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On 7 Jul 2008 17:16:26 -0400, Jim Henderson <nos### [at] nospam com> wrote:
>On Mon, 07 Jul 2008 22:08:24 +0100, Stephen wrote:
>
>> Ke-wrist, Someone did that to me once but it involved acid and bad trips
>> Ho, ho, ho!
>
>He does that sort of thing sober.
>
Ho, ho, ho! :)
>> Strange, as you get older these jokes seem less funny.
>
>I s'pose that could happen. He's not a spring chicken, though.
>
Pray that it does before he becomes an old broiler :)
>Yeah, I think that might've been a concern if he hadn't been on a stage.
>I think that separation from the audience helped avoid something like
>that - though kinda surprising nobody called 911....
>
Yeah that would have served him right. Especially if he got the bill from the
emergency services ;)
Maybe I'm being to hard on him. It is good to have a sense of humour, so I'm
told :)
>>>He also makes an *excellent* chili. :-)
>>
>> As you say, from my part of the world :)
>
><g> We have a party at my house each year (the show is in town) after
>the show's done, and he usually cooks for us. He's taken a class in
>Oxaca recently (and going to take another this October), so it evolves a
>little bit each year. We stayed with him and his wife both times we went
>to the UK, and that's how it got started - he made it for dinner one
>night.
Is that Mexican food? I've never tried it. I liked the Cuban food when I was in
Florida but not when I went to Havana for a weekend.
--
Regards
Stephen
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On Wed, 09 Jul 2008 14:49:52 +0100, Stephen wrote:
>>He does that sort of thing sober.
>>
> Ho, ho, ho! :)
Not that he doesn't like a pint or a glass (or three) of wine when the
time is right. :-)
>>I s'pose that could happen. He's not a spring chicken, though.
>>
> Pray that it does before he becomes an old broiler :)
I think the phrase "young at heart" applies - but yes, hoping for a good
long life for him.
> Yeah that would have served him right. Especially if he got the bill
> from the emergency services ;)
Here in the US, it wouldn't have been much. I think the company would
probably have covered it. But also, there is perhaps an irony that
people may not have dialed for fear of causing more harm. ;-)
> Maybe I'm being to hard on him. It is good to have a sense of humour, so
> I'm told :)
A little, there again, you had to be there....
>><g> We have a party at my house each year (the show is in town) after
>>the show's done, and he usually cooks for us. He's taken a class in
>>Oxaca recently (and going to take another this October), so it evolves a
>>little bit each year. We stayed with him and his wife both times we
>>went to the UK, and that's how it got started - he made it for dinner
>>one night.
>
> Is that Mexican food? I've never tried it. I liked the Cuban food when I
> was in Florida but not when I went to Havana for a weekend.
Yeah; I'm pretty sure I spelt the name wrong (or as Shaun would say,
"wrongly".) Good Mexican food is actually quite bland - the hot stuff
tends to be Tex-Mex instead. I spent a couple weeks in Mexico back in
the early 80's and *really* liked the food I had there. There's a couple
places in Salt Lake City that do good Mexican food as well, might have to
track some down for lunch today. Mmmm...
Jim
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Invisible wrote:
>>> OK, so I can't spell very well. I think we've established that one. It's
>>> actually a miracle I can read or write at all! :-P
>>
>> Given the subject of this thread, shouldn't you try to better yourself
>> at this instead of just using the "I can't spell very well" defense? ;)
>
> And who says I'm not trying?
>
> Either way, laughing at me isn't helping anything, is it? :-P
>
You have an important point there. What if those kids who say "I'm not good
at math" were once laughed at when giving a silly math answer?
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scott wrote:
> It makes things later in life, assuming you go on with studying maths, a
> lot
> easier. At university I used to cringe at the people who used their
> calculators so often to do what they really should be doing in their head.
> When you are tackling a large problem it is a huge advantage do be able to
> do at least the basics very quickly in your head.
And it's very easy to lose practice. I used to be good at maths at school.
Now (like 2 years later), I sometimes take a few seconds to solve those
silly "6+7" anti-spam tests. Probably I wouldn't take too much time to get
my brain in practice again, but...
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Invisible wrote:
>>> It is not immediately clear to me that the quotient of two
>>> polynomials is necessarily a polynomial.
>>
>> Nope, same as dividing two integers doesn't always give an integer
>> answer, you sometimes end up with fractional parts in the answer.
>
> Right. So there should also be a polymonial modulus operator?
Indeed there is. It is a useful operation in abstract algebra,
particularly in extending sets of numbers so that they contain elements
with useful algebraic properties. For instance, if you consider
polynomials over the real numbers modulo X^2-1, you get the same
structure as the complex numbers (you can do a similar thing for
quaternions).
In terms of real-world uses, iirc the polynomial modulus is valuable in
in information coding. For example modular multiplication of
polynomials is a step in the AES cypher and a CRC checksum is
essentially a polynomial modulus.
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>> Right. So there should also be a polymonial modulus operator?
>
> Indeed there is.
Ooo...
> For instance, if you consider
> polynomials over the real numbers modulo X^2-1, you get the same
> structure as the complex numbers (you can do a similar thing for
> quaternions).
Aren't quaternions non-associative?
> In terms of real-world uses, iirc the polynomial modulus is valuable in
> in information coding. For example modular multiplication of
> polynomials is a step in the AES cypher and a CRC checksum is
> essentially a polynomial modulus.
...and *this* is why I can't implement AES or CRC.
--
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*
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Invisible wrote:
>> For instance, if you consider polynomials over the real numbers modulo
>> X^2-1, you get the same structure as the complex numbers (you can do a
>> similar thing for quaternions).
>
> Aren't quaternions non-associative?
Non-commutative you mean, but yes. The quaternion representation using
modular polynomials considers polynomials of two variables (x and y) and
real coefficients where multiplication between x and y is
non-commutative. So the polynomials inherit their non-commutativity
from their variables.
Note that it's not actually important to say what sorts of numbers x and
y would actually represent -- we're just dealing with polynomials in
their purest form, where *all* we know about x and y are they have
operators like addition, subtraction, and multiplication which satisfy a
few basic mathematical properties:
http://en.wikipedia.org/wiki/Ring_(mathematics)
> ...and *this* is why I can't implement AES or CRC.
It's not so difficult as you might think, it's just takes some getting
used to the math. If you're interested in the implementation rather
than the math end of things, Wikipedia has a typically verbose article
on the subject: http://en.wikipedia.org/wiki/Computation_of_CRC
There's also some C code dealing with the polynomials used in AES at the
bottom of this page: http://en.wikipedia.org/wiki/Finite_field_arithmetic
Also forgot to mention that (if I recall correctly) ideas related to
this play an important role in some of the operations in computer
algebra systems.
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>> Aren't quaternions non-associative?
>
> Non-commutative you mean, but yes.
Oh. Maybe I'm thinking of hypercomplex numbers then? I recall that it's
not possible to generalise the complex field to 3 or 4 dimensions
(although I have no idea why), so all the generalisations people have
come up with fail to be fields.
By complete coincidence, I spent my afternoon today reading about magmas
and groups and rings and quasigroups and semigroups and groupoids and
rings and skew rings and splitting fields and fields of factors and
polynomial rings and that whole zoo of other things... Apparently
"vector field" doesn't mean what you'd expect at all! :-D
I've spent a while looking at group theory. It's not "hard", it's just
that there are lots and lots of terms to learn, mostly with unintuitive
names, and the whole system of groups constructed from groups
constructed from groups gets confusing quite fast. Similarly, the stuff
with rings and fields and so forth isn't "hard", there's just lots of
stuff to remember...
> Also forgot to mention that (if I recall correctly) ideas related to
> this play an important role in some of the operations in computer
> algebra systems.
Mathematica directly uses the commutivity and associativity of various
operators, not to mention distributivity. Indeed there are built-in
methods for telling Mathematica that your new function is associative
and/or commutative. (Distributive requires explicit transformation rules...)
--
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*
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Orchid XP v8 wrote:
> I've spent a while looking at group theory. It's not "hard", it's just
> that there are lots and lots of terms to learn, mostly with unintuitive
> names, and the whole system of groups constructed from groups
> constructed from groups gets confusing quite fast. Similarly, the stuff
> with rings and fields and so forth isn't "hard", there's just lots of
> stuff to remember...
Yeah, once you get used to it it's not to bad. I think that proving
things in it is a bit trickier though, as it seems to involve deep leaps
of logical intuition more than other areas such as analysis. It's sort
of fun like that though, as you can actually do interesting math without
having to acquire a ton of background knowledge.
>> Also forgot to mention that (if I recall correctly) ideas related to
>> this play an important role in some of the operations in computer
>> algebra systems.
>
> Mathematica directly uses the commutivity and associativity of various
> operators, not to mention distributivity. Indeed there are built-in
> methods for telling Mathematica that your new function is associative
> and/or commutative. (Distributive requires explicit transformation
> rules...)
bases (a useful took for computer algebra systems), but indeed being
careful about how you deal with the basic algebraic properties like
commutativity and associativity is more fundamental to such programs.
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Orchid XP v8 wrote:
> By complete coincidence, I spent my afternoon today reading about magmas
> and groups and rings and quasigroups and semigroups and groupoids and
> rings and skew rings and splitting fields and fields of factors and
> polynomial rings and that whole zoo of other things... Apparently
> "vector field" doesn't mean what you'd expect at all! :-D
And I've yet to figure out how to solve t out of
(X0 + Xd * t - Xc)^2 + (Y0 + Yd * t - Yc)^2 + (Z0 + Zd * t - Zc)^2 = Sr^2
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