Subject: Vietmath War: Field for Euclidean 3-Space
From: Archimedes.Plutonium@dartmouth.edu (Archimedes Plutonium)
Date: 8 Jan 1997 03:07:39 GMT
In article <5asef3$ale$1@dartvax.dartmouth.edu>
Archimedes.Plutonium@dartmouth.edu (Archimedes Plutonium) writes:
> Now, let us look at Euclidean geometry since we know it well. Do the
> points as Reals+i+j do those points form a Field? Can some of them
> form a Ring? Can some of them form a Group? But the highest algebraic
> structure of Euclidean 3-space is Euclidean geometry itself.
I am focused on questions here.
(1) To make the Reals complete in plane geometry, we append i to
them. I call this Reals+i.
Question: Consider Euclidean 3-Space. I have the suspicion that
Reals+i+j are what complete 3-Space. I am not sure of this because
Hamilton has Reals+i+j+k for his quaternions.
I am not sure because I do not see why the k is needed. I do not see
this because to me, i is a number which is a rotation itself, same with
j. But why is k needed when the Reals append i and append j suffice to
make Euclidean 3 space?
I have the feeling that Reals+i+j is the whole of Euclidean Geometry
and that the k is superfluous baggage. Is there something in another
subject of mathematics that would lend further support that Reals+i+j
are the good guys and that Reals+i+j+k are the bad guys?
(2) Question I harped around with a lot some years ago. I was looking
for the completion of the prime-adics conjoined with the
composite-adics. I conjectured that these All-adics formed Riemannian
geometry. What if any numbers to append to the All-Adics to make them
complete? Are the Reals the only set of numbers that need have appended
(imaginary numbers) new numbers to complete them? Or do the All-adics
need some appended numbers? Perhaps the All-adics can stand by
themselves and same also the Doubly Infinites.
(3) Anyone ever heard of a concept of number inside a number? A
concept of a point of geometry having internal points? Sounds
ludicrous. But consider the prime and composite adics of this number
....000002
Consider the 3-adic ....0002. Consider the 5-adic ....0002 and the
15-adic
....00002. They are all different , yet they are all the same. I
think the future mathematicians will be working with a new subject area
of mathematics that deals with this question. And the idea is that if
you look at Euclidean geometry say 3-Space as filled with points. Each
point in Euclidean geometry is a dimensionless entity. But a point in
Riemannian geometry is not a dimensionless entity, but rather it is
more like nested other things. Where the 15-adic point of ....0002 has
the 3-adic ...002 and the 5-adic ...002 inside of itself. This is very
important for physics, for a elementary particle such as the photon or
electron or whatever has internal nested parts and the Riemannian
geometry is the mathematics which pictures this.
Question: Is algebraic completeness related to geometry? I think not
from the above that we have 3 and only 3 geometries of Eucl, Riem, and
Loba and it appears that completeness is only a characteristic of
Euclidean geometry.
Subject: Re: matrix inversion with random numbers
From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik)
Date: 7 Jan 1997 15:43:36 -0500
In article <21nAjDAQ7T0yEwE4@gocomp.demon.co.uk>,
Richard H Gould wrote:
>In article <32D002F9.1BE9@math.okstate.edu>, David Ullrich
> writes
>> What makes you think there
>>exist matrix-inversion algorithms that use random numbers?
>>
>I assumed he meant that the matrix elements had no pattern - not
>integers, matrix not sparse, etc.
>
>Again, since no-one has answered, if my interpretation of the question
>is correct, look up the L-U decomposition method. It's quite elegant
>and easy to code if you want to write a program for matrix inversion.
>--
>Richard H Gould
>rhgould@gocomp.demon.co.uk
Reference for Monte Carlo inversion (basically for matrices A close to
identity in the sense that spectral radius(abs(I-A)) < 1 ):
"Matrix Inversion by Monte Carlo Methods" , by Florence Jeanne Oswald;
article included as Ch. 6 in
"Mathematical Methods for Digital Computers", edited by A. Ralston and
H. Wilf,, John Wiley & Sons, Inc., New York 1960
It seems that the first study in this direction is (as quoted by Oswald):
G.E. Forsythe and R.A. Leibler: Matrix Inversion by a Monte Carlo Method,
MTAC, vol. 4, 1950, pp. 127-129.
(I don't know what MTAC stands for.)
Cheers, ZVK (Slavek).
Subject: Re: Probability and Wheels: Connections and Closing the Gap
From: bm373592@muenchen.org (Uenal Mutlu)
Date: Wed, 08 Jan 1997 01:30:25 GMT
On Tue, 07 Jan 1997 13:23:44 -0700, Karl Schultz wrote:
>> So, the interessting question is: why should one ever play the
>> 168 ticket wheel and not simply 54 randomly choosen different single
>> tickets? IMHO mathematically spoken both cases should offer nearly
>> the same assurance and probability for hitting once or more at least
>> 3 correct numbers. (this maybe not 100% correct, but you can imagine
>> what I mean).
>
>Because the 168-wheel is a 100% guarentee. If you absolutely,
>positively want to ensure at least a 3-win, the best way to do it is
>to use this wheel.
>
>I see what you are getting at, saying that 54 tickets would give you
>a darned good chance, but the price to pay for the 100% certain 3-win
>is the larger wheel.
...
>> Personally, I would recommend the interessted player to go with the
>> 54 tickets, and recommend the wheel designers and researchers in
>> Design Theory (Covering Designs etc.) to also take into consideration
>> such probability calculations. Any comments?
>
>Any approach where you play more tickets helps, if you really must
>play at all.
It would be useful if we had a simulation software which for example
looks something like the following:
LOTSIM v k b nruns fFixedTickets fFixedDraw ...
v = total numbers (ie. 49)
k = nbrs per ticket (ie. 6)
b = nbr of random tickets (>= 1)
nruns = nbr of random drawings (simulation) (>= 1)
fFixedTickets = randomly fill tickets once OR refill each time (0/1)
fFixedDraw = randomly draw once and keep OR redraw each time (0/1)
...
Output:
m ep en rn rp dn dp ...
---------------------------
k .. .. .. .. .. .. .
. .. .. .. .. .. .. .
. .. .. .. .. .. .. .
0 .. .. .. .. .. .. .
---------------------------
Sum: .....
m = matching nbrs (0..k)
ep = expected theoretic probability
en = expected theoretic frequency
rn = simulated real frequency
rp = simulated real probability
dn = +/- diff frequency
dp = +/- diff probability
...
If there is already a similar publicly available program I would
like to hear about it. I think of programming such a thing too.
Further comments/options/ideas to include in the program are welcome.
IMHO such a program would be helpful in 'practically' answering some
still outstanding problems like in the case of the above 54 tickets
in question.
Subject: Re: The Inferiority of Cydonian "Math" in Comparison to the Earthly Science-Art
From: ba137@lafn.org (Brian Hutchings)
Date: Tue, 7 Jan 1997 22:43:17 GMT
In a previous article, euryale@earthlink.net (Judith Stroud) says:
Judith, *must* you quote *all* of that **** ??... anyway, yes;
at any rate, they wanted to *be* the antichrist, or one
of his marcher-lords. well, not so sure about Nero, and Napolean,
although they were both insane emporors, but
the Nazis had a training-grounds on the island of Capri,
specifically because of its antichristian lore/history.
>essence of numbers and numerology is that everything can be reduced or
>extrapolated into anything else - reference the 666 in Nero, Hitler and
>Napoleon's name. Were they all the antichrist? I can calculate the
>circumference of the earth with my left breast and the right formula.
>This is merely evidence that numbers permeate the universe, not that
>anyone happened to do it on purpose...
--
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)
Subject: Re: Non reversable mathematics?
From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik)
Date: 7 Jan 1997 19:54:36 -0500
In article ,
Stephen Montgomery-Smith wrote:
:>John Edser (edser@ans.com.au) wrote:
:>: Are all the assumptions made by mathematics reversable?
:
:>: Is it possible to create a mathemeatics whereby,
:>: the the movement across the "=" is only one way?
:
:>: From a strictly "non mathematician",
:
:>: John Edser
:>: edser@ans.com.au
:
:There is a situation when = is one way. As x tends to zero one has that
:
: O(x) = O(x^2)
:
:but NOT
:
: O(x^2) = O(x).
:
::-)
That's because the casually used "is" can mean "equals" or "belongs to",
"has the property (with appropriate grammatical construction) or "is
included among" (or perhaps many other relations), and in everyday
language, physical/social circumstances prevent us, for example, from
confusing "politicians are crooks" with "crooks are politicians".
This casual irreversible use crept into mathematical notation, and the
temptation to use the "=" sign was too strong (and many would read it
aloud as "is" rather than "equals" anyway).
ZVK (Slavek)
Subject: Re: Vietmath War: war victims; blinded victims
From: jpb@iris8.msi.com (Jan Bielawski)
Date: Wed, 8 Jan 1997 01:38:44 GMT
In article <5aedhv$po1@dartvax.dartmouth.edu> Archimedes.Plutonium@dartmouth.edu (Archimedes Plutonium) writes:
< In article
< jpb@iris8.msi.com (Jan Bielawski) writes:
<
< > In article <141421356237310@einsteinium.universe> Carl Friedrich Socrates Einsteinium writes:
< > < In article <32BEC8B2.794BDF32@clipper.ens.fr>
< > < David A. Madore writes:
< > <
< > < > example, is one allowed to ask the question of whether Finite Integers
< > < > satisfy Fermat's Last Theorem?
< > <
< > < This is as ridiculous as asking how many angels can stand on
< > < a pin head, or how many letters will fit in a cubic centimeter.
< >
< > Why is it ridiculous? Tell us, does the number ...00001 exist?
< > What about ...0002 ? If you can conceive such numbers (I have no
< > doubt that you can) then there CAN'T be anything "ridiculous" about
< > asking whether such numbers satisfy a^n + b^n = c^n .
<
< Say Jan, I used to live in Coronado and could use some (free)
< pictures for my website autobio, of that Naval station situated across
< from Hotel Del, a little ways down south of Del where I went to SWOS
< school.
Is this the one on Silver Strand?
< I would love a picture of the waters from that Natl Monunment
< there? Cabrillo? Also a picture of the Naval Base where all the ships
< are docked. If possible, send to Archimedes Plutonium, c/o Dartmouth
< College, Hanover NH 03755 and it will get to me.
Any P.O. Box or something?
<
< Good to see that you Jan are not one tracked, like a RR but instead
< can lift yourself up from off the track and go another direction, as
< per, Democritus Uranium and cubic cm.
Any reason why you didn't answer my question?
--
Jan Bielawski
Molecular Simulations, Inc. )\._.,--....,'``. | http://www.msi.com
San Diego, CA /, _.. \ _\ ;`._ ,. | ph.: (619) 458-9990
jpb@msi.com fL `._.-(,_..'--(,_..'`-.;.' | fax: (619) 458-0136
#DISCLAIMER******************************************************************#
+Unless stated otherwise, everything in the above message is personal opinion+
+and nothing in it is an official statement of Molecular Simulations Inc. +
#****************************************************************************#
Subject: Re: Speed of Light
From: "Jan Zumwalt"
Date: 8 Jan 1997 04:19:25 GMT
I'm sorry if I am way off base and intruding. I only caught a small glimpse
of your discussion but...
I saw an article about 5 years ago in a scientific magazine written by a
NASA mathematician paid to consider the problems encountered in near light
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 slow
for connections in access of 300ft. Based on your discussion I think this
Conflicts to one of your opinions but I'm not sure.
Of course the author is human and fallible too. I have always awed at how
people consider written works a Holy grail!
--
Hope this helps!
----------------------------------------------
Jan W. Zumwalt - Engineer
Computer Information Systems
zumwalt@alaska.net
http://www.alaska.net/~zumwalt
*****************************************************************
* Beware the man of one book. - Chinese proverb *
* The only real equality is found in a cemetary. - German *
* The one thing not so common, is COMMON SENSE! -USA *
* Beware the opinion of a person who has used only one OS! -JZ *
*****************************************************************
Zdislav V. Kovarik wrote in article
<5aubr7$b12@mcmail.CIS.McMaster.CA>...
: In article ,
wrote:
: :In article <32D17201.A85@sqruhs.ruhs.uwm.edu>, electronic monk
: :donniet@sqruhs.ruhs.uwm.edu> writes: >
: :third, if you did decide to shine
: :>a flashlight toward the front of the ship, you sould see the light
creep
: :>forward, it would be going the speed of light, and since your ship is
: :>getting really close to the speed of light, the light from the
: :>flashlight would just barely be going faster then you.
: :
: :No, this is wrong. being on the ship you'll see the light moving at c
: :relative to you.
:
: To add to this, if no one did it yet: The formula for the composition z
of
: two velocities v, w is
:
: z = (v + w) / (1 + v * w / c^2)
:
: For example, if you fly at the speed (3/4)*c, as observed from an Earth
: station, and if you launch a rocket forward so that it reaches the speed
: (3/4)*c, as observed by you, then (check it out) the resulting speed of
: the rocket, as observed from the station, is (24/25)*c.
:
: The composition of velocities is a commutative group operation on the
: interval (-c,c) -- left as an exercise.
:
: (We are not surprised that, in a not-too-different situation, the angles
: do not compose the same way as theis slopes: if you have a board at a
: slope s1, put down on a road of slope s2 so that it makes the board
: steeper, then the resulting slope is not (s1 + s2) (provided the
: slopes are small enough, so that s1 *s2 < 1) but
:
: (s1 + s2) / (1 - s1 * s2)
:
: (remember the addition formula for tangents?))
:
: Cheers, ZVK (Slavek).
:
Subject: FOURIER TRANSFORM (Discrete)
From: abian@iastate.edu (Alexander Abian)
Date: 8 Jan 97 05:31:18 GMT
Dear Emma (at your and others request I am continuing my previous posting
of Fourier Interpolation to The Fourier Transform (discrete)
Let me recall that the basic Fourier Interpolation (discrete)(by f*) scheme of
a function f defined at x = -1, 0, 1 was given in my previous posting as:
/w 1 w^-1\ /w^x\
| | | |
(9) f*(x)/s = s(f(-1), f(0), f(1)) |1 1 1 | | 1 |
| | | |
where | w^-1 1 w | |w^-x |
\ / \ /
(10) w = (2pi/3)i
and
(11) s = 1/sqrt 3
Now, we can stare at (9) till dooms day and not see what to extract from it.
In no books, in no lecture notes, no one has mentioned what key element
is hidden in (9). I claim, that the hidden element is as obvious as
the sun in a cloudless sky - but it takes .... brain and eyes of m....
caliber person to bring it out.
This is how to use (9) "of course after I say it -it becomes obviously
elementary and kindergartenish".
Looking at (9), I wish it could be written as
/ \
|w^x |
| |
(13) (g(-1),g(0),g(1)) | 1 |
| |
|w^-x|
\ /
Based on (9) and (11), we DEFINE the function g (of course , as
usual at x = -1, 0, 1) given by
/w 1 w^-1\
| |
(14) (g(-1), g(0), g(1)) = s(f(-1), f(0), f(1)) |1 1 1 |
| |
|w^-1 1 w |
\ /
as the FOURIER TRANSFORM (discrete) of f.
It can be readily verified that the inverse of the 3 by 3 matrix appearing
in (14) is
/w^-1 1 w \
| |
(15) 1/3 | 1 1 1 |
| |
| w 1 w^-1|
\ /
Clearly, using (15) we can solve (14) for (f(-), f(0), f(1)) and obtain
the formula for INVERSE FOURIER TRANSFORM (discrete) given by
/w^-1 1 w^ \
| |
(16) (f(-1), f(0), f(1)) = s(g(-1), g(0), g(1)) | 1 1 1 |
| |
|w 1 w^-1|
\ /
REMARK. It is worth noticing how (14) and (16) are interrelated: the
matrices are inverse of each other and the roles of functions
f and g are interchanged.
PS. It is midnight and am tired and have to stop now. I hope I did not
make some obvious mistakes. Tomorrow I hope I will finish this
topic by introducing the Convolution product of functions and
proving the Convolution Theorem of Fourier Transforms
PSS I made some minor corrections in my previous posting FOURIER
INTERPOLATION. In reading the present posting please consult my
lastest FOURIER INTERPOLATION posting!
--
--------------------------------------------------------------------------
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: Infinitude of Primes in P-adics
From: dik@cwi.nl (Dik T. Winter)
Date: Wed, 8 Jan 1997 02:23:21 GMT
In article <5asb6a$ma6$1@dartvax.dartmouth.edu> Archimedes.Plutonium@dartmouth.edu (Archimedes Plutonium) writes:
> In article
> dik@cwi.nl (Dik T. Winter) writes:
>
> > > Let me recap here some facts :
> > > p-adics form a field
> >
> > No. Obviously, if there *are* primes you do not have a field.
>
> 3-adics form a field, 5-adics form a field
> any p-adic where p is prime forms a field
Eh? Did you read what I wrote after the sentence above? If there *is*
a prime, there is no field. In the 3-adics 3 is the only prime (%), but
because there is a prime there is no field. In a field every element
except 0 has a multiplicative inverse; there is no inverse of 3 in the
3-adics. There is no inverse of p in the p-adics. In the rationals for
instance 3 is *not* prime.
--
% Prime in the sense of prime ideals. Not in the sense of: there are
no a and b such that a.b = 3; because there are such a and b
(for instance: 2 * ...11111111120 = 10 = 3 in the 3-adics). My previous
proof was a bit wrong however. If there is an element with a multiplicative
inverse there is no prime in the traditional sense. That was what the
original proved, but in that sense 3 is not even a prime in the 3-adics.
However, an element without multiplicative inverse can still generate a
prime ideal (as is the case with p in the p-adics), but in a field there
are no such elements.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
Subject: Re: Marilyn's Ping-Pong balls
From: mlerma@math.utexas.edu (Miguel Lerma)
Date: 8 Jan 1997 03:11:13 GMT
BNJohnson (bnjohnson@aol.com) wrote:
> Is anyone but me a little disturbed by Marilyn vos Savant's column in this
> past
> Sunday's Parade? Someone asked what is the minimum number of ping pong
> balls required to construct a square-based pyramid, with equilateral
> triangles
> for each of the four sides.
>
> She answered that, surprisingly, the minimum and *only* number of ping
> pong
> balls which would work is 4900. She then said a word or two about
> pyramidal
> numbers.
I was also puzzeld by that claim, but I believe that she probably
meant that 4900 is squared and pyramidal at the same time. It is
a square because 4900 = 70^2, and it is pyramidal because 4900 =
1^2 + 2^2 + 3^2 + 4^2 + ... + 24^2 (each layer of the pyramid is
a square of side 1, 2, 3,..., 24). Since 1^2 + 2^2 + ... + n^2 =
1/6 (2 n^3 + 3 n^2 + n), the problem is finding positive integer
solutions to the equation:
2 x^3 + 3 x^2 + x = 6 y^2
Some search reveals a couple of solutions: (1,1) and (24,70)
(the latter being the one mentioned by Marylin). Since the former
(1,1) is obvious, perhaps (24,70) can be considered as the only
"non trivial" one. At least that is the claim in Beiler: "Recreations
in the Theory of Numbers", Dover, 1966, p.196.
Miguel A. Lerma
Subject: Re: Why can't 1/0 be defined???
From: Darrell Ryan
Date: Tue, 07 Jan 1997 20:43:22 -0600
electronic monk wrote:
>
>
> like i said before, we can think of zero having levels in the same way
> that we can think of infinity as having levels. if we think of zero
> only as a limit, then it will make sense. x^2 > x for all x>0,
Huh? Since when is (1/2)^2 > 1/2 ?????
> and then we could say 1/(x^2) > 1/x for all x>0.
Huh? Since when is 1/(2^2) > 1/2 ?????
>so, lim 1/(x^2) as x-->0+ > lim 1/x as x-->0+.
Not so. lim x-->0+ [1/(x^2)] = +infinity (the limit does not exist.
*Even if* lim x-->0- = +infinity, which it does)
and lim x-->0+ (1/x) = +infinity (the limit does not exist)
Remember, when we write +infinity or -infinity as an answer to a limit
problem, we are really saying, "The limit does not exist." The infinity
part tells us the reason *why* the limit does not exist.
Of course, none of this really has anything to do with why 1/0 is
undefined. What number can you multiply by 0 and get an answer of 1?
Infinity? Nada. Infinity is not even a number. 'Nuff said.
____________________________________________________________
Darrell Ryan
e-mail dryan@edge.net
personal website http://edge.edge.net/~dryan
company website http://www.edge.net/stmc
Subject: Are exponentials linearly independent?
From: dean@math.math.ucdavis.edu (Dean Hickerson)
Date: 8 Jan 1997 05:28:43 GMT
Sherman Stein asked me to post the following question:
Suppose that b(0), ..., b(n) are distinct complex numbers. Then it is
well-known and easy to show that the exponential functions
b(0) z b(n) z
e , ..., e are linearly independent over C. What if n is
infinite? If b(0), b(1), ... are distinct complex numbers and a(0), a(1),
... are complex numbers such that
b(0) z b(1) z
a(0) e + a(1) e + ... converges to 0 for all complex numbers
z, must a(0), a(1), ... all be equal to 0?
Dean Hickerson
dean@ucdmath.ucdavis.edu
Subject: Re: Q: Matrix logarithms and linear dynamics
From: delliott@Glue.umd.edu (David L. Elliott)
Date: 7 Jan 1997 17:14:57 -0500
In article ,
wrote:
>given that the solution of a linear system dxdt = A.x in R^n is
>
>(*) x(t) = exp(A.t).x(0)
>
>is the time "t" to reach x(t) from x(0) well defined for all A? seems
>that t would be given by the matrix natural log: nl(A). at least this
>works for scalar systems and i imagine that if A can be diagonalized,
>then it should be just as easy in higher dimensions.
True.
Given a matrix M = exp(At) the time "t" is well defined as ln(exp(At))
(ln is natural log)
except when the system is periodic (imaginary eigenvalues).
However, given the vectors x(t) and x(0),
you must first solve for M in the equation x(t)= M x(0),
which is easy if A is diagonalized first.
This problem becomes interesting if you realize that -- knowing A, x(0)
and x(t)-- you only want t. That means that numerically you can search
for the real number t, for which you probably have a good initial guess,
in an optimization problem
min ||x(t)- exp(At)x(0)||
t
>
>intuitively, the time between two points in the state space - provided
>they are connected by a flowline - is well defined; it could be found
>by numerical integration for instance. however, i don't know how this
>is related to the matrix logarithm.
>
See the above remark.
>for example if A = {{-1,0},{0,-2}} then the (uncoupled) solutions are:
>
> x(t) = exp(-t).x(0)
> y(t) = exp(-2t).y(0)
>
>where {x(0), y(0)}, {x(t), y(t)} corresponding to the initial and
>final points in R^2 are given. the eigenvalues of A are {-1,-2}.
>however, doesn't the matrix log of a matrix with negative eigenvalues
>have complex entries? how is this matrix log computed to begin with?
>(matlab has a function to compute it numerically)
The eigenvalues in question are exp(-1) and exp(-2) which are positive,
so the log is real.
>
>can someone email/post some insight or point to the literature?
>thanks for the info,
>
> +---------------------------------+
> | Alan Calvitti |
> | Control Engineering |
> | Case Western Reserve University |
> +---------------------------------+
--
David L. Elliott, Vis. Sr. Res. Sci.
Institute for Systems Research/ A.V. Williams Building
University of Maryland/ College Park, MD 20742
(Prof. Emeritus, Washington University) delliott@ISR.umd.edu
Subject: Re: x^3+y^3=z^3+w^3
From: dean@math.math.ucdavis.edu (Dean Hickerson)
Date: 8 Jan 1997 10:41:38 GMT
dennis@netcom.com (Dennis Yelle) wrote:
>In article <5aqjv8$vij@news.alaska.edu> fthg@aurora.alaska.edu (Hannibal
>Grubis) writes:
>>'m curious about whether or not there is a pattern to numbers
>>expressible as the sum of two cubes in two different ways that are NOT of
>>the form (12k)^3 + k^3 = (10k)^3 + (9k)^3
>> i.e. Ramanujan's famous number 1729
>>Would it be difficult to write a program to develop a list of these
>>numbers?
>
>Probably quite difficult to write a program to list all of them.
>
>Here are the ones I found when I limited the search
>to those with 1 <= x,y,z,w <= 100:
>
> 1729 = 1^3 + 12^3 = 9^3 + 10^3
Bob Silverman (numtheor@tiac.net) replied:
>Why do people always compute first and think later??
It's often the most efficient way of attacking a problem. If the solution
of a problem isn't immediately obvious, then a few minutes of computing may
give you the data you need to see a pattern which will help you to solve it.
>It is *trivial* to write a program to list all of them since if you had
>bothered to consult a Number Theory book which discusses Diophantine
>equations you would learn that a complete parameterization of ALL solutions
>is known.
Bob, could you clarify your statement? I've consulted such a book and I
don't see how it helps.
A complete parameterization of all *rational* solutions is known: Theorem 235
of "An Introduction to the Theory of Numbers" by Hardy and Wright states
that, except for the trivial solutions x=y=0, u=-v and x=u, y=v, the
general rational solution of x^3+y^3 = u^3+v^3 is given by
2 2 2 2
x = L (1 - (a-3b)(a +3b )), y = L ((a+3b)(a +3b ) - 1),
(0)
2 2 2 2 2 2
u = L ((a+3b) - (a +3b ) ), v = L ((a +3b ) - (a-3b)),
where L, a, and b are any rational numbers with L != 0.
But generating just the integer solutions is tricky, since x, y, u, and v
may be integers even if L, a, and b are not. For example, the solution
x=1, y=12, u=9, v=10 comes from L=-361/42, a=10/19, b=-7/19.
One approach is to write a=r/t and b=s/t where r, s, and t are integers
with gcd(r,s,t)=1. Then we find that
4 2 2 2 2 4
x = t - t (r-3s)(r +3s ), y = t (r+3s)(r +3s ) - t ,
(1)
3 2 2 2 2 2 2 3
u = t (r+3s) - (r +3s ) , v = (r +3s ) - t (r-3s),
is an integer solution. Dividing x, y, u, and v by their gcd and then
multiplying by an arbitrary integer gives the general integer solution
(except for the trivial solutions mentioned above).
But suppose we want to list all solutions with the absolute values of x, y,
u, and v less than some bound M. Then the solution above is ineffective
unless we can give a useful bound on r, s, and t. Of course we can find a
bound on them, using the fact (obtained from Hardy and Wright's proof) that
r 2 u x - v x - u y + 2 v y s v x - u y
- = ------------------------- and - = -----------------. (2)
t 2 2 t 2 2
2 (x - x y + y ) 2 (x - x y + y )
Thus |r|, |s|, and |t| are all less than 6M.
So to list all solutions with |x|, ... less than M, we could run through
roughly (12M)^3 triples (r,s,t), compute x, y, u, and v from (1), divide
x, y, u, and v by their gcd, and see if the resulting solution is as
small as we want it to be.
But that's less efficient than just testing about (2M)^3 values of x, y,
and u to see if x^3+y^3-u^3 is a cube. So this method is worse than
the obvious brute force attack.
Can anyone suggest a better approach?
Dean Hickerson
dean@ucdmath.ucdavis.edu
Subject: Re: Marilyn's Ping-Pong balls
From: hetherwi@math.wisc.edu (Brent Hetherwick)
Date: 8 Jan 1997 05:12:40 GMT
BNJohnson (bnjohnson@aol.com) wrote:
:
: Is she wrong (heresy)? Was she badly edited? Did she simply not include
: enough details? Any thoughts?
I was too chicken to post a question when her column first appeared: One
must not trifle with Marilyn, at least until one is absolutely positively
undeniably the-world's-gonna-end-if-I'm-wrong sure. Do you recall the
fiasco over the Monty Hall paradox? Many bright mathematicians were quite
embarassed by that one. I have a photocopy of a column where she
publishes letters by quite a few "prominent" mathematicians who dared call
her an idiot. I thought that she was quite the sharp cookie until I read
her little paperback on Fermat.
I now believe that she "gets all her mathematics from books", and this
latest column supports my hypothesis. Wouldn't anyone who spent any time
thinking about the problem indicate the restrictions on the solution which
exclude the trivial, "five ball solution"? Particularly when one is
writing a column addressed mainly to people who don't make a living
estimating densities of optimal sphere-packings and who would naturally
rule out trivial interpretations of this problem.
If you haven't read her book on Fermat's Last Theorem (_The World's Most
Famous Math Problem_, or something like that), do so immediately. The
depth of her ignorance of mathematics knows no nameable bound; this fact
is revealed in excruciating detail there. In it, she criticizes Wiles's
proof, while knowing < 0 about it. Among her justifications are included
criticisms of induction (which she confuses with _empirical_ induction)
and hyperbolic geometry (which she claims may be inconsistent), for their
roles in Wiles's proof.
Thus, it seems more than likely that Martin Gardner has quite a few
"friends" on Marilyn's bookshelf, which would explain her cryptic answer
to the pyramid problem.
$$$ 666 $$$ 666 $$$ 666 $$$ 666 $$$ 666 $$$ 666 $$$ 666 $$$ 666 $$$ 666
hetherwi@math.wisc.edu
$$$ 666 $$$ 666 $$$ 666 $$$ 666 $$$ 666 $$$ 666 $$$ 666 $$$ 666 $$$ 666
Subject: Adding two Gaussians
From: Michael Hucka
Date: 08 Jan 1997 00:35:17 -0500
I have a situation in my work in which I need to add Gaussian curves
together, but can't figure out how to achieve a certain goal. In its
simplest form, the situation is as follows. I am using Gaussians of the form
2 2 2
-2 Pi sigma x
g(x) = e
Suppose I take the sum of two such Gaussians, both with the same sigma, but
one with a spatial offset:
s(x) = g(x) + g(x - x0)
When function s is graphed with increasing values of x0 > 0, the two
Gaussians are spaced farther and farther apart, and you get a range of shapes
like this:
_ __ _ _
/ \ / \ / \/ \
/ \ / \ / \
/ \ / \ / \
--' `-- --' `-- --' `--
Fig. 1 Fig. 2 Fig. 3
x0 = 0 x0 small x0 larger
What I need to do is to find the largest value that x0 can take without the
dip appearing in the top of the sum in Figure 3.
I've tried to approach this by trying to find a way to express the conditions
in the form of a function in x0, so that I can then look for the maximum of
that function, and thereby determine the maximum permissible value of x0.
But I seem to be unable to express the conditions in a usable form. (My math
skills are extremely rusty.) Maybe a different approach is needed
altogether?
I would very much appreciate help with this problem.
--
Mike Hucka hucka@umich.edu
Ph.D. candidate, computational models of human visual processing, U-M AI Lab
UNIX admin & programmer/analyst, EECS Dept., University of Michigan
Subject: Re: Adding two Gaussians
From: dc@cage.rug.ac.be (Denis Constales)
Date: Wed, 08 Jan 1997 11:34:54 +0200
In article <823ewc686i.fsf@eagle.eecs.umich.edu>, Michael Hucka
wrote:
> Suppose I take the sum of two such Gaussians, both with the same sigma, but
> one with a spatial offset:
>
> s(x) = g(x) + g(x - x0)
>
> When function s is graphed with increasing values of x0 > 0, the two
> Gaussians are spaced farther and farther apart, and you get a range of shapes
> like this:
> _ __ _ _
> / \ / \ / \/ \
> / \ / \ / \
> / \ / \ / \
> --' `-- --' `-- --' `--
> Fig. 1 Fig. 2 Fig. 3
> x0 = 0 x0 small x0 larger
>
> What I need to do is to find the largest value that x0 can take without the
> dip appearing in the top of the sum in Figure 3.
It's more helpful to choose a symmetric formulation, like
s(x) = exp(-(x-a)^2) + exp(-(x+a)^2)
because then the possible dip occurs at x=0. To investigate it,
take the second derivative; it's a bad dip if that value is positive.
The condition turns out to be a <= 1/sqrt(2) for no bad dip to occur.
Maple V.4 computation:
> s := exp(-(x-a)^2) + exp(-(x+a)^2);
2 2
s := exp(-(x - a) ) + exp(-(x + a) )
> sxx := diff(s,x,x);
2 2 2
sxx := -2 exp(-(x - a) ) + (-2 x + 2 a) exp(-(x - a) )
2 2 2
- 2 exp(-(x + a) ) + (-2 x - 2 a) exp(-(x + a) )
> factor(subs(x=0,sxx));
2 2
4 exp(-a ) (-1 + 2 a )
For your actual problem, where there's 2 Pi^2 sigma^2 in the exponential,
the condition will be, by rescaling and in terms of x0 = 2 a:
x0 <= sqrt(2) / (sqrt(2) * Pi * sigma) = 1 / (Pi * sigma).
--
Dr. Denis Constales - dcons@world.std.com - http://cage.rug.ac.be/~dc/
Subject: Re: x^3+y^3=z^3+w^3
From: ikastan@alumnae.caltech.edu (Ilias Kastanas)
Date: 8 Jan 1997 10:37:35 GMT
In article ,
Dennis Yelle wrote:
>In article <5aqjv8$vij@news.alaska.edu> fthg@aurora.alaska.edu (Hannibal Grubis) writes:
>>'m curious about whether or not there is a pattern to numbers
>>expressible as the sum of two cubes in two different ways that are NOT of
>>the form (12k)^3 + k^3 = (10k)^3 + (9k)^3
>> i.e. Ramanujan's famous number 1729
>>Would it be difficult to write a program to develop a list of these
>>numbers?
>
>Probably quite difficult to write a program to list all of them.
>
>Here are the ones I found when I limited the search
>to those with 1 <= x,y,z,w <= 100:
>
> 1729 = 1^3 + 12^3 = 9^3 + 10^3
> 4104 = 2^3 + 16^3 = 9^3 + 15^3
> 13832 = 2^3 + 24^3 = 18^3 + 20^3
...
> 1016496 = 47^3 + 97^3 = 66^3 + 90^3
If x, y; s, t are a solution to x^3 + y^3 = s^3 + t^3, then
another solution is
x(x^3 + 2y^3)(s^3 - t^3), t(t^3 + 2s^3)(x^3 - y^3);
y(y^3 + 2x^3)(s^3 - t^3), s(s^3 + 2t^3)(x^3 - y^3).
Ilias
