mm-395
mm-395
Subject: Job opening: member of technical staffThe
following job is available immediately:Job title: member of
technical staff.Job Description:Developing efficient algorithm
for electromagnetic field scatteringoffsemiconductor devices.
Assist in the enhancement and improvement ofcurrent thin film
algorithms.Qualifications:Ph. D in computational physics or
numerical analysis or equivalent,with morethan 5 years
experience of demonstrated proficiency in solving
complexphysicsproblems and numerical programming in C++,
Fortran 90/95.Thermawave is located in Fremont,
California.Please contact via
email:hchu@thermawave.com===Subject: Random generation of
covariance matricesHello there,I am looking for a possibility
to generate covariance matrices (pos.semidefinite, symmetric)
randomly. The diagonal elements (variances)should be uniform
distributed in [0,1]. Off-diagonal elements(covariances)
should be so that the resulting correlations are
uniformdistributed in [-1,1].My basic idea is as follows:1.
Generate random (n x n) correlation matrix.2. Generate random
uniform distributed (n) vector with the variancesin [0,1].3.
Calculate the resulting covariances.2. and 3. shouldn't be a
problem. But how can I generate a random (n xn) correlation
matrix with all diagonal entries being 1 andoff-diagonal
elements being uniform distributed in [-1,1] ensuringthat
everything stays positive semidifinite? Could
theBendel/Mickey-algorithm be of any help (to my knowledge it
does onlyensure uniform Eigenvalues which does not mean
uniform distributedentries).If anybody has a Matlab code for
this, it would be extremely helpfulto me... Otherwise I am
also grateful for any advice.Best wishes.Peter===Subject: Re:
Random generation of covariance matrices> Hello there,> I am
looking for a possibility to generate covariance matrices
(pos.> semidefinite, symmetric) randomly. The diagonal
elements (variances)> should be uniform distributed in [0,1].
Off-diagonal elements> (covariances) should be so that the
resulting correlations are uniform> distributed in [-1,1].This
doesn't look like a natural distribution of covariancematrices.
It is more likely that your application requiresthat you
generate them according to a Wishart distribution.Arnold
Neumaier===Subject: Probability of choosing N from M -
extended. by support1.mathforum.org (8.11.6/8.11.6/The Math
Forum, $Revision: 1.9 primary) id i2GEjVd19977;I can't quite
work out the formula for the following. Can anyonehelp? - I
have a bag containing B blue balls, R red balls and W
whiteballs. - I draw N balls from the bag at random, without
replacement, where 0< N < (B+R+W) - What is the probability
that I have drawn b blue balls, r red balland w white balls,
where obviously N=b+r+w, as a function of r, b, w,R, B and
W?===Subject: Help with Number Pattern by
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id i2GElIC20600;What is the next
number in this set?? 1 11 21 1112 3112 211213 312213 212223
11421331121314????????===Subject: Who are You Dennis ? by
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id i2GEjZo20145;>My calculus class was
challenged to determine and prove which is the>greater of the
two: e^pi or pi^e. I have worked on this for quitesome>time,
but cannot prove my conclusion. Any
suggestions?>Dennis===Subject: Re: Who are You Dennis ?Find
which is bigger on your calculator?>My calculus class was
challenged to determine and prove which is the>greater of the
two: e^pi or pi^e. I have worked on this for quite> some>time,
but cannot prove my conclusion. Any
suggestions?>Dennis===Subject: Re: Who are You Dennis ?> Find
which is bigger on your calculator?>My calculus class was
challenged to determine and prove which is the>greater of the
two: e^pi or pi^e. I have worked on this for quite> some>time,
but cannot prove my conclusion. Any suggestions?>DennisNot a
valid procedure, because there's no guarantee that one is
_really_bigger than the other (even though a difference of 3%
is well within therange of the calculator's error).To _prove_
it using a calculator you'd find the bigger one, round
bothnumbers _down_ at some precision and find the resulting
lower bound. Thentake the smaller one and round both numbers
_up_ (not necessarily at thesame precision) for an upper
bound. If x.u is a guaranteed upper bound ofx, and y.l is a
guaranteed lower bound of y then showing that x.u <
y.lguarantees that x < y.Alternately you could try to do some
sort of an error-bounds proof, but itwould be more work to
come up with a conclusive proof, and would requireknowing
details about the particular calculator you're using, etc.,
etc.===Subject: Re: Who are You Dennis ?> Find which is bigger
on your calculator?>My calculus class was challenged to
determine and prove which is the>greater of the two: e^pi or
pi^e. I have worked on this for quite> some>time, but cannot
prove my conclusion. Any suggestions?>Dennis> Not a valid
procedure, because there's no guarantee that one is _really_>
bigger than the other (even though a difference of 3% is well
within the> range of the calculator's error).> To _prove_ it
using a calculator you'd find the bigger one, round both>
numbers _down_ at some precision and find the resulting lower
bound. Then> take the smaller one and round both numbers _up_
(not necessarily at the> same precision) for an upper bound.
If x.u is a guaranteed upper bound of> x, and y.l is a
guaranteed lower bound of y then showing that x.u < y.l>
guarantees that x < y.> Alternately you could try to do some
sort of an error-bounds proof, but it> would be more work to
come up with a conclusive proof, and would require> knowing
details about the particular calculator you're using, etc.,
etc.Only a mathematician could make something difficult out of
this. Both e andPi are known to any number of decimal places.
My cheapo calculator, Casiofx-991MS, works to about 10
significant figures. It givese^Pi = 23.1406...Pi^e =
22.4591...Since the numbers differ to only two significant
figures I fail to see howthere could possibly be any doubt as
to which is larger.Am I missing something?===Subject: Re:
Matlab ebooks by support1.mathforum.org (8.11.6/8.11.6/The
Math Forum, $Revision: 1.9 primary) id i2GEjaQ20170;>Matlab
ebooks>Elementary Mathematical and Computational Tools for
Electrical and>Computer Engineers Using MATLAB>for more .nfo,
and for the complete list of 12,000 approx CDs, please>send
e-mail -> astra@===Subject: implementation of digital filter
on tms 320vc5402 kit by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
i2GEjad20193;i need source code for a digital filter of any
type in C FORIMPLEMENTATION ON TMS KIT USING CC
STUDIO===Subject: linking matlab functions with a front end by
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id i2GEjaO20184;dear friends, i have
made individual functions in MATLAB which require user togive
various filter parameters and design digital filters. i need
tolink these programs to a front end made in vb or html such
that thefront end takes up all the filter parameters and then
delivers them tomatlab at the backend. finally the result
should be displayed. i amnot finding any means to d o so since
matlab files are not .exe files===Subject: Re: linking matlab
functions with a front endWhy don't you use Java?The Matlab
GUI is written in Java and Matlab already has lots of
functions available to integrate with the JVM. The person
using your tool then only has to run Matlab to start the UI
with some Matlab command..>dear friends,> i have made
individual functions in MATLAB which require user to>give
various filter parameters and design digital filters. i need
to>link these programs to a front end made in vb or html such
that the>front end takes up all the filter parameters and then
delivers them to>matlab at the backend. finally the result
should be displayed. i am>not finding any means to d o so
since matlab files are not .exe files> ===Subject: Re: PCA
utility fro MS Excel or so on by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
i2GEjbl20238;>I am looking for an Excel add-in for performing
Principal ComponentAnalysis.>Anyone can hel
me?>Thank's>-----------== Posted via Deja News, The Discussion
Network==---------->http://www.dejanews.com/
Search, Read, Discuss, or Start Your Own ===Subject: Gauss -
Kronrod ( I need help!)Hi!I've to do something for school
about numerical integration.But I could really need some help
understanding the method of Gauss-Kronrod(or at least the
Gauss quadrature).It would be very nice if someone could
explain it to me.Brieg===Subject: Re: Gauss - Kronrod ( I need
help!)>Hi!>I've to do something for school about numerical
integration.>But I could really need some help understanding
the method of Gauss-Kronrod>(or at least the Gauss
quadrature).>It would be very nice if someone could explain it
to me.>Brieggauss rule takes n nodes and n weights free and
then choses them as to maximize the degree of a all
polynomials which can be integrated exactly with it.the
outcome is exactness degree 2n-1 (sometimes called order
2n)for the nodes fixed the weights are always uniquely
determined by the requirementthat any polynomial of order n-1
should be integrated exactly with the rule.this is a simple
linear system with a vandermonde matrix as systems matrix.that
using the zeors z_i,n of the Legendrepolynomials as nodes
raises the degree of exactness(hence the order) by n can be
seen quite simply using Euklids algorithm:
p_{2n-1}(x)=p_(n-1}(x) prod_{i=1 to n} (x-z_i,n) + q_{n-1}(x)
applying the rule and the integral to both sides of this
decompositionand using the orthogonality property of the
Legendrepolynomials (all this on[-1,1]) transformation on
general [a,b] is obvious (since it is linear it doesn'tvchnage
polynomial degrtees hence order)the uniqueness of the gaussrule
follows similarly.now, for kronrod, add n further nodes to the
n given Gaussnodes and require that thenew rule integrates
polynomials of order 3n-1 exactly. again this gives a system
which can be solved uniquely. so you have two nested rules.
from the order relation one concludes that the error of the
gaussrule applied to an interval of length H is something like
H^{2n}*const*derivative of order 2n of the integrand
somewhereand for gauss-kronrod H^{3n}*const2*derivative of
order 3n ......hence for small H it is reasonable to assume
that the difference of thetwo formulae gives an error estimate
for the less precise (the Gauss)rule.without the need to
discuss behaviour of derivative 2n of the integrandsee
quadpack, there this is used
heavilyhttp://www.netlib.org/quadpackhthpeter===Subject: Re:
svd for complex matrices by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
i2GGa9102280;>HI,>I'm trying to write a c program that allows
me to perform svd of a>complex matrix so that
[Uc,Wc,Vc]=svd(Ac), Ac=UWV*. I have aprogram>that performs svd
for real matrix and i know that I must replceevery>z=x+iy by a
matrix (x -y; y x) so that i can transform my complex>matrix
into a real one and let my program give me Ur, Vr and
Wr2m*2n>and 2n*2n matrices. But what is the correlation
between>(Uc,Ur),(Vc,Vr) and (Wc,Wr).>Please I need a quick
reply.>multiply the complex version, writtem in real and
imaginary partsout.>then you will be able to compare real and
imaginary components withthe real>version. but it is not a
good idea to do this tranformation into areal problem>of
doubled dimension, since the complete algorithm can be
donedirectly in the >complex filed, replacing transposition by
taking the complexconjugate transpose.>hth>peterHi Sir,If I
write, as you mentioned, my matrix A= UDV'= X+iY, U=U1+iU2
andV=V1+iV2, I'll obtain X=U1DV1'+U2DV2' and Y=U2-U1V2'. So,
as far as Iknow this decomposition of X and Y is not unique
and we can notdetermine U,D and V by means of the SVD of X and
Y separately.So I still need help to deal with this
problem.Patrik===Subject: Re: eigenvalues of large, symmetric,
matrces with many zeroes.Actually, I need to compute things
likesum_i (1/n) frac{lambda_i}{a - blambda_i}where lambda_i,
i=1ldots n are the eigenvalues,and a and b are parameters.Any
idea ?>I need to compute the eigenvalues of a symmetric,>real,
nXn matrix A where:>1. each row of A is a vector of zeroes with
>at most 4 ones>2. n is very large, of order 10^5 or 10^6.>Any
suggestions are greatly appreciated.>Federico Echenique> I
assume you want only some ... (the largest , the smallest,
some in some> interval .... , otherwise you must proceed
completely different ...)>
http://www.ime.unicamp.br/~chico/arpack++>
http://www.caam.rice.edu/software/ARPACK> hth>
peter===Subject: Re: Continuity Between Triangular Bezier
Patches by support1.mathforum.org (8.11.6/8.11.6/The Math
Forum, $Revision: 1.9 primary) id i2GHpkR11131;Dear ,My name
is Michael Osofsky and I am an MIT student. I saw yourposting
and would like to talk to you briefly about a research
projectI am doing because it is very relevant.Thank
you.Sincerely,Michael Osofskymosofsky@sloan.mit.edu>Hello
there!>Triangular Bezier surface patches seem to be a
promising choice to>interpolate my scattered points because
they can be easily controledto>meet some specific constraints
stemming from my application. Ofcourse the>triangular patches
must be joined together smoothly. And this iswhere
my>questions begin.>(Two of the books I read about the topic
are:>- Beach (1991):An Introduction to the Curves and Surfaces
of>Computer-Aided Design.>- Farin (1988): Curves and Surfaces
for Computer Aided GeometricDesign, A>Practical Guide.>In the
following I will refer to these books, so if you have
them>accidentally by hand ...)>Let us have two adjacent
triangles (-> sharing one common edge). Letthe>Bezier surface
be of bicubic or higher degree. We use barycentric>coordinates
for each triangle; also the control points are
labeledwith>three indeces, where the indeces for each
individual point sum to 3in the>bicubic and 4 in the biquartic
case. Let the control points havingone>index = 0 be placed in
R3 so that their projection on the x/y-planelies>on the same
projection of the corresponding triangle edge
(morecasually:>the control points at the triangle boundaries
are on their edges (in>x/y)). The control points subdivide the
original parent triangle into>subtriangles; a subset of these
subtriangles have one of their edgeson>one of the edges of the
parent triangle.>To allow the two original triangles to join
smoothly the subtriangles>adjacent to the common edge of the
parent triangles have to bemutually>coplanar. As Farin states
in his book (footnote in chapter 18.7) this>condition is
necessary, but not sufficient (... it does not even>guarantee
a continuous tangent plane.) The continuity between
the>original triangles does not have to be C1, I would be
satisfied withG1.>Still, both, Beach and Farin, write in their
books that thecoplanarity of>the subtriangle pairs at the
common edge is not sufficient for this>condition. >To find a
sufficient constraint for G1-continuity, Farin introduces
a>point x(t) on the common edge. (t takes values from 0 to 1
on thisedge.)>Using the de-Casteljau-algorithm, he determines
two adjacentsubtriangles>corresponding to x(t). Farin now
states that for each t between 0 and1>that the points of the
two subtriangles must be coplanar. Heformulates>this as>l p(t)
+ (1-l) r(t) = m qb(t) + (1-m) qr (t),>where qb(t) and qr(t)
are the end points of the common edge of thetwo>subtriangles,
p(t) and r(t) the remaining corners for eachsubtriangle.>The
interesting values are l, and m: If there exist l and m so
thatthe>equation above becomes true, the two subtriangles are
coplanar foreach t.>Now, l and m can be function of t as well
(-> l = l(t), m = m(t)). Tofind>an equation for the
G1-constraint, Farin lets l(t) and m(t) belinear.>My question
now is: What function other than linear and constant
are>suited for l(t) and m(t) to allow for G1-continuity? Could
the linear>function for l(t) and m(t) be replaced by quadratic
or higher degree>polynomials? (If yes, Farins footnote of
chapter 18.7 would not holdtrue,>i.m.h.o.)>I have another,
more pragmatic question: My triangular
Bezier-patchesare>biquartic, so I have three control points
along each edge, and each>triangle holds three interior
points. I have a method to determinethe>points along the
edges, but how should I proceed to determine
theinterior>points the way all the resulting subtriangle pairs
are in factcoplanar?>Which book or paper could show me usefull
approaches?>Thank you very much for taking the time to read
this. Since I am nota>trained matematician the descriptions
above may be sort of hairy.Still I>hope you understood what my
problems are.>Any hints and comments are appriciated.>(If you
know of other relevant news-groups I could send thisquestions
to>just let me know.)>Greetings from Switzerland,
>-------------------------------------------------------------
-------------------> Schneider tel.: ++411 257 52 56>Dept. of
Geography fax.: ++411 312 52 27>University of Zurich
e-mail:benni@geo.unizh.ch>
http://www.geo.unizh.ch/
~benni----------------------------------------------------
----------------------------> New Tel.-No starting August 1st:
++411 635 52 56> New Fax-No starting August 1st: ++41 1635 68
48>-----------------------------------------------------------
--------------------->-- > Schneider>Dept. of
Geography>University of Zurich>Winterthurerstrasse 190>CH-8057
Zurich (Switzerland)>tel.: ++41 1 257 52 56>e-mail:
benni@geo.unizh.ch>web: www.geo.unizh.ch/~benni===Subject: Re:
How do i find the equation of this curve> Litres 1/4 1/2 3/4 1
1-1/4 1-1/2 > Time (minutes) 9.85 7.83 7.52 7.42 7.02 6.98For
Time as a function of Litres, my best SSQ function is:Time = a
* (Litres-b)^ca = 7.1283998290225998317737E+00b =
2.3667636558827243042913E-01c =
-7.4873017560034629824806E-02For Litres as a function of Time,
my best SSQ function is:Litres = a * Time^(b * Time) + ca =
7.4237430144951122201746E+02b = -4.7343146918086520535951E-01c
= 2.1593913295217040393403E-01These are from the Function
Finder at my website, http://zunzun.com James Phillips
http://zunzun.com===Subject: Re: How do i find the equation of
this curveIf this is homework, read your book first, do the
work on your own, youwill learn it better,paper and pencil,
not keyboard.If not, You should repeat the experiment to find
another set of numbers tosee what the variances are,If highly
repeatable (within 1%) then you need a complex function. One
wouldexpect that if you are measuring by hand, that you would
be doing extremelywell to get it within 5%, more likely 10%.
If so, then a simple functionshould be used.l do.Assume your
second measurement set isLitters 1/4 1/2 3/4 1 1-1/4 1-1/2
Time (minutes) 10.30 9.33 7.60 7.30 7.20 6.80Then the function
you would want to use should be simpler, simpler modelstend to
reflect nature better.You can use the functions in Excel to do
this too.> Litres 1/4 1/2 3/4 1 1-1/4 1-1/2> Time (minutes)
9.85 7.83 7.52 7.42 7.02 6.98===Subject: Re: How do i find the
equation of this curve> Litres 1/4 1/2 3/4 1 1-1/4 1-1/2> Time
(minutes) 9.85 7.83 7.52 7.42 7.02 6.98What process is
producing this data? If you know whatform the relation should
have it makes the jobof fitting your data to a curve much
easier and meaningful.RonL===Subject: Re: How do i find the
equation of this curve >> Litres 1/4 1/2 3/4 1 1-1/4 1-1/2 >>
Time (minutes) 9.85 7.83 7.52 7.42 7.02 6.98 >What process
is producing this data? If you know what >form the relation
should have it makes the job >of fitting your data to a curve
much easier and meaningful.looks much like time=
a+b*exp(c*litres) a=.7110110139291032D+01b=
.8641139049141408D+01c= -.4627978801499478D+01at least the
plot looks reasonable hthpeter ===Subject: Re: How do i find
the equation of this curve> Litres 1/4 1/2 3/4 1 1-1/4 1-1/2>
Time (minutes) 9.85 7.83 7.52 7.42 7.02 6.98If possible, get
more data. The more data you get, the more accurate anycurve
that you would fit. As it stands, a curve that fits your data
willlikely not really represent the system you are trying to
model.Tim===Subject: Re: How do i find the equation of this
curveThere are an infinite number of curves that solve your
problem.There are many interpolation techniques to find a
curve that fit (orapproximate) your data, each of which gives
a different curve.After a quick plot, I couldn't see any
trivial curve that fits thosepoints...FernandoCherese
escribi.97 en el mensaje> Litres 1/4
1/2 3/4 1 1-1/4 1-1/2> Time (minutes) 9.85 7.83 7.52 7.42 7.02
6.98===Subject: Re: R_E_A_D T_H_I_S> For example, the list of
x,y coordinates> -10 100> -8 64> -6 36> -4 16> -2 4> 0 0> 2 4>
4 16> 6 36> 8 64> 10 100> is an approximation of the parabola y
= x^2, and should be easily> plottable.Pulse Peak also fits
this data set fairly well:y = 4.0 * a * e^(-(x-b)/c) * (1.0 -
e^(-(x-b)/c)) + da = -9.9320725379321840591729E+04b =
-2.1855502538402916457017E+02c = 3.1514531636498367106469E+02d
= 9.9320737224414318916388E+04Here is auto-generated C++
code:#include and obtain the steady state probability.>but
now i got NINE dimensional markov chain. could you plz give
some>suggestions to solve the set of equations or
multidimensional markovchain>(dimension exceed my
imagination)?>-->ZHANG Yan===Subject: Re: solution of
non-linear dynamical system?Originator:
jeyadev@kaveri>greetings!>i'm a biologist interested in the
dynamical system u[t]=1-exp(u[t-1])>starting from u[0]=1, in
case it matters. >Yes it matters! What would've happened if
you'd started with u[0]=0?Nothing? (Couldn't resist!)Surendar
Jeyadev jeyadev@wrc.xerox.bounceback.com Remove 'bounceback'
for email address===Subject: help! power iteration for
eigenvalue approximationi've been looking around for answers
to this problem but i just couldn't geta clue:i am using power
iteration method to compute the first k dominanteigenvalues and
eigenvectors of a symmetric matrix A. here is what i do:1. use
power iteration to get the dominant eigenvalue k1 and
eigenvector v1of matrix A2. residueA = A - k1 * v1 *
v1.transpose3. use power iteration to get the dominant
eigenvalue k2 and eigenvector v2of residueA4. ... repeat the
above until i have enough approximation termsfor some reason i
don't know, the eigenvalue produced by this process is notin
order i want. for example,A=[ 0 0 0 0 1 2 3 4 0 0 0 1 2 3 4 3
0 0 1 2 3 4 3 2 0 1 2 3 4 3 2 1 1 2 3 4 3 2 1 0 2 3 4 3 2 1 0
0 3 4 3 2 1 0 0 0 4 3 2 1 0 0 0 0 ];using matlab, the first
three eigenvalues are (in order of absolute value)14.0759,
-9.4306, 4.5685but with the power iteration method, i get them
in order: 14.0759,4.5685, -9.4306here is the problem picked
out:the residue matrix after first iterationresidue1=[-0.6592
-0.9584 -1.2071 -1.3525 -0.3525 0.7929 2.04163.3408-0.9584
-1.3933 -1.7550 -0.9663 0.0337 1.2450 2.60672.0416-1.2071
-1.7550 -1.2106 -0.4767 0.5233 1.7894 1.24500.7929-1.3525
-0.9663 -0.4767 0.2251 1.2251 0.5233 0.0337 -0.3525-0.3525
0.0337 0.5233 1.2251 0.2251 -0.4767 -0.9663 -1.3525 0.7929
1.2450 1.7894 0.5233 -0.4767 -1.2106 -1.7550 -1.2071 2.0416
2.6067 1.2450 0.0337 -0.9663 -1.7550 -1.3933 -0.9584 3.3408
2.0416 0.7929 -0.3525 -1.3525 -1.2071 -0.9584 -0.6592];its
eigenvalues are -9.4306, 4.5685 ... but the power iteration
method givesme 4.5685, which is not the dominant eigenvalue.so
i am confused what the power iteration is really doing, because
in somecases it finds the eigenvalue with maximum ABSOLUTE
VALUE, while in othersit finds the one with maximum VALUE. did
i do anything wrong? please helpme!!!rui===Subject: Fast
eigenvalue/eigenvector solutionsI need a fast module for
determination of the eigenvalues andeigenvectors of a 4x4 real
symmetric matrix. (It needs to be fastbecause it will be
running many times on a slow machine.)Usually, only the
largest two eigenvalues will be needed, andoccasionally the
eigenvectors corresponding to one of those twoeigenvalues, but
my study of the literature suggests that the gain inspeed in
computing only some of the eigenvalues is minimal.Additional
requirements: We must be able to find and publish (with
ourdocumentation) a proof-of-correctness for the
algorithm.===Subject: Re: Fast eigenvalue/eigenvector
solutions>I need a fast module for determination of the
eigenvalues and>eigenvectors of a 4x4 real symmetric matrix.
(It needs to be fast>because it will be running many times on
a slow machine.)>Usually, only the largest two eigenvalues
will be needed, and>occasionally the eigenvectors
corresponding to one of those two>eigenvalues, but my study of
the literature suggests that the gain in>speed in computing
only some of the eigenvalues is minimal.>Additional
requirements: We must be able to find and publish (with
our>documentation) a proof-of-correctness for the algorithm.I
do not know if this is any faster than the usual
recentmethods, such as QR with diagonal modification.One can
rather quickly reduce the matrix to tridiagonalform, using two
Householder transformations (one squareroot for each). Then the
characteristic equation canbe obtained, and the quartic solved.
Finding the vectors from the roots is easy, but if the roots
wantedare close to each other, or to unwanted roots, there can
be considerable roundoff error. Transforming backis also quick.
BTW, when there is bad roundoff error here, any othermethod
will also have it. However, well-behaved functionscombining
roots and vectors behave better under those.QR does maintain
that.All of this is in the old literature, around 1950,except
for QR.This address is for information only. I do not claim
that these viewsare those of the Statistics Department or of
Purdue University.Herman Rubin, Department of Statistics,
Purdue Universityhrubin@stat.purdue.edu Phone: (765)494-6054
FAX: (765)494-0558===Subject: Re: Fast eigenvalue/eigenvector
solutionsI would suggest looking at LAPACK (www.netlib.org) or
CLAPACK if you preferC.> I need a fast module for determination
of the eigenvalues and> eigenvectors of a 4x4 real symmetric
matrix. (It needs to be fast> because it will be running many
times on a slow machine.)> Usually, only the largest two
eigenvalues will be needed, and> occasionally the eigenvectors
corresponding to one of those two> eigenvalues, but my study of
the literature suggests that the gain in> speed in computing
only some of the eigenvalues is minimal.> Additional
requirements: We must be able to find and publish (with our>
documentation) a proof-of-correctness for the
algorithm.===Subject: Re: Fast eigenvalue/eigenvector
solutions> I need a fast module for determination of the
eigenvalues and> eigenvectors of a 4x4 real symmetric matrix.
(It needs to be fast> because it will be running many times on
a slow machine.)Bring the matrix to tridiagonal form and
applythe standard QR algorithm with double shift. This is very
fast(globally linear with convergence factor 0.7 or so,locally
typically supercubic convergence) and very reliable(stable and
foolproof). It will be hard to find anything better,even if you
only need the largest two eigenvalues.Maybe you can save a
little time in the reduction to tridiagonalform, by exploiting
that you have a 4x4 matrix and working out thereduction
symbolically (instead of using fast Givens rotations).And you
may save a little more by specializing Table 18.5.1 in thebook
by Parlett to the case n=4.Arnold Neumaier===Subject: Re:
matrices of revolution> The OP said rotate a vector around
another vector some number of radians. > How is that not a
general rotation?> General rotations are often given as a
sequence of separate rotations > about separate axes of
rotation, or in other similarly complex ways, > rather that as
a single rotation of a single vector about a single fixed >
axis.Oh. I think of that as the thing people do when they
don't know about quaternions and of quaternions as the general
way to describe a 3D rotation. I get what you meant,
though.meerohIf this message helped you, consider buying an
itemfrom my wish list:
2My formula after
a bunch of algebra including squaring both sides(although this
is wrong):4n^2 + 4n + 1 = n^3 + 3n^2 + 2n + 1I've spent over 2
hours stewing on this and am having a mental block onhow next
to proceed. If anyone can lend a hint on what step(s) Ishould
take, I'd much appreciate it. I'm missing the boat somewhere
onthis.thanks!===Subject: Re: please give a hint on this> Show
by induction that:> 1 + 2 + 3 + 4 + .. n = (1^3 + 2^3 + 3^3 +
4^3 + .. n^3)^1/2Same as n(n+1)/2 = (1^3 + 2^3 + 3^3 + 4^3 +
.. n^3)^1/2so prove [n(n+1)/2]^2 = (1^3 + 2^3 + 3^3 + 4^3 + ..
n^3)===Subject: Re: please give a hint on this Adjunct
Assistant Professor at the University of Montana.>Hello:>I'm
completely stumped and thought I had this figured out but
realized I>did not.>Problem is>Show by induction that:>1 + 2 +
3 + 4 + .. n = (1^3 + 2^3 + 3^3 + 4^3 + .. n^3)^1/2>I tried to
show for (n + 1) and>ended up with a formula (had to square
both sides) which worked for n=0,>n=1, and n=2. However, it
fails once n > 2Try showing that 1^3 + 2^3 + 3^3 + ... + n^3 =
(1+2+3+...+n)^2.You have proven it for n=0,1,2. Assume true for
k; then1^3 + ... + k^3 + (k+1)^3 = [1^3 + ... + k^3] + (k+1)^3
= (1+2+...+k)^2 + (k+1)^3You want to show this is equal to
(1+2+...+k+(k+1))^2 =(1+2+...+k)^2 + 2(1+...+k)(k+1) + (k+1)^2
= (1+2...+k)^2 + (k+1)[2(1+...+k) + (k+1)]so the result will be
proven if you can show that2(1+...+k) + (k+1) = (k+1)^2.Do you
know a formula for (1+...+k)?===========Subject: finding
coordinates of points on a circle by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
i2GEjXt20080;how can i find coordinates of as many as possible
points knowing thecentre and radius? how could interpolation be
helpful in this case?===Subject: Re: finding coordinates of
points on a circle by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
i2GFNot25386;>how can i find coordinates of as many as
possible points knowing the>centre and radius? how could
interpolation be helpful in this case?if (xc,yc) is the center
of the circle and r is its radius,the points (x,y) on the
circle satisfy the relation(x-xc)^2 + (y-yc)^2 = r^2.So if you
fix x in [xc-r ; xc+r] and set y1 = yc + sqrt(r^2-(x-xc)^2) and
y2 = yc - sqrt(r^2-(x-xc)^2), then (x,y1) and (x,y2) both lie
on the circle. Best wishesTorsten. ===Subject: Please help
thx:) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id i2GEjVD19969;Hi guys! I need help
for a final in my math92 course tomorrow and I amstuck on one
study prob in particular, so if someone could give me
theanswer and maybe show me how they got that would be
awesome. Problem: A police department knows that city growth
and the number ofburglaries are related by a linear equation.
City records show 585burglaries were reported in a year when
the local pop. was 67500, and685 were reported when the pop.
was 77500. How many burglaries can beexpected when the pop.
reaches 100,000? ===Subject: Re: Please help thx:)> Hi guys! I
need help for a final in my math92 course tomorrow and I am>
stuck on one study prob in particular, so if someone could
give me the> answer and maybe show me how they got that would
be awesome.> Problem: A police department knows that city
growth and the number of> burglaries are related by a linear
equation. City records show 585> burglaries were reported in a
year when the local pop. was 67500, and> 685 were reported when
the pop. was 77500. How many burglaries can be> expected when
the pop. reaches 100,000?You have two coordinates with with to
determine your slope and y-intercept.Think about your word
problem. Your X axis is the population and your Yaxis is the
number of burglaries for that population.City records show 585
burglaries were reported in a year when the localpop. was
67500...So here you have your first coordinate:
(67500,585)...and 685 were reported when the pop. was
77500.Here's your second coordinate: (77500,685)Now, take
those two points and determine your slope, then take one of
thecoordinates with your slope and use the point/slope formula
and you'll haveyour linear equation.===Subject: Re: Please help
thx:) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id i2GFZYk26671;>Hi guys! I need help
for a final in my math92 course tomorrow and Iam>stuck on one
study prob in particular, so if someone could give
methe>answer and maybe show me how they got that would be
awesome. >Problem: A police department knows that city growth
and the number of>burglaries are related by a linear equation.
City records show 585>burglaries were reported in a year when
the local pop. was 67500, and>685 were reported when the pop.
was 77500. How many burglaries can be>expected when the pop.
reaches 100,000? the equation of line through (67500|585) and
(77500|685) is given by(y-585)/(x-67500) =
(685-585)/(77500-67500) or simplifiedy = 585 +
0.01*(x-67500).Now insert x=100000 to get y=910.Best
wishesTorsten. ===Subject: Re: Beware undergrads, copiers of
your posts>
http://homepages.cwi.nl/~dik/english/mathematics/jsh.html>
James HarrisI never would've seen his webpage if you didn't
run around pointing itout all the time.Kcin===Subject: Re:
Beware undergrads, copiers of your postsi dont think you need
to worry about your stuff daryl. in case you haventnoticed,
you're the alt.math JOKE. nobody would want to take that
legacyaway from you.===Subject: Re: Beware undergrads, copiers
of your posts> He lives in the Netherlands. I guess they don't
care about copyrights> there, which surprises me since it *is*
a European nation, but then Europe> isn't what it used to
be.I'm sure there are plenty of people willing to mirror Dik's
pages in the US,if that would make you happier. :-)===Subject:
Re: AIR> This is an open chat room for Academy At Ivy Ridge
students only. No, this is a Usenet newsgroup,
alt.math.undergrad. It is not owned by anyuniversity or
organization and is open to everyone.> For> math questions
only, not for chatting but for getting answeres.Though this
newsgroup is under the alt.math.* hierarchy, it's for
undergradsand surely can accomodate more than math-specific
discussions.===Subject: Re: AIR> What is 2xy+365+xy+zy-z=?
(simplified)Combine like elements, factor it out and you'll
have your answer. :)===Subject: Re: Math, Avoiding burnout,
study tips recommended> I forgot to mention:> I'm a firm
believer in studying a little bit everyday (even if it's just>
1-2 hours) rather than taking 2-3 days off at a time and
cramming a> 6-hour study session. The approach of spending 1-3
hours daily and not> doing *any* overnighter study
cram-sessions has gotten me good grades> over the years.I'm
with you there. My routine is to either doing an hour in the
morning oran hour during lunch every day. Cramming for hours
on end doesn't help youlearn, and if you're only studying to
get through the test...well, that'sanother story entirely, eh?
:) > The only question there is if it's beneficial to take at
least 1 day off> per week ? Right now, that's something I have
not done.Working full-time and going to school part time, as
well as being a father,I don't know what a day off is, really.
For me, having a day to go tothe library with my kids and
just read text while they read chapter books*is* a day off.
Perhaps for you, being a full time student, a day off is
definitely needed.Your day off from studies would be the
equivalent of my day off fromprogramming. It helps keep
perspective and keeps you from getting burnedout, which you're
fearing now.===Subject: New source of the math!!!I have come
across a source providing good support to thoseexperiencing
difficulties with doing homework in maths
includingarithmetic(http://www.bymath.com/studyguide/ari/form1
.htm),
geometry(http://www.bymath.com/studyguide/geo/pro/pro1/pro1.
htm),
algebra(http://www.bymath.com/studyguide/alg/pro/pro1/pro1.htm
), functionsand graphics
(http://www.bymath.com/studyguide/fun/pro/pro.htm),principles
of
analysis(http://www.bymath.com/studyguide/ana/pro/pro1/pro1.
htm),all this is also supported by a good deal of examples
andillustrations.===Subject: Re: Mupad syntax
<1Ko3c.1895$Or1.84@news.chello.at> sobald man statt a eine
Zahl einsetzt z.B. 2 das Ergeniss nicht richtig> formatiert
wird.> also> simplify(2^v/2) Bug: a^v/a is actually
a^v*a^(-1), but 2^v/2 is actually 2^v*1/2(where 1/2 is a
rational number of type DOM_RAT) and combine andsimplify don't
realize this can be simplified. I suggest posting abug report
at www.mupad.de/bugs.html. +--+ +--+| |+-|+ Christopher
Creutzig (ccr@mupad.de) +--+ Tel.: 05251-60-5525===Subject:
Re: Sum evaluation with Wilf-Zeilberger method by
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id i2CDZe210830;Thank you very
much,this was exactly what i needed to prove the Lemma.The
function G can be simplified toG
=(-1)^(n-k+1)*k*m*binomial(m-k,k)*binomial(m-2*k,n-k+1)/((n+1)
*(m-n-1))Frank Haefner frankdothaefner@web.de===Subject:
Maxima vs MuPAD - computation times by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
i2DDCD425568;Hello All,My comparison of Maxima and MuPAD
includes now also computation times.They show that Maxima is
very fast: it is over ten times faster thanMuPAD. For details
seehttp://homepage.mac.com/peso1Pekka Sorjonen===Subject: Re:
Maxima vs MuPAD - computation times On this kind of general
mix of stuff, Macsyma should not beespecially fast, and I
would expect Maple to be faster. For thethings that they can
compute (which would be a small subset ofWester's test cases),
programs like NTL would be faster.Since Macsyma is one of the
oldest systems, it would be interestingto understand why one
of the newest systems should be so much slower.That is, what
did the designers of MuPad fail to notice about thedesign of
Macsyma?If the MuPAD designers say oh, we never looked at
Macsyma, wedesigned MuPAD from scratch then perhaps they
should lookNOW at the design.Do you suppose that MuPAD fell
victim to Greenspun's 10th law,Any sufficiently complicated C
or Fortran program contains an ad-hoc, informally-specified,
bug-ridden, slow implementation of half of Common Lisp. ?>
Hello All,> My comparison of Maxima and MuPAD includes now
also computation times.> They show that Maxima is very fast:
it is over ten times faster than> MuPAD. For details see>
http://homepage.mac.com/peso1> Pekka Sorjonen===Subject: Re:
Maxima vs MuPAD - computation times by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
i2FFeei00966;Thank you for informative comments. AFAIK MuPAD
is written in C andC++, so the 10th law may affect it.BTW,
version 2.1 is out. It contains a MuPAD-solution of the
problem88, thanks to Stefan Wehmeier;
seehttp://homepage.mac.com/peso1Pekka Sorjonen===Subject: How
can this be correct?Consider the following 2nd order
non-homogenous diff'l equation:mu'' + cu' + ku = FoCos(wt).if
the damping constanct c = 0, then, mu'' + ku = FoCos(wt).Now,
for the case where wo is not equal to w ( w = frequency), then
y = C1Cos(wot) + C2 Sin(wot) + (FoCos(wt))/m(wo^2-w^2), [1] is
a solution.Now, if we treat wo equal to w, then the above
solution is the same with theexeption of the last term. The
last term is FotSin(wt)/2mw, [2]. This isknown as
Resonance.Anyway, what is interesting is that you can apply
L'Hopital's rule to thelast term of [1] and get [2]. Try it!
Then take the Limit as wo approachesw. Any ideas as to why
this works when in fact the last term in [1] is
NOTindeterminate?Gord Bramfield.===Subject: Probability event
subset notationI am working on Prob. assignment on event
Independence.Where can I find a description of the index
terminology used to describe subsets?Ai1 - how should this be
interpreted{Ai1, Ai2, ...,Aik) k>=2The concept seems clear but
I can't understand the notation.Joe===Subject: Re:
Recommendation between two programs...It may be difficult to
obtain a good comparison between Mathematica andMaple because
people tend to devote themselves to one or the other. In
anycase, I have used Mathematica fairly intensively for 8
years and I am verysatisfied with it.I am a retired person and
have used it to study some modern mathematics andphysics. For
this purpose it has satisfied my every need. Mathematica
hasvery good functional language facilities, and also nice
rule basedprogramming. The function language is a very
powerful and mathematical wayof looking at calculations.One
thing that I have learned is that to study any particular
field or workon any particular application you will probably
have to program some of yourown routines. The mathematical and
scientific world is just too vast andvaried for any symbolic
program to have all the require routines built in.So which
program has the best programming facilities, and easiest to
learn,is probably an important consideration.David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/>
Hello:> I am looking at getting a math software program for
school. I am> currently a college student, and I was just
introduced to Maple 9 by> my Calculus instructor. WOW! I never
knew such a product existed.> Thinking I would do a little more
research on Maple 9, I actually> discovered Mathematica, IDL,
and Matlab (actually I was introduced to> Matlab as well, but
we didn't tool around with it).> Here is a little about me: I
am working toward a degree in> Biochemistry; I want to take as
many forms of higher maths as I can> (Calc II, III, etc.); I
have a little computer knowledge; price> doesn't matter too
much (I'm not rich but I could justify spending> about two
hundred dollars if I could find a neat and useful software>
application); my only other math tools that I've used were a
TI-83+ SE> (and a slide rule for fun). I am looking at buying
this type of> software as an aide in my studies and to double
check my work.> Right now, I am looking at either Mathematica
or Maple 9. Any help or> advice you offer would be greatly
appreciated.> Styron from North Carolina===Subject: ANN: Rhine
Workshop March 25, 26 NINTH RHINE WORKSHOP ON COMPUTER ALGEBRA
----------------- University of Nijmegen Nijmegen The
Netherlands -----------------The ninth Rhine Workshop on
Computer Algebra will take place at theMathematics Department
of the University of Nijmegen, the Netherlands,The programme
will run from 9 am on Thursday to around 4:00 pm onFriday, and
will include lectures, demonstrations, and a specialevening
programme on Thursday.For all information and registration,
please consult the Conference web pages
http://www.math.kun.nl/rwca04/===Subject: Hot piece of news
from Maplesoft by support1.mathforum.org (8.11.6/8.11.6/The
Math Forum, $Revision: 1.9 primary) id
i2GEjUI19938;kogeddes@scg.math.uwaterloo.ca (Keith Geddes)
(University of KOG> Summary of Maple Software Development, by
version release dates:KOG> Maple 1.0 (January
1982)[skipped]KOG> Maple 8 (May 2002)Hurrah! Therefore, Maple
9 does exist!I am certain, my feelings are shared by the
rejoicing throng of3,000,000 Maple corporate and individual
users over the globe.Vibrant with eagerness, I was wondering
when you would going todeliver it to the Maple customers
instead of the beta releases Maple 9, Maple 9.01, Maple 9.02,
Maple 9.03 ?Also, I am quite positive, Maplesoft will not ask
me or any othercustomer to pay for the shipment/downloading of
this long hoped-forMaple 9 you have just mentioned?Best
wishes,Vladimir BondarenkoGEMM architectCo-founder, CEO,
Mathematical DirectorCyber Tester, LLC13 Dekabristov Str,
SimferopolCrimea 95000, Ukrainetel: +38-(0652)-447325tel:
+38-(0652)-230243tel: +38-(0652)-523144fax:
+38-(0652)-510700http://www.cybertester.com/http://
maple.bug-list.org/http://www.CAS-testing.org/................
....................................................===Subject
: approximately equal to by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
i2GGBbn31946;Why is there a almost equal to and approximately
equal to what isthe difference?se unicode Mathematical
Operatorshttp://www.unicode.org/charts/PDF/U2200.pdf===Subject
: Mathcad question about finding then polynomial coefficients
of numerator and denominator. I do a lot of symbolic math that
envolves transfers function that have apolynomial in the
numerator and denominator. I need a easy way to get
thecoefficients of the polynomials from both the numerator and
the denominator.At this time I copy the numerator and
denominator to separate equations anduse the
tools->polynomials command. However, this is tedious and
manual.Is there a function, command or trick that can do this
more automatically?It would be great of there was a numer()
and denom() function but I can fundone.Peter Nachtwey