXB <txb1...@gmail.com> wrote:
> > > I have a collection of vectors that span some n-dimensional space.
> > > These vectors are typically unit vectors of the dimensional space or the
> > > difference of two unit vectors. I want to find all the three-vector
> > > combinations that sum to zero. Currently I do this by selecting every
> > > possible 3 vector combination to see if a non-trivial linear combination
> > > exists that sums to zero. Specifically, I select a vector (v1) and perform
> > > a least-squares solution to find a and b so that a*v2 + b*v3 = v1. If the
> > > sum is zero (and a, b = +1 or -1), I record the grouping of (v1, v2, and
> > > v3) and continue. This is working, however, I feel like there is a simpler
> > > methodology that I am missing. Eventually I use this information to create
> > > an adjacency matrix for these vectors and display it as a graph. I know
> > > very little graph theory (I am an engineer, not a mathematician) so I am
> > > not sure if I could use any graph theory to help.
>
> > > Thanks.
>
> > > txb
>
> > The vectors will always be coplanar. This means that the cross product
> > of each combination will be colinear; the same value when normalized.
>
> > U( v1 x v2 ) = U( v1 x v3 ) .
>
> And how, pray tell, do you propose to define/compute the cross product
> in n-dimensional space, when n > 3?
Yeah, I overlooked that point. Still, the coplanar comment can stand.
Is it true that the condition is merely
v1 + v2 + v3 = 0 ?
Or is TXB looking for scaled versions as well? e.g. is
(U( 5, 1, 0, 0 )), ( 1, 0, 0, 0 ), ( 0, 1, 0, 0 )
where U is a unit vector scalar conversion in the criterion?
If the dimension is large then the chance of hitting a coplanar triple
is slight. There may be a dot product solution lurking here that is so
simple. But then, the sum itself seems so simple that I don't see much
puzzle anyway. It's a need for speed, eh? Use partial data seems a
good method of elimination, but depends on the qualities of the data.
The sum of the angles will always be 2 pi and so by computing these
angles another process of elimination can be had. This is simple data
without the dimensional complexities. Integerizing that angle could
get pretty fast results as type char or short since the resulting
angle can be coded to overflow or some such hardware condition. Also
no actual trig operation is necessary since a coarse LUT would be
adequate. All other angular sums will be smaller, but the condition
queried above will also be included within this criterion. Depending
on how many candidates are eliminated this method could be fast if the
list for final precision verification is short.
- Tim
Frederick Williams
> And how, pray tell, do you propose to define/compute the cross product
> in n-dimensional space, when n > 3?
It can be done when n = 7, and that, along with n = 1, is your lot.
--
Needle, nardle, noo.
Frederick Williams
>
> Gerry Myerson wrote:
>
> > And how, pray tell, do you propose to define/compute the cross product
> > in n-dimensional space, when n > 3?
>
> It can be done when n = 7, and that, along with n = 1, is your lot.
That's confused, isn't it? Either I should just have mentioned n = 7
since you wrote n > 3, or I should have included 3 and 1.
So: cross products may be defined in n-dimensional space for n = 1, 3,
7, but not for other n. I'm sure Gerry Myerson knows that, I'm just
saying so for my own benefit. :-)
--
Needle, nardle, noo.
Gerry Myerson
<13059615-753f-4932...@z28g2000yqh.googlegroups.com>,
"Tim Golden BandTech.com" <tttp...@yahoo.com> wrote:
You know, if you actually knew some mathematics, you'd be dangerous.
spudnik
http://homepages.wmich.edu/~drichter/octonions.htm
> So: cross products may be defined in n-dimensional space
> for n = 1, 3, 7, but not for other n.
thus:
yeah, no-one knows, you're not just pretending to be a dog,
on the net ... but
that wasn't a vote for Obama, so to say, per se;
it was a vote for you to consider how to approach a real math-guy,
and they seem to be, to me, in conversation (Mazur, last year;
Ribet, maybe over 10ya, BW [*]; both at UCLA
as distinguished lectureres or some thing). * Before Wiles,
came out of his Secret Attic Project!
on the other hand, considering that he was the private lawyer
for CCX, while a Senator, he could actually know some thing
about derivatives, and hence real calculus (because,
the derivatives on CO2-credits, and ordinary derivatives,
are actually "second derivatives," considering stocks or bonds
to be first derivatives .-)
and, CCX is really, really big, and possibly a major source
of political bucks; go figure!
thus:
it was a non-tenured position; still, I know,
it was a long article, so, you mayn't have noticed a few
of the nuances. (I have met Mary. I told her that,
both times that Gore's movie was shown on campus,
*no-one showed up* but me, possibly because
they already are total believers.)
because of the budget-cuts, it is my experience that
people in anything remotely climate-related are on tenterhooks
-- what ever that means. although, I have to say,
they ended the funding of Public Citizen via registration fees,
some time ago.
oh, and, on the wayside,
diesel gets much better milage that gasoline, though, and
their is quite a long-on-going "green diesel" programme ... also,
there is essentially no difference between "biodiesel" and
ordinary diesel, excepting the bouquet. so, there.
> Enstrom probably has more cash on hand than a Mexican drug lord.
thus: I have met Ribet; also Barry Mazur, and
both of them have results toward Wiles' alleged proof
(and, I rather believe them, that it works .-)
thus: Occam was a part of the oligarchical "science establishment,"
like Galileo and Newton, who issued a commonplace sentiment.
Arctic ice is simply & utterly fragile, evanescent & unstable,
to any changes in the differentials of weather/climate. anyway,
I have yet to see one of you, seriously
to consider Singer's very old metastudy on glaciers ... so,
there's no where, hereinat; thou refusest.
thus: oh, very well said. however, I'll wait for the report
from Netherlands U., about the problems.
the whole world is going a)
nuclear electric and b)
into space, and
we are still stuck in '50s and '60s techniques (like
the "new" reactors from GE and Toshiba/Westinghouse); I mean,
it might as well be what Elron Hubbard said
about USA 'Fifties "culture, implanted from the planet Marduk."
(of course, solar is gangbusters *in* space .-)
> the best way to think of wind energy is as a form of agriculture.
thus: upon skimming, it seemed to contain only one sentence
about Singer; would you like me to give the highlights
of his CV, or would you prefer to look it "up?"
I recall a *Popular Science* ([L.A.] Times Mirror Corp.) cover-
story,
where three mainstream "holers" were given beautiful portraits,
and Singer was given a mugshot, no CV.
I just missed the authors of the prohominem that you cited,
at the local library to sell it, perhaps sponsored by the city;
everything Greenpolitical in L.A., starts in SM,
largely because of the WAND Corp., I know it.
> http://www.americanscientist.org/bookshelf/pub/manufactured-ignorance
thus: I stopped at the part about what Heisenberg "says"
about the electron. it's just about the impossiblity
of two kinds of measurements, taken at the same time, and
it certainly applies to many "macro" situations (like,
Lord Bertie Was Lying, Then .-)
Schroedinger made an excellent joke about the Copenhagenskoolers,
and that cat stinks to high-Heaven, by now!
thus: size of sail is not really limited;
tack at a right angle to the wind.
> Yes and directly into the wind.
thus: ah, so; never done a proof by contradiction??
thus: wow, that is exemplary, dood. (of course,
it's in the M&M paper, that they most certainly did
n o t find "no results," however slight .-)
> You cannot have an aether that is
> at rest with any one body. If the earth
> were at rest with the aether at any
> point, then every other galaxy that is
> accelerating away from earth would
> experience an aether flow towards earth,
> and *their* light would be affected....
> Therefore the aether must be
> moving in all directions wrt ALL bodies equally....
> M and M simply disproved something
> that could not be possible in the first place.
thus: Bjorn's change-of-heart could've been predicted, since
1st espoused his views in a Holy Economist guest editorial
-- the only thing that is ever signed by an author in it. so,
naturally, he is a proponent of bpTM's old KyotoTM cap&tradeTM,
and my Rep. Waxman's and my California Assemblywoman's
(now Senator) cap&trade derivatives, a.k.a. "free-er trade
on the free market -- free beer &or freedom!"
and, of course, one of Bjorn's telltales is that
cold generally leads to more deaths than heat,
per annum.
thus: he stole that from Hooke, then burnt all
of his portraits -- "ahahaha,
on the shoulders of that clever little dwarf!" (viz,
Sir I., the plagiarist, Freemason, alchemist-
who-burnt-his-"Principles"-in-an-accident-and-
had-it-"reconstructed"-by-the-RS-with-the-dydx-rectangle
etc. ad vomitorium .-)
thus: Euclid's proof is so simple, that
it takes a truly linguistically challenged individual
to dyss it; after all, all
of math problems are, really, wordproblemmas!
thus: as for ordinary spatial finite elements,
you really need tetrahedronometry; eh?
--les ducs d'Enron!
http://tarpley.net
--Light, A History!
http://wlym.com/~animations/fermat/index.html
achille
How about a dot product based strategy like this:
Let's say you have N n-dim vectors v_1 .. v_N. First compute
the N(N+1)/2 inner products D_ij = v_i . v_j and store them
somewhere, then loop over all possible combinations of i<j<k:
If you want to test whether v_1 + v_2 + v_3 = 0
as vectors, you check:
D_11 + 2*D_12 + 2*D_13
+ D_22 + 2*D_23
+ D_33 = 0
If you want to test whether v_1, v_2, v_3 are coplanar,
you check:
det( D_ij ) = 0 ( 1 <= i,j <= 3 )
With a little bit more effort, you can implement
collinear test using D_ij in a similar manner.
In practice, due to round off error, above tests will
not be met exactly. However, you can use it as a fast
filter (once you have D_ij) before carrying out a more
accurate test on the vectors directly.
Tim Golden BandTech.com
wrote:
> In article
> <13059615-753f-4932-9a69-729a4047a...@z28g2000yqh.googlegroups.com>,
Well, Gerry, you should check out polysign numbers.
http://bandtechnology.com/PolySigned
That is how I come to have these strong opinions. What about
Frederick's claim of a 7D cross product? I have no idea what he is
talking about but it sure sounds interesting.
It is interesting that angle in high dimension is still well behaved.
Anytime you have an angle of pi you do have an inverse of your
original vector, regardless of the dimensional qualities of the
vector.
As I go over it again I will stand by the 2 pi claim. If you have a
falsification then please present it. I don't necessarily believe all
of mainstream mathematics so I find it acceptable that you do not
believe mine. Simply stated: the sum of the angles between three
coplanar vectors will always be 2 pi, in any dimension. The converse
holds: If the sum of the angles of three vectors is 2 pi, then the
vectors are coplanar. This is not a proof, but is an observation. I'm
open to falsification, but doubt that you will find one.
- Tim
Gerry
>
> How about a dot product based strategy like this:
>
> Let's say you have N n-dim vectors v_1 .. v_N. First compute
> the N(N+1)/2 inner products D_ij = v_i . v_j and store them
> somewhere, then loop over all possible combinations of i<j<k:
>
> If you want to test whether v_1 + v_2 + v_3 = 0
> as vectors, you check:
>
> D_11 + 2*D_12 + 2*D_13
> + D_22 + 2*D_23
> + D_33 = 0
>
> If you want to test whether v_1, v_2, v_3 are coplanar,
> you check:
>
> det( D_ij ) = 0 ( 1 <= i,j <= 3 )
>
> With a little bit more effort, you can implement
> collinear test using D_ij in a similar manner.
>
> In practice, due to round off error, above tests will
> not be met exactly. However, you can use it as a fast
> filter (once you have D_ij) before carrying out a more
> accurate test on the vectors directly.
You're looping over i, j, k, so you have a O(N^3) algorithm.
My idea of computing all v_i + v_j, sorting them, then
searching for -v_k in the list, is O(N^2 log N).
--
GM
Gerry
wrote:
> Well, Gerry, you should check out polysign numbers.
> http://bandtechnology.com/PolySigned
No, thanks, I've seen more than enough of your polysign numbers
in previous discussions here.
> That is how I come to have these strong opinions. What about
> Frederick's claim of a 7D cross product? I have no idea what he is
> talking about but it sure sounds interesting.
It is. Why don't you look into it, and report back to us?
> It is interesting that angle in high dimension is still well behaved.
> Anytime you have an angle of pi you do have an inverse of your
> original vector, regardless of the dimensional qualities of the
> vector.
>
> As I go over it again I will stand by the 2 pi claim. If you have a
> falsification then please present it. I don't necessarily believe all
> of mainstream mathematics so I find it acceptable that you do not
> believe mine. Simply stated: the sum of the angles between three
> coplanar vectors will always be 2 pi, in any dimension. The converse
> holds: If the sum of the angles of three vectors is 2 pi, then the
> vectors are coplanar. This is not a proof, but is an observation. I'm
> open to falsification, but doubt that you will find one.
I won't find one because it's trivially true. But it has very little
to do with the original problem. OP wants v_1 + v_2 + v_3 = 0. All
coplanarity will get you is a_1 v_1 + a_2 v_2 + a_3 v_3 = 0 for some
real coefficients a_1, a_2, a_3. That's very far from what OP wants.
--
GM
achille
In terms of computational complexity, your algorithm is better.
But the 'coefficient' in front of my O(N^3) part should be smaller.
The memory requirement is also less demanding: O(N^2) vs O(N^2 n).
One should use a different algorithm on different situations.
Frederick Williams
>
> What about
> Frederick's claim of a 7D cross product? I have no idea what he is
> talking about but it sure sounds interesting.
Look up Hurwitz's theorem. There is an especially readable account in
Ebbinghaus, et al, Numbers, Springer.
--
Needle, nardle, noo.
hagman
> I have a collection of vectors that span some n-dimensional space.
> These vectors are typically unit vectors of the dimensional space or the
> difference of two unit vectors.
Does that mean, given a standard basis e_1, e_2, ..., e_n,
your vectors are all of the form
v = e_i for some i
or
v = e_i - e_j for some i != j?
I assume that v = -e_i is also allowed as otherwise no vector of the
first
kind could be included in a sum v_1 + v_2 + v_3 = 0:
Just consider the scalar product of v_1+v_2+v_3 and e_1 + ... + e_n.
> I want to find all the three-vector
> combinations that sum to zero.
In the above setting, v_1 + v_2 + v_3 = 0 is only possible if
a)
{v_1, v_2, v_3} = {e_i - e_j, e_j - e_k, e_k - e_i}
for some different indices i,j,k
or
b)
{v_1, v_2, v_3} = {e_i - e_j, e_j, -e_i}
for some different indices i,j
If you really mean general unit vectors instead of base vectors,
I don't dee much that can be improved over the general setting
except that v_1 + v_2 must not be longer than sqrt(2) and the like
Ki Song
wrote:
>
> As I go over it again I will stand by the 2 pi claim. If you have a
> falsification then please present it. I don't necessarily believe all
> of mainstream mathematics so I find it acceptable that you do not
> believe mine. Simply stated: the sum of the angles between three
> coplanar vectors will always be 2 pi, in any dimension. The converse
> holds: If the sum of the angles of three vectors is 2 pi, then the
> vectors are coplanar. This is not a proof, but is an observation. I'm
> open to falsification, but doubt that you will find one.
>
> - Tim
I don't really understand what you mean by "sum of the angles between
three coplanar vectors will always be 2 pi." I'm assuming you meant
dimension greater than or equal to 2 when you said "in any dimension."
This statement is trivially false if you just meant "three vectors
that that lie in a 2 dimensional subspace."(Take any vector X. Then,
take 2X and 3X.) Even if you meant "three vectors that span a 2-
dimensional subspace," it is trivially false. (Take two linearly
independent vectors, X,Y. Choose X+Y as the third vector. Angle
between X&Y, X&(X+Y), and Y&(X+Y) will not add up to 2Pi.)
So obviously, you mean something else. Exactly what do YOU mean by
three co-planar vectors?
The converse of the statement is also trivially false. Look at a
tetrahedron constructed from 4 equilateral triangles. Imagine that
one of the vertices is the origin, and imagine that the three edges
are vectors in R^3. What are the angles between the vectors? What is
the sum of the angles of these vectors? Are these vectors "coplanar"?
I agree with Gerry that perhaps it would be wise for you to learn a
bit of conventional mathematics before rejecting it. If you know some
algebra, you'd know that your polysign numbers is just a example of a
quotient ring.
Gerry Myerson
<19b1efbb-3181-47ee...@x20g2000pro.googlegroups.com>,
achille <achil...@yahoo.com.hk> wrote:
> On Sep 2, 8:43 pm, Gerry <ge...@math.mq.edu.au> wrote:
> > On Sep 2, 5:16 pm, achille <achille ...@yahoo.com.hk> wrote:
> >
> >
> >
> >
> >
> > > How about a dot product based strategy like this:
> >
> > > Let's say you have N n-dim vectors v 1 .. v N. First compute
> > > the N(N+1)/2 inner products D ij = v i . v j and store them
> > > somewhere, then loop over all possible combinations of i<j<k:
> >
> > > If you want to test whether v 1 + v 2 + v 3 = 0
> > > as vectors, you check:
> >
> > > D 11 + 2*D 12 + 2*D 13
> > > + D 22 + 2*D 23
> > > + D 33 = 0
> >
> > > If you want to test whether v 1, v 2, v 3 are coplanar,
> > > you check:
> >
> > > det( D ij ) = 0 ( 1 <= i,j <= 3 )
> >
> > > With a little bit more effort, you can implement
> > > collinear test using D ij in a similar manner.
> >
> > > In practice, due to round off error, above tests will
> > > not be met exactly. However, you can use it as a fast
> > > filter (once you have D ij) before carrying out a more
> > > accurate test on the vectors directly.
> >
> > You're looping over i, j, k, so you have a O(N^3) algorithm.
> > My idea of computing all v i + v j, sorting them, then
> > searching for -v k in the list, is O(N^2 log N).
>
> In terms of computational complexity, your algorithm is better.
> But the 'coefficient' in front of my O(N^3) part should be smaller.
You are calculating O(N^2) dot products
and O(N^3) 3-by-3 determinants,
whereas I'm calculating O(N^2) sums
and sorting a list of length O(N^2).
I'm not convinced that your coefficient is smaller.
> The memory requirement is also less demanding: O(N^2) vs O(N^2 n).
> One should use a different algorithm on different situations.
--
Gerry Myerson
<c8fb106b-0f83-48ca...@y11g2000yqm.googlegroups.com>,
Ki Song <kiwis...@gmail.com> wrote:
> On Sep 2, 7:09 am, "Tim Golden BandTech.com" <tttppp...@yahoo.com>
> wrote:
> >
> > As I go over it again I will stand by the 2 pi claim. If you have a
> > falsification then please present it. I don't necessarily believe all
> > of mainstream mathematics so I find it acceptable that you do not
> > believe mine. Simply stated: the sum of the angles between three
> > coplanar vectors will always be 2 pi, in any dimension. The converse
> > holds: If the sum of the angles of three vectors is 2 pi, then the
> > vectors are coplanar. This is not a proof, but is an observation. I'm
> > open to falsification, but doubt that you will find one.
>
> I don't really understand what you mean by "sum of the angles between
> three coplanar vectors will always be 2 pi." I'm assuming you meant
> dimension greater than or equal to 2 when you said "in any dimension."
> This statement is trivially false if you just meant "three vectors
> that that lie in a 2 dimensional subspace."(Take any vector X. Then,
> take 2X and 3X.) Even if you meant "three vectors that span a 2-
> dimensional subspace," it is trivially false. (Take two linearly
> independent vectors, X,Y. Choose X+Y as the third vector. Angle
> between X&Y, X&(X+Y), and Y&(X+Y) will not add up to 2Pi.)
I guess what's needed is that there's no half=space containing all
three vectors. Equivalently, the origin is interior to the convex hull
of the three vectors.
> So obviously, you mean something else. Exactly what do YOU mean by
> three co-planar vectors?
>
> The converse of the statement is also trivially false. Look at a
> tetrahedron constructed from 4 equilateral triangles. Imagine that
> one of the vertices is the origin, and imagine that the three edges
> are vectors in R^3. What are the angles between the vectors? What is
> the sum of the angles of these vectors? Are these vectors "coplanar"?
Call the edges/vectors u, v, and w. The angle theta between two vectors
is defined by u dot v = (length of u)(length of v) cos theta. In this
case, each of the three angles is pi/3, the sum is pi, the vectors are
not coplanar.
Tim Golden BandTech.com
> On Sep 2, 7:09 am, "Tim Golden BandTech.com" <tttppp...@yahoo.com>
> wrote:
>
>
>
> > As I go over it again I will stand by the 2 pi claim. If you have a
> > falsification then please present it. I don't necessarily believe all
> > of mainstream mathematics so I find it acceptable that you do not
> > believe mine. Simply stated: the sum of the angles between three
> > coplanar vectors will always be 2 pi, in any dimension. The converse
> > holds: If the sum of the angles of three vectors is 2 pi, then the
> > vectors are coplanar. This is not a proof, but is an observation. I'm
> > open to falsification, but doubt that you will find one.
>
> > - Tim
>
> I don't really understand what you mean by "sum of the angles between
> three coplanar vectors will always be 2 pi." I'm assuming you meant
> dimension greater than or equal to 2 when you said "in any dimension."
Actually I thing this works on the real line as well, where the angle
between -1 and +1 is pi. Alright, maybe it doesn't work on 0D space.
But I'd have to think about it.
> This statement is trivially false if you just meant "three vectors
> that that lie in a 2 dimensional subspace."(Take any vector X. Then,
> take 2X and 3X.) Even if you meant "three vectors that span a 2-
> dimensional subspace," it is trivially false. (Take two linearly
> independent vectors, X,Y. Choose X+Y as the third vector. Angle
> between X&Y, X&(X+Y), and Y&(X+Y) will not add up to 2Pi.)
Yes, you are right. I made it overly general there. This is a fine
counterexample.
>
> So obviously, you mean something else. Exactly what do YOU mean by
> three co-planar vectors?
Well, it is the three vectors that sum to zero, but also arbitrary
scalars of these vectors. These are a subset of the coplanar vectors
and I failed to consider the vectors you chose, though there is the
interpretation even on those that
theta = 2 pi - theta
which could get me out of the failure, but I would not grovel along
that route. I like taking the smallest angle and accept your
criticism.
>
> The converse of the statement is also trivially false. Look at a
> tetrahedron constructed from 4 equilateral triangles. Imagine that
> one of the vertices is the origin, and imagine that the three edges
> are vectors in R^3.
> What are the angles between the vectors?
60 degrees
> What is the sum of the angles of these vectors?
180 degrees ( pi radians )
> Are these vectors "coplanar"?
no, and we don't have 2 pi radians, so this is not a counterexample.
Still, I am open to you presenting one.
>
> I agree with Gerry that perhaps it would be wise for you to learn a
> bit of conventional mathematics before rejecting it. If you know some
> algebra, you'd know that your polysign numbers is just a example of a
> quotient ring.
Ah yes, the quotient ring. Such a sensible construction...
And when I ask you whether the '+' in
a + b i
is a ring operator, where a and b are real, what will you say? My
opinion is that all of abstract algebra is a bastard child. There is
nothing wrong with bastard children, but when you tell them this that
was hidden from them then there is sure to be a rift. As
mathematicians we should not accept a subject because others have
accepted the subject. This procedure of mimicry ends the validity of
mathematics, and is consistent with the modern academic system.
I admit there is no better way to procede, but also that your closed
mind is not my problem. I further admit that my accusation of the
bastard child could be false, but then we get to proving it and here
we cannot just rely upon mimicing a text, no matter how many variants
exist. That I would call the biblical approach and do see that the
human has a tendency towards veneration that is not at all
mathematical. Because the ones that came before did not know what the
ones that come after know, then their work may be falsifiable. As I
err on this thread so too can anyone err. We are caught as humans. All
of us. This is a great place to make mistakes. Let's see who can own
there mistakes. You Ki have made an accusation, or two, that I believe
do deserve some followup. As to whether you would be capable of
admitting an individual error that is a group error, well, this is a
deep statement on human behavior. You all are a bunch of bastards.
Hah. I hope you get a healthy chuckle here.
- Tim
Tim Golden BandTech.com
wrote:
How about a direct link to something about a 7D cross product?
I've begun to search, but
http://en.wikipedia.org/wiki/Hurwitz's_theorem
shows lots of interesting stuff, but no bull's eye.
Indirection is not necessary. This medium is a fine detector through
its lack of censorship. Your own coherency is at stake, but then, so
is mine.
- Tim
Frederick Williams
> How about a direct link to something about a 7D cross product?
Googling "seven-dimensional cross product" immediately turns up
http://en.wikipedia.org/wiki/Seven-dimensional_cross_product, wherein
our own (if I may call him that) late, and oftentimes rather crotchety,
Pertti Lounesto gets a mention.
> ... Your own coherency is at stake, but then, so
> is mine.
Oh dear, what shall we do?
--
Needle, nardle, noo.
Tim Golden BandTech.com
wrote:
> "Tim Golden BandTech.com" wrote:
> > How about a direct link to something about a 7D cross product?
>
> our own (if I may call him that) late, and oftentimes rather crotchety,
> Pertti Lounesto gets a mention.
Well, crotchety is OK with me, though there should be some content to
go along with it.
e(i) X e(i+1) = e(i+3)
is a bizzarre way of putting it, and makes me wonder what on earth is
this thing doing in electromagnetics? Shouldn't we find that
electromagnetic phenomena are a part of a more universal principle?
Wouldn't this be cause for a general dimensional solution? Yet it does
not come along this way. Somehow the lack of generality of the cross
product makes it a sore point for physics. I suppose it is better to
be impressed than depressed, though the buttons on my keyboard don't
seem to give a damn, and they are operating under electromagnetic
principles regularly, even when repressed. This trilogy must not be
offended and I fear that another one is coming on, but if in the
expansion the multiplicity rises to arbitrary qualities, well, it's
really not for me to say here. Having staked my incoherency down I
submit to your 7D cross product, all the while wondering what happened
to the three? It is still there, but it is pressed so far as to have
lost its original detent within the cryptic nature of its meanderings.
To wander so far can hardly be considered an arrival, yet it is a
fascinating place.
Thanks Frederick for maintaining your own coherency, while I have
sacrificed some of mine, preferably for your entertainment.
- Tim
Tim Little
> Yet it does not come along this way. Somehow the lack of generality
> of the cross product makes it a sore point for physics.
Not really. Electromagnetism can be easily generalised by unifying
the electric and magnetic vectors into a single antisymmetric rank-2
tensor.
- Tim
Tim Golden BandTech.com
wrote:
> On Sep 2, 6:12 pm, Ki Song <kiwisqu...@gmail.com> wrote:
Kiiii...
Kii Soooonnggg...
Where arrre yoouu?