mm-

Subject: Adventures teaching number theory
I¹ve written a few programs in both UBasic and Blassic
which may be of interest to students in number theory courses.
The programs are available at:
http://www.geocities.com/dunric/games.html
Users of either UBasic or Blassic may also wish to
check out Dark Forest 2, which is a somewhat
simpliÞed adventure game written using both
programming languages. That game is available freely
from:
http://www.geocities.com/dunric/dforest.html
Common questions to consider are:
1) Given a n number of times, how often does a random number
between 1 and 35 have of falling between 5 and 10? 10 and 15?
15 and 25?
2) When a random number is seeded, what is the likelihood
that it will
fall below 20 or above 25? Below 10 or above 30?
3) Given a random number of n, what is the likelihood of a
monster
winning during combat? The player winning during combat?

Paul Allen Panks
dunric@gmail.com

panks@sdf.lonestar.org
SDF Public Access UNIX System - http://sdf.lonestar.org
===
Subject: Random number question?
How does HLA¹s random number generator differ from
traditional BASIC¹s
version? I¹ve long based much of my adventure gaming on
random numbers,
especially during player/monster Þghting. Does seeding a
number truly
make it random, or only quasi-random?
Let¹s say I have a number between 1 and 35. What guarantee do
I have that
the computer won¹t habitually (or accidentally) pick the same
range of
numbers twice? Or the same individual number twice?
1 CLEAR
5 CLS : PRINT Random Number test
10 FOR x = 1 TO 10
20 RANDOMIZE TIMER
30 i = INT(RND * 35) + 1
40 PRINT i
45 NEXT x
The program was run three separate times, with the following
results:
Test #1
Random Number test
27
25
17
31
24
18
27
34
23
22
Test #2
Random Number test
32
29
21
35
29
22
32
4
27
27
Test #3
Random Number test
21
18
10
24
17
11
21
28
16
16
Ah oh...the last two tests repeated the last two sets of
numbers not once
but TWICE! That¹s not good!
Now for the same program in HLA:
program random;
#include (console.hhf);
#include (stdlib.hhf);
#include (math.hhf);
static
i:int32:=0;
x:int32:=0;
begin random;
console.cls();
stdout.put(Random number test,nl);
start:
add(1,x);
rand.randomize();
rand.urange(1,35); // pick a random number, 1 through 35
mov(eax,i); // move it into i variable
mov(i,eax); // set i to eax value
stdout.put(i,nl);
if(x<10) then
jmp start;
endif;
end random;
The HLA version gives the following 3 results:
Test #1
Random number test
29
24
11
19
22
8
27
33
27
28
Test #2
Random number test
19
4
32
30
23
34
24
20
20
30
Test #3
Random number test
35
26
29
17
26
18
33
29
19
1
Some repeats, but not as bad as before.
Is there a way to truly limit the number of repeats during a
set of random
number generation? I can foresee a lot of random numbers in
my own mind,
but they have to be truly, truly random for the random number
generator to
be doing a good job.
Any ideas as to why both sets of random number generators
seem different
in functionality?
Paul Panks
dunric@gmail.com

panks@sdf.lonestar.org
SDF Public Access UNIX System - http://sdf.lonestar.org
===
Subject: Re: Random number question?
You might want to deÞne your requirements a little more
precisely. If
RND were a perfect random number generator (generating
absolutely
independent U(0,1) RV¹s), any pair of numbers generated by
your QBasic
program would be duplicates with probability 1/36, giving a
little less
than 25% probability that there will be at least one pair of
successive
numbers that are duplicates in the sequence of 10 numbers. If
you are
also concerned about non-consecutive duplicates, that
probability would
approach 75% in a sequence of 10 from a perfect RNG.
If you want to avoid duplicates altogether, you will have to
use the
generated random numbers to sample without replacement from
the
candidate values.
If your concern is to minimize autocorrelations of various
lags, so that
the rates of duplicates do not signiÞcantly exceed those of a
perfect
random number generator, then there is a large literature on
random
number generation. You might start with
http://www-personal.engin.umich.edu/~wagnerr/
MersenneTwister.html
http://www.isds.duke.edu/~rlw/marsaglia/cdmake.ps
http://lib.stat.cmu.edu/apstat/183
Jerry
> How does HLA¹s random number generator differ from
traditional BASIC¹s
> version? I¹ve long based much of my adventure gaming on
random numbers,
> especially during player/monster Þghting. Does seeding a
number truly
> make it random, or only quasi-random?
> Let¹s say I have a number between 1 and 35. What guarantee
do I have that

> the computer won¹t habitually (or accidentally) pick the
same range of
> numbers twice? Or the same individual number twice?
> 1 CLEAR
> 5 CLS : PRINT Random Number test
> 10 FOR x = 1 TO 10
> 20 RANDOMIZE TIMER
> 30 i = INT(RND * 35) + 1
> 40 PRINT i
> 45 NEXT x
> The program was run three separate times, with the
following results:
> Test #1
> Random Number test
> 27
> 25
> 17
> 31
> 24
> 18
> 27
> 34
> 23
> 22
> Test #2
> Random Number test
> 32
> 29
> 21
> 35
> 29
> 22
> 32
> 4
> 27
> 27
> Test #3
> Random Number test
> 21
> 18
> 10
> 24
> 17
> 11
> 21
> 28
> 16
> 16
> Ah oh...the last two tests repeated the last two sets of
numbers not once

> but TWICE! That¹s not good!
> Now for the same program in HLA:
> program random;
> #include (console.hhf);
> #include (stdlib.hhf);
> #include (math.hhf);
> static
> i:int32:=0;
> x:int32:=0;
> begin random;
> console.cls();
> stdout.put(Random number test,nl);
> start:
> add(1,x);
> rand.randomize();
> rand.urange(1,35); // pick a random number, 1 through 35
> mov(eax,i); // move it into i variable
> mov(i,eax); // set i to eax value
> stdout.put(i,nl);
> if(x<10) then
> jmp start;
> endif;
> end random;
> The HLA version gives the following 3 results:
> Test #1
> Random number test
> 29
> 24
> 11
> 19
> 22
> 27
> 33
> 27
> 28
> Test #2
> Random number test
> 19
> 32
> 30
> 23
> 34
> 24
> 20
> 20
> 30
> Test #3
> Random number test
> 35
> 26
> 29
> 17
> 26
> 18
> 33
> 29
> 19
> Some repeats, but not as bad as before.
> Is there a way to truly limit the number of repeats during
a set of random

> number generation? I can foresee a lot of random numbers in
my own mind,
> but they have to be truly, truly random for the random
number generator to

> be doing a good job.
> Any ideas as to why both sets of random number generators
seem different
> in functionality?
> Paul Panks
> dunric@gmail.com
===
Subject: Re: Question about Simple Taylor Series
> In the question below, ^ means to the power of, and gamma
is the
> complete gamma function (where, for an integer n,
gamma(n+1) = n!).
>
> If a is an integer, then
>
> e - summation{[for n = a to inÞnity] (x^n)/gamma(n+1)}
>
> equals
>
> summation{[for n = 0 to a - 1] (x^n)/gamma(n+1)}
>
> That¹s an easy enough problem, even for me. However, the
case where
> a is not an integer is not so easy, for me. Could someone
please
> stoop to helping me?
>
> In case it matters, I¹m trying to create better C code for
calculating
> the standard error function and percentiles corresponding to
> chi-squared statistics.
>
> David Sexton
>
> Your formulae don¹t look quite right.
>
> The probability density function for the standard gamma
distribution
> is exp(-x).x^(a-1)/gamma(a) for x > 0 and a > 0.
>
> The cumulative density function is obtained by integrating
the pdf.
> The integral can also be expressed in terms of various
summations,
> continued fractions, asymptotic expansions... A&S gives
some of these
> formulae in sections 6 & 26 (see
> Which method to use when is the real issue when producing
code to
> deliver accurate results.
>
> The chi-squared distribution and the normal distribution
can both be
> expressed in terms of the gamma distribution. Personally, I
use a
> specialised function to evaluate the cdf of the normal
distribution
> and then call this function when evaluating the asymptotic
expansion
> for the gamma distribution.
>
> If you want to calculate percentile points then you need to
be able to
> invert the cdf. I Þnd the easiest way to solve F(x;a)=p is
to use
> simple Þrst guess methods, based on Normal approximations
and the
> like, and then solve log(F(x;a))=log(p) using the
Newton-Raphson
> method. For small p, the logged equation is usually a much
more linear
> equation than the original F(x;a)=p.
>
> You are more than welcome to start from the Javascript code
in
> http://members.aol.com/iandjmsmith/myfunctions.js It can be
easily
> translated to C - I believe there are even free translators
for this
> purpose. The code is mainly intended to deliver accurate
answers over
> a wide range of parameters. It is pretty quick, but could
deÞnitely
> be made quicker and more compact if that¹s what you mean by
better.
>
> Calculators for the Normal, Chi-squared and the Gamma
distributions
> calculator examples demonstrate how to call the functions
available in
> myfunctions.js and the higher level error-handling required.
>
>
> Ian Smith
> on this page
> http://mathworld.wolfram.com/IncompleteGammaFunction.html
> and formulae 6.2.5 and 6.2.7 on page 217 of
> http://www.library.cornell.edu/nr/bookcpdf/c6-2.pdf
> I get the most accurate results with the MathWorld formula,
but it
> only works when a (of Gamma(a,x)) is an integer. I was
hoping to
> Þnd, at least, a similarly accurate formula for all the
cases when 2a
> is an integer. The Numerical Recipes formulae are fast, but
> disappointingly inaccurate when a is large. I am willing to
trade
> quite a bit of speed for accuracy.
> The problem with the NM formulae is the necessity of using
logs and
> antilogs to deal with the very large numbers involved.
There¹s a
> simple trick for preventing overþow with the MathWorld
formula that
> I can¹t use with the NM formulae.
> David Sexton
The Poisson cumulative distribution function with mean m, can
be
evaluated using the following equations
For i > m and m > 0
Pr(X>=i;m) = cf(i)*Pr(X=i;m)
where
cf(i) = 1 + cf(i+1)m/(i+1) {Summation Equation }
cf(i) = 1 + m/(i-m+1 + m/(i-m+2 + 2m/(i-m+3 + 3m/(i-m+4...)
{Continued
Fraction Equation due to ???}
When i is reasonably large and close to m the continued
fraction
equation can be used to sum the tail of the distribution
(i.e. we use
it to evaluate cf(i+x) for some x) and then the summation
equation
used to Þnd cf(i).
For 0 <= i and i <= m
Pr(X<=i;m) = cf(i)*Pr(X=i;m)
where
cf(i) = 1 + cf(i-1)i/m {Summation Equation }
cf(i) = 1 + i/(m-i+1 + (i-1)/(m-i+3 + 2(i-2)/(m-i+5 +
3(i-3)/(m-i+7...) {Continued Fraction Equation due to
Legendre}
When i is reasonably large and close to m the continued
fraction
equation can be used to sum the tail of the distribution
(i.e. we use
it to evaluate cf(i-x) for some x) and then the summation
equation
used to Þnd cf(i).
For accurate evaluation of Pr(X=i;m) see
http://members.aol.com/iandjmsmith/UseOfLogGamm.htm
These formulae can be slow for very large values of m & i. The
asymptotic approximation in
http://members.aol.com/iandjmsmith/PoissonApprox.htm ,
produces
relative errors of 6e-13 or less when i is >= 21 and m is
between 0.7i
and 1.3i.
Note that for very large values of m & i, the main
contributor to
error is the cancellation error in the evaluation of m-i.
For the Gamma distribution, the other case which is not dealt
by the
above formulae is where a -> 0 and x < 1. In this case the
integral of
exp(-x).x^(a-1)/gamma(a) can be evaluated by a Taylor series
expansion
of exp(-x) followed by term by term integration, giving
x^a/gamma(a+1).(1-a(x/(a+1)+x^2/2!/(a+2)-x^3/3!/(a+3...)).
We can also evaluate 1 -
x^a/gamma(a+1).(1-a(x/(a+1)+x^2/2!/(a+2)-x^3/3!/(a+3...))
accurately
via 1-
(1+(x^a-1))/(1+(gamma(a+1)-1)).(1-a(x/(a+1)+x^2/2!/(a+2)-x^3/
3!/(a+3...)).
x^a - 1 is just exp(a.log(x))-1 (you will need a special
function to
evaluate exp(x)-1 accurately) and gamma(a+1) - 1 can be
evaluated
accurately via Abramowitz & Stegun¹s series 6.1.33.
Ian Smith
===
I am looking for algorithm to generate random packing of
spherical
Something like this:
http://www.che.utexas.edu/~truskett/images/packed_spheres.jpg
===
> I am looking for algorithm to generate random packing of
spherical
> Something like this:
>
http://www.che.utexas.edu/~truskett/images/packed_spheres.jpg
Search Google for sphere packing problem. You will get lots
of hits,
for example:
http://mathworld.wolfram.com/SpherePacking.html.
Good luck,
OUP
===
> I am looking for algorithm to generate random packing of
spherical
>
> Something like this:
>
http://www.che.utexas.edu/~truskett/images/packed_spheres.jpg
>
> Search Google for sphere packing problem. You will get lots
of hits,

> for example:
> http://mathworld.wolfram.com/SpherePacking.html.
> Good luck,
> OUP
http://mathworld.wolfram.com/ is a gold mine of information.
But I still have not Þnd workable program for it.
Maybe some one could point me directly to program generated
coordinates of
the
spheres. I am going to use it for simulation of þow through
this type of
media.
===
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> listproc ktbuk
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> mailnews@dizum.com ktbuk
> likely ktbuk
> kneider@kornet.net ktbuk
> carried ktbuk
> daviddoherty@yahoo.com ktbuk
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> planeta@yahoogroups.com ktbuk
> whites@aol.com ktbuk
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> reffnera@Þndlay.edu ktbuk
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> bmagpam@yahoo.com ktbuk
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> harlem@aol.com ktbuk
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> electronic ktbuk
> huge ktbuk
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> ericvanlint@yahoo.com ktbuk
> jazeera ktbuk
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> insync ktbuk
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> hopwood ktbuk
> bmagrhea@yahoo.com ktbuk
> rolando@truthmail.com ktbuk
> sreal ktbuk
> stop ktbuk
> irisbrose@yahoo.de ktbuk
> gritty ktbuk
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> onto ktbuk
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> littletnbigdaddy@aol.com ktbuk
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> were ktbuk
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> vjeuck@aol.com ktbuk
> xaknuli ktbuk
> bitnet ktbuk
> habitat ktbuk
> seens ktbuk
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> psychicmedium@yahoo.com ktbuk
> bbradd ktbuk
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> charanjib@yapost.com ktbuk
> stem ktbuk
> taigamasuku@yahoo.com ktbuk
> tharkey ktbuk
> commit ktbuk
> java ktbuk
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> carmen ktbuk
> reading ktbuk
> arabia ktbuk
> mailboxes ktbuk
> dhellas ktbuk
> bigole ktbuk
> bignews@fasttrec.com ktbuk
> jhopkins@uiah.Þ ktbuk
> sour ktbuk
> forgot ktbuk
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> genel@yahoogroups.com ktbuk
> volume ktbuk
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> leonardr@segfault.org ktbuk
> arrogance ktbuk
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> borarslan@yahoo.com ktbuk
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> Þshkill@uþ.edu ktbuk
> played ktbuk
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> ljossifoff@hotmail.com ktbuk
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> saonah@ivillage.com ktbuk
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> dnewman@networktest.com ktbuk
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> goran ktbuk
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> postmaster@columbia.edu ktbuk
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> mature ktbuk
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> rojanapanya@yahoo.com ktbuk
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> paulsongizaho@eyou.com ktbuk
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> mboricevic@yahoo.com ktbuk
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> ragchew ktbuk
> higher ktbuk
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> also ktbuk
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> niet ktbuk
> bmagned ktbuk
> xnewsreader ktbuk
> scripts ktbuk
> wclistserve@fanciful.org ktbuk
> anymouse ktbuk
> dnewman ktbuk
> cartoons ktbuk
> derganc@didamail.com ktbuk
> porter ktbuk
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> member ktbuk
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> paulson@eyou.com ktbuk
> inputs ktbuk
> begins ktbuk
> Þlter ktbuk
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> borislav ktbuk
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> yasushicountry@yahoo.com ktbuk
> surgam@hotmail.com ktbuk
> yahoogroupes ktbuk
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> batesa@yahoo.com ktbuk
> arboriculture ktbuk
> twentythree ktbuk
> produced ktbuk
> parties ktbuk
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> secy@sbcglobal.net ktbuk
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> bmagwhitneyh@yahoo.com ktbuk
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> bmagsusan@yahoo.com ktbuk
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> aren ktbuk
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> alumni@Þndlay.edu ktbuk
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> denied ktbuk
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> poohsan@hotmail.com ktbuk
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> anybody ktbuk
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> bmagphipps@yahoo.com ktbuk
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> contrib ktbuk
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> archstl@sbcglobal.net ktbuk
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===
Subject: Evaluating an inÞnite series
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i5MBteQ20620;
to Þnd this sum:
Inf.
Sum 1/[(3k-2)^2] = ?
k=1
Any help would be appreciated!
===
Subject: Re: Evaluating an inÞnite series
> to Þnd this sum:
> Inf.
> Sum 1/[(3k-2)^2] = ?
> k=1
> Any help would be appreciated!
To evaluate Sum{k=0 to infty}{f(k)} by residues, look at the
contour integral
oint{ dz f(z) cot(z) }
on the circle of radius R = N + 1/2 .
You see that this will give you (up to trivial factors)
twice the sum you want, plus the residue at z=0 and at any
poles of f(z). This, of course, has to equal the actual
integral
on the contour.
I leave it to you as a HW problem to show that the integral
on the circle shown (which goes through none of the poles)
vanishes in the limit N to infty .
BTW, this is all found in standard books on functions of a
complex
variable, such as Copson or Whittaker & Watson, as well as in
many
books on mathematical methods of physics. I am thinking in
particular
of Riley, et al.

Julian V. Noble
Professor Emeritus of Physics
jvn@lessspamformother.virginia.edu
^^^^^^^^^^^^^^^^^^
http://galileo.phys.virginia.edu/~jvn/
For there was never yet philosopher that could endure the
toothache patiently.
-- Wm. Shakespeare, Much Ado about Nothing. Act v. Sc. 1.
===
Subject: Re: Evaluating an inÞnite series
> to Þnd this sum:
> Inf.
> Sum 1/[(3k-2)^2] = ?
> k=1
> Any help would be appreciated!
> To evaluate Sum{k=0 to infty}{f(k)} by residues, look at the
> contour integral
> oint{ dz f(z) cot(z) }
> on the circle of radius R = N + 1/2 .
> You see that this will give you (up to trivial factors)
> twice the sum you want, plus the residue at z=0 and at any
> poles of f(z). This, of course, has to equal the actual
integral
> on the contour.
> I leave it to you as a HW problem to show that the integral
> on the circle shown (which goes through none of the poles)
> vanishes in the limit N to infty .
> BTW, this is all found in standard books on functions of a
complex
> variable, such as Copson or Whittaker & Watson, as well as
in many
> books on mathematical methods of physics. I am thinking in
particular
> of Riley, et al.
Oops-- two corrections:
1. Integrate oint{ dz f(z) cot(z pi) } around |z| = N + 1/2
^^^
2. The result will give you the sums
sum{k=1 to N} left( { f(k) + f(-k) } right)
one of which is the sum you want and the other is one you have
to
evaluate by a trick.
But anyway, you get the general idea.

Julian V. Noble
Professor Emeritus of Physics
jvn@lessspamformother.virginia.edu
^^^^^^^^^^^^^^^^^^
http://galileo.phys.virginia.edu/~jvn/
For there was never yet philosopher that could endure the
toothache
patiently. -- Wm. Shakespeare, Much Ado about Nothing. Act v.
Sc.
1.
===
Subject: Covariance
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i5MBtgn20700;
I am reading this paper on Concurrent Mapping and
localisation and one
of the portion covers the its state covariance matrix as
below.
Lets say P be covariance, so Pab is Covariance matrix between
variables a and b.
a is a variable of 3 dimensions, b is of 2 dimensions. Eg:
a(1,2,3),b(1,2)
So Paa and Pbb are actually 3x3 and 2x2 matrices
respectively. I can
retrieve Paa and Pbb based on my dataset. Now i wish to Þnd
P, which
is of the following 5x5 matrix.
Paa Pab
transpose(Pab) Pbb
is there a way to derive Pab (3x2) from Paa and Pbb itself?
Or is
there supposed to be another way?
rdgs,
jax
===
Subject: Derivative of Psi (Digamma) Function
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i5MBtfN20625;
In trying to Þnd a way to evaluate the series on My last post,
Inf.
Sum 1/[(3k-2)^2] = 1 + 1/(4^2) + 1/(7^2) + ... = ?
k=1
Another good samaritan told Me that Mathematica gives the
answer as
1/9*(Polygamma(1,1/3)), but I would like to have the answer
in simpler
functions, similar to how Polygamma(0,a/b) can be written
using
Gauss¹s Digamma Theorem-- So, is it possible to get the Þrst
derivative of Digamma, (which would be Polygamma(1,a/b)) by
taking the
derivative of Gauss¹s Equation, and using that expression to
get a
formula in terms of trig, log, etc. for the series above?
===
Subject: who is the authority in meshless Þeld?
i am a freshman
===
Subject: Re: Estimating the condition of an upper triangular
matrix
>> given a small (10 times 10) matrix in upper triangular
form I am
>seeking
>> an
>> algorithm to estimate the condition (i.e. the ratio of the
largest
and
>> smallest singular value) of that matrix. I want to avoid
any kind of
>> overestimation. What is the state of the art on that Þeld?
>> cheers
>> Thomas
>> The eigenvalues of a triangular matrix are just the
diagonal elements
>> themselves. Is this what you have?
>> If so, the condition number is just the ratio of the
largest eigenvalue
>to
>> the smallest eigenvalue.
>> I *think* that¹s true anyway!
>> Mike
>I agree, indeed I thought I was missing something in the
post. Anyways
>isn¹t the ratio a measure of the stiffness of the matrix?
Condition wrt
to
>inversion or exponentiation might not be given by the ratio.
>--
>Ciao,
>Gerry T.
no, this is unfortunately false. condition number relative to
a norm is
norm(A)*norm(inverse(A)) and if you take norm=euclidean norm
this is
sqrt(maxeigenvalue of transpose(A)*A/min eigenvalue of
transpose(A)*A)
and hence is not easily obtained for triangular A. Indeed,
you can have
an arbitrarily illconditioned R with all eigenvalues =1
hth
peter
===
Subject: Re: Estimating the condition of an upper triangular
matrix
Peter Spellucci
<nospamspellucci@fb04373.mathematik.tu-darmstadt.de >
message
>> given a small (10 times 10) matrix in upper triangular
form I am
>seeking
>> an
>> algorithm to estimate the condition (i.e. the ratio of
the largest
and
>> smallest singular value) of that matrix. I want to avoid
any kind
of
>> overestimation. What is the state of the art on that
Þeld?
>> cheers
>> Thomas
>> The eigenvalues of a triangular matrix are just the
diagonal elements
>> themselves. Is this what you have?
>> If so, the condition number is just the ratio of the
largest
eigenvalue
>to
>> the smallest eigenvalue.
>> I *think* that¹s true anyway!
>> Mike
>I agree, indeed I thought I was missing something in
the post. Anyways
>isn¹t the ratio a measure of the stiffness of the matrix?
Condition wrt
to
>inversion or exponentiation might not be given by the
ratio.
>--
>Ciao,
>Gerry T.
> no, this is unfortunately false. condition number relative
to a norm is
> norm(A)*norm(inverse(A)) and if you take norm=euclidean
norm this is
> sqrt(maxeigenvalue of transpose(A)*A/min eigenvalue of
transpose(A)*A)
> and hence is not easily obtained for triangular A. Indeed,
you can have
> an arbitrarily illconditioned R with all eigenvalues =1
> hth
> peter
Peter,
There are some ambiguities in your post.
First, as Mike pointed out, the eigenvalues of a triangular
matrix are its
diagonal entries, no calculations required. If you are saying
that the max
eigenvalue/min eigenvalue is not the condition wrt inversion
as intimated
by Mike then I agree. As I remarked, that ratio is a measure
of the
stiffness of the triangular matrix and it is not the
condition number wrt
inversion. Second, the condition number is not necessarily
wrt inversion.
As I remarked, the OP might have meant condition wrt
exponentiation in
which case again the max eigenvalue/min eigenvalue is not the
condition
number which is given in some norm as the Fretchet derivative
of the
matrix.

HTH,
Gerry T.
===
Subject: Question on Tutte Polynomial (planar 3-connected
graphs)
Originator: bergv@math.uiuc.edu (Maarten Bergvelt)
literature about the Tutte polynomial of single-component
three-connected planar simple graphs. SpeciÞcally, if there
are two
such non-isomorphic graphs with identical vertex count, edge
count,
and Tutte polynomial, though I am interested in other results
as well.
Ralph Furmaniak
rfurmani@math uwaterloo ca
===
Subject: Re: This is NOT naive bayes rule. How to interpret
it reasonably?
Originator: bergv@math.uiuc.edu (Maarten Bergvelt)
>I know the naive bayes rule is
>P(A,B | C) -> P(A|C)*P(B|C).
It may be naive, but what does it have to do with Bayes?
When it¹s an equality, it means that A and B are conditionally
independent given C.
>I used -> instead of = to emphasize it is just an
approximation.
>Similarly,
>P(A|B,C) -> P(A|B) * P(A|C)
>does NOT work too bad under some circumstances, as it is in
some
>sense doing something like averaging, as this one:
>P(A|B,C) -> 0.5 (P(A|B) + P(A|C))
>I would like to Þgure out if there has been any research on
P(A|B,C)
>-> P(A|B) * P(A|C) ? In what sense is it a reasonable rule?
This makes no sense at all to me. For example, if A, B and C
are
independent it says P(A) -> P(A)^2. Well, it¹s reasonable if
P(A)
is 0 or 1, but that¹s about it.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Re: This is NOT naive bayes rule. How to interpret
it reasonably?
Originator: bergv@math.uiuc.edu (Maarten Bergvelt)
>I know the naive bayes rule is
>P(A,B | C) -> P(A|C)*P(B|C).
> It may be naive, but what does it have to do with Bayes?
> When it¹s an equality, it means that A and B are
conditionally
> independent given C.
>I used -> instead of = to emphasize it is just an
approximation.
>Similarly,
>P(A|B,C) -> P(A|B) * P(A|C)
>does NOT work too bad under some circumstances, as it is in
some
>sense doing something like averaging, as this one:
>P(A|B,C) -> 0.5 (P(A|B) + P(A|C))
>I would like to Þgure out if there has been any research on
P(A|B,C)
>-> P(A|B) * P(A|C) ? In what sense is it a reasonable rule?
> This makes no sense at all to me. For example, if A, B and
C are
> independent it says P(A) -> P(A)^2. Well, it¹s reasonable
if P(A)
> is 0 or 1, but that¹s about it.
> Robert Israel israel@math.ubc.ca
> Department of Mathematics http://www.math.ubc.ca/~israel
> University of British Columbia
> Vancouver, BC, Canada V6T 1Z2
I found that, instead of that ungrounded conjecture
P(A | B,C) ~= P(A | B) * P(A | C),
we can get something similar by real derivation:
P(A | B,C) = P(A | B) * P(A | C) / P(A)
where two assumptions are used: one is that B,C are
independent, the
other one is B,C are conditionally independent given A.
Here is the derivation, please correct me if I am wrong:
P(A | B,C) = NOMINATOR / DENOMINATOR, where
NOMINATOR = P( B,C | A) * P(A)
~= P(B | A) * P(C | A) * P(A)
= P(A | B)P(B) * P(A | C)P(C) / P(A)
DENOMIATOR = P(B,C) ~= P(B) * P(C)
HENCE
P(A | B,C) = P(A | B) * P(A | C) / P(A)
I¹m interested in knowing if this derivation has any important
implications? It looks a little bit similar to the conditional
independence rule P(A,B|C) = P(A|C)*P(B|C), but it isn¹t.
===
Subject: Re: bounding rational points on Curves
Originator: bergv@math.uiuc.edu (Maarten Bergvelt)
> Is there a generalized Mordell-Conjecture??
> in the following sense:
> Let K be a number Þeld and let C/K be a curve of genus g>1.
Denote by
> Iso(C) the set of curves C¹ deÞned over K such that C is
isomorphic
> to C¹ over bar{K}. Exists there a bound N such that
> #C¹(K) < N for all C¹ in Iso(C) ???
I doubt this is known, even for twist families of curves of
the form
C_a : ay^n = f(x),
where f(x) is Þxed and a varies in K^* (mod n¹th powers). In
the
characteristic 0 function case, there is a bound for #C_a(K)
for
twists of this sort in terms of the rank of the Jacobian of
C_a, but
probably(?) the rank of J_a(K) is unbounded as a varies.
On the other hand, if you¹re willing to accept a conjecture of
Bombieri and Lang,
B-L Conjecture:
K-rational points on a variety of general type cannot be
Zariski
dense,
then Caporaso, Harris, and Mazur proved something much
stronger than
what you are asking. They prove that if the B-L conjecture is
true,
then there is a constant N = N(K,g) so that for every (smooth
projective) curve C/K of genus g, one has the bound
#C(K) < N(K,g).
Since all of the curves in your Iso(C) have the same genus,
the result
you want is a very special case.
BTW, there are many things that one might call generalized
Mordell
conjectures. A better name for your type of question is a
uniform
quantitative Mordell conjecture, since you want uniformity in
the
parameter, and you¹re looking at quantitative results (bounds
for
number of points) as opposed to effective results (bounds for
the
height of points).
JS
===
Subject: Re; Question on Tutte Polynomial (planar 3-connected
graphs)
Originator: bergv@math.uiuc.edu (Maarten Bergvelt)
I should also specify that I am refering to pairs of graphs
that are
not dual to each other.
-- Ralph Furmaniak
===
Subject: Paper published by Algebraic and Geometric Topology
Originator: bergv@math.uiuc.edu (Maarten Bergvelt)
The following paper has been published:
Algebraic and Geometric Topology
URL:
http://www.maths.warwick.ac.uk/agt/AGTVol4/agt-4-22.abs.html
Title:
Embeddings of graph braid and surface groups in right-angled
Artin groups
and braid groups
Author(s):
John Crisp, Bert Wiest
Abstract:
We prove by explicit construction that graph braid groups and
most
surface groups can be embedded in a natural way in
right-angled Artin
groups, and we point out some consequences of these embedding
results. We also show that every right-angled Artin group can
be
embedded in a pure surface braid group. On the other hand, by
generalising to right-angled Artin groups a result of Lyndon
for free
groups, we show that the Euler characteristic -1 surface
group (given
by the relation x^2y^2=z^2) never embeds in a right-angled
Artin
group.
Secondary: 05C25
Keywords:
Cubed complex, graph braid group, graph group, right-angled
Artin group,
conÞguration space
Author(s) address(es):
Institut de Mathematiques de Bourgogne (IMB), UMR 5584 du CNRS
Universite de Bourgogne, 9 avenue Alain Savary, B.P. 47870
21078 Dijon cedex, France
and
IRMAR, UMR 6625 du CNRS, Campus de Beaulieu, Universite de
Rennes 1
35042 Rennes, France
Email: jcrisp@u-bourgogne.fr,
bertold.wiest@math.univ-rennes1.fr
===
Subject: crystographic root systems
Originator: bergv@math.uiuc.edu (Maarten Bergvelt)
Hi everybody,
Given is a irreducible (Þnite) crystographic root systems
(V,R,S),
with $V$ a Euclidean vector space containing R, S
a choice of simple roots, $Theta$ THE longest highest root,
and v and regular element in V. Now the assertion I am not
able to prove in a uniform manner is the following:
if s and t are elements of the Weylgroup W of R, and
s(v) - t(v) is a real multiple of $Theta$, then s = t
or t = s_{Theta} s, whereby s_{Theta} the (orthogonal)
reþection in the mirror given by the orthogonal complement
given by $Theta$.
With some easy calculations, i have been able to prove it
for each classical root systems A,B,C and D seperately.
For easthetical reasons, I think there must be a uniform
proof.
I would appreciate if respondends also send a copy to my
mail acount: brokensymmetrie at yahoo dot com.
Broken Symmetrie
===
Subject: Re: crystographic root systems
Originator: bergv@math.uiuc.edu (Maarten Bergvelt)
> Hi everybody,
> Given is a irreducible (Þnite) crystographic root systems
(V,R,S),
> with $V$ a Euclidean vector space containing R, S
> a choice of simple roots, $Theta$ THE longest highest root,
> and v and regular element in V. Now the assertion I am not
> able to prove in a uniform manner is the following:
> if s and t are elements of the Weylgroup W of R, and
> s(v) - t(v) is a real multiple of $Theta$, then s = t
> or t = s_{Theta} s, whereby s_{Theta} the (orthogonal)
> reþection in the mirror given by the orthogonal complement
> given by $Theta$.
If s(v)=t(v), then s=t by regularity of v.
If not, look at the (afÞne) line through s(v) and t(v). By
assumption,
this
line is perpendicular to the hyperplane of Þxed points of the
reþection
s_Theta, so s_Theta(s(v)) also lies on this line. But these
three
points
are at the same distance from the origin, so two of them must
coincide,
i.e. either s_Theta(s(v))=s(v) or s_Theta(s(v))=t(v). By
regularity
again, this implies s_Theta s=s or s_Theta s=t, the former
being
obviously impossible.

Christian Ohn
e-mail: Þrstname dot lastname at univ hyphen valenciennes dot
fr
===
Subject: Re: A Repunit Observation
Originator: bergv@math.uiuc.edu (Maarten Bergvelt)
apologies for my question about the asymptotic formula.
With a bit more hunting around, I found the following
equation (13) on
http://mathworld.wolfram.com/PrimeFormulas.html
p_n ~ n (ln n + ln ln n -1 + (ln ln n - 2)/ln n - (ln^2 (ln
n) - 6ln
ln n + 11)/(2ln^2 n) + ...).
It is due to Cipolla, Cipolla, M. La determinazione
assintotica dell¹
numero primo. Napoli Rend. 3, 132-166, 1902
and is the inverse of li(x) obtained by series reversion.
I hope that this info is of interest.
> I wish to thank-you for your counter-example. I failed to
include the
> condition that that n should be greater than 5 in A REPUNIT
> OBSERVATION.
> Dennis
> dennismkulvicki@starday.org
>>Let Rn denote the repunit of n digits, DeÞne s(n) as the
sum of the
>>exponents of the prime factors of Rn. Consider the
following REPUNIT
>>OBSERVATION: If s(n)is equal to less than 3, then n is
prime.
>The prime factorization of R4 = 1111 is 11^1 * 101^1 so s(4)
= 2 < 3,
>but 4 is not prime.
>This may be the only (base 10) exception, however. If n is
divisible
> by
>p, then Rn is divisible by Rp. Moreover, R(pq) > R(p) R(q)
for
> p,q>1.
>So s(n) >= 3 whenever n has at least two distinct prime
factors
>or is a k¹th power for k >= 3. Thus the only possible
exceptions
>to your observation are when n is the square of a prime.
>Now for n = p^2, Rn/Rp is a number of p^2 - p + 1 digits. In
order
> for
>s(n) < 3 we would need this to be prime. Although it may
have no
>obvious prime factors, heuristically the probability that a
number of
>this size is prime is on the order of 1/p^2. And since
>sum_{p prime} 1/p^2 is Þnite, we might expect that there are
only
>Þnitely many cases where R(p^2)/Rp is prime.
>Here are some other counterexamples in other bases:
>base n
> 2 2^2, 3^2, 7^2
> 3 3^2
> 4 2^2
> 5 7^2
> 6 2^2
> 8 3^2
>16 2^2
>20 3^2
>Robert Israel israel@math.ubc.ca
>Department of Mathematics <a
>
href=http://www.math.ubc.ca/~israel>http://www.math.ubc.ca/~
israel</a>
>University of British Columbia
>Vancouver, BC, Canada V6T 1Z2
===
Subject: Re: Are there better explicit bounds on the nth
prime than
Rosser¹s?
Originator: bergv@math.uiuc.edu (Maarten Bergvelt)
hi,
just took a look at Massias and Robin¹s paper.
They have the bound
p_k leq k(ln k + ln ln k - 1 + (ln ln k-1.8)/ln k).
Furthermore, I just did some computing and it seems that the
1.8 can
be replaced by something like 2.04 for k sufÞciently large
(note that
is just an observation, no claims to have proofs here).
Anyone have an idea what this term should really be?
Maybe 2+o(1)?
Can we derive it from the Prime Number theorem? Not the
explicit upper
bound itself, but maybe an asymptotic version?
What about the next term? What should that look like?
>
>
I¹m looking for good bounds on the nth prime, and the theta
function
> stuff, but since it¹s from 1941, there might be some
improvements...
> functions, please do share them with me.
What I really am looking for is an upper bound for the theta
function
> below e^95, which Rosser doesn¹t give (he does give
something, but it
> has a restriction that isn¹t good for me...). The lower
bound he
gives
> is great, and so are the those for the nth prime.
>
Dror Speiser
>
> Rosser¹s 1941 paper
>
> J. B. Rosser, Explicit bounds for some functions of prime
numbers,
> Amer. J. Math. 63 (1941), 211--232.
>
> have been improved in
>
> J. B. Rosser and L. Schoenfeld, Sharper bounds for the
Chebyshev
> functions Theta(X) and Psi(X) , Math. Comp. 29 (1975),
243--269.
> and
> L. Schoenfeld, Sharper bounds for the Chebyshev functions
Theta(X)
> and Psi(X). II, Math. Comp. 30 (1976), 337--360
>
> (see
http://www.ams.org/mcom/1996-65-213/S0025-5718-96-00669-2/
home.html
> )
>
> The best? currently known lower bound for P_n might be
>
> Dusart, P. The k_th Prime is Greater than k(ln k + ln ln k
- 1) for
k
>= 2. Math. Comput. 68, 411-415, 1999.
> (see http://mathworld.wolfram.com/RossersTheorem.html )
>
> Hugo Pfoertner
> More information on this problem was given by Peter Luschny
and Hermann
> Kremer in a thread in the (German) newsgroup
de.sci.mathematik, see
> Peter Luschny noted:
> Die obere Schranke p_k <= k(ln(k)+ln(ln(k)))
> Þndet sich bei Rosser und Schoenfeld, der Beweis scheint
> aber nicht einwandfrei zu sein, ist aber sp.8atestens
> seit Massias und Robin gesichert.
> freely translated
> An upper bound p_k <= k(ln(k)+ln(ln(k)))
> The remaining doubts were resolved by Massias and Robin.
> ( J.-P. MASSIAS & G. ROBIN, Bornes effectives pour certaines
> fonctions concernant les nombres premiers, Journal de
Th.8eorie
> des Nombres de Bordeaux 8 (1996), 215-242. MR 97g:11099 )
> Hugo
===
Subject: Can one foliate 3-space with skew lines?
X-mailer: epigone
Epigone-thread: streefrorzhem
Originator: bergv@math.uiuc.edu (Maarten Bergvelt)
Can one foliate space with straight lines that are pairwise
skew?
Rich Peterson CSU Sacramento
===
Subject: Re: Can one foliate 3-space with skew lines?
Originator: bergv@math.uiuc.edu (Maarten Bergvelt)
>Can one foliate space with straight lines that are pairwise
skew?
It looks to me offhand like this works, but I can¹t Þnd any
more scrap paper to check. Begin by exhausting 3-space by
the pairwise disjoint sets H(A) = {(x,y,z):
x^2+y^2=A^2(1+z^2)},
for A greater than or equal to 0. H(0) is the z-axis, and
for A greater than 0, H(A) is a hyperboloid of one sheet,
and thus a ruled surface, foliated (in two ways) by straight
lines that are pairwise skew. I *think* that it¹s both
clear and true that for A not equal to B, each generator
of the ruling on H(A) is skew to each generator of the
ruling on H(B). Then choosing the rulings consistently,
I think you get the desired foliation of 3-space.
Lee Rudolph
===
Subject: Re: Can one foliate 3-space with skew lines?
Originator: bergv@math.uiuc.edu (Maarten Bergvelt)
>>Can one foliate space with straight lines that are pairwise
skew?
> It looks to me offhand like this works, but I can¹t Þnd any
> more scrap paper to check. Begin by exhausting 3-space by
> the pairwise disjoint sets H(A) = {(x,y,z):
x^2+y^2=A^2(1+z^2)},
> for A greater than or equal to 0. H(0) is the z-axis, and
> for A greater than 0, H(A) is a hyperboloid of one sheet,
> and thus a ruled surface, foliated (in two ways) by straight
> lines that are pairwise skew. I *think* that it¹s both
> clear and true that for A not equal to B, each generator
> of the ruling on H(A) is skew to each generator of the
> ruling on H(B). Then choosing the rulings consistently,
> I think you get the desired foliation of 3-space.
It works! -- Here¹s an explicit parametrization:
To any point (x,y,0) in the plane
associate the line
t |--> (x-t*y, y+ t*x, t)
in three-space.
These lines are pairwise skew and reach every point in
three-space.
===
Subject: Welcome to sci.math.research
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===
Subject: Re: a linear algebra problem
Originator: edgar@math.ohio-state.edu (Gerald Edgar)
>...
> Let A_1, A_2, --- A_(n-1) be n-1 Linearly independent
skew-symmetric
> n x n matrices such that the rows of all these n-1
matrices, taken
> together, span whole R^n. Then can we say that there exists
a
> non-singular ( invertible ) n x n matrix R such that the
Þrst row of
> R(A_i)R¹ is e_(i+1) for all i =1,2,. (n-1) ??? where
> 1) e_i is a vector in R^n with i th component being 1 and
rest are
> zero.
> 2) R¹ denotes the transpose of R.
No. Take for instance n = 8. Let A_1 be invertible, say
( 0 I)
(-I 0).
Its rows already span R^8.
Now we can Þnd six linearly independent 4 x 4 skew matrices,
say B_2, ..., B_7. For i = 2, ..., 7, deÞne A_i to be
(B_i 0)
( 0 0).
Clearly A_1 is not in their span, so these seven are linearly
independent.
But for any R the matrices R(A_2), ..., (RA_7) will have their
last 4 columns equal to 0. In particular, the Þrst rows of
these
six matrices will have to be linearly dependent. And any
dependence
relation on them will remain true on the Þrst rows of the
products R(A_2)R¹, ..., R(A_7)R¹.
William C. Waterhouse
PEnn State
===
Subject: Converting table output and plotting
I have the following equation:

Table[{v, NSolve[{i == v/r2 + i1, i1==i1(r1+r2)/r2 +
a/r2*Log[i1/io +
1] - v/r2} /. {r2 -> 200, r1 -> .025, io -> 10^-9, a -> .05},
{i,
i1}]},{v, 1, 1.2, .1}]//ColumnForm


OUTPUT IS:

{1, {{i->.402, i1->.397}}}
{1.1{{i->1.611,i1->1.60}}}
{1.2,{{i->3.86, i1->3.85}}}


The Þrst terms 1, 1.1,1.2 are Voltage (v) and the second term
is
Current (i), the third term is Current1 (i1).

My objective is to vary my equation above and then create the
output
in such a form that I can directly plot v vs. i, that is 1st
term
vs. 2nd term. Unfortunately, I don¹t know how to do this
since the
outputs in the table aren¹t as I like.

Note: My original approach was to write the transcendental
equation in
a single form whereby, NSolve[ i == function (v, i) ]
Unfortunately,
Mathematica seems to choke on this one. By calculating an
additional
(uninteresting) value (namely, v1), it gives me the output
I¹m looking
for. Again, except that I don¹t know how to convert this into
a set of
numbers that I can plot.

Hopefully this makes sense. Any help would be greatly
appreciated.
Vince
===
Subject: Re: Converting table output and plotting
> I have the following equation:
>
> Table[{v, NSolve[{i == v/r2 + i1, i1==i1(r1+r2)/r2 +
a/r2*Log[i1/io +
> 1] - v/r2} /. {r2 -> 200, r1 -> .025, io -> 10^-9, a ->
.05}, {i,
> i1}]},{v, 1, 1.2, .1}]//ColumnForm
>
>
> OUTPUT IS:
>
> {1, {{i->.402, i1->.397}}}
> {1.1{{i->1.611,i1->1.60}}}
> {1.2,{{i->3.86, i1->3.85}}}
>
>
> The Þrst terms 1, 1.1,1.2 are Voltage (v) and the second
term is
> Current (i), the third term is Current1 (i1).
>
> My objective is to vary my equation above and then create
the output
> in such a form that I can directly plot v vs. i, that is
1st term
> vs. 2nd term. Unfortunately, I don¹t know how to do this
since the
> outputs in the table aren¹t as I like.
>
> Note: My original approach was to write the transcendental
equation in
> a single form whereby, NSolve[ i == function (v, i) ]
Unfortunately,
> Mathematica seems to choke on this one. By calculating an
additional
> (uninteresting) value (namely, v1), it gives me the output
I¹m looking
> for. Again, except that I don¹t know how to convert this
into a set of
> numbers that I can plot.
>
> Hopefully this makes sense. Any help would be greatly
appreciated.
> Vince
What you want is to invoke a replacement rule in order to
convert
the sets of rules to values. This is addressed in several
notes to
comp.soft-sys.math.mathematica as well as in the note below.
http://support.wolfram.com/mathematica/kernel/features/
usingrulesolutions.ht
ml
As for the solver method used above, I would instead recommend
FindRoot. For one, you really do not have control over how
Solve (and
NSolve) will form polynomials from your input, use inverse
functions, react to branch cuts, etc. For another, FindRoot
will use
methods better suited to Þnding one root fast, rather than
methods
appropriate to Þnding all solutions to a (possibly high
degree)
polynomial system.
One way to code the whole thing is as below. Note that I use
start
values for FindRoot that happen to work below, but in general
one
might need to give that aspect more thought.
current[v_, r1_, r2_, io_, a_] := Module[{i,i1},
i /. FindRoot[{i == v/r2 + i1,
i1 == i1*(r1+r2)/r2 + a/r2*Log[i1/io + 1] - v/r2}, {i,1},
{i1,1}]
]
For example:
In[40]:= current[1, .025, 200, 10^(-9), .05]
Out[40]= 0.402681
In[41]:= current[1.76,.025,200,10^(-9),.05]
Out[41]= 22.7168
If you want to Þx the parameter values and only do a modest
number of
evaluations to create an approximation function over some
range you
could do as follows.
currentfunc =
Interpolation[Table[{v,current[v,.025,200,10^(-9),.05]},
{v,1,2,1/10}], InterpolationOrder->5];
We¹ll test at one point (not a node in the interpolation)
that we get
good agreement between these.
In[52]:= InputForm[{current[1.76,.025,200,10^(-9),.05],
currentfunc[1.76]}]
Out[52]//InputForm= {22.716831023032384, 22.7168419297376}
It¹s not quite so good near the beginning.
In[53]:= InputForm[{current[1.04,.025,200,10^(-9),.05],
currentfunc[1.04]}]
Out[53]//InputForm= {0.7494431878242906, 0.7322606923864048}
But still the behavior is reasonable, and if need be you
could home in
with more interpolation points in that region.
To plot you could do
Plot[current[v,.025,200,10^(-9),.05], {v,1,2}]
or
Plot[currentfunc[v], {v,1,2}]
Daniel Lichtblau
Wolfram Research
===
Subject: Re: roots of polynomials
bassam@ahu.edu.jo.or.karzeddin@yahoo.co.uk (bassam mizied
karzeddin)
>Þnd out one real root for the following uneven degree
equation:
>f(x)=x^n+ax^m+b=0
>n is uneven positive integer
>m is positive integer less than n
>f(x) is a rational integeral function of x
>(a&b) are real rational cooefÞcients
> REPLY:
> DEAR ORGANIZER
> I found one real root for the case (a=b=1)& m is uneven
positive
> integer,here is the solution:
> x=-{1-1/n+(2m-n+1)/(2!n^2)-(3m-2n+1)(3m-n+1)/(3!n^3)
> +(4m-3n+1)(4m-2n+1)(4m-n+1)/(4!n^4)
> -(5m-4n+1)(5m-3n+1)(5m-2n+1)(5m-n+1)/(5!n^5)+...}
> In fact, the proof of this (above mentioned equation )is
too lengthy
> and it was part of my interest for so many years.
> The complete solutions for all cases are also obtainable.
> I dont know if i could convey this properly.
See http://mathworld.wolfram.com/QuinticEquation.html. Here
series expansion

is used to obtain closed form solutions for the Bring quintic
form,
x^5 + a x + b
in terms of hypergeometric functions in one variable. This
technique
yields solutions in closed form for any polynomial equation
which can be
written in the form x^n+ax^m+b. The method dates back to the
1860s ...
Paul

Paul Abbott Phone: +61 8 9380 2734
School of Physics, M013 Fax: +61 8 9380 1014
The University of Western Australia (CRICOS Provider No
00126G)

35 Stirling Highway
Crawley WA 6009 mailto:paul@physics.uwa.edu.au
AUSTRALIA http://physics.uwa.edu.au/~paul
===
Subject: Re: power of triangle
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i5TC3di25265;
>I would like to know the answer for the following question:
>Given a triangle of three diffrent sides,were :
>L denotes the large side
>M denotes the middle side
>S denotes the small side
>There is a single power call it Z such that
>L^z=M^z+S^z
>Are there,any expressions(in radicals or series form) for Z ?
>Do that satisfy (z=2)when the angle between(s&m)is 90 degree
or :
>(z=1)when the angle between (s&m)is 180 degree
dear news groub
dear tarawna
do you mean that every certain angle has a uniqe power valu?
is that related to fermate last theorm?
are the sides of triangle of complex ,real, rational or
integeral
length?
is it neccesary that( S,M,L) form a triangle?
why sides of triangle are not equal?
are there any related topics?
i have tried out some numerical examples,and fond that always
there is
a single power value that seems to be irrational!& grater
than one.
thank you.
===
Subject: Re: Maxima on Zaurus, CAS on palm
> One of the nice parts about Macsyma is the modularity.
> On the PDP-10 where it was developed in MacLisp,
> there was a decent sized kernel that was loaded.
> Additional packages, such as for SOLVE, INTEGRATE,
> etc. would load up as required. Can you implement
> such a method for the Palm?
Slightly off-topic... GCL (Gnu Common Lisp) which underlies
Maxima (and Axiom) runs on the Sharp Zaurus, a palm-like
handheld. If you want a portable Maxima the Zaurus will
Þt the bill. If you want a CAS on your palm try Yacas.
===
Subject: Is it safe to buy mathematica?
Is it safe to buy mathematica from somebody on the internet? I
understand that it is idividually licensed. Could you get
into trouble,
ie, not be able to use it, even if you bought an unopened box
from
somebody?
BTW does mathematica for various OS¹s come on different CD¹s
or
they put eevrything on one cd and your computer picks out the
right
version?
===
Subject: Re: Is it safe to buy mathematica?
> Is it safe to buy mathematica from somebody on the
internet? I
> understand that it is idividually licensed. Could you get
into trouble,
> ie, not be able to use it, even if you bought an unopened
box from
> somebody?
> BTW does mathematica for various OS¹s come on different
CD¹s or
> they put eevrything on one cd and your computer picks out
the right
> version?
My box had sets of separate CDs for the various distributions.
CD1-Mac OS, Mac OS X
CD2-Linux (PC, Alpha, PowerPC)
CD3-Windows
The CDs aren¹t numbered as such, and I don¹t think you can
run them
concurrently. You need to call up Wolfram Research and ask,
especially
since
the version of the license in the box you want to buy may not
apply to you
in which case I don¹t think they would give you the activator
code. You
need
to call them Þrst, before you plop any money down.
Greysky
===
Subject: Re: Is it safe to buy mathematica?
> My box had sets of separate CDs for the various
distributions.
> CD1-Mac OS, Mac OS X
> CD2-Linux (PC, Alpha, PowerPC)
> CD3-Windows
> The CDs aren¹t numbered as such, and I don¹t think you can
run them
> concurrently. You need to call up Wolfram Research and ask,
especially
since
> the version of the license in the box you want to buy may
not apply to
you
> in which case I don¹t think they would give you the
activator code. You
need
> to call them Þrst, before you plop any money down.
===
Subject: Re: Is it safe to buy mathematica?
> Is it safe to buy mathematica from somebody on the
internet? I
> understand that it is i[n]dividually licensed. Could you
get into
trouble,
> ie, not be able to use it, even if you bought an unopened
box from
> somebody?
> BTW does mathematica for various OS¹s come on different
CD¹s or
> they put eevrything on one cd and your computer picks out
the right
> version?
No Brainer--Check with Mathermatica about transfer...
===
Subject: Re: Is it safe to buy mathematica?
> No Brainer--Check with Mathermatica about transfer...
The ad says it is a sealed box, so never been registered.
===
Subject: Re: Is it safe to buy mathematica?
> No Brainer--Check with Mathematica about transfer...
> The ad says it is a sealed box, so never been registered.
Looks like a go...
===
Subject: Re: software for MDS cluster analysis
This is certainly jumping right into the deep end of the pool!
Especially since this is your Þrst time to use a package, and
assuming
that your time is the most expensive cost component, I
suggest you take
a look at SPSS. It has both MDS including INDSCAL and cluster
analysis. If you are at a college or university the chances
are they
already have it available.
You can also get a 30 day trial from SPSS.
For quality assurance purposes you are usually better off
entering the
raw data verifying it, then letting the software do the
transformations.
Are you sure you just want to scale the common (pooled,
aggregated)
matrix and ignore all individual differences.
Art
Art@DrKendall.org
Social Research Consultants
University Park, MD USA
(301) 864-5570
> participants. I have created the co-occurrence matrix and
the subject
> distance matrix. Now, I need to feed these matrixes into a
> statistical software to run MDS cluster analysis. I want to
see the
> clusters that emerged from the data, hopefully pictorially
and in
> numbers. I do not have experience using any stats software
(so I am
> learning about all these as I go) and would appreciate your
inputs on
> which software is the best (intuitive and easy to learn and
use) for
> my analysis. Also, is there a free (or at a reasonable
price)
> software package for this type of anlaysis.
>
Œ`¹`¹`¹`¹`¹`¹`¹`¹`¹`¹`¹`¹`¹¹`¹`¹`¹`¹`¹`¹`¹`¹`¹`¹`¹`¹`¹`¹`¹`¹`
¹`¹`¹`¹`¹`¹`
> sci.psychology.research is a moderated newsgroup.
> here bimonthly or the charter on the web at
http://psychcentral.com/spr/
> Submissions are acknowledged automatically.
Œ`¹`¹`¹`¹`¹`¹`¹`¹`¹`¹`¹`¹`¹¹`¹`¹`¹`¹`¹`¹`¹`¹`¹`¹`¹`¹`¹`¹`¹`¹`
¹`¹`¹`¹`¹`¹`
sci.psychology.research is a moderated newsgroup.
here bimonthly or the charter on the web at
http://psychcentral.com/spr/
Submissions are acknowledged automatically.
===
Subject: Re: 3d arc generator FORTRAN
>Message-id: <vB9Ec.73027$Ax1.486780@news20.bellglobal.com>
>>Message-id: <hrm7nj8a1nh2@legacy>
>>
>>Hey,
>>Given the radius, beginning and end coordinates in 3d of a
circular
>>arc, I would like to generate a code giving coordinates of
a speciÞed
>>number of intermediates points on the arc.
>>I¹m using FORTRAN.
>>Can anyone help?
>>Caroline
>> Sounds a lot like a sphere to me.
>> Dan ;-)
>What a Dim Dumb Dan from a desperate man.
>--
>You¹re Welcome,
>Gerry T.
Go away little Troll.
===
Subject: Integrating Bessel fcns, kelvin fcns, hankel
functions
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i5SBiWe30054;
Does anyone know how to integrate Bessel Functions with a
complex
argument? (or a kelvin or hankel function for that matter)
Namely
Integrate J_2(r Exp[(3 I Pi)/4]) dr from r=0 to r=InÞnity
Any help would be greatly appreciated
Matt Swingle
mss4@duke.edu
===
Subject: Re: Integrating Bessel fcns, kelvin fcns, hankel
functions
charset=iso-8859-1
> Does anyone know how to integrate Bessel Functions with a
complex
> argument? (or a kelvin or hankel function for that matter)
> Namely
> Integrate J_2(r Exp[(3 I Pi)/4]) dr from r=0 to r=InÞnity
> Any help would be greatly appreciated
> Matt Swingle
> mss4@duke.edu
Here¹s one obvious place to look: Abramowitz and Stegun,
Handbook of
Mathematical Functions. It may be rather long in the tooth -
full of
tables, and so on - but has several chapters on Bessel
functions. I don¹t
have my copy immediately to hand, but I think it may be able
to help you.
I¹ve just checked: I have a pointer to an online copy, at
http://jove.prohosting.com/~skripty/. I hope that this is of
use to you.
John
johnDOTmorrisonATtescoDOTnet

Ein Buch is wie ein Spiegel: wenn ein Affe hineinsieht, so
kann kein
Apostel herausgucken. - Georg Christoph Lichtenberg (1742 -
1799)
---
Outgoing mail is certiÞed Virus Free.
Checked by AVG anti-virus system (http://www.grisoft.com).
===
Subject: Re: Integrating Bessel fcns, kelvin fcns, hankel
functions
> Does anyone know how to integrate Bessel Functions with a
complex
> argument? (or a kelvin or hankel function for that matter)
> Namely
> Integrate J_2(r Exp[(3 I Pi)/4]) dr from r=0 to r=InÞnity
> Any help would be greatly appreciated
> Matt Swingle
> mss4@duke.edu
This has to be in G. N. Watson¹s *Treatise on Bessel
Functions*.
Phil Hobbs
===
Subject: Kronecker product times vector
I¹m attempting to calculate the product, A x, of a matrix, A,
and a
vector, x. A is given by its Kronecker factors, i.e.,
A = otimes limits_i A_i.
Take as example
A = I_p otimes A_1 otimes I_q otimes A_2 otimes I_r otimes A_3
otimes I_s,
where I_p is the identity matrix of dimension p, and the
A_i¹s are
squared matrices.
For obvious reasons I attempted to form the product without
explicitly
calculating A. In fact, using stride permutations I have
succeeded for
small problems (See for example
www.cs.duke.edu/~nikos/cps258/lectures/KP-FFT.ps). However,
as the
number of factors of A increases the coding gets increasingly
messy
and tedious. Hence, my questions: Are there any established
libraries
for calculating the matrix vector product of the type given
above?
Secondly, searching the net, I came across several
publications were
automatic code generation systems have been applied to
products of the
above type, especially in the context of DSP and FFT.
Unfortunately, I
could so far not get my hands on any of these? Any
suggestions?
Alex
===
Subject: Re: Is this problem solvable?
>I have a problem I am not sure is solvable. There is an
unknown
>formula that calculates a value based on 10 parameters. Of
these 10
>parameters, 1 is an integer and the other 9 are fractional.
For
>example:
>a = 18, b = 6.5, c = 2.7, d = 3.2, e = 6.4, f = 9.2, g =
1.1, h = 3.8,
>i = 5.0, j = 3.1
>f(a,b,c,d,e,f,g,h,i,j) = 3010
>I have about 20,000 data points of a,b,c,d,e,f,g,h,i = x.
There¹s a
>catch, though. The only data I am allowed to see is integer
values for
>each variable. For example:
>18 6 2 3 6 9 1 3 5 3 = 3010
>I believe the formula used to calculate x is exponential
because
>higher variables tend to produce signiÞcantly higher values.
Also, I
>know that b and c are weighted more than the other
variables, and a
>and j are weighted very little, if at all. Also, most of my
20,000
>data points have b = 6 and f = 7.
>What I would like to Þgure out is a method to determine the
>fractional part of each variable. Really, if I could only
Þnd the
>fractional value of f, that would be Þne too. So given the
input:
>18 6 2 3 6 9 1 3 5 3 = 3010
>I would like to be able to Þgure out that f = 9.2. I realize
there
>probably isn¹t a way to Þnd the fractional values with 100%
>conÞdence, but if I knew with 95% conÞdence that 9.1 <= f <=
9.3,
>that would be sufÞcient.
that means that not only you do not know what f(...) is
about, you even
don¹t
know the inputs exactly???
>Is this problem solvable? If so, what methods should I use?
I am not
>looking for someone to do this work for me, but rather to
guide me in
>Þnding an answer. I haven¹t taken many advanced math
classes, but I
>am willing to do research. I just don¹t know what method(s)
to
>research. Feel free to email any questions to mkoss917 {at}
yahoo
>(dot) c-o-m.
>Michael.
no this is unsolvable.
you have no idea how f works besides some growth properties.
if everything is in the rationals, then you may assume that
f is some rational function of the 10 variables. then you
might
look at the kind of dependency of f with only one input
position varying.
plot it, say in halþogarithmic scale in order to get an idea
which
(integer) exponent would be appropriate for it (may be a
negative one
if one rational decay can be observed).
furthermore you might check whether it is completely
separable or whether
intercation of the variables is to be assumed.
then you could try a mixed
rational / polynomial model of f, say
F(a,b,c,d,e,f,g,h,i)= z(1)*a^3+z(2)*b^{-2}+... + z(20)*a*b+
....
and Þt the coefÞcients z(1),z(2),... using least squares and
compute intervals of conÞdence for those parameters z(i).
but this is for this speciÞc model. whether this model is
correct or not
is
not part of this analysis. and how large the errors in the
inputs have
been, how
should one analyse this. say you have a two parameter case
f(a,b)=x
and have datasets (a(i),b(i),x(i)) with errors in a,b,x.
now you may be convinced for example by taking datasets
(a(i),a(i),x(i)), (a(i),0,x(i)), (0,a(i),x(i)),
(a(i),-a(i),x(i))
that
f(a,b)=z(1)*a*b.
then getting z(1) in the least squares sense is easy, and if
a(i),b(i),x(i)

are all rational, then also z(1) will be rational. but
if you have for example a dataset
10,10,300
then
z(1)=3
would Þt perfectly this point but
also
3*(100/11)*(110/10)
would, hence you can assume your true a,b, had been not 10,10
but 100/11 and 110/10 ????
hth
peter


===
Subject: Re: Berenger¹s PML in FDTD
> In my own research I found PML to be largely useless in the
time
> domain because of corner reþections. The application was
> elastomechanics (soil-structure interaction under seismic
load). The
> old-fashioned Lysmer silent boundary did better.
Prof. Felippa,
At the risk of seeming to toot my own horn, let me mention
that we have in
fact developed PML models for elastodynamics and implemented
them using
Þnite
elements, in both the frequency and time domains, and used
them to
successfully solve classical soil-structure interaction
problems that
involved
corner regions. Our results for these problems are noticeably
more accurate
than the Lysmer boundary model but obtained at similar cost.
The modelling
of
corner regions is trivial in PML - the choice of attenuation
functions in
the
corner blocks follows naturally from the theory.
The references are:
U. Basu and A. K. Chopra. Perfectly matched layers for
time-harmonic
elastodynamics of unbounded domains: theory and Þnite-element
implementation.
Computer Methods in Applied Mechanics and Engineering,
192(11-12):1337-1375,
U. Basu and A. K. Chopra. Perfectly matched layers for
transient
elastodynamics of unbounded domains. International Journal
for Numerical
The IJNME paper has a slight error for which we have
submitted a
correction. (The error does not affect the results)
I would be happy to send to PDF Þles for the papers and the
correction.
Recently, we have successfully used these PML models towards
dam-water-foundation rock analysis under earthquake
excitation - I would be
happy to send the report once it is Þnalised.
Ushnish

Ushnish Basu ubasu@ce.berkeley.edu
+1 510 644-1906 http://www.ce.berkeley.edu/~ubasu
===
Subject: Re: Berenger¹s PML in FDTD
> Hiya,
>
> I¹m in the process of implementing my own FDTD code in 3D
in c
language.
> I¹ve previously validated my code for the TEz, TMz, TEy and
TEx by
> computing
> the relative error in each instance. In 3D I¹ve computed a
reference
> solution over 230 timesteps which I then compare to my test
data.
> Unfortunately after about 160 timesteps reþections can be
seen. My
> excitation is always placed @ 10cells from the air/PML
interface. The
> reþections seem to occur @ the same time step for each
normal
interface.
> Can I conclude that I¹m having problems with my corners? If
so, how can
I
> go
> about testing them?
>
>
> In my own research I found PML to be largely useless in the
time
> domain because of corner reþections. The application was
> elastomechanics (soil-structure interaction under seismic
load). The
> old-fashioned Lysmer silent boundary did better.
I¹ve never heard of Lysmer silent boundary¹s being used with
FDTD. The
most performant FDTD ABC¹s appear to either Berrenger¹s PML or
Gedney¹s UPML.
===
Subject: Re: Berenger¹s PML in FDTD
> Hiya,
>
> I¹m in the process of implementing my own FDTD code in 3D
in c
language.
> I¹ve previously validated my code for the TEz, TMz, TEy and
TEx by
> computing
> the relative error in each instance. In 3D I¹ve computed a
reference
> solution over 230 timesteps which I then compare to my test
data.
> Unfortunately after about 160 timesteps reþections can be
seen. My
> excitation is always placed @ 10cells from the air/PML
interface.
The
> reþections seem to occur @ the same time step for each
normal
interface.
> Can I conclude that I¹m having problems with my corners? If
so, how
can I
> go
> about testing them?
>
>
>
> In my own research I found PML to be largely useless in the
time
> domain because of corner reþections. The application was
> elastomechanics (soil-structure interaction under seismic
load). The
> old-fashioned Lysmer silent boundary did better.
> I¹ve never heard of Lysmer silent boundary¹s being used
with FDTD. The
> most performant FDTD ABC¹s appear to either Berrenger¹s PML
or
> Gedney¹s UPML.
You are correct. My application was 3D elastodynamics, with
FEM in
space and FD in time.
===
Subject: Re: Complex arithmetic
Jacques <jacqueslen@sympatico.ca> a .8ecrit dans le message de
> I want to do arithmetic in rings such as Q]e^(Pi*I/n)], for
some n,
> exactly. For example, I would like to do E(23) + E(45) * 4
* E(10)
> in GAP (where E(n) = e^(Pi*I*n)). However, I have many, many
> calculations to do so I want to do them extra quickly. the
ring is
> known before the calculations are started(in fact, only 26
elements
> are needed to determine all the operations performed). So I
want to do
> this extra quickly. How would I go about doing this? I¹ve
thought
> about libPari, but it seems that pari is used for þoating
point
> calculations. If anyone would know how to do it in libpari
(or pari)
> --Jacques
The lcc-win32 compiler features 100 digits precision with the
qþoat
data type.
#include <qþoat.h>
#include <stdio.h>
int main(void)
{
qþoat q = -1.570796Q;
qþoat result = tan(q);
printf(tan is: %70.40qfn,result);
}
tan
-3060023.3061935646797771659336362687358523
If you can write your calculations in the C language it will
work very
fast.
===
Subject: Re: Complex arithmetic
I would like to do it symbolically, ie be able to retrieve
the answer
as a Þnite sum of the form sum{i}( a_i * e^(Pi*I*i))
> Jacques <jacqueslen@sympatico.ca> a .8ecrit dans le message
de
> I want to do arithmetic in rings such as Q]e^(Pi*I/n)], for
some n,
> exactly. For example, I would like to do E(23) + E(45) * 4
* E(10)
> in GAP (where E(n) = e^(Pi*I*n)). However, I have many, many
> calculations to do so I want to do them extra quickly. the
ring is
> known before the calculations are started(in fact, only 26
elements
> are needed to determine all the operations performed). So I
want to do
> this extra quickly. How would I go about doing this? I¹ve
thought
> about libPari, but it seems that pari is used for þoating
point
> calculations. If anyone would know how to do it in libpari
(or pari)
> --Jacques
> The lcc-win32 compiler features 100 digits precision with
the qþoat
> data type.
> #include <qþoat.h#include <stdio.hint main(void)
> qþoat q = -1.570796Q;
> qþoat result = tan(q);
> printf(tan is: %70.40qfn,result);
> tan
> -3060023.3061935646797771659336362687358523
> If you can write your calculations in the C language it
will work very
> fast.
===
Subject: representing kinematics
Hi!
I have a problem in the representation of kinematic joint
with 6 DoF.
The mathematical representation of single rotations and
translations
is no problem using homogeneous coordinates. Also the
calculation of a
chain of joints is no problem. However I do not know how to
predict
the Þnal point, if each angle in rotation and each
postponement in
translation is equipped with a probability distribution. How
can i
predict the distribution of the Þnal point. Does anyone know
an
analytical solution?
Simon
===
Subject: Re: Root Finder and Example
> DERIVING ALL ROOTS TO ANY POLYNOMIAL (with example)
I will eventually read this toughly. But for now is this a
symbolic
technique or a numeric technique?
===
Subject: Maple 6
Suppose that m, e and N are large integers. Why does
(m&^e)mod(N)
execute so much faster than (m^e)mod(N)?
John W. Adams
Lehigh University
===
Subject: Re: Maple 6
>Suppose that m, e and N are large integers. Why does
(m&^e)mod(N)
>execute so much faster than (m^e)mod(N)?
Because (m^e) mod N actually computes m^e (which can be huge)
Þrst
and then reduces mod N, while (m &^ e) mod N works with
integers mod N
throughout. See the help page ?mod.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Re: Maple 6
John Adams
Lehigh University
===
Subject: Re: Maple 6
> Suppose that m, e and N are large integers. Why does
(m&^e)mod(N)
> execute so much faster than (m^e)mod(N)?
Because Maple doesn¹t have to compute the power; &^ is an
inert operator.
Joe
===
Subject: Re: Integral of Bessel function multiplied by exp
> Hello everyone,
> does anybody know how to compute analytically (at least,
> approximately)
> such integral
> if the upper limit of your integral is infty
> then you are lucky
> int_0^{infty} exp(-a*x) * J_0(b*sqrt(x)) * dx =
> frac{1}{a} * exp( - frac{b^2}{4*a})
> (see eqn 24.94 in
> M.R.Spiegel, Handbook of Mathematics,
> SCHAUM¹s Outline Series, McGraw-Hill Book Company)
> with eqn 24.32 (you are near to the solution)
> I_n(x) = i^{-n} * J_n(i*x)
> you get
> int_0^y exp(-a*x_1) * J_0(i*sqrt(x_2)*sqrt(x_1)) * dx_1
> = int_0^y exp(-a*x_1) * J_0(c*sqrt(x_1)) * dx_1
> with
> c = i*sqrt(x_2)
> thus your integral gets
> int_0^{infty} exp(-a*x_1)*I_0(sqrt(c*x_1))dx_1
> = frac{1}{a} * exp( - frac{c^2}{4*a} )
> = frac{1}{a} * exp( - frac{i^2*sqrt(x_2)^2}{4*a} )
> = frac{1}{a} * exp( - frac{-1*x_2}{4*a} )
> = frac{1}{a} * exp( frac{x_2}{4*a} )
> and so on for the next integrals
> int_0^y exp(-a*x)*I_0(sqrt(b*x))dx,
> where I_0 is the modiÞed 0-th order Bessel function of the
Þrst
> kind?
> In fact, I need to compute
> int_0^y ...int_0^y
>
exp(-a*(x_1+...+x_n))*I_0(sqrt(x_1*x_2))*...*I_0(sqrt(x_{n-1}
*x_n))dx_n...dx_
1,
> where n is **very** big number.
> P. Trifonov
> here are some additional books:
> F. Oberhettinger,
> Tables of Bessel Transforms,
> Springer Publications
> I.S. Gradshteyn, I.M. Ryzhik
> Tables of Integrals, Series and Products
> Academic Press
> I hope this helps & happy computing
> Herbert
He might also want to consult Watson, A Treatise on the
Theory of
Bessel Functions.

Julian V. Noble
Professor Emeritus of Physics
jvn@lessspamformother.virginia.edu
^^^^^^^^^^^^^^^^^^
http://galileo.phys.virginia.edu/~jvn/
For there was never yet philosopher that could endure the
toothache
patiently. -- Wm. Shakespeare, Much Ado about Nothing. Act v.
Sc.
1.
===
Subject: Re: Reverse Linear Regression
> I was wondering if there is an algorithm that can be used
> to create a set of n linear (X,Y) data points with a deÞned
slope m,
> y intercept b, STD DEV m and STD DEV b.
> I could use it in making up lab questions
> i.e If a table of Voltage and Current data is given for
> an unknown resistor, plot the values Voltage vs Current and
Þnd the
> value of resistor and the error in the resistance. Also
determine how
> linear the data is. In some cases I have the more advanced
students
> use their calculators to analyze the data or sometimes I
have them use
> graph paper and eyeball it for the line of best Þt.
> I have done this by hand using my HP48GX calculator, but
> it would be nice to just type in the values for the slope,
etc specify
> the number of points wanted and have it churn out the
original
> data.
> I would like to do his in MAPLE if possible. I use it quite
> often, and my programming skills in other languages are
either
> extremely rusty or nonexistent.
> It would also be nice to generate non-linear data for log or
> exponential functions, since these functions also Þt
certain physical
> phenomena.
> Harold A. Climer
> Dept. Of Physics,Geology, and Astronomy
> U.T. Chattanooga
> 318 Grote Hall
> 615 McCallie Ave
> Chattanooga TN 37403
Any decent spreadsheet program (Excel, Quattro pro) will do
this for you.
Generate linear data and add (normally distributed) noise.
You might
want to add the noise to both independent and dependent
variables (of
course they will have different standard deviations and
amplitudes).
You can even add systematic error by choosing your noise to
have a mean
different from 0.

Julian V. Noble
Professor Emeritus of Physics
jvn@lessspamformother.virginia.edu
^^^^^^^^^^^^^^^^^^
http://galileo.phys.virginia.edu/~jvn/
For there was never yet philosopher that could endure the
toothache
patiently. -- Wm. Shakespeare, Much Ado about Nothing. Act v.
Sc.
1.
===
Subject: Re: Reverse Linear Regression
|> I was wondering if there is an algorithm that can be used
|>to create a set of n linear (X,Y) data points with a deÞned
slope m,
|>y intercept b, STD DEV m and STD DEV b.
By STD DEV m and STD DEV b, I assume you mean the standard
error of
estimate for m and b. If I haven¹t misread my statistics
references,
for STD DEV m = A and STD DEV b = B, with sum_i x_i = C, what
you need
are
sum_i y_i = C m + b n
sum_i x_i^2 = (B/A)^2 + C^2/n
sum_i x_i y_i = m C^2/n + b C + m (B/A)^2
sum_i y_i^2 = (n-2) B^2 + n b^2 + 2 C m b + ((B/A)^2 + C^2/n)
m^2
Now to produce n data points with given values of
sum_i x_i = C, sum_i y_i = D, sum_i x_i^2 = E, sum_i x_i y_i
= F,
sum_i y_i^2 = G, you can start with 2n random numbers u_i,
v_i of
arbitrary distributions, and take
x_i = p + q u_i
y_i = r + s u_i + t v_i
for suitable values of p,q,r,s,t. If everything works (and I
think it
should as long as n,m,b,A,B,C satisfy some consistency
conditions),
you should be able to solve for p,q,r,s,t.
Thus you can do something like this:
> gendata:= proc(n, m, b, A, B, C)
local D,E,F,G,u,v,i,p,q,r,s,t,su,sv,suu,suv,svv,sol;
D:= C*m + b*n;
E:= (B/A)^2 + C^2/n;
F:= m*C^2/n + b*C + m*(B/A)^2;
G:= (n-2)*B^2 + n*b^2 + 2*C*m*b + ((B/A)^2 + C^2/n)*m^2;
u:= RandomTools:-Generate(list(þoat(method=uniform),n));
v:= RandomTools:-Generate(list(þoat(method=uniform),n));
su:= add(u[i],i=1..n);
sv:= add(v[i],i=1..n);
suu:= add(u[i]^2,i=1..n);
suv:= add(u[i]*v[i],i=1..n);
svv:= add(v[i]^2,i=1..n);
sol:= fsolve({
n*p + q*su = C,
n*r + s*su + t*sv = D,
n*p^2 + 2*p*q*su+q^2*suu = E,
n*p*r + (p*s+q*r)*su+p*t*sv+q*s*suu + q*t*suv = F,
n*r^2 + s^2*suu+t^2*svv + 2*r*s*su+2*r*t*sv+2*s*t*suv = G});
if sol = NULL then error Could not Þnd transformation Þ;
subs(sol, map(U -> p+q*U, u)),
subs(sol, zip((U,V) -> r+s*U+t*V,u,v));
end;
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Re: Reverse Linear Regression
> I was wondering if there is an algorithm that can be used
> to create a set of n linear (X,Y) data points with a deÞned
slope m,
> y intercept b, STD DEV m and STD DEV b.
> I could use it in making up lab questions
> i.e If a table of Voltage and Current data is given for
> an unknown resistor, plot the values Voltage vs Current and
Þnd the
> value of resistor and the error in the resistance. Also
determine how
> linear the data is. In some cases I have the more advanced
students
> use their calculators to analyze the data or sometimes I
have them use
> graph paper and eyeball it for the line of best Þt.
> I have done this by hand using my HP48GX calculator, but
> it would be nice to just type in the values for the slope,
etc specify
> the number of points wanted and have it churn out the
original
> data.
> I would like to do his in MAPLE if possible. I use it quite
> often, and my programming skills in other languages are
either
> extremely rusty or nonexistent.
> It would also be nice to generate non-linear data for log or
> exponential functions, since these functions also Þt
certain physical
> phenomena.
Erm, I may be missing something here, as I¹m a mathematical
ninny. To
get your data, all you need to do is generate some perfect
data for
the given slope and intercept, and add some noise to it.
I don¹t use Maple, but in R (www.r-project.org), a statistical
language, this
can be done very easily.
[running R on the command line - I don¹t use any guis with it]
x <- runif( 100, 20, 40 )
y <- x * 5.43 + 7.8
noise <- rnorm( 100, 0, 3 )
y <- y + noise
x
y
In plain english, the Þrst line creates 100 uniformly
distributed
numbers between 20 and 40. These are the x values.
The second line creates the y values for a slope of 5.43, and
an
intercept of -7.8.
The third line creates some random noise with a mean of 0,
and a
standard deviation of 3.
The fourth line adds the noise to the y values, so the points
will no
longer exactly match the lines.
The Þfth and sixth lines just print out the values of x and y,
suitable for
cutting and pasting into documents.
The above is just for linear data, with normally distributed
errors.
You could of course replace the line calculating the perfect
data with
some other function:
y <- x ^ 0.3 + log2( x )
For different types of errors, you could pass the normally
distributed
errors through some function, or use some other probability
distribution to generate them.
While you asked for a method for Maple, R is free and runs on
a number
of operating systems.
Ross-c
===
Subject: Cumulative Density Function for simple Multinomial
Distribution
I have a simple multinomial distribution with C equi-probable
categories. [The probability of each category is 1/C]
I¹d like to compute a cumulative density function for N
trials (N = C *
Q) for all the combinations (X1,X2,X3,...,Xc) such that
MAX(Xi) <= M.
Was wondering is there is a quick way to get this value.
Values of interest C{4,8,16,32}, M{6,12,18,24},
Q{M/3,M/2,...,M}
come up with a solution.
I am not even sure how to compute the simple probability for
the uniform
result X1=X2=X3=...=Xc=Q, using Stirling¹s formula I get
sqrt(C)/sqrt((2*PI*Q)^(C-1))
Jean-Pierre Panziera
===
Subject: Geometrical analogies for distribution functions?
I just realized this in bed, but the normal distribution
corresponds to the
cross-sectional height of a parallelogram.
e.g. The histogram of the sum of two dice approximates the
normal
distribution. The more sides the two dice have, the closer
the histogram
will look to a normal distribution. If the dice have
inÞnitely many sides
(a continuous distribution), the histogram will be a normal
distribution.
Discrete case: you can plot the outcome of two dice on a
sheet of graph
paper. e.g. two six-sided dice:
Let the x-axis be the value of the Þrst die, the y-axis be
the sum of the
two dice. At x=1, we would have a bar that starts at (1,2)
extending to
(1,7). At x=2, we would have a bar that starts at (2,3)
extending to
(2,8).
... At x=6, we would have a bar that starts at (6,7)
extending to (6,7).
In
the end, we will have bars stacked roughly in the shape of a
parallelogram,
and the histogram will correspond to the cross-sectional
width of the
parallelogram at a given value of y. As we increase the
number of sides on
the dice to inÞnity, the stack of bars will converge to a
parallelogram,
whose height as a function of x will look like the normal
distribution.
I have not seen such a geometrical analogy mentioned in any
stat textbook
I¹ve read. Are there other geometrical analogies to other
kinds of
distributions?
===
Subject: Reverse Linear Regression question
I was wondering if there is an algorithm that can be used
to create a set of n linear (X,Y) data points with a deÞned
slope m,
y intercept b, STD DEV m and STD DEV b.
I could use it in making up lab questions
i.e If a table of Voltage and Current data is given for
an unknown resistor, plot the values Voltage vs Current and
Þnd the
value of resistor and the error in the resistance. Also
determine how
linear the data is. In some cases I have the more advanced
students
use their calculators to analyze the data or sometimes I have
them use
graph paper and eyeball it for the line of best Þt.
I have done this by hand using my HP48GX calculator, but
it would be nice to just type in the values for the slope,
etc specify
the number of points wanted and have it churn out the original
data.
I would like to do his in MAPLE if possible. I use it quite
often, and my programming skills in other languages are either
extremely rusty or nonexistent.
It would also be nice to generate non-linear data for log or
exponential functions, since these functions also Þt certain
physical
phenomena.
Harold A. Climer
Dept. Of Physics,Geology, and Astronomy
U.T. Chattanooga
318 Grote Hall
615 McCallie Ave
Chattanooga TN 37403
===
Subject: b not in sp(v_1..v_n), v_n in sp(v_1...v_n-1,b)
X-Enigmail-Version: 0.84.2.0
X-Enigmail-Supports: pgp-inline, pgp-mime
Hi all.
x_y, x_{y} - i mean that x is written normally
and y is in subscript.
1) { v_1 .... v_n } is a subgroup of V
2) b is in V
3) b is not in Sp { v_1 ... v_n }
4) v_n is in Sp { v_1 ... v_{n-1}, b }
Prove that { v_1 ... v_n } are dependent
My answer (different from book) is
pretty simple:
Assume (the opposite) that { v_1 ... v_n }
are independent. => (based on #3) { v_1 ... v_n, b }
is also independent. That contradicts #4.
The assumption is therefore wrong.
Is my answer correct? (I mean ok to write
it on exam)
P.S.
English is not my native tongue and the question
is translated - please be forgiving ;)
Ilya
===
Subject: Re: b not in sp(v_1..v_n), v_n in sp(v_1...v_n-1,b)
ŒLooks right to me.
BTW, is this still linear algebra, or is this more general
algebra...?
If it isn¹t linear algebra, well, my word has little meaning,
as I¹m
taking that (er, general algebra...) class this coming fall.
Also, just out of curiousity--what other proof does the book
have? The
proof you have seems pretty straight forward... enough to be
the
textbook answer. Do they prove it by direct methods (as in
showing
that v_n is in Sp{v_1,...,v_n-1})?
> Hi all.
> x_y, x_{y} - i mean that x is written normally
> and y is in subscript.
> 1) { v_1 .... v_n } is a subgroup of V
> 2) b is in V
> 3) b is not in Sp { v_1 ... v_n }
> 4) v_n is in Sp { v_1 ... v_{n-1}, b }
> Prove that { v_1 ... v_n } are dependent
> My answer (different from book) is
> pretty simple:
> Assume (the opposite) that { v_1 ... v_n }
> are independent. => (based on #3) { v_1 ... v_n, b }
> is also independent. That contradicts #4.
> The assumption is therefore wrong.
> Is my answer correct? (I mean ok to write
> it on exam)
> P.S.
> English is not my native tongue and the question
> is translated - please be forgiving ;)
> Ilya
===
Subject: Re: b not in sp(v_1..v_n), v_n in sp(v_1...v_n-1,b)
X-Enigmail-Version: 0.84.2.0
X-Enigmail-Supports: pgp-inline, pgp-mime
> ŒLooks right to me.
> BTW, is this still linear algebra,
Yep
> or is this more general algebra...?
> If it isn¹t linear algebra, well, my word has little
meaning, as I¹m
> taking that (er, general algebra...) class this coming fall.
> Also, just out of curiousity--what other proof does the
book have? The
> proof you have seems pretty straight forward... enough to
be the
> textbook answer. Do they prove it by direct methods (as in
showing
> that v_n is in Sp{v_1,...,v_n-1})?
The answer in the book:
(2) -> there are scalars alfa_1,...alpha_{n-1}, beta
such that v_n = alpha_1*v_1+....alpha_{n-1}*v_{n-1} + beta*b
If beta not equals to 0:
b = (1/beta)*v_n - (alpha_1/beta)*v_1 - ..
.. - (alpha_{n-1}/beta)*v_{n-1}
and that means that b is in Sp({v_1 ... v_n}) which
contradicts (3)
therefore beta must be zero.
v_n=alpha_1*v_1+...+alpha_{n-1}*v_{n-1}
alpha_1*v_1+...+alpha_{n-1}*v_{n-1} - v_n = 0
v_n¹s (factor? multiplier? - don¹t know how it translates)
is non-zero therefore { v_1 ... v_{n-1} , v_n } is dependent
... wow ;)
>>1) { v_1 .... v_n } is a subgroup of V
>>2) b is in V
>>3) b is not in Sp { v_1 ... v_n }
>>4) v_n is in Sp { v_1 ... v_{n-1}, b }
===
Subject: Re: arithmetic Robak 0,(9)+{1+}0=1 Time Theory
charset=iso-8859-1
ksRobak <ed_robak@interia.pl>
> ksRobak <ed_robak@interia.pl>

> arithmetic Robak -- Time Theory by ksRobak (continue)
> ~~~~~~~~~~~~~~~~~~~~~~~~~
> 1/3 = 1/3 * 10/10 = (10/3) / 10 _|_ 10/3 = X
> 1/3 = X/10 = 0,X
> 1/3 = 0,X = 9/30 + 1/30 = 3/10 + (10/3)/10^2 = 3/10 +
X/10^2 = 0,3X
> 0,X = 0,3X
> 1/3 = 0,X = 0,3X = 0,33X = 0,3333...X = 0,(3)X = 0,(3)(10/3)
> 0,(3)(10/3) - 0,(3) = (10/3) / 10^Re1
> Edward Robak ed_robak@wp.pl |/ re:
1+1+1+1+1+1+1+1+...(->oo)=Re1 <== const
Re1 + a/b = Re1 + a/b
Re1 * a/b = Re1 * a/b
1/Re1 = +0 <== PUNKT
x/Re1 = {x +}0
{x +}0 * Re1 = x
+0 * ab = {ab +}0
+0/x = {1/x +}0
Edward Robak ed_robak@wp.pl |/ re:
===
Subject: Re: arithmetic Robak 0,(9)+{1+}0=1 Time Theory
charset=iso-8859-1
ksRobak <ed_robak@interia.pl1+1+1+1+1+1+1+1+...(->oo)=Re1 <==
const
> Re1 + a/b = Re1 + a/b
> Re1 * a/b = Re1 * a/b
> 1/Re1 = +0 <== PUNKT
> x/Re1 = {x +}0
> {x +}0 * Re1 = x
> +0 * ab = {ab +}0
> +0/x = {1/x +}0
> Edward Robak ed_robak@wp.pl |/ re:
geometric Robak -- Time Theory by ksRobak (continue)
~~~~~~~~~~~~~~~~~~~~~~~~~
A B B C
o------------------o o------------------o
1 2 3...->oo Re1 1 2 3...->oo Re1
PUNKT={AB+}0 PUNKT={BC+}0
AB + BC = AC
A B B C
o------------------o o------------------o
A B B C
o------------------o o------------------o
A B B C
o------------------o o------------------o
A B C
o------------------o------------------o
A C
o------------------------------------o
PUNKT={AB+}0 + {BC+}0 = {AC+}0
~~~~~~~~~~~~~~~
{AC+}0 * Re1 = AC
~~~~~~~~~~~~~~~
Edward Robak ed_robak@wp.pl |/ re:
--
AGEOMETRETOS MEDEIS EISITO
===
Subject: Answer to this riddle? Please email me the answer.
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i68HkNB27580;
Here is the puzzle:

You may recall that, on Citrus Island, three tribes - the
Whites,
the Oranges, and the Lemons - have different standards of
veracity.
(They are otherwise indistinguishable.) Whites always tell
the truth;
Oranges always lie; Lemons, when asked a series of questions,
tell
the truth and lie alternately. A Lemon¹s Þrst answer in a
series may
be either true or otherwise.

A visitor was somewhat confused recently when he was
introduced to
three natives named White, Orange, and Lemon, who are - not
necessarily respectively - a White, an Orange, and a Lemon.
There
was also a fourth native named Yellow. The visitor asked each
of the
Þrst three natives (a) what his own tribe was, and (b) what
was Mr.
Yellow¹s tribe.

To these questions, Mr. White replied: I¹m not a White. Mr.
Yellow is an Orange.

Mr. Orange replied: I¹m not Orange. Mr. Yellow is a Lemon.

Mr. Lemon replied: I¹m not Lemon. Mr. Yellow is a White.

To which tribe does Mr. Yellow actually belong?
===
Subject: Re: Answer to this riddle? Please email me the
answer.
days. My association with the Department is that of an
alumnus.
[Cc.ed to poster]
>Here is the puzzle:
> You may recall that, on Citrus Island, three tribes - the
Whites,
>the Oranges, and the Lemons - have different standards of
veracity.
>(They are otherwise indistinguishable.) Whites always tell
the truth;
>Oranges always lie; Lemons, when asked a series of
questions, tell
>the truth and lie alternately. A Lemon¹s Þrst answer in a
series may
>be either true or otherwise.
> A visitor was somewhat confused recently when he was
introduced to
>three natives named White, Orange, and Lemon, who are - not
>necessarily respectively - a White, an Orange, and a Lemon.
There
>was also a fourth native named Yellow. The visitor asked
each of the
>Þrst three natives (a) what his own tribe was, and (b) what
was Mr.
>Yellow¹s tribe.
> To these questions, Mr. White replied: I¹m not a White. Mr.
>Yellow is an Orange.
> Mr. Orange replied: I¹m not Orange. Mr. Yellow is a Lemon.
> Mr. Lemon replied: I¹m not Lemon. Mr. Yellow is a White.
>To which tribe does Mr. Yellow actually belong?
Consider the statement I¹m not White. This cannot be made by
a white
(since it would be false), nor can it be made by an Orange
(since it
would be true). So I¹m not White can only be said by a Lemon,
who
would be telling the true. So Mr. White is a lemon, and Mr.
Yellow is
not Orange.
So Mr. Lemon and Mr. Orange are a white and an orange, in some
order. Consider the statement I¹m not a Lemon. This cannot be
said
by an Orange (since it would be true), so Mr. Lemon must be a
white. Therefore, Mr. Yellow is a white.
Note that this means that Mr. Orange must be orange, and
indeed, both
his answers are false.

It¹s not denial. I¹m just very selective about
what I accept as reality.
--- Calvin (Calvin and Hobbes)
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Friday the 13th: Max/Min on any given year
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i68J43402432;
I came across a cute problem in a recent math competition. The
problem asked to Þnd the maximum and minimum number of times
Friday
the 13th could occur in any given year. After giving it some
serious
thought, I realized that this problem is neither simple nor
quick to
solve.
Here¹s my approach and answers (with spoil space). Any
comments on
the approach and accuracy of the answers would be greatly
appreciated!
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
Month # of Days Range of day #s The 13th falls on day #
----- --------- --------------- -----------------------
J 31 1-31 13
F (*) 28 32-59 45
M 31 60-90 72
A 30 91-120 103
M 31 121-151 133
J 30 152-181 165
J 31 182-212 194
A 31 213-243 225
S 30 244-273 256
O 31 274-304 286
N 30 305-334 317
D 31 335-365 347
* this analysis only for non-leap years
Now we want to calculate for each month the value of
mod(13th day # - 13,7) ,
which yields the following results:
J F M A M J J A S O N D
{0, 4, 3, 6, 1, 5, 6, 2, 5, 0, 3, 5}
This implies that if Jan 13 were a Sunday, we¹d get 3
occurrences of
Friday 13 (since modulus would be 5) and they would occur in
June,
September, and December. This is the maximum number of
occurrences.
Likewise, if Jan 13 were a Thursday, we¹d only get one
occurrence of
Friday 13 and this would be in May.
===
Subject: Re: Friday the 13th: Max/Min on any given year
> I came across a cute problem in a recent math competition.
The
> problem asked to Þnd the maximum and minimum number of
times Friday
> the 13th could occur in any given year. After giving it
some serious
> thought, I realized that this problem is neither simple nor
quick to
> solve.
I just went through the calculation... took me about 5-10
minutes
(with paper and pencil, of course :), probably. Considering
that I
have never seen a similar problem, this seems to be in a
difÞculty
level ranging from very easy to slightly non-trivial in a math
competition (assuming that it¹s high school competition).
> Month # of Days Range of day #s The 13th falls on day #
> ----- --------- --------------- -----------------------
> J 31 1-31 13
> F (*) 28 32-59 45
> M 31 60-90 72
> A 30 91-120 103
> M 31 121-151 133
> J 30 152-181 165
> J 31 182-212 194
> A 31 213-243 225
> S 30 244-273 256
> O 31 274-304 286
> N 30 305-334 317
> D 31 335-365 347
> * this analysis only for non-leap years
I did a similar analysis which accounts for leap years as
well--your
answer is still valid, as, in a leap year, the maximum
possible
occurance of Friday, 13th is still three times (if you have
Friday,
13th in January), and minimum is still once. So, I suppose we
can¹t
avoid bad luck in any case (perhaps, if we tried the Chinese
calendar, we might be able to avoid it :)
> Now we want to calculate for each month the value of
> mod(13th day # - 13,7) ,
Here -13 is unnecessary, as you will be Þnding the mode of
your
distribution anyway. (I don¹t know the term for
least-frequently
occuring value--is there one?)
> which yields the following results:
> J F M A M J J A S O N D
> {0, 4, 3, 6, 1, 5, 6, 2, 5, 0, 3, 5}
> This implies that if Jan 13 were a Sunday, we¹d get 3
occurrences of
> Friday 13 (since modulus would be 5) and they would occur
in June,
> September, and December. This is the maximum number of
occurrences.
> Likewise, if Jan 13 were a Thursday, we¹d only get one
occurrence of
> Friday 13 and this would be in May.
BTW, while your analysis is valid, taking mod 7 Þrst will
make your
calculations much easier--and quicker :). Let January begin
from
number 0 (for convenience), take mod 7 of all the months but
December
(er.. can you see why you need mod 7 of January but not
December?),
then you should have values 0, 2, and 3--and 1, for February
in a leap
year. Add up all of them, writing down the subtotal after each
addition, so you end up with 12 numbers in an increasing
order, and
then take the modulus of each of them again.
Er... wait. I just saw an error in your analysis. For the nth
day on
which the 13th of the month falls, your answers for February
and June
are wrong. Fortunately for your Þnal answer, while the error
on
February takes one occurance away from what used to be the
mode of
your distribution, the error for June adds one occurance to
previously-same-day-of-the-week-occuring-twice group
consisting of
September and December (whew--that was a long adjective). But
the
correct analysis should show that the most Friday, 13th occur
(for a
non-leap year) when the 13th of January is a Tuesday.
Best wishes,
Andrew
PS. I¹m just curious that the error on February and June
didn¹t carry
over to the other months--how did that happen? Typo?
===
Subject: Re: Friday the 13th: Max/Min on any given year
[...]
There are 2 errors in the table below, for February and June:
> Month # of Days Range of day #s The 13th falls on day #
> ----- --------- --------------- -----------------------
> J 31 1-31 13
> F (*) 28 32-59 45
^^
44
> M 31 60-90 72
> A 30 91-120 103
> M 31 121-151 133
> J 30 152-181 165
^^^
164
> J 31 182-212 194
> A 31 213-243 225
> S 30 244-273 256
> O 31 274-304 286
> N 30 305-334 317
> D 31 335-365 347
> * this analysis only for non-leap years
> Now we want to calculate for each month the value of
> mod(13th day # - 13,7) ,
> which yields the following results:
> J F M A M J J A S O N D
> {0, 4, 3, 6, 1, 5, 6, 2, 5, 0, 3, 5}
The correct table is:
J F M A M J J A S O N D
{0, 3, 3, 6, 1, 4, 6, 2, 5, 0, 3, 5}
and for leap years:
J F M A M J J A S O N D
{0, 3, 4, 0, 2, 5, 0, 3, 6, 1, 4, 6}
> This implies that if Jan 13 were a Sunday, we¹d get 3
occurrences of
> Friday 13 (since modulus would be 5) and they would occur
in June,
> September, and December. This is the maximum number of
occurrences.
> Likewise, if Jan 13 were a Thursday, we¹d only get one
occurrence of
> Friday 13 and this would be in May.
The maximum number is indeed 3, but it occurs in February,
March
and November of non-leap years, and January, April and July of
leap years. The minimum is indeed 1, and occurs in May, June
or
August of a non-leap year, or May, June or October of a leap
year.
Note that the 13th always falls on a different day of the week
in each month of May through November.

Jim Heckman
===
Subject: law of sines
a=65,c=62,h=63.5,C=78 degrees
what is A?
If there is no solution, why?
===
Subject: Re: law of sines
> a=65,c=62,h=63.5,C=78 degrees
> what is A?
> If there is no solution, why?
The law of sines for a triangle with sides a.b and c and
opposite angles
, respectively of A,B and C, may be expressed as
sin(A)/a = sin(B)/b = sin(C)/c.
For your problem, it appears that you may ignore b and B and
use only
sin(A)/a = sin(C)/c,
so that
A = arcsin(a*sin(C)/c).
===
Subject: Re: Math Tutoring Site
Right now, I¹m just starting out. I might increase it in the
future, but
for now, I¹m just trying to Þnd a few students to start out
with.
Jeff
> x-no-archive: yes
> Dude--- Take some business classes ! The biggest mistake
people make
> is not charging enough. Is your time only worth $5/hr?
>> Just wanted to post for anyone interested...
>> This site offers tutoring for math students online for
$5/hour
>> http://www.yourmathguide.com
===
Subject: I think I¹ve got it.
I¹m sorry it¹s taken me so long to post again, I¹ve been away
from my
computer for a while.
compared to some of the personal emails I¹ve been getting.
The following are links to my derivation:
http://img59.photobucket.com/albums/v179/chrax/Brian/
1stone.jpg
http://img59.photobucket.com/albums/v179/chrax/Brian/
2ndone.jpg
I would appreciate feedback, as I still don¹t want to look at
any
resources. My three formulas at the end have given me right
answers so far
in all the tests I¹ve put them through, so I think I¹m right.
The derivations lacks comments or justiÞcation, other than
that of
the math itself. If you notice any þaws in my reasoning or
have any
questions please feel free to post a reply or send an email.
I¹ll probably respond to an email more quickly.
-=-
I noticed that my initial question primarily ilicited annoyed
responses.
(I even got a rather rude email from someone named Mathed
Man.) After I
explained myself a little further I got primarily
constructive responses.
In the Linux world there are FAQs with regarding how to ask
proper
questions are there any FAQs on how to ask a proper math
question?
If not, does anyone have any suggestions?
===
Subject: Re: maple: derivatives wrt. derivatives
>>So for instance is it possible to get somthing like
>>f:=diff(x(t),t)^2;
>>diff(f,diff(x(t),t));
>>This should give
>>2*diff(x(t),t)
> For example,
> y(t):=diff(x(t),t):
>>f:=y^2:
>>diff(f,y)(t);
> /d
> 2 |-- x(t)|
> dt /
> Alec Mihailovs
===
Subject: Re: SYMBOLIC SOFTWARE FOR FINANCE?? GOOD IDEA??
> programs are divided into symbolic and numerical.
Some programs are good for both symbolic and numeric
computations.
Mathematica and Maple are two such programs.
> My question is, if a person is taking a double major in
biology and
> Þnance (real estate), would they be able to get by with just
> mathematica?
I¹d recommend also learning a spreadsheet program like
Miscrosoft Excel in
addition to Mathematica.
> Everything I have read tells me that a program like
Œmathworks matlab¹
> is better for real estate formulas and the like.
Where did you read that? It should be trivial to program any
real-estate
formula into Mathematica.
===
THE FIFTH INTERNATIONAL CONFERENCE ON KNOWLEDGE BASED
COMPUTER SYSTEMS
The next international conference on Knowledge Based Computer
Systems
to act as a forum for promoting interaction among researchers
in the
Þeld of ArtiÞcial Intelligence in India and abroad. The
International
Conference on Natural Language Processing(ICON) will be held
concurrently
at the same venue.
____________________________________________________________
Submission Deadlines
____________________________________________________________
Papers are invited on substantial, original and unpublished
research on
all aspects of ArtiÞcial Intelligence, including, but not
limited to the
following:
Case Based Reasoning
Cognitive Modelling
Data Mining
Expert Systems
Foundations of AI
Fuzzy Logic
Genetic Algorithms
Intelligent Agents
Intelligent Tutoring Systems
Intelligent Information Retrieval
Knowledge Acquisition
Knowledge Management
Knowledge Representation
Machine Learning
Machine Translation
Neural Networks
Natural Language Processing
Planning and Scheduling
Reasoning
Robotics
Search Techniques
Soft Computing
Speech Processing
Theorem Proving
Uncertainty Handling
Vision
Full papers (abstracts alone will not be accepted) must be
submitted
electronically at the conference website (preferred) or by
e-mail. All
submissions will be individually acknowledged by e-mail.
Papers must
contain an abstract not exceeding 250 words. Papers must be
no more
than 10 pages in length, inclusive of all Þgures, tables,
references
and appendices. Papers must be submitted in PDF/single-Þle
HTML
format. Refereeing will be Œblind¹ and hence authors¹ names
and
afÞliations must be given on a separate cover sheet only.
Instructions regarding the format of submissions will are
available at
Language Processing area, without strong AI content may be
transferred
Advisory Committee
------------------
Aravind K. Joshi, Univ. of Pennsylvania
P.V.S Rao, Tata Infotech Ltd, Mumbai
R. Narasimhan, CMC, Bangalore
Programme Committee
------------------
K.S.R Anjaneyulu, HP Labs, Bangalore
Vivek Balaraman, TRDDC, Pune
Pushpak Bhattacharya, IIT Mumbai
P.P. Chakraborti, IIT Kharagpur
S. Kambhampati, Arizona State University
M. Narasimha Murty, IISc, Bangalore
Bernd Neumann, Univ. of Hamburg
Arun K. Pujari, Univ. of Hyderabad
S. Ramani, HP Labs, Bangalore
P.V.S Rao, Tata Infotech Ltd, Mumbai (Chair)
Durgesh D. Rao, DR Systems, Mumbai
P. Saint-Dizier, Univ. of Paul Sabatier, France
K. Samudravijaya, TIFR, Mumbai
R. Sangal, IIIT Hyderabad
M. Sasikumar, C-DAC Mumbai (formerly NCST) (co-chair)
S. Sengupta, Tata Infotech Ltd, Mumbai
R. Uthurusamy, GMR Labs, USA (co-chair)
Contact Address
---------------
C-DAC Mumbai (formerly NCST)
Raintree Marg, Near Bharati Vidyapeeth,
Opp. Kharghar Railway Station,
Sector 7, CBD Belapur,
Navi Mumbai 400614, India
Phone: +91-22-27560013 email: kbcs AT ncst.ernet.in
--
[ Jayprasad J. Hegde, http://staff.ncst.ernet.in/jjhegde/ ]
===
Subject: Re: arithmetic Robak 0,(9)+{1+}0=1 Time Theory
charset=iso-8859-1
ksRobak <ed_robak@interia.pl>
> ksRobak <ed_robak@interia.pl>

> arithmetic Robak -- Time Theory by ksRobak (continue)
> ~~~~~~~~~~~~~~~~~~~~~~~~~
> 1/3 = 1/3 * 10/10 = (10/3) / 10 _|_ 10/3 = X
> 1/3 = X/10 = 0,X
> 1/3 = 0,X = 9/30 + 1/30 = 3/10 + (10/3)/10^2 = 3/10 +
X/10^2 = 0,3X
> 0,X = 0,3X
> 1/3 = 0,X = 0,3X = 0,33X = 0,3333...X = 0,(3)X = 0,(3)(10/3)
> 0,(3)(10/3) - 0,(3) = (10/3) / 10^Re1
> Edward Robak ed_robak@wp.pl |/ re:
1+1+1+1+1+1+1+1+...(->oo)=Re1 <== const
Re1 + a/b = Re1 + a/b
Re1 * a/b = Re1 * a/b
1/Re1 = +0 <== PUNKT
x/Re1 = {x +}0
{x +}0 * Re1 = x
+0 * ab = {ab +}0
+0/x = {1/x +}0
Edward Robak ed_robak@wp.pl |/ re:
===
Subject: Re: arithmetic Robak 0,(9)+{1+}0=1 Time Theory
charset=iso-8859-1
ksRobak <ed_robak@interia.pl1+1+1+1+1+1+1+1+...(->oo)=Re1 <==
const
> Re1 + a/b = Re1 + a/b
> Re1 * a/b = Re1 * a/b
> 1/Re1 = +0 <== PUNKT
> x/Re1 = {x +}0
> {x +}0 * Re1 = x
> +0 * ab = {ab +}0
> +0/x = {1/x +}0
> Edward Robak ed_robak@wp.pl |/ re:
geometric Robak -- Time Theory by ksRobak (continue)
~~~~~~~~~~~~~~~~~~~~~~~~~
A B B C
o------------------o o------------------o
1 2 3...->oo Re1 1 2 3...->oo Re1
PUNKT={AB+}0 PUNKT={BC+}0
AB + BC = AC
A B B C
o------------------o o------------------o
A B B C
o------------------o o------------------o
A B B C
o------------------o o------------------o
A B C
o------------------o------------------o
A C
o------------------------------------o
PUNKT={AB+}0 + {BC+}0 = {AC+}0
~~~~~~~~~~~~~~~
{AC+}0 * Re1 = AC
~~~~~~~~~~~~~~~
Edward Robak ed_robak@wp.pl |/ re:
--
AGEOMETRETOS MEDEIS EISITO
===
Subject: Galois group for deg 12 polynomial
charset=Windows-1252
Can someone Þnd the Galois group of this polynomial?
x^12+7*x^11+13*x^10-13*x^9-62*x^8-27*x^7+75*x^6+55*x^5-39*x^4
-34*x^3+3*x^2+6
*x+1
What software (if any) can do this? This polynomial arose in
a plane
geometry problem of arranging 11 circles in a circle. The
answer is to
Jim Buddenhagen
===
Subject: Re: Galois group for deg 12 polynomial
> Can someone Þnd the Galois group of this polynomial?
x^12+7*x^11+13*x^10-13*x^9-62*x^8-27*x^7+75*x^6+55*x^5-39*x^4
-34*x^3+3*x^2+6

*x+1
> What software (if any) can do this? This polynomial arose
in a plane
> geometry problem of arranging 11 circles in a circle. The
answer is to
> Jim Buddenhagen
Michael Pohst¹s group can apparently compute the groups of
polynomials over
Q up to degree 15. See
http://www.math.tu-berlin.de/~kant/kash.html . The
program is free.
--Edwin Clark
===
Subject: DC Proof 1.0 -- Revised and expanded tutorial
DC Proof is a line-by-line proof checker with a professional,
easy-to-use
interface designed with non-specialist in mind. It allows you
to write
error-free, formal proofs in logic, set theory and number
theory. The
recently revised and expanded tutorial should serve as an
excellent
introduction to formal logic and proof for the beginner.
Exercises with
hints and full solutions are provided.
Download DC Proof 1.0 at my website:
http://www.dcproof.com
System requirements: MS Windows (Win95 or later) and Internet
Explorer
(version 4 or later).
Freeware download for personal use only. Enquire about
multi-user
agreements.
Dan Christensen
Toronto, Canada
dc@dcproof.com
===
Subject: Re: Base 12. was: Origin of 360 degrees in a Circle?
Base 12 is vastly superior to base 10. The problem is we¹re
stuck with base
10. For example, 6 is the smallest perfect number, it¹s
multiple 12, is the
Þrst abundant number, and sigma(12)=28, the second perfect
number. Also,
6=3!, 4*6=4!. Many common fractions have exact decimal in
base 12, 1/3=0.4,
2/3=0.8, 1/4=0.3, 1/8=0.16, 1/16=0.09. There¹s also Benford¹s
Law on the
probability distribution of Þrst nonzero digits, given by
log_b(1+1/d).
In
base 12 the probabilty of a Þrst nonzero digit being a
divisor>1 is then
43% while in base 10 it is 26%. I think this would have
implications in
supercomputing: if your doing trillions of calculations a
second and
converting between binary and (duo)decimal, would Benford¹s
Law imply that
base 12 is measurably faster? The fact that you have a
greater chance of
combining divisors of your base should have some effect.
The Pythagorean Triples have a transparent structure in base
12. There¹s no
apparent structure in base 10. If anyone¹s interested, I¹ll
email you a pdf
Þle.
One more point: go look through the Atlas of Finite Groups
and observe haw
many groups have order of the form 2^p*3^q*5^r *(everything
else) where
typically p is large, q<p but comparable to p, and r is much
smaller than
q.
That means the order written in base 12 has more zeros,
sometimes four or
more times as many in base 10.
Last point: go through your favorite book on Elliptic
Functions and convert
all of the integers you Þnd to base 12 and see what happens.
Walter Kehowski
> ...
> If you want divisibility by 11: either dream, or surprise
me!
> I seem to recall reading, in a discussion on the Duodecimal
Society
> (a group advocating the use of base 12), a mention of
different group
> advocating using a prime base (maybe 11). The suggestion in
the
> However, in Martin Gardner¹s _Mathematical Magic Show_,
Chapter 8, I
> found the following:
> Above 5, very few number systems have been based on primes.
> W.W. Rouse Ball, in _A Short Account of the History of
> Mathematics_ (fourth edition, 1908), cites only the 7-base
> system of the Bolas, a West African tribe, and the 11-base
> system of the early Maoris, although I cannot vouch for
> either assertion.
> Jay
===
Subject: Debunking physics
For the debunking of physics see
http://paradigm.blogharbor.com
===
Subject: Re: Debunking physics
> For the debunking of physics see
http://paradigm.blogharbor.com
> [Physics] assumes that its measurement (its quantities) and
mathematics
> represent the ultimate logic of the Universe, and impose
these upon its
> observations and experiments.
Wrong. Physicists have long wondered how come mathematics
provides
such a
good description of physical phenomena, as illustrated by
Wigner in a
famous paper on the subject.
In a nutshell, you¹ve got it back to front.
===
Subject: Re: roots of polynomials
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i6ACbf103044;
>Reply to Dr. G.A.Edgar
>this equation of an odd degree,therefore,it must intersects
x-axes at
>least at one point,that is what I mean by (Þnd out) that
point of
>intersection.now if i assume (a&b) are real cooefÞcients
rather than
>real rational coofÞcients ,would that make it uncofusiable.
>I saw a strange formula (on the same topic)from
mr.karzeddin.I have
>been exersizing that formula for (n=5 )& (m=1 or 3)&
surprizingly it
>gives me very close values of intersections(f(x)& x-axes) as
long as
I
>take more elements of that mentioned series
>I hoop,this would make it clear.
>sincerely yours
Is there a radical closed form solution for the following
equation
x^5 + (27 + 24i)*x +(-4 + 48i) = 0

where i = (-1)^(1/2)
===
Subject: Re: roots of polynomials
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i68Bxoe29130;
>>A very interesting formula. It comes, I think, from a
series in
powers
>>of t for a root of x^n + t x^m - 1. According to Maple:
>>[view in Þxed-width font]
> series(RootOf(x^n + t*x^m - 1, x), t);
>> 1 n - 1 - 2 m 2 (2 n - 1 - 3 m) (n - 1 - 3 m) 3
>>1 - - t - ----------- t - ----------------------------- t
>> n 2 3
>> 2 n 6 n
>> (3 n - 1 - 4 m) (n - 1 - 4 m) (2 n - 1 - 4 m) 4
>> - --------------------------------------------- t
>> 4
>> 24 n
>> (n - 1 - 5 m) (3 n - 1 - 5 m) (2 n - 1 - 5 m) (4 n - 1 - 5
m)
5
>> -
-------------------------------------------------------------
t
>> 5
>> 120 n
>> / 6
>> + Ot /
>I think this is best viewed as a special case of the Lagrange
Inversion
>Formula. See
><<a
href=http://mathworld.wolfram.com/
LagrangeInversionTheorem.html>http://m
athworld.wolfram.com/LagrangeInversionTheorem.html</a>>.
>Write the equation x^n + t x^m = s as z = s - t z^(m/n) = s
+ t p(z)
>where x = z^(1/n) = F(z) and p(z) = - z^(m/n). The Lagrange
Inversion
>Formula says
>x = F(z) = F(s) + sum_{k=1}^inÞnity t^k/k! (d/ds)^(k-1)
(p(s)^k
F¹(s))
>>I don¹t know about the convergence of this series.
>It will converge for sufÞciently small |t|. I¹m not sure
exactly
how
>small.
>Robert Israel israel@math.ubc.ca
>Department of Mathematics <a
href=http://www.math.ubc.ca/~israel>http://www.math.ubc.ca/~
israel</a>
>University of British Columbia
>Vancouver, BC, Canada V6T 1Z2
I did not get the lagrange inversion you explained,
Why do you add many variables like (z,F(z),p(z))
can you simplify the expression?
(d/ds)^(k-1) (p(s)^k F¹(s)), were , s is constant hrer.
Is there a direct formula for lagrange like the one
I have introduced earlier.
In fact,I have derived(in 1991) a direct formula for x
in terms of (n,m,a,b)only, for the equation:
f(x)=x^n+ax^m+b=0
with out any further complication.
===
Subject: Re: roots of polynomials
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i685mCA31364;
>> Þnd out one real root for the following uneven degree
equation:
>> f(x)=x^n+ax^m+b=0
>> n is uneven positive integer
>> m is positive integer less than n
>> f(x) is a rational integeral function of x
>> (a&b) are real rational cooefÞcients
>The Galois group for x^5+x+3 over the rationals is S5.
>What do you mean by Þnd out?
>>Eng.Osama Ababnah,had passed on to me yuor e-mail
>> question about the meaning of Þnd out?
>> as you see from the formula I added the dots notations ...
>>which, means that the root obtained by this series comes
>> out of inÞnte sums,as you suggest in your message.
>>the solution,therfore,for uniqe only root is always
irrational
number.
>I don¹t get your therefore. Just because something is the
sum of
>an inÞnite series doesn¹t mean it¹s irrational. For example,
>x^n + a x^m - (1+a) has a rational root.
>Robert Israel israel@math.ubc.ca
>Department of Mathematics <a
href=http://www.math.ubc.ca/~israel>http://www.math.ubc.ca/~
israel</a>
>University of British Columbia
>Vancouver, BC, Canada V6T 1Z2
I have not presented my solution to (x^n+ax^m+b=0)yet,
so you are true but:
My reply was to the case, when (a=b=1)
Can you provide one rational root for (x^n+x^m+1=0)
===
Subject: Re: roots of polynomials
>I have not presented my solution to (x^n+ax^m+b=0)yet,
>so you are true but:
>My reply was to the case, when (a=b=1)
>Can you provide one rational root for (x^n+x^m+1=0)
Of course not, because there isn¹t any (x would be an
algebraic
integer, and all algebraic integers that are rational are
ordinary
integers).
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2