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convergence of QR-algorithm
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Gottfried Helms
science forum Guru


Joined: 24 Mar 2005
Posts: 301

PostPosted: Fri Jun 30, 2006 12:54 pm    Post subject: Re: Best Fit Circle made Easy Reply with quote

Am 29.06.2006 09:28 schrieb Han de Bruijn:
Quote:
Peter Spellucci wrote:

In article <4ac1e$44a2739b$82a1e228$1666@news1.tudelft.nl>,
Han de Bruijn <Han.deBruijn@DTO.TUDelft.NL> writes:
http://hdebruijn.soo.dto.tudelft.nl/jaar2006/kromming.pdf
you seem to assume that R^2 approx a^2+b^2 ??

No approximations involved: R^2 = a^2 + b^2
if R = radius and (a,b) = midpoint .

where is a problem with circle fit?
[ ... ]
this is clearly a nonlinear least squares problem.
[ ... ]

While my formulation is _linear_ all over the place.

this should give at least a reasonable first guess for the nonlinear problem
and will in many cases be already a very satisfacting solution.

Yes. But the point is that I _hate_ nonlinear problems. :-(

...

Han de Bruijn

Hmm, another try which i did recently:


Let the coordinates of all N points yi,xi
and let yi',xi' be the coordinates of the set of points
translated to their mean.

denote
ßi = arctan(yi'/xi')

then the minimization means in terms of a projected circle with
center (0,0) and optimal radius r to estimate:

Sum (xi' - r*cos(ßi))²+(yi'-r*sin(ßi)² = min

Derivative:

Sum 2(r*cos²(ßi) -xi*cos(ßi) + 2(r*sin²(ßi) -yi*sin(ßi)) = 0

2r Sum(cos²(ßi)+sin²(ßi)) = 2 Sum (xi*cos(ßi)) + yi*sin(ßi))
denote

laei = sqrt(xi'² + yi'²)

then

2r Sum 1 = 2 Sum (xi'² / laei) + yi'²/laei)

Sum (xi'² + yi'²)/laei
r = ------------------------
N

Sum laei
r = --------
N

r = Mean (laei)

... which generalizes easily to more dimensions.

Gottfried Helms
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Richard Mathar
science forum beginner


Joined: 23 May 2005
Posts: 45

PostPosted: Sun Jul 02, 2006 6:33 pm    Post subject: Re: Elliptic integral of the seconnd kind with complex amplitude Reply with quote

In article <1hhrlv3.1xijemp15dioq3N%cbarron3@ix.netcom.com>,
cbarron3@ix.netcom.com (Carl Barron) writes:
Quote:
Gerald I. Evenden <gerald.evenden@verizon.net> wrote:

Is anyone aware of a technique for the solution of the incomplete
elliptic integral of the second kind with a complex amplitude:

E(x+iy,m)

m is small, in the range of 0.006.

If there is a library reference, is should be usable under either
a GNU or MIT license.

Thanks in advance.

There is an article on elliptic integrals and elliptic functions.

Bulirsch Numerishe Mathematik vol 7 p78-90,vol 7 p353-354,vol 13,
p 305-315.

Either the first most likely since it contain algol 60 code provides an
algorithm to compute first and second elliptic integrals with complex
argument and real modulus. If its the first the second looks like
errata, and the third I don;t recall seeing.

its at guttenburg(sp??) online library.


See also

@article{CarlsonNumMath33,
author = {B. C. Carlson},
title = {Computing Elliptic Integrals by Duplication},
journal = {Num.\ Math.},
volume = {33},
pages = {1--16},
year = 1979
}

@article{CarlsonMathComp53,
author = {B. C. Carlson},
title = {A Table of Elliptic Integrals: Cubic Cases},
journal = {Math.\ Comput.},
volume = {53},
number=187,
pages = {327--333},
year = 1989
}

@article{Carlsonarxiv94,
author = {B. C. Carlson},
title = {Numerical Computation of real or complex Elliptic Integrals},
journal = {arxiv:math.CA/9409227},
year = 1994
}
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sonnenrain17@gmx.de
science forum beginner


Joined: 02 Jul 2006
Posts: 2

PostPosted: Sun Jul 02, 2006 8:25 pm    Post subject: Re: Newton Raphson variants... Reply with quote

For equations having one unknown there are numerous variants which are
more efficient than Newton's method, the most well known ones mentioned
before. For systems of equations the situation changes, there are not
as many. In any case one has to compare carefully the number of
function evaluations and derivative(s) used for each iteration in order
to derive reasonable conclusions. E.g. there is a method having order
1+sqrt(2) (for "single" roots) which uses one evaluation of f and one
for its first derivative per step which works for systems too. The
computational costs are comparable to Newton's method if implemented
reasonably. See Numerische Mathematik, 32 (1979) and 38 (1982). As far
as I know there is no method using one value of f and one evaluation of
the jacobian per step which has higher order.
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Phil Carmody
science forum Guru Wannabe


Joined: 05 Jun 2005
Posts: 267

PostPosted: Sat Jul 08, 2006 12:53 am    Post subject: Re: Newton Raphson variants... Reply with quote

Peter Luschny <spamgrube@luschny.de> writes:
Quote:
Phil Carmody schrieb:
"Hugo Pfoertner" <nothing@abouthugo.de> wrote

And where did Ernst Schröder lose the German Umlaut "ö"? The only
acceptable transcription of the German Umlaut derived from "o" is "oe".
^^^^^^^^^^^^^

i.e. Schröder --> Schroeder

Schroeder's original article

Ignore Hugo, he's full of old-Europe huff and nonsense. In a
_German_ medium the only acceptable transliteration is as he
^^^^^^^^^^^^^^^
indicates. _Usenet is not a German medium_. It is the target that
decides the rules for transliteration therein, not the source
language.

Ignore Phil, he does not know the difference between
'transcription' ans 'transliteration'.

Given that the process being performed was precisely transliteration,
that's the word I used. And I also know how wide space characters are,
unlike you.

Phil
--
The man who is always worrying about whether or not his soul would be
damned generally has a soul that isn't worth a damn.
-- Oliver Wendell Holmes, Sr. (1809-1894), American physician and writer
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Titus Piezas III
science forum Guru Wannabe


Joined: 10 Mar 2005
Posts: 102

PostPosted: Tue Jul 18, 2006 5:54 am    Post subject: Re: a^2+b^2 = c^2+d^2 Reply with quote

Robert Israel wrote:
Quote:
In article <44bb3bc6$0$990$ba4acef3@news.orange.fr>,
Patrick Coilland <pcoilland@pcc.fr> wrote:

Robert,

According to the initial problem "let P be the number of primitive
solutions to the Pythagorean triple a^2+b^2 = c^2 with hypotenuse c less
than a bound N. Then P/N = 1/(2pi) as N -> infinity", I think theat the
OP question is :

"let P be the number of primitive solutions to the integer equation
a^2+b^2 = c^2+d^2 = z with z *less than* a bound N. Then does the ratio
p/Sqrt(N) approach a real constant as N -> infinity? Can it be expressed
as a rational multiple of pi?".

Hello Patrick and Robert,

My apologies for the confusion. I'll try to make it clearer. Given the
equation,

a^2+b^2 = c^2+d^2 = z

let p be the number of primitive solutions with *hypotenuse sqrt[z]
less than* a bound N.

Case 1:

For c=d=0, and z a square, Lehmer proved the function F(N): = p/N =
1/(2pi) as N -> infinity. Example, for N = 10, we only have the
solution (3,4,5) giving F(N) = 1/10 = 0.1. The sequence of F(N) for N =
10, 10^2, 10^3, 10^4,... is

F(N) = 0.1, 0.16, 0.158, 0.1593, 0.15919, ...

converging to 1/(2pi) = 0.159154...

Case 2:

However, let (a,b,c,d) all non-zero with z not necessarily a square.
(Hence we are looking for a pair of triangles with integral legs with
the same hypotenuse.) Example, for N = 10, there are only 3 primitive
solutions,

1^2+7^2 = 5^2+5^2 = 7.07^2
1^2+8^2 = 4^2+7^2 = 8.06^2
2^2+9^2 = 6^2+7^2 = 9.21^2

so F(N) = p/N = 3/10 = 0.3.

Questions: 1) What is the sequence of F(N) for this? 2) Does it
converge to some constant? 3) If yes, can it be expressed in terms of
pi, analogous to Lehmer's result?

I hope it is clearer now.

P.S. The sequence of F(N) for case 2 for N = 10, 10^2, 10^3,... can
easily be ascertained by anyone with programming skills which I hope
someone in this channel would be curious enough to do so.

--Titus
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Paul O
science forum beginner


Joined: 01 Feb 2006
Posts: 11

PostPosted: Tue Jul 18, 2006 7:03 pm    Post subject: Re: An Interesting Subject Reply with quote

theo3 wrote:
Quote:
This is a webpage that basically gives all answers that people are
looking for. If you are interested in looking at it please check it
out. For the German vrsion please click on the German flag.
http://www.theomonistik.net/ee/374.htm



Yawn...




Paul D Oosterhout
(from SAIC)
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