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Newsgroups: sci.math
From: Ludwig.Plutonium@dartmouth.edu  (Ludwig Plutonium)
Subject: Re: FermatÕs Last Theorem
Message-ID: 
Organization: Dartmouth College, Hanover, NH
References: <2728f8$51j@news.u.washington.edu>
  
<278vgj$pi2@paperboy.osf.org> 
 
Date: Fri, 17 Sep 1993 03:19:22 GMT
Lines: 44
In article 
Ludwig.Plutonium@dartmouth.edu  (Ludwig Plutonium)
>In article <278vgj$pi2@paperboy.osf.org> karl@dme3.osf.org (Karl 
>Heuer) writes:
>
>>Ludwig.Plutonium@dartmouth.edu  (Ludwig Plutonium) writes:
>>>The eventual arithmetic proof of FLT, I am confident, will come 
>>>from the counting numbers; P-triples are possible only in exp2 
>>>because 2+2=2x2=4.
>>
>>I have my doubts as to the connection between that equation and 
>>FLT; however, you may be interested to know that other solutions 
>>are possible if you allow those left-infinite decimal strings that 
>>we discussed earlier. When k=4, there is a unique nonzero solution 
>>to N+N+N+N = N*N*N*N = M. Here is the answer, worked out to 60 
>>decimal places. You can check it by doing the arithmetic yourself, 
>>right to left.
>>
>>  N = . . .8217568575974462578891103859665245689398767183
>>            82655349981184
>>  M = . . .2870274303897850315564415438660982757595068735
>>            30621399924736
>>
>>Karl Heuer   karl@osf.org
>
>   Karl Heuer double bless you to the infinite Fields of Elysium. I >would not mind if you discovered the worldÕs first valid proof of >FLT, instead of me.
>   Karl can you do the same thing for exp3 and exp5, i.e., a unique >solution?
    Karl I think the proof would then go like this. Take any exp
greater than 2, then when there are rational solutions to FLT those are
turned into infinite integers by just deleting the decimal point. Near
the end of the proof would be something that only with finite integers
is a Ptriple possible because only 2+2=2x2=4.
    Then again I could be all wrong and there in fact exists a
counterexample to FLT provided that one considers infinite integers are
no different from finite integers. That is, finite integers are
infinite integers with just infinite repetition of zeroes to the left.
WOULD THAT NOT BE THE SUPREME IRONY SO FAR IN THE HISTORY OF MATH. That
there is a counterexample to FLT. The whole world will laugh
hysterically if Wiles gets approval and Ludwig Plutonium comes up with
the counterexample. Which choice would you pick--- a 1000 page math
community accepted (fake) proof, or a counterexample? So far my
confidence in the math community is that they would prefer the 1000
page ordeal.
-------------------------------------------------------------
Newsgroups: sci.math
From: Ludwig.Plutonium@dartmouth.edu  (Ludwig Plutonium)
Subject: Re: FermatÕs Last Theorem
Message-ID: 
Organization: Dartmouth College, Hanover, NH
References: <2728f8$51j@news.u.washington.edu>
  
<278vgj$pi2@paperboy.osf.org> 
Date: Fri, 17 Sep 1993 03:44:56 GMT
Lines: 9
In article <278vgj$pi2@paperboy.osf.org> karl@dme3.osf.org (Karl Heuer)
writes:
>
> N = . . .8217568575974462578891103859665245689398767183
>            82655349981184
> M = . . .2870274303897850315564415438660982757595068735
>            30621399924736
   LET US FIND A CROP OF COUNTEREXAMPLES TO FLT.
Ha ha ha ha ha ha ha ha ha ha ha ha ha ha ha ha ha ha ha ha ha ha ha!
-------------------------------------------------------------
EMAIL
From: ÒTerry TaoÓ 
Date: Sat, 18 Sep 93 09:55:36 EDT
To: Ludwig.Plutonium@Dartmouth.EDU
Subject: Re: FermatÕs Last Theorem
Newsgroups: sci.math
In-Reply-To: 
References: <2728f8$51j@news.u.washington.edu>
  
<278vgj$pi2@paperboy.osf.org> 
 
Organization: Princeton University
Cc:
In article  you write:
>In article  
>Ludwig.Plutonium@dartmouth.edu  (Ludwig Plutonium)
>
>>In article <278vgj$pi2@paperboy.osf.org> karl@dme3.osf.org (Karl 
>>Heuer) writes:
>>
>>>Ludwig.Plutonium@dartmouth.edu  (Ludwig Plutonium) writes:
>>>>The eventual arithmetic proof of FLT, I am confident, will come 
>>>>from the counting numbers; P-triples are possible only in exp2 
>>>>because 2+2=2x2=4.
>>>
>>>I have my doubts as to the connection between that equation and 
>>>FLT; however, you may be interested to know that other solutions 
>>>are possible if you allow those left-infinite decimal strings that 
>>>we discussed earlier. When k=4, there is a unique nonzero solution 
>>>to N+N+N+N = N*N*N*N = M. Here is the answer, worked out to 60 
>>>decimal places. You can check it by doing the arithmetic yourself, 
>>>right to left.
>>>
>>>  N = . . .8217568575974462578891103859665245689398767183
>>>            82655349981184
>>>  M = . . .2870274303897850315564415438660982757595068735
>>>            30621399924736
>>>
>>>Karl Heuer   karl@osf.org
>>
>>   Karl Heuer double bless you to the infinite Fields of Elysium. I 
>>would not mind if you discovered the worldÕs first valid proof of 
>>FLT, instead of me.
>>   Karl can you do the same thing for exp3 and exp5, i.e., a unique 
>>solution?
No.
Theorem. The equation N+N+N=N*N*N has no solution in 10-adics, apart
from N=0.
Proof: consider the powers of 2 and 5 in N. Suppose 2 divides N a times
and 5 divides N b times. The lhs of the above equation has 2^a 5^b as
its factors of 2 and 5 (which are by the way the only primes in
10-adics), and the rhs has 2^3a 5^3b as its factors, hence a and b must
be 0.
But then, if neither 2 or 5 divides N, then N must be invertible,
unless N=0. Thus, dividing by N, we get N*N = 3. But comparing the
final digits of both sides, we see that this is impossible.
Similarly: The equation N+N+N+N+N=N*N*N*N*N has no solution in
10-adics, apart from N=0.
Proof. Suppose 2^a5^b are the prime factors of N, again. Then the lhs
has prime factors of 2^a 5^(b+1) and the rhs has prime factors of 2^5a
5^5b. But these can never match, hence there is no solution (unless
N=0; 0 is the only number that has non-unique prime factorization).
The fact that  N+N+N=N*N*N has no solutions in 10-adics, whereas there
ARE solutions of FLT in 10-adics for n=3 (see for example the post by
William Schneeberger), shows that there is no proof that ÒFLT is true
for n=3 => N+N+N=N*N*N for some non-zero NÓ unless you use a property
of the integers that the 10-adic integers do not have.
>
>    Karl I think the proof would then go like this. Take any exp 
>greater than 2, then when there are rational solutions to FLT those 
>are turned into infinite integers by just deleting the decimal point.
An important point here: the operation of turning rational numbers into
infinite integers by deleting the decimal point does NOT preserve
addition or multiplication. For example, in rationals
.33333...   x  .33333....  =  .11111....
whereas
....33333  x  ....33333  =  .....88889
and
.5555....  +  .4444...   = 1
whereas
....5555    +  ....4444    = ....9999
Thus, a rational solution of FLT does not automatically lead to a
10-adic solution of FLT.
>Near the end of the proof would be something that only with finite 
>integers is a Ptriple possible because only 2+2=2x2=4.
I would very much like to see a proof of this statement: if you can
prove this, then you have proved FLT. Then again, see an above point
that you would need to use a property of the integers that is not
shared by the 10-adic integers.
>   Then again I could be all wrong and there in fact exists a 
>counterexample to FLT provided that one considers infinite integers 
>are no different from finite integers.
What you mean here is that there exists a counterexample to FLT in
infinite integers. It is not quite correct to say that Òinfinite
integers are no different from finite integersÓ. Every finite integer
is a 10-adic integer, but not conversely. What is true is that
multiplication and addition are the same operation for both of them.
However, finite integers have several properties that 10-adic integers
do not have, for example, they are all finite. Another is that
induction works for finite integers, but not for 10-adic integers. 
(otherwise, you could prove that all 10-adic integers were finite by
induction).
>   That is, finite integers are infinite integers with just infinite 
>repetition of zeroes to the left. WOULD THAT NOT BE THE SUPREME 
>IRONY SO FAR IN THE HISTORY OF MATH. That there is a 
>counterexample to FLT.
The commonly accepted wording of FLT ends Ò... where a, b, c, n are
(finite) integersÓ (with the finite added for emphasis). If you remove
this last phrase, then the FLT that most mathematicians think of would
then have to be called ÒFLT for integersÓ. It is true that FLT is false
for p-adics, matrices, quaternions, and a lot of other number systems.
In this sense, there are counter examples to the general FLT. But there
is no counter example to FLT (integers): this was proved by Wiles.
>   The whole world will laugh hysterically if Wiles gets approval 
>and Ludwig Plutonium comes up with the counterexample. Which 
>choice would you pick--- a 1000 page math community accepted 
>(fake) proof, or a counterexample? So far my confidence in the math 
>community is that they would prefer the 1000 page ordeal.
There seems to be a point you keep missing. If you change the
definitions of terms (like integer, real, etc), then theorems change as
well. Thus,
ÒFLT for normal integersÓ (Wiles)
is a different theorem than
ÒFLT is not true for 10-adic integersÓ (proved by many people)
and both results (admittedly one is very long, the other very short),
are good mathematics and knowing one does not automatically get you the
other result. So there is no real irony, except that theorems that hold
for one number system need not hold for all number systems.
Of course, you may dispute that the commonly accepted definition of
ÒintegerÓ SHOULD be the commonly accepted definition. But even if you
replace the concept of integer, the ÒoldÓ concept of integer is still a
valid one, so you canÕt just blithely say (for example) Òwell, if I
redefine integers to be 10-adic, so the reals are now equal cardinality
to the integers, then there is no infinite set with smaller cardinality
than the reals anymoreÓ, because the ÒoldÓ notion of integer still
exists.
Terry
-------------------------------------------------------------
Newsgroups: sci.math
From: Ludwig.Plutonium@dartmouth.edu  (Ludwig Plutonium)
Subject: Re: Fermat's Last Theorem
Message-ID: 
Organization: Dartmouth College, Hanover, NH
References: 
<278vgj$pi2@paperboy.osf.org>  
<27glo6$elj@paperboy.osf.org> 
Date: Mon, 20 Sep 1993 15:39:46 GMT
Lines: 15
In article <27glo6$elj@paperboy.osf.org> karl@dme3.osf.org (Karl Heuer)
writes:
>In article  >Ludwig.Plutonium@dartmouth.edu  (Ludwig Plutonium) writes:
>>does this above monster 4N=N^4 repeat in a block like Rational >>numbers repeat
>
>No, it doesn't.
   Karl is this new number which you discovered (if you do not have a
name for it as yet, I suggest HeuerPu Numbers, but that is up to you)
analytic irrational or transcendental? Given that concepts of
transcendental can be translated to P-adic.
   Also, please tell me if there is a mirror reflection in the Reals of
HeuerPu Numbers. Is there a Real number between 0 and 1 which has
HeuerPu properties?
-------------------------------------------------------------
EMAIL
Date: Tue, 21 Sep 93 00:01:54 EDT
From: ÒKin ChungÓ 
To: Ludwig.Plutonium@Dartmouth.EDU
Subject: INFINITE INTEGERS
In-Reply-To: 
Organization: Princeton University
Cc:
Before you embrace the so-called Òinfinite integersÓ too closely,
consider this straightforward sum:
....9999999999999999999999999999999999999999999999999
+....0000000000000000000000000000000000000000000000001
___________________________________________
  ....000000000000000000000000000000000000000000000000
Using your identification of the (finite) integers with a subset of the
infinite integers, this shows that (-1) = ...9999. Do you see what IÕm
trying to get at?
-------------------------------------------------------------
From: karl@dme3.osf.org (Karl Heuer)
Newsgroups: sci.math
Subject: Re: Fermat's Last Theorem
Date: 21 Sep 1993 20:59:42 GMT
Organization: Open Software Foundation
Lines: 25
Message-ID: <27npvu$blc@paperboy.osf.org>
References: 
<27glo6$elj@paperboy.osf.org>  
In article 
Ludwig.Plutonium@dartmouth.edu  (Ludwig Plutonium) writes:
>(if you do not have a name for it as yet, I suggest HeuerPu Numbers, 
>but that is up to you)
I've been calling N^k=k*N the "LP equation". I don't think its
solutions need names of their own, but "LP numbers" will do for now.
>   Karl is this new number which you discovered analytic irrational or transcendental? Given that concepts of transcendental can be translated to P-adic.
It's easy to prove that it's irrational, because the rationals have the
same properties in the 10-adic numbers that they do in the reals. Since
it's a zero of the polynomial x^k-k*x, it's a (non-Real) irrational
algebraic number.
>Also, please tell me if there is a mirror reflection in the Reals
The LP equation has real solutions for all k; e.g. sqrt(3) for k=3. 
(As someone else already noted, these solutions will have magnitude
>1.)
There are similarities to the Reals, but it's not just a renaming. 1/3
exists as a (repeating) 10-adic integer, but it's not . . .3333; it's .
. .66667 instead. (Multiply it out: . . .66667 * 3 = . . .00001 no
matter how many places you carry it to.) Also, x^2 = 3 has a solution
in the Reals but not in the 10-adics; while x^2 = -31 has a solution in
the 10-adics but not in the Reals.
-------------------------------------------------------------
From: karl@dme3.osf.org (Karl Heuer)
Newsgroups: sci.math
Subject: Re: P-ADIC NUMBERS: RENAMED AS INFINITE INTEGERS
Date: 21 Sep 1993 21:14:12 GMT
Organization: Open Software Foundation
Lines: 12
Message-ID: <27nqr4$bpe@paperboy.osf.org>
References:  
In article 
Ludwig.Plutonium@dartmouth.edu  (Ludwig Plutonium) writes:
>   Let us rename the math subject "P-adic Numbers" to that of 
>"Infinite Integers."
I'd rather keep the existing terminology. There are several consistent
models of arithmetic that include objects that look "infinite" in some
sense: the Hyperintegers/Hyperreals, the Surintegers/Surreals, the
compact number line, the Riemann sphere, the transfinite ordinals, the
transfinite cardinals, etc.
The only thing that's "infinite" about the p-adic numbers is their
representation as a digit string, and that's analogous to the infinite
number of digits in a non-terminating Real.
-------------------------------------------------------------
Newsgroups: sci.math
From: Ludwig.Plutonium@dartmouth.edu  (Ludwig Plutonium)
Subject: Re: P-ADIC NUMBERS: RENAMED AS INFINITE INTEGERS
Message-ID:  
Organization: Dartmouth College, Hanover, NH
References:  
<27nqr4$bpe@paperboy.osf.org>
Date: Thu, 23 Sep 1993 16:34:43 GMT
Lines: 17
In article <27nqr4$bpe@paperboy.osf.org>
karl@dme3.osf.org (Karl Heuer)
>I'd rather keep the existing terminology. There are several >consistent models of arithmetic that include objects that look >"infinite" in some sense: the Hyperintegers/Hyperreals, the >Surintegers/Surreals, the compact number line, the Riemann >sphe,
the transfinite ordinals, the transfinite cardinals, etc.
>
>The only thing that's "infinite" about the p-adic numbers is their 
>representation as a digit string, and that's analogous to the infinite 
>number of digits in a non-terminating Real.
   How about TRANSFINITE INTEGERS? Any objections?
   I am trying to give a good name to these infinite strings for
another assault on CantorÕs claim that there are more than one type of
infinity.
-------------------------------------------------------------
From: "Terry Tao" 
Subject: Re: Wiles proof of FLT
To: Ludwig.Plutonium@Dartmouth.EDU (Ludwig Plutonium)
Date: Sun, 26 Sep 93 17:10:49 EDT
In-Reply-To: <5509284@blitzen.Dartmouth.EDU>; from "Ludwig Plutonium"
at Sep 26, 93 5:07 pm
>
> Terry tell me if all P-adic numbers have inverses. Can you prove it
>
If P is prime, then all numbers which are not multiples of P have
inverses. (in other words, all numbers whose last digit is not 0.)
If P is not prime, then all numbers which are coprime to P have
inverses, i.e. the last digit of that number is coprime to P.
To prove it, it is sufficient to show that you can invert the last N
digits, for each N. This is a standard exercise in modular arithmetic.
Terry
p.s. I would still like to hear your comment on my proof that there
must be a counter-example to FLT. Do you think my proof is flawed?
-------------------------------------------------------------
EMAIL
From: ÒTerry TaoÓ 
Date: Tue, 28 Sep 93 20:34:10 EDT
To: Ludwig.Plutonium@Dartmouth.EDU
Subject: Re: FermatÕs Last Theorem
Newsgroups: sci.math
In-Reply-To: 
References: <2728f8$51j@news.u.washington.edu>
  
<278vgj$pi2@paperboy.osf.org> 

Organization: Princeton University
Several observations.
(1). In my mind, the reason why 2 is exceptional in FLT is not because
2x2 = 2+2, but rather because 2 is even. If n is odd, then FLT can be
rewritten in the much more beautiful
u^n + v^n + w^n is never 0 unless uvw is 0 (where u,v,w are integers).
(2). P-adic counter-examples to FLT have been known for some time -
almost at the same time that they were discovered. P-adics are like
real numbers, in a sense: who's interested in a real number
counter-example to FLT?
(3). Wiles uses special properties of the finite integers that the
infinte integers do not have, one of which is that there are infinitely
many primes in the finite integers.
(4) Your statement "Wiles's proof contradicts the Fourier theorem" is
indirect non-existence - after all, that's what you said when I used
the same principle to show that FLT must be false for finite integers.
(5) FLT is true for finite integers, false for p-adic integers. Each
finite integer is a p-adic integer, but the set of finite integers is
only a SUBSET of the set of p-adic integers. They are different things,
and you have two different FLTs for two different number systems. FLT
is assumed to be over the finite integers unless otherwise specified,
so your statement that ÒAll proofs of FLT are fakeÓ is wrongly deduced.
However, you have made a true deduction in saying that no proof of FLT
can rely purely on algebraic manipulation, because of the p-adic
counter example. It must use a property that the finite integers have
but the infinite integers do not, for example
(a) induction;
(b) infinitude of primes;
(c) no zero divisors (WillÕs two numbers, a and b, multiply to 0)
etc.
Terry
-------------------------------------------------------------
EMAIL
From: ÒTerry TaoÓ 
Subject: Re: Fermat's Last Theorem
To: Ludwig.Plutonium@Dartmouth.EDU (Ludwig Plutonium)
Date: Tue, 28 Sep 93 22:22:46 EDT
In-Reply-To: <5569918@blitzen.Dartmouth.EDU>; from "Ludwig Plutonium"
at Sep 28, 93 10:18 pm
>
>--- You wrote:
>However, you have made a true deduction in saying that no proof of >FLT can rely purely on algebraic manipulation, because of the p-adic >counter example. It must use a property that the finite integers >have but the infinite integers do not, for
example
>--- end of quoted material ---
>Thanks that is important I needed that.
>
>Terry tell me if there is a Real analog of that number Karl produced. >Karl says it is greater than 1. Can you pinpoint it better.
>
the cube root of 4.
Terry
-------------------------------------------------------------
EMAIL
From: ÒWilliam SchneebergerÓ  
Subject: Re: your counterexample posting
To: Ludwig.Plutonium@Dartmouth.EDU (Ludwig Plutonium)
Date: Wed, 29 Sep 93 12:10:41 EDT
In-Reply-To: <5581074@blitzen.Dartmouth.EDU>; from ÒLudwig PlutoniumÓ
at Sep 29, 93 11:39 am
Sorry, I don't have a copy. But here's the deal:
We solve a == 0 (mod 5)
 a == 0 (mod 25)
 a == 0 (mod 125)
     .
     .
     .
and
 a == 1 (mod 2)
 a == 1 (mod 4)
 a == 1 (mod 8)
by the Chinese Remainder Theorem. Similarly we solve
 b == 0 (mod 5)
 b == 0 (mod 25)
 b == 0 (mod 125)
 b == 0 (mod 625)
     .
     .
     .
and
 b == 1 (mod 2)
 b == 1 (mod 4)
 b == 1 (mod 8)
     .
     .
     .
Now one can prove that a*a=a, b*b=b, a*b=0, a+b=1. This leads
immediately to the fact that, for all (finite natural numbers) n,
a^n + b^n = c^n where c == 1.
There are, however, much more interesting solutions to FLT in these
numbers. The above solution may well be considered trivial as abc == 0.
For the p-adic numbers (infinite integers in a prime base p) the above
solution does not exist. But I know that there do exist solutions for n
relatively prime to p(p-1).
Will
-------------------------------------------------------------
EMAIL
From: ÒTerry TaoÓ 
Subject: Re: Schneebergers post
To: Ludwig.Plutonium@Dartmouth.EDU (Ludwig Plutonium)
Date: Wed, 29 Sep 93 14:45:31 EDT
In-Reply-To: <5581425@blitzen.Dartmouth.EDU>; from "Ludwig Plutonium"
at Sep 29, 93 11:55 am
>
>Hi Terry. I lost Schneebergers post of counterexamples. Would you >have a copy? Please relay
>
I haven't got that post either, but here's how you can compute them:
The idempotents of the 10-adics are the solutions of a^2 = a. Thus
their first digit must be 6 or 5 (by considering the problem modulo 10)
- the idempotents 0 and 1 being discounted. Let us, say, consider the
one with last digit 5: they sum up to 1 anyway.
You can compute successive digits iteratively. If the next digit is a,
i.e. the last two digits are 10a+5, then the last two digits of the
square is 25, so a must be 2.
Similarly, if we let the next digit be b, so the last three digits are
100b + 25, then the last three digits of the square is 625, hence b =
6. And so on.
Terry
-------------------------------------------------------------
EMAIL
From: ÒTerry TaoÓ 
Subject: Re: Schneebergers post
To: Ludwig.Plutonium@Dartmouth.EDU (Ludwig Plutonium)
Date: Wed, 29 Sep 93 15:02:53 EDT
In-Reply-To: <5581425@blitzen.Dartmouth.EDU>; from "Ludwig Plutonium"
at Sep 29, 93 11:55 am
Actually, all you need to do is take 5 and keep squaring it. The powers
of 5 will converge in the 10-adic topology to one of Will's numbers.
(recall: whereas the metric in say the reals, is |x-y| for the distance
between x and y, the metric between two p-adics x and y is 1/p^n, where
n is the highest number of times that p divides x-y.)
Terry
-------------------------------------------------------------
EMAIL
From: ÒWilliam SchneebergerÓ  
Subject: Re: your counterexample posting
To: Ludwig.Plutonium@Dartmouth.EDU (Ludwig Plutonium)
Date: Wed, 29 Sep 93 15:21:43 EDT
In-Reply-To: <5581838@blitzen.Dartmouth.EDU>; from ÒLudwig PlutoniumÓ
at Sep 29, 93 12:15 pm
So, I guess, some solutions for exponent 3 are
              a == 1
              b == 10
              c == . . .52979382777667001
              a == 1
              b == 20
              c == . . .4437336001
              a == 1
              b == 30
              c == . . .4919009001
In fact for any finite n, a == 1, b a multiple of 10, we can solve the
equation of FLT.
But, look, all IÕve shown here is that in the 10-adic numbers there is
a solution to the equation. I have _not_ contradicted the statement of
FLT which says that there is no solution among the usual finite
integers. This is more of a problem.
Later.
-------------------------------------------------------------
Newsgroups: sci.math
From: Ludwig.Plutonium@dartmouth.edu  (Ludwig Plutonium)
Subject: Re: Wiles's proof of FLT
Message-ID: 
Organization: Dartmouth College, Hanover, NH
References: <27st80$asv@clipper.clipper.ingr.com>

Date: Wed, 29 Sep 1993 16:24:55 GMT
Lines: 15
In article 
Ludwig.Plutonium@dartmouth.edu  (Ludwig Plutonium) writes:
>FLT was outstanding because there is no proof of FLT in the general >case. The general theorem of FLT has no proof because transfinite >integers are just as real as finite integers. All attempts at a proof >of the general equation of FLT are doomed to
failure. 
   PROOF OF FLT. The general form of FLT where a^n+b^n=c^n are such
that the four numbers a,b,c,n could be transfinite integers as well as
finite integers. Hence a proof in the general case is impossible. The
counterexamples in the P-adics is the proof. Anything else would have
to restrict the four numbers a,b,c,n to finite number cases and show
that in those restrictions there are no P-triples. QED

In article <568bli$cu4@jerry.loop.net>, drum@loop.com (Dru Morgan) wrote:
> Is there a formula for finding the lenght of  a groove that goes around a 
> spiral (such as a record album)?  If you know the diameter of the record and 
> the distance (period?) between grooves, what would be the formula for the 
> length of the groove?  You would also have to subtract the part in the 
> middle where there is no groove.  Please make sure to cc-by-mail your
response 
> to me at drum@loop.com
> 
> Thanks.
> Dru Morgan
   This would work if it weren't for Dynatrack. This was a method of
increasing both record time and volume. The volume you get out of a record,
and consequently its signal to noise ratio, is dependent on the maximum
deviation of the groove. This would mean that a loud record would need to
have its grooves spaced further apart to prevent the grooves from cutting
into each other. This reduces recording length. 
   Old record cutting lathes could only be set to cut at a given groove
spacing, so that led to a tradeoff between signal to noise (volume) and
track length.(This is why 12" sound better than albums-they can be cut
'hotter' with really wide groove spaces.) Then came dynatrack, where the
lathe would take into account the volume of the music and automatically
increase the groove spacing to compensate. So the groove spacing actually
varies depending on the music's loudness.
-- 
Tohru sagt,"Spork!"

davk@netcom.com (David Kaufman) wrote:
>
>          For K-12 Students, Teachers And Others
>     Interested In Exploring Math, Science And Ethics
>   Through Collaboration For Enrichment And Achievement.
>------------------------------------------------------------
>        ----------------------------------------
>       |   Surf The Web For Free And Learn How  |
>       |   At The Science and Business Library  |
>        ----------------------------------------
>
>	The Science Library at 34th Street on Madison Ave in 
>NYC has 55 computers downstairs that are used to access the 
>Web with over 30 million sites. 
Even the bridge and tunnel people may find this a bit inconvenient.  
As a New Yorker by birth, I recommend going armed.
-- 
Alan "Uncle Al" Schwartz
UncleAl0@ix.netcom.com ("zero" before @)
http://www.ultra.net.au/~wisby/uncleal.htm
 (Toxic URL! Unsafe for children, Democrats, and most mammals)
"Quis custodiet ipsos custodes?"  The Net!
                  INTERNET SEARCH ENGINES, 2.03
Find anything at the top of the list, everything at the bottom.
Big Boppers
     http://www.search.com/
     http://pacific.discover.net/~dansyr/engines.html
     http://albany.net/allinone/
     http://www.webcom.com/webscout/
     http://www.probe.net/~niles/
     http://www.ARTECH.com/post.html
     http://www.tstimpreso.com/hotsheet/
     http://www.cnet.com/
     http://home.netscape.com/home/internet-search.html
Research It!
     http://www.cam.org/~psarena/it.html
240,000+ indexed and documented shareware packages
     http://www.jumbo.com/
     http://shareware.com/
     http://ftpsearch.unit.no/
     http://www.intbc.com/sleuth/
UU-decoding/viewing USENET binary posts
     http://shell.ihug.co.nz/~ijh/
Newsgroups
     http://www.dejanews.com/ (Usenet search engine)
     http://gagme.wwa.com/~boba/groups/
     http://www.speakeasy.org/%7Edbrick/newspage/root.html
     http://sunsite.unc.edu/usenet-i/hier-s/0top-1.html
Medline (8 million medical references)
     http://www.healthgate.com/
More search engines:
     http://www.hotbot.com/
     http://www.opentext.com/
     http://www.webcrawler.com/
     http://pointcom.com/
     http://www.cs.colorado.edu/wwww/
     http://www.earthlink.net/free/bigbee/webdocs/links.html
     http://www.pond.com/~justice/engine.html
     http://rama.poly.edu:1800/WWW.html
     http://www.lycos.com/
     http://cuiwww.unige.ch/meta-index.html
     http://pubweb.nexor.co.uk/
The Phone Book
     http://www.yellowpages.com/
     http://www.bigbook.com/
     http://www.bigyellow.com/
     http://www.four11.com/
     http://wyp.net/info/search/NA.html
     http://www.iaf.com/
     http://www.switchboard.com/
     http://www.telephonebook.com/
US Snail Mail ZIP codes
     http://www.usps.gov/
Thomas Register (all North American manufacture)
     http://www.thomasregister.com/
YAHOO (Web Index by topic),
     http://www.yahoo.com/
Archieplex (ARCHIE search front end)
     http://web.nexor.co.uk/archie.html
The List (3500 Internet Providers by area code)
     http://www.thelist.com/
Link hubs are homepages which provide hundreds of hypertext links to 
other Web sites.  Here are some of Uncle Al's haunts:
But first... something completely different
     http://www.pythonline.com/
     http://www.cheesesofnazareth.com/
     http://www.paranoia.com/coe/e-sermons/butcher.html
     http://www.student.nada.kth.se/~nv91-asa/mad.html
     http://www.us.mensa.org/
http://users.aol.com/rpollanen/  (massive and indexed)
http://www.bigeye.com/           (1000+ URLs)
http://cool.infi.net/            (Cool Site of the Day)   
http://www.hotwired.com/         (join, it's free!)
http://www.netlynews.com/        (truth free of Official truth)
http://www.suck.com/             (kewl)
http://www.firstsite.com/        (aggressive info)   
http://www.iguide.com/           (indexed guide to the net)
http://www.msnbc.com/            (Microsoft + NBC = something)
http://www.cnn.com/              (folks you might trust)
http://www.cyberzine.com/seeress/vision.html
http://kzsu.stanford.edu/uwi/reviews-l.html
http://www.vpm.com/tti/stick1-5.html#SURFSITES
http://gagme.wwa.com/~boba/spider1.html
http://www.oslonett.no/home/frodeni/odin/
http://www.gsfc.nasa.gov/NASA_homepage.html
http://www.ziff.com/~pcmag/websites.htm
http://www.bekkoame.or.jp/Users/user.home.page.html
http://www.whitehouse.gov/
Everything is everywhere.  Magic is loose in the world!

In article <56ao38$eml@news1.uk.pipeline.com>, richard_gain@uk.pipeline.com(Richard Gain) says:
>
>I have started a project to make a dodecahedral shaped sculpture from junk
>I have lying around but I can't proceed without a couple of pieces of
>information.  
> 
>  1)   What is the angle between two adjacent faces? 
> 
>  2)  What is the relationship between edge length of the pentagonal faces
>and diameter (face to face or vertex to vertex, whichever is easier) of the
>dodecahedron? 
> 
>Any help with solving either or both of these would be appreciated. 
> 
>
Generating a dodecahedron with 3D turtletalk: 
3 dimensional turtle talk has four commands: 
M for move n many steps forward (in the direction the turtle is currently pointing) 
T for turn d degrees 
R for repeat the preceding sequence m times 
X for rotate axially z degrees (imagine a airplane rolling) 
The syntax is MnTdRmXz etc. 
got that? 
The dodecahedron's description is: 
M40T72R5M40X63.435T288X296.565R5M40T72M40X63.435T288X296.565R4 
This from a web page http://www.servtech.com/public/jmb184/interests/mathematics/dodecahedron.html

I'm reading J.B.Paris' paper "A hierarchy of cuts in models
of arithmetic". I don't understand some of the basic notions.
First some notation.
Let P denote Peano's 1st order axioms for arithmetic.
Let I\Sigma_n be the same set of axioms but with induction
confined to \Sigma_n formulas.
Let N denote the standard model of arithmetic.
Let M denote a countable nonstandard model of I\Sigma_1.
Each element x of M is identified with the initial segment
consisting of all of the predecessors of x in M.
Definition: A subset I of M is called a cut if I contains 2,
is closed under multiplication and is an initial segment of M
other than M. 
Definition: Let I be a cut of M. Let A be a subset of I.
We say that A is coded in M if there is some element b of M
such that for all x in I, x belongs to A if and only if the
x-th prime of M belongs divides b. (Presumably one speaks of
b as being the code of A).
Here is what I don't understand.
(1) Paris writes: "Similarly we can talk of f:I->M being coded.
    Henceforth all subsets of I or functions on I
    etc. which are mentioned are assumed to be coded in M unless
    otherwise stated." Now, I can imagine how that might work,
    e.g. use a bijection between MxM and M to think of a function
    as a subset of M. But is that what is actually intended? Even
    if it is, I still don't understand the following:
(2) Paris writes: "Suppose f: I -> a is coded by e in M. Then by
    overspill there is a b > I [i.e. b > x for all x in I] such that
    e codes a map f': b->a with f' restricted to I = f Hence we
    have a continuation of f a little way above I."
My vague intuition is that what is going on in (2) is that
one can't actually define I in M, so whatever one can say about
f using the code e will have to define something other than I,
but necessarily containing I. According to that intuition,
(2) seems to make sense. However, I would feel better about
it if I could see a formal proof.
If you can clarify these points, please let me know.
Allan Adler
adler@pulsar.cs.wku.edu

Phil Fischer wrote:
> What a bunch of moronic blather. The most stringent test of the perihelion
> advance predicted by GR is the Taylor-Hulse pulsar. You might recall that
> the discoverers of this pulsar (Taylor and Hulse) were recently awarded Nobel
> prizes. This system has a much larger perihelion advance than
> mercury. Observation and analysis of pulsar timing has yielded fantastic
> agreement with GR. End of discussion.
Hahahahahahaha .....  end of discussion.
Hahahahahahaha .....  what an intellectual singularity.
Of course - I forgot ... everyone who is awarded a Nobel prize is
correct by default.   Certainly, if they were handing out such
awards in the days of Ptolemy, then he would have received a few.
Water joke ..... surf on .....
Verily verily I say unto you ....
   Those who are stuffed up proponents of the status quo have
   already received the reward of their labor.
I find sci.physics the most amusing newsgroup to read for this
very reason ... "Know_it_Alls" - Please stand up and be recognised.
The prizes have been awarded.
The books have already been written.
Nature is known to the scientific mind.
All that is left to do now
is to teach the children well ....
Pete Brown
--------------------------------------------------------------------
 BoomerangOutPost:       Mountain Man Graphics, Newport Beach, {OZ}
 Webulous Coordinates:   http://magna.com.au/~prfbrown/ancients.html
 QuoteForTheDay:         HERACLITUS
       born about 540 bc in Ephesus of royal family, 
       Heraclitus was a solitary, his words were obscure, 
       and he never disguised his contempt for mankind and 
       other "philosophers and poets" such as Pythagoras and Homer:
     "The rest of mankind are unaware of what they do while awake,
     just as they forget what they do while sleeping."
     Rebuking some for their unbelief, Heraclitus says:
     "Knowing neither how to hear nor how to speak"
     The opinions of mankind - "to be children's playthings".
     "What sense or mind have they?
     They put their trust in popular bards 
     and take the mobs for their teacher, 
     unaware that most men are bad, and the good are few.
     "Human nature has no insight, but divine nature has it."
     "Man is infantile in the eyes of a god, 
      as a child in the eyes of a man."
     "To God all things are fair, and good and just, 
      but men have supposed some unjust and some just."
     "One man is to me ten thousand, if he be the best."
     "The way up and the way down are the same"
     "Divine things for the most part escape recognition because of
unbelief."
     "The limits of the soul woudst thou not discover 
      though thou shoudst travel every road: so deep a logos has it."
     "What we see awake is death - what we see asleep is sleep."
     "The body is a tomb" ...... (Note: this is a standard Pythagorean
belief)
     "A man's character is the immortal 
      and potentially divine part of him" [Fr 115]
     "In the CIRCLE the beginning and the ending are common.
     "What he calls death is not utter annhilation, 
      but changes to another element" - [Plato on Heraclitus]
     Heraclitus called fire "Want and satiety"
     "For fire will come and judge and convict all things."
     "From all things one, and from one all things."
     "Immortal mortals, mortal immortals, 
      livingdeath of the others and dying their lives"
     (Guthrie: the transformation of opposites occur concurrently)
     "Everything is an exchange for fire" .... 
      fire is the arche of nature [Simplicius:Phys23:33-24]
     "Let us not make random conjectures about the greatest of matters."
     According to the writings of Macrobius, 
     Heraclitus describes the soul as ....
     "A spark of the substance of the stars."
             -  Heraclitus ..... (about 500 BC)
---------------------------------------------------------------------

Erik Max Francis  writes:
>Dean Povey wrote:
>> In AD gravitation, the perihelion advance for each planet is
>> proportional to the square root of the division of the solar mass by
>> the orbital radius power 3.
>> 
>>              Tp = sqrt(M / r^3)      [ditto: DGP]
>Care to derive this?
>> If the Mercury value is taken as 43" . . . .
>Do you _actually_ mean that Autodynamics can't predict Mercury's perhelion
>precession without being given it?  That's not very impressive.  Right
>there general relativity has a head start on you.
From what I can gather from the web pages, the AD equation uses a constant
which indicates the quantity of mass received from pico-gravitons 
per each gram of mass present, per second. This is a universal constant which
is the same for all celestial bodies.  Hence, the input of Mecury's perhelion
advance is merely a method to calculate this constant.  (You could predict
Mercury's perhelion advance by using accurate observations of another body to 
calculate the constant.)  
I don't see much wrong with this, you find constants throughout physics,
(eg. the GR equation uses G and pi).
For more information read the Autodynamics web page.
Dean. 

John Anonymous MacDonald wrote:
> Anyone have tips for keeping David Kaufman's droppings out of
> the newsgroups?
For that matter what about Persuter's?  Suggestion might be for all to
ignore them both and maybe they'll go away.  Nah, too young to take a
hint.
Curmudgeon

John Baez wrote:
> Now, spin is a form of angular momentum intrinsic to the electron,
> but there is another kind of angular momentum, namely orbital angular
> momentum, caused by how the electron (or whatever particle) is moving
> around in space.  It turns out that orbital angular momentum also
> has magnetic effects, but only causes diamagnetism.  The idea
> that when you apply a magnetic field to some material, it can also make
> the electrons in it tend to move in orbits perpendicular to the
> magnetic field, and the resulting current creates a magnetic field.
> But this magnetic field must *oppose* the external magnetic field.
> Ergo, diamagnetism.
> 
> Why does orbital angular momentum work one way, while spin works
> the other way? 
Actually  orbital moments and spins
do work the same way! The nonrelativistic Hamilton of
a particle in a magnetic field can be written:
H=p^2/2m + mu B (L+g S) + 1/2m e^2/c^2 B^2 r^2
The second expression shows, that a orbital momentum L=(r x p)
really acts the same way as
a spin. To decide this physically, look at an atom with a finite angular
moment 
in a magnetic field (which is paramagnetic with a similiar behavior as 
a free spin).
Diamagnetism can be traced back to the third expression!
The most "classical" way to explain diamagnetism is to look at the
 induced electric field, when (adiabatically) switching on the magnetic
field. By this electric field, currents are induced, which (according
to Lenz' rule) shield the magnetic field which do not decay fully in a
quantum-mecanical system - the best diamagnet is a superconductor. 
Does somebody know how this "classical" picture can be used to derive the
"diamagnetic term" in the hamilton?
Achim

Kenneth Lareau (darshan@squonk.net) wrote:
: I am wondering if anyone has ever bothered to create an actual nomenclature
: for the duodecimal system.  I made a half-hearted attempt several years ago,
: but never came up with anything that sounded remotely coherent... then a-
: gain, I'm not a linguist. :)
In addition to Mr. Edgar's list, here is a book, in French, that I found
in my University's math library.
Author: Essig, Jean, 1899-
Title:  Douze, notre dix futur; essai sur la nume'rotation duode'cimale
        et un syste`me me'trique concordant.
Publ:   Paris, Dunod, 1955.
Paul Guertin
guertinp@jsp.umontreal.ca

U Lange wrote:
> 
> Fredrik Sandstrom (fred@spider.compart.fi) wrote:
> : I've got a simple(?) question.  I came to think of it one day, and I
> : can't come to a conclusion.  What is
> :
> :  lim    0/x         ?
> : x -> 0
> :
> 
> Just use the definition of the limit and the fact that the function 0/x
> is defined on IR-{0}:
> 
> lim  0/x = a    iff for each epsilon>0 there is a delta>0 such that
> x->0                |0/x-a| 
> Obviously, a=0, since |0/x|=0 |x|<100000 (Or any other "delta" you fancy).
> 
> : One possible answer would perhaps be that lim_(x->0+) 0/x = 1 and
> : lim_(x->0-) 0/x = -1.  Other possibilites are +/- infinity, or perhaps
> : 0.  What do you think?
> 
> These are not possible answers. lim  0/x = 0 and nothing else.
>                                 x->0
> 
> --
> Ulrich Lange                       Dept. of Chemical Engineering
>                                    University of Alberta
> lange@gpu.srv.ualberta.ca          Edmonton, Alberta, T6G 2G6, Canada
Answer 1 zero which is zero.

In article <56ao38$eml@news1.uk.pipeline.com>,
richard_gain@uk.pipeline.com(Richard Gain) wrote:
> I have started a project to make a dodecahedral shaped sculpture from junk
> I have lying around but I can't proceed without a couple of pieces of
> information.  
>  
>   1)   What is the angle between two adjacent faces? 
>  
>   2)  What is the relationship between edge length of the pentagonal faces
> and diameter (face to face or vertex to vertex, whichever is easier) of the
> dodecahedron? 
>  
> Any help with solving either or both of these would be appreciated. 
>  
I think that *THE BOOK* for you is _Mathematical Models_, by H. Martyn Cundy
and A. P. Rollett.  Oxford University Press.
-- 
Christopher J. Henrich
chenrich@monmouth.com

In article , hbaker@netcom.com says...
>
>In article <01bbc6aa$3586fe80$65d2989e@pacificp>, "Richard Cowey"
> wrote:
>
>> Take a Lat and Long of a place, say 50 deg. 20 min. 18 sec. North and 1
>> deg. 3 min. 42 sec. West as point A
>> And point B is, 52.13.54. North, 2.46.23. West.
>> How do you get the distance and heading between the two.
>
>see
>
>ftp://ftp.netcom.com/pub/hb/hbaker/FAQ-lat-long.txt
As I recall, under about 600 miles it isn't necessary
to do the spherical trigonometry for practical navigation,
Subtract the latitudes and Longitudes to get the difference.
At the equator a degree is about 69.17 miles and at 52 degrees
north it is 68.9 miles so you are well within the range where 
you can just take the diagonal of a rectangle with sides as the 
difference of the latitude and longitude.
Take 52.13.54 and convert it to 51.73.54
subtract from that              50.30.18
get				 1.43.36
convert to seconds as 6216
Take 3.46.23 and convert it to 2.45.83
subtract from that 	       1.03.42
get 			       1.42.41
convert to seconds as 6161
pythagorean theorem hypotenuse =~ 8752 seconds
there are about 101 feet five 1/2 inches in a second
at the equator and 101 feet 3/4" in a second at 52 degrees
8752 seconds = 2.25.52 degrees =~167.5 miles
aproximately 884,447 feet 9 inches
steve

Jack W. Crenshaw wrote:
> 
> George Weisz wrote:
> >
> > Hi...
> > Could somebody please help.
> >
> > I am looking for the rotation matrix R derived from a quaternion
> > q()
> >
> >
> > d = 2acos(q0)
> >
> >          d            q1          q2           q3
> > s = sin ---     zx = ---    zy = ---     zz = ---
> >          2            s           s            s
> >
> >     +-            -+
> >     |  ?   ?   zx  |
> > R = |  ?   ?   zy  |
> >     |  ?   ?   zz  |
> >     +-            -+
> >
> > Any help or pointer will be much appreciated.
> >
> > Thanks
> >
> > George
> 
> George, the fragment of the rotation matrix you gave isn't correct.
> Here's the general form of the matrix:
> 
>  [a^2-b^2-c^2+d^2     2(ab-cd)         2(ac+bd)      ]
>  [  2(ab+cd)      -a^2+b^2-c^2+d^2     2(bc-ad)      ]
>  [  2(ac-bd)          2(bc+ad)     -a^2-b^2+c^2+d^2  ]
> 
> Better check the signs, but I think I got it right.
> 
> In your notation, q1=a, q2=b, q3=c, q0=d
> 
> Jack
Thanks Jack....
Your help was very welcome....
The signs as you have shown are correct.
Thanks again
George

Sylvestre Blanc wrote:
> 
> Aaron Birenboim wrote:
> >
> > I have come across a simple, finite series problem...  I'm not sure there
> > is an answer.
> >
> > I want to know if there might be a simple formula equivalent to :
> >
> >      1*1 + 2*2 + 3*3 + 4*4 + ... + n*n
> >
> > I'm looking for a simple answer like n*(n-1)/2 = 1+2+3+4+5+...+n
> >
> > --
> > Aaron Birenboim   |         aaron@ptree.abq.nm.us            | Albuquerque, NM
> > http://www.swcp.com/~aaron         RESUME> http://www.swcp.com/~aaron/res.html
> > PearTree Consulting (WWW, UNIX, Scientific Computing,...)
> 
> Try n*(n+1)*(2*n+1)/6
In the same way that the sum of the first n integers has a simple
graphical demonstration obtained from fitting two triangles together to
form an n by n+1 rectangle, eg for n=7:     
    000000X
    00000XX
    0000XXX
    000XXXX
    00XXXXX
    0XXXXXX
1+2+3+4+5+6+7 = (6*7)/2,
the sum of the first n squares given above can also be demonstrated by
representing the sum by stacking a cube on top of a square of 4 cubes on
top of a square of 9 cubes . . . on top of a square of n*n cubes, and
then fitting 6 of these shapes together to form an n*(n+1)*(2n+1)
cuboid.

In article <55p2a3$ctc@info-server.bbn.com>, David Karr  wrote:
>
>ObPuzzle: Suppose you don't want to do all this math.  What's a
>sure-fire algorithm that will allow you to stack the cans correctly
>without any extra calculation, and without any risk that you'll have
>to move a can that you've already placed because you were wrong about
>needing it there?  (That is, ensure that if you have, say, only 55
>cans, you don't foolishly put 11 cans in the bottom row.)
Stack the cans in a triangle, then start building up the sides of
the triangle one row at a time. This is the order you put down the
first ten cans in:
     10
    6   9
  3   5   8 
1   2   4   7
Can 11 will go to the right of can 7, can 12 will rest on cans
7 and 11, and so on.
ObFollowup: What if the cans were to be arranged in a tetrahedron?
-- 
__/\__  Jonathan S. Haas             | Jake liked his women the way he liked
\    /  jhaas@microsoft.com          | his kiwi fruit: sweet yet tart, firm-
/_  _\  Microsoft Corporation        | fleshed yet yielding to the touch, and
  \/    Don't Tread On Me            | covered with short brown fuzzy hair.

Salem Reyen writes:
>
>	Does anyone know how to obtain (aka purchase) 
>math books (translated from Russian to English) from US?
mit press and dover paperback both carry some such
translations.	
				-don davis, boston

Richard Gain wrote:
> 
> I have started a project to make a dodecahedral shaped sculpture from junk
> I have lying around but I can't proceed without a couple of pieces of
> information.
> 
>   1)   What is the angle between two adjacent faces?
> 
>   2)  What is the relationship between edge length of the pentagonal faces
> and diameter (face to face or vertex to vertex, whichever is easier) of the
> dodecahedron?
> 
> Any help with solving either or both of these would be appreciated.
> 
I figured out how to carve a dodecahedron out of a solid cube,
if that would be of any help.
The length of an edge of a dodecahedron is    1/(phi^2)
                                            = 1 - 1/phi
                                            = 0.38196...
times the distance between parallel edges.
(phi is the golden ratio or  (1 + sqrt(5)) / 2).
Whenever you involve pentagons, phi is bound to pop up.....
Three pairs of parallel edges can be constructed on the 
surface of the cube with a compass.  Laid-out edges
on intersecting faces must be skew to each other.
The other edges appear magically when you saw off the 12
wedges that planes containing one of the edges that
you laid out and an endpoint of one of the other edges you
laid out.
Hope I helped,
Spencer

	Below are 2 replies I received by e-mail  that state that measuring
the difference of Time 1E-5s Versus 1.4E-5s for the two crystals
5 cm by 1.5 cm as easy to test.  Even  5 mm by 1.5 mm crystals
could be easily tested, if such crystals were sold commercially.
	Thanks again to Mr. Don Taylor for his imput below that could
decide how energy and sound may travel through metal
atom structures, if no one out their knows already.
	Does anyone know if sound goes in straight lines or if it zig-zags 
from atom bond to atom bond when it moves perpendicular to the square
arranged atoms in their layers?  Does anyone know?
	Is anyone interested in doing the experiment to find out?
--------------------------------------------------------------------------
First e-mail reply follows:
--------------------------------------------------------------------------
From: Don Taylor 
>
>           For K-12 Students, Teachers And Others
>      Interested In Exploring Math, Science And Ethics
>    Through Collaboration For Enrichment And Achievement.
>------------------------------------------------------------------------
>
>       Can the experiment be made that measures the time sound
> travels 5 centimeters in a single aluminum crystal which can
> distinguish the prediction that it takes 1E-5 seconds
> (1/100,000 s) for a crystal oriented (110) and 1.4E-5 s
> for a crystal oriented (100) as shown below.
...
It would certainly be easy to make such a measurement.
Take any two channel oscilloscope with better than say 10Mhz bandwidth and DC
coupling.
Take any square wave signal generator and produce good quality square waves at
oh, say 1000Hz.  You want good clean rising edges on this.
Bond a piezo transducer to the front of each of your crystals and drive both
of the transducers with the square wave.  (Depending on the generator you may
need to match the impedance of the generator to the relatively low impedance
of the transducers.)
Similarly bond a piezo transducer to the back of each of your crystals and
use any reasonable scope probes to pick up these signals.
Trigger the scope on the rising edge of the square wave.  Set up the
sensitivity, offset, trigger level, etc. on the scope and display both
channels.  About 2 microseconds/div should display the leading edge of the
square wave and both received pulses on the back face.  Push your holdoff
out so you get consistent clean triggering and you should see your two
transit times cleanly displayed one above the other.
Now, does sound actually behave the way you predict it does in aluminum,
that I can't tell you.
>       If this above experiment is warranted, and if the
> difference in times is measurable, then the purchase of the
> 2 aluminum crystals (5 cm by 1.5 cm each) needed to do the
> experiment is about $1450 for each oriented single crystal.
And it sounds like your probably should know this before making an order ;}
(if all this description of oscilloscopes is hazy, show the description to
anyone over in the electronics department and see what they think.  But I
think that trying to show on a scope tube the difference between 10 and
and 14 microseconds ***under controlled conditions*** is a piece of cake.
----------------------------------------------------------------------------
Second e-mail reply follows:
----------------------------------------------------------------------------
On Tue, 12 Nov 1996, David Kaufman wrote:
>       If you don't object, I intend to post this reply of yours
> below and give you full credit for it.
I don't really care.  I don't need any credit for it, but it is up
to you.  If you were to approach almost anyone in an electronics
department and ask them the same question I would think you would
get the same answer.
>       Thanks again for your valuable information. I would also be
> interested in knowing just what are the limits for this kind of
> measurement difference. For example, could a 5 mm by 1.5 mm crystal pair
> be tested to show a difference between 1 and 1.4 microsecond?
There are a variety of limits involved here.
One is the oscilloscope used.  Relatively inexpensive oscilloscopes,
say costing $1000-$3000 could easily make this measurement and probably
make measurements between 1 and 1.05 microseconds fairly easily.  Good
laboratory quality oscilloscopes costing an order of magnitude more
money could make far more precise and accurate measurements.
Another is the characteristics of the transducers used to couple the
signal into and out of the crystals.  You want these to respond in a
consistent manner to each of tens of thousands of pulses into and out
of the crystal with timing that does not vary by more than a few percent
of what you are trying to see.  Thus the "bandwidth" of the transducers
is an issue and the bonding of the transducers to the crystal will be
important.  A carefully applied thin layer of adhesive will probably
do.
Another is the quality and reliability of the signal source used to put
the pulses into the crystal.  You would want this to have a very consistent
shape of the leading edge of the waveform.
--------------------------------------------------------------------------
In conclusion:
	Thanks again to Mr. Don Taylor for his imput that could
decide how energy and sound may travel through metal
atom structures, if no one out their knows already.
	Does anyone know if sound goes in straight lines or if it zig-zags 
from atom bond to atom bond when it moves perpendicular to the square
arranged atoms in their layers?  Does anyone know?
	Is anyone interested in doing the experiment to find out?
David Kaufman (davk@netcom.com) wrote:
:           For K-12 Students, Teachers And Others
:      Interested In Exploring Math, Science And Ethics
:    Through Collaboration For Enrichment And Achievement.
: ------------------------------------------------------------
: 	Can the experiment be made that measures the time sound
: travels 5 centimeters in a single aluminum crystal which can
: distinguish the prediction that it takes 1E-5 seconds 
: (1/100,000 s) for a crystal oriented (110) and 1.4E-5 s 
: for a crystal oriented (100) as shown below.
: 	Face Centered Cubic (FCC) Structure
:               /      \ /      \ /
:              O    O   O    O   O                     O
:             /  \     /  \     /  \                  /  \
:            /    \   /    \   /    \                /    \
:           /      \ /      \ /      \              /      \
:    Hit-->O   O    O   O    O   O    O Straight-->O   O    O
:    Atom  |\      /|\      /|\      /| Path       |\      /|
:          | \    / | \    / | \    / |            | \    / |
:          |  \  /  |  \  /  |  \  /  |            |  \  /  |
:          |   O    |   O    |   O    |            |   O    |
:          |   |    |   |    |   |    |            |   |    |
:          | O | O  | O | O  | O | O  |            | O | O  |
:          |   |    |   |    |   |    |            |   |    |
:          O   |O   O   |O   O   |O   O            O   |O   O
:           \  |   / \  |   / \  |   /              \  |   /
:          | \ |  /   \ |  /   \ |  /                \ |  / |
:          |  \| /     \| /     \| /                  \| /  |
:          |   O        O        O                     O    |
:          |                                                |
:          |<---------------------------------------------->|
:          |   5 Cm Single FCC Crystal Oriented (110)   |
:          I inferred this orientation needed, as (110) 
:                Please correct me, if necessary.
: Note: Atoms O in adjacent cube faces are adjacent atoms
:       bonded together also in a straight line from one
:       end of the 5 centimeter crystal to the other end.
: 	For aluminum, sound travels about 5000 meters per 
: second. Thus for a 5 centimeter (5/100 m) piece of 
: aluminum (as oriented above) it would require 1E-5 seconds 
: for sound to move in a straight line from atom bond to atom 
: bond at one end of the single crystal to the other end 
: calculated as follows.
:        s       5 m        1  
:     --------  ---  =  ----------- s = 1E-5 s 
:     5000 m    100      100,000
: 	Below is an aluminum crystal oriented (100), so 
: that if energy flows from adjacent atom to adjacent atom, 
: the sound will travel in a zig-zag direction. 
: 	The extra distance traveled in a zig-zag path is the 
: square root of 2 times the straight (direct) distance D, or 
: 2^.5 D = 1.414 D. 
: 	Thus the time to travel 5 cm in a zig-zag path is 
: 1.414E-5 seconds versus 1E-5 s for the straight path.
:            O________O________O________O         O________O
:            /        /        /        /|        /        /|
:           /   O    /   O    /   O    / |       /   O    / |
:          /        /        /        /  |      /        /  |
:   Hit-->O_______O/_______O/_______O/   |  -->O_______O/   |
:   Atom  |\     / |\     / |\     / | O | Path|\     / | O |
:         | \   /  | \   /  | \   /  |   /O    | \   /  |   /O
: Zig-zag |  \O/   |  \O/   |  \O/   |  /      |  \O/   |  /
:   Path  |        |        |        | /       |        | / 
:         |________|________|________|/        |________|/  
:         O        O        O        O         O        O   
:         |                                             |
:         |<------------------------------------------->|
:         | 5 Cm Single FCC Crystal Oriented (100)  |
:         I inferred this orientation needed, as (100) 
:                Please correct me, if necessary.
: Note: Atom adjacent bonds go from corner atom O to face
:       centered atom O.
: Remember: The above is only a prediction of how energy flows
:           in face-centered cubic (FCC) structures.
:           An alternate view is that energy will flow in a 
:           straight line also in the (100) oriented 
:           crystal rather than a zig-zag fashion as 
:           predicted.
:           What's your view and why?
: 	The problem is can the difference in time (for the 
: experiment outlined above) be detected to settle this 
: question of how energy and sound flows in atom structures.
: 	Does anyone know?
: Note: Their is also a straight path in the (100) oriented 
:       crystal. However, since the 5 cm long crystal is only 
:       1.5 cm wide, the straight path only reaches 1.5 cm 
:       along the 5 cm length. Therefore, energy traveling 
:       along this straight must also zig-zag the same 
:       distance to reach the end of the 5 cm sound path.
: 	If this above experiment is warranted, and if the 
: difference in times is measurable, then the purchase of the 
: 2 aluminum crystals (5 cm by 1.5 cm each) needed to do the 
: experiment is about $1450 for each oriented single crystal.
: Below between the 2 dashed lines is a repeat of the above 
: but from another perspective.
: ------------------------------------------------------------
: 	Does sound (or energy into the wall) travel like a 
: sphere from a hit wall atom?  Or does it travel from atom 
: bond to atom bond in straight lines in one direction, but at
: 45 degrees to this straight line direction, does sound 
: travel in a zig-zag manner as shown in the figure below?
: 	All numbers and letters in the figure below are atoms 
: in the face of cubes. Adjacent faces contain adjacent atoms
: that are bonded together.  
: 	Note the same bond to bond distance traveled of 2 paths
: from H the hit atom. But the overall distance traveled is 
: different for each path as follows:
:      ___________________________ 
:     /|       /|       /|       /|
:    / |  1   / |  3   / |  5   / |
:   /__|_____/__|_____/__|_____/  |
:   | H|_ _ _|_2|_ _ _|_4|_ _ _|_6| In Path H-1-2-3-4-5-6
:   |  /     |  |     |  /     |  / Energy from H Moves
:   | /   A  | /|     | /|     | /  6 Bond Lengths = 6 d.
:   |/_______|/_|_____|/_|_____|/    
:            | B|_ _ _|_ |          But distance if traveled
:            |  /     |  /          in a straight line from
:            | /   C  | /|          H to 6 = 3 (2)^.5 d 
:            |/_ _____|/ |          
:                     | D|_ _ _ _ _
:                     |  /        /| In Path H-A-B-C-D-E-F
:                     | /   E    / | Energy from H Moves
:                     |/_______ /  | 6 Bond Lengths = 6 d
:                               | F| But In A Straight line.
:                               |  /
:                               | /
:                               |/
: 	Can the difference in the time sound travels in the
: 2 paths above be detected? 
: 	If such a difference in the time sound travels were
: detected, it would prove that sound travels from bond to 
: bond in a zig-zag fashion. 
: 	Also the atoms in the crystal could be oriented because
: travel perpendicular to the square layer section takes 41.4%
: longer for sound to travel than the shortest straight line 
: path at a 45 degree angle to the square layers.
: ------------------------------------------------------------
: Thanks for joining this undertaking.
: 	Good luck on this exciting adventure to find useful 
: projects to explore and the tools to empower and to succeed 
: with.
: 	I offer this post to continue a useful discussion on 
: many valuable ideas about atoms that could become meaningful
: projects for students and others to undertake.
: ____________________________________________________________
:   Thanks to those who have offered constructive criticism.
:     
:              C by David Kaufman, Nov. 11, 1996
:                   Founder of the Cube Club
:    For Collaborative Math, Science and Ethics Excellence.
:          Be Good, Do Good, Be One, and Then Go Jolly.
:                  What else is there to do? 
: -- 
:                                              davk@netcom.com
-- 
                                             davk@netcom.com

ikastan@sol.uucp (ilias kastanas 08-14-90) writes:
|> 	As Bill's Dialogue Concerning Two New Sciences underscores,
Hah!  Very neat.   But I rather follow in Lakatos' footsteps there...
|> observing an infinite sequence of Bernoulli trials (i.e. a real number) is 
|> an abstraction.  All physical experiments and observations yield rationals; 
|> ...  ...  Forget P = 0; such a thing is simply
|> _impossible_... it cannot happen (!).  Never has, and never will.
Yep, that pretty well sums up the clincher.  Nicely put.
|> 	If we grant this abstraction, it is math, not physics. 
|>       ... ...
|>    and x is in B^w, "x can not occur" is logically false. 
Hmmm... well if we're in math-not-physics, the word "occur" shouldn't really
be there at all.  You'd just say, "x is not in the sample space", which is
false by definition.  So we agree.
|> 	The remarkable part is focusing on _one_ countable set, the definable
|>    (recursive) x's, and deeming it "more justified". 
Yeah right.  It's just laziness on our part.  We can only be bothered
talking about the things it's possible to talk about.  ;-)
|>       A little-known fact:
|>    coins have tiny cellular telephones and talk to each other;...
|>     ... They will not be caught
|>    doing anything recursive, or recursively enumerable... 
Or even definable?
|>  Clever little devils.
It is even LESS well known, (if that were possible), that all sets have
a CantorNet connection with auto-antigeneric parallel many-world devices
that prevent an infinite number of unspecified choices being made 
from their members.
DAMN cunning...
-------------------------------------------------------------------------------
             Bill Taylor           W.Taylor@math.canterbury.ac.nz
-------------------------------------------------------------------------------
                "Open the internet connection, please,  HAL."
                "I'm sorry Bill; I'm afraid I can't do that."
-------------------------------------------------------------------------------

Mike McCarty  suggests:
> Here's an approach: [...] prove that there are arbitrarily 
>large n such that [...]  1/(n sin n) > 1
Good luck!  A lower bound of   sqrt(5)/Pi (about 0.7) follows 
from general facts about continued fractions.  Improving on this 
would amount to proving that the continued fraction expansion for 
Pi has many terms larger than 1; but little or nothing along
those lines has been proved.

In article ,
Jamie Dreier  wrote:
>ikastan@alumnae.caltech.edu (Ilias Kastanas) wrote:
>
>[...]
>
>>         Loeb's theorem is actually equivalent to the Incompleteness theorem
>>    (it easily implies it... but, less obviously, it is also implied by it).
>
>In what sense? They are both theorems; EVERYTHING implies each of them!
	Of course; they are outright provable in PA (or PRA).  The point is 
that one implies the other in some weak theory... whichever happens to provide
what you use in the proof!  Such equivalences sometimes involve the empty theory
... i.e. "just logic".
For Loeb => G you just plug in a contradiction for formula phi.
>I know that there is a very elegant proof of the second incompleteness
>theorem using Lob's theorem, although the only version I have seen of this
>proof seems wrong to me (but probably this is a fault in me rather than in
>the proof).
	I posted that proof in sci.logic some time ago; I'll see if I can
find a copy.
							Ilias

Vladimir A. Pertsel  wrote:
>"Dmitri V. Papichev"  wrote in relcom.rec.puzzles:
>(Translation only)
>[...]
>49 absolutely identical insulated wires cross the river under the 
>water. The ends of each wire are on the opposite banks of the river, 
>disconnected  initially. There is an electricity source on one of 
>the banks. An electrician with a tester (*) has to label all the 
>wires (0 - 48) (Each wire should have the same label on its both ends).
>A boatman charges 1 rouble for each crossing of the river.
>What is the minimal sum of money, sufficient for the electrician to 
>fulfill the task?
>
>(*) The tester allows to determine whether the wire is
>    connected to the electricity source, when You touch 
>    the wire by the tester.
Seen it. Good one, though. Spoiler below...
He can do it in two crossings of the river.
First, he should make sure all the wires are not touching each other.
He should tie 48 of the wires together into 24 pairs. The extra
wire, he labels as 0. He connects wire 0 to the power source.
Then he crosses the river once. Using the tester, he locates wire 0.
He picks a random wire, labels it 1, and connects it to 0. Wire 1
must be paired with another wire, and since wire 1 is connected to
wire 0, which is connected to the power source, wire 1's partner
should be live. The electrician uses the tester to locate wire 1's
partner, and labels it 2. He picks a random wire, labels it 3,
and connects it to wire 2. He locates wire 3's partner, labels it
4, and connects it to a random wire, labeled 5. He labels 5's
partner 6, and so on.
Then he crosses back. He disconnects all the connections he made,
but leaves the formerly-paired wires wrapped together by their
insulation, so he can tell which wires were paired before.
He locates the wire that is now paired with 0, and labels it 1.
1's former partner gets labeled 2, and 1 and 2 get reconnected,
so power flows to 2's current partner, labeled 3. 3's former
partner is 4, 4's current partner is 5, and so on.
You can do this without a power source, if you have a continuity
tester (which can tell when two wires are connected.)
-- 
__/\__  Jonathan S. Haas             | Jake liked his women the way he liked
\    /  jhaas@microsoft.com          | his kiwi fruit: sweet yet tart, firm-
/_  _\  Microsoft Corporation        | fleshed yet yielding to the touch, and
  \/    Don't Tread On Me            | covered with short brown fuzzy hair.

DATA NETWORKS by DIMITRI BERTSEKAS/ROBERT GALLAGER
I am looking for a book that covers the basic theory of Data Networks
and it's application in today's environment. This book, which I have
only skimmed, is my only example of applied mathematics in this field.
I am sure there are many good books out on this subject.
I will endevor to return the favor to anyone who takes the time to
respond.
DTS
dtsmith@hiwaay.net

Erik Max Francis  wrote in article
<328A06AE.2F74F680@alcyone.com>...
> Dean Povey wrote:
> 
> > In AD gravitation, the perihelion advance for each planet is
> > proportional to the square root of the division of the solar mass by
> > the orbital radius power 3.
> > 
> >              Tp = sqrt(M / r^3)      [ditto: DGP]
> 
> Care to derive this?
> 
> > If the Mercury value is taken as 43" . . . .
> 
> Do you _actually_ mean that Autodynamics can't predict Mercury's
perhelion
> precession without being given it?  That's not very impressive.  Right
> there general relativity has a head start on you.
I'm more confused than ever now. If M = the solar mass then precession is
independent of the mass of the object. 
This also implies that the orbits are circular? which they are not or
precession would not exist.
And where does this 43" come in there's no place for it in the equation
unless the text says one thing and M is  the Mercury value.
I make high school students show their work.
> 
> > [These] values are equal to Hall's empirical values and close to the
> > expected values calculated by Newcomb.
> 
> Empirical values and expected values?  I don't see observational values.
> 
> -- 
>                              Erik Max Francis | max@alcyone.com
>                               Alcyone Systems |
http://www.alcyone.com/max/
>                          San Jose, California | 37 20 07 N 121 53 38 W
>                                  &tSftDotIotE; | R^4: the 4th R is respect
>          "But since when can wounded eyes see | If we weren't who we
were"
> 

In article <56e74c$ebq@Starbase.NeoSoft.COM>,
positron@Starbase.NeoSoft.COM (Jonathan Haas) wrote:
> ObFollowup: What if the cans were to be arranged in a tetrahedron?
A concise algorithm I sent David for the triangular version is:
     while there are cans left to stack
         place a can as high as possible
     end-while
Conveniently enough, this also solves the tetrahedral version.
mag
p.s.  I only answered because I wanted to use "tetrahedral" in a sentence.  :-)
-- 
.---o  Tom Maglierygry, Research Programmer                      .---o
`-O-.  NCSA, 605 E. Springfield                  (217) 333-3198  `-O-.
o---'  Champaign, IL 61820         O-         mag@ncsa.uiuc.edu  o---'

G. A. Edgar wrote:
> As I recall from long ago, one way that was used had the two
> extra digits: X, pronounced "dek", and a backward 3, pronounced "el".
> Of course 10 is not pronounced "ten" but "dozen".
Well that sure explains the "Muliplication Rock" piece on
"Little Twelve Toes" who counts
1, 2, 3, 4, 5, 6, 7, 8, 9, "dek", "el", "do" (long "o" sound)
And, although I haven't checked, I do believe that
the symbols you described were the ones that were
used for "dek" and "el"

binklmj5@wfu.edu (Mathew J. Binkley) wrote:
> n
>d L
>---    with n not an integer.
>  n
>dx
Take the Fourier transform, then multiply the resulting function
by (iw)^n, then transform back.
n can be fractional if you feel like it.
That only differentiates a periodic function, but any function can be
made periodic with a veeeeeery long period. The trouble with that
method is, it might destroy the property that a derivative is a purely
local operation. I haven't played with it theoretically or numerically
to know. It would be nice if the fractional derivative were still a
purely local operation.
Offhand, the Fourier method is the simplest way I can think of to
_define_ a fractional derivative.
I don't know if there is some deeper mathematics which requires
that it be a certain way.
Simon

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Suppose you have 12 coins. 1 is lighter or heavier than the rest.
Determine the bad coin and whether it is heavy or light using at most 3
weighings. Your descion tree should have 24 outcomes at the bottom.

Paul Mulvey  wrote:
>How to calculate the dimensions of a square whose diagonal is 35mm
>longer than the sides
Let side = x, then diagonal = sqrt(2)*x
       sqrt(2)*x = x + 35
or   sqrt(2)*x - x = 35
(sqrt(2)-1)*x = 35
x = 35/(sqrt(2)-1)

magix (magix@dibe.unige.it) wrote:
> Does anybody know the closed form of:
> (LATEK) \frac{d^m}{dx^m}(e^{-x^2})
> (Visual)
> .             2
> .        m  -x
> .       d  e
> .      --------
> .           m
> .        d x
> for any m in N?
             2
        m  -x                      2
       d  e            m         -x
      --------  =  (-1)  H_m(x) e 
           m
        d x
Where H_m(x) = m-th Hermite polynomial.
Miguel A. Lerma

In a previous article, fc3a501@AMRISC04.math.uni-hamburg.de (Hauke Reddmann) says:
I think, this has been called "fractorial" notation;
I saw it implimented in an article in Byte magazine.  it might be nice
to use primorials, but I don't think that it's trivial!
>: The 'all-ary' representation of x in [0,1) is sum_{n=1}^infinity d_n/n!
>: with d_n in { 0, 1, ..., n-1 }.  The first N digits then allow you to represent
>: k/N! for any k in {0, 1, ..., N!-1} (just count the number of different 
>: [d_1,d_2,...,d_N] and note that the map is 1-1).  Since every rational in [0,1)
>: can be written in the form k/N!, its representation is terminating.
>: 
>Another possibility: Egyptian fractions. E.g., 0.1001
>would represent 1/2+1/5.
-- 
You *don't* have to be a rocket scientist.  (College Career Counselor
					     to me, again )
There is no dimension without time.  --RBF (Synergetics, 527.01)

Matthew P Wiener wrote:
> Some consider that cheating.  "Real" cross products, the two vectors at a
> time sort, work in dimensions 3 *and* 7.  The familiar 3-dimensional one
> can be thought of as the purely imaginary part of quaternion multiplication.
> Similarly, the purely imaginary part of octonion multiplication can be read
> as a 7-dimensional cross product.  Most of the familiar identities hold in
> both cases.
Do cross products only work in 3 & 7 dimensions, or will they work
anytime
there are 2^n - 1 dimensions (where n > 1 and an integer)?

> Richard Gain wrote:
> >   1)   What is the angle between two adjacent faces?
> >   2)  What is the relationship between edge length of the pentagonal faces and diameter (face to face or vertex to vertex, whichever is easier) of the dodecahedron?
I wrote: 
> 1) 116,57 degrees
> 2) face to face: 2,270 times the edge length
As Robert A. Moeser has pointed out, this is wrong. The last figure
should read: 2,2270. Sorry for the copying error. H. Oe.

> Richard Gain wrote:
> >   1)   What is the angle between two adjacent faces?
> >   2)  What is the relationship between edge length of the pentagonal faces and diameter (face to face or vertex to vertex, whichever is easier) of the dodecahedron?
> H. Oelschlaeger wrote: 
> 1) 116,57 degrees 
> 2) face to face: 2,270 times the edge length
Erratum: as Robert A. Moeser has pointed out, it should read: 2,2270.
Sorry about this copying error.         H. Oe.

need by 6am nov 14
(f-p)sqr + (n+p)sqr =(q+p)sqr
need to solve for p

> Archimedes Plutonium (Archimedes.Plutonium@dartmouth.edu) wrote:
> : 
The redoubtable AP comments---
> :    You have failed to see my point and my message.
Well, *sigh* what I actually failed to see were the several email messages
warning me (too late) not to ever answer one of your posts.  Now I
understand why.
I saw your point and your message---I happen to disagree, which, last time
I checked, was my right and was not a green light for condescension and
derision from you.  
It is unfortunate that instead of using proper logical argument to attempt
to prove your "points,"  you chose to descend to a personal attack on me.
To quote Dr. Foakes-Jackson (1855-1941) of Cambridge ..."It's no use
trying to be clever, we are *all* clever here.  Just try to be kind."
Darla
---who hereby thanks her brave champions, both public and private.

James Hannum wrote:
> 
> Matthew P Wiener wrote:
> 
> > Some consider that cheating.  "Real" cross products, the two vectors at a
> > time sort, work in dimensions 3 *and* 7.  The familiar 3-dimensional one
> > can be thought of as the purely imaginary part of quaternion multiplication.
> > Similarly, the purely imaginary part of octonion multiplication can be read
> > as a 7-dimensional cross product.  Most of the familiar identities hold in
> > both cases.
> 
> Do cross products only work in 3 & 7 dimensions, or will they work
> anytime
> there are 2^n - 1 dimensions (where n > 1 and an integer)?
One can define a 4-d cross product which is very useful in quaternion 
math.  It's analogous to the 3-d one.  Using it, the product of two 
quaternions becomes something like
	q = q1 x q2
and the derivative of q becomes
	q dot = (1/2)omega x q
where omega is the angular velocity, and treated as though it were a 4-d 
vector with zero fourth component.
Jack

> Richard Gain wrote:
> >   1)   What is the angle between two adjacent faces?
> >   2)  What is the relationship between edge length of the pentagonal faces and diameter (face to face or vertex to vertex, whichever is easier) of the dodecahedron?
I wrote: 
> 1) 116,57 degrees
> 2) face to face: 2,270 times the edge length
As Robert A. Moeser has pointed out, this is wrong. The last figure
should read: 2,2270. Sorry for the copying error. H. Oe.