A46
==
Try in this manner, I think that it
is right:In[1]:= matr = {{0, 1, 2}, {1, 2, 3}}In[2]:= f = x m
+ pOk?***********************************************Stefano
FricanoDip. di Scienze Fisiche ed AstronomicheUniversit.88
degli StudiVia Archirafi, 36Palermo
90123*********************************************
====
Dear
Narasimham,Try in this manner, I think that it is
right:In[1]:= matr = {{0, 1, 2}, {1, 2, 3}}In[2]:= f = x m +
pOk?***********************************************Stefano
FricanoDip. di Scienze Fisiche ed AstronomicheUniversit.88
degli StudiVia Archirafi, 36Palermo
90123*********************************************
====

Johannes,I took a brief look at your problem, and I have a
couple comments. I agreethat there is no reason that FindRoot
should be slower in Mathematica thanin any other program, as
the functionality of FindRoot is not complex.First, I
couldn't get your function Probit to work with your data 
set.
Ihave not tried to figure out the problem here.Second, and 
more
importantly, your Probit function uses the nongradient formof
FindRoot, since it provides two starting points for each
variable insteadof just one. With two starting points for
each variable, FindRoot will nottry to evaluate the gradient,
and will instead use a root finding techniquewhich does not
rely on gradient information. Of course, using
FindRootwithout gradient information is much slower than
using FindRoot withgradient information, and I believe this
is the root of your problem.I'm guessing that you used the
two starting point, nongradient form ofFindRoot because
Mathematica is unable to find the gradient of your functionby
itself, and so Mathematica returned an error when you tried
usingFindRoot with only one starting point. However, when
using the two startingpoint, nongradient form of FindRoot,
gradient information is not needed, sosupplying a gradient to
FindRoot does nothing. Try supplying a gradientwhile using
only one starting point for each variable.Carl WollPhysics
DeptU of Washington
====
please check syntax for upper limit I
did not use it before this way.st1 = Solve[((Integrate[#1,
{t, 0,  cannot be followed by [Tau]}] & ) /@ .
====
fix
this with an option.]This forces Export to use a different
version of the W3C Universal style default because we have
seen problems with this stylesheet on some
browsers.-Dale
====
[...]Let's first define your 
special
product:In[34]:= comp[a_, b_] := Thread[Unevaluated[a*b],
List, -1]In[35]:= comp[{1, 2, 3}, {4, 5, 6}]Out[35]= {{4, 8,
12}, {5, 10, 15}, {6, 12, 18}}To decompose, my suggestion
is:In[36]:=decomp[m_] := Module[{a, b = GCD @@@ m, d}, d =
m/b; a = First[d]; Catch[ b = If[And @@ SameQ @@@ #, #[[All,
1]], Throw[$Failed]] &[a/# & /@ d]* b; {a, b}]]Catch and
Throw here only is to get a decent error. If you can assert
thedecomposability you may dispense with that. Let's
try:In[37]:= decomp[{{-4, -8, 12}, {5, 10, -15}, {6, 12,
-18}}]Out[37]= {{-1, -2, 3}, {4, -5, -6}}verify:In[38]:= comp
@@ %Out[38]= {{-4, -8, 12}, {5, 10, -15}, {6, 12,
-18}}In[39]:= decomp[{{-4, -8, 12}, {5, 10, 15}, {6, 12,
-18}}]Out[39]= $FailedThis one was not decomposable.You might
like to take the steps:In[40]:= m = {{-4, -8, 12}, {5, 10,
-15}, {6, 12, -18}};In[41]:= b = GCD @@@ mOut[41]= {4, 5,
6}Not quite the second factor, but we get the missing signs
if we divideIn[42]:= d = m/bOut[42]= {{-1, -2, 3}, {1, 2,
-3}, {1, 2, -3}}All elements of this list should be the same
except for a sign ±1. We arbitrarily take the first 
element
as the first factor (this gives theother solution of your
example).In[43]:= a = First[d]Out[43]= {-1, -2, 3}We want to
get the signs:In[44]:= a/# & /@ dOut[44]= {{1, 1, 1}, {-1,
-1, -1}, {-1, -1, -1}}In each element there should be all 1
or all -1. (This was tested above.)In[45]:= %[[All,
1]]Out[45]= {1, -1, -1}In[46]:= %*bOut[46]= {4, -5, -6}This
is now the second factor of your special product.--Hartmut
Wolf
====
I want to give notice, that I made available an new
version of my oo systemfor Mathematica on my web
site:www.schmitther.de.The following features were added to
the system:- the possibility to store objects on external
media(serialization/deserialization).- the possibility to
create classes in the programs.- additional functions, which
make available information about classes andobjects.Hermann
Schmitt
====
Yves,There are some cases where responders might
wish to respond privately or askfor more information and it
is very slow doing this day by day through thenewsgroup.The
regular Convert routine will simplify your
expression.Needs[Miscellaneous`Units`]Needs[Miscellaneous`
PhysicalConstants`]Sqrt[x*Meter^2]Convert[%, Meter]
gives...Sqrt[Meter^2*x]Meter*Sqrt[x]To use prefixes, you must
leave a space between the prefix and the unit.Convert[1500.
Milli Meter, Yard]1.64042 YardUnfortunately, when one goes
the other way, Mathematica sorts the symbols togive an
unnatural order.Convert[0.05Yard, Milli Meter]45.72 Meter
MilliIn using units, I think that it is often best to never
put the units in theequations. Put the units in the data.
Then you can manipulate and solve yourequations symbolically
without units ever showing, and substitute the dataWITH the
units at the end. Here is a very simple example. What height
will astone reach in 20 seconds if dropped from 2
miles.Without introducing units...v[t_] = Integrate[-g, {t,
0, t}]-g tx[t_] = x0 + Integrate[v[t], {t, 0, t}]-((g*t^2)/2)
+ x0Now we can introduce our data WITH the units attached to
obtain the answer.x[t] /. dataConvert[%, Mile] giving-1961.33
Meter + 2 Mile0.781286 MileThere are two packages at my web
site below, V4ExtendUnits andV4ReducedUnits, that extend the
Units package. Its ToUnit function is alittle more convenient
to use than Convert. One can also format the outputat the same
time as converting it and there is a new output form
calledPrefixForm. It automatically picks a proper 
prefix and
puts the output in aHoldForm to maintain the customary order.
So in the example
above...Needs[Miscellaneous`V4ExtendUnits`]45.7200 Milli
MeterThere are many other useful features. With the regular
Units package youcan't convert ßoating point quantities 
to
Degrees because Mathematica madethe mistake of making Degrees
both a constant and a unit. With ExtendUnitsyou can convert to
degrees. There is a BaseSI command that convertseverything to
base SI units. The SI command does not convert everything
tocommon units. You can convert to multiple units at the same
time, and alsoparse a quantity into descending compatible
units (for example, Day, Hour,Minute, Second). It will handle
units in DiracDeltas, UnitStep and ArcTan.To do numerical work
or plotting it is necessary to remove units one way oranother.
The package has a Deunitize function that will deunitize
tostandard unit systems. There is also a more controlled
function that allowsone to specify the implied input and
output units.The package also allows you to install your own
units, (GeV for GigaElectronVolt, or mps for Meter/Second,
say). The ReducedUnits package allowsthe introduction of
unit systems where certain physical constants are takenas 1.
These would include such systems as geometrized units or
atomic units.David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/Am I
missing something? Is there a way to get the units
properlysimplified?2) There is Meter and Centimeter
predefined. Where have the Millimetersgone? I did not succeed
in prepending any of the SI unit prefixes. Could I simply use
MilliMeter? Did not really work for me...Yves
====
First of
all, thanks to David and Bill!David Park schrieb:I agree. My
yk@despammed.com is totally functional, though. Really nice
spam filtering service there (Until they 
havenÇt stopped this
weekend, IÇll check this... no, works just 
fine).Again,
agreed:-) I tried to do so anyway, but got annoyed by
unsimplified units in the results.Again, thanks! I will check
out your packages...Yves
====
Narasimham,Here is one
method...Needs[Graphics`Colors`]Here is an array of slopes
and y-axis intercepts.mc = {{0, 0.2, 0.4, 0.6}, {-0.1, 0.1,
0.2, 0.3}};The following statement generates a table of
functions for the lines.functions = MapThread[#1 x + #2 &,
mc]{-0.1, 0.1 + 0.2*x, 0.2 + 0.4*x, 0.3 +
0.6*x}Plot[functions // Evaluate, {x, -1, 1},David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/
Sender: steve@smc.vnet.net
====
Hi all,Does anyone why f@a_=2 a
gives a proper answer for f[1]and not in the case of
Prefix[s[a_]]= 2 a eventhough that
Prefix[s[a_]]=f@a_???In[1]:=f@a_=2 af[1]Out[1]=2
aOut[2]=2In[3]:=Prefix[s[a_]]Prefix[s[a_]]=2
as[1]Out[3]=s@a_Set::write: Tag Prefix in s@a_ is
Protected.Out[4]=2 aOut[5]=s[1]Reply-To:
kuska@informatik.uni-leipzig.de
====
Hi, becausePrefix[] *print*
the function f[expr] in the form f @ expr.And printing an
expression is quite the oposite of parsing/reading a
expression. Jens
====
In a fit of inspiration/boredom, I just
made a palette that some of you might like. The buttons on it
are as follows:Clear All -- executes
ClearAll[Global`*]Remove All -- executes
Remove[Global`*]Quiet Spell Warnings -- executes
Off[General::spell, General::spell1] (By the way, why the
heck are there two kinds of spell messages?)Conserve Memory
-- executes $HistoryLength=10 and Share[]Unshadow -- executes
a version of Ulrich Jentschura's unshadowing functionLoad
Graphics Packages -- executes <<Graphics`Load Stat Packages
-- executes <<Statistics`and finally...Kern Text -- this scans
the current notebook and in each text cell replaces fi and 
ß
with ligatures and puts thin negative spaces between pairs of
letters such as ff, To, Vo, We, and so on. (ff should be a
ligature too, but for some reason I couldn't convince
Mathematica of its existence.)If you're interested, download
the palette here: http://www.appliedsymbols.com/Toolbar.nb .
I'll appreciate any suggestions about additional buttons 
that
would fit in.Selwyn Hollis
====
If I understand what you
wantPlot[Evaluate[b.{x,1}],{x,0,10}] should work.Here, I am
assuming b is a matrix of real numbers with the first column
containing the slope and the second the intercept.Perhaps a
bit nicer would beThis will plot each line in a different
color ranging fron red to blue on a light gray
background.
====
You wouldn't want Mathematica to simplify
Sqrt[x^2} to give x. Why not -x?Unless, of course, you are
really sure that x is positive. In this case useTomas
GarzaMexico City----- Original Message -----
====
According to
WRI tech support a bug exists in
Developer`SparseLinearSolve[].I recently contacted WRI about
another problem I was working on in
whichDeveloper`SparseLinearSolve[] was giving different
results to LinearSolve.They stated that a bug existed in the
current version however in thedevelopment version it is
apparently corrected.I don't know how widely this error will
effect other peoples problems butfor the previous problem I
was working on (which was an asymmetricalmatrix) it certainly
made using it impossible. In my problem an instabilitysomehow
arose. I was solving an equation where the output was the
input atthe next time step. Within a few time steps instead
of returning valuesbetween 0. and 1. I was getting 10^6 and
so on.LinearSolve gave good results but slow because of the
large matrix size.Mike
====
a={2,3,5}c={3,5,3}t={-1,3,5}I am
using Table[a*k+b*c+v*t,{k,1,2},{b,1,2},{v,1,2}]to generate
some points. After this I need to plot the points by using
3Dcommands like Scatter3DPlot.How can I get the points from
Table in the form of a list so that I can
plotthem?.Reply-To:
kuska@informatik.uni-leipzig.de
====
Hi,Show[Graphics3D[{
AbsolutePointSize[4], Point /@ Flatten[Table[a*k + b*c + v*t,
{k, 1, 2}, {b, 1, 2}, {v, 1, 2}],2]} ]]it is Faltten[_,2] that
help you. Jens
====
The following code fragment will plot the
points for you:a = {2, 3, 5};c = {3, 5, 3};t = {-1, 3, 5};x =
Table[a*k + b*c + v*t, {k, 1, 2}, {b, 1, 2}, {v, 1,
2}];Show[Graphics3D[{PointSize[0.03],Map[Point,Flatten[x,2]]}
]];The Flatten[x,2] part of this is the list that you
want.--Steve LuttrellWest Malvern, UKplot
====
The last time I
needed to work with an animated gif, I used a utilitycalled
GifBuilder (developed by Yves Piguet) with good results. This
wasfor a Mac, pre-OS X.Ken LevasseurMath. Sci.UMass
Lowell
====
Alexander,In your definition of p and p1, there are
parameters a and m, which you haven't specified. 
What should
those be?---Selwyn Hollis
====
just some
numbersa=1,m=23,e=0.1,el=30;
====
John,Why is this patch
specifically for Mathematica 4.2? Or does it patch 4.1 as
well?---Selwyn Hollis
====
Ashraf,I think it is because Prefix
is a formatting function that affects how theexpression is
displayed. But the prefix header is still 
there.Prefix[s[a_]]%
// FullForms@a_Prefix[s[Pattern[a, Blank[]]]]So when you try
to make an assignment with it, Mathematica objects
becausePrefix is protected.On the other hand...s@a_% //
FullForms[a_]s[Pattern[a, Blank[]]]and you can make an
assignment because s is not protected.I have to admit that
this is a little tricky and confused me at first.David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/2
aOut[2]=2In[3]:=Prefix[s[a_]]Prefix[s[a_]]=2
as[1]Out[3]=s@a_Set::write: Tag Prefix in s@a_ is
Protected.Out[4]=2 aOut[5]=s[1]
====
Studentversion 4.2,KDEI
generated a Plot,right-clicked into it,and tryed some of the
conversionswhen suddenly the only reaction of my systemwas to
mouse-MOVEMENTS .Not even Cntrl-Alt-Del got me a rebootof my
machine,I had to use the resetbutton.That's not very
nice.
====
Searches of past discussion yielded a couple of
threads on the topic ofwaterfall plots a couple of years ago,
but nothing too helpful.I'm just looking for hints as to 
which
fundamental plot command to startfrom.Reply-To:
kuska@informatik.uni-leipzig.de
====
Hi,can you tell us what a
waterfall plot is ? Jens
====
I'd like, in a .m file for a
package, to be able to know what directorythe .m file is being
loaded from, so I can search in the same place(or
subdirectories) for related files. I've looked 
through
everythingto do with directories I can find in The Book, but I
can't find anythingthat looks as though it would 
help.
John
====
Many thanks, this is exactly what I was
hoping/looking for. John
====
John,Although I don't recall ever
having problems with it, I am sure it notcompletely reliable.
The usage message makes this
clear:packageSourceDirectory[package] returns the list of
directories where thesource of the first currently loaded
package containing package in its context string might
be located.If you are interested you can go
tohttp://jowett.home.cern.ch/jowett/Mathematica/and download
the notebook defining the BumpEtc utility package.Perhaps
someone can improve on this function ?John Jowett
====
Your
*.m file normally is saved and resides in what ever
specificdirectory your notebook was opened and resides in. One
way to tell is to goto open the notebook in question and do a
file save as and you should beable to tell what location
you are working in from the displayed
pathinformation.shippee@jcs.mil
====
Sure, but my question was
not how *I* can find out, but how the .m file 
Johngobe
====
Maybe
this isn't what you want, but it seems to me the quickest
thingto do would be to make a Table which evaluates your
function at your 5
====
Yaw,All you need is two levels of
Flattening.Needs[Graphics`Colors`]Needs[Graphics`Animation
`]a = {2, 3, 5}c = {3, 5, 3}t = {-1, 3, 5}pts =
Flatten[Table[a*k + b*c + v*t, {k, 1, 2}, {b, 1, 2}, {v, 1,
2}], 2];plot1 = Show[Graphics3D[ {AbsolutePointSize[4], Point
/@ pts}],SpinShow[plot1]SelectionMove[EvaluationNotebook[],
All, GeneratedCell]FrontEndTokenExecute[OpenCloseGroup];
Pause[0.01];David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/
====

Bruce,Picking up from Allan Hayes
suggestion...http://forums.wolfram.com/mathgroup/archive/1998
/May/msg00424.html...you could do something like this. First
we extract the graphicsprimitives for a surface without
displaying it.surface = First[ParametricPlot3D[{x, y,
Cos[x]}, {x, 0, [Pi]/2}, {y, 0, [Pi]/2},Then we display
the surface with a blank EdgeForm to get rid of the
polygonedges and then extract the Polygons from the surface
and draw the lines fromtwo of the sides.Show[Graphics3D[
{EdgeForm[], surface, surface /.With the DrawGraphics package
from my web site, we could do this in onestatement and add a
watery color to the
surface.Needs[DrawGraphics`DrawingMaster`]Draw3DItems[
{SurfaceColor[ColorMix[LightSkyBlue, PaleGreen][0.3]],
EdgeForm[], surface = ParametricDraw3D[{x, y, Cos[x]}, {x, 0,
[Pi]/2}, {y, 0, [Pi]/2}], surface /. NeutralLighting[0.3,
0.7, 0.0, -30Á],David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/Sender:
steve@smc.vnet.net
====
I'm interestied in using Mathematica to
do something similar to this:
http://www.mapleapps.com/powertools/logos/appess3.gifDoes
anyone have a guess at what kind of rotation is being
used?And how one might get a similar shadow effect? (Without
resorting to a different rendering engine.)SelwynReply-To:
kuska@informatik.uni-leipzig.de
====
Hi,a) computing the true
shadows of the 3 light sourceswould requite the construction
of the shadow volumesand take an half hour per frameb) fake a
single shadow is easy
withNeeds[Graphics`Polyhedra`]Needs[Graphics`Shapes`]
toShadow[gray_?MatrixQ, {x1_, x2_}, {y1_, y2_}, z_] :=
Module[{n, m, dx, dy, points, cgraph}, {m, n} =
Dimensions[gray]; dx = (x2 - x1)/(n); dy = (y2 - y1)/(m);
points = Table[{x1 + dx*i, y1 + dy*j, z}, {j, 0, m}, {i, 0,
n}]; poly = Drop [#, -1] & /@ Drop[Transpose[{points,
RotateRight[points], RotateLeft /@ RotateRight[points],
RotateLeft /@ points},{3, 1, 2}], -1]; {EdgeForm[],
Transpose[{Map[SurfaceColor[GrayLevel[#]] &, gray,{2}],
Map[Polygon, poly, {2}]}, {3, 1, 2}]} ]makeShadow[t_,
opts___?OptionQ] := Module[{sh}, sh = DensityPlot[ 1 -
Exp[-(3 + Sin[t/2])*(x^2 + y^2)], {x, -2, 2}, {y, -2, 2},
toShadow[sh[[1]], {-2, 2}, {-2, 2}, -2] ]obj =
Table[RotateShape[Icosahedron[], t, t/2, t/3], {t, 0, 8,
1/10}];MapIndexed[obj];but that is *not* the shadow that the
three colored light sources wouldproduce -- that would be
much more expensive.If some one else would like to have
(hard) shadows I can include it into Jens Ashraf,the question
is: what behaviour did you expect? An answer to that
mightsolve your problem.First, Prefix[s[a_]] is *not* the same
as f@a_ In[1]:= f@a_ := 2 aIn[2]:= ?f Global`f f[a_] := 2
aIn[3]:= f[1]Out[3]= 2 Compare this to In[4]:=
Prefix[s[a_]]Out[4]= s@a_In[5]:= % //
FullFormOut[5]//FullForm= Prefix[s[Pattern[a, Blank[]]]]So
Prefix is a wrapper for printing purposes (only). Such you
cannot makedefinitions for it (as it is protected), but also,
it is a special form notsubmitted to the standard evaluation
sequence, see:In[7]:= Prefix[s[aaa___]] ^= s[aaa]Out[7]=
s[aaa]In[8]:= s[a_] := 2 aIn[9]:= ?s Global`s
Prefix[s[aaa___]] ^= s[aaa] s[a_] := 2 aIn[10]:=
Prefix[s[2]]Out[10]= Prefix[4]So even Upvalues 
won't work. You
may however doIn[12]:= Prefix[Unevaluated[s[2]]]Out[12]= 4The
very question, however, is whether this is what you intended,
or say,what did you want to achieve at all? --Hartmut
Wolf
====
That is more than just a shadow. There appears to be
diffraction around theedges as well. But it is pretty
cool.Kevin
====
One way (not necessarily the best) would
beFlatten[Transpose[Flatten[r,1]],1]where r =
Table[a*k+b*c+v*t,{k,1,2},{b,1,2},{v,1,2}]
====
how to
calculate Spherical Harmonic expansion Coefficients?I am
looking for an algorithm possibly for mathematica.please
helpthanks,AReply-To:
kuska@informatik.uni-leipzig.de
====
Hi,of what ? Of an
analytical known function
?Sum[a[l,m]*SphericalHarmonicY[l,m,th,phi],{l,0,Infinity},{m,-
l,l}]with a[l,m]=Integrate[
Conjugate[SphericalHarmonicY[l,m,th,phi]]f[th,phi]*Sin[th],
{th,0,Pi},{phi,0,2Pi} ] Jens
====
Perhaps if you were to give
an example of the type of problem you areinterested in
solving we could better help you.Kevin
====
Raster would be a
good starting point for waterfall
plots.records=50;samples=128;w=2. Pi 5; (* fundamental
angular frequency of made up data *)Clear[f];f[t_]:=Cos[w
t]+Cos[3 w t]/3+Cos[5 w t]/5+Random[Real,t]/60.; (* some
harmonic data with noise added *)samplerate=100; (* sampling
rate Hz
*)data=Partition[Table[f[i/samplerate],{i,0,samples*records-1
}],samples]; (* time series data partitioned into records.We
could have windowed here if desired *)(* Plot the time series
data next*)(* Then make a list of power spectra, one of each
record:
*)powerspectra=Map[Take[#,samples/2]&,Map[Abs[Fourier[#]]&,
data]];(* Only half the spectra is taken to remove the folded
over portion *)(* here is a simple frozen waterfall plot
*)You could then add windowing to the data records and a
proper frequency scale in Hz, and insert Logs and
normalization to get the intensity in greyscale or color
keyed to a decibel scale. Next you can animate this, so you
see the waterfall growing in height on your screen, using the
animation ideas discussed in this forum over the last year or
so, and you would have what you seek.startChristopher
PurcellDRDC-Atlantic, 9 Grove St., PO Box 1012,Dartmouth, NS
Canada B2Y 3Z7chris.purcell@drdc-rddc.gc.ca
====
John,There is
a nice trick for this. It works on every operating system and
all versions of Mathematica from at least 4.0 onward (and
probably earlier). It uses the undocumented function
System`Private`FindFile, but if that ever goes away we'll
have to add something to take its place because J/Link uses
this exact technique to find its application resources.Put a
line like this in your .m file: thisPackageDirectory =
DirectoryName[System`Private`FindFile[$Input]]The variable
thisPackageDirectory will get the directory name in which the
.m file resides no matter how the user reads the package, 
e.g.:
<<MyPackage.m <</full/path/to/MyPackage.m <<MyPackage`
Needs[MyPackage`] etc.This trick also works if you use the
standard Mathematica application directory structure (highly
recommended), wherein you have a directory named for your
package (say MyApp): Applications/ MyApp/ MyApp.m ... other
files and dirs ... init.m <-- this file reads 
simply:
Get[MyApp`MyApp`]This standard application layout allows
users to enter simply Needs[MyApp`]to load your MyApp.m
file. The thisPackageDirectory variable would then get Todd
GayleyWolfram Research
====
How to perturb the constant f in
parametric plots? f= 1 ;' for single f 
value'xy= { 2 f
Sin[t]^2 ,Sinh[f t]+f Log[t] };ParametricPlot[ xy ,
{t,0,2}];An attemt to Map did not work.' f={ 
.5,1,1.5};''trj
= Map [ xy &, f ];'--Reply-To:
kuska@informatik.uni-leipzig.de
====
Hi,something
like:13];NotebookPut[nb] Jens
====
Simon,Two ways:Firstly-
select the text that you want to set to font size 13.This
will cause the text to be 13 point on screen and in print.You
can select all text cells at once by holding down the Alt key
andclicking on one of them.SecondlyTo automatically print in
13 point while keeping to the default 12 point onscreen
:environment PrintOut. - set the font size of this cell to
13 as done above. - close the style
sheet.--Allan---------------------Allan HayesMathematica
Training and ConsultingLeicester
UKwww.haystack.demon.co.ukhay@haystack.demon.co.ukVoice: +44
(0)116 271 4198thesisReply-To:
kuska@informatik.uni-leipzig.de
====
Hi,number.You say it
already use Simplify[]/FullSimplify[] with a second
argument,i.e.,Unprotect[Simplify]Protect[Simplify]Convert[1
Milli Meter, Meter]Meter/1000and Notice the space between
Milli and Meter Jens
====
Mathematica treats the units as
simply another symbol which does not have an assigned value.
Mathematica does not simplify Sqrt[x^2] to x since there is
no way to know which branch is desired.There are a couple of
ways around this. Instead of Simplify you could use
PowerExpand as inPowerExpand[Sqrt[x Meter^2]] or Sqrt[x
Meter^2]//PowerExpand.However, Mathematica outputs Meter
Sqrt[x] probably not exactly what you want.A better choice
would be to use Ted Ersek's Units2.m package that can be
found on the Mathsource site. Alternatively, David Park as
written a units package that extends the package in the
standard distribution. David's package offers a different
feature set than Ted's. For my use, I prefer 
Ted's
package.Millimeter is there but needs to be expressed as
Milli Meter (note the space). The units package predefines
basic units in both the SI and CGS systems. Other units are
expressed as by using the appropriate pre-fix in front of the
basic unit. Millimeter is a derived unit in both the CGS and
SI systems so is not pre-defined in the same way as
Centimeter.Reply-To:
kuska@informatik.uni-leipzig.de
====
Hi,Export[img.gif,{
graph1,graph2,..},??Look in the additional information part
of the Online documentation ofExport[] Jens
====
There is a
MacOS X version called GifBuilder carbon. I occasionally
use version 1.0. It does not seem to have it's own web site
but one can easily find it by doing a Google search.Andrzej
KozlowskiYokohama,
Japanhttp://www.mimuw.edu.pl/~akoz/http://
platon.c.u-tokyo.ac.jp/andrzej/
====
Dear Mr. Woll,I tried to
follow your advice to make the jacobianfaster. Below you find
the optimized code. (You haveto past it into Mathematica.
Otherwise it's unreadable).Now I would be really surprised 
if
it could be improvedfurther, since the code is really short
and makes extensiveuse of the listable attribute and matrix
algebra operators.Nevertheless, evaluation time has increased
only by a smallamount.BUT MY ORIGINAL QUESTION IS OPEN:
MATHEMATICA STILLREMAINS to be slower by a factor of 12 than
its competitors. Johannes!((pdf = Function[{x},
Exp[(-(x^2/2))]/@(2
[Pi])];)[IndentingNewLine] (cdf = Function[{x},
0.5 ((1. + Erf[x/Sqrt[2.0]]))];)[IndentingNewLine]
SetAttributes[pdf, Listable]; SetAttributes[cdf,
Listable];)!(LogLikJacC = Compile[{{y, _Integer, 1},
{x, _Real, 2}, {b, _Real, 1}}, Module[{p, c, z = Map[If[#
[Equal] 1, 1, (-1)] &, y] x . b},
[IndentingNewLine]p = pdf[z]; c = cdf[z]; (-Transpose[ x]
. ((x (( p c z)(+)(p^2)(
))/c^2)))]])Johannes LudsteckEconomics
DepartmentUniversity of RegensburgUniversitaetsstrasse
3193053 Regensburg
====
To whom may concern,I have a problem
with Mathematica. I have installed Mathematica 4.2 on
Windows XP. I need to run a program in Mathematica, best
regard.Hung Viet.
====
Dear group, I have m balls shared in n
box. What is the probability that one box has r o more
balls?If m = 4, n=3 and r = 2 the solution will be (?): In[1]
:= Needs[DiscreteMath`Combinatorica`]In[2] := list =
Compositions[4,3];In[3] :=
(Count[list,2,2]+Count[list,2,3]+Count[list,2,4])/(3
Length[list])The question is that this way works only when m
and n are small , but for values sush as m=20, n=11 and r=5
the computer is out of memory. Any
idea?GuillermoSanchez----------------------------------------
-----This message was sent using Endymion MailMan.
====
Hey all
- I'm taking a basic calculus course that uses Mathematica. 
We
have been studying limits and I have been using the Limit
functionto check if my answers are correct.We were given the
following function and asked to determine a limit:(3E^(-x) -
E^(-3x)) / (E^(-3x) + E^(-x))Usually the approach is to
select the dominant terms, factor and thendetermine the
limit. My initial reason had me select -E^(-3x) in
thenumerator and E^(-3x) in the denominator. Factoring the
terms wouldand the real answer is somewhere near 3.I then
tried to use the Limit function which is not producing
ananswer (perhaps I'm not sure of the usage).Any help is
greatly appreciated, David Seruyange
====
HiHow can I tell
Mathematica to pull a particular factor out of an
expression.For example in this expression:{(a -
2*wone*Sin[a*t])/4, (a - 2*wone*Sin[a*t])/4, (a +
2*wone*Sin[a*t])/4,(a + 2*wone*Sin[a*t])/4, (-a - 2*Sqrt[a^2
+ wone^2*Sin[a*t]^2])/4, (-a - 2*Sqrt[a^2
+wone^2*Sin[a*t]^2])/4, (-a + 2*Sqrt[a^2 +
wone^2*Sin[a*t]^2])/4, (-a + 2*Sqrt[a^2
+wone^2*Sin[a*t]^2])/4}I would like to bring the a out to
the front of the expression, evenoutside the matrix.How can I
ask Mathematica to do this, and how can I stop the a
justßoating back into the main expression?YoursChris
RodgersSt John's Collegehttp://rodgers.org.uk/
====
 ...But
again it is wrong! It only is correct if a and b happento be
real quantities, which is nowhere stated!So the main question
still is: Why is Mathematica makingthese very silly errors?
One could expect it from an earlyversion of a product, but
that is not what Mathematica 4.xis!So could it be, perhaps,
that Mathematica has now evolved sofar that it has its own
kind of humor, and is making theselittle jokes to entertain
us? :-))-- Jo
====
 I'm sorry. Not ÔAssumption' but
ÔAssumptions' Out[2]=0.679391 This work using 
version 3.0 for
Windows-- marian otremba
====
Hi Alex:The notebook below may be
of some help. Just copy and paste into a Mathematica
notebook.hw
====
===================Notebook[{Cell[
CellGroupData[{Cell[BoxData[ (TraditionalForm`Expansion
in Spherical Harmonics)], Subtitle,Cell[BoxData[
(TraditionalForm`Basics)],
Subsubtitle],Cell[<Spherical harmonics are a set of
complex orthonormal functions defined on the surface of the
unit sphere. Mathematica furnishes a built-in function for
their symbolic and numerical computation. Let [Theta] and
[CurlyPhi] be the usual spherical polar and azimuthal
angles, respectively, which sweep the entire surface for (0
< [Theta] < [Pi]) and (0 < [CurlyPhi] < 2[Pi]).
These functions may be specified in Mathematica by
Cell[BoxData[ (TraditionalForm`(Y(l_,
m_))[InvisibleApplication] (([Theta]_,
[CurlyPhi]_)) := (Y_l%m)([Theta],
[CurlyPhi]))], Input],Cell[<Here the index l is
any positive integer zero or greater while the second index
m is restricted, for a given value of l, to one or another
of the values -l <= m <= l. Spherical harmonics are
customarilydefined so that the complex congugate of Y[l,m]
is given by (-1)^m Y[l, -m], and to have the orthogonality
property that Integrate[(-1)^m Y[l ,
-m][[Theta],[CurlyPhi]] Y[l', m'] 
Sin[[Theta]]
{[Theta], 0, [Pi]}, {[CurlyPhi], 0, 2[Pi]}] gives
zero unless both conditions l = l' and m = m' 
hold. If
these conditions hold the integral yields 1. These are the
conditions of orthonormality. After first executing the
preceding function to load the program that computes a
spherical harmonic, you might then verify the properties
described above by executing the following
cell.Text],Cell[CellGroupData[{Cell[BoxData[
(TraditionalForm`Expansion of a Function)],
Subsubtitle],Cell[<The properties given above permit
a given function of the spherical polar angles to be
represented as a (generally infinite) sum of of spherical
harmonics, The {l, m)th expansion coefficients an arbitrary
function f may be calculated byCell[BoxData[
(TraditionalForm`(coef(l_,
m_))[InvisibleApplication] ((f_)) :=
[Integral]_0%[Pi]([Integral]_0%(2
[Pi])(((-1)))^m (Y( l,
(-m)))[InvisibleApplication] (([Theta],
[CurlyPhi])) f (sin([Theta]))
[DifferentialD][CurlyPhi]
[DifferentialD][Theta]))], Input],Cell[BoxData[
(TraditionalForm`2 l of set The + a
coefficients for given is l such then)],
Text],Cell[BoxData[ (TraditionalForm`(coefset(
l_))[InvisibleApplication] ((f_)) := Table[(coef(l,
m))[InvisibleApplication] ((f)), {m, (-l),
l}])], Input],Cell[<For example, the expansion set
for the function Cos[[Theta]]^2 corresponding to l = 0,
1, and 2 areCell[BoxData[ (TraditionalForm`TableForm[
Table[(coefset( l))[InvisibleApplication]
(((cos^2)([Theta]))), {l, 0, 3}]])],
Input],Cell[<You can check for yourself that these
coefficients for this particular function are 
sufficient to
furnish the original function exactly. This is certainly
not true in general. Try some other f[[Theta],
[CurlyPhi]]. }, Open ]]}, Open ]]},]
====
Since I don't
really expect the right answer here, let me pose the
situation as it looks from here: I have to pay for an upgrade
in order to use a patch to fix a bug. Way to go!---Selwyn
Hollis
====
I am trying to use mathematica to define moment
generating function ofa multivariate normal distribution with
mean 0 and variance T,[N(0,T)]. I would like to be able to 
find
the nth moment for thisdistribution.For simplicity I would
like define m = exp[1/2 b'T b]= f(b), where bis 
a q*1 vector
with elements (b1 b2 b3 b4 ... bq) (b' is 1*q), and Tis a 
q*q
matrix with elementst11 t12 t13 ... t1qt21 t22 t23 ... t2q . .
. . . . . . . . . .tq1 tq2 tq3 tqq I have looked through the
online manual and have only found out how todefine a matrix or
vector of specific integer size.I have found the 
first moment
which disapears at b=0 to be m*b'*T.I have also found the
second moment to be m*T + m*T*b*b'T.I am having a very hard
problem finding the next moment (let alone thenext ten). I
would like to use mathematica to get all the momnets Icare to
look at. I would be greatful for any help!!Chris
Johnsoncjque@umich.edu
====
Hi Nigel,thanks for your comment.
Indeed, your assumption is the crux. Thedistribution of the
points is not even about the circle. Therefore your{x1,y1} is
not the centrepoint of the circle. It is a good assumption for
astartpoint for an iterative procedure. Up to now, we are
using this point tocalculate mean and standard deviation as
you, then we move {x1,y1}a littlebit in any direction and
calculate mean and standard deviation again. Is thestandard
deviation better as before, we move again in the same
direction.Otherways we move back and then in an orthogonal
direction. This is taken aslong as the change in the
standard deviaton is greater then a limit. When wehave a
look at the fitted circel then we think it is not the best 
fit
whatwe calculate in this way.Dieter--Dieter
Palmedl7udpdistributed about the circle
====
The title subject
was the source of perplexing behavior in a long script of
over 3000 lines (a complete FEM program). It tooks hours to
trace. Only the relevant portion is illustrated:c=1/2;
exp=Cos[psi]/2-2*Sin[psi];Print[exp//InputForm];
Print[exp//FullForm];Cos[psi]/2 -
2*Sin[psi]Plus[Times[Rational[1,2],Cos[psi]],Times[-2,Sin[psi
]]]Here it comes:Print[exp//InputForm];
Print[exp//FullForm];3*Cos[psi] -
2*Sin[psi]Plus[Times[3,Cos[psi]],Times[-2,Sin[psi]]]the
scenes. OK, program mistake. I should have cleared c, part of
a long replacement list left over from legacy code).
Conclusion: a constant can be silently replaced by
another,warning. Here the intent is obvious. But when the
constantis hidden behind a symbol, this can be the source of
bugs difficult to find, akin to self-modifying 
code. Is there a
way to automatically mark all constant subexpressions as
protected against /. or //.? Sort of a automatic const
declaration. That would enhance program safety.
====
First of
all, you are using = both in the meaning of Set (=) and Equal
(==). This is not a good idea.Secondly, Mathematica cannot
perform semantic pattern matching and will not search for
ways to do the replacement with b. So your best bet is
something like:Sin[1 + x] + x*Sin[1 + x] /. Solve[b == 1+x,
x][[1]]Sin[b] + (-1 + b)*Sin[b]Andrzej KozlowskiYokohama,
Japanhttp://www.mimuw.edu.pl/~akoz/http://
platon.c.u-tokyo.ac.jp/andrzej/
====
Use ReplaceAll,
ieIn[4]:=Out[4]=b Sin[b]See the help browser for further
information on ReplaceAll and HoldPattern.Sseziwa
====
In[1]:=a
= (1 + x)*Sin[1 + x]; In[2]:=Out[2]=b*Sin[b]In[3]:=a = Sin[1 +
x] + x*Sin[1 + x]; In[4]:=Out[4]=Sin[b] + x*Sin[b]Tomas
GarzaMexico City----- Original Message ----- 
====
On 12/27/02
at 2:15 AM, noplace@nowhere.no (Diligent Sp*m Avoider)One way
to accomplish this would be to use a replacement rule as
in:This will clearly handle more complex expressions in x.
But it doesn't assign a value to b
====
There is one basic
thing to remember. Mathematica replaces matching patterns but
not algebraic expressions:2 + x^(-2)2 + a^(-1)2 + x^(-2)2 +
a^(-1)2 + x^(-4)2 + y^(-2)To understand why some work but
others dont you have to look at the way expressions are
internally stored, using FullForm.Sometimes //. (replace
repeated) helps:1/Sqrt[1 - x^2] + y1/Sqrt[y] + yIdeally there
should be an algebraic replacement operatordifferent from /.
or //. The technology to do it is there:optimizing compilers
have used it since the late 1950s.
====
Jan,One technique is to
use a function I programmed called ImplicitSolve.Basically,
ImplicitSolve takes a set of N equations in N+1 unknowns,
where Nof the unknowns are dependent variables and one of
the unknowns is anindependent variable, and then solves for
the N dependent variables interms of the independent
variable. I use quotes because the choice ofwhich variable is
the independent variable is up to you. The output
ofImplicitSolve is a set of Interpolating functions, one for
each depedentvariable. In your question, you have two
equations and 3 unknowns. So, wecould use ImplicitSolve as
follows:In[7]:=Out[7]=Notice that one has to supply a root as
a starting point, since in generalthere can be many different
solutions, and ImplicitSolve needs to know whichone to
pick.Now that we have interpolating functions for y and z, it
is a simple matterto use ParametricPlot3D to plot your
curve.ParametricPlot3D[{t, sol[[1, 2]][t], sol[[2, 2]][t]},
{t, 0, 0.5}]I haven't included the generated plot, 
obviously.
You will just need to doso yourself.Now all you need is the
definition of ImplicitSolve, which I give
below:Clear[ImplicitSolve]:= If[x0 == {},
Message[ImplicitSolve::badroot, x]; Return[], x0
=x0[[1]]];x0], Evaluate[Sequence @@ Transpose[{y, y0}]],
opts][[All,2]], Null, Message[ImplicitSolve::incomplete];
Return[]];Thread[(#1[x0] & ) /@ y == y0]}], y, {x, min, max},
opts,xmax}, opts] finds a solution to the implicit equations
eqns for the functions y, ... with the independent variable
x in the range xmin toxmax. The root {x0,y0,...} should
satisfy the equations, or should provide a good starting
point for finding a solution when usingFindRoot. Options which
can be supplied with opts include those which can be
provided to FindRoot or NDSolve.;ImplicitSolve::badroot =
Supplied root is missing value for
`1`;ImplicitSolve::incomplete = Supplied root is
incomplete;ImplicitSolve::inaccurate = Supplied root is
inaccurate, using FindRoot toimprove accuracy;If you
commonly use StandardForm, then when pasting the above
function intoMathematica, it is a good idea to change the
default input format toInputForm first, and afterwards change
the default back to StandardForm.Then, convert the cell to
StandardForm. This prevents Mathematica frominterpreting
spaces at the end of lines as multiplications, causing a
wholebunch of Nulls to ouput, and possibly creating
errors.Carl WollPhysics DeptU of Washingtona space curve
defined as the intersection of the surfaces x^2 +y^2 +z^2 ==
1and z == Sin[x y].Jan M. (~_~)
====
There's nothing general
about it, but this may help reveal thecomplexity of such
problems (and the structure of the one you
mention):Off[Solve::svars, Root::deg]{x^3 - 3x y^2 - x^2
y^4 == 1, 2 x^3 - 5y z^4 + 3 z^2 == 2};solns = Solve[%, {x,
y}];solns2 = solns //. {solns[[1, 1, -1, -1, -1, -1, 1, 1,
xValues = solns2[[All, 2]]There are 14 solutions, each
involving y as the same function of x butwith x egual to one
of 14 roots of a certain polynomial withcoefficients depending
on z. (14th degree in x, and 16th degree in z)When z = 1, here
are the x values:N@(Root[Evaluate@solns[[1, 1, -1, -1, -1, -1,
1, 1, 1]] &, #] /. {{-1.4675791391058204 -
0.16275444114403523*I, -1.4675791391058204 +
0.16275444114403523*I, -0.983729467938907 -
1.1441614374058868* I, -0.983729467938907 +
1.1441614374058868*I, -0.513798649597158 - 0.6816748778475*I,
-0.513798649597158 + 0.6816748778475*I, 0.14830009529459443 -
1.3072652114458407*I, 0.14830009529459443 +
1.3072652114458407*I, 0.5076727899147009 -
1.4571487512553851*I, 0.5076727899147009 +
1.4571487512553851*I, 0.9900958185680836 -
0.35714916082429093*I, 0.9900958185680836 +
0.35714916082429093*I, 1.3190385528645066 -
0.7907735960845482*I, 1.3190385528645066 +
0.7907735960845482*I}None of these are real, so that's not a
good value of z. When z = 100,there are two real
solutions:Cases[N@(Root[Evaluate@solns[[1, 1, -1, -1, -1, -1,
1, 1, 1]] &, #] /. zValue) & /@ Range[14],_Real];5.80015}}Have
fun with it.Bobby-----Original Message-----Please help! Jan M.
(~_^)In general I believe it will be difficult to solve for 
the
intersection.Butin your specific case it is easy to make the
plot. Substitute yoursecondequation into the first equation to
obtain...eqn1 = x^2 + y^2 + z^2 == 1;x^2 + y^2 + Sin[x*y]^2 ==
1Now we use Mathematica's ImplicitPlot command to plot the
curve in thexy-plane.Needs[Graphics`ImplicitPlot`]plot1
=ImplicitPlot[eqn2, {x, -1, 1}, {y, -1, 1}];We convert the
resulting ContourGraphics to Graphics, take the
firstpart,which contains all the Line primitives, and extract
all the lines withtheCases statement.lines2d =
Cases[First[Graphics[plot1]], Line[_], Infinity];We then
convert all the 2D points to 3D points using a rule and
yoursecondequation to obtain the z coordinate, and display
the resulting curve in3D.lines2d =
Cases[First[Graphics[plot1]], Line[_],
Infinity];Show[Graphics3D[This can be done a little more
easily with the DrawGraphics package atmyweb site below
becuase the ImplicitDraw routine automataically
extractstheprimitive graphics and there is an explicit
RaiseTo3D
command.Needs[DrawGraphics`DrawingMaster`]Draw3DItems[{
ImplicitDraw[x^2 + y^2 + Sin[x y]^2 == 1, {x, -1, 1}, {y, -1,
1}] //RaiseTo3D[Sin[#1 #2] &]}];David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/
====
i
posted this on the student forum for quite a while but unable
to get anyresponds.filesize is getting too large when i have
imported few (relatively small, ~200Kb per picture) diagrams
or pictures.and comes the BIG problem, its my college acc
only allows 5Mb of personalstorage, and i cant work with
mathematica because the notebook has todecompressed first. AND
it would becomes 22Mb after decompression.i wonder is there
any command to reduce or compress the notebook file sizewithin
mathematica, or perhaps, alternative ways of attaching
images(without getting a LARGE notebook)here is the original
post:Hi, anyone ever come across this, after importing few
graphic files (jpg ortif), the notebook size becomes almost
10~20 times of it original size.i have one notebook with
around 15 pictures, it takes around 22Mb!! (thoseare just
pictures to go along with the theories, and it aint big, only
1/4of A4 paper and the original size is around 200kB per
file).i have tried to convert the picture into bitmap format
(within Mathematica,i found that it is originally in
postscript format after been imported), itdoes helps to
reduce the notebook file size (tremendously!), but the
graphicresolution turns out to be so dreadful till it
couldnt even worth a look onthe hardcopy.please advise.
====
[I
was not previously subscribed to this newsgroup, so apparently
I can'tattach to a thread more than a month old.Response to
Andrzej Kozlowski - OOP in Mathematica - 5 Nov 2002as well
as Orestis Vantzos - Re: OOP in Mathematica - 6 Nov
2002][Andrzej] -(one is the Object-Oriented Graph Theory in
Chapter III of Grey's MasteringMathematica),but all of
them can be programmed with no more effortusing the natural
Mathematica functional-pattern based style.Andrzej, that
is quite a claim.I do not wish to join a meaningless argument
of opinion.I would like to see some evidence.*** I would be
very happy to learn natural Mathematica ways to
accomplishvarious OOP objectives.If you can actually help me
and others do that, rather than just spoutingoff (like Mr. Jen
does) about how superior the Mathematica way is, then
thatwould be a great contribution.I will be presenting small
snippets of code from Java or Eiffel, anddiscussing how to
accomplish the result in Mathematica.I am well aware that
doing so may involve a complete re-arranging of how Ithink
about programming.If so, help me see it that
way.---------------Background:I have many years experience
with OOP in C++, Eiffel, and Java.I now have a practical
project that I would like to accomplish usingMathematica:I
wish to use Mathematica as a software modeling environment,
somewhatlike the software industry has standardized on as UML
(Unified ModelingLanguage).But rather than using UML, I would
like to use / help develop amathematically rigorous modelling
and simulation environment in Mathematica,which is intended
to ultimately spit out code in Java, Eiffel, C++, and/orC#.(I
won't ignore UML where relevant - but UML has gotten rather
unwieldy. Iwish to build up from a few basic principles -
which we can discuss as thisproject unfolds.)The reasons for
spitting out code in those other languages:* performance.*
tight binding to existing large class libraries or other
existing code.* many companies ARE going to program in those
languages, rightly orwrongly - it would be a great benefit to
the industry to have amathematically valid way to develop
software that ultimately ends up inthose languages.I am
starting with Java, because of a 3D application which I am
developing.I am suggesting an Open Source OOP modelling and
simulation environment bedeveloped on Mathematica.I would not
presume to host such a development, until I understand
whatother projects have been done / are being contemplated,
by others.However, if no one else steps forward, I will take
that on. Presumably withcode at SourceForge.net.I have used
Mathematica previously on other projects, but am just
nowbeginning to explore how to represent OOP concepts within
it.I am about to look at the two code examples I see
references to. (RomanMaeder) (Hermann
Schmitt).-----------------------------------------------
Orestis - I see your post of 4 Nov 2002, OOP in Mathematica
- A concreteexample.With your delegation implementation of
Objects. You accomplish things I didnot know how to do in
Mathematica, so I will study closely.[Orestis] - I will be
happy to discuss issues like
Inheritance,Encapsulation,Dynamic Binding, etc. in future
posts.I very much wish to discuss those issues.Bertrand
Meyer's Design By Contract is central to my methodology, so
Iwish to discuss how to implement that as well.-- Steve S.
Shaw (a.k.a. ToolmakerSteve)
====
Steve,Well, it took me a few
days to catch on to how to extrapolate list data fromthe
table output. I was finally able to take your program, and
apply it tothe Cayley Table for A_5.To keep the size to a
minimum for the newsgroup post, I have pasted mynotebook file
to export the Cayley Table for A_4.The notebook command looks
like:Export[C:cayley4table.csv, {{1, 2, 3, 4,
5, 6, 7, 8, 9, 10, 11, 12}, {2, 3, 1,
7, 9, 8, 10,  12, 11, 4, 5, 6}, {3, 1,
2, 10, 11, 12 , 4, 6, 5, 7, 9, 8}, {4,
6, 5, 1, 3, 2, 11,10 , 12, 8, 7, 9},
{5, 4, 6, 8, 7, 9,  1, 2, 3, 11, 12,
10}, {6, 5, 4, 11, 12, 10,  8, 9, 7,
1, 3, 2}, {7, 8, 9, 2, 1, 3,  5, 4,
6, 12, 10, 11}, {8, 9, 7, 5, 6, 4,
12,11 , 10, 2, 1, 3}, {9, 7, 8, 12,
10, 11 , 2, 3, 1, 5, 6, 4}, {10, 12,
11, 3, 2, 1, 9, 7, 8, 6, 4, 5}, {11,
10, 12, 6, 4, 5 , 3, 1, 2, 9, 8, 7},
{12, 11, 10, 9, 8, 7,  6, 5, 4, 3, 2,
1}}, CSV]Dianaexample,
====
Hi,I wish to define a function
as a table lookup. Here's an example: myFunc[i_Integer] =
Table[n, {n, 1, 10}][[i]]so when I invoke it with an index
argument i: myFunc[3]it returns the 3rd element in the
(precomputed) table (n.b.,precomputed, due to defining
function with Set(=) rather thanSetDelayed(:=) ).This works,
but nonetheless, upon defining the function, Mathematica
returnsthe error message: Part::pspec: Part
specification i is neither an integer nor alistof integers.I
suppose I'll simply ignore this message, but is there some
bettermeans of defining a table lookup function?
====
>
NDSolve seems to have difficulties with solving integral
equation.> n = 5; NDSolve[{D[[Sigma]norm[z, t], t] ==
3*z*Integrate[[Sigma]norm[z,> t]^n*z, {z, 0, 1}] -
[Sigma]norm[z, t]^n,> [Sigma]norm[z, 0] == 1.5*z,
[Sigma]norm[0, t] == 0}*[Sigma]norm[z,> t], {z, 0.01,
1}, {t, 0.01, 2}]> Mathematica returns a message>
NDSolve::deql: The first argument must have both an
equation and an > initial condition.> which I cannot
understand.> Can anybody tell what's wrong with my attempt?>
> -Toshi> I am not really sure that your question isn't a
joke.NDSolve solves differential equations, not integral
equations.Although there are relations between these topics,
they are certainlynot the same.Alois-- Vienna University of
Technology,
====
evaluating the following gives you a sample
x=x+1 button:NotebookPut@Notebook[{Cell[BoxData[
ButtonBox[(x = (x + 1)), RuleDelayed[ButtonFunction,
CompoundExpression[If[Not[ ValueQ[x]], Set[x, 0]], Set[x,
Plus[x, 1]]]], Rule[ButtonEvaluator, Automatic]] ],
NotebookDefault, PageBreakAbove -> True, CellTags ->
GeneratedButtonBoxx=x+1]}, ClosingAutoSave -> True,
Editable -> False, WindowToolbars -> {}, PageWidth -> 299.5,
WindowSize -> {89., 29.}, WindowMargins -> {{92., Automatic},
{Automatic, 56.}}, WindowFrame -> Palette, WindowElements ->
{}, WindowFrameElements -> CloseBox, WindowClickSelect ->
False, ScrollingOptions -> {PagewiseScrolling -> True},
ShowCellBracket -> False, CellMargins -> {{0., 0.},
{Inherited, 0.}}, Active -> True, CellOpen -> True,
ShowCellLabel -> False, ShowCellTags -> False, ImageMargins
-> {{0., Inherited}, {Inherited, 0.}}, Magnification -> 1.5](*
********************************* *)Now, how do you create
such a button in less than a minute? ... :One way is to just
create a section cell and the underlying ButtonFunction
code as input cells, i.e., type interactively such that you
get something like:NotebookPut[Notebook[ {Cell[CellGroupData[
{Cell[x=x+1, Section], Cell[If[!ValueQ[x], x=0],
Input], Cell[x=x+1, Input]}, Open]]}]]Then hit the
F2B (function to Button) button in ButtonTools.nb ( my
freewarebutton tools from http://www.mertig.com/mathdepot )
and you get the button.With the HP and VP you can easily
and quickly generate (horizontally orverically) palettes.
Check out the Help button, or also the source code.It
basically is all straightforward and there is actually
documentation about all those ButtonFunction features
somewhere. I agree that the whole Button-design could have
been made better, but up to a point is quite useful. Of
course the world is used to better GUI's these days but if
you really need nice GUI's and buttons, use Java and JLink (
and there are also simple examples in the JLink manual of how
to do this ).If you don't like Java, go with VBA and use the
nice Mahematica for Active Xproduct from
http://www.episoft.comRolf MertigMertig
Consultinghttp://www.mertig.com
====
tryIn[1]:=Clear[a,b,c,d,x,
y]x= {{0,0,0,1},{1,0,0,1},{0,1,0,1},{1,1,1,1}};y = {a,b,c,d};
LinearSolve[x,y]s01=
LinearSolve[x,y][[1]]Out[4]={-a+b,-a+c,a-b-c+d,a}Out[5]=-a+
bIn[6]:=g[a_,b_] =s01Out[6]=-a+bIn[7]:=g[1,3]Out[7]=2*NEVER*
use capital letters at the beginning of a variable's name!
Never!Matthias BodeSal. Oppenheim jr. & Cie.
KGaAKoenigsberger Strasse 29D-60487 Frankfurt am
MainGERMANYMobile: +49(0)172 6 74 95 77Internet:
http://www.oppenheim.de-----Ursprí.b9ngliche
Nachricht-----Gesendet: Freitag, 23. August 2002 06:25An:
mathgroup@smc.vnet.netBetreff: := Does not assign variables
properly.Why?Here's a piece of a conversion I had with
Mathematica.Why is a[A_,B_] := LinearSolve[X,Y][[1]] not
givingme the function I expect?In[261]:= X =
{{0,0,0,1},{1,0,0,1},{0,1,0,1},{1,1,1,1}};In[262]:= Y =
{A,B,C,D};In[263]:= LinearSolve[X,Y]Out[263]=
{-A+B,-A+C,A-B-C+D,A}In[264]:= LinearSolve[X,Y][[1]]Out[264]=
-A+BIn[265]:= a[A_,B_] := LinearSolve[X,Y][[1]]In[266]:=
a[1,3]Out[266]= -A+BThe output above is not what I want. I
want 2. Here'swhat I expect:In[267]:= a[A_,B_] :=
-A+B;In[268]:= a[1,3]Out[268]= 2This output is what I expect.
What is the difference betweenthe two?
====
First of all, just
look at your own posting below. You clearly have a *(Times)
where a , (comma) should be. Presumably your input ought to
be:n = 5; NDSolve[{D[[Sigma]norm[z, t], t] ==
3*z*Integrate[[Sigma]norm[z,t]^n*z, {z, 0, 1}] -
[Sigma]norm[z, t]^n, [Sigma]norm[z, 0] == 1.5*z,
[Sigma]norm[0, t] == 0},[Sigma]norm[z,t], {z, 0.01, 1},
{t, 0.01, 2}]However, even in the corrected version the
equation can't be solved. First of all you will get the
complaint:NDSolve::bcedge: Boundary conditions must be
specified at the edge of the spatial domain.In other words
Mathematica wants a boundary condition for [Sigma]norm[z,
0.1] or alternatively you should use {z,0,1} in NDSolve. But
actually I do not think this equation is solvable by any
numerical scheme even if you could provide the initial
conditions at the edge of the boundary that Mathematica
requests. To evaluate the integral in your equation NDSolve
needs to know the values of [Sigma]norm[z, t] for all z
between 0 and 1 and a given t, but this knowledge is not
available at any stage of the evaluation. I am not really an
expert, but this seems to me a clear example of an equation
that is not solvable by any numerical means.By the way, the
fact that you know a solution to a differential equation, and
even the fact that the solution is very simple does not imply
that the equation can be solved by any known method, except
of course guessing, which computer programs generally do not
use. Andrzej KozlowskiToyama International UniversityJAPANOn
Friday, August 23, 2002, at 05:25 AM, Toshiyuki ((Toshi))
Meshii >> NDSolve seems to have difficulties with solving
integral equation.>> n = 5; NDSolve[{D[[Sigma]norm[z, t],
t] == > 3*z*Integrate[[Sigma]norm[z,> t]^n*z, {z, 0, 1}] -
[Sigma]norm[z, t]^n,> [Sigma]norm[z, 0] == 1.5*z,
[Sigma]norm[0, t] == > 0}*[Sigma]norm[z,> t], {z, 0.01,
1}, {t, 0.01, 2}]>> Mathematica returns a message>>
NDSolve::deql: The first argument must have both an
equation and an > initial condition.>> which I cannot
understand.> Can anybody tell what's wrong with my 
attempt?>>
-Toshi>
====
>Here's a piece of a conversion I had with
Mathematica.>Why is a[A_,B_] := LinearSolve[X,Y][[1]] not
giving>me the function I expect?>>In[261]:= X =
{{0,0,0,1},{1,0,0,1},{0,1,0,1},{1,1,1,1}};>In[262]:= Y =
{A,B,C,D};>In[263]:= LinearSolve[X,Y]>Out[263]=
{-A+B,-A+C,A-B-C+D,A}>In[264]:=
LinearSolve[X,Y][[1]]>Out[264]= -A+B>In[265]:= a[A_,B_] :=
LinearSolve[X,Y][[1]]>In[266]:= a[1,3]>Out[266]= -A+B>>The
output above is not what I want. I want 2. Here's>what I
expect:>>In[267]:= a[A_,B_] := -A+B;>In[268]:=
a[1,3]>Out[268]= 2>>This output is what I expect. What is the
difference between>the two?X={{0,0,0,1},{1,0,0,1},
{0,1,0,1},{1,1,1,1}};Y={A,B,C,D};a[A_,B_]:=
LinearSolve[X,Y][[1]];?aGlobal`aa[A_, B_] :=
LinearSolve[X,Y][[1]]Note that the RHS of the stored
definition is not a function of the arguments. Now add
Evaluate to RHSa[A_,B_]:=
Evaluate[LinearSolve[X,Y][[1]]];?aGlobal`aa[A_, B_] := -A +
Ba[1,3]2Bob HanlonChantilly, VA USA
====
All of this looks like
a mistake to me because it seems far too easy. But anyway,
here is the solution that makes almost no use of Mathematica.
First of all, your equation is not a differential equation so
there is no point using DSolve.Secondly the use of z in
Integrate[[Sigma]norm[z]^n*z, {z, 0, d}] is deceptive,
since you are integrating over z, so let's replace it by
something else, say s. So your equation
is:(3*z)/d^3)*Integrate[[Sigma]norm[s]^n*s, {s, 0, d}]
==[Sigma]norm[z]^nwhich is supposed to hold true for every
z>0. Re-write it asIntegrate[[Sigma]norm[s]^n*s, {s, 0,
d}]/d^3 =[Sigma]norm[z]^n/3zfor all z. However, the left
hand side is a function of d, independent of z, so we can
write:[Sigma]norm[z_]:=(3z*g[d])^(1/n)Let's take this as a
definition and substitute in the original
equationIn[2]:=Simplify[((3*z)*Integrate[[Sigma]norm[s]^n*s
, {s, 0, d}])/d^3 == [Sigma]norm[z]^n, {d > 0, n > 0, z >
0}]Out[2]=TrueThat means you can take g to be an arbitrary
function of d.Andrzej KozlowskiToyama International
UniversityJAPANOn Friday, August 23, 2002, at 05:25 AM,
Toshiyuki ((Toshi)) Meshii > How can I solve the following
integral equation?> Mathematica seems not to work.> Is there
any way?>> DSolve[((3*z)/d^3)*Integrate[[Sigma]norm[z]^n*z,
{z, 0, d}] ==> [Sigma]norm[z]^n, [Sigma]norm[z], z]>>
note: z>0 & n>1>> I know that the answer is simple and>
$B&R(Bnorm[z_] = (1 + 1/(2*n))*(z/d)^(1/n)>>
-Toshi>
====
But suddenly the following Errormessage
appears:/usr/bin/local/mathematica: file or directory not
found(Maybe its more a UNIX-Problem or has to do
withMandrake-updates ... but so far everything exceptof
Mathematica seems to work fine.)Of course I checked for the
File (it is indeed there) and (a hint from a Unix-usegroup)
the needed libs libc.so.5, libm.so.5 are in /lib/ too.Had
anyone had, ore better solved, a similar problem?greetings
Detlef
====
How can I create a Mathematica thing (function?
program?) that wouldautomatically open a browser page, and
give my username and pwd to log meinto a https:.... site?
The server's login page has a script 
resettingfields: <script
language=javascript> function ResetFormFields() {
document.Login.username.value = ;
document.SearchUserForm.search.value = ;
document.FindCourseForm.searchText.value = ; } </script>
<body . . . ONLOAD=ResetFormFields();
document.Login.username.select();
document.Login.username.focus();>and the relevant input
cells are<input type=text name=username . . . ><input
type=password name=password . . . ><input type=submit
name=login value=Login>Nicholas
====
Howdy,I'm trying to
figure out the correct syntax to do the following. I havesome
function with three arguments, and I want to syntactically
describe thesingle-argument function that holds two of those
arguments constant (i.e.without creating that single-argument
function).More specifically, I have defined
Machine[radix_,multiplier_,state_] := Module [{c,s}, c =
Floor[state/base]; s = Mod[state,base]; multiplier*s + c
]where I have a generalize Ômachine', 
defined by the radix and
multiplier,which converts one state into another state. So 
I'd
like to be able to dosomething like this:
NestList[Machine[10,7,#], 3, 22]to get the series of states
that the radix-10 multiplier-7 machine runsthrough (starting
with state 3). However, this syntax doesn't seem to dowhat I
want.I hope that description makes sense. It seems like there
must be a syntaxto describe the function
Machine[10,7,#].Anyone have any ideas?Bob H
====
>>There is a
following problem with Mathematica 4.2:>when i try to load
Help Browser, i get a message:>building help browser index
(first time only)>scanning index file>and 
Mathematica stops
responding: you have to>kill process. There was no such a
problem with>version 4.1.>>Is there is a solution to 
that?I'd
start with the following
FAQ.http://support.wolfram.com/mathematica/interface/
helpbrowser/howrebuildindex.html-Dale
====
>I'd start with the
following
FAQ.>>http://support.wolfram.com/mathematica/interface/
helpbrowser/howrebuildindex.html>>-DaleSorry that i failed
say it immediately in a first place, but of coursei did try it
FAQ at first, and both tried to delete cache and rebuildindex,
but results where the same - whenever i try to invoke help
browser (or rebuild index, for that matter), mathematica
stops responding (i did read your answer to the same question
asked a week ago before - actually that is why i turned to the
FAQ).
====
People encounter this all the time. It is because
SelectionEvaluate does not do what you think. It does not
work like ToExpression, which causes immediate kernel
evaluation. Instead it works like when you press Shift-Enter,
which selects a cell for evaluation after all current
evaluations have finished.See
http://support.wolfram.com/mathematica/kernel/interface/
selectionevaluate.html-Dale>>Trying to manipulate notebooks
from the kernel I found>an unexpected bahavior with
Mathematica. First I tried>>the following commands one by one
(they are not int>the same cell, and they are not selected at
same time>for evaluation)>>nb = NotebookCreate[];>i = 0;>(*
Purpose is a Do loop here *)>NotebookDelete[nb]
(*1*)>NotebookWrite[nb, ++i, All]
(*2*)>SelectionEvaluate[nb] (*3*)>SelectionMove[nb, All,
Cell] (*4*)>>If I repeat evaluating coomands (*1,2,3,4*) one
by one>what is shown in the created notebook is an
animation>of the index i in the same cell.>Naturally a loop
must do the job. However when I>intent to collect (*1,2,3,4*)
in the same cell the>result is not the same even whitout the
Do loop (I>mean just evaluating this cell several times)>>I
would be grateful if some can explain what's going>on here 
or
if there is something wrong with
my>machine.>>Cesar>__________________________________
________________>Do You Yahoo!?>Yahoo! Finance - Get
real-time stock quotes>http://finance.yahoo.com
====
>Here's a
piece of a conversion I had with Mathematica.>Why is
a[A_,B_] := LinearSolve[X,Y][[1]] not giving>me the
function I expect?>>In[261]:= X =
{{0,0,0,1},{1,0,0,1},{0,1,0,1},{1,1,1,1}};>In[262]:= Y =
{A,B,C,D};>In[263]:= LinearSolve[X,Y]>Out[263]=
{-A+B,-A+C,A-B-C+D,A}>In[264]:=
LinearSolve[X,Y][[1]]>Out[264]= -A+B>In[265]:= a[A_,B_] :=
LinearSolve[X,Y][[1]]>In[266]:= a[1,3]>Out[266]= -A+B>>The
output above is not what I want. I want 2. Here's>what I
expect:>>In[267]:= a[A_,B_] := -A+B;>In[268]:=
a[1,3]>Out[268]= 2>>This output is what I expect. What is the
difference between>the two?This is a common misconception
about what := does. What it does is set up a variable
replacement for the unevaluated expression, not the evaluated
expression. Soa[A_,B_] := LinearSolve[X,Y][[1]]a[1,3]is
similar to doingReleaseHold[ Hold[ LinearSolve[X,Y][[1]] ] /.
{A->1, B->3}]which means that only explicit instances of A and
B are replaced. What you are attempting is done with =. This
assigns the function to the evaluated
expression.In[5]:=a[A_,B_] =
LinearSolve[X,Y][[1]]Out[5]=-A+BIn[6]:=a[1,3]Out[6]=2There
are certain situations where you get a different result if
you evaluate with the values or replace the values in the
symbolic evaluation. (This example is not one of them.) Using
= does the later. If you want to do the former, you should use
:= and have the right hand side use A and B explicitly or you
could use a Block.In[7]:=a[b_,c_] := Block[{A=b, B=c},
LinearSolve[X,Y][[1]]
]In[8]:=a[1,3]Out[8]=2---------------------------------------
-----------------------Omega ConsultingThe final answer to
your Mathematica needsSpend less time searching and more
time
finding.http://www.wz.com/internet/Mathematica.html
====
>
Normally I don't feel so stupid, but I'm 
trying to create an
interactive>Mathematica notebook, and I'm stuck at square
one.>>Specifically, how do I do something like create a button
to add a number (or>perform other mathematical functions) and
then display or otherwise>manipulate the result (eventually I
want to be able to click a button that>increments/decrements
an angle and animate the resulting transformation of
a>vector, with an eye to finally simulating a simple robotic
arm complete with>simple controls to manipulate the arm).>>I
want to have a variable (or matrix or whatever) defined as a
global>variable x, and then perform x = x+1 when a button is
clicked. I've been>using and programming computers for 
almost
25 years and I can't follow>Wolfram's 
documentation. Is it
just me, or is he always this obtuse when>explaining things?
I mean, given the amazing power of the Mathematica>system, a
sample list of buttons in a notebook that you could select
and>examine how they were implemented, would have been nice.
I can't find such a>list, and this seems to be 
par for the
course for the rest of the>documentation as well.Button
programming can be very confusing at first. It has a whole
series of quirks that make it in many ways unique to
Mathematica and programming in general. This causes a lot of
head-scratching, but once you understand what's going on
things get easier.The key is to create a button that uses the
kernel (by default it only uses the front end). Here's a
simple example to help you get started.In[1]:=
x=1;In[2]:=ButtonBox[Increment x,Active->True,
ButtonEvaluator->Automatic, ButtonFunction:>Print[x = ,++x]
]//DisplayFormSome random things to note:-
ButtonEvaluator->Automatic. This says use the kernel to
implement the ButtonFunction.- Buttons only create
side-effects. They generate no output. What you see when you
press the button is the result of 2 side effects. One from
++, which changes the value of x. The other from Print, which
creates a cell.- Print is, in general, a poor side-effect to
use in a button. It's difficult to control where 
the Print
cell is placed. It is worth your while to learn how to use
other front end side-effect functions (such as NotebookRead
and NotebookWrite) when using buttons.Here are some further
resources you might find
helpful:http://support.wolfram.com/mathematica/interface/
buttons/http://library.wolfram.com/conferences/devconf99/
hinton/Buttons19991022.nbhttp://library.wolfram.com/
conferences/devconf2001/horton2/horton2.nb-------------------
-------------------------------------------Omega
ConsultingThe final answer to your Mathematica needsSpend
less time searching and more time
finding.http://www.wz.com/internet/Mathematica.html
====
How do
I change format used for tick mark labels in a 2D plot? I
wouldlike to use DigitBlock->3 option of NumberForm to format
large numbersappearing as tick mark labels on a
histogram.Alexander
====
As Daniel Lichtblau pointed out, the
statement below about vertices is nonsense. Consider two
overlapping rectangles arranged as a cross. You need to
compute intersections and test them instead of vertices.Begin
forwarded message:>> Begin forwarded message:>> Dear
colleagues,>> any hints on how to implement a very fast
routine in Mathematica for> testing if two rectangles have an
intersection area?> Frank Brand> Here is one approach.>>
Given three points {x1,y1},{x2,y2},{x3,y3}, switch to
homogenous> coordinates a={1,x1,y1}, b={1,x2,y2},
c={1,x3,y3}. Then> Sign[Det[{a,b,c}]] is +1 if and only if
the point a lies on your left as> you walk along the line
though b and c in the direction from b to c.> ( If the result
is zero, then a lies on the line.)>> The value of the
determinant is x2y3-x3y2-x1y3+x3y1+x1y2-x2y1, and the> speed
of the algorithm depends essentially on how fast this
quantity can> be computed. Suppose we write a function
LeftSide[a,{b,c}] that computes> the sign of the
determinant.>> Now let {p1,p2, . . ., pn} be a list of
vertices (pi={xi,yi}) of a> convex polygon traced
counterclockwise. Then a lies within or on the> boundary of
the polygon if and only if none of the numbers>
LeftSide[a,{pi,p(i+1)}] are -1. That is, if -1 does not
appear in the> list LeftSide[a,#]&/@Partition[{p1,p2,. .
.,pn,p1},2,1].>> Now use the fact that if two convex
polynomials overlap, then some> vertex of one of them must
lie inside or on the boundary of the other.>> If an overlap
of positive area is required, then the check is that only> +1
appears--not that -1 does not appear.>> For two rectangles (
or parallelograms) this approach requires the> evaluation of
16 determinants, so it may be a bit expensive. If the> points
have rational coordinates, then (positive) denominators may
be> cleared in the homogeneous coordinates and the
computations can be done> in integer arithmetic, at the cost
of at least three more> multiplications per
determinant.>>Garry HelzerDepartment of MathematicsUniversity
of MarylandCollege Park, MD
20742301-405-5176gah@math.umd.edu>>
====
While playing arounf
with patterns and substitutions, I came across thefollowing
behavior which I didn't expect:z := SomeHead[{{1, 2}, {3,
4}}]z /. {SomeHead[q_] -> Flatten[q]}While this _does_ yield
the desired result {1,2,3,4}, Mathematica
complains:Flatten::normal: Nonatomic expression expected
at position 1 inFlatten[q].........as if it is trying to
evaluate Flatten[q], with q not bound to{{1,2},{3,4}}.Could
anybody explain why this happens? Sidney Cadot
====
>
............... I came across the> following behavior which I
didn't expect:>> z := SomeHead[{{1, 2}, {3, 4}}]> z /.
{SomeHead[q_] -> Flatten[q]}>> While this _does_ yield the
desired result {1,2,3,4}, Mathematicacomplains:>>
Flatten::normal: Nonatomic expression expected at position
1 in> Flatten[q].>............> Could anybody explain why
this happens?Sidney,The clue is that, before any replacing is
done, z is evaluated to SomeHead[{{1, 2}, {3, 4}}];
SomeHead[q_] is evaluated to SomeHead[q_] Flatten[q] is
evaluated to Flatten[q] and the message is generatedsaying
essentially nothing to ßatten.*Then* the replacement
SomeHead[{{1, 2}, {3, 4}}] /. {SomeHead[q_] -> Flatten[q]}is
performed.It would be better here to use RuleDelayed, :>,
instead of Rule, ->, sinceFlatten[q] would not then be
evaluated. z:=SomeHead[{{1,2},{3,4}}]
z/.{SomeHead[q_]:>Flatten[q]}
{1,2,3,4}--Allan---------------------Allan HayesMathematica
Training and ConsultingLeicester
UKwww.haystack.demon.co.ukhay@haystack.demon.co.ukVoice: +44
(0)116 271 4198
====
Never mind my question, I should have used
a delayed rule there (:> insteadof ->).Bit silly of
me.
====
Suppose f is a function of n real variables, and
returns avector of n real variables. What is the correct
syntax to find a rootof f using FindRoot?For instance, the
following works with n=7, but I would need to changethe code
if I changed the value of n:f[p_] := Table[fk[p, k],
{k,n}];pn = {p1,p2,p3,p4,p5,p6,p7};f1 := f[pn] [[1]];f2 :=
f[pn] [[2]];f3 := f[pn] [[3]];f4 := f[pn] [[4]];f5 := f[pn]
[[5]];f6 := f[pn] [[6]];f7 := f[pn] [[7]];theRoot =
FindRoot[{f1==0,f2==0,f3==0,f4==0,f5==0,f6==0,f7==0}, {p1,
1/n},{p2, 1/n},{p3, 1/n},{p4, 1/n},{p5, 1/n}, {p6, 1/n},{p7,
1/n}];-- John MacCormick Systems Research Center, HP Labs,
1501 Page Mill Road,
====
On 8/23/02 at 12:25 AM,
meshii@mech.fukui-u.ac.jp (Toshiyuki (Toshi)>NDSolve seems to
have difficulties with solving integral equation.>n = 5;
NDSolve[{D[[Sigma]norm[z, t], t]
==>3*z*Integrate[[Sigma]norm[z, t]^n*z, {z, 0, 1}] -
[Sigma]norm[z,>t]^n, [Sigma]norm[z, 0] == 1.5*z,
[Sigma]norm[0, t] ==>0}*[Sigma]norm[z, t], {z, 0.01, 1},
{t, 0.01, 2}]>Mathematica returns a message>>NDSolve::deql:
The first argument must have both an equation and an>
initial condition.A general solution to a differential
equation includes a constant of integration. It isn't
possible to arrive a numerical solution without providing
sufficient information to determine this constant. Specifying
the initial condition provides this information.Look at the
examples using the help browser to see what NDSolve
needs.
====
>I'm trying to figure out the correct syntax to 
do
the following. I have>some function with three arguments, and
I want to syntactically describe>the>single-argument function
that holds two of those arguments constant (i.e.>without
creating that single-argument function).>>More specifically, I
have defined>> Machine[radix_,multiplier_,state_] := Module
[{c,s},> c = Floor[state/base]; s = Mod[state,base];>
multiplier*s + c> ]>>where I have a generalize 
Ômachine',
defined by the radix and multiplier,>which converts one state
into another state. So I'd like to be able to>do>something
like this:>> NestList[Machine[10,7,#], 3, 22]>>to get the
series of states that the radix-10 multiplier-7 machine
runs>through (starting with state 3). However, this syntax
doesn't seem to>do>what I want.>>I hope that description
makes sense. It seems like there must be a syntax>to describe
the function Machine[10,7,#].>I assume that you want base
rather than radix in the definition (or vice
versa).Machine[base_, multiplier_, state_]:= Module[{c, s},
c=Floor[state/base]; s=Mod[state, base];
multiplier*s+c];NestList[Function[Machine[10,7,#]],3,22]{3,
21, 9, 63, 27, 51, 12, 15, 36, 45, 39, 66, 48, 60, 6, 42, 18,
57, 54, 33, 24, 30, 3}The abbreviation for Function[body] is
body&%==NestList[Machine[10, 7, #]&, 3, 22]TrueBob
HanlonChantilly, VA USA
====
>While playing arounf with
patterns and substitutions, I came across the>following
behavior which I didn't expect:>>z := SomeHead[{{1, 2}, {3,
4}}]>z /. {SomeHead[q_] -> Flatten[q]}>>While this _does_
yield the desired result {1,2,3,4}, Mathematica
complains:>>Flatten::normal: Nonatomic expression expected
at position 1 in>Flatten[q].>>........as if it is trying to
evaluate Flatten[q], with q not bound
to>{{1,2},{3,4}}.>>Could anybody explain why this
happens?z:=SomeHead[{{1,2},{3,4}}]The error is associated
with defining the Rule not executing it. Note that you get the
error with{SomeHead[q_]->Flatten[q]};Use
RuleDelayedz/.{SomeHead[q_]:>Flatten[q]}{1, 2, 3, 4}Bob
HanlonChantilly, VA USA
====
Can anyone help me with this
problem. If I have an n-element list, (say where each element
is itself alist), such as {{a,b}, {a,b}, {a,b}}is there a way
to strip off the outermost nesting of the list toobtain just
a sequence of of these n elements, that is{a,b},{a,b},{a,b}
so that I can use this for input for some function.I would
like to do something likeOuter[SomeFunction, Table[{a,b},{N}
]] where I can enter Ndynamically.The problem, of course, is
that the output of the Table command is onebig listand Outer
is expecting a sequence of N separate lists
afterSomeFunction. 
====
some function with three arguments,
and I want to syntactically describe thesingle-argument
function that holds two of those arguments constant
(i.e.without creating that single-argument function).More
specifically, I have defined 
Machine[radix_,multiplier_,state_]
:= Module [{c,s}, c = Floor[state/base]; s = Mod[state,base];
multiplier*s + c ]where I have a generalize 
Ômachine', defined
by the radix and multiplier,which converts one state into
another state. So I'd like to be able to dosomething like
this: NestList[Machine[10,7,#], 3, 22]to get the series of
states that the radix-10 multiplier-7 machine runsthrough
(starting with state 3). However, this syntax doesn't seem 
to
dowhat I want.I hope that description makes sense. It seems
like there must be a syntaxto describe the function
Machine[10,7,#].Anyone have any ideas?Bob H
====
Sidney,With a
direct rule Mathematica tries to Flatten the symbol q
immediately.You want a delayed rule to avoid the error
message.z /. SomeHead[q_] :> Flatten[q]David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/{{1,2},
{3,4}}.Could anybody explain why this happens? Sidney
Cadot
====
Lucas,I hope that your question will provoke a
number of replies because I thinkit is an interesting
topic.First, it would be nice if Mathematica had a
ShowPrecedence statement toquickly retrieve a precedence
number for any command. It is a bit timeconsuming to search
through the table in Section A.2.7.Next, it would be nice if
the user could set the precedence for operatorsthat don't
have built-in definitions.Since it appears that you 
can't do
that, is it possible to switch themeanings of Plus and
CirclePlus in your theory? Are you depending on thenumerical
behavior of Plus? Go with the precedence that Mathematica
givesyou. Or you could use something like VerticalBar or
RightTee which havelower precedence than Plus, but perhaps
don't have the look that you want.You could try to use the
Notation package for your CirclePlus sum. But Ialways find the
Notation package difficult to use and like to useStandardForm
as much as possible. One possibility
is...!((([Sum]+(i = 1)%5 x[i])) /. Plus ->
CirclePlus)These are weak answers to your question, but
maybe they will bring in morediscussion.David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/desired
effect.Also, I would like to introduce a notation like
N[BigCirclePlus] x[[i]] --> x[[1]] [CirclePlus] x[[2]]
[CirclePlus].... i=0analagous to summation, but mathematica
does not appear to offer theCirclePlus in a large format. to
relate this to the case above, x[1]= (a + b) andx[2] = (c +
d), so each indexed element is a subexpression.Finally, I
would like to be able to set up the CirclePlus operatorsuch
that the following algebraic relations hold:Sum
BigCirclePlus E = BigCirclePlus Sum E i j ij j i ijd
d-- BigCirclePlus E = BigCirclePlus -- Edx j j j dx
j-Lucas Scharenbroich-MLS Group / JPL
====
You are quite right,
Mathematica does evaluate Flatten[q] before substituting {{1,
2}, {3, 4}}. It then issues the error message and returns
Flatten[q]. Only now {{1, 2}, {3, 4}} is substituted for q,
Flatten[{{1, 2}, {3, 4}}] is evaluated and you get the right
answer. To avoid all this just use RuleDelayed instead of
Rule:In[1]:=z := SomeHead[{{1, 2}, {3, 4}}];In[2]:=z /.
{SomeHead[q_] :> Flatten[q]}Out[2]={1,2,3,4} Andrzej
KozlowskiToyama International UniversityJAPAN>> While playing
arounf with patterns and substitutions, I came across the>
following behavior which I didn't expect:>> z :=
SomeHead[{{1, 2}, {3, 4}}]> z /. {SomeHead[q_] ->
Flatten[q]}>> While this _does_ yield the desired result
{1,2,3,4}, Mathematica > complains:>> Flatten::normal:
Nonatomic expression expected at position 1 in> Flatten[q].>>
........as if it is trying to evaluate Flatten[q], with q not
bound to> {{1,2},{3,4}}.>> Could anybody explain why this
happens?> Sidney Cadot>>
====
> [...] Y = {A,B,C,D};
[...]> In[265]:= a[A_,B_] := LinearSolve[X,Y][[1]]> In[266]:=
a[1,3]> Out[266]= -A+B> The output above is not what I want.
I want 2.> [...]You are mixing the (global (*)) symbols A
and B with the (local (*)) patterns A_ and B_. If you want to
replace the (global) solution with local values you should
writeIn[4]:= a[AA_, BB_] := LinearSolve[X, Y][[1]] /. {A ->
AA, B -> BB}; a[1, 3]Out[5]= 2________________(*) as far we
can say that in Mathematica-- Rainer Gruber
====
>
NestList[Machine[10,7,#], 3, 22]You almost have
it:NestList[Machine[10,7,#]&, 3, 22]More elegantly, you can
defined Machine this way:Machine[radix_,multiplier_][state_]
:= ...Then you can writeNestList[Machine[10,7], 3, 22]Tom
Burton
====
> z := SomeHead[{{1, 2}, {3, 4}}]> z /.
{SomeHead[q_] -> Flatten[q]}> While this _does_ yield the
desired result {1,2,3,4}, Mathematica complains:>
Flatten::normal: Nonatomic expression expected at position
1 in> Flatten[q].The -> operator is prompt: It permits
evaluation of the RHS Flatten[q]immediately, before it is
used in the substitution. Evidently, q wasundefined before you
tried this, so Flatten[q]'s evaluation yielded itself.Later,
in the process of substitution, with q set as you intended,
Flattenevaluated again, this time yielding the desired
result.Try it again after setting q to something. Then you
get not only thecomplaint but also the wrong answer.To avoid
these issues, use the delayed operator :> instead of the
promptoperator ->. A rule of thumb is: Use delayed operators
:= and :> when theLHS (SomeHead[q_] in your case) contains a
blank (_).Tom Burton
====
 Mark:In: ?KOut: K is a default
generic name for a summation index in a symbolic sum.Turns
out that there are seven single-letter symbols.In:
ClearAll[Global`*]; Select[Names[*], (StringLength[#] ==
1) &]Out: {C, D, E, I, K, N, O}K is a little weird, because
it's not Protected.In: Attributes[{C, D, E, I, K, N, O}]Out:
{{NHoldAll, Protected}, {Protected, ReadProtected},
{Constant, Protected, ReadProtected}, {Locked, Protected,
ReadProtected}, {}, {Protected}, {Protected,
ReadProtected}}----Selwyn Hollis> I just learned that K is a
System` Symbol:> Information[K]> Context[K]> I learned
this due to an error message:> Block[{K = 1}, Sum[j, {j,
i}]]> The same message can be generated by the following:>
> K := 1> Sum[j, {j, i}]> The message can be eliminated by
Removing K:> Remove[K]> Block[{K = 1}, Sum[j, {j, i}]]> K
:= 1> Sum[j, {j, i}]> Surely, this is not intentional.>
--Mark.> 
====
Dear MathgroupI have convert pde to ode like
thisdu_i/dt = 1-4 u_i + .02 (u_i-1 .9a 2 u_i + u i+1)/(delta
(x))^3 + (u_i)^3 v_idv_i /dt = 3 u_i + .02 (v_i-1 .9a 2 v_i +
u i+1)/(delta (x))^3 - (u_I)^3 v_idelta(x)=(i)/(N+1)x_i=
(i)/(N+1)Boundary conditionu_0 (t) = 1 = u_N+1(t)v_0(t) = 3 =
v_N+1(t),Initial conditiou_i(0) = 1+sin (2 pi x_i )v_i = 3For
i = 1,á., NTime Interval =[t_o, t_end] = [0,10]Could i get
code in Mathematica (by using Euler of 4 Runge - Kutta..)to
solve this ordinary differential equation.I am very happy if
you give me
helpKhaled___________________________________________________
______________MSN Photos is the easiest way to share and
print your photos:
http://photos.msn.com/support/worldwide.aspx
====
 I am a
newbie to mathematica. I have a 14 functions which are
thefunction of r,theta and phi. I want to do some
mathematical operationover them. How can I do? Can it be
possible to call them in Do or Forloop with some index?
Please suggest. Raj
====
Is there an easy (elegant?) way to
generate the set of all k-tuplestaking values from some set
(list) S? I want the arguments of thefunction to be k (the
length of the tuples) and the set S. That is,KTuples[3,{a,b}]
should
produce{{a,a,a},{a,a,b},{a,b,a},{a,b,b},{b,a,a},{b,a,b},{b,b,
a},{b,b,b}}.
====
Does anyone know if there are any
publications that describe theintegration of above?In
particular I am working with a 4 gaussian process with joint
pdfp(x1,y1,x2,y2) transformed to-> p(r1,r2,theta1,theta2),
the processeshave non-zero mean and I would like to integrate
the joint pdf twice(over r1 and r2) and obtain an expression
for the pdf of the phaseangles between p(theta1,theta2). I
could assume that the means are thesame (to make life
easier). Has anyone come accross this - it doesn'tseem that
unusual to me. I am asking because the integrals arebecoming
Ôinteresting'.thankyou for any tips.
====
Start Excel and go to
Tools -> Macro -> Security..., then on the SecurityLevel
tab click on Medium and then click OK.----- Original
Message ----->> My original version of MathLink for Excel
(MLX.xla) was for Excel 97,and> It worked fine. Then I
upgraded the package to Excel 2000 by unloadingthe> macro
from the Wolfram site. It continued working perfectly on my
old> environment (Windows 98 + Mathematica 4.1).>> But
I recently upgraded to Mathematica 4.2 under Windows 2000;
and when> trying to load the macro (MLX.xla) for Excel2000,
I got (in Excel 2000)> the> following message: This
workbook contains a type of macro (Microsoft> Excel>
version 4.0 macro) that cannot be disable nor signed.
Therefore, this> workbook cannot be opened under high
security level.>> Does any one have any experience on
this?> Emilio Martin-Serrano>
====
Download the
newest patch from wolfram research - this should
theoreticallyhelp.> Group,>> My original version of MathLink
for Excel (MLX.xla) was for Excel 97, and> It worked fine.
Then I upgraded the package to Excel 2000 by unloading the>
macro from the Wolfram site. It continued working perfectly
on my old> environment (Windows 98 + Mathematica 4.1).>> But
I recently upgraded to Mathematica 4.2 under Windows 2000;
and when> trying to load the macro (MLX.xla) for Excel2000, I
got (in Excel 2000)the> following message: This workbook
contains a type of macro (MicrosoftExcel> version 4.0 macro)
that cannot be disable nor signed. Therefore, this> workbook
cannot be opened under high security level.>> Does any one
have any experience on this?> Emilio
Martin-Serrano>
====
>I'm sorry for that my question is not
clear,I have correct below.> I have a very interesting math
problem:If I have a scales,and I>> have 40 things that their
mass range from 1~40 which each is a nature>> number,and now
I can only make 4 counterweights to measure out each>> mass
of those things.Question:What mass should the
counterweights>> be???>> The answer is that 1,3,9,27 and I
wnat to use mathematica to solve>> this problem.>> In fact,I
think that this physical problem has various>>
answer,ex.2,4,10,28>> this way also work,because if I have a
thing which weight 3 , and I>> can measure out by comparing
2<3<4 . But,If I want to solve this math>> problem:>>
{x|x=k1*a+k2*b+k3*c+k4*d}={1,2,3,4,,,,,,40} where a,b,c,d is
nature numbers.>> and {k1,k2,k3,k4}={1,0,-1}>> How to solve
it ?? >> mathematica solving method. appreciate any idea
sharing>> sincerely >> bryan>Just use brute
force.Needs[DiscreteMath`Combinatorica`];var = {a, b, c,
d}; n = Length[var];s = Outer[Times, var, {-1, 0, 1} ];f =
Flatten[Outer[Plus, Sequence@@s]];Since the length of f is
just 3^n then the range of numbersto be covered should be
{-(3^n-1)/2, (3^n-1)/2}.Consequently, the largest of the
weights can not exceed(3^n-1)/2 - (1+2+...+(n-1)) or ((3^n-1)
- n(n-1))/234Thread[var->#]& /@ (First /@ Select[{var,f} /.
Thread[var->#]& /@ KSubsets[Range[((3^n-1) - n(n-1))/2], n],
Sort[#[[2]]] == Range[-(3^n-1)/2,(3^n-1)/2]&]){{a -> 1, b ->
3, c -> 9, d -> 27}}Bob HanlonChantilly, VA USA
====
Can we
please get a response from WRI?i.e. regarding:In[1]:= Limit[
(Log[x]^Log[Log[x]])/ x , x->Infinity]Out[2]:= 
InfinityIt
should be 0Jonathan Rockmannmtheory@msn.com
====
I'm finding
that the ImageSize option in Export has no effect when
exporting Cell or Notebook objects. For instance, the
following two commands produce precisely the same graphic:
Export[image1.jpg, Cell[Some cell contents, Text,
FontSize -> 100]] Export[image2.jpg, Cell[Some cell
contents, Text, FontSize -> 100], ImageSize -> {576,
288}]Has anyone encountered this problem before?(This is with
Mathematica 4.1.5 and Mac OS X.)----Selwyn
Hollisslhollis@mac.com
====
Does anyone know how to get the
JavaPlot window (or any windows of this type)which can be
seen at
http://www.wolfram.com/products/mathematica/newin42/
java.htmlthat WRI advertises comes with
4.2?Jonathanmtheory@msn.com
====
How can I get mathematica to
display the inverse of functions like:f(x) = x^2 - 7*x +
10orf(x) = cos(3*x + 1/2*pi)orf(x) = (x - 3) / (x + 2)I'm
having trouble getting the syntax right.
====
G'day,Looks like
you forgot to define base in your Machine function and
whenusing pure functions, don't forget the
ampersand.ThusNestList[Machine[10,7,#]&, 3, 22]returns{3, 8,
4, 2, 1, 7, 10, 5, 9, 11, 12, 6, 3, 8, 4, 2, 1, 7, 10, 5, 9,
11,12}with base = 2.Yas> Howdy,>> I'm trying to 
figure out the
correct syntax to do the following. I have> some function with
three arguments, and I want to syntactically describe the>
single-argument function that holds two of those arguments
constant (i.e.> without creating that single-argument
function).>> More specifically, I have defined>>
Machine[radix_,multiplier_,state_] := Module [{c,s},> c =
Floor[state/base]; s = Mod[state,base];> multiplier*s + c>
]>> where I have a generalize Ômachine', 
defined by the radix
and multiplier,> which converts one state into another state.
So I'd like to be able to do> something like this:>>
NestList[Machine[10,7,#], 3, 22]>> to get the series of
states that the radix-10 multiplier-7 machine runs> through
(starting with state 3). However, this syntax doesn't seem 
to
do> what I want.>> I hope that description makes sense. It
seems like there must be a syntax> to describe the function
Machine[10,7,#].>> Anyone have any ideas?>> Bob
H>
====
>I'm sorry for that my question is not clear,I have
correct below.>>ajvp7h$ibk$1@smc.vnet.net>...> I have a very
interesting math problem:If I have a scales,and I> have 40
things that their mass range from 1~40 which each is a
nature> number,and now I can only make 4 counterweights to
measure out each> mass of those things.Question:What mass
should the counterweights> be???> The answer is that
1,3,9,27 and I wnat to use mathematica to solve> this
problem.> In fact,I think that this physical problem has
various> answer,ex.2,4,10,28> this way also work,because
if I have a thing which weight 3 , and I> can measure out
by comparing 2<3<4 . But,If I want to solve this math>
problem:> {x|x=k1*a+k2*b+k3*c+k4*d}={1,2,3,4,,,,,,40} where
a,b,c,d is nature>numbers.> and {k1,k2,k3,k4}={1,0,-1}>
How to solve it ?? > mathematica solving method. appreciate
any idea sharing> sincerely > bryan>>Just use brute
force.>>Needs[DiscreteMath`Combinatorica`];>var = {a, b,
c, d}; n = Length[var];>s = Outer[Times, var, {-1, 0, 1}
];>f = Flatten[Outer[Plus, Sequence@@s]];>Since the
length of f is just 3^n then the range of numbers>to be
covered should be {-(3^n-1)/2, (3^n-1)/2}.>Consequently, the
largest of the weights can not exceed>(3^n-1)/2 -
(1+2+...+(n-1)) or >>((3^n-1) -
n(n-1))/2>34>Thread[var->#]& /@> (First /@
Select[{var,f} /. Thread[var->#]& /@>
KSubsets[Range[((3^n-1) - n(n-1))/2], n],> Sort[#[[2]]] ==
Range[-(3^n-1)/2,(3^n-1)/2]&])>{{a -> 1, b -> 3, c -> 9, d
-> 27}}>A modification to my earlier response. Since you are
try to cover all of the values up to (3^n-1)/2 then you can
speed up the brute force method by requiring that a+b+c+d ==
(3^n-1)/2Needs[DiscreteMath`Combinatorica`];var =
{a,b,c,d}; n = Length[var]; m = (3^n-1)/2;s = Outer[Times,
var, {-1,0,1} ];f = Flatten[Outer[Plus, Sequence@@s]];k=
Select[KSubsets[Range[m - n(n-1)/2], n], (Plus @@ #) ==
m&];Thread[var->#]& /@ (First /@ Select[{var,f} /.
Thread[var->#]& /@ k, Sort[#[[2]]] == Range[-m,m]&]){{a -> 1,
b -> 3, c -> 9, d -> 27}}Bob HanlonChantilly, VA USA
====
Some
functions with vector arguments work as expected and some
giveerrors. I would appreciate if someone can clarify
why.First an example of a function definition that works 
fine
:f[u_,v_] := u.vCalling it with f[{x1,y1},{x2,y2}] gives the
expected result x1 x2 +y1 y2.Now an example that does not
work :g[u_,v_] := u-vCalling it with g[{x1,y1},{x2,y2}] one
would expect the answer {x1-x2,y1-y2} but instead one gets
error messages.Why ? And how do I fix g (i.e write a function
that outputs thedifference of 2 vectors).
====
Garry, Here's a
solution using your LeftSide concept; it works perfectly
but takes twice as much time as my solution. Both solutions
look at every vertex of both rectangles, but mine uses two
sides from each and yours requires looking at all four sides
of each rectangle. I'd think yours should be a triße 
faster
than this, though. There may be efficiencies I'm 
missing (in
both solutions). ClearAll[cis, rect, pickRect, extent,
cannotIntersect, intersects, daveRect] cis[t_] := {Cos@t,
Sin@t} rect[{pt : {_, _}, angle_, {len1_, len2_}}] :=
Module[{pt2}, {pt, pt2 = pt + len1 cis[angle], pt2 - len2
cis[angle - Pi/2], pt - len2 cis[angle - Pi/2]}] daveRect :=
{{Random[], Random[]}, Random[] + Pi/2, {Random[], Random[]}}
pickRect := rect@daveRect extent[r1_, r2_] := {Min@#, Max@#} &
/@ ((Take[r1, 2] - r1[[{2, 3}]]).Transpose@r2)
cannotIntersect[{{min1_, max1_}, {min2_, max2_}}] := max2 <
min1 || min2 > max1 intersects[r1_, r2_] := Catch[
If[cannotIntersect[#], Throw[False]] & /@
Flatten[Transpose[Outer[extent,  {r1}, {r1, r2},
1]~Join~Outer[extent, {r2}, {r2, r1}, 1], {1, 3, 2}], 1];
Throw[True]] ClearAll[leftSide,leftIntersects,sides]
sides[a_List]:=Partition[Join[a,{First@a}],2,1]
leftSide[{a_,b_},{{c_,d_},{e_,f_}}]:=-b c+a d+b e-d e-a f+c
f>0 leftSide[a:{{_,_}..},b:{{_,_},{_,_}}]:=leftSide[#,b]&/@a
leftSide[a_List,b:{{{_,_},{_,_}}..}]:=leftSide[a,#]&/@b
leftIntersects[a_,b_]:=!Or@@(And@@#&/@leftSide[a,sides@b])&&!
Or@@(And@@#&/@leftSide[b,sides@a])
davePairs={daveRect,daveRect}&/@Range[10000];
rectanglePairs=Map[Reverse@rect[#]&,davePairs,{2}];
Timing[right=intersects[Sequence@@#]&/@rectanglePairs;]
Timing[test=leftIntersects[Sequence@@#]&/@rectanglePairs;]
right[Equal]test {3.187999999999999*Second, Null}
{6.765000000000001*Second, Null} True Bobby Treat
-----Original Message----- intersection > Begin forwarded
message: > Dear colleagues, > any hints on how to
implement a very fast routine in Mathematica for > testing if
two rectangles have an intersection area? > Frank Brand >
Here is one approach. > Given three points
{x1,y1},{x2,y2},{x3,y3}, switch to homogenous > coordinates
a={1,x1,y1}, b={1,x2,y2}, c={1,x3,y3}. Then >
Sign[Det[{a,b,c}]] is +1 if and only if the point a lies on
your left as > you walk along the line though b and c in the
direction from b to c. > ( If the result is zero, then a lies
on the line.) > The value of the determinant is
x2y3-x3y2-x1y3+x3y1+x1y2-x2y1, and the > speed of the
algorithm depends essentially on how fast this quantity can >
be computed. Suppose we write a function LeftSide[a,{b,c}]
that computes > the sign of the determinant. > Now let
{p1,p2, . . ., pn} be a list of vertices (pi={xi,yi}) of a >
convex polygon traced counterclockwise. Then a lies within or
on the > boundary of the polygon if and only if none of the
numbers > LeftSide[a,{pi,p(i+1)}] are -1. That is, if -1 does
not appear in the > list LeftSide[a,#]&/@Partition[{p1,p2,. .
.,pn,p1},2,1]. > Now use the fact that if two convex
polynomials overlap, then some > vertex of one of them must
lie inside or on the boundary of the other. > If an overlap
of positive area is required, then the check is that only > +1
appears--not that -1 does not appear. > For two rectangles (
or parallelograms) this approach requires the > evaluation of
16 determinants, so it may be a bit expensive. If the >
points have rational coordinates, then (positive)
denominators may be > cleared in the homogeneous coordinates
and the computations can be done > in integer arithmetic, at
the cost of at least three more > multiplications per
determinant. > Garry Helzer Department of Mathematics
University of Maryland College Park, MD 20742 301-405-5176
gah@math.umd.edu 
====
I would like to have a list of all the
directives that can be used forcommand in the Windows Start
Menu). I am especially interested in adirective with which
you can appoint a file which has to be evaluated by thekernel
after it has been launched. I imagine that such a directive
lookslike -f filename, but I can not find a list 
of the correct
directives.The Mathematica version I have is 4.1 on a Windows
NT system.Rene Klaver
====
> 1) Get data from excel into a
coordinate > list {x,y,z}> e.g. Node 1, {x1,y1,z1}> Node 2,
{x2,y2,z2}> etc...Why not just use data = ReadList[filename,
Table[Number, {4}]];Alternatively, data = ReadList[filename,
Number, RecordLists->True]; data = Partition[data, 4]; cord =
{#[[1]], #[[2]], #[[3]]}& /@ data;> 2) Convert from
Rectangular to > Cylindrical (maybe)Coordinate tranforms live
in Calculus`VectorAnalysis`. It's straightforwardto write a
function that uses > 3) Plot3D the dataThis will be tricky --
ListSurfacePlot3D plots f[x, y]. ListContourPlot3Dwhich plots
f[x, y, z] can be quite slow.> 4) Generate a harmonic
bessel function for that > plot3D/graph> 5) Find the
equation(s) that spits out > these harmonic > bessel
functions (which I think might be in the general form > of
Hankel > Function solutions to the Helmholtz equation which
shows > cylinder harmonics > of order v)Maybe I
misunderstand ... isn't this the same as fitting 
a bessel
functionto your data? For this, you can use
Statistics`NonlinearFit`. > I can figure out step 5 if I can
get steps 1 through 4 > figured out. If > anyone can write a
recipe for me to follow that would be > great, or even some >
tips and clues...Anything!!!Dave.
====
You've gotten some very
helpful responses so I'm just going to add a coupleof
comments. The following is an example of yet another type of
button that you can useas a model. Copy and paste the
following code into a notebook. When youevaluate it, it will
create a button within the same notebook which whenpressed
evaluates the entire notebook. Sort of like selecting from
the menuis something in the notebook already to evaluate. So
input 1+2 (and don'thit enter) somewhere in the notebook so
that you can see whatthe effect of the button
is.DisplayForm[Button[Evaluate*Notebook,ButtonFunction:>
FrontEndExecute[FrontEndToken[EvaluateNotebook]],
ButtonEvaluator->None,Active->True]]It has the nice feature
that it doesn't get stuck in a loop creating the button over
and over again. However, with some enthusiasm you can make
thishappen too.Personally, I am fascinated by the Java
integration in version 4.2. But I wouldn't recommend it to
anyone who is just starting to program in
Mathematica.Further, there is hardly enough material
contained in the Help Browser to make sense of much of it.It
is always a good idea to have several basic constructs of
efficient, usefulprograms that can serve as models when
programming. If you look atsome of the MathGroup archived
threads you will find similar complaints that the Mathematica
documentation is confusing.Jonathan
Rockmannmtheory@msn.com----- Original Message -----I want to
have a variable (or matrix or whatever) defined as a
globalvariable x, and then perform x = x+1 when a button is
clicked. I've beenusing and programming computers for almost
25 years and I can't followWolfram's 
documentation. Is it
just me, or is he always this obtuse whenexplaining things? I
mean, given the amazing power of the Mathematicasystem, a
sample list of buttons in a notebook that you could select
andexamine how they were implemented, would have been nice. I
can't find such alist, and this seems to be par 
for the course
for the rest of thedocumentation as well.
====
This was REALLY
interesting. Here's a solution that looks only at the7,560
relevant combinations. It first chooses three numerators.
Thenit chooses two denominators for the first fraction. Then
twodenominators for the second fraction. The last fraction is
determinedat that point.<<
DiscreteMath`Combinatorica`ClearAll[f, g, h, j]r = Range[1,
9];f = KSubsets[#1, #2] &;g[r_List, n_Integer, {}] := f[r,
n]g[r_List, n_Integer, e_?VectorQ] := Join[e, #] & /@
f[Complement[r, e],n]g[r_List, n_Integer, e : {__?VectorQ}]
:= Flatten[g[r, n, #] & /@ e, 1]h[r_List, e : {__?VectorQ}]
:= Join[#, Complement[r, #]] & /@ ej[{a_, b_, c_, d_, e_, f_,
g_, h_, i_}] := a/(d e) + b/(f g) + c/(h i)Timing[Select[h[r,
Fold[g[r, #2, #1] &, {}, {3, 2, 2}]], j@# == 1 &]]{0.532
Second, {{1, 5, 7, 3, 6, 8, 9, 2, 4}}}Hence the only solution
is1/(3*6)+5/(8*9)+7/(2*4)Bobby Treat-----Original
Message-----University Professor of Philanthropy and the
LawDirector, National Center on Philanthropy and the LawNew
York University School of LawRoom 206A110 West 3rd StreetNew
York, N.Y. 10012-1074-----Original
Message-----byran____________________________________________
____________________________service. For more information on
a proactive anti-virus service workingaround the clock,
around the globe, visit
http://www.messagelabs.com___________________________________
_____________________________________
====
Garry,Here's a
solution using your LeftSide concept; it works perfectly
buttakes twice as much time as my solution. Both solutions
look at everyvertex of both rectangles, but mine uses two
sides from each and yoursrequires looking at all four sides
of each rectangle. I'd think yoursshould be a triße 
faster
than this, though. There may be efficienciesI'm 
missing (in
both solutions).ClearAll[cis, rect, pickRect, extent,
cannotIntersect, intersects,daveRect]cis[t_] := {Cos@t,
Sin@t}rect[{pt : {_, _}, angle_, {len1_, len2_}}] :=
Module[{pt2}, {pt, pt2 = pt + len1 cis[angle], pt2 - len2
cis[angle - Pi/2], pt - len2 cis[angle - Pi/2]}]daveRect :=
{{Random[], Random[]}, Random[] + Pi/2,
{Random[],Random[]}}pickRect := rect@daveRectextent[r1_, r2_]
:= {Min@#, Max@#} & /@ ((Take[r1, 2] -
r1[[{2,3}]]).Transpose@r2)cannotIntersect[{{min1_, max1_},
{min2_, max2_}}] := max2 < min1 || min2 > max1intersects[r1_,
r2_] := Catch[ If[cannotIntersect[#], Throw[False]] &
/@Flatten[Transpose[Outer[extent, {r1}, {r1, r2},
1]~Join~Outer[extent, {r2}, {r2, r1}, 1], {1, 3, 2}],1];
Throw[True]]ClearAll[leftSide,leftIntersects,sides]sides[a_
List]:=Partition[Join[a,{First@a}],2,1]leftSide[{a_,b_},{{c_,
d_},{e_,f_}}]:=-b c+a d+b e-d e-a f+c
f>0leftSide[a:{{_,_}..},b:{{_,_},{_,_}}]:=leftSide[#,b]&/@
aleftSide[a_List,b:{{{_,_},{_,_}}..}]:=leftSide[a,#]&/@
bleftIntersects[a_,b_]:=!Or@@(And@@#&/@leftSide[a,sides@b])&&
!
Or@@(And@@#&/@leftSide[b,sides@a])davePairs={daveRect,
daveRect}&/@Range[10000];rectanglePairs=Map[Reverse@rect[#]&,
davePairs,{2}];Timing[right=intersects[Sequence@@#]&/@
rectanglePairs;]Timing[test=leftIntersects[Sequence@@#]&/@
rectanglePairs;]right[Equal]test{3.187999999999999*Second,
Null}{6.765000000000001*Second, Null}TrueBobby
Treat-----Original Message-----intersection>> Begin forwarded
message:>> Dear colleagues,>> any hints on how to implement a
very fast routine in Mathematica for> testing if two
rectangles have an intersection area?> Frank Brand> Here is
one approach.>> Given three points {x1,y1},{x2,y2},{x3,y3},
switch to homogenous> coordinates a={1,x1,y1}, b={1,x2,y2},
c={1,x3,y3}. Then> Sign[Det[{a,b,c}]] is +1 if and only if
the point a lies on your leftas> you walk along the line
though b and c in the direction from b to c.> ( If the result
is zero, then a lies on the line.)>> The value of the
determinant is x2y3-x3y2-x1y3+x3y1+x1y2-x2y1, and the> speed
of the algorithm depends essentially on how fast this
quantitycan> be computed. Suppose we write a function
LeftSide[a,{b,c}] thatcomputes> the sign of the
determinant.>> Now let {p1,p2, . . ., pn} be a list of
vertices (pi={xi,yi}) of a> convex polygon traced
counterclockwise. Then a lies within or on the> boundary of
the polygon if and only if none of the numbers>
LeftSide[a,{pi,p(i+1)}] are -1. That is, if -1 does not
appear in the> list LeftSide[a,#]&/@Partition[{p1,p2,. .
.,pn,p1},2,1].>> Now use the fact that if two convex
polynomials overlap, then some> vertex of one of them must
lie inside or on the boundary of the other.>> If an overlap
of positive area is required, then the check is thatonly> +1
appears--not that -1 does not appear.>> For two rectangles (
or parallelograms) this approach requires the> evaluation of
16 determinants, so it may be a bit expensive. If the> points
have rational coordinates, then (positive) denominators may
be> cleared in the homogeneous coordinates and the
computations can bedone> in integer arithmetic, at the cost
of at least three more> multiplications per
determinant.>>Garry HelzerDepartment of MathematicsUniversity
of MarylandCollege Park, MD
20742301-405-5176gah@math.umd.edu>>
====
Garry,Also note your
solution requires rectangle points to be in clockwiseorder
(mine doesn't), but yours works for arbitrary convex 
polygons
aswritten.Bobby-----Original Message-----ClearAll[cis, rect,
pickRect, extent, cannotIntersect,
intersects,daveRect]cis[t_] := {Cos@t, Sin@t}rect[{pt : {_,
_}, angle_, {len1_, len2_}}] := Module[{pt2}, {pt, pt2 = pt +
len1 cis[angle], pt2 - len2 cis[angle - Pi/2], pt - len2
cis[angle - Pi/2]}]daveRect := {{Random[], Random[]},
Random[] + Pi/2, {Random[],Random[]}}pickRect :=
rect@daveRectextent[r1_, r2_] := {Min@#, Max@#} & /@
((Take[r1, 2] -
r1[[{2,3}]]).Transpose@r2)cannotIntersect[{{min1_, max1_},
{min2_, max2_}}] := max2 < min1 || min2 > max1intersects[r1_,
r2_] := Catch[ If[cannotIntersect[#], Throw[False]] &
/@Flatten[Transpose[Outer[extent, {r1}, {r1, r2},
1]~Join~Outer[extent, {r2}, {r2, r1}, 1], {1, 3, 2}],1];
Throw[True]]ClearAll[leftSide,leftIntersects,sides]sides[a_
List]:=Partition[Join[a,{First@a}],2,1]leftSide[{a_,b_},{{c_,
d_},{e_,f_}}]:=-b c+a d+b e-d e-a f+c
f>0leftSide[a:{{_,_}..},b:{{_,_},{_,_}}]:=leftSide[#,b]&/@
aleftSide[a_List,b:{{{_,_},{_,_}}..}]:=leftSide[a,#]&/@
bleftIntersects[a_,b_]:=!Or@@(And@@#&/@leftSide[a,sides@b])&&
!
Or@@(And@@#&/@leftSide[b,sides@a])davePairs={daveRect,
daveRect}&/@Range[10000];rectanglePairs=Map[Reverse@rect[#]&,
davePairs,{2}];Timing[right=intersects[Sequence@@#]&/@
rectanglePairs;]Timing[test=leftIntersects[Sequence@@#]&/@
rectanglePairs;]right[Equal]test{3.187999999999999*Second,
Null}{6.765000000000001*Second, Null}TrueBobby
Treat-----Original Message-----intersection>> Begin forwarded
message:>> Dear colleagues,>> any hints on how to implement a
very fast routine in Mathematica for> testing if two
rectangles have an intersection area?> Frank Brand> Here is
one approach.>> Given three points {x1,y1},{x2,y2},{x3,y3},
switch to homogenous> coordinates a={1,x1,y1}, b={1,x2,y2},
c={1,x3,y3}. Then> Sign[Det[{a,b,c}]] is +1 if and only if
the point a lies on your leftas> you walk along the line
though b and c in the direction from b to c.> ( If the result
is zero, then a lies on the line.)>> The value of the
determinant is x2y3-x3y2-x1y3+x3y1+x1y2-x2y1, and the> speed
of the algorithm depends essentially on how fast this
quantitycan> be computed. Suppose we write a function
LeftSide[a,{b,c}] thatcomputes> the sign of the
determinant.>> Now let {p1,p2, . . ., pn} be a list of
vertices (pi={xi,yi}) of a> convex polygon traced
counterclockwise. Then a lies within or on the> boundary of
the polygon if and only if none of the numbers>
LeftSide[a,{pi,p(i+1)}] are -1. That is, if -1 does not
appear in the> list LeftSide[a,#]&/@Partition[{p1,p2,. .
.,pn,p1},2,1].>> Now use the fact that if two convex
polynomials overlap, then some> vertex of one of them must
lie inside or on the boundary of the other.>> If an overlap
of positive area is required, then the check is thatonly> +1
appears--not that -1 does not appear.>> For two rectangles (
or parallelograms) this approach requires the> evaluation of
16 determinants, so it may be a bit expensive. If the> points
have rational coordinates, then (positive) denominators may
be> cleared in the homogeneous coordinates and the
computations can bedone> in integer arithmetic, at the cost
of at least three more> multiplications per
determinant.>>Garry HelzerDepartment of MathematicsUniversity
of MarylandCollege Park, MD
20742301-405-5176gah@math.umd.edu>>
====
Try
this:Off[Remove::rmnsm]Remove[Global`p@, Global`p@@]n =
7;pn = Unique[p] & /@ Range[10]f[p_] = Array[fk[p, #] &,
n]f[p]fEq[p_] = MapThread[Equal, {f[p], Array[0 &, n]}]Bobby
Treat-----Original Message-----f2 := f[pn] [[2]];f3 := f[pn]
[[3]];f4 := f[pn] [[4]];f5 := f[pn] [[5]];f6 := f[pn]
[[6]];f7 := f[pn] [[7]];theRoot =
FindRoot[{f1==0,f2==0,f3==0,f4==0,f5==0,f6==0,f7==0}, {p1,
1/n},{p2, 1/n},{p3, 1/n},{p4, 1/n},{p5, 1/n}, {p6, 1/n},{p7,
1/n}];-- John MacCormick Systems Research Center, HP Labs,
1501 Page Mill Road,
====
the problem, but I duplicated it just
now, so I'm puzzled. I'm usingWindows XP Home, 
and
System`Private`$BuildNumber is 168634.4.2 for Microsoft
Windows (June 5, 2002)f[n_] :=
Log[n]^Log[Log[n]]Limit[f[n]/n, n -> 
Infinity]InfinityBobby
Treat -----Original Message----- By the way, Mathematica gets
the following limit wrong:f[n_] :=
Log[n]^Log[Log[n]]Limit[f[n]/n, n -> 
Infinity]InfinityThat
limit is zero. For the Sieve to be polynomial, we only need
thesequence to be BOUNDED (for some power of n in the
denominator).Bobby Treat
====
Garry,No, you don't have to
compute intersections, and yes, you can testvertices only. I
haven't coded it yet, but the LeftSide idea seemslike a
good one.It is sufficient to test whether all vertices of one
convex polygon areon the left (out) side of some side of the
second polygon (both polygonsin clockwise order). If that
happens for any side of either polygon,the polygons don't
intersect. In the cross example, some vertices areto the
right for every side you try.Bobby Treat-----Original
Message-----intersection>> Begin forwarded message:>> Dear
colleagues,>> any hints on how to implement a very fast
routine in Mathematica for> testing if two rectangles have an
intersection area?> Frank Brand> Here is one approach.>>
Given three points {x1,y1},{x2,y2},{x3,y3}, switch to
homogenous> coordinates a={1,x1,y1}, b={1,x2,y2},
c={1,x3,y3}. Then> Sign[Det[{a,b,c}]] is +1 if and only if
the point a lies on your leftas> you walk along the line
though b and c in the direction from b to c.> ( If the result
is zero, then a lies on the line.)>> The value of the
determinant is x2y3-x3y2-x1y3+x3y1+x1y2-x2y1, and the> speed
of the algorithm depends essentially on how fast this
quantitycan> be computed. Suppose we write a function
LeftSide[a,{b,c}] thatcomputes> the sign of the
determinant.>> Now let {p1,p2, . . ., pn} be a list of
vertices (pi={xi,yi}) of a> convex polygon traced
counterclockwise. Then a lies within or on the> boundary of
the polygon if and only if none of the numbers>
LeftSide[a,{pi,p(i+1)}] are -1. That is, if -1 does not
appear in the> list LeftSide[a,#]&/@Partition[{p1,p2,. .
.,pn,p1},2,1].>> Now use the fact that if two convex
polynomials overlap, then some> vertex of one of them must
lie inside or on the boundary of the other.>> If an overlap
of positive area is required, then the check is thatonly> +1
appears--not that -1 does not appear.>> For two rectangles (
or parallelograms) this approach requires the> evaluation of
16 determinants, so it may be a bit expensive. If the> points
have rational coordinates, then (positive) denominators may
be> cleared in the homogeneous coordinates and the
computations can bedone> in integer arithmetic, at the cost
of at least three more> multiplications per
determinant.>>Garry HelzerDepartment of MathematicsUniversity
of MarylandCollege Park, MD
20742301-405-5176gah@math.umd.edu>>
====
> Dear All:> I want to
make an animation that simulate the Doppler Effect, just 2D>
circles travel out one by one, and at the same time,the
origin of the> wave also moves toward one direction. I have
no idea to make the speed> of the wave origin and the speed
of traveling wave independent.Is> sincerely bryancircles, but
the second one includes two lines which approximate theshock
waves, I believe. The arguments of the functions are vx &
vy(velocities in x & y directions) and tstart, tend, tstep
whichdetermine how many circles to draw and the spacing of
the circles. Ijust used the Do loop to make the pictures, &
then you can select thecells & pick
Animate.dopp[vx_,vy_,tstart_,tend_,tstep_]:=Show[Graphics[
Table[{Hue[t/tend],Circle[{vx*t,vy*t},tend-t]},{t,tstart,
tend,tstep}]],AspectRatio->Automatic,Axes->True,PlotLabel->
Mach
Sqrt(vx^2+vy^2)];doppcone[vx_,vy_,tend_,tstep_]:=Show[
Graphics[Table[{Hue[t/tend],Circle[{vx*t,vy*t},tend-t]},{t,0,
tend,tstep}]],Graphics[Line[{{-tend*Sin[If[vx==0,0,ArcTan[vy
/vx]]],tend*Cos[If[vx==0,0,ArcTan[vy/vx]]]},{Sign[vx]*Cos[If[
vx==0,0,ArcTan[vy/vx]]]*(tend-tstep)*sqrt[vx^2+vy^2]-
tstep*Sin[If[vx==0,0,ArcTan[vy/vx]]],Sin[If[vx==0,0,ArcTan[
vy/vx]]]*Sign[vx]*(tend-tstep)*sqrt[vx^2+vy^2]+tstep*Cos[If[
vx==0,0,ArcTan[vy/vx]]]}}]],Graphics[Line[{{tend*Sin[If[vx==
0,0,ArcTan[vy/vx]]],-tend*Cos[If[vx==0,0,ArcTan[vy/vx]]]},{
Sign[vx]*Cos[If[vx==0,0,ArcTan[vy/vx]]]*(tend-tstep)*sqrt[vx^
2+vy^2]+tstep*Sin[If[vx==0,0,ArcTan[vy/vx]]],Sign[vx]*Sin[If
[vx==0,0,ArcTan[vy/vx]]]*(tend-tstep)*srt[vx^2+vy^2]-tstep*
Cos[If[vx==0,0,ArcTan[vy/vx]]]}}]],AspectRatio->Automatic,
Axes->True];Do[dopp[i,0,0,10,1],{i,0,1.4,0.2}];Hope I did
that without typos - I can't see how to post the
notebook
====
It would take me a lot of time to understand
howfunctions like SelectionEvaluate works...Cesar.> People
encounter this all the time. It is because> SelectionEvaluate
does > not do what you think. It does not work like>
ToExpression, which causes > immediate kernel evaluation.
Instead it works like> when you press > Shift-Enter, which
selects a cell for evaluation> after all current >
evaluations have finished.> See
>http://support.wolfram.com/mathematica/kernel/interface/
selectionevaluate.html> -Dale>
__________________________________________________Do You
Yahoo!?Yahoo! Finance - Get real-time stock
quoteshttp://finance.yahoo.com
====
I have a set of code that
takes about 5 minutes to run and udpate a varietyof
databases.I would like to add some code that will time the
total lenght of time to runthe code and display it in a
message box.Any idea how to do this?Long range goal would be
to dump the times into excel.This would be so I could compare
the lenght of time to run the code based oncode changes (ie
alter query statements, alter loops, etc)Anyone have
experience with this?
====
this will display the time
difference in minutes!see datediff in MSDN for more
infobefore the code executes datStartTime = nowafter the code
ends datEndTime = now msgbox Code ran from  & datStartTime &
 to  & datEndTime &vbnewline & Total Time to Execute was: 
& datediff(n,datEndTime,datStartTime)> I have a set of code
that takes about 5 minutes to run and udpate avariety> of
databases. I would like to add some code that will time the
total lenght of time torun> the code and display it in a
message box. Any idea how to do this? Long range goal would
be to dump the times into excel. This would be so I could
compare the lenght of time to run the code basedon> code
changes (ie alter query statements, alter loops, etc) Anyone
have experience with this?
====
boxes.When I hit ÔEnter' on any
of the text boxes the form unloads.Howcan I avoid this?What
may be the problem? Help required Plz.Netharam.
====
do you
have a command button to unload the form? if you do, it
sounds likeit may be set as default! if this is the case then
set the default valuefor the command button to false!MikeJust
a thought!! may need more info!> boxes.When I hit 
ÔEnter' on
any of the text boxes the form unloads.How> can I avoid
this?What may be the problem? Help required Plz.
Netharam.
====
It's likely that on one of your command buttons
you have the Enter propertyset to True. I would trace through
the code and find out where the form isunloading. Check the
call stack to see what started it all.--David GugickIntrinsic
Design, Inc.Coefficient - Database Analysis for Microsoft SQL
Serverhttp://www.idisoft.com> boxes.When I hit 
ÔEnter' on any
of the text boxes the form unloads.How> can I avoid this?What
may be the problem? Help required Plz.
Netharam.
====
windows.i.e only the search list of the window.
I have a databse inaccess & I wish to display the fileds of it
in text boxes placedadjacently & when the user resizes one the
adjacent textbox's leftshould automatically be adjusted.I 
aslo
wish to have a common scrollbar to scroll thro' them.I have 
a
vague idea of how this can be done,anyway need some
help.Netharam.
====
You may want to do some reading about the
ListView (or was thatList-View?) control, using the Report
View setting. You can usemultiple columns, optional headers,
it allows column resizing, andtakes care of horizontal and
vertical scroll bars, if they're needed.HTH,R/mi.>
windows.i.e only the search list of the window. I have a
databse in> access & I wish to display the fileds of it in
text boxes placed> adjacently & when the user resizes one the
adjacent textbox's left> should automatically be adjusted.I
aslo wish to have a common scroll> bar to scroll thro' 
them.>
> I have a vague idea of how this can be done,anyway need some
help.> Netharam.
====
the fileds of an access database
table.Actually I wish to print abill,I tried using printform
of the form object,but that was notsatisfactory,so I Ôve
decided to create a report with the detailsstored in arrays
i.e 
ÔitemID','qty',[Ca
pitalOTilde]rate' ,'bill No.' 
etc & link itto the
report dynamically.Now I wish to print this report as a
bill.Is this possible?If so Plz. help me with some
coding.Netharam.
====
What are you asking? How to print?> the
fileds of an access database table.Actually I wish to print a>
bill,I tried using printform of the form object,but that was
not> satisfactory,so I Ôve decided to create a report with
the details> stored in arrays i.e 
ÔitemID','qty',[Ca
pitalOTilde]rate'
,'bill No.' etc & link it> to the report 
dynamically.Now I
wish to print this report as a bill. Is this possible?If so
Plz. help me with some coding. Netharam.
====
Hey all!I just
created a small app that adds a custom number of bands to a
Coolbarcontrol and in each of these bands I place a Toolbar
control with a customnumber of buttons....how can I trap the
events for these custom buttons?The code I'm using is
below...Dim i As Integer ÔGeneral counter Dim j As Integer
ÔGeneral counter Dim ControlName As String ÔName 
of custom
control For i = 1 To 3 CoolBar1.Bands.Add
CoolBar1.Bands(CoolBar1.Bands.Count).Caption = Group # &
Format(i,0) ControlName = tb & Format(i, 0)
Controls.Add MSComctlLib.toolbar, ControlName, CoolBar1
With Controls(ControlName) .TextAlignment = tbrTextAlignRight
.Wrappable = False For j = 1 To 3 .Buttons.Add , , Button #
& Format(j, 0) .Buttons(.Buttons.Count).Style =
tbrButtonGroup Next j .Buttons(1).Value = tbrPressed End With
Set CoolBar1.Bands(CoolBar1.Bands.Count).Child =
Controls(ControlName) Next i
====
Hey all!When a user
double-clicks a Toolbar control, the Customize Toolbar
windowpops up, which is great, but how can I add the Text
Options and Icon Optionscombo boxes to appear on this
window?
====
I am working with Win 2000 and MS Office XP pro and
each time I start orclose Word or Excel or Outlook the
followin message appers:Compile error in hidden module:
AutoExecWhat to do?Nenad
====
tocompile a program so that I can
execute it on the server (IIS,win2000server) regulary. And
that program has to get some data from anAccess (yeah I know,
I shouldn't use Access...) -database and save itin a
Text-file.My big problem is that I only know VB-script. And if
there's as littlecommon between VB and VBscript as between
Java and Javascript I don'tknow how to do it. Is there an
easy way doing what I want?Maybe I can stick to VBscript and
still get the server to excecute thescript regulary (and that
would mean Im in the wrong forum aswell..<;)Well..I bet one of
you gurus out there have the answer to me?///hjalmar
====
I
would like to write some code thatOpens a rs (a query in
access)Loop thru each record in table oneFor each record in
table 1, I want to compare a field (CS which is a numberCross
section) to another table (table2) which has a range values
for CSTable 2:M-100 0 100M-200 100 200M-300 200 300etcI
assume I would need to loop thru each record in TABLE 2 to
find the rangeThen I want to update a field in 
TABLE1 with the
value from TABLE2.Can someone help me write this? Am I
approaching this correctly?
====
ave schrieb:> I would like
to write some code that> Opens a rs (a query in access)>
Loop thru each record in table one> For each record in
table 1, I want to compare a field (CS which is a number>
Cross section) to another table (table2) which has a range
values for CS> Table 2:> M-100 0 100> M-200 100 200> M-300
200 300> etcFirst of all: This might cause wrong data.Which
one of the M-x00 values will be assigned, if CS is exactly
100 or 200 ?> I assume I would need to loop thru each record
in TABLE 2 to find the range> Then I want to update a 
field
in TABLE1 with the value from TABLE2.> Can someone help me
write this? Am I approaching this correctly?Maybe. But why
don't you use SQL ?Let's assume, you try to 
update the field
NewValue in Table1 with the Value of Ref in Table2.Try
this:DoCmd.RunSQL UPDATE Table1 SET NewValue =
[Table2].[Ref]  & This is one instruction only: Quick, easy,
clean, fast, nearly no VBA programming and running on each SQL
Database.Ralf.
====
Does anyone have any idea how to set IE as
the default browser in a script? Dom Ciccotto
====
Quick
question...If one changes the (name) property of a form or an
object on a form is there an easy way to reßect those changes
to the underlying code? I can't imagine everyone names their
object correctly the first pass through! :)Maybe 
I'm on drugs,
but I could have sworn that changes to the (name) of an
object were reßected in the code. I guess I've been away
from VB too long!
====
The answer is no. The code does not
change when an object name changes such as a form or command
button, however, to fix the issue simply use the Replace
function from the Edit Menu in VB. You must have a module or
CODE window open to use the function.It must be noted though
that the predifined EVENTS (when you add them by selecting
them from the combo box at the top of the code page) will
change with object name (But only if you add them again,
existing routines will still be there with old or wrong
object name attached to Form_Unload, etc.Matt> Quick
question...> If one changes the (name) property of a form
or an object on a form is > there an easy way to reßect those
changes to the underlying code? I can't > imagine everyone
names their object correctly the first pass through! :)>
Maybe I'm on drugs, but I could have sworn that changes to
the (name) of an > object were reßected in the code. I guess
I've been away from VB too long!> 
====
Download mztools
(http://www.mztools.com) and let it do all the work for
you.Martin> If one changes the (name) property of a form
or an object on a form is > there an easy way to reßect those
changes to the underlying code? I can't > imagine everyone
names their object correctly the first pass through!
:)>
====
Is there anywhere I can find a license-free library of
statisticalfunctions? In particular, I would like to find an
implementation of theerf() function and the inverse normal
function.In general, is there any one place to go when
hunting for shared code? Ioften find myself in need of math
functions not supported by Visual Basic.Typically, I'll 
spend
days implementing the functions myself, even thoughI'm 
certain
someone else has already done so and is probably quite
willingto share their work.-TC
====
I'm still a VB6 programming
beginner, so this may be a stupid questions, butif I wanted
to make the next letter a user typed to automatically
capitalizeitself, how would I do that?
====
Yes, there is. But
you'll need to be more specific about whaat 
you're doing.For
instance, you can write code for a textbox which will
automaticallyproper case a person's name as 
it's typed. But,
we'll need to know more tohelp...> I'm still a 
VB6
programming beginner, so this may be a stupid questions,but>
if I wanted to make the next letter a user typed to
automaticallycapitalize> itself, how would I do that?
====
I
want to be able to capitalize the first word of every sentence
like Worddoes. I already have code that checks for periods,
question marks, andexclamation points. It then checks for a
space after that. It checks to besure a space follows the
punctuation or it resets the function until thenext period,
etc. It's set not to count double spaces, in case people
preferto use double spaces between sentences. Now I just
need to know what ittakes to turn the next letter they type
(the first letter of the first word> Yes, there 
is. But you'll
need to be more specific about whaat 
you'redoing.> For
instance, you can write code for a textbox which will
automatically> proper case a person's name as 
it's typed.
But, we'll need to know more to> help... I'm 
still a VB6
programming beginner, so this may be a stupid questions,>
but> if I wanted to make the next letter a user typed to
automatically> capitalize> itself, how would I do
that?
====
Joe,I would suspect that Ascii or Replace with UCase
will do the trick for you. I assume that you can detect when a
new sentence is started and thus, the first word of the
sentence. You can use two approaches:'Approach 1:Function
ReplaceFirstChar(Word As String) As String ÔDeclare a local
storage string Dim S As String ÔTrap for no string sent If
Word =  Then ReplaceFirstChar =  Exit Function End If
ÔSelect only the first character S = Left(Word, 1) 
ÔDetermine
if a change is neccessary If Asc(S) > 96 And Asc(S) < 123
Then S = Chr(Asc(S) - 32) End If ÔDecide if there is more to
tack on to the string or not Ôand return the results If
Len(Word) > 1 Then ReplaceFirstChar = S & Right(Word,
Len(Word) - 1) Else ReplaceFirstChar = S End IfEnd
Function'Approach 2:Function ReplaceFirstChar2(Word As
String) As String ÔDeclare a local storage string Dim S As
String ÔTrap for no string sent If Word =  Then
ReplaceFirstChar2 =  Exit Function End If ÔSelect only the
first character S = Left(Word, 1) ÔReplace the 
character in
the word Ôand return the results ReplaceFirstChar2 =
Replace(Word, S, UCase(S), 1, 1, vbTextCompare)End
FunctionNote that both approaches accomplish the same goal. I
prefer to use approach 2 as it is less code and simpler to
use. However I felt that, because you said you are VB
Beginner you should know about the two functions ASC & CHR as
well. They both are handy when you using loops (for/next,
while/wend, do/loop, etc.) to cycle through strings or to
cycle through numbers that will be used to represent strings.
ASC returns the numeric ASCII code of the character, and CHR
returns the ASCII code equivalent (character) of an integer
from 0 to 255. I encourage you to look up all the functions
(ASC, CHR, Replace, Left, Right, Mid, UCase, LCase) and get
familiar with them. They will become your best friends.Hope
that helps,Matt> I want to be able to capitalize the first
word of every sentence like Word> does. I already have code
that checks for periods, question marks, and> exclamation
points. It then checks for a space after that. It checks to
be> sure a space follows the punctuation or it resets the
function until the> next period, etc. It's set not to count
double spaces, in case people prefer> to use double spaces
between sentences. Now I just need to know what it> takes to
turn the next letter they type (the first letter of the 
first
word>>Yes, there is. But you'll need to be more
specific about whaat you're> doing.>>For 
instance, you can
write code for a textbox which will automatically>>proper case
a person's name as it's typed. But, 
we'll need to know more
to>>help...>I'm still a VB6 programming beginner, so this
may be a stupid questions,>>but>if I wanted to make the
next letter a user typed to
automatically>>capitalize>itself, how would I do that?>

====
I am having a problem using the PrintDialog API, maybe
someone can give mesome help.The code is all pretty much the
standard API call, however, when setting theprinter for the
second time (i.e., setting the printer once, then using
thesame function to change something about the setup), I get
the error Theobject invoked has disconnected from its
clients on the line For eachobjPrinter in Printers. Even
stranger, it only does this when settingcertain printers...it
works fine for most printers.fully qualifying the Printers
collection will fix the problem. Though Ihaven't 
tried it...I
couldn't figure out how to fully qualify 
printers (isit,
VB.Printers?).Has anyone ever had this problem? Any ideas how
to fix or work around
it?http://support.microsoft.com/default.aspx?scid=kb;en-us;
q319832
====
HalloCan you take a look onI have problems to get
base working. Has anyone idea what`s wrong withconnection to
base. I tested the base on my private server. There it`s
workperfectly well.Slavomir
====
I did see the following
error.Microsoft OLE DB Provider for ODBC Drivers error
Ô80004005' [Microsoft][ODBC Microsoft Access 
Driver]General
error Unable to open registry key ÔSOFTWAREODBCBrazos
volatile counter'. This error occurs when you read a value
from the registry and do not have the right permissions.Use
registry editor (regedt32.exe) and check the permission for
your IIS user to see if he has the right permission.I have
seen this error when you have a user DSN and system DSN with
the same name.Make sure you do not have a user DSN with the
same name as the one you use.Hope this helps!Ramesh
Thyagarajan, MCSD,MCDBAMicrosoft Developer supportThis
posting is provided AS IS with no warranties, and confers
no rights.Are you secure? For information about the
Microsoft Strategic Technology Protection Program and to
order your FREE Security Tool Kit, please visit
http://www.microsoft.com/security.
====
> When I try to create
schema.ini-file (I tried many different values),> it just
generates error message> Text file specification 
field
separator matches decimal separator or> text delimiter.Found
the problem: Comma is set by default (here in Europe)
torepresent both separator and decimal symbol.I added
DecimalSymbol=. to schema.ini and all the problems
weregone.Damn these cultural/regional differencies!
-Eero
====
I usully use this codeDim cn As ADODB.ConnectionDim
Rs As ADODB.RecordsetSet cn = New
ADODB.Connectioncn.CursorLocation=adUseClientcn.Open
Provider=Microsoft.Jet.OLEDB.4.0;Data Source=D:;Extended
Properties=Text;Set Rs = New
ADODB.RecordsetRs.CursorLocation = adUseClientRs.Open select
* from [products.csv], cn, adOpenStatic, adLockOptimisticSet
DataGrid1.DataSource = RsScot Rose, MCSDMicrosoft Visual Basic
Developer SupportWant to know more? Check out the MSDN at
msdn.microsoft.com or the Microsoft Knowledge Base at
support.microsoft.comThis posting is provided .8bAS IS.8a, with
no warranties, and confers no
rights.--------------------east.nntpserver.com!nntpserver.com
!newsfeed1.easynews.com!easynews.com!easynews!sn-xit-02!
sn-xit-01!sn-xit-08!sn-xit-04!>>I'm trying to connect to a
csv-file, but it does not work properly. I>can connect to 
file
like:>cn.open Driver={Microsoft Text Driver (*.txt;
*.csv)};Dbq= & thermo>& ;Extensions=asc,csv,tab,txt;>>and
open recordset like:>rs.open SELECT * FROM dummy.csv,
cn>>When I try to display results, text driver somehow finds
only one>field(rs.Fields.Count = 1)? Also when I try to
connect similar ODBC>connection with WinSQL, it only shows
one field, First Name.>When I try to create 
schema.ini-file
(I tried many different values),>it just generates error
message>Text file specification field 
separator matches decimal
separator or>text delimiter.>>Here are the first few lines
from the dummy csv
file:John,Doe,555-1234>Jane,Doe,555-5678>>Has
anyone had similar problems? -Eero>
====
app via ODBC, and Im
experiencing the following:Everything has worked fine until I
yesterday got into trouble. There is no problem connecting to
the database, executing procedures. But when I run one 
specific
query(stored procedure), the system hangs and I get the MSSQL
ODBC Driver message Timeout Expired. The procedure reads
from a table which at the time holds ca 600 rows. If I delete
10 of them, the problem dissapears-no matter where in the
table they are placed. It seems to me as if it has reached
its max-limit. If I execute the procedure from Query
Analyser there is no problem at all. This is my query from
VB:dbDBase.rsdbo DBase.Open (SourceString)Anyone who knows
what this is, IÇd be greatfull!!Lars
====
How long does it take
until it timesout? 30 seconds? How long did the query take in
Query Anylyzer? What version of ADO is on the system? What
kind of query are you running? Scot Rose, MCSDMicrosoft
Visual Basic Developer SupportWant to know more? Check out
the MSDN at msdn.microsoft.com or the Microsoft Knowledge
Base at support.microsoft.comThis posting is provided .8bAS
IS.8a, with no warranties, and confers no
rights.-------------------->Sender: Lars Bendix
<lars@frisak.no>>Thread-Index:
AcJI8ucrfYB5gDOXR+mPJJnCjAuC3w==>>app via ODBC, and Im
experiencing the following:>Everything has worked fine until I
yesterday got into >trouble. There is no problem connecting to
the database, >executing procedures. But when I run one
specific query>(stored procedure), the system hangs and I get
the MSSQL >ODBC Driver message Timeout Expired. >The
procedure reads from a table which at the time holds >ca 600
rows. If I delete 10 of t