Newsgroup sci.math.num-analysis 29352

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: My view is that math is largely independant from reality,
Number one: Math IS reality.  It really DOES exist.  Of course, I
am allowing ABTRACTIONS to exist.  It exists regardless of whether
anyone is there to think it.  I assume that this is what you mean by
"independent from reality".
Number two: ANY other subject (that might qualify as a major at a 
university, for instance) ALSO using abstractions to describe it.
It's called LANGUAGE.
Number three: Math is, by far, infinitely better suited at describing
ALL of reality, not just part of it, than all the subjects referred to in
"Number two".
: it is best applied to the abstractions and mental models we 
: made of real things.
That's because that's the only thing math CAN be applied to: the
abstractions and mental models we make of things.  Again, EVERY
human subject, with its corresponding language and terminology,
can only be applied to an abstraction in our heads.
 This is good because even the simplest things
: from reality are so complex that we wouldn't be able to cope with
: them.
Yes.  Just as doctors and lawyers and judges make simplifying models
in their mind to achieve an end such as reaching a decision.
 Reality is by nature three dimensional, so I don't it is
: possible to build a 2 dimensional figure in reality, the lines you
: draw will be made of atoms which are three dimensional.
: I don't think it is possible to build even ideal polyhedra.
Of course not.  NO one can create EXACTLY what is in their mind.
The greatest sculptors and artists cannot do it.  What they create 
is their very best approximation to what's in their minds.   "Best"
means "optimal subject to the condition that they wish to reach a 
decision in a finite amount of time" (and effort).

In my posting on the Afternotes, I failed to give the complete ftp
address.  It should read thales.cs.umd.edu.  I apologize for any
inconvenience.
Pete Stewart

Robin Becker wrote:
> 
> I would appreciate any pointers to formulae for conversion of calendar
> dates to integers which are invertible.
> 
>         ie
>         yyyy/mm/dd --> integer and the inverse
> 
>         integer    --> yyyy/mm/dd
> 
> I'm not terribly interested in dates pre 1901
> --
> Robin Becker
Algorithms for date conversions can be found in any astronomical
handbook. I suggest you:
    Jean Meeus: Astronomical Algorithms
Good Luck.
Hayo
-- 
***********************************************************************
* Hayo Hase                                        Tel: 09941-603-0   *
* Institut fuer Angewandte Geodaesie               Tel: 09941-603-104 *
* Fundamentalstation Wettzell                      Fax: 09941-603-222 *
* D-93444 Koetzting                        Net: hase@wettzell.ifag.de *
***********************************************************************

Given: A set of threedimensional coordinates identifying points on the
surface of the spheroid with the radius r.
Wanted: The maximum volume of the sum of tetraeders formed by three
surface points and the center point of the spheroid (= origin of
coordinate system).
Restriktion: Each volume element must contribute only once! With
infinite points on the surface the volume must result in the spheroidal
volume!
Problem: It exists many combinations of tetraeders. Which strategy or
algorithm will lead to the maximized volume of all tetraeders?
Example of the problem:
Choose four surface points plus the origin. With these five points two
sets of two different tetraeders can be formed. Which set has a larger
volume?
   Ok. You might say: just calculate it and you'll know it.
That's true. But if an additional point appears I obtain much more
possibilities of tetraeders. Where do I start? Which possibility leads
to the maximum volume?
   ---???
Exactly, that is my problem.
Hope to get an answer from the sci.math community.
Please mail me: hase@wettzell.ifag.de
-- 
***********************************************************************
* Hayo Hase                                        Tel: 09941-603-0   *
* Institut fuer Angewandte Geodaesie               Tel: 09941-603-104 *
* Fundamentalstation Wettzell                      Fax: 09941-603-222 *
* D-93444 Koetzting                        Net: hase@wettzell.ifag.de *
***********************************************************************

In article <32cc24f6.7396859@news>, Greg Ratzel 
writes
>On Thu, 2 Jan 1997 16:07:04 +0000, Gary Hampson
> wrote:
>>1) The function you describe as a time/space domain operation is simply
>>some weighting function of f(t). Similar to say Hamming or Hanning
>>window functions. On that basis its frequency domain equivalent is a
>>convolution with its fourier transform. 
>
>I disagree.  g(sigma) is more complicated than a window function.
>Window functions are multiplied by the time-domain function, so the
>corresponding frequency domain operation is convolution.
On more than 3 seconds reflection I agree, as they say "engage brain
before opening mouth", gone off half cocked.... and all that stuff.
-- 
Gary Hampson

In article <32C90B5C.64B1@public.ibercaja.es>, benigno
 writes
>Hi,
>       I need to handle Big matrix of around 800 x 800 to implement
>       some optimization algorithms, I would use C++ libraries if
>       possible to use on BorlandC++ 4.5, but if there is any shareware
Unless the matrix has some structure which allows many short cuts and
reduced storage (eg Toeplitz), then just get on with coding it. 800*800
is not that big (unless of course in the optimisation you need to
evaluate A.x many times, or you have some time critical conditions.
-- 
Gary Hampson

Hi
I have square matrices of floating point numbers and I would like to
have a criterion to decide if their determinants are zero or not. But,
assuming that the coefficients are not exact, the computed determinant
will obviously be non null, thus some threshold have to be used.
As this problem is fairly usual, I assume that lots of work has been
done on that topic. Of course I know that testing the condition number
of the matrix can be an answer to my problem, but I'm wondering if some
other methods have been studied.
Now more important is the choice of the threshold. Are there any work
studying how to choose a reasonnable value for the threshold when having
some knowledge of the noise on the coefficients.
    Thank's a lot....
	Theo Papadopoulo.

You've defined a non-linear finction g() which is
applied to the time-domain functin.
There are some things you can transfer between the time and frequency
domains. If you multiply two things in the time domain, it's like
convolving their DFTs. If you convolve two things in the time domain,
it's like multiplying their DFTs. This is the reason that a lot of
convolution gets done in the frequency domain instead of in the
time domain: multiplying is a lot quicker than convolving.
But now you want a non-linear function of the time-domain sequence and you
want to find the equivalent in the frequency domain. Tricky. I don't
know if it's possible in the general case but I can give you a
good approximation.
First, let g(x) = x^2.  That's exactly like multiplying two functions,
(which happen to be identical) so you need to convolve their DFTs.
You can do it with a simple loop over the components of your DFT but
it's more effort than doing the multiplication the simple way.
Secondly, let g(x) = polynomial of x:   A0 + A1x + A2x^2 etc.
Each term can be added inside or outside the DFT, so treat them separately.
Each term is a multiplication, so it comes out as a convolution inside
the DFT. It only takes one convolution to go from x^n to x^(n+1)
because you haven't thrown away your original DFT of x, have you?
so for an nth-degree polynomial, you must do n convolutions in the
frequency domain. This is all still exact, no approximations.
                  ^^^^^^^^^^^^^^^^^^^^^^^
                  ***********************
Thirdly, approximate your real g() as a polynomial. Depending how
large your signal is, you might get reasonable approximations with
a low-order approximation. Each successive term in the polynomial
will apply a correction.
The above really is a lot of effort. It's so much more effort than
just transforming back to the time domain and applying g(), even
if your Fourier transform is not a fast one.
Even one convolution in the frequency domain would be more simply
done by transforming back to the time domain where you can multiply.
However, there might be one or two insights into what g() is
doing to your spectrum if you input a pure sine wave and watched
the resulting spectrum as its amplitude got larger and larger. There
would be all sorts of distortion harmonics developing, rather
reminiscent of an amplifier which is being fed a signal which is
too large. Hmm. That wouldn't be your application, would it? It
seems to me that g() resembles an amplifier which is initially
linear, but which saturates at higher amplitude. Mind you, lots
of physical systems behave like this. Since you use k (wavenumber)
rather than f or omega, I'm inclined to think there are some
physical waves in a medium and that medium saturates somehow.
[===>>YOUR QUESTION<<===]
    Can I apply, say, the Fourier transform of g(\sigma) to each
    point of F(k), i.e.,
            if DFT[g(\sigma)]  =  G(k), then is  IDFT[G(F(k))] = g(f(t)),
            where DFT is the discrete Fourier transform and IDFT
            is the inverse discrete Fourier transform?  (*)
[===>>YOUR QUESTION<<===]
g(sigma) is a time-domain function. t goes in, amplitude comes out.
because you've said that sigma is a finction of time.
G(k) is a frequency-domain function. k goes in, amplitude comes out.
F(k) is similarly a frequency-domain function. Amplitude comes out.
You can't take G() of anything which is not a frequency. G() just
won't operate on amplitude, which is what you're asking it to do
when you write G(F(k)) .
Simon

Try 
Griffiths, P. and I.D. Hill, 1985, Applied Statistics Algorithms
	Chichester, Ellis Horwood
-- 
Gerry Middleton
Department of Geology, McMaster University
Tel: (905) 525-9140 ext 24187 FAX 522-3141

Has anybody and  good references? Most Numerical Analysis
book seem to concentrate on quadrature and interpolation :(
Specifically I want to concentrate on polys of degree < 10 
(I have implemented up to fourth degree analytically)
Also, most of the time just the least positive root is
required, not all roots (a la Sturm sequence)
Thanks,
Steve.

Another suggestion:
Maybe you don't care how precise your sigmoid g(x) is.  There are lots
of simpler saturation functions that may work and be easier to
analyze, depending on your application.  For example, in artificial
neural nets where a sigmoid response is called for, a piecewise-linear
saturation curve is often substituted.
Greg Ratzel

Angel Garcia (bp887@FreeNet.Carleton.CA) writes:
>      Well, TETET-96 is already past history and a good one.
> It is time to start thinking for the next TETET-97.
> For those who follow a bit the MAYUSCULE discoveries currently
> being made in that, seemingly chaotic, terrain of Mars:
> 
>     The next obvious question is about these tiny mounds
> near-by Ac, Ec, Mc, etc. in my IMAGE-2, -etc...
> Such set of mounds were 'noticed' long time ago by Hoagland
> and correctly started to be decrypted (1995) by Prof. Dr. Horace
> Crater. See the latest 'news' in URL of Prof. McDaniel, who
> labels them A, E, B, etc. (the same first letter as I do in
> my URL... but I add second letter ..c to distinguish from
> labels A, E which ARE fundamental and already published in 1993.
> 
>      Of course the decryption prsented in TETET-96 (see summary in
> my URL) is correct (exactly as intended by the martian Designer)
> or the majestic conundrums would never had appeared. One notices
> that mound G in McDaniel's is not mentioned in TETET-96: it
> simply means that nobody has so far found a good conundrum for it
> and yet, according to Crater and McDaniel, it is in perfect
> rectangle with Ec, Ac, Mc:
> 
>                                  * Nc
>                                / |
>                              /   |
>                           /      |
>             Gc *-----------------* Mc
>                |        /        |
>                |     /           |
>                |  /              |
>             Ec /-----------------* Ac     
>               /
>             /
> 
          /
         /
        ~
       ~
   A~*
> With (Ac,Mc)=(Mc,Nc)= (Ec,Ac)/2^(1/2) = 0.21998877608..
> 
> What the heck this Gc means ?... any other mounds around Gc ?.
> Any simple conundrum related to Gc ?. Clearly mounds A~ (Angel's mark)
> and Ec are in PERFECT alignment with point Nc (at nose of beast in
> Fortress): any numbers in these EXTREMELY simple triangles which
> can produce glamorous conundrums ?.... because it seems pretty obvious
> that such Gc of Crate-McDaniel and possibly others near-by HAVE to be
> in there for a reason !!!.
> --
     In such unique decryption the psychology is to "tune" with our
"martian brothers" who obviously spent years in building that grand
parade of monuments and mounds with embedded mathematics of so superior
class. 
A) They use the ONLY possible language: geometry-numbers and some fine-Art.
B) They are "talking" about the most general and fundamental topics:
    Cosmogony and evolution of planets and life in the solar system.
C) They are perfectly honest: they don't lie. They are a bit cryptic: not
   too much, probably because They want to be cautious with our feelings
   and our reasonable pride and our preconceived ideas.
D) ...
   It is NOT that hard to decrypt it: just trial-error on what has already
been decrypted. Consider the "process" in TETET-96:
      McDaniel and Crater had presented mounds Ac, Bc, Mc which being in
perfect alignment are pointing to exact center S of D&M; (decrypted in
TETET-92 already). Thus Lahoz measured distance from Ac to S and found that
it was very very close to 3*(Ac,Mc) and said:
   Voila They ate making here a perfect 'stright line' divided in 5 equal
segments: (Nc,S)= 5*(Ac,Mc) !!!.  The last point is NOT marked with 'mound'
but just the "NOSE of that terrific work-of-fine-Art: a felida at Fortres"
IS the intended final mark of the 5 segments.
Then Lahoz calculated roughly the numbers for these lengths and found:
      Ac, Mc  = 0.219988776... and  (Ec,Ac)= 2^(1/2)*0.219988..=0.3111111...
and said:
Voila: a conundrum 0.311111111...  we are in the right track !!!. And very
soon he added: because (5/2)*0.31111111... = 0.7777777777...it is
logical that THEY stress the conundrum via geometry of triangles to
include trivial factor 5/2: thus there MUST be a mark (or 'mound') at
any of the four end-sides of  5-string (Nc,S): so I looked around at
these 'predicted' places and VOILA there it was my mark A~ (Angel's
mark).
    Perpendicular from A~ to line (Nc,S) and the whole book was ON. !
  So simple that anybody could have done it !. 
--
Angel, secretary of Universitas Americae (UNIAM). His proof of ETI at
Cydonia and complete Index of new "TETET-96: Faces on Mars.." by Prof.
Dr. D.G. Lahoz (leader on ETI and Cosmogony) can be studied at URL:
     http://www.ncf.carleton.ca/~bp887    ***************************

On 3 Jan 1997, Steve wrote:
> 
> Has anybody and  good references? Most Numerical Analysis
> book seem to concentrate on quadrature and interpolation :(
                              ^^^^^^^^^^^^^^^^^^^^^^^^^^^^
I'm not sure what those are (hopefully not what I'm suggesting).
How about Newton's method?  It has quadratic convergence.
Given what you wrote below you have an idea of the starting value:
> Also, most of the time just the least positive root is
> required, not all roots (a la Sturm sequence)
Tom.
--
Tom Szymkiewicz (mrtom@wpi.edu)
WPI - Worcester Polytechnic Institute
USA
WWW: http://www.wpi.edu/~mrtom
PHYSICS!