Bug 48 et 49
gibbons
'1/15-2/30', la division renvoie 0, ce qui est évidemment faux, tant sur la
48 que sur la 49.
Homer Simpson
'1/15-2/30'), the division returns 0, which is of course wrong, on the 48 as
well as on the 49.
Maybe you should go back to school a little. Zero divided by non-null-number
is indeed zero.
It doesn't only work on the HPs, also on the TIs, Casios, Intels, whatever!
Homer Simpson
http://zap.to/hsimpson
lar...@ns.ulatina.ac.cr
Well, it's nice.....
But til I know 0/K, is always 0. For K not iqual to 0.
Also, in the 49, when working in exact mode, If you make:
0 [SPC] 0 [/] you get +infinite
And in aprox mode you get an error that says:
/ Error: Infinite result.
I mean, in exact mode you must get the same error, because infinit as a
symbol represents a specific number.
(1/2)
/ 3
| 2 3 x 3.14 (1/2)
| z dz cos (--------) = ln(e )
/ 1 9
Sent via Deja.com http://www.deja.com/
Before you buy.
Keith Farmer
Homer Simpson wrote:
> Maybe you should go back to school a little. Zero divided by non-null-number
> is indeed zero.
You may have missed the fact that he is dividing by a null constant --
0/(1/15-2/30) = 0/0, which I've seen addressed only as a limit (which I
figure in my head to be 0, in this case anyway).
Steen S. Schmidt
>0/(1/15-2/30) = 0/0, which I've seen addressed only as a limit (which I
>figure in my head to be 0, in this case anyway).
Well.....actually by applying l'Hospital on this gives the limít as
undefined. Differentiating denominator and numerator leaves '0/0', which is
undefined.
Regards
Steen
Keith Farmer
> And graphing 0/x, x->0 brings oppositely-signed asymptotes. I was
> thinking abs(0/x)
"Graphing 0/x" .. you know what I mean! :P
Keith Farmer
Homer Simpson
>0/(1/15-2/30) = 0/0, which I've seen addressed only as a limit (which I
>figure in my head to be 0, in this case anyway).
Oh... Then it will be hard to say it's a normal behavior. My apologies for
having replied too quickly.
I wonder whether there is a good explanation for this... If there is no
explanation, it's the second ten-years old bug we discover in a week (with
the monadic plus).
Homer Simpson
http://zap.to/hsimpson
Joėl Bourquard
Jean-Yves Avenard <aven...@epita.fr> wrote in message
news:385f9...@isoit370.bbn.hp.com...
> A ma connaissance
> 0/expression renvoie toujours zero...
>
> C'est exactement ce qu'il se passe sur la 49
>
> Jean-Yves
Oui, sauf si l'expression est nulle.. Dans ce cas, manque de bol : 0 / 0
!= 0 ...
Joėl
Jean-Yves Avenard
0/expression renvoie toujours zero...
C'est exactement ce qu'il se passe sur la 49
Jean-Yves
"gibbons" <gib...@infonie.fr> wrote in message
news:945754932.435907@news2...
Steen S. Schmidt
>symbol represents a specific number.
Infinite isn't a number as such.
Regards
Steen
Nate Eldredge
> (1/2)
> / 3
> | 2 3 x 3.14 (1/2)
> | z dz cos (--------) = ln(e )
> / 1 9
:)
I've seen this limerick too. But the math looks wrong. According to
the HP49 (just to keep this on topic), the left side is
'(-1+3*sqrt(3))/6', while the right side is '1/2'.
For extra credit: make one that is correct and still has the correct
rhyme/meter. Other forms of poetry might be interesting as well
(haiku, or for the really ambitious, a sonnet).
--
Nate Eldredge
neld...@hmc.edu
Richard Helps
expression in English ("Integrate", "Integral", "Take the integral of"
etc.) Please give the English language version of this expression. To
keep this on-toipic, I will then enter it into my HP49 text editor ;-b
Richard Helps
Nate Eldredge
> I'm sorry I don't see the limerick. There are too many ways to say this
> expression in English ("Integrate", "Integral", "Take the integral of"
> etc.) Please give the English language version of this expression. To
> keep this on-toipic, I will then enter it into my HP49 text editor ;-b
The version I heard went something like:
Integral z-squared d z
From one to the square root of three
Times the cosine
Of three pi over nine
Equals lawn of the square root of e.
(Credit: A book of science humor entitled _Absolute Zero Gravity_, by
Joel Cohen et al).
> Richard Helps
>
> Nate Eldredge wrote:
>
> > lar...@ns.ulatina.ac.cr writes:
> >
> > > (1/2)
> > > / 3
> > > | 2 3 x 3.14 (1/2)
> > > | z dz cos (--------) = ln(e )
> > > / 1 9
> >
> > :)
> >
> > I've seen this limerick too. But the math looks wrong. According to
> > the HP49 (just to keep this on topic), the left side is
> > '(-1+3*sqrt(3))/6', while the right side is '1/2'.
> >
> > For extra credit: make one that is correct and still has the correct
> > rhyme/meter. Other forms of poetry might be interesting as well
> > (haiku, or for the really ambitious, a sonnet).
>
--
Nate Eldredge
neld...@hmc.edu
Sophus Lefouque (P)
>
> >You may have missed the fact that he is dividing by a null constant --
> >0/(1/15-2/30) = 0/0, which I've seen addressed only as a limit (which I
> >figure in my head to be 0, in this case anyway).
>
> undefined. Differentiating denominator and numerator leaves '0/0', which is
> undefined.
l'Hôpital's rule is irrelevant here, as we are not taking any
limits at all. 0/0 is undefined, period.
BTW: '0/0 form' often mentioned in l'Hôpital's rule does not mean
the number zero divided by the number zero, it represents a
quotient of functions f(x)/g(x) in which both f(x) and g(x) tend
to zero.
Sophus Lefouque (P)
...
> I wonder whether there is a good explanation for this... If there is no
> explanation, it's the second ten-years old bug we discover in a week (with
> the monadic plus).
plain oversight, possibly ignorance. perhaps the development team
needs a Maths consultant in addition to its team of engineers.
Let F be a field, which includes the case F=R (real numbers)
and F=C (complex numbers). Every non-trivial field contains
at least the two elements 0 and 1. The fields R and C
contain infinitely many elements, in fact.
-------------------------------------------------------
0/0 is undefined.
-------------------------------------------------------
Why? (reducio ad absurdum):-
Suppose 0/0 = k, where k is in F.
Then, 0 = k·0
However all members k in F satisfy this
equation and there are at least two of
them. Hence it is impossible to define 0/0
BECAUSE IT WOULD NOT BE UNIQUE. ###
-------------------------------------------------------
n/0 (where n ‡ 0) is undefined.
-------------------------------------------------------
Why? (reducio ad absurdum):-
Suppose n is in F with n ‡ 0.
Suppose n/0 = k, where k is in F.
Then, n = k·0
The left-hand side is non-zero, while
the right-hand side is zero, no matter
what k is. This contradiction implies that
there is no such k. Hence it is impossible
to define n/0 BECAUSE NO SUCH VALUE EXISTS
that would fit the definition. ###
Therefore,
-------------------------------------------------------
n/0 is undefined for all n
-------------------------------------------------------
If n = 0, it cannot be defined as due to
NON-UNIQUENESS.
If n ‡ 0, it cannot be defined as due to
NON-EXISTENCE. ###
some people use a calculus approach to prove that division
by zero gives problems. while this is not wrong, it cannot
be applied to algebraic structures which do not come with
a built-in "calculus", for example finite (prime) fields
like F3.
furthermore, the EXISTENCE-UNIQUENESS style of argument
demonstrated above can be used to more generally to explain,
for example, why we cannot divide with vectors (whether it's
meant as opposite of cross product or dot product).
Steen S. Schmidt
>limits at all. 0/0 is undefined, period.
No it isn't, since the original expression was 0/(1/15-2/30). It's obvious
that 1/15-2/30 evaluates to zero, but when deriving a result from an
expression by inserting a value, and you get a undefined result, you need to
find the limit instead - which involves the use of l'Hôpital's rule on the
original expression in this case. That this particular case is so trivial
that it is indeed undefined just stresses the fact that l'Hôpital's rule has
to be applied in less simple cases;
An example would be 'COS(x)/SIN(x+pi/2)', where x=pi/2. If this were
evaluated as 'pi/2' COS (which yields 0) 'pi/2' DUP + SIN /, you'd get zero
as result, where it is in fact 1.
That's why Mr. Gibbons is incorrect in his assumption that the calc is
buggy - it's rather him who has to understand what he's making it do.
l'Hôpital's rule fits nicely here.
>BTW: '0/0 form' often mentioned in l'Hôpital's rule does not mean
>the number zero divided by the number zero, it represents a
>quotient of functions f(x)/g(x) in which both f(x) and g(x) tend
>to zero.
Yes. In this case f(x)=0 and g(x)=1/15-2/30.
Regards
Steen
Homer Simpson
Steen S. Schmidt a écrit dans le message
<83vuv1$23vk$1...@news.cybercity.dk>...
>>l'Hôpital's rule is irrelevant here, as we are not taking any
>>limits at all. 0/0 is undefined, period.
>>BTW: '0/0 form' often mentioned in l'Hôpital's rule does not mean
>>the number zero divided by the number zero, it represents a
>>quotient of functions f(x)/g(x) in which both f(x) and g(x) tend
>>to zero.
>Yes. In this case f(x)=0 and g(x)=1/15-2/30.
It isn't very useful to use this rule to "solve" such problems! So, let:
f:x -> 0
g:x -> 1/15-2/30
It wouldn't be false at all (in no case!) to say that g(x)=0, but let's
proceed:
diff(f(x),x)=0
diff(g(x),x)=0
Wonderful! Now you know that you can write 0/0 instead of 0/(1/15-2/30).
That was imo a complicated way to find out that the expression is
undefined...
Homer Simpson
http://zap.to/hsimpson
Keith J. Farmer
news:386370F0...@letterbox.com...
> plain oversight, possibly ignorance. perhaps the development team
> needs a Maths consultant in addition to its team of engineers.
They already have a professor (instructor, in fact), and it's the retraining
of the engineer-thinking user base that is creating some of the problems
with the 49 -- some of their favorite shortcuts no longer work, and some of
the forms of the expressions are unfamiliar to them.
> for example, why we cannot divide with vectors (whether it's
> meant as opposite of cross product or dot product).
Regardless of not having a single-valued output, it's still useful to have a
function return the family of functions if possible. Sound strange? It
shouldn't -- we do as much when we undo differentiation.
It'd be interesting, but I'm not sure it's entirely feasible -- set up a
calculator with a modifiable algebra just to see the consequences. But for
now, I'd be just as happy with tensors and operators.
Sophus Lefouque (P)
>...
it is very obvious that you are an amateur, as far as Mathematics
is concerned. you may be the world's best engineer for all i care
but your grasp of Mathematical concepts is atrocious. you are
confusing constants, constant functions, the limit of a function
at a point, the value of a function at a point, and the value of
a function at a point with singularity removed.
Steen S. Schmidt
You are free to think what you want, but I did at no point say that to see
if 0/(1/15-2/30) was defined or undefined you _had_ to use l'Hōpital's rule.
I said that you _could_ use it to check it, and the reason I did this was
that for any fraction which implies undefinity, you can use this rule - also
in this case, although it would be to cross the river to get water.
I guess you would argue that my example function 'COS(x)/SIN(x+pi/2)', where
x=pi/2, is undefined too, if you write it as 'COS(pi/2)/SIN(pi/2+pi/2)'?
This would yield 0/0, which is undefined, but I think we agree that the
result is actually 1?
Regards
Steen
Carlos Marangon
Excuse me, but I need to say you that in a discussion
we never attack the person of our interlocutor.
We need always attack the enclosed ideas, never the person.
In article <3863AA95...@letterbox.com>,
v...@letterbox.com wrote:
> Steen S. Schmidt wrote:
> >...
>
> it is very obvious that you are an amateur, as far as Mathematics
> is concerned. you may be the world's best engineer for all i care
> but your grasp of Mathematical concepts is atrocious. you are
> confusing constants, constant functions, the limit of a function
> at a point, the value of a function at a point, and the value of
> a function at a point with singularity removed.
>
>
--
Carlos Marangon
Area48 - HP48/49 for beginners!
http://www.area48.comIn article <3863AA95...@letterbox.com>,
v...@letterbox.com wrote:
> Steen S. Schmidt wrote:
> >...
>
> it is very obvious that you are an amateur, as far as Mathematics
> is concerned. you may be the world's best engineer for all i care
> but your grasp of Mathematical concepts is atrocious. you are
> confusing constants, constant functions, the limit of a function
> at a point, the value of a function at a point, and the value of
> a function at a point with singularity removed.
>
>
--
Carlos Marangon
Area48 - HP48/49 for beginners!
http://www.area48.com
Carlos Marangon
Sophus Lefouque (P)
...
> Regardless of not having a single-valued output, it's still useful to have a
> function return the family of functions if possible. Sound strange? It
> shouldn't -- we do as much when we undo differentiation.
no, not strange, but do it with caution.
for example, (§ means INTEGRATE)
§ 2 sin x cos dx = (sin x)² + C
§ 2 sin x cos dx = -(cos x)² + D
these answers seem to be at odds with each other, especially to a
student who just "plugs and plays" the calculator without
understanding what is going on.
Sophus Lefouque (P)
...
sorry, i got overheated.
> I guess you would argue that my example function 'COS(x)/SIN(x+pi/2)', where
> x=pi/2, is undefined too, if you write it as 'COS(pi/2)/SIN(pi/2+pi/2)'?
> This would yield 0/0, which is undefined, but I think we agree that the
> result is actually 1?
'COS(x)/SIN(x+pi/2)' if taken literally, would be undefined at x=pi/2.
however, the point of discontinuity at x=pi/2 is a removable
discontinuity, and the same is true for all x's where SIN(x+pi/2)=0.
so if we interpret 'COS(x)/SIN(x+pi/2)' as representing the extended
function which has the discontinuities patched-up (by defining it equal
to its limit at the points where SIN(x+pi/2)=0), then the value is 1
at x=pi/2.
f(x) = COS(x)/SIN(x+pi/2) for x s.t. SIN(x+pi/2)‡0
g(x) = COS(x)/SIN(x+pi/2) for x s.t. SIN(x+pi/2)‡0
= lim f(t) for x s.t. SIN(x+pi/2)=0
t->x
f(pi/2) is undefined.
g(pi/2) = 1.
Keith J. Farmer
> § 2 sin x cos dx = (sin x)² + C
> § 2 sin x cos dx = -(cos x)² + D
> these answers seem to be at odds with each other, especially to a
Such as the fact that constants are present and make up for the difference
between sin2 and -cos2, if D is taken as 1+C to make the relationship
explicit.
Sophus Lefouque (P)
...
> It isn't very useful to use this rule to "solve" such problems! So, let:
if you have only a hammer, every problem looks like a nail.
[if you own only a hospital, you want everybody to follow your
hospital's rules. :-P]
...
> Wonderful! Now you know that you can write 0/0 instead of 0/(1/15-2/30).
> That was imo a complicated way to find out that the expression is
> undefined...
...
i remember even books like those in the Schaum's series do caution
against the indiscriminate use of l'Hôpital's rules.
'0/(1/15-2/30)' is not even a limit to begin with. where is the
'lim'? you only use l'Hôpital's rules with limits and even then,
under special conditions.
ok, what about removable discontinuities where you redefine a
function to it's limit? let's try to interpret '0/(1/15-2/30)'
as the limit of a function at some point x=a with it's
"discontinuities removed". over what domain? ok, let's be
generous. assume the domain to be big enough to contain some
interval I containing a, like [a-1,a+1].
Let
f : x |———> 0 x in I
g : x |———> 1/15-2/30 x in I
Yes, actually g(x)=0 identically, but let's try to take
lim f(x)/g(x)
x->a
If one knows one's calculus concepts well, one would realise that
such a limit depends on the behaviour of f(x)/g(x) in the
neighbourhood of a (excluding a i.e. don't care about the value
of f(a)/g(a) or whether it is defined). So what is f(x)/g(x)
in the neighbourhood of a? Well, it is everywhere just as
undefined as f(a)/g(a)! so the discontinuity is unremovable
and the limit is nonsense. the whole neighbourhood has terminal
illness, even planting a hospital in the neighbourhood is of
no use. the expression '0/(1/15-2/30)' is doomed right from the
beginning.
More examples of abuse of l'Hôpital's rules:-
3 3 (d/dx)(3) 0
lim —— = —— (CORRECT) = lim —————————— = lim — = 0 (WRONG)
x->2 5x 10 x->2 (d/dx)(5x) x->2 5
[numerator and denominator don't both tend to zero, in fact
neither does.]
sin x cos x 1
lim ————— = 0 (CORRECT) = lim ————————— = ——— (WRONG)
x->0 cos x x->0 - sin x 0
[numerator and denominator don't both tend to zero]
8 4 (d/dx)(8) 0
——— = ——— (CORRECT) = lim —————————— = lim ——— (WRONG)
14 7 x->0 (d/dx)(14) x->0 0
[no lim!!!]
---
Here's a joke for those who like to apply Maths without thinking:-
Q: Solve x^2 = 4.
S: Diff both sides wrt x,
2x = 0
Therefore x = 0.
Substituting back to check,
0^2 = 4
0 = 4.
(where's the error? why?)
Steen S. Schmidt
>[if you own only a hospital, you want everybody to follow your
>hospital's rules. :-P]
Ok.....I deserved that one ;-)
>> Wonderful! Now you know that you can write 0/0 instead of 0/(1/15-2/30).
>> That was imo a complicated way to find out that the expression is
>> undefined...
[.....snip....]
>i remember even books like those in the Schaum's series do caution
>against the indiscriminate use of l'Hôpital's rules.
>
>'0/(1/15-2/30)' is not even a limit to begin with. where is the
>'lim'? you only use l'Hôpital's rules with limits and even then,
>under special conditions.
Yes, I was too fast, moved too quickly, and worst of all - fell into the
grave grave (grave^2?) of taking things for granted. Things that I cannot
take for granted unless I'm discussing these kinds of problems with my
fellow collegues - And yes, you're right; electrical engineering is the
field, not theoretical math. We do it the quick and dirty way, the way that
works in real life. I'm sorry that I got out of detail - I _know_ that
detail matter. I myself get very frustrated when people neglect detail and
take things for granted that _absolutely_ needs further explanation.
>Here's a joke for those who like to apply Maths without thinking:-
>Q: Solve x^2 = 4.
>S: Diff both sides wrt x,
> 2x = 0
> Therefore x = 0.
> Substituting back to check,
> 0^2 = 4
> 0 = 4.
>(where's the error? why?)
I hope I'll never be caught in error like that! Solving by differentiating?
Well.........I seem to ever neglect the fact that l'Hôpital's rule only
applies to limits (I again took something for granted :-)
Regards
Steen
Steen S. Schmidt
It's ok, no pun taken (except maybe a lesson in getting out of the sticky
old ways :-).
>'COS(x)/SIN(x+pi/2)' if taken literally, would be undefined at x=pi/2.
[snip]
Yes, I understand what you're saying. I have surely given this some thought
over the last couple of hours, no doubt about it.
The net result is that you're of course right in your deductions - I'm
sloppy at my math. Not that I don't know simple math, I think of myself as
rather adept at these kinds of things.
The problem, I think, is that when working with this, you (I) do certain
things to avoid making things more complicated than nescessary. Your field
happen to be theoretical math (it seems), and my field is applied math. In
this way I may seem as an amateur to you - understandably. My way of
thinking merely implies things that I take for granted and don't specify -
such as that l'Hôpital's rule only applies to limits (my answer to the
original question was to treat the solution as a limit, but I didn't make
that explicit). This also lead to your statement that I was confusing limits
with actual values at a point. I was, seemingly. I'm actually so used to
this behaviour (use), because in the digital/electronic world, there's no
such thing as a undefined value - just think of the SINC function - heavily
abbused by us :-)
All in all; People like you have just made me yank my old books of of the
shelves again - now we are going to have some nice healthy discussions at
work - no doubt about it.
To calm minds who may think that engineers are erratic people by now:
The things that we do make tend to work. It's actually only our deductions
that are somewhat lacking - we usually end at the right place anyways. I
don't know why ;-)
Merry christmas.
Steen Schmidt
Homer Simpson
>'lim'? you only use l'Hôpital's rules with limits and even then,
>under special conditions.
Yes, there's no limit here, but that is what is computed when one enters
'1/0', for example. We were looking for the 'right solution', thus trying to
find what limit should be computed to find the result '0/null' should
returns (obviously here it is '?').
>Here's a joke for those who like to apply Maths without thinking:-
>Q: Solve x^2 = 4.
>S: Diff both sides wrt x,
> 2x = 0
> Therefore x = 0.
> Substituting back to check,
> 0^2 = 4
> 0 = 4.
When from are we allowed to solve eqns by differentiating both sides? Afaik,
there is no (simple) relation between the points where a functions cancels
itself and where its derivative does...
Homer Simpson
http://zap.to/hsimpson
Homer Simpson
>> explanation, it's the second ten-years old bug we discover in a week
(with
>> the monadic plus).
[conditions]
>-------------------------------------------------------
> 0/0 is undefined.
>-------------------------------------------------------
[demonstration]
>-------------------------------------------------------
> n/0 (where n á 0) is undefined.
>-------------------------------------------------------
[demonstration]
>Therefore,
>-------------------------------------------------------
> n/0 is undefined for all n
>-------------------------------------------------------
But in fact there is an explanation.
When 1/0 returns infinity (unsigned), nobody complains, because it's pretty
intuitive, the result is ok. The assumption the calculator makes is that
when 1/0 is entered, what is expected is the result of limit(1/x, x=0).
The same way, 0/(null expr) can be interpreted as limit(0/x,x=0) which
equals zero. Only the case 0/0 is considered undefined, because the CAS
can't know what zero should be interpreted as a 'x -> 0'.
It's a choice of this CAS to returns something else than undefined when an
undefined calculation is performed (there is some kind of contradiction with
sign(0)=undefined if I'm not mistaken). Oh well.
Homer Simpson
http://zap.to/hsimpson
Veli-Pekka Nousiainen
news:s67pgku...@corp.supernews.com...
!> Sophus Lefouque (P) <vng....@letterbox.com> wrote in message
!> news:386370F0...@letterbox.com...
!> > plain oversight, possibly ignorance. perhaps the development team
!> > needs a Maths consultant in addition to its team of engineers.
!>
!> They already have a professor (instructor, in fact), and it's the
retraining
!> of the engineer-thinking user base that is creating some of the problems
!> with the 49 -- some of their favorite shortcuts no longer work, and some
of
!> the forms of the expressions are unfamiliar to them.
!>
!> > for example, why we cannot divide with vectors (whether it's
!> > meant as opposite of cross product or dot product).
!>
!> Regardless of not having a single-valued output, it's still useful to
have a
!> function return the family of functions if possible. Sound strange? It
!> shouldn't -- we do as much when we undo differentiation.
!>
!> It'd be interesting, but I'm not sure it's entirely feasible -- set up a
CASPAR ???
!> calculator with a modifiable algebra just to see the consequences. But
for
!> now, I'd be just as happy with tensors and operators.
Sophus Lefouque (P)
...
> Yes, there's no limit here, but that is what is computed when one enters
> '1/0', for example. We were looking for the 'right solution', thus trying to
> find what limit should be computed to find the result '0/null' should
> returns (obviously here it is '?').
re my comments about a super-human calculator with discretion.
> When from are we allowed to solve eqns by differentiating both sides? Afaik,
...
it's just a teaser. ;-)
--
Very Nice Guy (Pacific Version), Singapore
* Ah <sname> ah, <sname> : <sname> = short name or surname
e.g. "Ah- TanŻ ah_, TanŻ, your phone"
... Singlish database [1662 entries]
----- winners beget winners, losers stick with losers -----
Sophus Lefouque (P)
...
> But in fact there is an explanation.
> When 1/0 returns infinity (unsigned), nobody complains, because it's pretty
there is a way to extend the arithmetic outside the realm of real
numbers. we're working with an extended set R' = R u {oo}. i think
the best way to do so is by decree. i.e. make definitions like
k/0 = oo, oo + k = oo, ... etc. (where k is a finite real number)
any system is fine, as long as it is consistent.
some people prefer to work with the set R" = R u {+oo, -oo}.
what is HP-49's stand on infinity? it seems that HP-49 uses oo
to mean +oo or -oo.
> intuitive, the result is ok. The assumption the calculator makes is that
> when 1/0 is entered, what is expected is the result of limit(1/x, x=0).
...
> The same way, 0/(null expr) can be interpreted as limit(0/x,x=0) which
the limits approach can sometimes be used to explain things, but i
feel it's a bit dangerous. limits can be a pretty tricky concept.
consider the following limits:-
1
lim — = oo = unbounded or no limit (no pun intended)
x->0 x
unsigned infinity - in the sense that for every R>0, there exists
a d>0 such that whenever |x|<d we have |1/x|>R.
1
lim — = undefined
x->0 x
in the sense that the left limit and the right limit are unequal
1 1
lim — = +oo lim — = -oo
x->0+ x x->0- x
the limit on the left means for every R>0, there exists
a d>0 such that whenever 0<x<d we have 1/x>R.
the limit on the right means for every R>0, there exists
a d>0 such that whenever -d<x<0 we have 1/x<-R.
x²-4
lim ———— = 4
x->2 x-2
sin x x
lim ————— = 1 lim ————— = 1
x->0 x x->0 sin x
-x 2x 3x
lim —— = -1 lim —— = 2 lim —— = 3
x->0 x x->0 x x->0 x
x e^x
lim ——— = 0 lim ——— = oo
x->0 e^x x->0 x
2x²-4x 2
lim —————— = —
x->oo 5x²+3x 5
2x²-4x 4
lim —————— = - —
x->0 5x²+3x 3
2x²
lim —————— = 0
x->0 5x²+3x
2x²-4x
lim —————— = oo
x->0 5x²
note that limits of the form 0/0 can be anything (depending on the
expression and the value x tends to).
--
Very Nice Guy (Pacific Version), Singapore
* Ah <sname> ah, <sname> : <sname> = short name or surname
e.g. "Ah- Tan¯ ah_, Tan¯, your phone"
Sophus Lefouque (P)
...
> between sin2 and -cos2, if D is taken as 1+C to make the relationship
> explicit.
HP-49 gives
§ 2 sin x cos dx = (sin x)²
TI-89 gives
§ 2 sin x cos dx = -(cos x)²
both without the constant. an unenlightened student would have
thought it's
§ 2 sin x cos dx = (sin x)² + C according to HP-49
§ 2 sin x cos dx = -(cos x)² + C according to TI-89
and would have thought they are different. of course both are
correct but the C's are different. i made it easier to clean up
the mess by using C and D.
the bottom line is: using a computing tool is not a substitute for
thinking and understanding.
Sophus Lefouque (P)
...
> happen to be theoretical math (it seems),
nah! i'm just a cyber-bummer.
anyway the Applied vs Pure dichotomy usually has no problems. "leave
enough well alone" or "live and let live". but when you design a
calculator to cater to everybody, some hard thinking needs to be done.
you need a calculator that gives reliable, authoritative and rigorous
answers and yet be flexible to creatively interpret certain
"non-sensical" expressions as their limits. AND YOU NEED TO DO SO
CONSISTENTLY. you need to endow the calculator with an almost
super-human quality of discretion. this is an issue for the
designers. certainly no easy task.
--
Very Nice Guy (Pacific Version), Singapore
* Ah <sname> ah, <sname> : <sname> = short name or surname
e.g. "Ah- TanŻ ah_, TanŻ, your phone"
Tom Gutman
news:8457aj$1dk$1...@front2m.grolier.fr...
[SNIP]
>
> >Here's a joke for those who like to apply Maths without thinking:-
> >Q: Solve x^2 = 4.
> >S: Diff both sides wrt x,
> > 2x = 0
> > Therefore x = 0.
> > Substituting back to check,
> > 0^2 = 4
> > 0 = 4.
>
>
Afaik,
> itself and where its derivative does...
>
> Homer Simpson
> http://zap.to/hsimpson
>
>
What do you use as the meaning of equality? As I remember it, it is meant
to mean that the two sides are just different ways of writing the same
thing. If you do the same thing to the same thing, you again get the same
thing. The trap here is a little bit deeper. Note that if f(x) = g(x)*h(x)
is it quite legitimate to differentiate both sides to get f'(x) = g'(x)*h(x)
+ g(x)*h'(x).
--
Tom Gutman
Tom Gutman
> Homer Simpson wrote:
[SNIP]
>
> furthermore, the EXISTENCE-UNIQUENESS style of argument
> demonstrated above can be used to more generally to explain,
>
--
Tom Gutman
Keith Farmer
> And complex numbers are not vectors? Nor quartenions?
You may be able to write them with the same notation, but they're not
the same beast:
C=a+bi; i=scalar
V=a*I + b*J; I, J=vector
Remember that I and J both have associated with them a characteristic
'direction', whereas i does not. That is, you can have a magnitude i,
but not a magnitude J.
Homer Simpson
>thing.
In general. The "thing" you do has to be some kind of bijection. We all know
that << + 1 >> is this kind of function (inverse function: << 1 - >>). But
<< 0 * >> isn't (inverse function: good luck!).
The same way, the function << 'X' der >> _seems_ to have << 0 'X' ROT 'X'
int >> as its inverse function, but it isn't right (different functions can
have the same derivative blah blah blah)...
Thanks for pointing this out. Before reading this post, I thought that the
above method was complete nonsense, but it does make sense somehow (hum...
stupid thing).
Homer Simpson
http://zap.to/hsimpson
Tom Gutman
function or even something mathematical) and you apply a deterministic
operation on it you get another thing. If you take that same first thing
and apply the same operation you get the same second thing. It doesn't have
to be a bijections or invertable.
The problem in the example is a bit different. When we write X^2=4, X^2 is
not a function, but just a number (or a finite set of numbers). You cannot
differentiate a number.
Even the example I gave requires some care. (Now missing a proper extended
alphabet.) If we say that for all x f(x)=g(x)*h(x), then f(x) is proper a
function, and we can procede to differentiate both sides an possibly find
some interesting relations in the result. But if we say that there exists
an x such that f(x)=g(x)*h(x) then in that context f(x) is no longer a
function, but a single value and we can no longer differentiate it.
Another way of considering it is that if we wish to apply an operation (such
as differentiation) that depends not on the value at a specific point but on
the values in a compact set then the equality must hold not just at an
isolated point but over a compact set.
--
Tom Gutman
Tom Gutman
news:386ACA3F...@pacbell.net...
They are the same beast. Complex numbers and quartenions are quite
definitely vector spaces. They form an additive group, and multiplication
by reals is defined and distributitive.
Looking at the complex numbers, the common basis elements (1 and i) can be
mapped into a euclidian space witt directions. The common convention is
that 1 goes from right to left on the paper and i goes from bottom to top.
If you prefer the definition of a vector as something with a length and
direction (the physicist's definition rather than the mathematician's), note
that complex numbers are frequently written in polar notation, explicitly
giving its length and direction.
--
Tom Gutman
Sophus Lefouque (P)
...
> Thanks for pointing this out. Before reading this post, I thought that the
> above method was complete nonsense, but it does make sense somehow (hum...
> stupid thing).
but where's the mistake?
--
Very Nice Guy (Pacific Version), Singapore
Keith J. Farmer
news:84hr3q$g86$2...@ssauraaa-i-1.production.compuserve.com...
> If you prefer the definition of a vector as something with a length and
> direction (the physicist's definition rather than the mathematician's),
note
There is very little difference between the two. The bachelor in physics is
maybe 30 credits from a bachelors in mathematics, if not less. The only
difference I've seen, really, is that Elementary Classical Analysis, and the
very deep algebra are not required of the physics major (who probably taught
himself half of diffeq by the time he took the actual math course).
And yet you miss the point that the basis i is a scalar, wheras [1 0] and [0
1] are not. *How* you graph a thing is not important; what is important is
what information is inherent in that thing. Certainly, you can map the
complex numbers onto a cartesian space, [x y] = re(Z)*[1 0] + im(Z)*[0 1],
but that doesn't mean that the complex numbers are themselves vectors,
obeying the multiplicative properties that separate vectors from scalars.
Veli-Pekka Nousiainen
What use of them?
VPNnews:84eduh$u1$3...@ssauraaa-i-1.production.compuserve.com...
!> "Sophus Lefouque (P)" <vng....@letterbox.com> wrote in message
!> news:386370F0...@letterbox.com...
!> > Homer Simpson wrote:
!> [SNIP]
!> >
!> > furthermore, the EXISTENCE-UNIQUENESS style of argument
!> > demonstrated above can be used to more generally to explain,
!> > for example, why we cannot divide with vectors (whether it's
!> > meant as opposite of cross product or dot product).
!> >
!> And complex numbers are not vectors? Nor quartenions?
!>
!> --
!> Tom Gutman
!>
!>
!>
Tom Gutman
"Keith J. Farmer" <far...@pacbell.net> wrote in message
news:s6p3s7...@corp.supernews.com...
> > If you prefer the definition of a vector as something with a length and
> > direction (the physicist's definition rather than the mathematician's),
> note
>
> There is very little difference between the two. The bachelor in physics
is
> maybe 30 credits from a bachelors in mathematics, if not less. The only
> difference I've seen, really, is that Elementary Classical Analysis, and
the
> very deep algebra are not required of the physics major (who probably
taught
> himself half of diffeq by the time he took the actual math course).
Perhaps. But they have a very different outlook on life.
Differential equations is an interesting case in point. The mathematician
is quite satisfied to prove that a solution exists, and move on to the next
problem. The representation of the solution in a closed form or even a
procedure for a numerical solution are of no interest.
>
> And yet you miss the point that the basis i is a scalar, wheras [1 0] and
[0
> 1] are not. *How* you graph a thing is not important; what is important
is
> what information is inherent in that thing. Certainly, you can map the
> complex numbers onto a cartesian space, [x y] = re(Z)*[1 0] + im(Z)*[0 1],
> but that doesn't mean that the complex numbers are themselves vectors,
> obeying the multiplicative properties that separate vectors from scalars.
>
>
But it does. Exactly what property is needed for a vector space that is
lacking in the complex numbers? As I remember, all you need for a vector
space is an underlying field of scalars (the real numbers, in this case), an
additive group (the complex numbers form an additive group), and scalar by
vector multiplication with the proper associative and distributive
properties (normal complex multiplication restricted to a real by a complex
fits this need). As with any vector space, one can use any pair of
independent elements as a basis - for the complex numbers 1 and i are
traditionally used.
I seem to remember writing complex numbers as (a,b), with 1 defined as (1,0)
and i defined as (0,1). This is an exposition of complex numbers that
starts with their properties as a vector space with a chosen basis, and then
an additional multiplication operator is defined to make it a field.
The concept of vectors are having an inherent direction is not a
mathematical one. It is that of a physicist, who thinks in terms of real
world quantities and finds that low dimensional vector spaces are a good way
to represent quantities that have a magnitude and direction.
So is i a scalar or a vector. Considering the complex numbers as a vector
space over the real numbers, then i (as any complex number) is a vector.
But the complex numbers also have an additional operation (in addition to
the vector properties) that make them a field. As a field, they can be used
as the scalars for defining vector spaces over the complex numbers. Used
this way, i would be seen as a scalar.
A rather interesting question is why, of all the vector spaces over the
reals, only the two and four dimensional cases can be extended to form a
field. I know why the particular construction for complex numbers and
quartenions (essentially the same) only works in these two cases, but why no
other ways to do a multiplicative group?
--
Tom Gutman
Tom Gutman
news:84ihqd$e96$9...@tron.sci.fi...
> quartenions? [ x y z w ] or ( (x, y) , (z w) ) or ???
> What use of them?
> VPN
between the behaviour of the first coordinate of of the other coordinates.
Historically they have been interesting because they and the complex numbers
represent the only cases of a vector space over the reals also being a
field. (hmmm.... I wonder if that is quite right - possibly if the
multiplication is not required to be an extension of real number
multiplication there might be other solutions - any real mathematicians
around to comment?)
I have recently run into quartenions used in doing 3D geometry.
> Tom Gutman <Tom_G...@compuserve.com> wrote in message
> news:84eduh$u1$3...@ssauraaa-i-1.production.compuserve.com...
> !> "Sophus Lefouque (P)" <vng....@letterbox.com> wrote in message
> !> news:386370F0...@letterbox.com...
> !> > Homer Simpson wrote:
> !> [SNIP]
> !> >
> !> > furthermore, the EXISTENCE-UNIQUENESS style of argument
> !> > demonstrated above can be used to more generally to explain,
> !> > for example, why we cannot divide with vectors (whether it's
> !> > meant as opposite of cross product or dot product).
> !> >
> !> And complex numbers are not vectors? Nor quartenions?
> !>
> !> --
> !> Tom Gutman
> !>
> !>
> !>
>
>
--
Tom Gutman
Veli-Pekka Nousiainen
Please, more !! Can quartenions be used in the 49G?
VPN
Tom Gutman <Tom_G...@compuserve.com> wrote in message
news:84kcq7$lfd$1...@ssauraac-i-1.production.compuserve.com...
!> "Veli-Pekka Nousiainen" <v...@fcs.fi> wrote in message
!> news:84ihqd$e96$9...@tron.sci.fi...
!> > quartenions? [ x y z w ] or ( (x, y) , (z w) ) or ???
!> > What use of them?
!> > VPN
!> generally [w x y z], but sometimes [w (x y z)], since their is a
difference
!> between the behaviour of the first coordinate of of the other
coordinates.
!> Historically they have been interesting because they and the complex
numbers
!> represent the only cases of a vector space over the reals also being a
!> field. (hmmm.... I wonder if that is quite right - possibly if the
!> multiplication is not required to be an extension of real number
!> multiplication there might be other solutions - any real mathematicians
!> around to comment?)
!>
!> I have recently run into quartenions used in doing 3D geometry.
!>
!> > Tom Gutman <Tom_G...@compuserve.com> wrote in message
!> > news:84eduh$u1$3...@ssauraaa-i-1.production.compuserve.com...
!> > !> "Sophus Lefouque (P)" <vng....@letterbox.com> wrote in message
!> > !> news:386370F0...@letterbox.com...
!> > !> > Homer Simpson wrote:
!> > !> [SNIP]
!> > !> >
!> > !> > furthermore, the EXISTENCE-UNIQUENESS style of argument
!> > !> > demonstrated above can be used to more generally to explain,
!> > !> > for example, why we cannot divide with vectors (whether it's
!> > !> > meant as opposite of cross product or dot product).
!> > !> >
!> > !> And complex numbers are not vectors? Nor quartenions?
!> > !>
!> > !> --
!> > !> Tom Gutman
Tom Gutman
> Hi, Tom !
>
> Please, more !! Can quartenions be used in the 49G?
> VPN
>
Sure - but you would have to program the operations. Using user RPL, I
think the (w (x y z)) form would be easiest and fastest. In ML (and
possibly SYSRPL) the (w x y z) is probably better. Since your questions
lead me to think that you are not familiar with quartenions, I will use this
as an excuse to provide an elementary (my limits) exposition on quartenions,
generalized to include the complex numbers.
A means of extending a vector space over the reals to be a field, and why it
only works for dimensions 1 (trivially reconstructs the reals), 2 (complex
numbers), and 4 (quartenions).
Start with a vector space over the reals, with a basis so that the vectors
can be represented as a sequence of real numbers (a,b,c, ... ). Now pull
out the first (or any) dimension and represent the N dimensional vector as a
real number and an N-1 dimensional vector. (a,b,c,...) -> (a, (b,c,...))
We will now call that first dimension the scalar or real part of the number
and the rest of it as the vector part. For notational convenience, and
where context prevents ambiguities (or the ambiguities don't matter) we will
call vectors of the form (a, 0) scalars and write them as just a. Vectors
of the form (0, V) will be called vectors and written as just V. To help
distinguish vector components from scalar components (particularly in the
case of two dimensions, where the vector component is just one dimensional
and looks very much like a scalar) we will use symbols for the basis set and
write vectors as a linear combination of basis elements. In the two
dimensional case the single basis element is traditionally written a i
(except by EEs who prefer j), in the four dimensional case the basis
elements are traditionally written as I, J, and K.
Now we write all vectors as the sum of scalar and vector components - a+V.
This should be interpreted as (a,0)+(0,V). Since the scalar component is
one dimensional it acts much like a real, we impose on it all the behaviours
of reals, including real by vector multiplication. Vectors already come
with addition, and the existing vector addition is compatible with our new
notation.
Now how to multiply? Consider (a+V)x(b+W). Applying normal distributive
laws, we get four terms of three types. The first is a scalar by scalar
multiplication. That is no problem, and yields a scalar term. There are
then two terms of scalar by vector multiplication. That is also no problem,
and yields a vector term. That leaves the vector by vector multiplication.
Should be use the dot product? Or the Cross product? Or something else?
Not being dicisive nor inventive, we just use both the dot product and the
cross product. But with a twist. We take the cross product (yielding a
vector term) as is, but we take the negative of the dot product (yielding a
scalar term). So the result of multiplying a+V by b+W ends up as (ab-V.W) +
(aW+bV+VxW).
Now we see why this works only for one, two and four dimensions. These
yield zero, one, and three dimensional vector components, and these are the
only dimensions where the cross product works. For zero dimensions, there
is only one vector, and the cross product trivially is that unique vector.
In the one dimensional case, all vectors are parallel, hence the cross
product is always zero, which is a suitable one dimensional vector. The two
dimensional case fails because while we can define the cross product, it
does not lie in the original space - we need to embed the two dimensional
space in a three dimensional space to provide the required vectors
orthogonal to both of the original vectors. In three dimensions cross
products work, and are actually non-trivial. In higher dimensions, cross
products are not well defined as there are too many orthogonal directions.
It is trivial (i.e. I don't remember exactly how to do it and it is too much
work to work out) to show that this multiplication follows the necessary
associative and distributive laws. Note that since three dimensional cross
products are not commutative, quartenion multiplication is not commutative.
But since one dimensional cross products are (trivially) commutative,
complex multiplication is commutative.
To make a field, we do need to show a multiplicative inverse (trivially 1 is
the multiplicative identity). For a+V consider the conjugate a-V. The
result is a^2+V.V. The two scalar by vector components cancel out, and the
cross product of a vector with itself (or its additive invers) is zero.
Since we are working over the real numbers and V.V is a sum of squares,
a^2+V.V is positive, unless both a and V are zero. Hence we can take its
inverse. So (a+V)x((a-V)/(a^2+V.V)) is one, and (a-V)/(a^2+V.V) is the
multiplicative inverse of a+V. Hence we have a field.
Note that the complex numbers are embedded (many ways) in quartenions. If
we take any unit vector U and consider quartenions of the form a+bU (a,b
real) the result is isomorphic to the complex numbers. So any program that
provides for quartenion arithmetic also provides for complex arithmetic.
--
Tom Gutman
Veli-Pekka Nousiainen
Thaks for the brief introduction into the realms of quartenions
but are they used on any computation and could they be
inplemented in HP 49G (R->Q, Q->R, C->Q, Q->C) math?
Flag -4 as a Quartenion mode flag, etc...
Should we persuade Parisse to implement Q's in CAS?
??
--
Regards, VPN
_________________________________________________________
Veli-Pekka Nousiainen ; e-mail= v...@fcsolutions.com
Sokinsuontie 3 A 1, FIN-02760 Espoo, Finland
TEL, WORK= +358 (9) 859 2025 ; (WORK2= +358 (3) 4728 300)
Future Computing Solutions Oy ; URL= http://www.eiffel.fi
_________________________________________________________
HP25,HP28S,HP41CX,HP48SX,HP48GX,HP49G,HP71B,HP75C & TI89
Vote for the "82484A Curve Fit for HP71B" => HP49G !!!
Tom Gutman <Tom_G...@compuserve.com> wrote in message
news:84lr49$cu1$1...@ssauraab-i-1.production.compuserve.com...
!> "Veli-Pekka Nousiainen" <v...@fcs.fi> wrote in message
!> news:84llpe$n9h$4...@tron.sci.fi...
!> > Hi, Tom !
!> >
!> > Please, more !! Can quartenions be used in the 49G?
!> > VPN
!> >
!> [SNIP]
!>
!> Sure - but you would have to program the operations. Using user RPL, I
!> think the (w (x y z)) form would be easiest and fastest. In ML (and
!> possibly SYSRPL) the (w x y z) is probably better. Since your questions
!> lead me to think that you are not familiar with quartenions, I will use
this
!> as an excuse to provide an elementary (my limits) exposition on
quartenions,
!> generalized to include the complex numbers.
!>
!> A means of extending a vector space over the reals to be a field, and why
it
!> only works for dimensions 1 (trivially reconstructs the reals), 2
(complex
!> numbers), and 4 (quartenions).
!>
!> Start with a vector space over the reals, with a basis so that the
vectors
!> can be represented as a sequence of real numbers (a,b,c, ... ). Now pull
!> out the first (or any) dimension and represent the N dimensional vector
as
a
!> real number and an N-1 dimensional vector. (a,b,c,...) -> (a, (b,c,...))
!> We will now call that first dimension the scalar or real part of the
number
!> and the rest of it as the vector part. For notational convenience, and
!> where context prevents ambiguities (or the ambiguities don't matter) we
will
!> call vectors of the form (a, 0) scalars and write them as just a.
Vectors
!> of the form (0, V) will be called vectors and written as just V. To help
!> distinguish vector components from scalar components (particularly in the
!> case of two dimensions, where the vector component is just one
dimensional
!> and looks very much like a scalar) we will use symbols for the basis set
and
!> write vectors as a linear combination of basis elements. In the two
!> dimensional case the single basis element is traditionally written a i
!> (except by EEs who prefer j), in the four dimensional case the basis
!> elements are traditionally written as I, J, and K.
!>
!> Now we write all vectors as the sum of scalar and vector components -
a+V.
!> This should be interpreted as (a,0)+(0,V). Since the scalar component is
!> one dimensional it acts much like a real, we impose on it all the
behaviours
!> of reals, including real by vector multiplication. Vectors already come
!> with addition, and the existing vector addition is compatible with our
new
!> notation.
!>
!> Now how to multiply? Consider (a+V)x(b+W). Applying normal distributive
!> laws, we get four terms of three types. The first is a scalar by scalar
!> multiplication. That is no problem, and yields a scalar term. There are
!> then two terms of scalar by vector multiplication. That is also no
problem,
!> and yields a vector term. That leaves the vector by vector
multiplication.
!> Should be use the dot product? Or the Cross product? Or something else?
!> Not being dicisive nor inventive, we just use both the dot product and
the
!> cross product. But with a twist. We take the cross product (yielding a
!> vector term) as is, but we take the negative of the dot product (yielding
a
!> scalar term). So the result of multiplying a+V by b+W ends up as
(ab-V.W)
+
!> (aW+bV+VxW).
!>
!> Now we see why this works only for one, two and four dimensions. These
!> yield zero, one, and three dimensional vector components, and these are
the
!> only dimensions where the cross product works. For zero dimensions,
there
!> is only one vector, and the cross product trivially is that unique
vector.
!> In the one dimensional case, all vectors are parallel, hence the cross
!> product is always zero, which is a suitable one dimensional vector. The
two
!> dimensional case fails because while we can define the cross product, it
!> does not lie in the original space - we need to embed the two dimensional
!> space in a three dimensional space to provide the required vectors
!> orthogonal to both of the original vectors. In three dimensions cross
!> products work, and are actually non-trivial. In higher dimensions, cross
!> products are not well defined as there are too many orthogonal
directions.
!>
!> It is trivial (i.e. I don't remember exactly how to do it and it is too
much
!> work to work out) to show that this multiplication follows the necessary
!> associative and distributive laws. Note that since three dimensional
cross
!> products are not commutative, quartenion multiplication is not
commutative.
!> But since one dimensional cross products are (trivially) commutative,
!> complex multiplication is commutative.
!>
!> To make a field, we do need to show a multiplicative inverse (trivially 1
is
!> the multiplicative identity). For a+V consider the conjugate a-V. The
!> result is a^2+V.V. The two scalar by vector components cancel out, and
the
!> cross product of a vector with itself (or its additive invers) is zero.
!> Since we are working over the real numbers and V.V is a sum of squares,
!> a^2+V.V is positive, unless both a and V are zero. Hence we can take its
!> inverse. So (a+V)x((a-V)/(a^2+V.V)) is one, and (a-V)/(a^2+V.V) is the
!> multiplicative inverse of a+V. Hence we have a field.
!>
!> Note that the complex numbers are embedded (many ways) in quartenions.
If
!> we take any unit vector U and consider quartenions of the form a+bU (a,b
!> real) the result is isomorphic to the complex numbers. So any program
that
!> provides for quartenion arithmetic also provides for complex arithmetic.
Tom Gutman
> Hi, Tom !
>
> Thaks for the brief introduction into the realms of quartenions
> but are they used on any computation and could they be
> inplemented in HP 49G (R->Q, Q->R, C->Q, Q->C) math?
> Flag -4 as a Quartenion mode flag, etc...
> Should we persuade Parisse to implement Q's in CAS?
> ??
The only place I've seen them used is in 3D graphics. Just as 2D graphics
are often done by extending the 2D vectors with a constant one (to allow
translations to be incorporated into the same transformation matrix as
rotations and scaling), 3D vectors can be incorporated into quartenions with
a scalar of one. A quartenion can be used to represent a rotation, with the
vector part providing the direction and the scalar part the amount of
rotation, and one can implement the rotation with some simple quartenion
expression (which I don't remember, and am not sure I could reconstruct).
I don't know that quartenions are used enough to warrent having them built
into the calculator. But a prime feature of the HP49 is the use of flash
memory, and reasonably large amount of it. What would make sense is to put
hooks into the basic system to allow packages to be included that extend the
functions. Libraries already do that for explicitly selected functions.
What would be nice here is a way to define additional data types, and have
the existing overloaded operators (+, *, /, etc.) be able to be extended to
these new types.
I think Parisse has his hands full with just real and complex calculus. I'm
not even sure how much algebra and calculus survives the quartenion
extension - the loss of commutativity in multiplication should result in the
loss of a fair bit of algebra. Even going to the two dimensions of complex
numbers, analytic functions have been reduced to a very few (a countable
number, I think). How do they survive going to the four dimensions of
quartenions?
--
Tom Gutman
Veli-Pekka Nousiainen
OK, thanks !
I think I have gasped something about quartenions.
Maybe - when the "59G" is released (with 3D graphics)
we will have some use of quartenions?
Tell me more if it still pleases you, others might enjoy
of it, too.
--
Regards, VPN
_________________________________________________________
Veli-Pekka Nousiainen ; e-mail= v...@fcsolutions.com
Sokinsuontie 3 A 1, FIN-02760 Espoo, Finland
TEL, WORK= +358 (9) 859 2025 ; (WORK2= +358 (3) 4728 300)
Future Computing Solutions Oy ; URL= http://www.eiffel.fi
_________________________________________________________
;-) We R TiBORG - Res. is Futile - Ur 59 will be Assimilated !
Tom Gutman <Tom_G...@compuserve.com> wrote in message
news:84ocnv$gml$1...@ssauraac-i-1.production.compuserve.com...
!> "Veli-Pekka Nousiainen" <v...@fcs.fi> wrote in message
!> news:84o59e$cp0$1...@tron.sci.fi...
!> > Hi, Tom !
!> >
!> > Thaks for the brief introduction into the realms of quartenions
!> > but are they used on any computation and could they be
!> > inplemented in HP 49G (R->Q, Q->R, C->Q, Q->C) math?
!> > Flag -4 as a Quartenion mode flag, etc...
!> > Should we persuade Parisse to implement Q's in CAS?
!> > ??
!>
!> The only place I've seen them used is in 3D graphics. Just as 2D
graphics
!> are often done by extending the 2D vectors with a constant one (to allow
!> translations to be incorporated into the same transformation matrix as
!> rotations and scaling), 3D vectors can be incorporated into quartenions
with
!> a scalar of one. A quartenion can be used to represent a rotation, with
the
!> vector part providing the direction and the scalar part the amount of
!> rotation, and one can implement the rotation with some simple quartenion
!> expression (which I don't remember, and am not sure I could reconstruct).
!>
!> I don't know that quartenions are used enough to warrent having them
built
!> into the calculator. But a prime feature of the HP49 is the use of flash
!> memory, and reasonably large amount of it. What would make sense is to
put
!> hooks into the basic system to allow packages to be included that extend
the
!> functions. Libraries already do that for explicitly selected functions.
!> What would be nice here is a way to define additional data types, and
have
!> the existing overloaded operators (+, *, /, etc.) be able to be extended
to
!> these new types.
!>
!> I think Parisse has his hands full with just real and complex calculus.
I'm
!> not even sure how much algebra and calculus survives the quartenion
!> extension - the loss of commutativity in multiplication should result in
the
!> loss of a fair bit of algebra. Even going to the two dimensions of
complex
!> numbers, analytic functions have been reduced to a very few (a countable
!> number, I think). How do they survive going to the four dimensions of
!> quartenions?
!>
!> > --
!> > Regards, VPN
!> > _________________________________________________________
!> > Veli-Pekka Nousiainen ; e-mail= v...@fcsolutions.com
!> > Sokinsuontie 3 A 1, FIN-02760 Espoo, Finland
!> > TEL, WORK= +358 (9) 859 2025 ; (WORK2= +358 (3) 4728 300)
!> > Future Computing Solutions Oy ; URL= http://www.eiffel.fi
!> > _________________________________________________________
!> > HP25,HP28S,HP41CX,HP48SX,HP48GX,HP49G,HP71B,HP75C & TI89
!> > Vote for the "82484A Curve Fit for HP71B" => HP49G !!!
!> >
!> > Tom Gutman <Tom_G...@compuserve.com> wrote in message
!> > news:84lr49$cu1$1...@ssauraab-i-1.production.compuserve.com...
!> > !> "Veli-Pekka Nousiainen" <v...@fcs.fi> wrote in message
!> > !> news:84llpe$n9h$4...@tron.sci.fi...
!> > !> > Hi, Tom !
!> > !> >
!> > !> > Please, more !! Can quartenions be used in the 49G?
!> > !> > VPN
!> > !> >
!> > !> [SNIP]
!> > !>
!> > !> Sure - but you would have to program the operations. Using user
RPL, I
!> > !> think the (w (x y z)) form would be easiest and fastest. In ML (and
!> > !> possibly SYSRPL) the (w x y z) is probably better. Since your
!> questions
!> > !> lead me to think that you are not familiar with quartenions, I will
use
!> > this
!> > !> as an excuse to provide an elementary (my limits) exposition on
!> > quartenions,
!> > !> generalized to include the complex numbers.
!> > !>
!> > !> A means of extending a vector space over the reals to be a field,
and
!> why
!> > it
!> > !> only works for dimensions 1 (trivially reconstructs the reals), 2
!> > (complex
!> > !> numbers), and 4 (quartenions).
!> > !>
!> > !> Start with a vector space over the reals, with a basis so that the
!> > vectors
!> > !> can be represented as a sequence of real numbers (a,b,c, ... ). Now
!> pull
!> > !> out the first (or any) dimension and represent the N dimensional
vector
!> > as
!> > a
!> > !> real number and an N-1 dimensional vector. (a,b,c,...) -> (a,
!> (b,c,...))
!> > !> We will now call that first dimension the scalar or real part of the
!> > number
!> > !> and the rest of it as the vector part. For notational convenience,
and
!> > !> where context prevents ambiguities (or the ambiguities don't matter)
we
!> > will
!> > !> call vectors of the form (a, 0) scalars and write them as just a.
!> > Vectors
!> > !> of the form (0, V) will be called vectors and written as just V. To
!> help
!> > !> distinguish vector components from scalar components (particularly
in
!> the
!> > !> case of two dimensions, where the vector component is just one
!> > dimensional
!> > !> and looks very much like a scalar) we will use symbols for the basis
!> set
!> > and
!> > !> write vectors as a linear combination of basis elements. In the two
!> > !> dimensional case the single basis element is traditionally written a
i
!> > !> (except by EEs who prefer j), in the four dimensional case the basis
!> > !> elements are traditionally written as I, J, and K.
!> > !>
!> > !> Now we write all vectors as the sum of scalar and vector
components -
!> > a+V.
!> > !> This should be interpreted as (a,0)+(0,V). Since the scalar
component
!> is
!> > !> one dimensional it acts much like a real, we impose on it all the
!> > behaviours
!> > !> of reals, including real by vector multiplication. Vectors already
!> come
!> > !> with addition, and the existing vector addition is compatible with
our
!> > new
!> > !> notation.
!> > !>
!> > !> Now how to multiply? Consider (a+V)x(b+W). Applying normal
!> distributive
!> > !> laws, we get four terms of three types. The first is a scalar by
!> scalar
!> > !> multiplication. That is no problem, and yields a scalar term.
There
!> are
!> > !> then two terms of scalar by vector multiplication. That is also no
!> > problem,
!> > !> and yields a vector term. That leaves the vector by vector
!> > multiplication.
!> > !> Should be use the dot product? Or the Cross product? Or something
!> else?
!> > !> Not being dicisive nor inventive, we just use both the dot product
and
!> > the
!> > !> cross product. But with a twist. We take the cross product
(yielding
!> a
!> > !> vector term) as is, but we take the negative of the dot product
!> (yielding
!> > a
!> > !> scalar term). So the result of multiplying a+V by b+W ends up as
!> > (ab-V.W)
!> > +
!> > !> (aW+bV+VxW).
!> > !>
!> > !> Now we see why this works only for one, two and four dimensions.
These
!> > !> yield zero, one, and three dimensional vector components, and these
are
!> > the
!> > !> only dimensions where the cross product works. For zero dimensions,
!> > there
!> > !> is only one vector, and the cross product trivially is that unique
!> > vector.
!> > !> In the one dimensional case, all vectors are parallel, hence the
cross
!> > !> product is always zero, which is a suitable one dimensional vector.
!> The
!> > two
!> > !> dimensional case fails because while we can define the cross
product,
!> it
!> > !> does not lie in the original space - we need to embed the two
!> dimensional
!> > !> space in a three dimensional space to provide the required vectors
!> > !> orthogonal to both of the original vectors. In three dimensions
cross
!> > !> products work, and are actually non-trivial. In higher dimensions,
!> cross
!> > !> products are not well defined as there are too many orthogonal
!> > directions.
!> > !>
!> > !> It is trivial (i.e. I don't remember exactly how to do it and it is
too
!> > much
!> > !> work to work out) to show that this multiplication follows the
!> necessary
!> > !> associative and distributive laws. Note that since three
dimensional
!> > cross
!> > !> products are not commutative, quartenion multiplication is not
!> > commutative.
!> > !> But since one dimensional cross products are (trivially)
commutative,
!> > !> complex multiplication is commutative.
!> > !>
!> > !> To make a field, we do need to show a multiplicative inverse
(trivially
!> 1
!> > is
!> > !> the multiplicative identity). For a+V consider the conjugate a-V.
The
!> > !> result is a^2+V.V. The two scalar by vector components cancel out,
and
!> > the
!> > !> cross product of a vector with itself (or its additive invers) is
zero.
!> > !> Since we are working over the real numbers and V.V is a sum of
squares,
!> > !> a^2+V.V is positive, unless both a and V are zero. Hence we can
take
!> its
!> > !> inverse. So (a+V)x((a-V)/(a^2+V.V)) is one, and (a-V)/(a^2+V.V) is
the
!> > !> multiplicative inverse of a+V. Hence we have a field.
!> > !>
!> > !> Note that the complex numbers are embedded (many ways) in
quartenions.
!> > If
!> > !> we take any unit vector U and consider quartenions of the form a+bU
!> (a,b
!> > !> real) the result is isomorphic to the complex numbers. So any
program
!> > that
!> > !> provides for quartenion arithmetic also provides for complex
!> arithmetic.
!> > !>
!> > !> --
!> > !> Tom Gutman
!> > !>
!> >
!> >
!>
!> --
!> Tom Gutman
!>
!>
!>
!>
!>
Keith Farmer
> But it does. Exactly what property is needed for a vector space that is
I was not talking spaces, merely vectors as commonly used.
Extend it upward toward matrices, and you loose commutivity, etc. For
those reasons, I can not consider complex numbers to be the same as
vectors in the common sense -- they do not behave the same
mathematically. As was pointed out earlier, Z^-1 can exist, whereas
V^-1 does not. I'll stipulate they are both tensors, but not both
vectors, barring abstraction toward Hermitian functions and the like.
> I seem to remember writing complex numbers as (a,b), with 1 defined as (1,0)
> and i defined as (0,1). This is an exposition of complex numbers that
> starts with their properties as a vector space with a chosen basis, and then
> an additional multiplication operator is defined to make it a field.
But, as you can test, the definition of multiplication is different for
the two.
> A rather interesting question is why, of all the vector spaces over the
> reals, only the two and four dimensional cases can be extended to form a
Actually, I rather like the exercise that shows that stable orbits
obeying r^-2 appear only in 4D spacetime.
Tom Gutman
news:387130C3...@pacbell.net...
> Tom Gutman wrote:
> > But it does. Exactly what property is needed for a vector space that is
>
> I was not talking spaces, merely vectors as commonly used.
Ahh, yes. The physicists viewpoint (and for this purpose I include
engineers with physicists). You consider the vector to be the physical
quantity with magnitude and direction and identify that with the
mathematical vectors (which are just elements of vector spaces). But
mathematical vectors are rather different and much more general. As is
common with mathematics, low dimensional vector spaces can be mapped into
the physical quantities with the vector operations having useful
interpretations in the physical world.
>
> Extend it upward toward matrices, and you loose commutivity, etc. For
> those reasons, I can not consider complex numbers to be the same as
> vectors in the common sense -- they do not behave the same
> mathematically. As was pointed out earlier, Z^-1 can exist, whereas
> V^-1 does not. I'll stipulate they are both tensors, but not both
> vectors, barring abstraction toward Hermitian functions and the like.
>
> > I seem to remember writing complex numbers as (a,b), with 1 defined as
(1,0)
> > and i defined as (0,1). This is an exposition of complex numbers that
> > starts with their properties as a vector space with a chosen basis, and
then
> > an additional multiplication operator is defined to make it a field.
>
> But, as you can test, the definition of multiplication is different for
> the two.
Plain vector spaces do not have an proper multiplication (in a group
theoretical sense). The dot product has a range that is separate from the
domain - as such even the concepts of associativity and an identity element
do not exist (and you cannot have inverses without an identity element).
The cross product only exists for three dimensional vectors, and it also
fails to have an identity element (or associativity).
It is possible to add additional operations to a vector space - this in no
way detracts from the properties, or usefulness, of the original vector
space. In the case of two and four dimensions you can add a multiplication
operation that makes the vector space a field. You can add multiplication
operations with identies and associativity for other dimensionalities (for
dimensions of the form n^2, matrix multiplication does the job), but these
generally fail to have inverses for all elements. Whether any such
operation has any useful physical interpretation is a separate question.
Complex numbers are a two dimensional vector space, and can be used to
represent euclidean two space. It also turns out to be a very convenient
way to do two dimensional geometry and physics. That is because complex
muliplication has a useful physical meaning - scaling and rotation. An
example:
Uniform circular motion around the origin can be represented by the
parametric equation X=Re^(ikt). This can be differentiated to get
V=X'=ikRe^(ikt). Since multiplication by i is a 90 degree rotation, this
immediately shows the velocity at right angles to the radius.
Differentiating again, we get a=V'=-(k^2)Re^(ikt).
Many geometric problems get reduced to algebraic ones with a single variable
(as contrasted to the two variables using just cartesian coordinates).
The ordinary vector operation can even be recovered from the field
operations. The dot product of A and B is just Re(AxCon(B)). I think, but
have not verified, that the magnitude of the cross product is Im(AxCon(B)).
It's rather interesting that while the cross product only exists in
three-space, the magnitude of the cross product exists in any finite
dimensional space.
>
> > A rather interesting question is why, of all the vector spaces over the
> > reals, only the two and four dimensional cases can be extended to form a
>
> Actually, I rather like the exercise that shows that stable orbits
> obeying r^-2 appear only in 4D spacetime.
--
Tom Gutman
Keith Farmer
Tom Gutman wrote:
> Ahh, yes. The physicists viewpoint (and for this purpose I include
> engineers with physicists). You consider the vector to be the physical
"For this purpose" is a good way to state it -- even we're different beasts.
> quantity with magnitude and direction and identify that with the
No, I don't consider them to be (necessarily) physical, but I do
consider the bases to have an existance similar to that for units. EG:
the vector [1 0 0] differs from 1, in the same way that 1_m differs from
both 1 and 1_s.
> The cross product only exists for three dimensional vectors, and it also
> fails to have an identity element (or associativity).
Actually, on this point, the vector analysis class I took (it was under
the MATH department, should you balk at the following), put forth the
generic definitions for dot and cross products in n dimensions. If I
had the book at my desk, I would give you the reference, but
unfortunately it's about 1600 miles away.
> way to do two dimensional geometry and physics. That is because complex
> muliplication has a useful physical meaning - scaling and rotation. An
> example:
And complex number have other, deeper, uses. I'm not denying their
usefulness, merely the blanket statement that they're just like vectors.
> operations. The dot product of A and B is just Re(AxCon(B)). I think, but
AxCon() is defined as?
On a tangent -- a friend in college was discussing a seminar he'd
attended covering fractional dimension geometry. I've not seen any
references otherwise: you have a precis of the concept?
Tom Gutman
>
>
> Tom Gutman wrote:
> > Ahh, yes. The physicists viewpoint (and for this purpose I include
> > engineers with physicists). You consider the vector to be the physical
>
> "For this purpose" is a good way to state it -- even we're different
beasts.
>
> > quantity with magnitude and direction and identify that with the
>
> No, I don't consider them to be (necessarily) physical, but I do
> consider the bases to have an existance similar to that for units. EG:
> the vector [1 0 0] differs from 1, in the same way that 1_m differs from
> both 1 and 1_s.
Good analogy. Physical quantities exist independently of units (a moving
object has a velocity), but when we want to represent the quantities we must
provide units. Similarly vectors exist independently of a basis set, but
when we want to represent the individual vectors, we need some coordinate
system - for conventional representations that is provided by the basis set.
The real number 1 and the vector [1 0 0] and the complex number (1,0) are
all different (as even the HP49 knows). But mathematicians (to some extent)
and physicists (more so) and engineers (even more than that) like to abuse
(and simplify) notation. All of them are generally quite happy to use the
symbol "1" for both the real number 1 and the complex number (1,0) (or
1+0*i, if you prefer that notation), letting the context make the
distinction (if, indeed, it makes a difference).
Note that even as physical quantities, the basis set for vectors is not
really inherent in the vectors. The size of the basis vectors is set by our
(arbitrary) choice of units. the direction of the basis vector subject to
the usual orthogonality constraints) is completely arbitrary. It is usual
in a real problem to choose a coordinate system that simplifies the geometry
or the physics.
Operations differ as to how much they depend on a coordinate system. The
operations that define a mathematical vector space, addition and scalar
multiplication, do not require a coordinate system. The dot product depends
on the scaling of the basis vectors (i.e. requires the definition of the
unit sphere) but is invariant with respect to rotations and reflections of
the coordinate system. This is similar to working with physical quantities,
where we can do things like doubling a velocity without concern for the
units, but calculating an area depends on the units. A cross product is
invariant with respect to rotation of the coordinate system, but not
reflection. The field multiplication added to vectors for complex numbers
and quartenions depends on a specific coordinate system (at least up to
selecting the "scalar" component).
>
> > The cross product only exists for three dimensional vectors, and it also
> > fails to have an identity element (or associativity).
>
> Actually, on this point, the vector analysis class I took (it was under
> the MATH department, should you balk at the following), put forth the
> generic definitions for dot and cross products in n dimensions. If I
> had the book at my desk, I would give you the reference, but
> unfortunately it's about 1600 miles away.
Dot products exist for any number of dimensions, but since the result is a
scalar, not a vector, it cannot have an identity element. I don't see how
the full cross product can exist in other than three dimensions. The cross
product (as a vector) is required to be orthogonal to both the original
vectors. In one and two space, there simply is nothing orthogonal to two
arbitrary vectors, and in four and higher space there are too many of them.
As I noted, the magnitude of the cross product can be defined (and
calculated and is useful), but this is a scalar, not a vector. Regardless,
the cross product cannot have an identity element - as a vector, a vector is
not orthogonal to itself, and as a scalar it is not even a vector.
>
> > way to do two dimensional geometry and physics. That is because complex
> > muliplication has a useful physical meaning - scaling and rotation. An
> > example:
>
> And complex number have other, deeper, uses. I'm not denying their
> usefulness, merely the blanket statement that they're just like vectors.
Ah, the deep philosphy ... where to start?
"just like vectors". Now there's a loaded phrase. My claim is that complex
numbers are vectors. They obey all of the rules for vectors and can be used
for all purposes that plain two dimensional vectors can be used for. But
they are not "just" vectors. They are vectors+. The + is a multiplication
operation that makes them a field (as well as a vector space).
I also note that this multiplication has geometric meaning, and is quite
useful in geometric problems. This is not always the case for
multiplications imposed on vector spaces. For instance, 3x3 matrices form a
nine dimensional vector space. But I don't think that the normal matrix
multiplication has any intresting geometric interpretation in nine space.
But in the case of two dimensions, complex multiplication should not be
ignored when considering two dimensional geometry (and physics base on
that).
And certainly the complex numbers have many uses and interpretations. And
for many of them their geometric properties are not relevant. Again, while
they are vectors, they are not just vectors.
Actually, I was not really objecting to your statement that the dot and
cross products do not have inverses. Only to the implication that vectors
cannot be enhanced so as to have inverses (in some useful way). That is a
bit too limiting - although it is natural with a purely geometric view of
vectors.
>
> > operations. The dot product of A and B is just Re(AxCon(B)). I think,
but
>
> AxCon() is defined as?
Sorry - bad notatioin with a limited character set - I was using the x for
multiplication. Try A*Con(B), where Con is the complex conjugate.
>
> On a tangent -- a friend in college was discussing a seminar he'd
> attended covering fractional dimension geometry. I've not seen any
> references otherwise: you have a precis of the concept?
I've seen a little bit. One concept of dimensionality is based on measure.
If you take a simple one dimensional curve (embedded in whatever dimensional
space you like) its measure (length) varies linearly with the size of your
ruler (i.e. halve the length of the basis vectors, and the length doubles).
If you take a surface, the measure turns up as the area, and varies as the
square of the scale - so a surface is two dimensional. There are objects (I
think they are all created as limits of one sort or another) which scale
with fractional exponents. In that sense they can be said to have a
fractional dimensionality. And that is about all I actually know and
understand about that.
--
Tom Gutman
Keith Farmer
Tom Gutman wrote:
> vectors. In one and two space, there simply is nothing orthogonal to two
> arbitrary vectors, and in four and higher space there are too many of them.
If a vector is intended as a result, I'll agree. If, however, we allow
higher-order constructs -- planes, spaces, hyperspaces -- to be returned
from an n-dimensional cross product (A x B x C x ...) then where does
that lead?
> Ah, the deep philosphy ... where to start?
Maxwell's a good place, I hear, although I'm still overcoming my trauma
over that part of E&M.
> they are not "just" vectors. They are vectors+. The + is a multiplication
> operation that makes them a field (as well as a vector space).
I'll accept them as an extension. Altering/Adding operations will
change the behavior to differentiate between them.
> Sorry - bad notatioin with a limited character set - I was using the x for
> multiplication. Try A*Con(B), where Con is the complex conjugate.
No prob -- I gave up on x in frustration myself.
> square of the scale - so a surface is two dimensional. There are objects (I
> think they are all created as limits of one sort or another) which scale
> with fractional exponents. In that sense they can be said to have a
> fractional dimensionality. And that is about all I actually know and
> understand about that.
That makes sense -- I wasn't entirely sure how they were intending the
idea of a dimension to be carried. Physics and cosmology, thankfully,
seem limited to integer dimensions. Actually, modern physics as a whole
seems limited to integers of some group of factors, which really does
make things a whole lot easier to deal with than, say, Newtonian.
Veli-Pekka Nousiainen
Your article gave me an idea of putting units in a vector (or matrix).
Using lists and AXL I managed to do something useful.
ACO: Better support is needed !
--
Regards, VPN
_________________________________________________________
Veli-Pekka Nousiainen ; e-mail= v...@fcsolutions.com
Sokinsuontie 3 A 1, FIN-02760 Espoo, Finland
TEL, WORK= +358 (9) 859 2025 ; (WORK2= +358 (3) 4728 300)
Future Computing Solutions Oy ; URL= http://www.eiffel.fi
_________________________________________________________
HP25,HP28S,HP41CX,HP48SX,HP48GX,HP49G,HP71B,HP75C & TI89
Vote for the "82484A Curve Fit for HP71B" => HP49G !!!
Tom Gutman <Tom_G...@compuserve.com> wrote in message
news:850i0b$ltv$1...@ssauraac-i-1.production.compuserve.com...
<SNIP>
> Note that even as physical quantities, the basis set for vectors is not
> really inherent in the vectors. The size of the basis vectors is set by
our
> (arbitrary) choice of units. the direction of the basis vector subject to
> the usual orthogonality constraints) is completely arbitrary. It is usual
> in a real problem to choose a coordinate system that simplifies the
geometry
> or the physics.
>
<SNIP>
Tom Gutman
> In haste --
>
> Tom Gutman wrote:
two
> > arbitrary vectors, and in four and higher space there are too many of
them.
>
> higher-order constructs -- planes, spaces, hyperspaces -- to be returned
> from an n-dimensional cross product (A x B x C x ...) then where does
> that lead?
Let's see .... As usual when extending something that is essentially
unique, one must decide which aspects to keep and which to give up. The
cross product is a dyadic function. Its magnitude has a simple geometric
interpretation as the area of a parallelogram. The result, while not
exactly a vector, is close enough that it can be treated as a vector and
further vector operations performed.
Returning spaces (of whatever dimensions) does not seem appropriate. That
loses any concept of having a magnitude as a result. While the concept of
the space spanned by a set of vectors, and the space orthogonal to the
spanning space, is a useful one, I don't see it as a generalization of the
cross product.
Trying to obtain a specific vector in the orthogonal space could be of
interest. In the case of four space, to get a vector in the orthogonal two
space would require somehow generating two magnitudes that are somehow
related to the cross product concept. I am at a complete loss to see how
that would be done.
Giving up the dyadic nature does seem interesting. An n-way cross product
would presumably have a magnitude representing the hypervolume of an
n-dimensional parallelopiped. In an n+1 dimensional space, the n-cross
product could again be represented as an orthogonal vector back in the
original space.
Calculus seems quite happy with functions of n variables. But I have not
seen any extensions to algebra (group theory and its extensions) beyond
dyading operations. I suppose someone must have worked on them somewhere.
One of the interesting aspects of cross products is that they do seem to
have useful physical interpretations (angular momentum comes to mind). If
we consider physics in four dimensions, we can take three spacial vectors,
combine them to get a volume, and represent that volume along the time
dimension. While this sounds intriguing, it also seems extrememly odd - is
time actually volume????
>
[SNIP]
--
Tom Gutman
Keith Farmer
> the space spanned by a set of vectors, and the space orthogonal to the
> spanning space, is a useful one, I don't see it as a generalization of the
> cross product.
It depends on what you're putting in, and what you're expecting out.
One way of looking at the 3-space cross product is that you give it
characteristic vectors defining two families of 2-spaces (planes), and
finding a vector characteristic of a family of 2-spaces orthogonal to
the inputs (if it exists). A 4-space cp would take three families of
cubes and return the family of cubes orthogonal, etc. Generalizing, the
n-space cp would take various (n-1)-hedrons and return the orthogonal
(n-1)-hedron. In none of this are we actually having to specify a
particular point -- just a member of a family.
> Trying to obtain a specific vector in the orthogonal space could be of
> interest. In the case of four space, to get a vector in the orthogonal two
> space would require somehow generating two magnitudes that are somehow
> related to the cross product concept. I am at a complete loss to see how
> that would be done.
I see the problem as similar to getting solutions to differential
equations -- we obtain the general family of solutions, and then (if
necessary) apply constraints if we want a particular solution. But, as
in mathematics, theoretical physics is less concerned with particular
solutions than with general solutions
> Giving up the dyadic nature does seem interesting. An n-way cross product
> would presumably have a magnitude representing the hypervolume of an
> n-dimensional parallelopiped.
I believe that was mentioned in my vector analysis text. It's certainly
consistent, dimensionally.
> dimension. While this sounds intriguing, it also seems extrememly odd - is
> time actually volume????
I'm certainly not Hawking, but I suspect the analogue would represent
some sort of momentum within spacetime (one such quantity, momenergy, is
a constant). The derivation of momenergy, though, doesn't involve any
cross-product I know of. But is there a higher-order cross product
(akin to position x velocity) that will remain a constant? I don't
think I've seen anything mentioned.