Subject: Re: (-2) * (-3) = (- 6) ; why not ?
From: cjm@purdue.edu (Chad J McQuinn)
Date: Thu, 16 Jan 1997 18:16:01 -0500
In article <32d698a2.134401@aplnews>, mfein@aplcomm.jhuapl.edu (Matt
Feinstein) wrote:
...
> It -does- seem unnecessarily complicated..
Yes, it does, and I'm afraid I can't do a lot to simplify your argument,
but try this (somewhat) more intuitive approach.
Get the person to agree that (-1)*(-1) has absolute value 1. Then, if
(-1)*(-1)=-1, we have
(-1)=1 by division, a contradiction. Hence it must be that (-1)*(-1)=1.
Now, a*b*(-1)*(-1)=a*b, so (-a)*(-b)=a*b.
Hope this helps,
Chad
Subject: Re: Fibonacci
From: rjc@maths.ex.ac.uk
Date: Fri, 17 Jan 1997 05:51:04 -0600
In article ,
tram@olympic.seas.ucla.edu (Tri Tram) wrote:
>
> I am wondering what is wrong with this proof:
> Given the recurrence relation for Fibonacci numbers F(n)=F(n-1)+F(n-2). We
> considered the function G(n)=G(n-1)+G(n-2)+1. It seems obvious that G(n)>F(n)
> We can prove that G(n)=F(n)-1 by induction.
> We assume that G(k)=F(k)-1 for all k such athat 1<=k<=n and we consider
> G(n+1):
> G(n+1)=G(n)+G(n-1)+1 =F(n)-1+F(n-1)-1+1 = F(n+1)-1.
This is correct PROVIDED that your initial values of F and G also satisfy
G(n) = F(n) - 1. Otherwise your induction can't start. If you set
F(1) = F(2) = 1 and G(1) = G(2) = 0, then F(n) = G(n) + 1 always. But if
you had started with G(1) = G(2) = 1, then F(n) = G(n) + 1 is false, but
you can easily show by induction that G(n) > F(n) for all n > 2.
Robin J. Chapman "... needless to say,
Department of Mathematics I think there should be
University of Exeter, EX4 4QE, UK more sex and violence
rjc@maths.exeter.ac.uk on television, not less."
http://www.maths.ex.ac.uk/~rjc/rjc.html J. G. Ballard (1990)
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Subject: Re: Why can't 1/0 be defined???
From: pausch@electra.saaf.se (Paul Schlyter)
Date: 17 Jan 1997 08:57:47 +0100
In article ,
David Kastrup wrote:
> In computer "real" math those laws *are* already broken. Putting
> infinity into computer numbers does not do much harm anymore, thus.
> Or a special number NAN or so (what would be the typical result for
> 0/0).
0/0 should be computed as INDEF(inite). NAN (not-a-number) is
reserved for "values" which truly cannot be real numbers, such as
the sqrt or the log of a negative number.
Yes, there is a difference: if you add, subtract, or multiply an
INDEF with an INF, you'll get an INF. But if you do the same with
a NAN and an INF, you'll get a NAN.
--
----------------------------------------------------------------
Paul Schlyter, Swedish Amateur Astronomer's Society (SAAF)
Grev Turegatan 40, S-114 38 Stockholm, SWEDEN
e-mail: pausch@saaf.se psr@net.ausys.se paul@inorbit.com
WWW: http://www.raditex.se/~pausch/ http://spitfire.ausys.se:8003/psr/
Subject: Re: Cute Proofs...
From: rjc@maths.ex.ac.uk
Date: Fri, 17 Jan 1997 02:51:03 -0600
In article <5bh6ps$f5k@gap.cco.caltech.edu>,
zare@cco.caltech.edu (Douglas J. Zare) wrote:
>
> NCC-1701-E wrote:
> >[...]
> >Prob : prove that C(2n, n) is divisible by (n+1).
> >
> >Note : C(a,b) refers to the Binomial coefficient of x^b in (1+x)^a.
> >
> >Proof: C(2n,n)/(n+1) = 2C(2n, n) - C(2n+1, n+1) q.e.d
>
> I think that this statement assumes too much. The quotients are the
> Catalan numbers, which count, among other things, the number of lattice
> paths from (0,0) to (n,n) which do not go above the main diagonal. The
> following is in Graham, Knuth, and Patashnik's _Concrete Mathematics_ and
> Richard Stanley told me it was known to Cayley, though I have seen it
> published as new in a combinatorics journal in 1993:
>
> There are C(2n,n) lattice paths from (0,0) to (n,n). Let us construct
> equivalence classes of size n+1 such that each equivalence class contains
> precisely one path which does not go above the main diagonal. For each
> lattice path, write the sequence of horizontal and vertical steps, and add
> a horizontal step at the end. Then repeat this sequence of length 2n+1.
> n+1 lattice paths will give this one (one for each horizontal step which
> could have been added). There is a unique highest point per period where a
> line of slope n/n+1 can rest on the path. It is followed by a horizontal
> step, and this is the added step of the unique member of the equivalence
> class which does not go above the diagonal.
G/K/P attribute this to a 1959 paper of Raney. Also Conway & Guy use this
in their recent "The Book of Numbers". They give about half a dozen other
interpretations of Catalan numbers.
> I would like to have a similarly combinatorial proof that
> (3a! * 2)/(a! a+1! a+2!) is an integer.
So would I...
Robin J. Chapman "... needless to say,
Department of Mathematics I think there should be
University of Exeter, EX4 4QE, UK more sex and violence
rjc@maths.exeter.ac.uk on television, not less."
http://www.maths.ex.ac.uk/~rjc/rjc.html J. G. Ballard (1990)
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Subject: Re: Topology problem.
From: ikastan@alumnae.caltech.edu (Ilias Kastanas)
Date: 17 Jan 1997 08:49:20 GMT
In article <32DD1ADE.54FD@acs.ucalgary.ca>,
Bogdan Georgescu wrote:
>This should make it clear once and for all.
>
>~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
>Define a perfect continuous function as being a function
>
>f:[0,1]->R(the set of real numbers)
> ^ the compact[0,1]
>
>and for all x in [0,1] there is a sequence {x_n} which converges to x
>(but it is not constant (=x) starting from any n) such that f(x_n)=f(x)
>
>Prove:
>
>The sum of two perfect continuous functions is a perfect continuous
>function.
>~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
It is not true.
Let [0,1] -> [0,1]^2 : x --> (f(x), g(x)) be Peano's space-filling
arc, e.g. as defined by Hilbert, via successive subdivision into squares.
In particular 0 --> (0, 0) and 1 --> (1, 0). Then f and g are conti-
nuous and perfect; f(x) = 0 for continuum-many x, the arc points on the
square's left side, and so on. But f+g is not perfect; consider when
does f(x) + g(x) = 0.
Ilias
Subject: Re: I challenge anyone to solve this one
From: Vincent Johns
Date: Fri, 17 Jan 1997 02:48:03 -0600
Justin wrote:
>
> In article <5bgu7h$1m2a@inst.augie.edu>, (Augie) wrote:
>
> * "Nathan Crowder" wrote:
> *
> * >How many real solutions does the equation sin(x)=(x/100) have?
>
> X15.5, 6.34, 3.11, 0, -3.11, -6.34, -15.5
>
> --
> JZS 3=)
Well, he asked "how many", not "please list a few of them".
Other solutions include 9.33 and -9.33.
OTOH, if the argument x is measured in degrees, the number
of solutions is much smaller (I'd say about 3). ;-)
--
-- Vincent Johns
Please feel free to quote anything I say here.
Subject: Re: 1 / 2^.5 or 2^.5 / 2?
From: Vincent Johns
Date: Fri, 17 Jan 1997 03:14:01 -0600
(posted & emailed)
meron@cars3.uchicago.edu wrote:
>
> In article <32DD16E9.1BE1@cnas.smsu.edu>, Shyang Hwang writes:
> >Dave Seaman wrote:
> >>
> >> I have read that the original purpose of BASIC was to teach beginning
> >> programming students about concepts of assembly language, [...]
> >
> >Are you sure? I don't see how the concepts of assembly language can
> >ever be taught by teaching BASIC. Rather, I think BASIC was developed
> >so that people did not have to learn the concepts of assembly language
> >and still be able to program a computer by using an English-like
> >language.
> >
> No, that's not quite the reason, since English-like languages already
> existed by the time BASIC was developed. The specific rationale
> behind BASIC was the need for an "English-like" language which can be
> used on machines with very limited memory. Thus all features which
> already existed, in FORTRAN for example, but were not considered
> absolutely essential, were sacrificed to save space.
I don't claim to have any special insight into John Kemeny's
reasoning in implementing BASIC, but one huge advantage of using
it was the ability to run programs from terminals that might be
in several rooms around the campus, or at the opposite end of the
country. People used it both ways. There was no need to buy
separate computers; Teletype terminals were a lot cheaper than
a computer (though a terminal was more expensive than a PC is today)
and put computing power into the hands of many people who could not
otherwise afford it. Long-distance phone bills could eat up one's
budget, though.
BTW, limited memory is no reason not to use FORTRAN, etc.; I used
FORTRAN on a Control Data 160 computer with 6K bytes of memory
augmented by some virtual memory (in the form of punched paper
or Mylar tape). It wasn't blazingly fast, but it produced useful
results.
--
-- Vincent Johns
Please feel free to quote anything I say here.
Subject: Re: Please solve: x^3-x^2=1, X^4-x^3=1, or x^5-x^4=1
From: pmontgom@cwi.nl (Peter L. Montgomery)
Date: Fri, 17 Jan 1997 12:59:58 GMT
In article bruck@pacificnet.net (Ronald Bruck) writes:
>In article , jeriley@azstarnet.com wrote:
>
>:What are the solutions to these problems? I've had a math block and need a
>:clue.
>
(solutions to cubic and quartic deleted)
>I haven't asked Mathematica to solve the quintic, because (although I
>haven't checked the Galois group) it's probably not solvable in radicals.
>Numerically, the roots are approximately
> {{x -> -0.662358978622373 - 0.5622795120623012*I},
> {x -> -0.662358978622373 + 0.5622795120623012*I},
> {x -> 0.5 - 0.8660254037844385*I},
> {x -> 0.5 + 0.8660254037844385*I},
> {x -> 1.324717957244746}}
Two of the complex roots seem to have real part exactly 1/2.
The quadratic equation satisfied by those roots seems to be
x^2 - x + 1 = 0. Sure enough, the original quintic is reducible:
x^5 - x^4 - 1 = (x^2 - x + 1) * (x^3 - x - 1)
The quadratic and cubic can be solved using radicals if desired.
QUESTION: Is x^(n+1) - x^n - 1 irreducible over Z
unless n == 4 (mod 6)?
--
Peter L. Montgomery pmontgom@cwi.nl San Rafael, California
Will the bridge to the 21st century tolerate the floods of 1997?
Subject: Re: Proof of infinite prime #'s
From: jcea@argo.es
Date: Fri, 17 Jan 1997 03:51:14 -0600
> A caveat: the number
>
> a := (1 x 2 x 3 x 5 x 7 x 11 x 13) + 1
>
> is one more than a multiple of 2
> is one more than a multiple of 3
> is one more than a multiple of 5
> is one more than a multiple of 7
> is one more than a multiple of 11
> is one more than a multiple of 13
>
> but is *not* a prime. In fact, a = 59 x 509. Check it out!
59 and 509 are primes. If you consider that 1, 2, 3, 5, 7, 11 and
13 are the only primes, then 1*2*3*5*7*11*13+1 is:
a) A prime number
or
b) A composite number with factors beyond 13
Now the proof has been "fixed" ;-). Prime numbers are infinite.
Any comment?
--
Jesus Cea Avion
jcea@argo.es
"El amor es poner tu felicidad en la felicidad de otro" - Leibnitz
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Subject: Re: small rings
From: rjc@maths.ex.ac.uk
Date: Fri, 17 Jan 1997 04:01:06 -0600
In article <5bmqt7$8q2@gap.cco.caltech.edu>,
bbrock@pepperdine.edu wrote:
>
> Typically in a course that covers groups one eventually
> encounters a table of the number of nonisomorphic groups
> of small order, e.g.
>
> 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
> 1 1 1 2 1 2 1 5 2 2 1 5 1 2 1 14 1
>
> I wanted to make a similar table for rings (with multiplicative
> identity) of small order, and I came up with
>
> 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
> 1 1 1 4 1 1 1 ? ? 1 1 ? 1 1 1 ? 1
>
> It appears that there's only 1 if the order is squarefree.
> The number of rings of order 8 appears to be >10.
> In particular I think 8 is the size of the smallest
> noncommutative ring. Before I waste time finishing
> order 8, has anyone seen such a table before?
>
> Also, what is the size of the smallest ring that
> does not have a multiplicative identity?
I haven't seen such an enumeration before. If the order is square-free the
additive group is cyclic, and can be identified with the integers modulo n.
The multiplication is then given by x*y = axy for some a ( = 1*1). If a
is invertible, then the ring has an identity and is isomorphic to Z_n
as a ring. Any 4-element ring with an identity is either cyclic as
an additive group, and so Z_4 as a ring by the previous argument, or
of characteristic 2, and so equal to {0, 1, x, x+1}. The multiplication
is determined by x^2, and the ring is commutative. Thus the smallest
non-commutative ring has 8 elements, and such exist, e.g., the upper triangular
2x2 matrices over the field F_2.
If one drops the assumption of an identity, then one can get a ring
from any additive group by setting the multiplication to be identically zero.
So the smallest ring without identity has 2 elements. The smallest with
non-trivial multiplication have order 4, e.g., the integers mod 4 with
x*y = 2xy, or the non-commutative example of the 2x2 matrices over F_2 with
zero second row.
Incidentally, each finite ring is a direct product of rings of prime power
order, and so there are 4 unital rings of order 12.
Best of luck with order 8, I'm sure there are plenty of examples!
Robin J. Chapman "... needless to say,
Department of Mathematics I think there should be
University of Exeter, EX4 4QE, UK more sex and violence
rjc@maths.exeter.ac.uk on television, not less."
http://www.maths.ex.ac.uk/~rjc/rjc.html J. G. Ballard (1990)
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Subject: Heronšs formula
From: gwhite@jetlink.net (Glenn White)
Date: Wed, 15 Jan 1997 12:04:16 -0800
I was trying to figure the area of a triangle using Heronšs formula which
is area = sqrt((a+b+c)(c+b)(a+c)(b+C)).
The points I was trying to use were:
origin (0,0)
y-axis (0,2)
x-axis (4,0)
Using both the vertex formula and the basic (1/2)bh formula I come up with
4. I can not get the Heronšs formula to work out. I tried the following:
a = height
b = base
c = hypotenuse
This did not work. Can anybody give me some insight as to what a, b, and c
should be or is the formula in the book wrong?
...Glenn
Subject: Re: Just for fun...
From: David Kastrup
Date: 17 Jan 1997 11:49:12 +0100
rosentha@classic36.rz.uni-duesseldorf.de () writes:
> JRANCK@ix.netcom.com wrote:
> : My high school physics teacher asked this question the other day, just
> : for fun. Anybody know how to solve it?
>
> : You have 12 silver balls that look identical. However, one is either
> : slightly heavier or slightly lighter than the others. You also have a
> : balance, which you may use only 3 times to find the ball that is
> : different. How do you find it?
>
> : I can find it with 8 balls, but not 12. Anybody know?
> : Mike
> Hint: It does not work if you start with 2 sets of 6 balls.
Two sets of fours are fine. If they weigh equal, the offender is in
the rest 4 balls, and we have 8 reference balls of correct weight.
We then weigh 2 in the offending group against one clean and one from
the offending group. If they weigh equal, the non-weighed ball from
the offending group is faulty, and we can find out whether it is
lighter or heavier by one weighing.
If they don't, we weigh the two-group of offending balls against one
another. If they weigh equal, the one-group of offending balls is
really at fault, and we know in which direction. If they weigh
non-equal, the one weighing just like the two-group before is at
fault.
Ok, now let's suppose the first weighing turned out *unequal*. Then
things get complicated. I'll just mention that the next weighing
weighs some of the possibly faulty balls, some neutral ones, and
switches the side of some possibly faulty balls.
That's the really complicated thing, and so will be left as an
exercise to the reader. But I've heard that it is given in the FAQ of
rec.puzzles or so...
--
David Kastrup Phone: +49-234-700-5570
Email: dak@neuroinformatik.ruhr-uni-bochum.de Fax: +49-234-709-4209
Institut f=FCr Neuroinformatik, Universit=E4tsstr. 150, 44780 Bochum, Germa=
ny
Subject: Re: computational infeasibility and absolute noncomputability
From: Ariel Scolnicov
Date: 17 Jan 1997 15:35:49 +0200
harden writes:
> When I think about the issues of NP-completeness and the unsolved
> problem about whether or not P=NP, I notice many similarities between
> issues of computational infeasibility( if, indeed, P does not equal NP )
> and absolute noncomputability. For instance, take the issue of easy
> cases of the problems: for easy cases of noncomputable decision
> problems, there is an algorithm( like determining whether or not a
> Diophantine equation has a solution if either the number of
> indeterminates or the degree is 1 ), and for ( probably )computationally
> infeasible problems, the easy case is computationally feasible( like the
> partitions problem for a set of integers whose sum is odd ). If a
> decision problem cannot be done in P, then, to any machine that performs
> only a polynomial( in the number of bits of input ) number of operations
> will see a sequence of 1's and 0's formed by answering a( computable )
> decision problem which is not in P as noncomputable, with all the
> similarities between what it sees, with its limited view, and actual
> noncomputability. Could this somehow be employed to prove, if this is
> the case, that P does not equal NP( and that cryptographers will rejoice
> and use systems based on NP-complete problems )?
I think this is probably a question about whether we can use
diagonalisation methods to solve P=NP. Let me clarify: To show some
questions don't have a computable solution, you take, say, the halting
problem, and derive a contradiction from the assumption that some
program solves halting questions. The contradiction comes from feeding
(some variant of) the program to itself.
We know this kind of argument won't work for P=NP. This is because of
computations with "oracles". A computer-with-oracle (actually Turing
machine-with-oracle, if you're into that sort of thing) is just like
your ordinary computing device, EXCEPT that at any stage you can ask
whether something is a member of some language L. The language L,
fixed in advance, is the oracle: it is essentially a "black box" which
solves some problem for you. For instance, if L is the language of
pairs (k,n), where n is the longest time a terminating program of
length k may run on the empty input, then a
computer-with-(this)-oracle can solve the halting problem.
Denote by P^L the set of languages which a computer-with-oracle-L can
accept in polynomial time, and define NP^L similarly. It can be shown
that there are languages A and B for which P^A=NP^A and P^B!=NP^B. Now
suppose you had some diagonalisation proof that P!=NP. The same proof
would work for the case of P^A and NP^A, except you already know
they're equal. What the existence of these languages A and B really
means is that to prove P=NP or P!=NP you'll need to take a long hard
look and P and NP. This is quite different from the case for recursive
(=computable) functions, where you can just hook up some black boxes
to get a proof.
Better newsgroups for this sort of thing are sci.logic and comp.theory
(you'll also find people who know more about this than I do there, and
probably also on this group).
--
Ariel Scolnicov
Subject: Re: Why can't 1/0 be defined???
From: ags@seaman.cc.purdue.edu (Dave Seaman)
Date: 17 Jan 1997 09:33:32 -0500
In article ,
Travis Kidd wrote:
>ags@seaman.cc.purdue.edu (Dave Seaman) writes:
>>One nit. The expression 0^0 does indeed have a value, which has
>>nothing to do with limits. This is discusses in the sci.math FAQ,
>>which gives several reasons why 0^0 = 1, at least in the case where the
>>exponent is considered to be an integer.
>You cannot give reasons in mathematics, except for reasons why (actually
>how) we know something to be true. 0^0 is indeterminate--possibly on
>all the reals, possibly only between 0 and 1.
The adjective "indeterminate" applies only to limit expressions. I was
not discussing a limit expression. Since 0^0 = 1 is a theorem of ZF
and can be found in textbooks on axiomatic set theory, I consider that
to be a reason for saying that 0^0 = 1 just as surely as 2 + 2 + 4, as
long as you are talking about arithmetic on the natural numbers.
>>My favorite reason: 0^0 is the cardinality of the class of functions
>>mapping the empty set to itself, which is one.
>That's certainly one of the uses of 0^0. But not the only one.
Another "use" of 0^0 lies in the fact that if x is an element of any
monoid G, then x^0 = the identity element of G. The exponent 0 in this
case is the natural number 0. Since the real numbers are a monoid, it
follows that 0^0 = 1, where the first zero is a real number, the second
zero is a natural number, and the result is a real.
--
Dave Seaman dseaman@purdue.edu
++++ stop the execution of Mumia Abu-Jamal ++++
++++ if you agree copy these lines to your sig ++++
++++ see http://www.xs4all.nl/~tank/spg-l/sigaction.htm ++++
Subject: Re: Problomatic Teacher
From: D&A; Klinkenberg
Date: Tue, 14 Jan 1997 04:17:49 -0500
Hello,
Keith Pitcher wrote: (in part)
>
> Hello,
> recently my sister's math teacher asked a question, and did not accept
> her
> logical answer (He gave an "F" to the poor girl). I will be meeting with
> the teacher this week to discuss this matter, and am seeking support to
> show the teacher the error in his ways.
> His question for his standard 7th grade math class, in verbatim, was as
> follows:
>
> Q) Take a square piece of paper. Fold it in half. Do it again. Repeat 25
> times. How many sheets thick is the final folded piece of paper.
>
> My sister realized that this was a trick question, as she knew a piece
> of paper can
> not be folded that many times in half, and so far every question had
[snip]
To me this Q) is at least very ambiguous. I think it is hard to
interpret what step(s)are to be repeated. I agree that paper cannot be
folded that many times but I am finding fault with another issue. To
avoid the question of how many times a physical piece of paper can be
folded, let me change the teacher's problem to "repeat 3 times" like
this:
Step 1) Take a square piece of paper
Step 2) Fold it in half
Step 3) Do it again
Step 4) Repeat 3 times
I would need to ask some questions to see what is being requested. I
would want to know whether in step 3), "it" refers to step 2) alone or
to both steps 1) and 2). I would also want to know whether in step 4),
"repeat" refers to a) step 2) alone; b) all 3 steps; or c) some other
choice of steps.
If we assume that steps 3) and 4) each refer to step 2) alone, then
does step 4) mean "a total of 3 folds" or does it mean "3 more folds"
for a total of 5 folds? Other assumptions lead to similar questions.
Dan
Subject: Re: Random Associative Functions
From: WayneMV@LocalAccess.Com (Wayne M. VanWeerthuizen)
Date: Fri, 17 Jan 1997 10:09:30 GMT
On 14 Jan 97 13:00:10 GMT, Simon Read wrote:
>On Sat, 04 Jan 1997 22:55:27 GMT, WayneMV@LocalAccess.Com (Wayne M.
>VanWeerthuizen) wrote:
>>>
>>>I know the function is cummulative if the matrix is symetrical along
>>>the diagonal.
>symmetrical along diagonal means f(x,y) = f(y,x) which is...
>
> COMMUTATIVE.
Yes, sorry.
>You imply that is is easy to check if a matrix is symmetrical.
>I think you need to check every pair of elements either side of
>the leading diagonal. For an n x n matrix, this is approximately
>n^2/2 pairs. It's not an "instant" comparison. There may be a
>conceptually simple algorithm to _generate_ such a matrix, ie
>reflect it in the leading diagonal, but even this takes n^2/2
>operations.
>
>To check associativity, you need to check f(x,f(y,z)) and see if it
>equals f(f(x,y),z) To me this looks like n^3/2 comparisons.
Why are you dividing by two in this case?
>I can't, offhand, think of an easy way to generate an associative
>matrix, although there is a possibility of an n^2 algorithm, which
>would be a lot quicker than the n^3-type test of associativity.
>
>ZimoZ
>
Okay, I'll remember that the word is "commutative". And I'll remember
to spell "associative" with two S's and one C. ;-)
Your last paragraph is along the lines of what I was thinking, "Is
there an O(n^2) algorithm to create a random associative matrix?"
Subject: Re: Speed of Light
From: David Kastrup
Date: 17 Jan 1997 13:36:26 +0100
vit writes:
> > speed travel. The gist of the article included a computer program that
> > showed what objects would look like at various speeds.
> >
> > The most interesting aspect to me professionally was the observation of
> > electrical circuit response at those speeds. For instance at about .75c
> > computer response of electronic equipment would become prohibitively sl=
ow
> > for connections in access of 300ft. Based on your discussion I think th=
is
> > Conflicts to one of your opinions but I'm not sure.
> >
> -snip-
> I don't think the article you read was a very serious one. As stated in
> the postulats of theory of relativity, there's no way to distinguish
> between any two inertial systems. If the computer is not moving with
> respect to an inertial frame of refference, it has to perform the same
> as on Earth or anywhere else. No slowing down or whatever.
Well, viewed from outside, the computer would have serious problems
reacting to something approaching it because the info would run slowly
through the wires to the front.
Viewed from inside, the problem stays the same, just looks different.
From the inside it looks as
a) the approaching thing is indeed much nearer than from viewed
outside, so we have less time to react to it
b) the light from the approaching thing is not much faster than the
thing itself, so we have even less time to react (if we consider the
speed with which the *information* reaching us about the approach
increases, it can be definitely *more* than the speed of light, even
viewed from inside).
While the internals on this ship will work perfectly normally, the
interaction with the outside could be rather unfriendly, so to speak.
Relativistic effects make intergalactic speeding even more dangerous
than it would be anyway.
--
David Kastrup Phone: +49-234-700-5570
Email: dak@neuroinformatik.ruhr-uni-bochum.de Fax: +49-234-709-4209
Institut f=FCr Neuroinformatik, Universit=E4tsstr. 150, 44780 Bochum, Germa=
ny
Subject: Re: Average value of a function
From: candy@mildred.ph.utexas.edu (Jeff Candy)
Date: 17 Jan 1997 14:52:48 GMT
gareth :
|> I'm sure that this is a trivial problem, but I get so far in solving it
|> and then can't deal with the expressions i come up with. Here goes:
|> I'm trying to find an analytical way of finding the average value of a
|> function. For exmple, if you have the function f(x) = x^2 Then what is
|> and expression for the mean value of the function between x=a and x=b?
|> The actual function I am trying to do this for is f(x) =
|> +sqrt(r^2-x^2) (the euation of the semi-circle), However I would be
|> interested in a more general solution as well.
The average of f on the interval [a,b], denoted by , is:
b
1 /
= --- | f(x) dx
b-a /
a
If you grasp the concept of integration, the formula should
probably be intuitively obvious.
For your example f(x)=x^2 on [a,b], the answer, after some
simplification, is:
2 2
a + ab + b
= -------------
3
-------------------------------------------------------------------
Jeff Candy The University of Texas at Austin
Institute for Fusion Studies Austin, Texas
-------------------------------------------------------------------
Subject: Re: Why can't 1/0 be defined???
From: David Kastrup
Date: 17 Jan 1997 13:19:46 +0100
tkidd@hubcap.clemson.edu (Travis Kidd) writes:
> You cannot give reasons in mathematics, except for reasons why
> (actually how) we know something to be true. 0^0 is
> indeterminate--possibly on all the reals, possibly only between 0
> and 1.
Wrong. An expression involving only constants is *never*
indeterminate. It either has a value, or is undefined.
The function x^y is indeterminate at (0,0). That's an entirely
different thing. It means that lim (x,y)->(0,0) x^y does not exist.
This is commonly called an indeterminate limit.
An expression involving limits can be undefined because the function
the limit is taken of is undeterminate at the limit.
Fine distinction.
But for a function to have limits, it needs to have values. And where
it has values is the mathematicians' decision. It turns out that
0^0=3D1 has decidedly more important reasons supporting it than other
choices have which is why today's general consent is that it is the best
definition. See the appropriate section of the FAQ for more info.
--
David Kastrup Phone: +49-234-700-5570
Email: dak@neuroinformatik.ruhr-uni-bochum.de Fax: +49-234-709-4209
Institut f=FCr Neuroinformatik, Universit=E4tsstr. 150, 44780 Bochum, Germa=
ny
Subject: Re: Idle query: how good are math and science teaching outside the U.S.?
From: mert0236@sable.ox.ac.uk (Thomas Womack)
Date: 17 Jan 1997 11:34:42 GMT
M.LJoyce (martin@kimmi.demon.co.uk) wrote:
: Anyway, I still can't figure the inverse Laplace transform of
: s^2/(s+a)(s+b), any help ?
: P.S. Is there a DERIVE fan club out there, or has windoze got you all
: MATLAB ?
I'm a MAPLE zealot. It's available on the Massive Great Server here.
>readlib(laplace):
>invlaplace(s^2/(s+a)*(s+b),s,t);
3 2 2
- a exp(- a t) + Dirac(2, t) - a Dirac(1, t) + a Dirac(t) + b a exp(- a t)
+ b Dirac(1, t) - b a Dirac(t)
(which I think can be simplified to
a^2(b-a)exp(-at) + Dirac(2,t) - (a-b)Dirac(1,t) + a(a-b)Dirac(t) )
[where Dirac(n,t) is the n'th derivative of Dirac's delta function evaluated
at t].
: --
: Martin@kimmi.demon.co.uk
: "the crux of the biscuit is the apostrophe"F.Z.
--
Tom
The Eternal Union of Soviet Republics lasted seven times longer than
the Thousand Year Reich
Subject: Re: Geometry Problem
From: "OD"
Date: 17 Jan 1997 13:57:18 GMT
Alex Papazoglou wrote in article
...
> Here's a little problem I came across. Can someone give
> me its proof?
>
> Assume a (finite) number of points on the same plane:
> A1, A2, A3, ..., An (not necassarily in that order) with
> the following property: On the line connecting two
> of the n points at least one more of them can be placed.
> Prove that they are all points of the same line.
>
> Alex
>
>
This is obviously true for n<=2. Now suppose it is true for n=k>=2, and
consider a set of k+1 points. By hypothesis, the first k points lie on the
same line L. If the last point were not on the line L, then the line L'
joining the point A(k+1) to A1 would not contain other points of the set.
Indeed L and L' are different because A(k+1) belongs to L but not to L',
hence they have at most one point in common. Consequently A(k+1) belongs to
L, and the proposition is proved.
Subject: Re: paradox
From: owl@rci.rutgers.edu (Michael Huemer)
Date: 17 Jan 1997 08:36:59 -0500
This is the first message in this thread that I've seen (because the
first that's appeared in sci.philosophy.meta), so I'm here risking
repeating things that have already been said. Nevertheless...
>: > : > Thus all of the non-black things you find which
>: > : > aren't ravens (your red coat, the white ceiling, etc.)
>: > : > also support your generalization that "all ravens are black".
>: >
>: > Hm, isn't there a name for this paradox? 'Hempel paradox' or
Yes, it's called the Ravens Paradox.
>: This is not a paradox. The statement "all non-black things aren't
>: ravens" is the contrapositive of the hypothesis "all raven's are black.
>: Mr. Allen is correct; the two statements are logically equivalent.
Yes, they are logically equivalent, but yes, it is still a paradox.
(see below)
>: I'd not call the ornithologist lazy, however. In order to prove the
>: hypothesis (all ravens are black) by demonstrating the contrapositive,
>: he'll have to examine each and every non-black item and show that
>: none of them are ravens. I'd call that a more daunting task even than
>: checking up on all the ravens.
I think you've misunderstood the paradox. Let me explain it. First,
the paradox is about 'confirmation.' Confirmation is a relationship
between propositions. "A confirms B" means that A provides *some*
evidence, some support, for B. Note that it does not mean either of
the following: (a) that it absolutely proves B (i.e. entails B), or
(b) that it provides enough support that we should believe B. Rather,
A confirms B just means that A provides *at least some* support,
however small, to B. And of course, if you get *enough* confirmation,
then you get a justified belief (so the other two relations I
mentioned are *species* of confirmation).
Now, most of our knowledge, including all oour scientific knowledge,
is based on evidence which confirms but does not entail it. The
concept "confirmation" is clearly central both to philosophy of
science and to epistemology generally.
It would be nice to hae a theory of confirmation (something analogous
to the systems of deductive logic that we already have) - something
that would tell us when we have confirmation and when we don't. Of
course, the best thing would be if we could precisely measure
confirmation (e.g., "there is 57% confirmation between A and B
here..."), but the least we could hope for is a *qualitative* account
of when A confirms B.
Here's a start at that. Here are some intuitively plausible
principles that ought to govern the 'confirmation' relation:
1. The observation of an A that is B confirms "All A's are B."
2. The observation of an A that is non-B disconfirms "All A's are B."
3. The observation of a non-A is irrelevant to (neither confirms nor
disconfirms) "All A's are B."
4. If P is logically equivalent to Q, then whatever confirms P
confirms Q.
(The first three principles are collectively called "Nicod's
criterion".)
The Ravens Paradox results because we see that these 4 principles,
which at least appear obviously true, are inconsistent. For consider
the observation of a white shoe. This object is a non-black
non-raven. Therefore, by (1), it confirms "All non-black things are
non-ravens." But "All non-black things are non-ravens" is logically
equivalent to "All ravens are black." Therefore, by (4), the
observation of a white shoe confirms "All ravens are black." However,
by (3), the observation of a white shoe is irrelevant to whether all
ravens are black.
Thus, one of these principles has to go. Which one?
--
^-----^
Michael Huemer / O O \
http://www.rci.rutgers.edu/~owl | V |
\ /
Subject: Re: Evidence for God's Existence - TRY Math
From: David Kastrup
Date: 17 Jan 1997 13:03:24 +0100
geo@3-cities.com writes:
> llockhar@Direct.CA (Lloyd Lockhart) wrote:
>
> >Hmmmm... another intelligent one here....
>
> After reading a few of your buttnuggets, I can only presume you are
> another fundie-idiot.
You shouldn't mistake "buttnuggets" for Braille.
> If atheism ever becomes a capital offense,
> then I want to be regarded as the Charles Manson
> of atheism. - Geo
You appear more like the Lucy van Pelt of it. And I'm still
flattering.
--
David Kastrup Phone: +49-234-700-5570
Email: dak@neuroinformatik.ruhr-uni-bochum.de Fax: +49-234-709-4209
Institut f=FCr Neuroinformatik, Universit=E4tsstr. 150, 44780 Bochum, Germa=
ny
Subject: Just what do you want anyway?
From: "John P. Boatwright"
Date: Fri, 17 Jan 1997 05:55:25 +0000
John Sanger wrote:
> Nothing mentioned in your collection of fictions and fables has ever
> come to pass..... That collection has no validity as you cannot provide
> the proof for the existence of your diety....
Well, Eze 37 did and remains to finish out. Also Daniel, Isaiah,
Jeremiah, ect... I guess you guy's don't read the multitude of
bibles you hoard (Ask the newsgroup atheists: Dumb, Six, Catapilla,
Parklady, Dill, Goe, ect...), do you? Best wishes.
Subject: Re: Understanding Number Magnitudes
From: dfilpus@bnr.ca (David Filpus)
Date: 17 Jan 1997 15:19:15 GMT
In article <32DF2F16.149B@unpsun1.cc.unp.ac.za>, "Matthew C. Clarke" writes:
|> Joseph (JaKe) Kisenwether wrote:
|> >
|> > I'm trying to design a project to give K-12 students an understanding of
|> > the magnitudes of numbers. Too often they hear things link "The Earth is
|> > 4 1/2 billion years old." and dutifully memorize them without having any
|> > idea how large a billion is. I would like to come up with several
|> > examples of things which are measured by numbers of each order of
|> > magnitude. Let me give you a couple of examples:
|>
|> [snip]
|>
|> The best presentation of this I have seen is "Powers of Ten". This is
|> available in both video and book form, but I'm sorry I don't have a more
|> detailed reference or source. I did have the book but loaned it out and
|> didn't get it back.
|>
|> The video starts with an aerial view of a family having a picnic and
|> gradually zooms outwards so that you see the park they are lying in, the
|> city, the continent, the planet, the solar system etc. It then zooms
|> back in again, back to the starting point and beyound -- to the man's
|> hand, the skin cells, the molecules within a cell, the atomic structure
|> etc. The accompanying commentary emphasises the relative sizes of things
|> in terms of magnitudes of ten.
|>
|> Matt.
"Powers of Ten" is shown as an exhibit in the NASA part of the
Virginia Air and Space Museum in Hampton, VA. It is shown on a circular
screen in the floor of a darkened room. You truly feel like you are
moving up (or down) through the magnitudes, because you are above the screen
lookin down into it.
--
Dave Filpus | Opinions in this post are my own
NORTEL-PCN Technology | and do not reflect those
RTP, NC | of NORTEL Public Carrier Networks Technology
dfilpus@nortel.ca |
Subject: Re: Update inverse of a covariance matrix.
From: shm4@msg.ti.com (Jim Shima)
Date: Fri, 17 Jan 1997 15:45:47 GMT
longam@hotmail.com wrote:
>I need to find the inverse of a Hermitian covariance matrix. The
>covariance matrix is formed by averaging the sample covariances of the
>form: R = XX' Each sample covariance is formed by multiplying a sample
>column vector by its transpose.
I am also interested in finding an efficient way to solve for an
inverse covariance matrix. Most of the straight forward methods are
N^3 in complexity, and I would like to hear from anyone you has
encountered and tackled this problem. Can Woodbury's identity be used
here?
Also, does anybody know why using an adaptive filter (such as a
transversal FIR using LMS) to solve for the inverse is not a good
solution? I thought that since the solution to the matrix equations
is just the well-known Wiener-Hopf equations, one can find the optimal
filter weights iteratively instead of taking a huge inverse of a
matrix.
Any help is greatly appreciated.
Regards,
Jim
Jim Shima
Advanced Signal Processing
Texas Instruments, Inc.
email: shima@ti.com
*********************************************************
* Opinions stated are mine and not representative of TI *
*********************************************************
