Subject: Re: Random Associative Functions
From: Jim Hunter
Date: Fri, 10 Jan 1997 15:43:08 -0500
Wayne M. VanWeerthuizen wrote:
>
> On Sat, 04 Jan 1997 22:55:27 GMT, WayneMV@LocalAccess.Com (Wayne M.
> VanWeerthuizen) wrote:
>
> >
> >I'm looking for an efficient way to tell if an arbitrary function is
> >associative. The function is represented in an N by N matrix, which
> >represents f(x,y). All the values of the function are between 0 and
> >N-1. (Such as a table for modular addition.)
> >
> >I know the function is cummulative if the matrix is symetrical along
> >the diagonal. I looking for a simple rule of thumb to check for
> >associativity, without having to try all possible triples of values.
> >
> >I'm also looking for a fast way to generate a random function that is
> >associative. This seems to be a harder problem than merely testing.
> >
>
> I have not yet seen any responces to my question. Does anyone have an
> answer? Or was my question not well explained?
It's a tough problem isn't it?
Associativity is a Popeye property (I yam what I yam). It's something
trivial to state, yet defies being broken down the way you need it to
be to solve this problem. I've been waiting for replies on this myself
since I've been working on a similiarly-stated theoretical problem.
> ...
> How can I prove that f(f(x,y),z) = f(x,f(y,z)), without having to do
> an exhaustive test of every triple?
Well, I think your stuck here. Check all triples.
> What further restrictions must I consider when placing random numbers
> into the matrix if I want the resulting function to be assosiative?
>
I might be able to help you a little with this.
You should first tighten-up the problem by saying "I would like
the operator to be A and ?". Otherwise it's trivial for just A,
because a constant function on any set is A. ? should also be something
other than C to constrain the solutions from being trivial.
It would be very hard to generate these operators from scratch,
so start with some seed matrices known to be A.
There are two general methods I know of that can generate more matrices
from these in a random sense.
The first is by similarity transformation:
Let f be a binary operator on a set S and u be a bijective function
from S to itself. If f is assoc., then g(x,y) = u^-1(f(u(x),u(y)) is
assoc.
The second uses translates of the operator:
If f is assoc. and comm. and c is fixed, then g(x,y) = f(f(x,y),c) is
assoc. and comm.
For your stated problem, the set of bijective functions is the
set of N permutations. So similarity transforms have the
form, A* = p^-1(P'AP); where A is a seed matrix, P is a permutation
matrix, and p is the function that generates P. p(M) is applied
element-wise.
The second method generates: A* = A|c; where | is the operator defined
by A, and this is applied element-wise to A for some c in {0,1,...,N-1}.
---
Jim
Subject: Re: Why do stars collapse?
From: Jeff Wilson
Date: Fri, 10 Jan 1997 12:46:05 -0600
Ken Fischer wrote:
>
> How will you be able to tell if it is a black hole? :-)
> Only one star in 10,000 has greater than two solar masses,
> do I read that right? Does that mean by the time they
> burn all available fuel, they may be close to the mass of
> the Sun, and then a supernova would leave essentially
> not enough to make a black hole?
>
> Ken Fischer
Sounds like you might be wanting what's known as the Chandresekhar (sp)
limit. This defines the amount of mass in the *core* of a star that
determines the ultimate fate (white dwarf, neutron star, black hole).
Note that I said the *core* of the star. If the core mass is below
1.44 solar mass, it will evolve to a white dwarf. A core mass greater
than 1.44 but smaller than about 3.5 solar masses will become a
neutron star. Anything greater will become a black hole. There's
also the possibility that the remnant will completely destroy itself
leaving nothing.
The core size really doesn't imply anything about the pre-death size
of the star itself. It could be just about anything (but must, of
course, be larger than the core size). A star may undergo mass loss
of significant amounts at times before ending the fusion process.
Therefore, the star mass blow *many* solar masses of material away,
but it just depends on the stability of the star before the SN event.
I also doubt that you could even say that two different cores of the
same mass had similar masses before death due to the mass loss stages.
--
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
| Jeff Wilson | In space, no one can |
| jdwilson@nortel.com | hear you scream!! |
| Richardson, TX - my opinions are...MINE. | |
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Subject: Re: Complex Question !
From: tleko@aol.com
Date: 10 Jan 1997 20:28:21 GMT
:In article <5al7jn$241@gap.cco.caltech.edu
:Ilias Kastanas wrote:
:I'm glad you realized there is no need for arctan.
:>
:>In article <19970104143700.JAA18764@ladder01.news.aol.com> tleko@aol.com
:>(Tleko) wrote:
:> Not for e^z which is analytic (has no zeros).
:>
:>In article <19970106165000.LAA29625@ladder01.news.aol.com> tleko@aol.com
:>writes:
:> e^z=(sqrt(R^2+I^2))*(e^(i*atan(I/R)))=(e^x)*(e^(i*atan(tan(y)))
:> =(e^x)*(e^(i*y))=(e^x)*(cos(y)+i*sin(y))
:>>
:>>In article dik@cwi.nl (Dik T. Winter) wrote:
:>>Where do you find that atan(tan(y)) = y? Make a plot with MATLAB for
:>>y ranging from say -10 to 10.
:>>
:>>In article dik@cwi.nl (Dik T. Winter)wrote:
:>>I said: make a plot with MATLAB for y ranging from say -10 to 10.
:>>Do it, look at the plot and see a sawtooth function. You did it
:>>from -pi/2 to pi/2 which does not show the sawtooth.
Thank you very much for you reference to the sawtooth function.
I am plesed that you are using MATLAB.
What makes the function e^z distinct from most of the others is the
absence of zeros of polynomial varieties. Here is the program regarding
the zeros.
x=-6:.06:6;
y=-6:.06:6;
[X,Y]=meshgrid(x,y);
R=exp(X).*cos(Y);
I=exp(X).*sin(Y);
A=sqrt(R.^2+I.^2).*(-sin(atan(I./R))+cos(atan(I./R)));
c=contour(x,y,A,0:100:200);
clabel(c,'manual');
tleko@aol.com
Subject: Vitali nonmeasurable
From: abian@iastate.edu (Alexander Abian)
Date: 10 Jan 1997 23:05:09 GMT
In article <5b3i2a$5u4$1@nntp.ucs.ubc.ca>,
Robert Israel wrote:
>In article , abian@iastate.edu (Alexander Abian) writes:
[the word "SOME" is added by Abian (as explained by underlined sentence)
>> Referring to the classical example of Vitali's Lebesgue nonmeasurable
>> subset V of the real unit interval - is it the case that the outer
>> measure of SOME (instead of mistakingly originally typed "any") such
>> V is equal to 1 ? ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Abian answers:
Dear Mr. Israel,
Probably you did not notice that very soon after my posting that you
have quoted above, I posted another one (which is still posted) by
correcting the typo which I indicated in your quote above.
..............................................................
Isreal continues:
>................you can always take these representatives to be in any
>given interval [a,b] with 0 <= a < b <= 1, so the outer measure of V is
>at most b-a.
>What's a bit more challenging, I think, is to find a V whose outer measure is
>equal to 1. You can do it as follows:
>The family F of open subsets of [0,1] with measure < 1 has cardinality c.
>So it can be well-ordered in such a way that every member of F has fewer than
>c predecessors. Since the complement of each member of F has cardinality
>c, there is a function f: F -> [0,1] such that for each A in F, f(A) is not
>a member of A and is not in {f(B)+q: B a predecessor of A, q in Q}. Thus
>{ f(A): A in F } contains at most one representative of each coset. Complete
>V by putting in representatives of all cosets not already represented.
>Then the outer measure of this V is 1.
>
Abian answers:
Your answer seems correct.
The reason for my posting the question was that I had to present an example
of V with outermeasure 1 and I wanted to see if there is an example
simpler than the one I have devised.
My devised example is based on the fact that " If a subset S of [0,1]
has a nonempty intersection with every closed subset of positive (Lebesgue)
measure of [0,1] then the outermeasure of S is equal to 1.
This result can be found in my paper
"A SIMPLEST EXAMPLE OF A NONMEASURABLLE SET"
Simon Stevin Math Journal, September 1976 (vol 2?) pp. 101-102
Based on the above, I recently gave a construction of V with outer measure
equal 1.
Moreover, based on a variant of the same result, I recently gave examples
of V with outer measure precisely equal to b - a and not only "at most"
for a an b as mentioned by you. All one has to do is to consider
the set of all closed subsets of [a,b] which have positive Lebesgue)
measure, then pick up a point from each of them to include in the
construction of a V by appropriate addition of appropriate points not
used from [a, b].
--
--------------------------------------------------------------------------
ABIAN MASS-TIME EQUIVALENCE FORMULA m = Mo(1-exp(T/(kT-Mo))) Abian units.
ALTER EARTH'S ORBIT AND TILT - STOP GLOBAL DISASTERS AND EPIDEMICS
ALTER THE SOLAR SYSTEM. REORBIT VENUS INTO A NEAR EARTH-LIKE ORBIT
TO CREATE A BORN AGAIN EARTH (1990)
Subject: Re: (-2) * (-3) = (- 6) ; why not ?
From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik)
Date: 10 Jan 1997 16:43:11 -0500
In article <32D68320.41C67EA6@imag.fr>,
Jean-Christophe Janodet wrote:
>It's easy to explain why the sum of two negative
>numbers is negative, using the example of a barometer.
>Does anybody have a similar example to justify that
>the product of two negative integers is positive ?
>Or else, what are the theoretical reasons ?
>
>Sincerely, Jean-Christophe.
Step one: Suppose you agree that
(-2) * 3 = -6
(for example, if you owe 2 cows and your neighbour owes 3 time as many,
what is your neighbour's debt?).
Step two: Observe that 3 + (-3) = 0 and that any number times 0 is 0
(at least I hope you agree with that).
Step three: It looks reasonable that the distributive law should extend
from positive numbers to all numbers:
(a + b) * c = a * c + b * c
Combine these: I want
(-2) * (-3) + (-2) * 3 = (-2) * (-3 + 3) = (-2) * 0 = 0
so that
(-2) * (-3) + (- 6) = 0
and inevitably (-2) * (-3) = -(-6) = 6
whether I like it or not.
Hope it helps, ZVK (Slavek).
Subject: Re: Speed of Light
From: "Risto Lankinen"
Date: 10 Jan 1997 20:56:44 GMT
Hi!
Jason Lee wrote in article
<5ak1go$v1s@cymbal.aix.calpoly.edu>...
> And then Richard Mentock quoth:
> >Since the meter is defined in terms of the speed of light
> >http://math.ucr.edu/home/baez/physics/speed_of_light.html
> >the speed of light is always 299,792,458 meters/second.
> >
> >So, this is a value that won't change (the length of the
> >meter might!)
>
> If the meter is defined in terms of the speed of light with specifically
> that constant, then it can't change, since the speed of light will always
be
> the speed of light.
Doesn't this also mean that the definition of meter is at most as
accurate as the definition of second? Isn't it even possible that
both definitions are hugely inaccurate, as long as they are off to
the same direction (since speed is a *ratio* of distance and time)?
How is the second defined nowadays, anyway?
terv: Risto L.
Subject: Re: (-2) * (-3) = (- 6) ; why not ?
From: John
Date: Fri, 10 Jan 1997 20:08:44 -0500
Matt Feinstein wrote:
>
> John wrote:
>
> Jean-Christophe Janodet wrote:
> >
> > It's easy to explain why the sum of two negative
> > numbers is negative, using the example of a barometer.
> > Does anybody have a similar example to justify that
> > the product of two negative integers is positive ?
> > Or else, what are the theoretical reasons ?
> >
> > Sincerely, Jean-Christophe.
>
> A non-mathematical friend once asked me the same question, and this
> was the argument I came up with---
>
> We agree that zero times any number is zero and a number plus its
> negative is zero-- so, for example,
>
> 0 * (-3) = ( 2 + (-2)) *(- 3) = 2*(-3) + (-2)*(-3) = 0
>
> Therefore, 2*(-3) must be the negative of (-2)*(-3) since they sum to
> zero.
>
> A similar argument
>
> 0 * 3 = ( 2 + (-2)) * 3 = 2*3 + (-2)*3= 0
>
> shows that (- 2)*3 must be the negative of 2*3 since they sum to zero.
>
> Putting the two conclusions together, since 2*3 is positive, 2*(-3)
> must be negative; and since (- 2)*3 = 2*(-3) must be negative,
> (-2)*(-3) must be postive..
>
> It -does- seem unnecessarily complicated..
>
> Matt Feinstein
> mfein@aplcomm.jhuapl.edu
Hi Matt,
You are in good company. That looks similar to the argument Euler used.
(See the same reference, same page)
That is probably a necessary but not sufficient condition and hence not
a complete proof. This "rule of signs", according to the ref., cannot be
proven within the other rules of arithmetic. The "rule of signs" in
arithmetic is like one of the axioms of Euclidean geometry.
Regards,
John
Subject: Re: Cute Proofs...
From: mareg@csv.warwick.ac.uk (Dr D F Holt)
Date: 10 Jan 1997 21:20:24 -0000
In article <5b5gpm$66j@node.fwi.uva.nl>,
mdekker@wins.uva.nl (Martijn Dekker) writes:
>sci60065@leonis.nus.sg (NCC-1701-E) wrote:
>
>:Godfrey Degamo (gotd@jimmy.harvard.edu) wrote:
>:How about this one :
>:
>: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
>
>bad example. I see no proof here, only another statement from which
>the answer follows. Now prove this last identity.
Well, I have to disagree. I find this an excellent example, and I
think most mathematical journals would accept such a proof.
Proofs are intended to communicate ideas, and by including too many routine
details they often only succeed in frustrating this aim.
In this case it would tedious and unnecessary to prove the identity since
the reader might just as well do it.
Derek Holt.
Subject: Re: Probability and Wheels: Connections and Closing the Gap
From: bm373592@muenchen.org (Uenal Mutlu)
Date: Fri, 10 Jan 1997 21:24:56 GMT
On Thu, 09 Jan 1997 17:25:10 -0700, Karl Schultz wrote:
>2) Generates 54 drawings to represent the player playing 54 tickets.
I'm not sure about this point, because in my posting I had asked
Mr. Sharkey about the interchangeability of these two things, ie.
>>I would say that the probability for both the below events is equal:
>> - playing the same 1 ticket in 54 consecutive draws
>> - playing 54 different tickets in 1 drawing
>>
>>Isn't it?
>
>No, it's not. Not quite.
He gave also detailed examples (thanks), but since I can't play cards :-)
I have a hard imagination about them, so especially such examples with
cards confuse me more than they help me (I know it is my problem, but
if possible, I would like to see another example without cards etc.;
pure math and/or a formula would be ok.)
Another confusing thing (for me) is also the difference (if any)
between 'probability' and 'chance', for example in the following
excerpt of a posting of again Mr. Sharkey:
>If odds of AT LEAST 3 is 1/54 then chance of getting a 3 match in 54 is
> 1 - [ (1- 1/54)^54] or about 0.6355
(here, it's meant in 54 consecutive drawings (not 54 tickets in 1 drawing),
if I got it right).
But, what does this exactly mean? Here, we are using 1 ticket. Which of
the below is true:
- 63.55 % chance of winning "at least 3" in the next drawing
- 63.55 % chance of winning "at least 3" after participating in 54
consecutive drawings
- Does the above apply only if the same 1 ticket (ie. fixed nbrs) is used?
- Does it matter (for the above formula) if in every drawing different
(ie. randomly chosen) numbers are played (on 1 ticket).
If one plays in 54 consecutive drawings using the same 1 ticket:
- is then the probability for "at least 3" constant or does it change?
- is then the chance of winning "at least 3" constant or does it change?
(Sorry if I used any wrong terms (ie. probability vs. chance etc.), but
IMHO should be clear what it's meant)
Subject: Re: UNSOLVED MATHS PROBLEM
From: electronic monk
Date: Fri, 10 Jan 1997 15:51:53 -0800
Ang Ngee Hiong wrote:
>
> Is anyone free to solve this
>
> Q: A straight horizontal track is marked by stones A0, A1, A2,
A3, ...
> equally spaced along it, so that the distance A0A1 = A1A2 = ... = c.
The
> summit T of a hill on one side of the track is at a height h above
the
> level of the track, and the angle of elevation of T from An is
denoted by
> µn (n = 0, 1, 2, 3 ...). Show that:
^
|
how did you make that cool symbol?
> i) mod(h*cot µ0 - h*cot µ1) < c < h*cot µ0 + h*cot µ1
> ii) (cot µn)^2 = n*(cot µ1)^2 + n*(n-1)*(c/h)^2
just wondering.
electronic monk
Subject: Re: Best way to study maths?
From: rbrito@ime.usp.br (Rogerio Brito)
Date: Fri, 10 Jan 1997 18:57:50 -0500
Peter wrote:
>Hi, everybody.
>
>Could anyone give me the answer of my subject?
It's not easy to say what's the best way to learn
Mathematics. But I'll put here some of the methods that
I've been following.
As you may have noted, Math is a science that is built up
of conclusions, like a building. This means that if you
don't understand something well enough to give another
step, then you probably won't get very far. Math
requires a lot of attention and many definitions have
very subtle differences (the continuous and uniformily
continuous mappings definitions are what comes to mind).
It's not rare to catch me spending up to 30 minutes
to read one page of a book. If you don't previously know
the subject, you may want to check that every definition
makes sense, what is its importance and so on.
When reading theorems I try to pay a lot of attention to
what I'm reading and then I see if I can reproduce what
I've just read (often, you need piles of paper). If yes,
then maybe I've understood it. If not, I re-read the
whole page and try to find where are my problems (this
is, IMO, the hardest part, but it's where I see if I have
really learned something).
And I really try to do the most interesting exercises.
Well, this is the way I've been trying to learn
Mathematics.
[]s, Roger...
--
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
Rogerio Brito - rbrito@ime.usp.br - http://www.ime.usp.br/~rbrito
"Master, Master, where's the dreams that I've been after?
Master, Master, you promised only lies!
Laughter, laughter, all I hear or see is laughter.
Laughter, laughter, laughing at my cries."
James Hetfield (Metallica), Master of Puppets
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
Subject: Conferences/Announcements @ I'meister-Ukrainian
From: pyz@panix.com (Pan Tofli)
Date: 10 Jan 1997 14:05:27 -0800
[ Article crossposted from soc.culture.ukrainian ]
[ Author was Pan Tofli (pyz@panix.com) ]
[ Posted on 10 Jan 1997 12:24:05 -0500 ]
Greetings,
We've set up a new section at Infomeister-Ukrainian for
Conferences/Announcements. The first one which we have listed is:
"Symmetry in Nonlinear Mathematical Physics, Second International
Conference"
-- July 7-13, 1997, Kyiv, Ukraina
For more information on this conference please use one of the following
web coordinates:
http://www.osc.edu/ukraine.html#CONF
or
http://www.osc.edu/ukraina.html#CONF (for Ukrainian KOI8-speaking web
browsers)
Similar types of notices will gladly be accomodated. Much thanks to Prof.
Roman Andrushkiw for providing this one.
Max Pyziur
pyz@panix.com
Subject: Re: Why can't 1/0 be defined???
From: electronic monk
Date: Fri, 10 Jan 1997 16:03:35 -0800
Paul Schlyter wrote:
>
> In article <32D587B4.343@sqruhs.ruhs.uwm.edu>,
> electronic monk wrote:
> >BlackTopPilot wrote:
> >
> > > eg: (the limit of one over x, as x gets as large as you
wish)*zero =
> >er,
> > > uh, z e r o .....
> >
> >actualy,
> >
> >lim (x * 1/x) = oo * 0 = 1
> >x-->oo
>
> 1 = oo * 0 = oo*(0 + 0) = oo*0 + oo*0 = 1 + 1 = 2
>
> Q.E.D.
what i mean is that:
lim (x * 1/x) = 1
x->oo
lim (x * 1/x) = 0 * oo
x->oo
so 0*oo can equal one.
>
> --
> ----------------------------------------------------------------
> 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/
electronic monk
Subject: Re: Why can't 1/0 be defined???
From: electronic monk
Date: Fri, 10 Jan 1997 16:09:33 -0800
Dik T. Winter wrote:
>
> In article <32D587B4.343@sqruhs.ruhs.uwm.edu>
donniet@sqruhs.ruhs.uwm.edu writes:
> > > eg: (the limit of one over x, as x gets as large as you
wish)*zero =
> > er,
> > > uh, z e r o .....
> >
> > actualy,
> >
> > lim (x * 1/x) = oo * 0 = 1
> > x-->oo
>
> Better read correctly, he was writing about:
> lim (1/x * 0)
> x-->oo
you mean as x-->0 or you mean lim [x-->oo] (x * 0). but true enough,
that was what he was saying, and that equals zero. the problmem with
that is that he should of expressed 0 as a limit. this is because since
we can never get to infinity, we must also never be able to reach zero.
i expressed the limit this way.
> --
> dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland,
+31205924131
> home: bovenover 215, 1025 jn amsterdam, nederland;
http://www.cwi.nl/~dik/
electronic monk
Subject: Re: A question
From: rbrito@ime.usp.br (Rogerio Brito)
Date: Mon, 06 Jan 1997 22:34:29 -0500
mlerma@pythagoras.ma.utexas.edu (Miguel Lerma) wrote:
>HC Lee (hclee@axps1.phy.cuhk.hk) wrote:
>> There is a so called Jensen's inequality for random variables. Could
>> somebody tell me the details ?
>
>If E[X] exists, and if f(x) is a convex cup function, then
>
> E[f(X)] >= f(E[X])
>
>(C.R.C. Standard Mathematical Tables and Formulae, p.579).
>
>E means "expected value". I guess "convex cup" means just
>"convex", i.e., f((1-r) x + r x) <= (1-r) f(x) + r f(x) for
>0 <= r <= 1.
Hi.
Since this thread is very old, I'm e-mailing it to you,
Miguel, as well as posting it to the newsgroup (something
that I usually don't do).
The first time I have seen a definition to a "convex cup"
function instead of only "convex" was, maybe, two or
three weeks ago. I was surfing the net (well, who said
Internet isn't culture? :) and then I found some lecture
notes (BTW, very well written by Mr. Chee K. Yap) and it
has this footnote:
"We say a real function f(x) is convex cap if for all 0 <
\alpha < 1, f(x) + f(y) <= 2 f(\alpha x + (1 - \alpha)y).
For completeness, we say f(x) is convex cup if for all 0
< \alpha < 1, f(x) + f(y) >= 2 f(\alpha x + (1 -
\alpha)y)."
>Miguel A. Lerma
Hope this helps, Roger...
--
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
Rogerio Brito - rbrito@ime.usp.br - http://www.ime.usp.br/~rbrito
"Master, Master, where's the dreams that I've been after?
Master, Master, you promised only lies!
Laughter, laughter, all I hear or see is laughter.
Laughter, laughter, laughing at my cries."
James Hetfield (Metallica), Master of Puppets
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
Subject: Re: recurring sequence 1,3,6,1,6,3,1,9,9
From: mathar@qtp.ufl.edu (Richard Mathar)
Date: Fri, 10 Jan 1997 17:21:56 EST
jrupe@advtech.uswest.com (Jason Rupe) writes:
|>
|> While driving through Nebraska, I came across an interesting number result. I haven't worked out much detail around this because I would quess that this is a common result in mathematics that I just haven't run into (I have an engineering OR background). Here is what I have observed:
|>
|> Begin with the positive reals 1 2 3 4 5 .... Then take the partial sums from 1 to n for each n to yield 1 3 6 10 15 ... Then, sum all digits in each n so that, for example, 10 becomes 1+0=1, 15 becomes 1+5=6 and 148 would become 1+4+8=13 become 1+3=4, and such. This seems to result (should be easy to prove) in a recurring sequence of 1 3 6 1 6 3 1 9 9.
|>
While I am here sitting behind my desk in Florida, I see that
the sum of the first n integer numbers is n(n+1)/2 . The total of the
digits of the product is equal to the product of the
total of the factors (I used to use this prior to
the invention of pocket calculators). So the first
"n" in the product represents one of the numbers
between 0 and 8 (because the total of the digits is
the rest after division by 9), the second factor
"n+1" represents the next number modulo 9. Hence
the result is periodical with n with a period of 9.
--
mathar@qtp.ufl.edu
Subject: Re: Science Versus Ethical Truth.
From: Jim Hunter
Date: Fri, 10 Jan 1997 18:04:07 -0500
David Kastrup wrote:
>
> Just to give an example: a few years ago in San Francisco, some
> policemen broke into the home of some suspect thinking nobody would be
> there (they had a search warrant or somethiung like that).
> Unfortunately, both the tenant of the flat *and* another male were
> there, involved in, ugh, some action. They were arrested for I don't
> know what (something like indecent behaviour or whatever) and actually
> persecuted and sentenced. I might add that both were considerably of
> age.
>
This is hard to believe. San Francisco? I have to think a lawyer and
and both clients came away with a lot of money after the law suit.
---
Jim
