A8
==
It is a pleasure write to you again. I have
three questions to ask. Alexandre Costa Question One: The
below code is working fine but I think there are more elegant
ways of doing so.Any suggestions are VERY WELCOME.<<
DiscreteMath`Permutations`individual1 =
RandomPermutation[12]individual2 =
RandomPermutation[12]CutPosition = 3 ;CutLength = 4
;CutSequence = Take[Drop[individual1, CutPosition],
CutLength]SonOne = individual2;Do[ If[MemberQ[SonOne,
CutSequence[[i]]], SonOne = Delete[SonOne, Position[SonOne,
CutSequence[[i]]]]], {i, 1, CutLength}];SonOneSonOne =
RotateLeft[SonOne, 1]Do[SonOne = Insert[SonOne,
Reverse[CutSequence][[i]], CutPosition + 1], {i, 1,
CutLength}]SonOneQuestion Two:How can I change points
properties (such as Size and Color) for the plot below? The
PlotStyle Option does not work for this LabeledListPlot<<
Graphics`Graphics`listcities = {{1, 5}, {4, 6}, {7, 5}, {5,
4}, {9, 4}, {2, 3}, {4, 2}, {6, 2}, {1, 1}, {5, 1}, {3, 0},
{9, 0}};LabeledListPlot[listcities, Axes -> None, Frame ->
True, DisplayFunction -> $DisplayFunction]Question Three:
Why the Goto statement below is not working?q =
2;Label[start];q = 3;Label[begin];Print[q];q += 1; If[q
< 6, Goto[begin], Goto[start]]Reply-To:
murray@math.umass.edu
====
I presume packagename is a context
name, e.g., Graphics`Colors` . Then evaluating the
expression Names[packagename*]containing the wildcard
symbol * will probably do what you want.It may tell you more
than you want to know unless the package was decently
written. By decently I mean that names not needing to be
known to the outside world have been enclosed within a
Private context within the package's context.> After I load
a package using> <<packagename> Is there a way to figure out
what the package exports without editing> the source file?>
-- Murray Eisenberg murray@math.umass.eduMathematics &
Statistics Dept.Lederle Graduate Research Tower phone 413
549-1020 (H)University of Massachusetts 413 545-2859 (W)710
North Pleasant StreetAmherst, MA 01375
====
Jose,I
generally use the Needs statement to load a package. It
avoids doubleloading and all the error messages that
result. SayNeeds[Graphics`Graphics`]Then to obtain
information on the routines in the package just
use...?Graphics`Graphics`*names in the package. If you click
on any of the names you obtain the usagemessage for the
name. (On earlier Mathematica versions you only obtain
alist of the names and you have to use ?name to obtain the
usage message.)More link. If you click on that it will
bring you to the relevant Helppage. (Private packages may
not have that feature.)David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/
Sender: steve@smc.vnet.netApproved: Steven M. Christensen
<steve@smc.vnet.net>, ModeratorReply-To:
<drbob@bigfoot.com>
====
Maybe there's a faster way -- I hope
so! -- but here's my answer:n = 2^26Timing[RealDigits[N[Pi
- 3, n], 10, 20, 19 - n]]67108864{15027.922*Second, {{3, 3,
8, 6, 3, 2, 2, 0, 8, 9, 6, 2, 2, 3, 4, 0, 9, 8, 0,
3},-67108844}}hrMinSec[15027.922]{4*hours, 10*minutes,
27.92200000000048*seconds}Bobby Treat-----Original
Message-----
====
Could you tell me the CPU you used and its
speed etc...i am curious,other programs out there.> Maybe
there's a faster way -- I hope so! -- but 
here's my answer:>
n = 2^26> Timing[RealDigits[N[Pi - 3, n], 10, 20, 19 - n]]>
67108864> {15027.922*Second,> {{3, 3, 8, 6, 3, 2, 2, 0, 8, 9,
6, 2, 2, 3, 4, 0, 9, 8, 0, 3},> -67108844}}>
hrMinSec[15027.922]> {4*hours, 10*minutes,
27.92200000000048*seconds}> Bobby Treat> -----Original
Message-----> 4 for me?> Could someone calculate the number
Pi to 67,108,864 (2^26) decimal> places> I made the
calculation in another program and would like to verify its>
> 
====
> Could you tell me the CPU you used and its speed
etc...i am curious,> other programs out there.I used one
processor of a dual 1GH Mac and got the same answer with
thefollowing speed:4.2 for Mac OS X (June 4, 2002)oldmax =
$MaxPrecision 61. 10$MaxPrecision = 
InfinityInfinityWith[{n
= 2^26}, Timing[ pd = RealDigits[N[Pi, n + 1], 10, 20, 19 -
n]; ]]{28794.1 Second, Null}MaxMemoryUsed[]512055204pd{{3,
3, 8, 6, 3, 2, 2, 0, 8, 9, 6, 2, 2, 3, 4, 0, 9, 8, 0, 3},
-67108844}Tom Burton-- 
====
So would it take about the same
amont of time for the complete printoutof digits? Of course
it would take a few additional seconds to formatthe
output...Or does Mathematica take alot less time when it
truncates the output?> Could you tell me the CPU you used
and its speed etc...i am curious,> other programs out there.>
I used one processor of a dual 1GH Mac and got the same answer
with the> following speed:> 4.2 for Mac OS X (June 4, 2002)>
oldmax = $MaxPrecision> 6> 1. 10> $MaxPrecision = Infinity>
Infinity> With[{n = 2^26}, Timing[> pd = RealDigits[N[Pi, n +
1], 10, 20,> 19 - n]; ]]> {28794.1 Second, Null}>
MaxMemoryUsed[]> 512055204> pd> {{3, 3, 8, 6, 3, 2, 2, 0, 8,
9, 6, 2, 2, 3,> 4, 0, 9, 8, 0, 3}, -67108844}> Tom
Burton
====
Here's an answer to Question 1:individual1 =
RandomPermutation[12]individual2 =
RandomPermutation[12]cutPosition = 3;cutLength =
4;cutSequence = Take[Drop[individual1, cutPosition],
cutLength]sonOne = Complement[individual2,
cutSequence]sonOne = RotateLeft[sonOne, 1]sonOne =
Join[Take[sonOne, cutLength - 1], cutSequence, Drop[sonOne,
cutLength - 1]]Bobby Treat-----Original Message-----<<
DiscreteMath`Permutations`individual1 =
RandomPermutation[12]individual2 =
RandomPermutation[12]CutPosition = 3 ;CutLength = 4
;CutSequence = Take[Drop[individual1, CutPosition],
CutLength]SonOne = individual2;Do[ If[MemberQ[SonOne,
CutSequence[[i]]], SonOne = Delete[SonOne, Position[SonOne,
CutSequence[[i]]]]], {i, 1, CutLength}];SonOneSonOne =
RotateLeft[SonOne, 1]Do[SonOne = Insert[SonOne,
Reverse[CutSequence][[i]], CutPosition + 1], {i, 1,
CutLength}]SonOneQuestion Two:How can I change points
properties (such as Size and Color) for the plotbelow? The
PlotStyle Option does not work for this LabeledListPlot<<
Graphics`Graphics`listcities = {{1, 5}, {4, 6}, {7, 5}, {5,
4}, {9, 4}, {2, 3}, {4, 2}, {6, 2}, {1, 1}, {5, 1}, {3, 0},
{9, 0}};LabeledListPlot[listcities, Axes -> None, Frame ->
True, DisplayFunction -> $DisplayFunction]Question Three:
Why the Goto statement below is not working?q =
2;Label[start];q = 3;Label[begin];Print[q];q += 1; If[q
< 6, Goto[begin], Goto[start]]
====
I would like to build my
Mathematica notebooks in manner which allows me to carry out
two overlapping purposes: 1) prototype an algorithm
completely within Mathematica, 2) use Mathematica as a
partial evaluator to splice derived expressions into a C
language version of the algorithm (with the aid of the
Format package from MathSource). The key complication in
doing this is that for the Mathematica prototype I want
everything to be evaluated, while for splicing I want the
results from some of the functions to be retained as data in
temporary variables.A first step towards these purposes is
simple: write my component functions so they operate on
lists (the natural form of the prototype data), then provide
lists of arrays for arguments which I would like to supply as
data in the resulting C language output forms.An
example:--------------------------------------------------In
[1]:= t[x_] := {x[[4]]-x[[3]], x[[2]]-x[[1]]}In[2]:=
f[x_,y_] := {y, 1}.t[x]In[3]:= f[{1,2,3,4},2]Out[3]=
3In[4]:= f[Array[x,4],2]Out[4]= -x[1] + x[2] + 2 (-x[3] +
x[4])In[5]:= CAssign[fcnval, f[Array[x,4],2],
AssignToArray->{x}]Out[5]//OutputForm=
fcnval=-x[1]+x[2]+2.*(-x[3]+x[4]);---------------------------
-----------------------For various reasons, however, I would
like certain expressions of my symbolic derivation to be
treated as data for the C language version. In the above
example, for instance, I would like the array that function
t produces to be a data array. The modification of the above
example:--------------------------------------------------In[
6]:= tc[x_] := Array[tt,2]In[7]:= fc[x_,y_] := {y,
1}.tc[x]In[8]:= CAssign[tt, t[Array[x,4]],
AssignToArray->{x}]Out[8]//OutputForm= tt[0]=-x[3]+x[4];
tt[1]=-x[1]+x[2];In[9]:= CAssign[fcnval, fc[Array[x,4],2],
AssignToArray->{tt}]Out[9]//OutputForm=
fcnval=2.*tt[1]+tt[2];---------------------------------------
-----------But I don't want to carry around two versions of
everything, nor do I really want to thread all of the
supporting functions through my definitions in order to
choose the correct function for the purpose I would rather
define the C language alternatives only for functions like t
and tc in the example, with f using the appropriate one
depending on some evaluation ßag that I set at the highest
level.For example, it would be nice to be able to sprinkle
in the C language alternatives with a construct
likeIn[10]:= t[x_] := {x[[4]]-x[[3]],
x[[2]]-x[[1]]}In[11]:= DefineTemporaryForm[t[x_]] :=
Array[tt,Length[x]/2]then be able to writeIn[8]:=
CAssign[tt, t[Array[x,4]],
AssignToArray->{x}]Out[8]//OutputForm= tt[0]=-x[3]+x[4];
tt[1]=-x[1]+x[2];In[9]:= CAssign[fcnval,
UseTemporaryForm[f[Array[x,4],2]],
AssignToArray->{tt}]Out[9]//OutputForm=
fcnval=2.*tt[1]+tt[2];I would appreciate any pointers on a
good, clean and hopefully simple way to do this within
Mathematica!Alex
====
I've been using Mathematica 4.1 on
Win98 as a word processor formath-related documents, but
often people that need to see the documentsdon't have
Mathematica, and for whatever reason on my computer the
HTMLsaves don't work at all. I'd like to 
export to PDF
format. I can exportimages to PDF format no problem using,
for exampleExport[c:docsplot3.pdf,
Plot[Sin[x],{x,-2Pi,2Pi}]],and I can export cells correctly
to GIF, JPEG, and WMF formats (probablymore, those are the
only ones I tested) using, for
exampleExport[c:docscell4.gif, Cell[ <<...(copied
cell data from Edit->CopyAs->Cell Expression)...> ]]When
I change the filename to a .PDF and evaluate the cell, the
programdisplays ÔRunning...' for a second and 
gives the
ÔOut[n] = c:docscell4.pdf' message as if a 
file was
created, but no file is created anywhere with anyname that
I could find with Start->Find->[All files and 
folders created
inthe previous day].Is there a limitation to PDF exporting
I don't know about? Do I need toupgrade to 4.2? Am I doing
something wrong with the Export[] command? Do Ineed a
faster computer? A patch? Something else?
====
I recently
communicated with technical support about precisely this
issue. Here's their reply:``PDF and AI export use psrender,
which is a MathLink program that interfaces with the kernel.
Since the kernel has no knowledge of how cells are formatted,
export cannot generate PDF and AI for cells, just
Graphics.Note that this is contrary to the documentation,
which says:``All graphics formats in Export can handle any
type of 2D or 3D Mathematica graphics. ... They can also
handle Notebook and Cell objects.---Selwyn Hollis> I've
been using Mathematica 4.1 on Win98 as a word processor for>
math-related documents, but often people that need to see the
documents> don't have Mathematica, and for whatever reason
on my computer the HTML> saves don't work at all. 
I'd like
to export to PDF format. I can export> images to PDF format
no problem using, for example>
Export[c:docsplot3.pdf, Plot[Sin[x],{x,-2Pi,2Pi}]],>
and I can export cells correctly to GIF, JPEG, and WMF
formats (probably> more, those are the only ones I tested)
using, for example> Export[c:docscell4.gif, Cell[
<<...(copied cell data from Edit->Copy> As->Cell
Expression)...> ]]> When I change the filename to a .PDF and
evaluate the cell, the program> displays 
ÔRunning...' for a
second and gives the ÔOut[n] = c:docscell4.pdf> Ô
message as if a file was created, but no file is 
created
anywhere with any> name that I could find with
Start->Find->[All files and folders created in> the previous
day].> Is there a limitation to PDF exporting I don't know
about? Do I need to> upgrade to 4.2? Am I doing something
wrong with the Export[] command? Do I> need a faster
computer? A patch? Something else?> 
====
Here's a smoother
animation, taking into account the period and cuttingthe
step size in half (without using more frames):Do[Show[curve,
Graphics[Disk[{ xx[t], Cosh[xx[t]]}, 0.035]], PlotRange ->
{{-1.2, 1.2}, {0.9,1.65}}, AspectRatio -> Automatic, Axes ->
None], {t, 0, 2.3, 0.05}]SelectionMove[EvaluationNotebook[],
All,
GeneratedCell]FrontEndTokenExecute[OpenCloseGroup]
FrontEndTokenExecute[SelectionAnimate]Bobby
Treat-----Original Message-----and the equation of motion
is diffeq = Simplify[ D[D[L, x'[t]], t] ] == Simplify[ D[L,
x[t]] ]Now solve and animate ... xx[t_] = x[t]/. First[
NDSolve[{diffeq, x[0] == -1, x'[0] == 0}, x[t], {t, 0, 5}]]
curve = Plot[Cosh[x], {x, -1, 1}] Do[ Show[curve,
Graphics[Disk[{xx[t], Cosh[xx[t]]}, 0.025]], PlotRange ->
{{-1.2, 1.2}, {0.9, 1.65}}, AspectRatio -> Automatic,
Axes->None], {t, 0, 5, 0.1}]----Selwyn Hollis> Dear
Colleagues,> I intend to make an animation in which > ball A
rolls down on an inclined plane from the left whilst> ball B
- starting from the same height - rolls down Cosh[t]'s
pathfrom the> right.> x-axis is time t, y-axis is height
h.> Ball A is fine; ball B - which should arrive at h=0 before
A - isbeyond my> means.> Matthias Bode> Sal. Oppenheim
jr. & Cie. KGaA> Koenigsberger Strasse 29> D-60487 Frankfurt
am Main> GERMANY> Mobile: +49(0)172 6 74 95 77> Internet:
http://www.oppenheim.de> 
====
I would like to build my
Mathematica notebooks in manner which allows me to carry out
two overlapping purposes: 1) prototype an algorithm
completely within Mathematica, 2) use Mathematica as a
partial evaluator to splice derived expressions into a C
language version of the algorithm (with the aid of the Format
package from MathSource). The key complication in doing this
is that for the Mathematica prototype I want everything to
be evaluated, while for splicing I want the results from
some of the functions to be retained as data in temporary
variables.A first step towards these purposes is simple:
write my component functions so they operate on lists (the
natural form of the prototype data), then provide lists of
arrays for arguments which I would like to supply as data in
the resulting C language output forms.An
example:--------------------------------------------------In
[1]:= t[x_] := {x[[4]]-x[[3]], x[[2]]-x[[1]]}In[2]:=
f[x_,y_] := {y, 1}.t[x]In[3]:= f[{1,2,3,4},2]Out[3]=
3In[4]:= f[Array[x,4],2]Out[4]= -x[1] + x[2] + 2 (-x[3] +
x[4])In[5]:= CAssign[fcnval, f[Array[x,4],2],
AssignToArray->{x}]Out[5]//OutputForm=
fcnval=-x[1]+x[2]+2.*(-x[3]+x[4]);---------------------------
-----------------------For various reasons, however, I would
like certain expressions of my symbolic derivation to be
treated as data for the C language version. In the above
example, for instance, I would like the array that function
t produces to be a data array.The modification of the above
example:--------------------------------------------------In
[6]:= tc[x_] := Array[tt,2]In[7]:= fc[x_,y_] := {y,
1}.tc[x]In[8]:= CAssign[tt, t[Array[x,4]],
AssignToArray->{x}]Out[8]//OutputForm= tt[0]=-x[3]+x[4];
tt[1]=-x[1]+x[2];In[9]:= CAssign[fcnval, fc[Array[x,4],2],
AssignToArray->{tt}]Out[9]//OutputForm=
fcnval=2.*tt[1]+tt[2];---------------------------------------
-----------But I don't want to carry around two versions of
everything, nor do I really want to thread all of the
supporting functions through my definitions in order to
choose the correct function for the purpose I would rather
define the C language alternatives only for functions like t
and tc in the example, with f using the appropriate one
depending on some evaluation ßag that I set at the highest
level.For example, it would be nice to be able to sprinkle
in the C language alternatives with a construct
likeIn[10]:= t[x_] := {x[[4]]-x[[3]],
x[[2]]-x[[1]]}In[11]:= DefineTemporaryForm[t[x_]] :=
Array[tt,Length[x]/2]then be able to writeIn[8]:=
CAssign[tt, t[Array[x,4]],
AssignToArray->{x}]Out[8]//OutputForm= tt[0]=-x[3]+x[4];
tt[1]=-x[1]+x[2];In[9]:= CAssign[fcnval,
UseTemporaryForm[f[Array[x,4],2]],
AssignToArray->{tt}]Out[9]//OutputForm=
fcnval=2.*tt[1]+tt[2];I would appreciate any pointers on a
good, clean and hopefully simple way to do this within
Mathematica!Alex
====
To find out what loaded contexts are
related to the package,
use:Contexts[Integrate`*]{Integrate`,Integrate`Elliptic`}
To see the symbols exported by the context,
execute:Names[Integrate`*]{(* too many to list here
*)}Names[Integrate`*`*]{Integrate`Elliptic`Elliptic}To
find out what context a symbol comes
from:Context[Integrate]System`Bobby Treat-----Original
Message-----Sender: steve@smc.vnet.netApproved: Steven M.
Christensen <steve@smc.vnet.net>, ModeratorReply-To:
<drbob@bigfoot.com>
====
Oops! That should beindividual1 =
RandomPermutation[12]individual2 =
RandomPermutation[12]cutPosition = 3;cutLength =
4;cutSequence = Take[Drop[individual1, cutPosition],
cutLength]sonOne = DeleteCases[individual2,
_?(MemberQ[cutSequence, #] &)]sonOne = RotateLeft[sonOne,
1]sonOne = Join[Take[ sonOne, cutLength - 1], cutSequence,
Drop[sonOne, cutLength - 1]]Complement returns a sorted
result, and you didn't want that.Bobby Treat-----Original
Message-----sonOne = Join[Take[sonOne, cutLength - 1],
cutSequence, Drop[sonOne, cutLength - 1]]Bobby
Treat-----Original Message-----<<
DiscreteMath`Permutations`individual1 =
RandomPermutation[12]individual2 =
RandomPermutation[12]CutPosition = 3 ;CutLength = 4
;CutSequence = Take[Drop[individual1, CutPosition],
CutLength]SonOne = individual2;Do[ If[MemberQ[SonOne,
CutSequence[[i]]], SonOne = Delete[SonOne, Position[SonOne,
CutSequence[[i]]]]], {i, 1, CutLength}];SonOneSonOne =
RotateLeft[SonOne, 1]Do[SonOne = Insert[SonOne,
Reverse[CutSequence][[i]], CutPosition + 1], {i, 1,
CutLength}]SonOneQuestion Two:How can I change points
properties (such as Size and Color) for the plotbelow? The
PlotStyle Option does not work for this LabeledListPlot<<
Graphics`Graphics`listcities = {{1, 5}, {4, 6}, {7, 5}, {5,
4}, {9, 4}, {2, 3}, {4, 2}, {6, 2}, {1, 1}, {5, 1}, {3, 0},
{9, 0}};LabeledListPlot[listcities, Axes -> None, Frame ->
True, DisplayFunction -> $DisplayFunction]Question Three:
Why the Goto statement below is not working?q =
2;Label[start];q = 3;Label[begin];Print[q];q += 1; If[q
< 6, Goto[begin], Goto[start]]
====
I would appreciate help
with these problems:1. I'm plotting several thousand points,
which I can doeither with something like this (this is a
test):w = Table[Point[{Random[], Random[]}], {i, 1,
4096}];x = Table[Point[{Random[], Random[]}], {i, 1,
4096}];y = Table[Point[{Random[], Random[]}], {i, 1,
4096}];z = Table[Point[{Random[], Random[]}], {i, 1,
4096}];dw = Graphics[{PointSize[0.01], RGBColor[ 1, 0, 0],
w}];dx = Graphics[{PointSize[0.01], RGBColor[.8, .8, .8],
x}];dy = Graphics[{PointSize[0.01], RGBColor[ 0, .5, .9],
y}];dz = Graphics[{PointSize[0.01], RGBColor[.8, .8, 0],
z}];Show[dw, dx, dy, dz, AspectRatio -> Automatic, PlotRange
-> {{0, 1}, {0, 1}}, Axes -> Automatic, Frame -> True,
Background -> GrayLevel[.026],This gives me dots in 4 colors
for distinguishing differentkinds of points in my real
application. This works fine butneeds the Point structure.
Or, I can do (this is for onekind of point),t =
Table[{Random[], Random[]}, {i, 1, 1024}];ListPlot[t,
AspectRatio -> Automatic, Axes -> Automatic, Frame -> True,
Background -> GrayLevel[.026] ];which seems simpler and may
fit into the rest of the programmore easily.1. How do I get
the RGBColor Rule or the equivalent intothe latter? The
RGBColor[] call is not a rule.2. In the latter case I also
want to plot 4 types of points.How do I get ListPlot to put
down 4 plots superimposed?Or can't I?3. In either case, I
need to make the whole plot area abouttwice as big. That
is, it now occupies about a 4 square. Tosee details better
in my real plot, and because with 16k pointsthe small plot
just looks almost like a solid blur, I want tomake it more
like 8 square, or as big as will fit the screen(without
changing the plot range or anything else). Theremust be a
scale factor somewhere.
====
Steve,You could try something
like this:Needs[Graphics`Colors`]Clear[t];Do[t[i] =
Table[{Random[], Random[]}, {i, 1, 1024}], {i,
4}];Show[Graphics[ {PointSize[0.005], RoyalBlue, Point /@
t[1], OrangeRed, Point /@ t[2], SpringGreen, Point /@ t[3],
CadmiumLemon, Point /@ t[4]}], AspectRatio -> Automatic,
Frame -> True, PlotRegion -> {{0.02, 0.96}, {0, 1}},
Background -> IvoryBlack, ImageSize -> 700];The overall size
of the plot can be controlled with the ImageSize
option.ListPlot is more a hindrance than a help - throw it
in the ash can. Just MapPoint onto the lists of point
coordinates. Give the color before each set ofpoints. I
adjusted the PlotRegion to obtain some black margin on all
sidesof the frame.David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/dy
= Graphics[{PointSize[0.01], RGBColor[ 0, .5, .9], y}];dz =
Graphics[{PointSize[0.01], RGBColor[.8, .8, 0], z}];Show[dw,
dx, dy, dz, AspectRatio -> Automatic, PlotRange -> {{0, 1},
{0, 1}}, Axes -> Automatic, Frame -> True, Background ->
GrayLevel[.026],This gives me dots in 4 colors for
distinguishing differentkinds of points in my real
application. This works fine butneeds the Point structure.
Or, I can do (this is for onekind of point),t =
Table[{Random[], Random[]}, {i, 1, 1024}];ListPlot[t,
AspectRatio -> Automatic, Axes -> Automatic, Frame -> True,
Background -> GrayLevel[.026] ];which seems simpler and may
fit into the rest of the programmore easily.1. How do I get
the RGBColor Rule or the equivalent intothe latter? The
RGBColor[] call is not a rule.2. In the latter case I also
want to plot 4 types of points.How do I get ListPlot to put
down 4 plots superimposed?Or can't I?3. In either case, I
need to make the whole plot area abouttwice as big. That
is, it now occupies about a 4 square. Tosee details better
in my real plot, and because with 16k pointsthe small plot
just looks almost like a solid blur, I want tomake it more
like 8 square, or as big as will fit the screen(without
changing the plot range or anything else). Theremust be a
scale factor somewhere.
====
Bobby,Your ODE is different from
both mine and Borut's. You do however get the same as mine
with your approach, if you differentiate the total energy
rather than the Lagrangian. KE = Simplify[(x'[t]^2 +
D[f[x[t]], t]^2)/2]; PE = g*f[x[t]]; totalE = KE + PE; treat
= D[totalE, t] == 0 x''[t] - 
(x''[t] /. First[Solve[treat,
x''[t]]]//Apart) == 0Borut's 
ODE differs only in his factor
of 2 in the first-order term. I suspect that might have been
a typo... Borut?By the way, some nice animations of
mechanical systems (some based on the same sort of approach)
can be seen here:
http://www.math.armstrong.edu/faculty/hollis/DEmovies/--
Selwyn> The two of you derived slightly different ODE's. I
think Selwyn's is> correct, but I only had one physics
course, 30 years ago.> Here's a notebook expression showing
how Selwyn's approach can be used> to derive the ODE for
y[t] = f[x[t]], with arbitrary f. It shows that> solution in
a form that is easily compared with Borut's ODE, and then 
it>
shows my own solution (same as Borut's, by a slightly 
simpler
method).> Bobby Treat<snip> KE = Simplify[(x'[t]^2 +
D[f[x[t]], t]^2)/2]; > PE = g*f[x[t]]; > L = KE - PE; > treat
= D[L, t] == 0<snip>
====
I've modified Selwyn's solution 
to
make it more general. In particular,the height can be
specified (up to about 35 meters). The differentialequation
is solved for t=0 to 5 first. The quarter-period is
computedby finding a zero; then the differential equation is
solved again fort=0 to the quarter-period, and the solution
is extended using reßectionand periodicity. This yields a
higher-precision solution. Then I graphthe solution, with a
stepsize equal to period/40, from t=0 to 
Ôperiod',labeling
each frame with the values of t, x[t], and y[t]. Using a
stepsize that divides period/4 guarantees the lowest point
is reached ON aframe when appropriate.Here's the solution
as a notebook
expression:Notebook[{Cell[CellGroupData[{Cell[Borut L's
solution:, Subsubtitle],Cell[TextData[StyleBox[Having
noticed your statement  ... BEYOND MY MEANS  I thing
you aren't yetnfamiliar with Lagrangian formalism. 
It's
quite easy to derive a generalnequation of motion for a
point mass, subjected to gravity and to moving on ancurve f
= y(x) (i.e. f = Cosh[#]&).nn1) I'll leave re-deriving
equation to you, here is what I've got (just copynpaste
it).:, FontFamily->Courier New, FontSize->10,
CharacterEncoding->WindowsANSI]], Text],Cell[BoxData[
RowBox[{selwyn, =, RowBox[{First, [, RowBox[{
RowBox[{ RowBox[{ RowBox[{x, ''}], [, t, ]}],
/., RowBox[{Solve, [, RowBox[{diffeq, ,, RowBox[{
RowBox[{x, ''}], [, t, ]}]}], ]}]}], //,
Simplify}], ]}]}]], Input],Cell[BoxData[
RowBox[{borut, =, RowBox[{First, [, RowBox[{
RowBox[{ RowBox[{x, ''}], [, t, ]}], /.,
RowBox[{Solve, [, RowBox[{ RowBox[{getEq, [,
Cosh, ]}], ,, RowBox[{ RowBox[{x, ''}], [,
t, ]}]}], ]}]}],  ]}]}]], Input],Cell[BoxData[
RowBox[{ RowBox[{getEq, [, f_, ]}], :=,
[IndentingNewLine], RowBox[{Simplify, [, RowBox[{
RowBox[{ RowBox[{ RowBox[{x, ''}], [, t, ]}],
+, RowBox[{ SuperscriptBox[ RowBox[{ RowBox[{x, '}],
[, t, ]}], 2],  , FractionBox[ RowBox[{2,  ,
RowBox[{ RowBox[{f, '}], [, RowBox[{x, [, t,
]}], ]}],  , RowBox[{ RowBox[{f, ''}], [,
RowBox[{x, [, t, ]}], ]}]}], RowBox[{1, +,
SuperscriptBox[ RowBox[{ RowBox[{f, '}], [,
RowBox[{x, [, t, ]}], ]}], 2]}]]}], +,
FractionBox[ RowBox[{g,  , RowBox[{ RowBox[{f, '}],
[, RowBox[{x, [, t, ]}], ]}]}], RowBox[{1,
+, SuperscriptBox[ RowBox[{ RowBox[{f, '}], [,
RowBox[{x, [, t, ]}], ]}], 2]}]]}],==,
0}], ]}]}]], Input],Cell[TextData[{ StyleBox[2)
Next, you integrate it .:, FontFamily->Courier New,
FontSize->10, CharacterEncoding->WindowsANSI], }],
Text],Cell[BoxData[ RowBox[{ RowBox[{getSol, [,
RowBox[{f_, ,, RowBox[{h0_, ?, Positive}], ,,
RowBox[{x0_, ?, Positive}]}], ]}], :=,
RowBox[{Module, [, RowBox[{ RowBox[{{, tStop,
}}], ,, [IndentingNewLine], RowBox[{First, @,
RowBox[{NDSolve, [, [IndentingNewLine], RowBox[{
RowBox[{{, RowBox[{ RowBox[{getEq, [, f, ]}],
,, RowBox[{ RowBox[{x, [, 0, ]}], [Equal],
h0}], ,, RowBox[{ RowBox[{ RowBox[{x, '}], [,
0, ]}], [Equal], 0}]}], }}], ,,
[IndentingNewLine], x, ,, [IndentingNewLine],
RowBox[{{, RowBox[{t, ,, 0, ,, 10}], }}],
,, [IndentingNewLine], RowBox[{MaxStepSize,
[Rule], RowBox[{1, /, 100}]}], ,,
RowBox[{StoppingTest, [RuleDelayed], RowBox[{If,
[, RowBox[{ RowBox[{h0, <, 0}], ,, RowBox[{x,
>, 0}], ,, RowBox[{x, <, 0}]}], ]}]}]}],
[IndentingNewLine], ]}]}]}], [IndentingNewLine],
]}]}]], Input],Cell[<3) Here follows animation
code, specially for linear versus cosh case, apply my
initial conditions (below)>, Text],Cell[BoxData[
RowBox[{ RowBox[{makeDuo, [, RowBox[{f_, ,, g_,
,, h0_}], ]}], :=, RowBox[{Module, [, RowBox[{
RowBox[{{, RowBox[{ RowBox[{solf, =,
RowBox[{getSol, [, RowBox[{f, ,, h0}], ]}]}],
,, RowBox[{solg, =, RowBox[{getSol, [,
RowBox[{g, ,, h0}], ]}]}], ,, tf, ,, tg,
,, maxT, ,, minT}], }}], ,,
[IndentingNewLine], RowBox[{ RowBox[{tf, =,
RowBox[{solf, [, RowBox[{[, RowBox[{ 1, ,, 2,
,, 1, ,, 1, ,, 2}], ]}], ]}]}], ;,
[IndentingNewLine], RowBox[{tg, =, RowBox[{solg,
[, RowBox[{[, RowBox[{ 1, ,, 2, ,, 1, ,, 1,
,, 2}], ]}], ]}]}], ;, [IndentingNewLine],
RowBox[{maxT, =, RowBox[{Max, [, RowBox[{{,
RowBox[{tf, ,, tg}], }}], ]}]}], ;,
[IndentingNewLine], RowBox[{minT, =, RowBox[{Min,
[, RowBox[{{, RowBox[{tf, ,, tg}], }}], ]}]}],
;, [IndentingNewLine], RowBox[{Do, [, RowBox[{
RowBox[{Plot, [, RowBox[{ RowBox[{{, RowBox[{
RowBox[{f, @, x}], ,, RowBox[{g, @, x}]}],
}}], ,, RowBox[{{, RowBox[{x, ,, 0, ,,
RowBox[{ArcCosh, @, h0}]}], }}], ,,
RowBox[{AspectRatio, [Rule], Automatic}],  ,,
RowBox[{Frame, [Rule], True}], ,,
RowBox[{Axes, [Rule], False}], ,,
RowBox[{Epilog, [Rule], RowBox[{{, RowBox[{ RowBox[{
AbsolutePointSize, [, 10, ]}], ,, RowBox[{Hue,
[, 0, ]}], ,, RowBox[{Point, [, RowBox[{
RowBox[{{, RowBox[{ RowBox[{x, [, t, ]}], ,,
RowBox[{f, @, RowBox[{x, [, t, ]}]}]}],
}}], /., solf}], ]}], ,,  RowBox[{Hue, [,
.6, ]}], ,, RowBox[{Point, [, RowBox[{
RowBox[{{, RowBox[{ RowBox[{x, [, t, ]}], ,,
RowBox[{g, @, RowBox[{x, [, t, ]}]}]}],
}}], /., solg}], ]}]}], }}]}]}], ]}], ,,
RowBox[{{, RowBox[{t, ,, 0, ,, minT, ,,
minT}], }}]}], ]}]}]}], ]}]}]],
Input],Cell[TextData[{ StyleBox[4) My initial
conditions. In my opinion, you weren't true on this.
Saying nSTARTING FROM THE SAME HEIGHT  is not enough
- you should specify x asnwell, thus specifing starting
POINT and not just height y., FontFamily->Courier New,
FontSize->10, CharacterEncoding->WindowsANSI], }],
Text],Cell[BoxData[{ RowBox[{makeDuo, [, RowBox[{
RowBox[{ RowBox[{ RowBox[{Cosh, [, 1., ]}],  ,
#}], &}], ,, RowBox[{ RowBox[{Cosh, [, #,
]}], &}], ,, 23}],  ]}], [IndentingNewLine],
RowBox[{SelectionMove, [, RowBox[{
RowBox[{EvaluationNotebook, [, ]}], ,, All, ,,
GeneratedCell}], ]}], n, RowBox[{
FrontEndTokenExecute, [, <OpenCloseGroup>,
]}], n, RowBox[{ FrontEndTokenExecute, [,
<SelectionAnimate>, ]}]}], Input]}, Open
]]},ScreenRectangle->{{0, 1024}, {0,
711}},WindowSize->{815, 569},WindowMargins->{{0,
Automatic}, {Automatic, -1}},ShowSelection->True]Bobby
Treat-----Original Message-----and the equation of motion
is diffeq = Simplify[ D[D[L, x'[t]], t] ] == Simplify[ D[L,
x[t]] ]Now solve and animate ... xx[t_] = x[t]/. First[
NDSolve[{diffeq, x[0] == -1, x'[0] == 0}, x[t], {t, 0, 5}]]
curve = Plot[Cosh[x], {x, -1, 1}] Do[ Show[curve,
Graphics[Disk[{xx[t], Cosh[xx[t]]}, 0.025]], PlotRange ->
{{-1.2, 1.2}, {0.9, 1.65}}, AspectRatio -> Automatic,
Axes->None], {t, 0, 5, 0.1}]----Selwyn Hollis> Dear
Colleagues,> I intend to make an animation in which > ball A
rolls down on an inclined plane from the left whilst> ball B
- starting from the same height - rolls down Cosh[t]'s
pathfrom the> right.> x-axis is time t, y-axis is height
h.> Ball A is fine; ball B - which should arrive at h=0 before
A - isbeyond my> means.> Matthias Bode> Sal. Oppenheim
jr. & Cie. KGaA> Koenigsberger Strasse 29> D-60487 Frankfurt
am Main> GERMANY> Mobile: +49(0)172 6 74 95 77> Internet:
http://www.oppenheim.de> 
====
Try this:<< Graphics`Colors`n
= 4096;data = {w, x, y, z} = Array[Random[] &, {4, n,
2}];g@{a_, b_} := ListPlot[a, PlotStyle ->
{PointSize[0.005], b}, DisplayFunction -> Identity,Show[g /@
Transpose[{data, {Red, Blue, Yellow, White}}], PlotRange ->
{{0, 1}, {0, 1}}, Axes -> Automatic, Frame -> True,
DisplayFunction -> $DisplayFunction, Background ->
GrayLevel[.026], ImageSize -> 500];Bobby Treat-----Original
Message-----dx = Graphics[{PointSize[0.01], RGBColor[.8, .8,
.8], x}];dy = Graphics[{PointSize[0.01], RGBColor[ 0, .5,
.9], y}];dz = Graphics[{PointSize[0.01], RGBColor[.8, .8,
0], z}];Show[dw, dx, dy, dz, AspectRatio -> Automatic,
PlotRange -> {{0, 1}, {0, 1}}, Axes -> Automatic, Frame ->
True, Background -> GrayLevel[.026],This gives me dots in 4
colors for distinguishing differentkinds of points in my
real application. This works fine butneeds the Point
structure. Or, I can do (this is for onekind of point),t =
Table[{Random[], Random[]}, {i, 1, 1024}];ListPlot[t,
AspectRatio -> Automatic, Axes -> Automatic, Frame -> True,
Background -> GrayLevel[.026] ];which seems simpler and may
fit into the rest of the programmore easily.1. How do I get
the RGBColor Rule or the equivalent intothe latter? The
RGBColor[] call is not a rule.2. In the latter case I also
want to plot 4 types of points.How do I get ListPlot to put
down 4 plots superimposed?Or can't I?3. In either case, I
need to make the whole plot area abouttwice as big. That
is, it now occupies about a 4 square. Tosee details better
in my real plot, and because with 16k pointsthe small plot
just looks almost like a solid blur, I want tomake it more
like 8 square, or as big as will fit the screen(without
changing the plot range or anything else). Theremust be a
scale factor somewhere.
====
The same command worked for me,
insofar as creating the plot isconcerned.I did get an
error message opening the file, saying that Adobe Acrobatwas
Unable to find or create the font 
ÔMathematica1Mono-Bold'.
Somecharacters may not display or print correctly. The
plot looks fine,though.Bobby Treat-----Original
Message-----Export[c:docscell4.gif, Cell[
<<...(copied cell data from Edit->CopyAs->Cell
Expression)...> ]]When I change the filename to a .PDF and
evaluate the cell, the programdisplays 
ÔRunning...' for a
second and gives the ÔOut[n] =c:docscell4.pdf' 
message
as if a file was created, but no file is created 
anywhere
withanyname that I could find with Start->Find->[All 
files
and folders createdinthe previous day].Is there a
limitation to PDF exporting I don't know about? Do I need
toupgrade to 4.2? Am I doing something wrong with the
Export[] command? DoIneed a faster computer? A patch?
Something else?
====
I'm trying to export a table from
Mathematica 4.0 to Notepad, for examples,or something
similar, but it's impossible for me! It always exports the
cellwith all its rubbish, or an image (if I use
Word).How can I export just plain
_numbers_?ThanxFip
====
You can useExport[file.dat,
expr] where expr is a two-dimensional array.Or for more
ßexibility study the following.Assuming data has n rows and
2 columns.file = OpenWrite[file.dat];Map[ ( (* the 
N may
be redundant haven't tested the code w/o it *) str =
ToString[PaddedForm[N[#[[1]]],{10,6}]]<> <>
ToString[PaddedForm[N[#[[2]]],{10,6}]]; (* make numbers e
notation: the *^ notation has given me fits, so I just put
this in as a precaution *) str = StringReplace[str, *^ ->
e]; WriteString[file, str, n] )&,
data];Close[file]Hope this helps,Lawrence> I'm 
trying to
export a table from Mathematica 4.0 to Notepad, for
examples,> or something similar, but it's impossible for me!
It always exports the cell> with all its rubbish, or an
image (if I use Word).> How can I export just plain
_numbers_?> Thanx> Fip> -- Lawrence A. Walker
Jr.http://www.kingshonor.com
====
> I'm trying to export a
table from Mathematica 4.0 to Notepad, for examples,> or
something similar, but it's impossible for me! It always
exports the cell> with all its rubbish, or an image (if
I use Word).> How can I export just plain _numbers_?Try with
this:WriteSimpleTableForm[file_String, data_List, opts___] :=
Module[{str}, str = OpenWrite[file]; WriteString[str,
ToString[TableForm[data, opts]]]; Close[str]]This is my
small solution for this problem.marek
====
I put the
following in
.../.../KeyEventTranslations.tr:Item[KeyEvent[<,Modifiers->
{Control}], FrontEndExecute[{
FrontEnd`NotebookWrite[FrontEnd`SelectedNotebook[],
[LeftDoubleBracket][RightDoubleBracket],After]}]],to
make it easy to insert the double-brackets. This works fine,
but Iwant the cursor to be placed between the brackets not
after them. Howcan this be done?Gru137
Peter--=--=--=--=--=--=--=--=--=--=--=--=--=--=
http://home.t-online.de/home/phbrf~~~
====
>I've been using
Mathematica 4.1 on Win98 as a word processor
for>math-related documents, but often people that need to
see the documents>don't have Mathematica, and for whatever
reason on my computer the HTML>saves don't work at all. 
I'd
like to export to PDF format. I can export>images to PDF
format no problem using, for
example>Export[c:docsplot3.pdf,
Plot[Sin[x],{x,-2Pi,2Pi}]],>and I can export cells correctly
to GIF, JPEG, and WMF formats (probably>more, those are the
only ones I tested) using, for
example>Export[c:docscell4.gif, Cell[ <<...(copied
cell data from Edit->Copy>As->Cell Expression)...> ]]>When
I change the filename to a .PDF and evaluate the cell, the
program>displays ÔRunning...' for a second and 
gives the
ÔOut[n] = c:docscell4.pdf>' message as if a 
file was
created, but no file is created anywhere with any>name that
I could find with Start->Find->[All files and 
folders created
in>the previous day].>Is there a limitation to PDF exporting
I don't know about? Do I need to>upgrade to 4.2? Am I doing
something wrong with the Export[] command? Do I>need a
faster computer? A patch? Something else?This is a
limitation of PDF export. The mechanism for exporting GIF,
JPEG, and other raster formats is completely different than
the system used for PDF export. Because of this PDF export
is limited to graphics expressions. Cells and Notebooks
cannot be converted via Export.However, there is a way to
generate PDFs using the frontend.
Seehttp://support.wolfram.com/mathematica/graphics/export/
convertpdfghostscript.htmlhttp://support.wolfram.com/
mathematica/graphics/export/convertpdfdistiller.html-Dale=
=== Dear Mathematica friends,how can Mathematica 4.1 be used
to combine sound and graphics?In particular, I would like to
prepare a demo video aboutdifferential equations. I can
Plot the solution and I canPlay the solution. How to
combine the 2 results into a single file that can be played
back using xine or DivX,like ordinary video can?Best wishes
from Prague-- Pavel PokornyMath Dept, Prague Institute of
Chemical
Technologyhttp://staffold.vscht.cz/mat/Pavel.Pokorny
====

Hoi,it depends a bit what you want to do.E.g.: If you want
to write applications for clients then go with thatsystems
your clients use (probably Windows).If you just use it for
yourself, for development: use what you like more.anymore,
likethere were in 3.0 times.The copy and paste problems
are gone if you switch off the KDE Klipper.On the other
hand, there are a few OS- (or better
Window-manager-specific)1.: you cannot rotate text (i.e.,
FrameLabel settings will look weird (vertically arranged
horizontal letters), you have to use RotateLabel -> False
generally, or play with the Fonts settings such that
horizontal tick marks still fit)2.: If you work with bigger
graphics in notebooks I suggest Windows (or MacOS X)
sinceleast on my XFree 4.2 installation with a not too
modern graphics card). Also I find resizing of larger
notebooks somewhat slow.3.: If you like to work with keyboard
shortcuts: Windows is better, clearly.4.: There are a couple
of Font issues which are better on Windows since not all
fontsengels, nederlands, duits of spaans)Rolf
MertigMertig ConsultingBerlin
====
I'd like to add a
JLink animation of the rolling ball based on
Selwyn'ssolution: UseFrontEndForRendering =
False;createWindow[] := Module[{frame}, frame =
JavaNew[com.wolfram.jlink.MathFrame, Doppler
Animation]; drawArea =
JavaNew[com.wolfram.jlink.MathCanvas];
drawArea@setUsesFE[UseFrontEndForRendering];
drawArea@setSize[800,
600];JavaBlock[frame@setLayout[JavaNew[
java.awt.BorderLayout]]; frame@add[drawArea,
ReturnAsJavaObject[BorderLayout`CENTER]]; frame@pack[];
frame@setSize[800, 600]; frame@setLocation[200, 200];
JavaShow[frame]];frame] drawRoll[t_] := Show[curve,
Graphics[{RGBColor[1, 0, 0], Disk[{xx[t], Cosh[xx[t]]},
0.05]}], PlotRange -> {{-1.2, 1.2}, {0.9, 1.65}}, AspectRatio
-> Automatic, Axes -> None, DisplayFunction -> Identity];
times = Range[0, 5, .05]; LoadJavaClass[java.lang.Thread];
AnimationPlot[t_List] := JavaBlock[Block[{frm},frm =
createWindow[];Map[(obj = drawRoll[#];
drawArea@setMathCommand[obj];drawArea@repaintNow[];
Thread@sleep[5];) &, t]]]AnimationPlot[times]jerry
blimbaum panama city, ß-----Original Message-----and the
equation of motion is diffeq = Simplify[ D[D[L, x'[t]], t] ]
== Simplify[ D[L, x[t]] ]Now solve and animate ... xx[t_] =
x[t]/. First[ NDSolve[{diffeq, x[0] == -1, x'[0] == 0}, 
x[t],
{t, 0, 5}]] curve = Plot[Cosh[x], {x, -1, 1}] Do[ Show[curve,
Graphics[Disk[{xx[t], Cosh[xx[t]]}, 0.025]], PlotRange ->
{{-1.2, 1.2}, {0.9, 1.65}}, AspectRatio -> Automatic,
Axes->None], {t, 0, 5, 0.1}]----Selwyn Hollis> Dear
Colleagues,> I intend to make an animation in which > ball A
rolls down on an inclined plane from the left whilst> ball B
- starting from the same height - rolls down Cosh[t]'s path
fromthe> right.> x-axis is time t, y-axis is height h.>
Ball A is fine; ball B - which should arrive at h=0 before A -
is beyondmy> means.> Matthias Bode> Sal. Oppenheim jr. &
Cie. KGaA> Koenigsberger Strasse 29> D-60487 Frankfurt am
Main> GERMANY> Mobile: +49(0)172 6 74 95 77> Internet:
http://www.oppenheim.de> 
====
I noticed that n Mathematica
3.0 , IntegerDigits function is giving wrong results. This
problem is not found in Mathematica 4.1. Whether any body
else has also noted any such problem.For example
IntegerDigits[10^18+7] will give the digits 0 and 7 ,
omitting 1.Shyam Sunder Gupta
====
You're right; I wasn't
exporting a Cell. I misunderstood his post andsuccessfully
executed this:Export[c:docsplot3.pdf,
Plot[Sin[x],{x,-2Pi,2Pi}]]but that was working for him,
already.Sorry for the confusion.Bobby-----Original
Message-----Sender: steve@smc.vnet.netApproved: Steven M.
Christensen <steve@smc.vnet.net>, Moderator
====
> w =
Table[Point[{Random[], Random[]}], {i, 1, 4096}];> x =
Table[Point[{Random[], Random[]}], {i, 1, 4096}];> y =
Table[Point[{Random[], Random[]}], {i, 1, 4096}];> z =
Table[Point[{Random[], Random[]}], {i, 1, 4096}];> 2. In the
latter case I also want to plot 4 types of points.> How do I
get ListPlot to put down 4 plots superimposed?> Or can't
I?You can, but you have to do it via
Graphics`MulitpleListPlot`. Load the requisite packages
Needs[Graphics`Colors`]Needs[Graphics`MultipleListPlot
`]Define cols = {Black, Red, Green, Blue}; pnts =
{PlotSymbol[Triangle], PlotSymbol[Box],
PlotSymbol[Diamond],PlotSymbol[Star]};Thengrf =
MultipleListPlot[w, x, y, z, PlotStyle->cols,
SymbolShape->pnts, SymbolsStyle->cols ];Documentation shows
how to incorporate legends and to define other
symbols.Dave.
====
====================================== Dr.
David Annetts EM Modelling Analyst Australia
David.Annetts@csiro.au
====
==================================

====
=====>Dear Group,> It is a pleasure write to you again. I
have three questions to ask.> Alexandre Costa>Question
Two:>How can I change points properties (such as Size and
Color) for the plot>below? The PlotStyle Option does not
work for this LabeledListPlot><<
Graphics`Graphics`>listcities = {{1, 5}, {4, 6}, {7, 5}, {5,
4}, {9, 4}, {2, 3}, {4, 2},>{6, 2}, {1, 1}, {5, 1}, {3, 0},
{9, 0}};>LabeledListPlot[listcities, Axes -> None, Frame ->
True,> DisplayFunction -> $DisplayFunction]It's not a
built-in feature of LabeledListPlot, so you'll have to do it
manually.gr=LabeledListPlot[listcities, Axes -> None, Frame
-> True, DisplayFunction -> $DisplayFunction]You can add
graphics directives to the points and text with a replacement
rule.Show[gr/.x_Point|x_Text->{RGBColor[1,0,0],x}]>Question
Three: Why the Goto statement below is not working?>q =
2;>Label[start];>q = 3;>Label[begin];>Print[q];>q += 1; If[q
< 6, Goto[begin], Goto[start]]First, let me say that noone
should ever use Goto. You should always use a loop or some
other process instead. With that said . . .When you type
semicolon separated input into the frontend, each command is
treated as a separate input (as if they were in separate
input cells). So the Labels and the Gotos are evaluated
separately and there's no way to jump from one to the other.
Instead, the commands need to be within the same expression.
This can be done by wrapping the command in a
CompoundExpression( q = 2; Label[start]; q = 3;
Label[begin]; Print[q]; q += 1; If[q < 6, Goto[begin],
Goto[start]])Or some other expression, like a
Module.------------------------------------------------------
--------Omega ConsultingThe final answer to your
Mathematica needsSpend less time searching and more time
finding.http://www.wz.com/internet/Mathematica.html
====
>I
would appreciate help with these problems:>1. I'm plotting
several thousand points, which I can do>either with
something like this (this is a test):>w =
Table[Point[{Random[], Random[]}], {i, 1, 4096}];>x =
Table[Point[{Random[], Random[]}], {i, 1, 4096}];>y =
Table[Point[{Random[], Random[]}], {i, 1, 4096}];>z =
Table[Point[{Random[], Random[]}], {i, 1, 4096}];>dw =
Graphics[{PointSize[0.01], RGBColor[ 1, 0, 0], w}];>dx =
Graphics[{PointSize[0.01], RGBColor[.8, .8, .8], x}];>dy =
Graphics[{PointSize[0.01], RGBColor[ 0, .5, .9], y}];>dz =
Graphics[{PointSize[0.01], RGBColor[.8, .8, 0], z}];>Show[dw,
dx, dy, dz, AspectRatio -> Automatic,> PlotRange -> {{0, 1},
{0, 1}},> Axes -> Automatic,> Frame -> True,> Background ->
GrayLevel[.026],>This gives me dots in 4 colors for
distinguishing different>kinds of points in my real
application. This works fine but>needs the Point structure.
Or, I can do (this is for one>kind of point),>t =
Table[{Random[], Random[]}, {i, 1, 1024}];>ListPlot[t,
AspectRatio -> Automatic,> Axes -> Automatic,> Frame ->
True,> Background -> GrayLevel[.026]> ];>which seems simpler
and may fit into the rest of the program>more easily.>1. How
do I get the RGBColor Rule or the equivalent into>the
latter? The RGBColor[] call is not a rule.>2. In the latter
case I also want to plot 4 types of points.>How do I get
ListPlot to put down 4 plots superimposed?>Or can't I?>3. In
either case, I need to make the whole plot area about>twice
as big. That is, it now occupies about a 4 square. To>see
details better in my real plot, and because with 16k
points>the small plot just looks almost like a solid blur, I
want to>make it more like 8 square, or as big as will fit the
screen>(without changing the plot range or anything else).
There>must be a scale factor somewhere.You can plot more
than one list with MultipleListPlot.<<Graphics`If you don't
like the default symbols used by MultipleListPlot, then you
can create your own.ColorPoint[color_] :=
MakeSymbol[{color,Point[{0,0}]}]Create the
data.w=Table[{Random[],Random[]},{i,1,1024}];x=Table[{
Random[],Random[]},{i,1,1024}];y=Table[{Random[],Random[]},{
i,1,1024}];z= Table[{Random[],Random[]},{i,1,1024}];And
plotMultipleListPlot[w,x,y,z,
AspectRatio->Automatic,Axes->Automatic,Frame->True,
SymbolShape->{ColorPoint[RGBColor[1,0,0]],
ColorPoint[RGBColor[.8, .8, .8]],ColorPoint[RGBColor[ 0, .5,
.9]], ColorPoint[RGBColor[.8, .8, 0]]},
SymbolStyle->PointSize[0.01],Background->GrayLevel[.026]]----
----------------------------------------------------------
Omega ConsultingThe final answer to your Mathematica
needsSpend less time searching and more time
finding.http://www.wz.com/internet/Mathematica.html
====
>I
have to use a graphics of mathematica with powerpoint for a
little>Y-axes and line of function) are bold or more
visible: how can I do>that?Use the PlotStyle and AxesStyle
options as
inPlot[x,{x,0,5},PlotStyle->Thickness[0.015],AxesStyle->
Thickness[0.015]]
====
 Dear Mathematica friends,how can
Mathematica 4.1 be used to combine sound and graphics?In
particular, I would like to prepare a demo video
aboutdifferential equations. I can Plot the solution and I
canPlay the solution. How to combine the 2 results into a
single file that can be played back using xine or DivX,like
ordinary video can?Best wishes from Prague-- Pavel
PokornyMath Dept, Prague Institute of Chemical
Technologyhttp://staffold.vscht.cz/mat/Pavel.Pokorny
====

I'd like to add a JLink animation of the rolling ball based
on Selwyn'ssolution: UseFrontEndForRendering =
False;createWindow[] := Module[{frame}, frame =
JavaNew[com.wolfram.jlink.MathFrame, Doppler
Animation]; drawArea =
JavaNew[com.wolfram.jlink.MathCanvas];
drawArea@setUsesFE[UseFrontEndForRendering];
drawArea@setSize[800,
600];JavaBlock[frame@setLayout[JavaNew[
java.awt.BorderLayout]]; frame@add[drawArea,
ReturnAsJavaObject[BorderLayout`CENTER]]; frame@pack[];
frame@setSize[800, 600]; frame@setLocation[200, 200];
JavaShow[frame]];frame] drawRoll[t_] := Show[curve,
Graphics[{RGBColor[1, 0, 0], Disk[{xx[t], Cosh[xx[t]]},
0.05]}], PlotRange -> {{-1.2, 1.2}, {0.9, 1.65}}, AspectRatio
-> Automatic, Axes -> None, DisplayFunction -> Identity];
times = Range[0, 5, .05]; LoadJavaClass[java.lang.Thread];
AnimationPlot[t_List] := JavaBlock[Block[{frm},frm =
createWindow[];Map[(obj = drawRoll[#];
drawArea@setMathCommand[obj];drawArea@repaintNow[];
Thread@sleep[5];) &, t]]]AnimationPlot[times]jerry
blimbaum panama city, ß-----Original Message-----and the
equation of motion is diffeq = Simplify[ D[D[L, x'[t]], t] ]
== Simplify[ D[L, x[t]] ]Now solve and animate ... xx[t_] =
x[t]/. First[ NDSolve[{diffeq, x[0] == -1, x'[0] == 0}, 
x[t],
{t, 0, 5}]] curve = Plot[Cosh[x], {x, -1, 1}] Do[ Show[curve,
Graphics[Disk[{xx[t], Cosh[xx[t]]}, 0.025]], PlotRange ->
{{-1.2, 1.2}, {0.9, 1.65}}, AspectRatio -> Automatic,
Axes->None], {t, 0, 5, 0.1}]----Selwyn Hollis> Dear
Colleagues,> I intend to make an animation in which > ball A
rolls down on an inclined plane from the left whilst> ball B
- starting from the same height - rolls down Cosh[t]'s path
fromthe> right.> x-axis is time t, y-axis is height h.>
Ball A is fine; ball B - which should arrive at h=0 before A -
is beyondmy> means.> Matthias Bode> Sal. Oppenheim jr. &
Cie. KGaA> Koenigsberger Strasse 29> D-60487 Frankfurt am
Main> GERMANY> Mobile: +49(0)172 6 74 95 77> Internet:
http://www.oppenheim.de> 
====
>I've been using Mathematica
4.1 on Win98 as a word processor for>math-related documents,
but often people that need to see the documents>don't have
Mathematica, and for whatever reason on my computer the
HTML>saves don't work at all. I'd like to 
export to PDF
format. I can export>images to PDF format no problem using,
for example>Export[c:docsplot3.pdf,
Plot[Sin[x],{x,-2Pi,2Pi}]],>and I can export cells correctly
to GIF, JPEG, and WMF formats (probably>more, those are the
only ones I tested) using, for
example>Export[c:docscell4.gif, Cell[ <<...(copied
cell data from Edit->Copy>As->Cell Expression)...> ]]>When
I change the filename to a .PDF and evaluate the cell, the
program>displays ÔRunning...' for a second and 
gives the
ÔOut[n] = c:docscell4.pdf>' message as if a 
file was
created, but no file is created anywhere with any>name that
I could find with Start->Find->[All files and 
folders created
in>the previous day].>Is there a limitation to PDF exporting
I don't know about? Do I need to>upgrade to 4.2? Am I doing
something wrong with the Export[] command? Do I>need a
faster computer? A patch? Something else?This is a
limitation of PDF export. The mechanism for exporting GIF,
JPEG, and other raster formats is completely different than
the system used for PDF export. Because of this PDF export
is limited to graphics expressions. Cells and Notebooks
cannot be converted via Export.However, there is a way to
generate PDFs using the frontend.
Seehttp://support.wolfram.com/mathematica/graphics/export/
convertpdfghostscript.htmlhttp://support.wolfram.com/
mathematica/graphics/export/convertpdfdistiller.html-Dale=
===I noticed that n Mathematica 3.0 , IntegerDigits function
is giving wrong results. This problem is not found in
Mathematica 4.1. Whether any body else has also noted any
such problem.For example IntegerDigits[10^18+7] will give
the digits 0 and 7 , omitting 1.Shyam Sunder
Gupta
====
You're right; I wasn't exporting a Cell. I
misunderstood his post andsuccessfully executed
this:Export[c:docsplot3.pdf,
Plot[Sin[x],{x,-2Pi,2Pi}]]but that was working for him,
already.Sorry for the confusion.Bobby-----Original
Message-----Sender: steve@smc.vnet.netApproved: Steven M.
Christensen <steve@smc.vnet.net>, Moderator
====
> w =
Table[Point[{Random[], Random[]}], {i, 1, 4096}];> x =
Table[Point[{Random[], Random[]}], {i, 1, 4096}];> y =
Table[Point[{Random[], Random[]}], {i, 1, 4096}];> z =
Table[Point[{Random[], Random[]}], {i, 1, 4096}];> 2. In the
latter case I also want to plot 4 types of points.> How do I
get ListPlot to put down 4 plots superimposed?> Or can't
I?You can, but you have to do it via
Graphics`MulitpleListPlot`. Load the requisite packages
Needs[Graphics`Colors`]Needs[Graphics`MultipleListPlot
`]Define cols = {Black, Red, Green, Blue}; pnts =
{PlotSymbol[Triangle], PlotSymbol[Box],
PlotSymbol[Diamond],PlotSymbol[Star]};Thengrf =
MultipleListPlot[w, x, y, z, PlotStyle->cols,
SymbolShape->pnts, SymbolsStyle->cols ];Documentation shows
how to incorporate legends and to define other
symbols.Dave.
====
====================================== Dr.
David Annetts EM Modelling Analyst Australia
David.Annetts@csiro.au
====
==================================

====
=====>Dear Group,> It is a pleasure write to you again. I
have three questions to ask.> Alexandre Costa>Question
Two:>How can I change points properties (such as Size and
Color) for the plot>below? The PlotStyle Option does not
work for this LabeledListPlot><<
Graphics`Graphics`>listcities = {{1, 5}, {4, 6}, {7, 5}, {5,
4}, {9, 4}, {2, 3}, {4, 2},>{6, 2}, {1, 1}, {5, 1}, {3, 0},
{9, 0}};>LabeledListPlot[listcities, Axes -> None, Frame ->
True,> DisplayFunction -> $DisplayFunction]It's not a
built-in feature of LabeledListPlot, so you'll have to do it
manually.gr=LabeledListPlot[listcities, Axes -> None, Frame
-> True, DisplayFunction -> $DisplayFunction]You can add
graphics directives to the points and text with a replacement
rule.Show[gr/.x_Point|x_Text->{RGBColor[1,0,0],x}]>Question
Three: Why the Goto statement below is not working?>q =
2;>Label[start];>q = 3;>Label[begin];>Print[q];>q += 1; If[q
< 6, Goto[begin], Goto[start]]First, let me say that noone
should ever use Goto. You should always use a loop or some
other process instead. With that said . . .When you type
semicolon separated input into the frontend, each command is
treated as a separate input (as if they were in separate
input cells). So the Labels and the Gotos are evaluated
separately and there's no way to jump from one to the other.
Instead, the commands need to be within the same expression.
This can be done by wrapping the command in a
CompoundExpression( q = 2; Label[start]; q = 3;
Label[begin]; Print[q]; q += 1; If[q < 6, Goto[begin],
Goto[start]])Or some other expression, like a
Module.------------------------------------------------------
--------Omega ConsultingThe final answer to your
Mathematica needsSpend less time searching and more time
finding.http://www.wz.com/internet/Mathematica.html
====

Mario:Use, for example, PlotStyle->Thickness[0.01] for plot
andAxesStyle->Thickness[0.01].Other way is using
AbsoluteThickness instead of Thickness.See The Mathematica
Book: Section 2.9.3.Germ.87n Buitrago A.----- Original
Message -----Sender: steve@smc.vnet.netApproved: Steven M.
Christensen <steve@smc.vnet.net>, Moderator
====
Mario:Use,
for example, PlotStyle->Thickness[0.01] for plot
andAxesStyle->Thickness[0.01].Other way is using
AbsoluteThickness instead of Thickness.See The Mathematica
Book: Section 2.9.3.Germ.87n Buitrago A.----- Original
Message -----Sender: steve@smc.vnet.netApproved: Steven M.
Christensen <steve@smc.vnet.net>, Moderator
====
I have to use
a graphics of mathematica with powerpoint for a
littleY-axes and line of function) are bold or more
visible: how can I dothat?
====
>I have to use a graphics of
mathematica with powerpoint for a little>Y-axes and line
of function) are bold or more visible: how can I
do>that?Use the PlotStyle and AxesStyle options as
inPlot[x,{x,0,5},PlotStyle->Thickness[0.015],AxesStyle->
Thickness[0.015]]
====
>I would appreciate help with these
problems:>1. I'm plotting several thousand points, which I
can do>either with something like this (this is a test):>w =
Table[Point[{Random[], Random[]}], {i, 1, 4096}];>x =
Table[Point[{Random[], Random[]}], {i, 1, 4096}];>y =
Table[Point[{Random[], Random[]}], {i, 1, 4096}];>z =
Table[Point[{Random[], Random[]}], {i, 1, 4096}];>dw =
Graphics[{PointSize[0.01], RGBColor[ 1, 0, 0], w}];>dx =
Graphics[{PointSize[0.01], RGBColor[.8, .8, .8], x}];>dy =
Graphics[{PointSize[0.01], RGBColor[ 0, .5, .9], y}];>dz =
Graphics[{PointSize[0.01], RGBColor[.8, .8, 0], z}];>Show[dw,
dx, dy, dz, AspectRatio -> Automatic,> PlotRange -> {{0, 1},
{0, 1}},> Axes -> Automatic,> Frame -> True,> Background ->
GrayLevel[.026],>This gives me dots in 4 colors for
distinguishing different>kinds of points in my real
application. This works fine but>needs the Point structure.
Or, I can do (this is for one>kind of point),>t =
Table[{Random[], Random[]}, {i, 1, 1024}];>ListPlot[t,
AspectRatio -> Automatic,> Axes -> Automatic,> Frame ->
True,> Background -> GrayLevel[.026]> ];>which seems simpler
and may fit into the rest of the program>more easily.>1. How
do I get the RGBColor Rule or the equivalent into>the
latter? The RGBColor[] call is not a rule.>2. In the latter
case I also want to plot 4 types of points.>How do I get
ListPlot to put down 4 plots superimposed?>Or can't I?>3. In
either case, I need to make the whole plot area about>twice
as big. That is, it now occupies about a 4 square. To>see
details better in my real plot, and because with 16k
points>the small plot just looks almost like a solid blur, I
want to>make it more like 8 square, or as big as will fit the
screen>(without changing the plot range or anything else).
There>must be a scale factor somewhere.You can plot more
than one list with MultipleListPlot.<<Graphics`If you don't
like the default symbols used by MultipleListPlot, then you
can create your own.ColorPoint[color_] :=
MakeSymbol[{color,Point[{0,0}]}]Create the
data.w=Table[{Random[],Random[]},{i,1,1024}];x=Table[{
Random[],Random[]},{i,1,1024}];y=Table[{Random[],Random[]},{
i,1,1024}];z= Table[{Random[],Random[]},{i,1,1024}];And
plotMultipleListPlot[w,x,y,z,
AspectRatio->Automatic,Axes->Automatic,Frame->True,
SymbolShape->{ColorPoint[RGBColor[1,0,0]],
ColorPoint[RGBColor[.8, .8, .8]],ColorPoint[RGBColor[ 0, .5,
.9]], ColorPoint[RGBColor[.8, .8, 0]]},
SymbolStyle->PointSize[0.01],Background->GrayLevel[.026]]----
----------------------------------------------------------
Omega ConsultingThe final answer to your Mathematica
needsSpend less time searching and more time
finding.http://www.wz.com/internet/Mathematica.html
====
For
the life of me I am not sure why the following is not working
in my v. 4.2:ru[a]=a->x;f[x_]:=(a+b) /. ru[a];Why do I
getf[c] = b+xand notf[c] = b+c?What gives?Lawrence--
Lawrence A. Walker Jr.http://www.kingshonor.comReply-To:
kuska@informatik.uni-leipzig.de
====
because te right hand
side of SetDelayed[] is notevaluate.Tryru[a] = a ->
x;f[x_] := (a + b) /. ru[a];f1[x_] := Evaluate[(a + b) /.
ru[a]];and f1[] does what you expectIn[]:={f[c],
f1[c]}Out[]={b + x, b + c} Jens> For the life of me I am
not sure why the following is not working in my> v. 4.2:>
ru[a]=a->x;> f[x_]:=(a+b) /. ru[a];> Why do I get> f[c] =
b+x> and not> f[c] = b+c?> What gives?> Lawrence> -->
Lawrence A. Walker Jr.>
http://www.kingshonor.com
====
Sir,we are having FreeBSD
Server,In this we connected a heavy duty Dot matrix printer
locally.While on taking outputs, username and file name are
printing as BANNER TYPE.Instead of this, we would like to
take the print out as username and file name should be
printed as header through out the file.if u are having any
scripts like that please send usRaj MohanSystem
Administrator Reply-To: jmt@dxdydz.net
====
I would like to
add :Some characters, at least with a french keyboard, are
not directly available typed through their Mathematica
entities : [EHat] , [OHat], etc.While this is not an
issue when programming, it can become quite painful when
writing documentation.> Hoi,> it depends a bit what you want
to do.> E.g.: If you want to write applications for clients
then go with that> systems your clients use (probably
Windows).> If you just use it for yourself, for development:
use what you like more.> anymore, like> there were in 3.0
times.> The copy and paste problems are gone if you switch
off the KDE Klipper.> On the other hand, there are a few OS-
(or better Window-manager-specific)> 1.: you cannot rotate
text (i.e., FrameLabel settings will look weird > (vertically
arranged> horizontal letters), you have to use> RotateLabel
-> False generally, or play with the Fonts settings > such
that horizontal> tick marks still fit)> 2.: If you work with
bigger graphics in notebooks I suggest Windows (or > MacOS X)
since> least on my XFree 4.2> installation with a not too
modern graphics card).> Also I find resizing of larger
notebooks somewhat slow.> 3.: If you like to work with
keyboard shortcuts: Windows is better, clearly.> 4.: There
are a couple of Font issues which are better on Windows since
> not all fonts> engels, nederlands, duits of spaans)> Rolf
Mertig> Mertig Consulting> Berlin> 
====
Hoi,it depends
a bit what you want to do.E.g.: If you want to write
applications for clients then go with thatsystems your
clients use (probably Windows).If you just use it for
yourself, for development: use what you like more.anymore,
likethere were in 3.0 times.The copy and paste problems
are gone if you switch off the KDE Klipper.On the other
hand, there are a few OS- (or better
Window-manager-specific)1.: you cannot rotate text (i.e.,
FrameLabel settings will look weird (vertically arranged
horizontal letters), you have to use RotateLabel -> False
generally, or play with the Fonts settings such that
horizontal tick marks still fit)2.: If you work with bigger
graphics in notebooks I suggest Windows (or MacOS X)
sinceleast on my XFree 4.2 installation with a not too
modern graphics card). Also I find resizing of larger
notebooks somewhat slow.3.: If you like to work with keyboard
shortcuts: Windows is better, clearly.4.: There are a couple
of Font issues which are better on Windows since not all
fontsengels, nederlands, duits of spaans)Rolf
MertigMertig ConsultingBerlin
====
I am considering the
following integralW[m_,n_]:=Integrate[BesselJ[m,
x]*BesselJ[n, x], {x, 0, Infinity}]where m,n are reals >=0.
With Mathematica 4.1 I obtain:If[Re[m+n]>-1,
-Cos[(m-n)Pi/2]/(2 Pi)*(2 EulerGamma + Log[4] + PolyGamma[0,
1/2(1 + m - n)] + PolyGamma[0, 1/2(1 - m + n)] +
2PolyGamma[0, 1/2(1 + m + n)])and so using this answer as a
definition I obtain W[0,0]=-(2 EulerGamma + Log[4] + 4
PolyGamma[0, 1/2])/(2 Pi)=0.84564I suspect that these
integrals are divergent (*). So I try the
numericalintegration:NW[m_,n_]:=NIntegrate[BesselJ[m,
x]*BesselJ[n, x], {x, 0, Infinity}]so
thatNW[0,0]=11.167Othe couples areW[1,0]=Indeterminate
NW[1,0]=0.597973W[0,1.5]=0.537095
NW[0,1.5]=-5.79306W[1,1]=0.20902
NW[1,1]=17.5425W[2,0]=0.427599
NW[2,0]=-6.83464W[2,1]=Indeterminate NW[2,1]=4.69013(*) The
integral is a particular case of the Weber-Schafheitlin
integrals(Abramowitz, 11.4.33).Any explanation about the
analytical expression will be gratefully
accepteed.Roberto.Roberto BrambillaCESIVia Rubattino
5420134 Milanotel +39.02.2125.5875fax
+39.02.2125.5492rlbrambilla@cesi.it
====
WRI Tech Support
sent me an answer to the font problem. Copying theType 1
(not True Type) fonts made the error go away:A workaround
for the font-related issue that you encountered, is
toplacecopies of the Mathematica Type 1 fonts
fromC:Program
FilesWolframResearchMathematica4.2SystemFiles
FontsType1intoC:Program FilesAdobeAcrobat
5.0ResourceFontSincerely,George
KambouroglouTechnical SupportWolfram Research,
Inc.support@wolfram.com-----Original Message-----I've been
using Mathematica 4.1 on Win98 as a word processor
formath-related documents, but often people that need to
see the documentsdon't have Mathematica, and for whatever
reason on my computer the HTMLsaves don't work at all. 
I'd
like to export to PDF format. I can exportimages to PDF
format no problem using, for
exampleExport[c:docsplot3.pdf,
Plot[Sin[x],{x,-2Pi,2Pi}]],and I can export cells correctly
to GIF, JPEG, and WMF formats (probablymore, those are the
only ones I tested) using, for
exampleExport[c:docscell4.gif, Cell[ <<...(copied
cell data from Edit->CopyAs->Cell Expression)...> ]]When
I change the filename to a .PDF and evaluate the cell, the
programdisplays ÔRunning...' for a second and 
gives the
ÔOut[n] =c:docscell4.pdf' message as if a 
file was
created, but no file is created anywhere withanyname that
I could find with Start->Find->[All files and 
folders
createdinthe previous day].Is there a limitation to PDF
exporting I don't know about? Do I need toupgrade to 4.2?
Am I doing something wrong with the Export[] command?
DoIneed a faster computer? A patch? Something
else?
====
SSG> I noticed that n Mathematica 3.0 ,
IntegerDigits function isSSG> giving wrong results. This
problem is not found in MathematicaSSG> 4.1. Whether
anybody else has also noted any such problem.SSG> For
example IntegerDigits[10^18+7] will give the digits 0SSG>
and 7, omitting 1.Having made totally 4,000,000 attemtps,I
was not able to reproduce your example.What version of
Mathematica do you
use?......................................................
4.2 for Microsoft Windows (February 28, 2002)4.1 for
Microsoft Windows (November 2, 2000)4.0 for Microsoft
Windows (April 21, 1999)Microsoft Windows 3.0 (April 25,
1997)Windows 387 2.2 (April 9, 1993)IntegerDigits[10^18
+ 7]IntegerDigits[10^18 + 7]IntegerDigits[10^18 +
7]IntegerDigits[10^18 + 7]IntegerDigits[10^18 +
7]{1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,7}{
1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,7}{
1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,7}{
1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,7}{
1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,7}.......................
...............................Best wishes,Vladimir
BondarenkoMathematical DirectorSymbolic Testing
GroupWeb : http://www.CAS-testing.org/
http://maple.bug-list.org/VER2/ (under tuning)
http://maple.bug-list.org/VER3/ (under tuning)
http://maple.bug-list.org/VER1/ (under tuning)Voice :
(380)-652-447325 Mon-Fri 9 a.m.-6 p.m.Mail : 76 Zalesskaya
Str, Simferopol, Crimea, Ukraine
====
Just tested it with no
errors, version 3.0John A. Velling-----Original
Message-----
====
I would appreciate any info on these
issues.1. I have a symbol slback. I was gettingtag
times lback protected  (or something) errors.(A typically
uninformative error message.)I suspected there was a
separation between the sand l, and they looked too far
apart, but when Ispaced past them with the arrow keys,
there didnot appear to be anything like a space in
between.I retyped the symbol and all was ok. This
hashappened before. What's going on ???2. I'm 
using Raster
to plot data pointsin a 256x256 array. Two problems:
First,regardless of the ImageSize setting (I need
thedisplay as big as possible), the individual data
cellswhen examined closely at 300% vary in size byalmost
2:1. This makes detailed inspection of thedata values
difficult. Surely there is some settingof something which
would make each data cell anexact number of screen pixels?
Second, I'musingShow[ Graphics[Raster[ rescol,
ColorFunction -> Hue]], AspectRatio -> Automatic, ImageSize
-> 700];but the Hues don't allow enough easily
distinguishableshades to visually recognize even 6 data
values easily.(I'm slightly colorblind.) It would be nice to
have blackand white available for two of the data values,
but Huedoes not allow this. I don't understand what
thedocumentation says about RasterArray.3. I wanted to get
this raster image into a formatwhere I could dissect it
with Photoshop or equivalent.After much fooling around, I
found that I can export theselection as an html file, read
it into Navigator, do File>Edit Page, which brings up
Netscape Composer, rightclick the image which, allows
saving it as a GIF, whichI can finally work on with a photo
editor. Maybethere is an easier way, or maybe this
description willhelp someone with the same need.Reply-To:
kuska@informatik.uni-leipzig.de
====
> I would appreciate any
info on these issues.> 1. I have a symbol slback. I was
getting> tag times lback protected  (or something)
errors.> (A typically uninformative error message.)> I
suspected there was a separation between the s> and l, and
they looked too far apart, but when I> spaced past them with
the arrow keys, there did> not appear to be anything like a
space in between.> I retyped the symbol and all was ok. This
has> happened before. What's going on ???As long as you
don't send us the *exact* input we can'thelp 
you. You
should also send the mathematica versionyou are using. But
typical this error comes from a equationa*b==cwhere the
user has mixed up Equal[] and Set[]> 2. I'm using Raster to
plot data points> in a 256x256 array. Two problems: First,>
regardless of the ImageSize setting (I need the> display as
big as possible), the individual data cells> when examined
closely at 300% vary in size by> almost 2:1. This makes
detailed inspection of the> data values difficult. Surely
there is some setting> of something which would make each
data cell an> exact number of screen pixels? Second, I'm>
using> Show[ Graphics[Raster[ rescol,> ColorFunction ->
Hue]],> AspectRatio -> Automatic,> ImageSize -> 700];> but
the Hues don't allow enough easily distinguishable> shades
to visually recognize even 6 data values easily.> (I'm
slightly colorblind.) The most humans can distinguish 160
gray levels>It would be nice to have black> and white
available for two of the data values, but Hue> does not
allow this.And something like:mycolor[i_] :=
Switch[Round[i], 0, RGBColor[0, 0, 0], 1, RGBColor[1, 0, 0],
2, RGBColor[1, 1, 0], 3, RGBColor[0, 1, 0], 4, RGBColor[0, 1,
1], 5, RGBColor[0, 0, 1], _, RGBColor[1, 1, 1]]Show[Graphics[
Raster[Table[Random[Integer, {0, 6}], {16}, {16}],
ColorFunction -> mycolor, ColorFunctionScaling ->
False]]]does not help ?> I don't understand what the>
documentation says about RasterArray.If you can't be more
specific *what* you not understandwe can not help you.> 3. I
wanted to get this raster image into a format> where I could
dissect it with Photoshop or equivalent.And ? waht does
Export[] do ? it write the expressionin a desired format,
TIFF, PNG, PPM are all losslesscompressed bitmap formats,
that Mathematica can exportand that can be imported into
PhotoShop> After much fooling around, I found that I can
export the> selection as an html file, read it into
Navigator, do File>> Edit Page, which brings up Netscape
Composer, right> click the image which, allows saving it as
a GIF, which> I can finally work on with a photo editor.
Maybe> there is an easier way, or maybe this description
will> help someone with the same need.May be that thhis
description help someone who can't readthe fancy documation
on Import[] and Export[]. Jens
====
[snip]> 2. I'm using Raster
to plot data points> in a 256x256 array. Two problems:
First,> regardless of the ImageSize setting (I need the>
display as big as possible), the individual data cells> when
examined closely at 300% vary in size by> almost 2:1. This
makes detailed inspection of the> data values difficult.
Surely there is some setting> of something which would make
each data cell an> exact number of screen pixels?GRAY: (No
answer received.) I can get around thisby making the images
bigger but this is not a completesolution.> Second, I'm
using> Show[ Graphics[Raster[ rescol,> ColorFunction ->
Hue]],> AspectRatio -> Automatic,> ImageSize -> 700];> but
the Hues don't allow enough easily distinguishable> shades
to visually recognize even 6 data values easily.> (I'm
slightly colorblind.)> The most humans can distinguish 160
gray levels>It would be nice to have black> and white
available for two of the data values, but Hue> does not
allow this.> And something like:> mycolor[i_] :=>
Switch[Round[i],> 0, RGBColor[0, 0, 0],> 1, RGBColor[1, 0,
0],> 2, RGBColor[1, 1, 0],> 3, RGBColor[0, 1, 0],> 4,
RGBColor[0, 1, 1],> 5, RGBColor[0, 0, 1],> _, RGBColor[1, 1,
1]]> Show[Graphics[> Raster[Table[Random[Integer, {0, 6}],
{16}, {16}],> ColorFunction -> mycolor, ColorFunctionScaling
-> False]]]GRAY: Sounds good. I'll try it. Interesting that
the help does notcontain this in a form I could easily
find.> 3. I wanted to get this raster image into a format>
where I could dissect it with Photoshop or equivalent.> And ?
what does Export[] do ? it write the expression> in a desired
format, TIFF, PNG, PPM are all lossless> compressed bitmap
formats, that Mathematica can export> and that can be
imported into PhotoShopGRAY: I foolishly thought something
like Export would be underthe File menu.> After much
fooling around, I found that I can export the> selection as
an html file, read it into Navigator, do File>> Edit Page,
which brings up Netscape Composer, right> click the image
which, allows saving it as a GIF, which> I can finally work
on with a photo editor. Maybe> there is an easier way, or
maybe this description will> help someone with the same
need.GRAY: I just found that simply Copying the image and
Pastingit into Paint Shop Pro (or no doubt lots of other
bitmapeditors) works. For outputting a Raster
noninteractively, there is Export,but I haven't tried it
yet. Jens, thank you for your reply.
====
explain to me why I
get this behavior and what I can do to fix it.When I run
Mathematica in X with the graphical user interface by typing
mathematica,I can run the following commands and get the
expected output, namely a plot with the plot label in a big
font. However, when I run the math command and get the
text interface and run the same code I do not get the label
in a big font. cc = Plot[Sin[x], {x, 0, Pi}, {PlotLabel ->
StyleForm[Label, FontSize -> 60]}]Export[test.eps,
cc, eps]I believe that this is due to the different
method in which the math and mathematica commands setup
fonts. Is there anyway to get the behavior that I want? As a
side note when I use the old syntax of
cc=Plot[Sin[x],{x,0,Pi},{PlotLabel->FontForm[Label,{
Courier,60}]}]I get it to work. However, I would like not
to use this syntax as ithas several limitations.Any
suggestions would be greatly appreciated.
====
I have a dual
processor Dell computer...unfortunately, I can only access
oneof the processors...I purchased Wolfram's parallel
processing toolkit which, because of the very poor
documentation, never did me any good...i'm toldthat another
option for accessing the dual processors is to write C
codeetc...I can't C program, so i'm wondering 
this...duzz
anyone have C code,both source and binary, that they could
give me for accomplishing this....inaddition, i would like
to access this C code with JavaNativeCode, etc...cananyone
help me with this? thanks...jerry blimbaum NSWC panama city,
ßReply-To: kuska@informatik.uni-leipzig.de
====
you can have
a parallel implicit Runge-Kutta methodprogrammed in C *and*
with the Parallel Computing Toolkitboth on the base of the
MathLink protocol.Since I have a SGI I can't help you with
the binaryfor a Dell what ever computer but if you like
the source ...You can also have a native MPI source of the
code but you willneed a running MPI for your computer.
Jens> I have a dual processor Dell computer...unfortunately,
I can only access one> of the processors...I purchased
Wolfram's parallel processing toolkit which> , because of
the very poor documentation, never did me any good...i'm
told> that another option for accessing the dual processors
is to write C code> etc...I can't C program, so 
i'm wondering
this...duzz anyone have C code,> both source and binary, that
they could give me for accomplishing this....in> addition,
i would like to access this C code with JavaNativeCode,
etc...can> anyone help me with this? thanks...> jerry
blimbaum NSWC panama city, ß
====
I have a set of
inequalities that I solve with InequalitySolve. But thenit
gives a complete set of solutions, but not in the way I would
like itto be! :-) For example, the simple following
calculation will give:In[1]:= ineq = {y4 >= -1, y5 >= -1, y6
+ y4 >= y5 - 1, y5 >= y6, y6 >= -1};
InequalitySolve[ineq,{y4,y6,y5}]Out[1]:= y4 == -1 && y6 >=
-1 && y5 == y6 || y4 > -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 +
y6the result is good, but I would like it to be in the
simpler butequivalent form y4 >= -1 && y6 >= -1 && y6 <=
y5 <= 1 + y4 + y6How can I tell InequalitySolve to do that?
It is simple for this example,but for a large set of simple
inequalities InequalitySolve gives lines andlines of
results instead of a simple result.Vincent
Bouchard
====
f[c] == (c+b)/.ru[c] == (c+b)/.c->x == x+b ==
b+xBobby Treat-----Original Message-----f[c] = b+c?What
gives?Lawrence-- Lawrence A. Walker
Jr.http://www.kingshonor.com
====
Sorry; disregard my
earlier answer. ru[c] isn't defined, so 
thatobviously wasn't
the correct sequence of events. Here's the right one:f[c] ==
HoldPattern[(a+b)/.ru[a]]/.x->c == (a+b)/.ru[a] ==
(a+b)/.a->x == x + b == b + xThe substitution of c for x
occurs before the rule ru[a] is evaluated,so there's no x in
the expression to replace. Instead, there's an 
Ôa'to replace
with x.If the other sequence had been correct, you have no
rule for evaluatingru[c], so it would have remained just
that -- ru[c]. When the kerneltried to apply it as a rule
to (a+b), there would have been an error.That didn't happen,
so that wasn't the sequence of events.Bobby
Treat-----Original Message-----f[c] = b+c?What
gives?Lawrence-- Lawrence A. Walker
Jr.http://www.kingshonor.com
====
You really should read
about the difference between := (SetDelayed) and =
(Set).When you enter your definitionf[x_]:=(a+b) /.
ru[a]the right hand side is not evaluated. So next when you
evaluate f[c] you get(a+b)/.ru[a] and only now ru[a] is
evaluated, thus giving you (a+b)/.a->x which is b+xThere
are several ways to get what you want. One is to use =
instead of :=, another to force evaluation of the right hand
side withf[x_]:=Evaluate[(a+b) /. ru[a]]yet another to
insert the actual rule in the definition of f:f[x_]:=(a+b) /.
a->xAndrzej KozlowskiToyama International
UniversityJAPANOn Wednesday, September 25, 2002, at 02:50
PM, Lawrence A. Walker Jr. > For the life of me I am not sure
why the following is not working in my> v. 4.2:>
ru[a]=a->x;> f[x_]:=(a+b) /. ru[a];> Why do I get> f[c] =
b+x> and not> f[c] = b+c?> What gives?> Lawrence> -- >
Lawrence A. Walker Jr.>
http://www.kingshonor.com
====
Well,I have written the
following notebook with
MathematicaIn[78]:=a=Sqrt[4*Pi/Sqrt[3]]In[79]:=fcom[k_,
mu_]:=((
1+Exp[-I*mu*2*Pi/a]+Exp[-I*mu*4*Pi/a])*(1-Exp[-I*k*(a-mu)])/
(I*k)+ Exp[-I*mu*2*
Pi/a]*((1-Exp[-I*(k+2*Pi/a)*(a-mu)])/(I*(k+2*Pi/a))+(
1-Exp[-I*(k-2*Pi/a)*(a-mu)])/(I*(k-2*Pi/a)))+ Exp[-I*mu*4*
Pi/a]*((1-Exp[-I*(k+4*Pi/a)*(a-mu)])/(I*(k+4*Pi/a))+(
1-Exp[-I*(k+2*Pi/a)*(a-mu)])/(I*(k+2*Pi/a)))+(
1-Exp[-I*(k-2*Pi/a)*(a-mu)])/(I*(k-2*Pi/a))+(
1-Exp[-I*(k-4*Pi/a)*(a-mu)])/(I*(k-4*Pi/a)))/(3*a)In[80]:=
f0[k_,mu_]:=((1+Exp[-I*mu*2*Pi/a]+Exp[-I*mu*4*Pi/a])*(a-mu)+
Exp[-I*mu*2* Pi/a]*((1-Exp[-I*2*Pi/a*(a-mu)])/(I*(2*Pi/a))+(
1-Exp[-I*(-2*Pi/a)*(a-mu)])/(I*(-2*Pi/a)))+ Exp[-I*mu*4*
Pi/a]*((1-Exp[-I*(4*Pi/a)*(a-mu)])/(I*(4*Pi/a))+(
1-Exp[-I*(2*Pi/a)*(a-mu)])/(I*(2*Pi/a)))+(
1-Exp[-I*(-2*Pi/a)*(a-mu)])/(I*(-2*Pi/a))+(
1-Exp[-I*(-4*Pi/a)*(a-mu)])/(I*(-4*Pi/a)))/(3*a)In[81]:=fp1
[k_,mu_]:=((
1+Exp[-I*mu*2*Pi/a]+Exp[-I*mu*4*Pi/a])*(1-Exp[-I*2*Pi/a*(
a-mu)])/( I*2*Pi/a)+
Exp[-I*mu*2*Pi/a]*((1-Exp[-I*(4*Pi/a)*(a-mu)])/(I*(4*Pi/a))+
(a-mu))+ Exp[-I*mu*4*
Pi/a]*((1-Exp[-I*(6*Pi/a)*(a-mu)])/(I*(6*Pi/a))+(
1-Exp[-I*(4*Pi/a)*(a-mu)])/(I*(4*Pi/a)))+(
a-mu)+(1-Exp[-I*(-2*Pi/a)*(a-mu)])/(I*(-2*Pi/a)))/(3*a)In[82
]:=fm1[k_,mu_]:=(( 1+Exp[-I*mu*2*Pi/a]+Exp[-I*mu*4*Pi/a])*(
1-Exp[I*2*Pi/a*(a-mu)])/(-I*2*Pi/a)+
Exp[-I*mu*2*Pi/a]*((a-mu)+(1-Exp[-I*(-4*Pi/a)*(a-mu)])/(I*(-
4*Pi/a)))+
Exp[-I*mu*4*Pi/a]*((1-Exp[-I*(2*Pi/a)*(a-mu)])/(I*(2*Pi/a))+
(a-mu))+( 1-Exp[-I*(-4*Pi/a)*(a-mu)])/(I*(-4*Pi/a))+(
1-Exp[-I*(-6*Pi/a)*(a-mu)])/(I*(-6*Pi/a)))/(3*a)In[83]:=fp2
[k_,mu_]:=((
1+Exp[-I*mu*2*Pi/a]+Exp[-I*mu*4*Pi/a])*(1-Exp[-I*4*Pi/a*(
a-mu)])/( I*4*Pi/a)+ Exp[-I*mu*2*
Pi/a]*((1-Exp[-I*(6*Pi/a)*(a-mu)])/(I*(6*Pi/a))+(1-Exp[-I*( 
2*Pi/a)*(a-mu)])/(I*(2*Pi/a)))+ Exp[-I*mu*4*
Pi/a]*((1-Exp[-I*(8*Pi/a)*(a-mu)])/(I*(8*Pi/a))+(
1-Exp[-I*(6*Pi/a)*(a-mu)])/(I*(6*Pi/a)))+(
1-Exp[-I*(2*Pi/a)*(a-mu)])/(I*(2*Pi/a))+(a-mu))/(3*a)In[84]:
=fm2[k_,mu_]:=(( 1+Exp[-I*mu*2*Pi/a]+Exp[-I*mu*4*Pi/a])*(
1-Exp[I*4*Pi/a*(a-mu)])/(-I*4*Pi/a)+ Exp[-I*mu*2*
Pi/a]*((1-Exp[-I*(-2*Pi/a)*(a-mu)])/(I*(-2*Pi/a))+(
1-Exp[-I*(-6*Pi/a)*(a-mu)])/(I*(-6*Pi/a)))+ Exp[-I*mu*4*
Pi/a]*((a-mu)+(1-Exp[-I*(-2*Pi/a)*(a-mu)])/(I*(-2*Pi/a)))+(
1-Exp[-I*(-6*Pi/a)*(a-mu)])/(I*(-6*Pi/a))+(
1-Exp[-I*(-8*Pi/a)*(a-mu)])/(I*(-8*Pi/a)))/(3*a)In[85]:=Lp[
0|N[0],mu_]:=f0[0,mu]In[86]:=Lp[2*Pi/a|N[2*Pi/a],mu_]:=fp1[
2*Pi/a,mu]In[87]:=Lp[-2*Pi/a|-N[2*Pi/a],mu_]:=fm1[2*Pi/a,mu
]In[88]:=Lp[4*Pi/a|N[4*Pi/a],mu_]:=fp2[4*Pi/a,mu]In[89]:=
Lp[-4*Pi/a|-N[4*Pi/a],mu_]:=fm2[-4*Pi/a,mu]In[90]:=Lp[k_,mu
_]:=fcom[k,mu]In[91]:=Ll[k_,mu_]:=0/;mu>=aIn[92]:=Ll[k_,
mu_]:=0/;mu<=-aIn[93]:=Ll[k_,mu_]:=Lp[k,mu]/;0<=mu<aIn[
94]:=Ll[k_,mu_]:=Exp[I*k*mu]*Conjugate[Lp[k,mu]]/;-a<mu<0In
[95]:=Ft[k_,mu_]:=(Exp[-2*k^2/3-mu^2/2+k*mu/Sqrt[3]]*Abs[Ll[
k,mu]]^2-1)/( 2*Pi*Sqrt[4*k^2/3+mu^2-2*k*mu/Sqrt[3]])At this
point I need to compute the (numerical) integration of both Ll
and, more important, Ft in all the real plane
({k,-Infinity,Infinity},{mu,-Infinity
,Infinity}).As  a first
attempt I am trying with the following statement:NIntegrate[
Abs[Ll[k,mu]],{k,-10,-4*Pi/a,-2*Pi/a,0,2*Pi/a,4*Pi/a,10},{mu,
-10,-a,0,a,10},
Method->MonteCarlo,MaxPoints->100000000,Compiled->False]but
this is not enough to ensure convergence of the integration.
Notice that  I have inserted some points in the integration
path in order to avoid problems with numerical divergences
which Mathematica detects in fcom[k,mu] (but these
divergences do not really exist, analitically) Does 
somebody has a smart suggestion to perform this
computation?Fabio
====
In a presentation I wish to use Plot
to generate a sequence of frames andthen animate them. The
problem is that the audience sees the animationtwice. Once
when the frames are being generated and then again after I
haveclosed the group and double clicked on the top graphic.
However, the firstshowing during generation is enough (but
uncontrolled). Is it possible to tidy up the generation of
the graphic so that it becomesthe animation?I have tried
the followingDo[Plot[Sin[t]*Sin[x], {x, 0, Pi}, PlotRange
-> {{0, Pi}, {-1, 1}},ImageSize -> 400];
SelectionMove[EvaluationNotebook[], All,
GeneratedCell];FrontEndExecute[{FrontEnd`SelectionAnimate[
0.1]}]; FrontEndExecute[{FrontEndToken[Clear]}], {t, 0,
15, 0.1}]This works but the cell dividing line ßashes on and
off spoiling theanimation and if there is anything in the
cell below this jumps up and down.Is there a proper way of
doing this?Hugh Goyder
====
I need to program the following
recursion scheme for time seriesforecasting (The Innovation
Algorithm).I will write it in pseudo-Mathematica notation.
K is the given m x mautocovariance (numerical) matrix of the
processv[0] = K[[1,1]];H[n,n-k] = (v[k])^(-1) (K[[n+1,k+1]]
- Sum[(H[k,k-j] H[n,n-j] v[j]),{j,0,k-}]) for
k=0,1,...,n-1v[n] = K[[n+1,n+1]] -Sum[(H[n,n-j])^2
v[j],{j,0,n-1}]The scheme should be solved in the order
v[0], H[1,1], v[1], H[2,2],H[2,1], v[2], H[3,3], H[3,2],
H[3,1], ...I have already tried to program it in a
straght-forward way, but ashave no experience with
recursive functions with two variables, itdoesn't seem to
work properly and is also very slow. Any help would
beRobertReply-To: murray@math.umass.edu
====
I don't
understand the problem. Say I enter the 2D form for the
reciprocal of the square root of a^2 + b^2, but I enter it
in Standard Form (i.e., using the Control key with / to form
the fraction, and the Control key with 2 to get the square
root symbol).If then I click anywhere within the
subexpression inside the square root, then with 2 to three
clicks the entire expression under the square root will be
highlighted. I can then do a copy-and-paste to put that into
any cell, in the usual way -- with keyboard key combinations,
right-click context menu, or Mathematica Edit menu.Isn't
that GUI enough?Would you REALLY want to be able to DRAG
the highlighted expression to a new place? Think of how much
of a mess this could cause through inadvertent movement of
the mouse after highlighting some subexpression. In fact, I
hate Microsoft Word's doing just that!>>I'm a 
poor physicist
trying to figure out how to sort out the>physical from the
non-physical solutions to a problem. To do>>that, I need
to be able to look at an expression and pick out
a>>subexpression, the part under the radical.> GRAY:> This
points up the need, which I've been aware of for years, to
be> able to select any part of an expression and drag it to
a new line for> further processing. Mathematica and the
other CAS I'm familiar with are still> pretty much stuck
with a command line interface. They need a true> GUI with
extensive interaction.> When I can see on the screen exactly
what I want to do, why> should I have to type a bunch of
stuff in to access what I want?> Maybe we'll see this in
version 5? 6? Never?> -- Murray Eisenberg
murray@math.umass.eduMathematics & Statistics Dept.Lederle
Graduate Research Tower phone 413 549-1020 (H)University of
Massachusetts 413 545-2859 (W)710 North Pleasant
StreetAmherst, MA 01375
====
> I'm a poor physicist trying to
figure out how to sort out the> physical from the
non-physical solutions to a problem. To do> that, I need
to be able to look at an expression and pick out a>
subexpression, the part under the radical.GRAY: This points
up the need, which I've been aware of for years, to beable
to select any part of an expression and drag it to a new line
forfurther processing. Mathematica and the other CAS I'm
familiar with are stillpretty much stuck with a command
line interface. They need a trueGUI with extensive
interaction. When I can see on the screen exactly what I want
to do, whyshould I have to type a bunch of stuff in to
access what I want? Maybe we'll see this in version 5? 6?
Never?Reply-To: murray@math.umass.edu
====
I get the same
result in 4.2 (under Windows). Not sure why. But you could
try this modification: PlaySeq[L_, dur_] :=
Do[PlayTone[L[[i]], dur]; Pause[dur], {i, Length[L]}]Then
PlaySeq[{224, 256, 384}, 3]seems to work as expected,
prolonging the notes before the last one to their full
expected duration.> PlayTone[F_, dur_] := Play[Sin[2*Pi*F*t],
{t, 0, dur}]> allows me to play a sine wave of frequency F and
duration dur> PlaySeq[L_, dur_] := Do[PlayTone[L[[i]], dur],
{i, Length[L]}]> should allow me to play a sequence of
tones> with a given list of frequencies L> and all the same
duration dur.> It does not work at least in 4.1> all tones but
the last one are much shorter than dur.> any help?> --
Murray Eisenberg murray@math.umass.eduMathematics &
Statistics Dept.Lederle Graduate Research Tower phone 413
549-1020 (H)University of Massachusetts 413 545-2859 (W)710
North Pleasant StreetAmherst, MA 01375
====
this is almost
the solution.but it creates some artificial rests.when the
sounds are created, there is some silence between the
different notes.when after creation one clicks the play
sound icon of the group with all the sounds, these rests
disappear.is there a way of Play creating ghte waves, but
NOT playing them,and then, doubleclicking the sound
icon,hearing the sound without rests immediately?> I get the
same result in 4.2 (under Windows). Not sure why. But you >
could try this modification:> PlaySeq[L_, dur_] :=
Do[PlayTone[L[[i]], dur];> Pause[dur], {i, Length[L]}]>
Then> PlaySeq[{224, 256, 384}, 3]> seems to work as
expected, prolonging the notes before the last one to > their
full expected duration.>>PlayTone[F_, dur_] :=
Play[Sin[2*Pi*F*t], {t, 0, dur}]>>allows me to play a sine
wave of frequency F and duration dur>PlaySeq[L_, dur_] :=
Do[PlayTone[L[[i]], dur], {i, Length[L]}]>should allow me to
play a sequence of tones>>with a given list of frequencies
L>>and all the same duration dur.>It does not work at least
in 4.1>all tones but the last one are much shorter than
dur.>any help? > -- --Erich Neuwirth, Computer Supported
Didactics Working GroupVisit our SunSITE at
http://sunsite.univie.ac.at
====
PlayTone[F_, dur_] :=
Play[Sin[2*Pi*F*t], {t, 0, dur}]allows me to play a sine
wave of frequency F and duration durPlaySeq[L_, dur_] :=
Do[PlayTone[L[[i]], dur], {i, Length[L]}]should allow me to
play a sequence of toneswith a given list of frequencies
Land all the same duration dur.It does not work at least
in 4.1all tones but the last one are much shorter than
dur.any help?-- --Erich Neuwirth, Computer Supported
Didactics Working GroupVisit our SunSITE at
http://sunsite.univie.ac.at
====
using Set (=) instead of
SetDelayed (:=) in f[x_]:=(a+b) /. ru[a]; will give you
what you desire.Matthias Bode.-----Ursprí.b9ngliche
Nachricht-----Gesendet: Mittwoch, 25. September 2002
07:51An: mathgroup@smc.vnet.netBetreff: Strange ReplaceAll
behaviorFor the life of me I am not sure why the following
is not working in my v. 4.2:ru[a]=a->x;f[x_]:=(a+b) /.
ru[a];Why do I getf[c] = b+xand notf[c] = b+c?What
gives?Lawrence-- Lawrence A. Walker
Jr.http://www.kingshonor.com
====
If you want to do it with
the mouse, it's not that difficult. I 
clickwithin the part I
want and double-click until the part is selected.(Each
double-click expands the selection by one level of nesting.)
Thenpush Ctrl-C, click where you want to put the result,
and push Ctrl-V.Does that help any?Bobby-----Original
Message-----able to select any part of an expression and
drag it to a new line forfurther processing. Mathematica
and the other CAS I'm familiar with arestillpretty much
stuck with a command line interface. They need a trueGUI
with extensive interaction. When I can see on the screen
exactly what I want to do, whyshould I have to type a bunch
of stuff in to access what I want? Maybe we'll see this in
version 5? 6? Never?
====
I met a critical problem for running
Mathematica 4.1 on MacOS X 10.2.(OSX native version) I hope
such a problem is suitable to be posted tothis ML.What
happens is the following. When Mathematica 4.1 installed in
HDDis started, a window quickly appears saying that A
serious error has occured while Mathematica was starting up.
Mathematica will probably not function properly until this
problem is resolved. ... The Mathematica fonts are not
properly installed in your system. Without these fonts,
typeset mathematical expressions cannot be displayed
properly.This message is correct. If I chose Continue
Anyway, mathematicalexpressins are totally broken.What is
strange is that when I start Mathematica 4.1 by double
clickingMathematica in CDROM, it works without any problem.
Further more,Once Mathematica 4.1 is correctly started in
this way, I can restart Mathematicadirectly from double
clicking Mathematica in HDD, which is the installed
one.Once CDROM is ejected, the error appears again.It is
clearly font related problem but I do not know how to solve.
If someoneknow how to manage, please let me
know.-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-I
met a critical problem for running Mathematica 4.1 on MacOS X
10.2.(OSX native version) I hope such a problem is suitable to
be posted tothis ML.What happens is the following. When
Mathematica 4.1installed in HDDis started, a window quickly
appears saying that A serious error has occured while
Mathematica was starting up. Mathematica will probably not
function properly until this problem is resolved. ...
The Mathematica fonts are not properly installed in your
system. Without these fonts, typeset mathematical expressions
cannot be displayed properly.This message is correct. If I
chose Continue Anyway, mathematicalexpressins are
totally broken.What is strange is that when I start
Mathematica 4.1 by double clickingMathematica in CDROM, it
works without any problem. Further more,Once Mathematica 4.1
is correctly started in this way, I can
restartMathematicadirectly from double clicking
Mathematica in HDD, which is theinstalled one.Once CDROM
is ejected, the error appears again.It is clearly font
related problem but I do not know how to solve.
Ifsomeoneknow how to manage, please let me
know.-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=---
Apple-Mail-2-977368608--
====
Toshinao,Try reinstalling
Mathematica from the CD.---Selwyn Hollis> I met a critical
problem for running Mathematica 4.1 on MacOS X 10.2.> (OSX
native version) I hope such a problem is suitable to be
posted to> this ML.> What happens is the following. When
Mathematica 4.1 installed in HDD> is started, a window
quickly appears saying that> A serious error has occured
while Mathematica was starting up.> Mathematica will probably
not function properly until this problem is> resolved.>
...> The Mathematica fonts are not properly installed in
your system.> Without these fonts, typeset mathematical
expressions cannot be> displayed properly.> This message is
correct. If I chose Continue Anyway, mathematical>
expressins are totally broken.> What is strange is that when
I start Mathematica 4.1 by double clicking> Mathematica in
CDROM, it works without any problem. Further more,> Once
Mathematica 4.1 is correctly started in this way, I can
restart > Mathematica> directly from double clicking
Mathematica in HDD, which is the > installed one.> Once CDROM
is ejected, the error appears again.> It is clearly font
related problem but I do not know how to solve. If >
someone> know how to manage, please let me know.>
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-> I
met a critical problem for running Mathematica 4.1 on MacOS X
10.2.> (OSX native version) I hope such a problem is suitable
to be posted to> this ML.> What happens is the following.
When Mathematica 4.1installed in HDD> is started, a window
quickly appears saying that> A serious error has occured
while Mathematica was starting up.> Mathematica will probably
not function properly until this problem is> resolved.>
...> The Mathematica fonts are not properly installed in
your system.> Without these fonts, typeset mathematical
expressions cannot be> displayed properly.> This message
is correct. If I chose Continue Anyway, mathematical>
expressins are totally broken.> What is strange is that
when I start Mathematica 4.1 by double clicking> Mathematica
in CDROM, it works without any problem. Further more,> Once
Mathematica 4.1 is correctly started in this way, I can
restart> Mathematica> directly from double clicking
Mathematica in HDD, which is the> installed one.> Once CDROM
is ejected, the error appears again.> It is clearly font
related problem but I do not know how to solve. If>
someone> know how to manage, please let me know.>
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=->
--Apple-Mail-2-977368608--> 
====
The reason why
InequalitySolve returns it's answer in what sometimes turns
out to be unnecessarily complicated form is that the
underlying algorithm, Cylindrical Agebraic Decomposition
(CAD) returns its answers in this form. Unfortunately it
seems to me unlikely that a simplification of the kind you
need can be can be accomplished in any general way. To see
why observe the following. First of
all:In[1]:=FullSimplify[x > 0 || x == 0]Out[1]=x >=
0This is fine. However:In[2]:=FullSimplify[x > 0 && x < 2
|| x == 0 && x < 2]Out[2]=x == 0 || 0 < x < 2Of course
what you would like is simply 0 <= x < 2. One reason why you
can't get it is that while Mathematica can perform a
LogicalExpand, as in:In[3]:=LogicalExpand[(x > 0 || x
== 0) && x < 2]Out[3]=x == 0 && x < 2 || x > 0 && x <
2There i no LogicalFactor or anything similar that would
reverse what LogicalExpand does. if there was then you could
perform the sort of simplifications you need
for:In[4]:=FullSimplify[(x > 0 || x == 0) && x <
2]Out[4]=0 <= x < 2However, it does not seem to me very
likely that such logical factoring can be performed by a
general enough algorithm (though I am no expert in this
field). In any case, certainly Mathematica can't 
do this.I
also noticed that Mathematica seems unable to show that the
answer it returns to your problem is actually equivalent to
your simpler one. In fact this looks like a possible bug in
Mathematica. Let's first try the function 
ImpliesQ from the
Experimental context:<< Experimental`Now Mathematica
correctly gives:In[6]:=ImpliesQ[y4 >= -1 && y6 >= -1 && y6
<= y5 <= 1 + y4 + y6, y4 == -1 && y6 >= -1 && y5 == y6 || y4
> -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 +
y6]Out[6]=TrueHowever:In[7]:=ImpliesQ[y4 == -1 && y6 >=
-1 && y5 == y6 || y4 > -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 +
y6, y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 +
y6]Out[7]=FalseThat simply means that ImpliesQ cannot
show the implication, not that it does not hold. ImpliesQ
relies on CAD, as does FullSimplify. Switching to
FullSimplify we see that:In[8]:=FullSimplify[y4 == -1 && y6
>= -1 && y5 == y6 || y4 > -1 && y6 >= -1 && y6 <= y5 <= 1 + y4
+ y6, y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 +
y6]Out[8]=TruewhileIn[9]:=FullSimplify[y4 >= -1 && y6
>= -1 && y6 <= y5 <= 1 + y4 + y6, y4 == -1 && y6 >= -1 && y5
== y6 || y4 > -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 +
y6]Out[9]=y4 >= -1 && y6 <= y5 <= 1 + y4 + y6On the other
hand, taking just the individual summands of Or as
hypotheses;In[10]:=FullSimplify[y4 >= -1 && y6 >= -1 && y6
<= y5 <= 1 + y4 + y6, y4 > -1 && y6 >= -1 && y6 <= y5 <= 1 +
y4 + y6]Out[10]=TrueIn[11]:=FullSimplify[y4 >= -1 && y6
>= -1 && y6 <= y5 <= 1 + y4 + y6, y4 == -1 && y6 >= -1 && y5
== y6 ]Out[11]=TrueIn fact FullSimplify is unable to use
Or in assumptions, which can be demonstrated on an abstract
example:In[12]:=FullSimplify[C,(A||B)&&(C)]Out[12]=True
In[13]:=FullSimplify[C,LogicalExpand[(A||B)&&(C)]]Out[13]=
CThis could be fixed by modifying
FullSimplify:In[14]:=Unprotect[FullSimplify];In[14]:=
FullSimplify[expr_,Or[x_,y__]]:=Or[FullSimplify[expr,x],
FullSimplify[exp r,y]];In[15]:=Protect[FullSimplify];Now
at least we get as before:In[16]:=FullSimplify[y4 == -1 &&
y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1 && y6 <= y5 <= 1
+ y4 + y6, y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 +
y6]Out[16]=Truebut also:In[17]:=FullSimplify[y4 >= -1
&& y6 >= -1 && y6 <= y5 <= 1 + y4 + y6, y4 == -1 && y6 >= -1
&& y5 == y6 || y4 > -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 +
y6]Out[17]=TrueThis seems to me a possible worthwhile
improvement in FullSimplify, though of course not really
helpful for your problem.Andrzej KozlowskiToyama
International UniversityJAPAN> I have a set of
inequalities that I solve with InequalitySolve. But > then>
it gives a complete set of solutions, but not in the way I
would like > it> to be! :-) For example, the simple
following calculation will give:> In[1]:= ineq = {y4 >= -1,
y5 >= -1, y6 + y4 >= y5 - 1, y5 >= y6, y6 >= > -1};>
InequalitySolve[ineq,{y4,y6,y5}]> Out[1]:= y4 == -1 && y6 >=
-1 && y5 == y6 ||> y4 > -1 && y6 >= -1 && y6 <= y5 <= 1 + y4
+ y6> the result is good, but I would like it to be in the
simpler but> equivalent form> y4 >= -1 && y6 >= -1 && y6 <=
y5 <= 1 + y4 + y6> How can I tell InequalitySolve to do that?
It is simple for this > example,> but for a large set of
simple inequalities InequalitySolve gives lines > and> lines
of results instead of a simple result.> Vincent Bouchard
====

I have been trying to integrate the following :
Integrate[Cosh[2 Abs[x-y]] 2 y, {y,0,1/2},
Assumptions->{Im[x]==0,x>0}] However, Mathematica chokes and
simply returns the integral as it is. However, if I split
up the integral into two portions, it quickly gives me an
answer for the parts. Is there something implicit that I am
missing in the Assumptions ?MS.-- 12:02am up 5:06, 1 user,
load average: 0.54, 0.22, 0.08
====
The modification to
FullSimplify that I sent earlier works correctly only for
assumptions of the form Or[a,b] (and even then not is not
always what one would like). For what it's worth here is a
better (but slow)
version:In[1]:=Unprotect[FullSimplify];In[2]:=
FullSimplify[expr_, x_ || y__] := FullSimplify[
FullSimplify[expr, x] || FullSimplify[expr,
Or[y]]];In[3]:=Protect[FullSimplify];For
example:In[4]:=FullSimplify[Sqrt[(x - 1)^2] + Sqrt[(x -
2)^2] + Sqrt[(x - 3)^2], x > 1 || x > 2 || x > 3]Out[4]=-1 +
x + Abs[-3 + x] + Abs[-2 + x] || -3 + 2*x + Abs[-3 + x] ||
3*(-2 + x)Andrzej KozlowskiToyama International
UniversityJAPAN> The reason why InequalitySolve returns
it's answer in what sometimes > turns out to be 
unnecessarily
complicated form is that the underlying > algorithm,
Cylindrical Agebraic Decomposition (CAD) returns its >
answers in this form. Unfortunately it seems to me unlikely
that a > simplification of the kind you need can be can be
accomplished in any > general way. To see why observe the
following. First of all:> In[1]:=> FullSimplify[x > 0 || x ==
0]> Out[1]=> x >= 0> This is fine. However:> In[2]:=>
FullSimplify[x > 0 && x < 2 || x == 0 && x < 2]> Out[2]=> x
== 0 || 0 < x < 2> Of course what you would like is simply 0
<= x < 2. One reason why you > can't get it is that while
Mathematica can perform a LogicalExpand, > as in:>
In[3]:=> LogicalExpand[(x > 0 || x == 0) && x < 2]> Out[3]=>
x == 0 && x < 2 || x > 0 && x < 2> There i no
LogicalFactor or anything similar that would reverse what
> LogicalExpand does. if there was then you could perform
the sort of > simplifications you need for:> In[4]:=>
FullSimplify[(x > 0 || x == 0) && x < 2]> Out[4]=> 0 <= x <
2> However, it does not seem to me very likely that such
logical > factoring can be performed by a general enough
algorithm (though I am > no expert in this field). In any
case, certainly Mathematica can't do > this.> I also noticed
that Mathematica seems unable to show that the answer > it
returns to your problem is actually equivalent to your
simpler one. > In fact this looks like a possible bug in
Mathematica. Let's first try > the function 
ImpliesQ from the
Experimental context:> << Experimental`> Now Mathematica
correctly gives:> In[6]:=> ImpliesQ[y4 >= -1 && y6 >= -1 &&
y6 <= y5 <= 1 + y4 + y6,> y4 == -1 && y6 >= -1 && y5 == y6 ||
y4 > -1 && y6 >= -1 && y6 <= y5 > <= 1 + y4 + y6]> Out[6]=>
True> However:> In[7]:=> ImpliesQ[y4 == -1 && y6 >= -1 && y5
== y6 || y4 > -1 && y6 >= -1 &&> y6 <= y5 <= 1 + y4 + y6, y4
>= -1 && y6 >= -1 && y6 <= y5 <= 1 + > y4 + y6]> Out[7]=>
False> That simply means that ImpliesQ cannot show the
implication, not that > it does not hold. ImpliesQ relies on
CAD, as does FullSimplify. > Switching to FullSimplify we see
that:> In[8]:=> FullSimplify[y4 == -1 && y6 >= -1 && y5 == y6
|| y4 > -1 && y6 >= -1 &&> y6 <= y5 <= 1 + y4 + y6, y4 >= -1
&& y6 >= -1 && y6 <= y5 <= 1 + > y4 + y6]> Out[8]=> True>
while> In[9]:=> FullSimplify[y4 >= -1 && y6 >= -1 && y6 <=
y5 <= 1 + y4 + y6,> y4 == -1 && y6 >= -1 && y5 == y6 || y4 >
-1 && y6 >= -1 && y6 <= y5 > <= 1 + y4 + y6]> Out[9]=> y4 >=
-1 && y6 <= y5 <= 1 + y4 + y6> On the other hand, taking just
the individual summands of Or as > hypotheses;> In[10]:=>
FullSimplify[y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 +
y6,> y4 > -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6]>
Out[10]=> True> In[11]:=> FullSimplify[y4 >= -1 && y6 >= -1
&& y6 <= y5 <= 1 + y4 + y6,> y4 == -1 && y6 >= -1 && y5 == y6
]> Out[11]=> True> In fact FullSimplify is unable to use Or
in assumptions, which can be > demonstrated on an abstract
example:> In[12]:=> FullSimplify[C,(A||B)&&(C)]> Out[12]=>
True> In[13]:=> FullSimplify[C,LogicalExpand[(A||B)&&(C)]]>
Out[13]=> C> This could be fixed by modifying FullSimplify:>
In[14]:=> Unprotect[FullSimplify];> In[14]:=>
FullSimplify[expr_,Or[x_,y__]]:=Or[FullSimplify[expr,x],
FullSimplify[ex > pr,y]];> In[15]:=> Protect[FullSimplify];>
Now at least we get as before:> In[16]:=> FullSimplify[y4 ==
-1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1 &&> y6 <=
y5 <= 1 + y4 + y6, y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + >
y4 + y6]> Out[16]=> True> but also:> In[17]:=>
FullSimplify[y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 +
y6,> y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1
&& y6 <= y5 > <= 1 + y4 + y6]> Out[17]=> True> This seems to
me a possible worthwhile improvement in FullSimplify, > though
of course not really helpful for your problem.> Andrzej
Kozlowski> Toyama International University> JAPAN> I have
a set of inequalities that I solve with InequalitySolve. But
> then> it gives a complete set of solutions, but not in the
way I would like > it> to be! :-) For example, the simple
following calculation will give: In[1]:= ineq = {y4 >= -1, y5
>= -1, y6 + y4 >= y5 - 1, y5 >= y6, y6 >= -1};>
InequalitySolve[ineq,{y4,y6,y5}] Out[1]:= y4 == -1 && y6 >=
-1 && y5 == y6 ||> y4 > -1 && y6 >= -1 && y6 <= y5 <= 1 + y4
+ y6 the result is good, but I would like it to be in the
simpler but> equivalent form y4 >= -1 && y6 >= -1 && y6 <=
y5 <= 1 + y4 + y6 How can I tell InequalitySolve to do that?
It is simple for this > example,> but for a large set of
simple inequalities InequalitySolve gives > lines and> lines
of results instead of a simple result.> Vincent
Bouchard
====
>-----Original Message----->Sent: Wednesday,
September 25, 2002 7:51 AM>For the life of me I am not sure
why the following is not >working in my >v.
4.2:>ru[a]=a->x;>f[x_]:=(a+b) /. ru[a];>Why do I get>f[c] =
b+x>and not>f[c] = b+c?>What gives?>Lawrence>-- >Lawrence
A. Walker Jr.>http://www.kingshonor.comLawrence, in your
definition of f, x doesn't show up explicitely. 
So, in
theevaluation sequence, when the definition for f[c] is
applied, no x appearsat rhs i.e.(a + b) /. ru[a] and such c
cannot be inserted. The result is the same asdirectly
executingIn[11]:= (a + b) /. ru[a]Out[11]= b + xIf you
don't like this, you have to make explicit the Value of 
ru[a]
in thedefiniton of f. One way to do so is to use Set instead
of SetDelayed:In[9]:= f[x_] = (a + b) /. ru[a]Out[9]= b +
xIn[10]:= f[c]Out[10]= b + cThe drawback of this that
not only the value of ru[a] is inserted but alsothe whole
expression including ReplaceAll is evaluated. If this is
notwanted, you have to insert the value of ru[a] into the
unevaluated rhs atthe definition. The general means for this
are function application, With orReplace:In[7]:= (g[x_] :=
(a + b) /. #) &[ru[a]]In[8]:= g[c]Out[8]= b + cIn[16]:=
Clear[g]In[20]:=Unevaluated[g[x_] := (a + b) /. rule] /.
rule -> ru[a]In[21]:= g[c]Out[21]= b + cHere we have to
prevent evaluation of the defintion before our rule
isinserted, this is achieved by Unevaluated.With is a bit
more complicated, since the scoping rules for SetDelayed
wouldnot allow the substition of an expression at rhs
containing a patternvariable (the pattern variable is
renamed in this case). A simple answer tothis is to also
substitute the argument variable (the pattern): In[31]:=
Clear[g]In[32]:=With[{rule = ru[a], arg = x_}, g[arg] := (a
+ b) /. rule]In[33]:= g[c]Out[33]= b + c--Hartmut
Wolf
====
Hartmut,I add an explicit illustration to your
ingenious solution using With.Hartmut's solution: Clear[g];
ru[a] = a -> x; With[{rule = ru[a], arg = x_}, g[arg] := a +
b /. rule]; g[ c] b + cWhy is arg = x_ needed?Without it
we get Clear[g]; With[{rule=ru[a]}, g[x_]:=a+b/.rule]; g [c]
b+xThe reason for this shows in ?g Global`g g[x$_] := a +
b /. a -> xThe x in x_ has been changed to x$ and there is
no x$ on the right side.This is a general feature of
scoping. Taking it further we get Clear[g];
With[{rule=ru[a]},g[x_]:=a+x/.rule]; g [c] c+x ?g Global`g
g[x$_] := a + x$ /. a ->
x--Allan---------------------Allan HayesMathematica
Training and ConsultingLeicester
UKwww.haystack.demon.co.ukhay@haystack.demon.co.ukVoice
: +44 (0)116 271 4198>-----Original Message----->Sent:
Wednesday, September 25, 2002 7:51 AM>For the life of me I
am not sure why the following is not>working in my>v.
4.2:>ru[a]=a->x;>f[x_]:=(a+b) /. ru[a];>Why do I get>f[c] =
b+x>and not>f[c] = b+c?>What gives?>Lawrence>-->Lawrence
A. Walker Jr.>http://www.kingshonor.com>> Lawrence,> in your
definition of f, x doesn't show up explicitely. 
So, in the>
evaluation sequence, when the definition for f[c] is applied,
no x appears> at rhs i.e.> (a + b) /. ru[a] and such c
cannot be inserted. The result is the same as> directly
executing> In[11]:= (a + b) /. ru[a]> Out[11]= b + x> If
you don't like this, you have to make explicit the Value of
ru[a] inthe> definiton of f. One way to do so is to use Set
instead of SetDelayed:> In[9]:= f[x_] = (a + b) /. ru[a]>
Out[9]= b + x> In[10]:= f[c]> Out[10]= b + c> The drawback
of this that not only the value of ru[a] is inserted but
also> the whole expression including ReplaceAll is
evaluated. If this is not> wanted, you have to insert the
value of ru[a] into the unevaluated rhs at> the definition.
The general means for this are function application,
Withor> Replace:> In[7]:= (g[x_] := (a + b) /. #)
&[ru[a]]> In[8]:= g[c]> Out[8]= b + c> In[16]:= Clear[g]>
In[20]:=> Unevaluated[g[x_] := (a + b) /. rule] /. rule ->
ru[a]> In[21]:= g[c]> Out[21]= b + c> Here we have to
prevent evaluation of the defintion before our rule is>
inserted, this is achieved by Unevaluated.> With is a bit
more complicated, since the scoping rules for
SetDelayedwould> not allow the substition of an expression
at rhs containing a pattern> variable (the pattern variable
is renamed in this case). A simple answerto> this is to
also substitute the argument variable (the pattern):>
In[31]:= Clear[g]> In[32]:=> With[{rule = ru[a], arg = x_},
g[arg] := (a + b) /. rule]> In[33]:= g[c]> Out[33]= b + c>
--> Hartmut
Wolf
====
TryIn[1]:=M={{1,2,3},{4,5,6},{7,8,9}};In[2]:=
Export[test.csv,M,CSV];Also, avoid using names with
capital initial, since these are usuallyreserved for
Mathematica functions.Tomas GarzaMexico City-----
Original Message -----> M={{1,2,3},{4,5,6},{7,8,9}};>
TableForm[M]> 1 2 3> 4 5 6> 7 8 9> The commands like Put or
Write create files with a standard Mathematica> matrix
notation with braces, I get the the same result with Save
As/Copy> Robert
====
I would appreciate if someone could
tell me how to put my output forexample a matrix in
TableForm as a space-delimited ASCII file inMathematica
3.0.for example:M={{1,2,3},{4,5,6},{7,8,9}};TableForm[M]1
2 34 5 67 8 9The commands like Put or Write create files with
a standard Mathematicamatrix notation with braces, I get the
the same result with Save As/CopyRobert
====
Much longer. At
least 10 times as long. Not sure because I quit. As I
readBobby Treat's response, one of my processors had been
laboring for more thanthree days to generate all the
digits. (I had not considered to display orprint all of
them. Why on Earth would I want to do that?) So I started
over,asking RealDigits to skip most of the digits.Tom
Burton> So would it take about the same amont of time for
the complete printout> of digits? Of course it would take a
few additional seconds to format> the output...> Or does
Mathematica take alot less time when it truncates the
output?> Could you tell me the CPU you used and its speed
etc...i am curious,> other programs out there.> I used one
processor of a dual 1GH Mac and got the same answer with
the> following speed:> 4.2 for Mac OS X (June 4, 2002)>
oldmax = $MaxPrecision> 6> 1. 10> $MaxPrecision = Infinity>
Infinity> With[{n = 2^26}, Timing[> pd = RealDigits[N[Pi, n +
1], 10, 20,> 19 - n]; ]]> {28794.1 Second, Null}>
MaxMemoryUsed[]> 512055204> pd> {{3, 3, 8, 6, 3, 2, 2, 0, 8,
9, 6, 2, 2, 3,> 4, 0, 9, 8, 0, 3}, -67108844}> Tom Burton>

====
Often posters to MathGroup copy and paste in the
complete cell expression,including the In and Out numbers,
when posting to MathGroup.I wonder if this is the best
method because one can't then just copy out allthe
statements and paste them into a Mathematica notebook. All
the statementnumbers have to be edited out and if there
are many statement definitionsthis is an extended task for
any responder. This, of course, decreases thechances for a
response. A better method is for the poster to just copy
andpaste the CONTENTS of each cell. This is more work for
the poster, but itmay pay off in better responses.David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/===
=David,I agree with you that it makes responding and using
easier if the In and Outnumbers are not included.An
easier way to do this, rather than copying the contents of
the cells, isUsually a little manual reformating is needed
to distinguish text, input andoutput --- I try to use one
tab indent for input and two tabs indent foroutput, plus
some blank line adjustment.I wonder if anyone has a way of
automatically achieving this
reformating.--Allan---------------------Allan
HayesMathematica Training and ConsultingLeicester
UKwww.haystack.demon.co.ukhay@haystack.demon.co.ukVoice
: +44 (0)116 271 4198> Often posters to MathGroup copy and
paste in the complete cell expression,> including the In and
Out numbers, when posting to MathGroup.> I wonder if this is
the best method because one can't then just copy outall>
the statements and paste them into a Mathematica notebook.
All thestatement> numbers have to be edited out and if
there are many statement definitions> this is an extended
task for any responder. This, of course, decreases the>
chances for a response. A better method is for the poster to
just copy and> paste the CONTENTS of each cell. This is
more work for the poster, but it> may pay off in better
responses.> David Park> djmp@earthlink.net>
http://home.earthlink.net/~djmp/
====
copying and pasting is
straightforward. I always run my notebooks this way,because
showing the In[x], Out[x] does not seem to provide me with
muchbeyond clutter.Kevin> Often posters to MathGroup copy
and paste in the complete cell expression,> including the In
and Out numbers, when posting to MathGroup.> I wonder if this
is the best method because one can't then just copy outall>
the statements and paste them into a Mathematica notebook. All
thestatement> numbers have to be edited out and if there
are many statement definitions> this is an extended task for
any responder. This, of course, decreases the> chances for
a response. A better method is for the poster to just copy
and> paste the CONTENTS of each cell. This is more work for
the poster, but it> may pay off in better responses.> David
Park> djmp@earthlink.net>
http://home.earthlink.net/~djmp/
====
>In a presentation I wish
to use Plot to generate a sequence of frames and>then animate
them. The problem is that the audience sees the
animation>twice. Once when the frames are being generated
and then again after I have>closed the group and double
clicked on the top graphic. However, the first>showing during
generation is enough (but uncontrolled).>Hugh GoyderThis
creates a graphics cell from a graphics
expression.GraphicCell[graphics_] :=
Cell[GraphicsData[PostScript,
DisplayString[graphics]],Graphics]cellgroup.Block[{$
DisplayFunction=Identity, graphs}, graphs =
Table[GraphicCell[ Plot[Sin[t]*Sin[x], {x, 0, Pi}, PlotRange
-> {{0, Pi}, {-1, 1}}, ImageSize -> 400]], {t,0,15,.1}];
NotebookWrite[EvaluationNotebook[],Cell[CellGroupData[graphs,
Closed]]]; SelectionMove[EvaluationNotebook[], All,
GeneratedCell];
FrontEndExecute[{FrontEndToken[EvaluationNotebook[],
SelectionAnimate]}]
]------------------------------------------------------------
--Omega ConsultingThe final answer to your Mathematica
needsSpend less time searching and more time
finding.http://www.wz.com/internet/Mathematica.html
====

Actually, the reason why ImpliesQ (and FullSimplify) fail
toprove the implication is not that the hypothesis is a
disjunction.To use the cylindrical algebraic decomposition
algorithm theyneed to know that the assumptions imply that
all variables arereal. The assumptions mechanism infers
variable domains in a purelysyntactical way, i.e. v is
assumed to be real if there is an Element[v, Reals]
statement or v appears in an inequality.It does not attempt
to analyze assumptions further, to figureout that, say y6 >=
-1 implies that y6 is real, and then ifwe have y5 == y6 then
y5 must be real too. Doing such an analysisin general would
require solving the assumptions over complexnumbers, and
then finding out which variables need to be real.This would
be in general too time consuming to do, but analyzing linear
dependencies like the ones in your example is a
possiblefuture improvement.ImpliesQ cannot prove the
implication here, because it knows onlythat y6 is
real.In[1]:= <<Experimental` In[2]:= ImpliesQ[y4 == -1 &&
y6 >= -1 && y5 == y6, y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1
+ y4 + y6] Out[2]= False If we add an explicit assumption
that y4 and y5 are real, ImpliesQ (and FullSimplify) can
prove this implication, and the full version of your
example.In[3]:= ImpliesQ[Element[y4|y5, Reals] && y4 == -1
&& y6 >= -1 && y5 ==y6, y4 >= -1 && y6 >= -1 && y6 <= y5 <=
1 + y4 + y6] Out[3]= True In[4]:= ImpliesQ[Element[y4|y5,
Reals] && (y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6
>= -1 && y6 <= y5 <= 1 + y4 + y6), y4 >= -1 && y6 >= -1 && y6
<= y5 <= 1 + y4 + y6] Out[4]=True In[5]:= FullSimplify[y4
>= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6,Element[y4|y5,
Reals] && (y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6
>= -1 && y6 <= y5 <= 1 + y4 + y6)] Out[5]=True Adam
StrzebonskiWolfram Research > On second thoughts I
realized that there seems to be an inherent> ambiguity about
what one coudl mean by using alternatives (statements> joned
by Or) assumptions. In fact it now seems to me that the>
reasonable intertpretation for ImpliesQ and FullSimplify
ought to> perhaps be different. It seems to me that
ImpliesQ[Or[a,b],c] ought to> return True if aand only if
ImpliesQ[a,c] and ImpliesQ[b,c] both return> True. If so
this could be acomplished by adding the rule>
ImpliesQ[Or[a,b],c] = And[ImpliesQ[a,c],ImpliesQ[b,c]]. That
could then> be used in proving that the two answers to the
system of inequalities> that of Vincent's original posting
are equivalent. On the other hand> probably FullSimplify[a,
Or[p,q]] ought to return>
Or[FullSimplify[a,p],FullSimplify[a,q]] (or do nothing as it
doe snow).> The first approach would seem to be consistent
with the way> FullSimplify works with domain specifications
but would however have> the strange effect of returning True
if just one of the alternatives> were true and the other
false. So perhaps after all it is best to> leave
FullSimplify as it is. However, it seems to me that
ImpliesQ> shoud be able to handle such cases (?)> Andrzej
Kozlowski> Toyama International University> JAPAN> The
modification to FullSimplify that I sent earlier works
correctly> only for assumptions of the form Or[a,b] (and
even then not is not> always what one would like). For what
it's worth here is a better (but> slow) version:> In[1]:=>
Unprotect[FullSimplify];> In[2]:=> FullSimplify[expr_, x_ ||
y__] := FullSimplify[> FullSimplify[expr, x] ||
FullSimplify[expr, Or[y]]];> In[3]:=> Protect[FullSimplify];>
For example:> In[4]:=> FullSimplify[Sqrt[(x - 1)^2] + Sqrt[(x
- 2)^2] +> Sqrt[(x - 3)^2], x > 1 || x > 2 || x > 3]>
Out[4]=> -1 + x + Abs[-3 + x] + Abs[-2 + x] ||> -3 + 2*x +
Abs[-3 + x] || 3*(-2 + x)> Andrzej Kozlowski> Toyama
International University> JAPAN> The reason why
InequalitySolve returns it's answer in what sometimes> turns
out to be unnecessarily complicated form is that the
underlying> algorithm, Cylindrical Agebraic Decomposition
(CAD) returns its> answers in this form. Unfortunately it
seems to me unlikely that a> simplification of the kind you
need can be can be accomplished in any> general way. To see
why observe the following. First of all:> In[1]:=>
FullSimplify[x > 0 || x == 0]> Out[1]=> x >= 0> This is fine.
However:> In[2]:=> FullSimplify[x > 0 && x < 2 || x == 0 && x
< 2]> Out[2]=> x == 0 || 0 < x < 2> Of course what you would
like is simply 0 <= x < 2. One reason why> you can't get it
is that while Mathematica can perform a> LogicalExpand,
as in:> In[3]:=> LogicalExpand[(x > 0 || x == 0) && x < 2]>
Out[3]=> x == 0 && x < 2 || x > 0 && x < 2> There i no
LogicalFactor or anything similar that would reverse>
what LogicalExpand does. if there was then you could perform
the sort> of simplifications you need for:> In[4]:=>
FullSimplify[(x > 0 || x == 0) && x < 2]> Out[4]=> 0 <= x <
2> However, it does not seem to me very likely that such
logical> factoring can be performed by a general enough
algorithm (though I> am no expert in this field). In any
case, certainly Mathematica can't> do this.> I also noticed
that Mathematica seems unable to show that the answer> it
returns to your problem is actually equivalent to your
simpler> one. In fact this looks like a possible bug in
Mathematica. Let's> first try the function 
ImpliesQ from the
Experimental context:> << Experimental`> Now Mathematica
correctly gives:> In[6]:=> ImpliesQ[y4 >= -1 && y6 >= -1 &&
y6 <= y5 <= 1 + y4 + y6,> y4 == -1 && y6 >= -1 && y5 == y6 ||
y4 > -1 && y6 >= -1 && y6 <= y5> <= 1 + y4 + y6]> Out[6]=>
True> However:> In[7]:=> ImpliesQ[y4 == -1 && y6 >= -1 && y5
== y6 || y4 > -1 && y6 >= -1 &&> y6 <= y5 <= 1 + y4 + y6, y4
>= -1 && y6 >= -1 && y6 <= y5 <= 1 +> y4 + y6]> Out[7]=>
False> That simply means that ImpliesQ cannot show the
implication, not that> it does not hold. ImpliesQ relies on
CAD, as does FullSimplify.> Switching to FullSimplify we see
that:> In[8]:=> FullSimplify[y4 == -1 && y6 >= -1 && y5 ==
y6 || y4 > -1 && y6 >= -1> &&> y6 <= y5 <= 1 + y4 + y6, y4 >=
-1 && y6 >= -1 && y6 <= y5 <= 1 +> y4 + y6]> Out[8]=> True>
while> In[9]:=> FullSimplify[y4 >= -1 && y6 >= -1 && y6 <=
y5 <= 1 + y4 + y6,> y4 == -1 && y6 >= -1 && y5 == y6 || y4 >
-1 && y6 >= -1 && y6 <= y5> <= 1 + y4 + y6]> Out[9]=> y4 >=
-1 && y6 <= y5 <= 1 + y4 + y6> On the other hand, taking just
the individual summands of Or as> hypotheses;> In[10]:=>
FullSimplify[y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 +
y6,> y4 > -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6]>
Out[10]=> True> In[11]:=> FullSimplify[y4 >= -1 && y6 >= -1
&& y6 <= y5 <= 1 + y4 + y6,> y4 == -1 && y6 >= -1 && y5 == y6
]> Out[11]=> True> In fact FullSimplify is unable to use Or
in assumptions, which can be> demonstrated on an abstract
example:> In[12]:=> FullSimplify[C,(A||B)&&(C)]> Out[12]=>
True> In[13]:=> FullSimplify[C,LogicalExpand[(A||B)&&(C)]]>
Out[13]=> C> This could be fixed by modifying FullSimplify:>
In[14]:=> Unprotect[FullSimplify];> In[14]:=>
FullSimplify[expr_,Or[x_,y__]]:=Or[FullSimplify[expr,x],
FullSimplify[e> xpr,y]];> In[15]:=> Protect[FullSimplify];>
Now at least we get as before:> In[16]:=> FullSimplify[y4 ==
-1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1> &&> y6 <=
y5 <= 1 + y4 + y6, y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 +>
y4 + y6]> Out[16]=> True> but also:> In[17]:=>
FullSimplify[y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 +
y6,> y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1
&& y6 <= y5> <= 1 + y4 + y6]> Out[17]=> True> This seems to
me a possible worthwhile improvement in FullSimplify,> though
of course not really helpful for your problem.> Andrzej
Kozlowski> Toyama International University> JAPAN> I
have a set of inequalities that I solve with InequalitySolve.
But> then> it gives a complete set of solutions, but not in
the way I would> like it> to be! :-) For example, the
simple following calculation will give:> In[1]:= ineq = {y4
>= -1, y5 >= -1, y6 + y4 >= y5 - 1, y5 >= y6, y6>= -1};>
InequalitySolve[ineq,{y4,y6,y5}]> Out[1]:= y4 == -1 && y6 >=
-1 && y5 == y6 ||> y4 > -1 && y6 >= -1 && y6 <= y5 <= 1 + y4
+ y6> the result is good, but I would like it to be in the
simpler but> equivalent form> y4 >= -1 && y6 >= -1 && y6 <=
y5 <= 1 + y4 + y6> How can I tell InequalitySolve to do that?
It is simple for this> example,> but for a large set of
simple inequalities InequalitySolve gives> lines and> lines
of results instead of a simple result.> Vincent Bouchard>
>
====
On second thoughts I realized that there seems to be an
inherent ambiguity about what one could mean by using
alternatives (statements joned by Or) assumptions. In fact
it now seems to me that the reasonable intertpretation for
ImpliesQ and FullSimplify ought to perhaps be different. It
seems to me that ImpliesQ[Or[a,b],c] ought to return True if
aand only if ImpliesQ[a,c] and ImpliesQ[b,c] both return
True. If so this could be acomplished by adding the
ruleImpliesQ[Or[a,b],c] = And[ImpliesQ[a,c],ImpliesQ[b,c]].
That could then be used in proving that the two answers to
the system of inequalities that of Vincent's original
posting are equivalent. On the other hand probably
FullSimplify[a, Or[p,q]] ought to return
Or[FullSimplify[a,p],FullSimplify[a,q]] (or do nothing as it
doe snow). The first approach would seem to be consistent with
the way FullSimplify works with domain specifications but
would however have the strange effect of returning True if
just one of the alternatives were true and the other false.
So perhaps after all it is best to leave FullSimplify as it
is. However, it seems to me that ImpliesQ shoud be able to
handle such cases (?)Andrzej KozlowskiToyama International
UniversityJAPAN> The modification to FullSimplify that I
sent earlier works correctly > only for assumptions of the
form Or[a,b] (and even then not is not > always what one
would like). For what it's worth here is a better (but >
slow) version:> In[1]:=> Unprotect[FullSimplify];> In[2]:=>
FullSimplify[expr_, x_ || y__] := FullSimplify[>
FullSimplify[expr, x] || FullSimplify[expr, Or[y]]];>
In[3]:=> Protect[FullSimplify];> For example:> In[4]:=>
FullSimplify[Sqrt[(x - 1)^2] + Sqrt[(x - 2)^2] +> Sqrt[(x -
3)^2], x > 1 || x > 2 || x > 3]> Out[4]=> -1 + x + Abs[-3 +
x] + Abs[-2 + x] ||> -3 + 2*x + Abs[-3 + x] || 3*(-2 + x)>
Andrzej Kozlowski> Toyama International University> JAPAN>
The reason why InequalitySolve returns it's answer in what
sometimes > turns out to be unnecessarily complicated form is
that the underlying > algorithm, Cylindrical Agebraic
Decomposition (CAD) returns its > answers in this form.
Unfortunately it seems to me unlikely that a > simplification
of the kind you need can be can be accomplished in any >
general way. To see why observe the following. First of all:
In[1]:=> FullSimplify[x > 0 || x == 0] Out[1]=> x >= 0 This
is fine. However: In[2]:=> FullSimplify[x > 0 && x < 2 || x ==
0 && x < 2] Out[2]=> x == 0 || 0 < x < 2 Of course what you
would like is simply 0 <= x < 2. One reason why > you can't
get it is that while Mathematica can perform a >
LogicalExpand, as in:> In[3]:=> LogicalExpand[(x > 0 || x
== 0) && x < 2] Out[3]=> x == 0 && x < 2 || x > 0 && x < 2
There i no LogicalFactor or anything similar that would
reverse > what LogicalExpand does. if there was then you
could perform the sort > of simplifications you need for:
In[4]:=> FullSimplify[(x > 0 || x == 0) && x < 2] Out[4]=> 0
<= x < 2 However, it does not seem to me very likely that
such logical > factoring can be performed by a general
enough algorithm (though I > am no expert in this field). In
any case, certainly Mathematica can't > do this. I also
noticed that Mathematica seems unable to show that the answer
> it returns to your problem is actually equivalent to your
simpler > one. In fact this looks like a possible bug in
Mathematica. Let's > first try the function 
ImpliesQ from the
Experimental context: << Experimental` Now Mathematica
correctly gives: In[6]:=> ImpliesQ[y4 >= -1 && y6 >= -1 && y6
<= y5 <= 1 + y4 + y6,> y4 == -1 && y6 >= -1 && y5 == y6 || y4
> -1 && y6 >= -1 && y6 <= y5 > <= 1 + y4 + y6] Out[6]=> True
However: In[7]:=> ImpliesQ[y4 == -1 && y6 >= -1 && y5 == y6 ||
y4 > -1 && y6 >= -1 &&> y6 <= y5 <= 1 + y4 + y6, y4 >= -1 &&
y6 >= -1 && y6 <= y5 <= 1 + > y4 + y6] Out[7]=> False That
simply means that ImpliesQ cannot show the implication, not
that > it does not hold. ImpliesQ relies on CAD, as does
FullSimplify. > Switching to FullSimplify we see that:
In[8]:=> FullSimplify[y4 == -1 && y6 >= -1 && y5 == y6 || y4
> -1 && y6 >= -1 > &&> y6 <= y5 <= 1 + y4 + y6, y4 >= -1 &&
y6 >= -1 && y6 <= y5 <= 1 + > y4 + y6] Out[8]=> True while
In[9]:=> FullSimplify[y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 +
y4 + y6,> y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >=
-1 && y6 <= y5 > <= 1 + y4 + y6] Out[9]=> y4 >= -1 && y6 <= y5
<= 1 + y4 + y6 On the other hand, taking just the individual
summands of Or as > hypotheses;> In[10]:=> FullSimplify[y4 >=
-1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6,> y4 > -1 && y6 >=
-1 && y6 <= y5 <= 1 + y4 + y6] Out[10]=> True In[11]:=>
FullSimplify[y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 +
y6,> y4 == -1 && y6 >= -1 && y5 == y6 ] Out[11]=> True In
fact FullSimplify is unable to use Or in assumptions, which
can be > demonstrated on an abstract example:> In[12]:=>
FullSimplify[C,(A||B)&&(C)] Out[12]=> True In[13]:=>
FullSimplify[C,LogicalExpand[(A||B)&&(C)]] Out[13]=> C This
could be fixed by modifying FullSimplify: In[14]:=>
Unprotect[FullSimplify]; In[14]:=>
FullSimplify[expr_,Or[x_,y__]]:=Or[FullSimplify[expr,x],
FullSimplify[e > xpr,y]]; In[15]:=> Protect[FullSimplify];
Now at least we get as before: In[16]:=> FullSimplify[y4 ==
-1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1 > &&> y6 <=
y5 <= 1 + y4 + y6, y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + >
y4 + y6] Out[16]=> True but also: In[17]:=> FullSimplify[y4
>= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6,> y4 == -1 && y6
>= -1 && y5 == y6 || y4 > -1 && y6 >= -1 && y6 <= y5 > <= 1 +
y4 + y6] Out[17]=> True This seems to me a possible
worthwhile improvement in FullSimplify, > though of course
not really helpful for your problem.> Andrzej Kozlowski>
Toyama International University> JAPAN I have a set of
inequalities that I solve with InequalitySolve. But > then>
it gives a complete set of solutions, but not in the way I
would > like it> to be! :-) For example, the simple
following calculation will give: In[1]:= ineq = {y4 >= -1, y5
>= -1, y6 + y4 >= y5 - 1, y5 >= y6, y6 >= -1};>
InequalitySolve[ineq,{y4,y6,y5}] Out[1]:= y4 == -1 && y6 >=
-1 && y5 == y6 ||> y4 > -1 && y6 >= -1 && y6 <= y5 <= 1 + y4
+ y6 the result is good, but I would like it to be in the
simpler but> equivalent form y4 >= -1 && y6 >= -1 && y6 <=
y5 <= 1 + y4 + y6 How can I tell InequalitySolve to do that?
It is simple for this > example,> but for a large set of
simple inequalities InequalitySolve gives > lines and> lines
of results instead of a simple result.> Vincent
Bouchard
====
Hallo!,My name is Nagesh and pursuing research
studies in Refrigeration. At present I am writing a Dynamic
Refrigeration System Simulation Package. I am using
Mathematica as a programming language for the same since last
one year. I don't have any programming experience before
this. I have following querries:-1. Is any body here have
expertise or information about the capability of Mathematica
as a system simulation tool?2. Is is possible to program a
user friendly interface for my system simulation package
with Mathematica or I have to use some other software?3. My
refrigeration system simulation package is likely to have
approximately 60 First order Differential equations. Is is
possible to solve these in Mathematica ? If yes then can
anybody here guide me about this further.I am explaining
below in short about the objectives I want to fulfill from
coding out of my main input file1. Example from Main Input
File ( this will contain about 200-250 variables which will
be entered by the user of this package)Below is examples of
two variables entered into this file, which will be used in
other analysis files for further evaluation.2. Example from
other analysis file ( there will be about 20-25 other such
component analysis files ) where the above mentioned
variables from main input file will be used for further
evaluations:-Below is one example from this file explaining
how the variables from main input file will be used in other
files.I hope that this short information will be useful for
guiding me to solve the following problems that I am facing.
I am facing follwing problems or objectives:-1. My 1st
Objective:- The user of this package must be able to change
only the value of the variable in the main input file but he
must not be able to change the name of the variable itself.
For example he must be able to change the value of the
variable   but he must not be able to change the name of
this variable itself.Here our problem is how to achieve or
program it so that our objective will be fullfilled.2. My
2nd Objective:- How I can program the main input file so that
it will be user friendly in terms of its visuals and
satisfying the constraint mentioned above in objective1.3.
My 3rd Objective:- How can I program the optional values for
each variable in the main input file ? so that there will be
always a value assigned to each variable listed in main
input file whenever the user opens up this file. 
If user want
to change the values of some variables then he can change
them and run the simulation otherwise the simulation run will
be done with optional values assigned to each variable in
the input file.4. My 4th Objective:- How can I program the
check for correctness of the input values supplied by the
package user ?I will be very greatful to you for helping and
suggesting some technique or method to solve some of my
above mentioned programming problems.Can anybody here guide
me about the above mentioned programming problems ?Nagesh
Rajepandhare
====
I want to add a further
suggestion:Probably each differential equation logically
belongs to one of the definedclasses in the sense, that it
helps to define the behaviour of thecomponent.In this case,
each differential equation should be stored in one of
thesesclasses as textstring in a form appropriate for the
execution inMathematica.Hermann Schmitt----- Original
Message -----> Is any body here have expertise or
information about the capability of> Mathematica as a system
simulation tool?> Mr. Kuska answers:> Since the most
system simulation tools are simply solving a system of>
ordinary differntial equations it is simple to do this with
NDSolve[].> My comment:> That is: He sees the simulation
system merely as a set of differential> equations.> The
question of Mr. Nagesh:> My 4th Objective:- How can I
program the check for correctness of the> input values
supplied by the package user ?> The answer of Mr. Kuska is:>
And @@ (NumericQ /@ {aListOfAllYourNumericParameters})> My
comment:> This is a nice command and shows the knowledge of
Mr.Kuska. But does Mr.> Nagesh understand it and is it
sufficient to check, if all inputs are> numerical?>
Additionally I think, it is not userfriendly to see the input
merely as a> set of 200-250 numbers.> My suggestion is, that
JLink is used, a suggestion Mr. Kusk takes into>
consideration, too.> But further I suggest, that classes are
defined in Java, which representthe> parts of the system.>
Constructors of the classes should build objects with default
values.> Graphical user interfaces> should give the
opportunity to change the data fields in the objects and>
check the input for correctness.> The system should give the
opportunity, to store the objects on harddisk>
(serialization).> accessed directly.> My name is Nagesh and
pursuing research studies in Refrigeration. At> present I am
writing a Dynamic Refrigeration System SimulationPackage.>
I> am using Mathematica as a programming language for the
same since last> one> year. I don't have any programming
experience before this. I have> following> querries:-> 1.
Is any body here have expertise or information about the
capability> of> Mathematica as a system simulation tool?>
Since the most system simulation tools are simply solving
a system of> ordinary differntial equations it is simple to
do this with NDSolve[].> 2. Is is possible to program a user
friendly interface for my system> simulation package with
Mathematica or I have to use some other> software?> Write a
MathLink or J/Link frontend that launch the kernel. But you>
should keep> in mind that the user interface is typical 80-90
% of your code.> If you just whant to solve some ode's it is
probably easyer to> include one of the excelent ode-solvers
from netlib in your C-code> than to call Mathematica to do
that. As long as you dont wish to change> the ode's very
often (than Mathematica is more ßexible) you should> not use
Mathematica.> 3. My refrigeration system simulation package is
likely to have> approximately 60 First order Differential
equations. Is is possible to> solve these in Mathematica ?>
Sure.> If yes then can anybody here guide me about> this
further.> Write down the equations and call NDSolve[].> I
am explaining below in short about the objectives I want to
fulfill> from> coding out of my main input file> 
1. Example
from Main Input File ( this will contain about 200-250>
variables> which will be entered by the user of this
package)> This sounds like a *very* userfiendly interface ;-)>
> Below is examples of two variables entered into this file,
which willbe> used in other analysis files for further
evaluation.> 2. Example from other analysis file ( there will
be about 20-25 other> such> component analysis files ) where
the above mentioned variables frommain> input file will be
used for further evaluations:-> Below is one example from
this file explaining how the variables from> main> input 
file
will be used in other files.> I hope that this short
information will be useful for guiding me to> solve> the
following problems that I am facing. I am facing follwing
problems> or> objectives:-> 1. My 1st Objective:- The user
of this package must be able to change> only> the value of
the variable in the main input file but he must not beable>
to> change the name of the variable itself. For example he
must be able to> change the value of the variable   but
he must not be able tochange> the> name of this variable
itself.> Here our problem is how to achieve or program it so
that our objective> will> be fullfilled.> Options with
defaulf values ? or something like>
{aParam,bParam}={ODEParameter1,ODEParameter2} /.> userRules
/.> {ODEParameter1->1,ODEParameter2->2}> 2. My 2nd
Objective:- How I can program the main input file so that
it> will> be user friendly in terms of its visuals and
satisfying the constraint> mentioned above in objective1.>
What is *userfiendly* in a file with 250 variables 
???> 3. My
3rd Objective:- How can I program the optional values for
each> variable in the main input file ? so that there will be
always a value> assigned to each variable listed in main
input file whenever the user> opens> up this file. 
If user
want to change the values of some variables then> he> can
change them and run the simulation otherwise the simulation
runwill> be> done with optional values assigned to each
variable in the input file.> See above.> 4. My 4th
Objective:- How can I program the check for correctness
ofthe> input values supplied by the package user ?> And @@
(NumericQ /@ {aListOfAllYourNumericParameters})>
Jens>
====
I have developed one package where it is included
a functions forsolving System of ODE. You can downloaded
from:http://web.usal.es/~guillerm/biokmod.The functions
are: SystemDSolve and SystemNDSolve Guillermo Sanchez> I
want to add a further suggestion:> Probably each differential
equation logically belongs to one of the defined> classes in
the sense, that it helps to define the behaviour of the>
component.> In this case, each differential equation should
be stored in one of theses> classes as textstring in a form
appropriate for the execution in> Mathematica.> Hermann
Schmitt> ----- Original Message -----> I agree with Mr.
Kuska, that the system Mr Nagesh describes is not>
userfriendly. But I think, the suggestions of Mr. Kuska do
not make it> more> userfriendly, rather the opposite is
true.> Mr. Nagesh asks> Is any body here have expertise or
information about the capability of> Mathematica as a system
simulation tool?> Mr. Kuska answers:> Since the most
system simulation tools are simply solving a system of>
ordinary differntial equations it is simple to do this with
NDSolve[].> My comment:> That is: He sees the simulation
system merely as a set of differential> equations.> The
question of Mr. Nagesh:> My 4th Objective:- How can I
program the check for correctness of the> input values
supplied by the package user ?> The answer of Mr. Kuska is:>
And @@ (NumericQ /@ {aListOfAllYourNumericParameters})> My
comment:> This is a nice command and shows the knowledge of
Mr.Kuska. But does Mr.> Nagesh understand it and is it
sufficient to check, if all inputs are> numerical?>
Additionally I think, it is not userfriendly to see the input
merely as a> set of 200-250 numbers.> My suggestion is,
that JLink is used, a suggestion Mr. Kusk takes into>
consideration, too.> But further I suggest, that classes are
defined in Java, which represent> the> parts of the
system.> Constructors of the classes should build objects
with default values.> Graphical user interfaces> should give
the opportunity to change the data fields in the objects and>
check the input for correctness.> The system should give the
opportunity, to store the objects on harddisk>
(serialization).> accessed directly.> My name is Nagesh
and pursuing research studies in Refrigeration. At> present
I am writing a Dynamic Refrigeration System Simulation>
Package.> I> am using Mathematica as a programming language
for the same since last> one> year. I don't have any
programming experience before this. I have> following>
querries:-> 1. Is any body here have expertise or information
about the capability> of> Mathematica as a system
simulation tool?> Since the most system simulation tools
are simply solving a system of> ordinary differntial
equations it is simple to do this with NDSolve[].> 2. Is is
possible to program a user friendly interface for my
system> simulation package with Mathematica or I have to
use some other> software?>> Write a MathLink or J/Link
frontend that launch the kernel. But you> should keep> in
mind that the user interface is typical 80-90 % of your
code.> If you just whant to solve some ode's it is probably
easyer to> include one of the excelent ode-solvers from
netlib in your C-code> than to call Mathematica to do that.
As long as you dont wish to change> the ode's very often
(than Mathematica is more ßexible) you should> not use
Mathematica.> 3. My refrigeration system simulation package
is likely to have> approximately 60 First order Differential
equations. Is is possible to> solve these in Mathematica ?>
Sure.> If yes then can anybody here guide me about> this
further.> Write down the equations and call NDSolve[].> I
am explaining below in short about the objectives I want to
fulfill> from> coding out of my main input file> 
1. Example
from Main Input File ( this will contain about 200-250>
variables> which will be entered by the user of this
package)> This sounds like a *very* userfiendly interface ;-)>
> Below is examples of two variables entered into this file,
which will> be> used in other analysis files for further
evaluation.> 2. Example from other analysis file ( there will
be about 20-25 other> such> component analysis files )
where the above mentioned variables from> main> input file
will be used for further evaluations:-> Below is one example
from this file explaining how the variables from> main>
input file will be used in other files.> I hope 
that this short
information will be useful for guiding me to> solve> the
following problems that I am facing. I am facing follwing
problems> or> objectives:-> 1. My 1st Objective:- The user
of this package must be able to change> only> the value of
the variable in the main input file but he must not be>
able> to> change the name of the variable itself. For
example he must be able to> change the value of the
variable   but he must not be able to> change> the> name
of this variable itself.> Here our problem is how to achieve
or program it so that our objective> will> be fullfilled.>
Options with defaulf values ? or something like>
{aParam,bParam}={ODEParameter1,ODEParameter2} /.> userRules
/.> {ODEParameter1->1,ODEParameter2->2}> 2. My 2nd
Objective:- How I can program the main input file so that
it> will> be user friendly in terms of its visuals and
satisfying the constraint> mentioned above in objective1.>
What is *userfiendly* in a file with 250 variables 
???> 3. My
3rd Objective:- How can I program the optional values for
each> variable in the main input file ? so that there will
be always a value> assigned to each variable listed in main
input file whenever the user> opens> up this file. 
If user
want to change the values of some variables then> he> can
change them and run the simulation otherwise the simulation
run> will> be> done with optional values assigned to each
variable in the input file.> See above.> 4. My 4th
Objective:- How can I program the check for correctness of>
the> input values supplied by the package user ?> And @@
(NumericQ /@ {aListOfAllYourNumericParameters})> Jens>
>
====
I agree with Mr. Kuska, that the system Mr Nagesh
describes is notuserfriendly. But I think, the suggestions
of Mr. Kuska do not make it moreuserfriendly, rather the
opposite is true.Mr. Nagesh asksIs any body here have
expertise or information about the capability ofMathematica
as a system simulation tool?Mr. Kuska answers:Since the
most system simulation tools are simply solving a system
ofordinary differntial equations it is simple to do this
with NDSolve[].My comment:That is: He sees the simulation
system merely as a set of differentialequations.The
question of Mr. Nagesh:My 4th Objective:- How can I program
the check for correctness of the input values supplied by the
package user ?The answer of Mr. Kuska is:And @@ (NumericQ
/@ {aListOfAllYourNumericParameters})My comment:This is a
nice command and shows the knowledge of Mr.Kuska. But does
Mr.Nagesh understand it and is it sufficient to check, if all
inputs arenumerical?Additionally I think, it is not
userfriendly to see the input merely as aset of 200-250
numbers.My suggestion is, that JLink is used, a suggestion
Mr. Kusk takes intoconsideration, too.But further I
suggest, that classes are defined in Java, which represent
theparts of the system.Constructors of the classes should
build objects with default values.Graphical user
interfacesshould give the opportunity to change the data
fields in the objects andcheck the input for
correctness.The system should give the opportunity, to store
the objects on harddisk(serialization).accessed directly.>
My name is Nagesh and pursuing research studies in
Refrigeration. At> present I am writing a Dynamic
Refrigeration System Simulation Package.I> am using
Mathematica as a programming language for the same since
lastone> year. I don't have any programming experience
before this. I havefollowing> querries:-> 1. Is any body
here have expertise or information about the capabilityof>
Mathematica as a system simulation tool?> Since the most
system simulation tools are simply solving a system of>
ordinary differntial equations it is simple to do this with
NDSolve[].> 2. Is is possible to program a user friendly
interface for my system> simulation package with Mathematica
or I have to use some othersoftware?> Write a MathLink or
J/Link frontend that launch the kernel. But you> should
keep> in mind that the user interface is typical 80-90 % of
your code.> If you just whant to solve some ode's it is
probably easyer to> include one of the excelent ode-solvers
from netlib in your C-code> than to call Mathematica to do
that. As long as you dont wish to change> the ode's very
often (than Mathematica is more ßexible) you should> not use
Mathematica.> 3. My refrigeration system simulation package is
likely to have> approximately 60 First order Differential
equations. Is is possible to> solve these in Mathematica ?>
Sure.> If yes then can anybody here guide me about> this
further.> Write down the equations and call NDSolve[].> I am
explaining below in short about the objectives I want to
fulfillfrom> coding out of my main input file> 1. 
Example
from Main Input File ( this will contain about
200-250variables> which will be entered by the user of this
package)> This sounds like a *very* userfiendly interface ;-)>
Below is examples of two variables entered into this file,
which will be> used in other analysis files for further
evaluation.> 2. Example from other analysis file ( there will
be about 20-25 othersuch> component analysis files ) where
the above mentioned variables from main> input file will be
used for further evaluations:-> Below is one example from
this file explaining how the variables frommain> input 
file
will be used in other files.> I hope that this short
information will be useful for guiding me tosolve> the
following problems that I am facing. I am facing follwing
problemsor> objectives:-> 1. My 1st Objective:- The user
of this package must be able to changeonly> the value of
the variable in the main input file but he must not be
ableto> change the name of the variable itself. For
example he must be able to> change the value of the variable
  but he must not be able to changethe> name of this
variable itself.> Here our problem is how to achieve or
program it so that our objectivewill> be fullfilled.>
Options with defaulf values ? or something like>
{aParam,bParam}={ODEParameter1,ODEParameter2} /.> userRules
/.> {ODEParameter1->1,ODEParameter2->2}> 2. My 2nd
Objective:- How I can program the main input file so that
itwill> be user friendly in terms of its visuals and
satisfying the constraint> mentioned above in objective1.>
What is *userfiendly* in a file with 250 variables 
???> 3. My
3rd Objective:- How can I program the optional values for
each> variable in the main input file ? so that there will be
always a value> assigned to each variable listed in main
input file whenever the useropens> up this file. 
If user
want to change the values of some variables thenhe> can
change them and run the simulation otherwise the simulation
run willbe> done with optional values assigned to each
variable in the input file.> See above.> 4. My 4th Objective:-
How can I program the check for correctness of the> input
values supplied by the package user ?> And @@ (NumericQ /@
{aListOfAllYourNumericParameters})>> Jens
====
See my
comments in the following text!Hermann Schmitt-----
Original Message -----> Is any body here have expertise or
information about the capability of> Mathematica as a
system simulation tool?> Mr. Kuska answers:> Since the
most system simulation tools are simply solving a system
of> ordinary differntial equations it is simple to do this
with NDSolve[].> My comment:> That is: He sees the
simulation system merely as a set of differential>
equations.> hmm, since the original poster write> My
refrigeration system simulation package is likely to have>
approximately 60 First order Differential equations.> it
seems not completly wrong to assume that the system consists
of> of ode's ..Yout should not ignore the word
merely.It is not enough to have a set 60 differential
equations and a set of200-250 numbers. That is not
simulation system, which can be used by userswith the
exception, perhaps, of the programmer of the system
himself.How does e.g. the user know what meaning a number in
the set has, ought hecount the numbers from the
beginning?Your nice command shows only, if there is an
input, which is not a number.But I think the user would like
to know, which of the 200 elements are notnumbers.The only
good of your command is, that it looks nice and shows
yourknowledge!> The question of Mr. Nagesh:> My 4th
Objective:- How can I program the check for correctness of
the> input values supplied by the package user ?> The
answer of Mr. Kuska is:> And @@ (NumericQ /@
{aListOfAllYourNumericParameters})> My comment:> This is a
nice command and shows the knowledge of Mr.Kuska. But does
Mr.> Nagesh understand it and is it sufficient to check, if
all inputs are> numerical?> It seems you have a deeper
knowlege about the things that Mr. Nagesh> understand. You
know him personally ? It is not very polite to> make
speculations *what* a other person understand.> And no it
is not sufficent to check that all parameters are numbers.>
Typical paramters described by intervals, where the
assumptions of> a model are valid. But for this checks one
needs more knowlege> about the meaning of the parameters.
And one needs the knowlege about> the differntial equations,
to find out the eigenvalues of the jacobian> ...> Additionally
I think, it is not userfriendly to see the input merely
asa> set of 200-250 numbers.> My suggestion is, that JLink
is used, a suggestion Mr. Kusk takes into> consideration,
too.> That will be fast as lightning !> But further I
suggest, that classes are defined in Java, which
representthe> parts of the system.> That is notable
nonsense! When the differntial equations should be> solved
with Mathematica, the parts can't be Java classes.
Mathematica's> NDSolve[] need a explicit expression to
integrate the equations.That's not nonsense, the Mathematica
program does not fetch the classes butthe numbers in the
classes (or better in the objects).the Java classes in
textform. They can then be fetched from the
Mathematicaprogram and transformed into expressions by the
Command ToExpression.The aim is, to have a clear
separation of the system into components, whichare
manageable and understandable.> OK you can call a Java class
member from Mathematica but this will> be incredible slow
when the right hand side is evaluated 200-300> times and
every evaluation make several callbacks to the Java source.>
Event handler of the Java main program (without some
modification in> the event loop) while it is evaluating an
other command.My idea is to fetch the values once from the
Java objects at the beginning.200-250 numbers is not so
much..> Constructors of the classes should build objects with
default values.> That's a great idea. If a simulation run
should be documented, one> always> need the full listing of
the Java source to find the actual parameter> settings.That
is not my opinion. I think not every user of the simulation
systemshould have to know Java and Mathematica.The user
must look for the values in the objects. And the values are
in theobjects, if they come from the constructor or from
the graphical userinterface.I think, there should be
listings of the objects including names of thevariables. In
the objects the values are in an meaningful environment.>
Graphical user interfaces> should give the opportunity to
change the data fields in the objects and> check the input
for correctness.> *and* what has a GUI for 200-250 variables
to do with Mathematica ?> BTW one has typical much less
variables because many parameters> are fixed and it make no
sense to change, for example, material> constants of
materials that can't exchanged> The system should give the
opportunity, to store the objects on harddisk>
(serialization).> accessed directly.>> Can you be so kind, to
explain *how* your posting help a person that> ask How can
I build a simulation system with Mathematica ?> You *can*
say Don't use Mathematica, use Java! but this has
nothing> to do with the question or with my reply.It is
you, who proposes to solve the problem with C/C++ and not to
useMathematica (see below!)My point of view is:Use
Mathematica, for what Mathemtica is good, and Java, for what
Java isgood.Mathematica is not so good as Java for data
entry and Java is better thanMathematica to represent the
simulation system (by objects).> But I still would suggest to
use C/C++ and a numerical> ode-solver, make a fancy
GUI/Script> interface and don't use Mathematica for such a
simple task.> The ode-solver is the smallest part of such a
simulation system.> Jens> My name is Nagesh and pursuing
research studies in Refrigeration. At> present I am writing
a Dynamic Refrigeration System SimulationPackage.> I> am
using Mathematica as a programming language for the same
sincelast> one> year. I don't have any programming
experience before this. I have> following> querries:-> 1.
Is any body here have expertise or information about
thecapability> of> Mathematica as a system simulation
tool?> Since the most system simulation tools are simply
solving a systemof> ordinary differntial equations it is
simple to do this with NDSolve[].> 2. Is is possible to
program a user friendly interface for my system> simulation
package with Mathematica or I have to use some other>
software?> Write a MathLink or J/Link frontend that launch
the kernel. But you> should keep> in mind that the user
interface is typical 80-90 % of your code.> If you just whant
to solve some ode's it is probably easyer to> include one of
the excelent ode-solvers from netlib in your C-code> than to
call Mathematica to do that. As long as you dont wish
tochange> the ode's very often (than Mathematica is more
ßexible) you should> not use Mathematica.> 3. My
refrigeration system simulation package is likely to have>
approximately 60 First order Differential equations. Is is
possibleto> solve these in Mathematica ?> Sure.> If yes
then can anybody here guide me about> this further.> Write
down the equations and call NDSolve[].> I am explaining
below in short about the objectives I want tofulfill>
from> coding out of my main input file> 1. Example from Main
Input File ( this will contain about 200-250> variables>
which will be entered by the user of this package)> This
sounds like a *very* userfiendly interface ;-)> Below is
examples of two variables entered into this file, whichwill
be> used in other analysis files for further evaluation.> 2.
Example from other analysis file ( there will be about
20-25other> such> component analysis files ) where the
above mentioned variables frommain> input file will be used
for further evaluations:-> Below is one example from this file
explaining how the variablesfrom> main> input file will be
used in other files.> I hope that this short information will
be useful for guiding me to> solve> the following problems
that I am facing. I am facing follwingproblems> or>
objectives:-> 1. My 1st Objective:- The user of this package
must be able tochange> only> the value of the variable in
the main input file but he must not beable> to> change the
name of the variable itself. For example he must be
ableto> change the value of the variable   but he must
not be able tochange> the> name of this variable itself.>
Here our problem is how to achieve or program it so that
ourobjective> will> be fullfilled.> Options with defaulf
values ? or something like>
{aParam,bParam}={ODEParameter1,ODEParameter2} /.> userRules
/.> {ODEParameter1->1,ODEParameter2->2}> 2. My 2nd
Objective:- How I can program the main input file so
thatit> will> be user friendly in terms of its visuals
and satisfying theconstraint> mentioned above in
objective1.> What is *userfiendly* in a file with 
250 variables
???> 3. My 3rd Objective:- How can I program the optional
values for each> variable in the main input file ? so that
there will be always avalue> assigned to each variable
listed in main input file whenever theuser> opens> up this
file. If user want to change the values of some
variablesthen> he> can change them and run the simulation
otherwise the simulation runwill> be> done with optional
values assigned to each variable in the inputfile.> See
above.> 4. My 4th Objective:- How can I program the check
for correctness ofthe> input values supplied by the
package user ?> And @@ (NumericQ /@
{aListOfAllYourNumericParameters})> Jens>Reply-To:
kuska@informatik.uni-leipzig.de
====
> I agree with Mr. Kuska,
that the system Mr Nagesh describes is not> userfriendly. But
I think, the suggestions of Mr. Kuska do not make it more>
userfriendly, rather the opposite is true.> Mr. Nagesh asks>
Is any body here have expertise or information about the
capability of> Mathematica as a system simulation tool?>
Mr. Kuska answers:> Since the most system simulation
tools are simply solving a system of> ordinary differntial
equations it is simple to do this with NDSolve[].> My
comment:> That is: He sees the simulation system merely as a
set of differential> equations.hmm, since the original
poster writeMy refrigeration system simulation package is
likely to have approximately 60 First order Differential
equations.it seems not completly wrong to assume that the
system consists ofof ode's ..> The question of Mr. Nagesh:>
My 4th Objective:- How can I program the check for
correctness of the> input values supplied by the package
user ?> The answer of Mr. Kuska is:> And @@ (NumericQ /@
{aListOfAllYourNumericParameters})> My comment:> This is a
nice command and shows the knowledge of Mr.Kuska. But does
Mr.> Nagesh understand it and is it sufficient to check, if
all inputs are> numerical?It seems you have a deeper
knowlege about the things that Mr. Nagesh understand. You
know him personally ? It is not very polite tomake
speculations *what* a other person understand.And no it
is not sufficent to check that all parameters are
numbers.Typical paramters described by intervals, where the
assumptions ofa model are valid. But for this checks one
needs more knowlege about the meaning of the parameters. And
one needs the knowlege aboutthe differntial equations, to
find out the eigenvalues of the jacobian...> Additionally I
think, it is not userfriendly to see the input merely as a>
set of 200-250 numbers.> My suggestion is, that JLink is
used, a suggestion Mr. Kusk takes into> consideration,
too.That will be fast as lightning ! > But further I
suggest, that classes are defined in Java, which represent
the> parts of the system.That is notable nonsense! When
the differntial equations should besolved with Mathematica,
the parts can't be Java classes. 
Mathematica'sNDSolve[] need
a explicit expression to integrate the equations.OK you can
call a Java class member from Mathematica but this willbe
incredible slow when the right hand side is evaluated
200-300times and every evaluation make several callbacks to
the Java source.Event handler of the Java main program
(without some modification inthe event loop) while it is
evaluating an other command.> Constructors of the classes
should build objects with default values.That's a great
idea. If a simulation run should be documented,
onealwaysneed the full listing of the Java source to find
the actual parametersettings. > Graphical user interfaces>
should give the opportunity to change the data fields in the
objects and> check the input for correctness.*and* what has
a GUI for 200-250 variables to do with Mathematica ?BTW one
has typical much less variables because many parametersare
fixed and it make no sense to change, for example,
materialconstants of materials that can't exchanged> The
system should give the opportunity, to store the objects on
harddisk> (serialization).> accessed directly.> Can you be
so kind, to explain *how* your posting help a person
thatask How can I build a simulation system with
Mathematica ?You *can* say Don't use Mathematica, use
Java! but this has nothingto do with the question or with
my reply. But I still would suggest to use C/C++ and a
numerical ode-solver, make a fancy GUI/Scriptinterface and
don't use Mathematica for such a simple task.The ode-solver
is the smallest part of such a simulation system. Jens> My
name is Nagesh and pursuing research studies in Refrigeration.
At> present I am writing a Dynamic Refrigeration System
Simulation Package.> I> am using Mathematica as a
programming language for the same since last> one> year. I
don't have any programming experience before this. I have>
following> querries:-> 1. Is any body here have expertise or
information about the capability> of> Mathematica as a
system simulation tool?> Since the most system simulation
tools are simply solving a system of> ordinary differntial
equations it is simple to do this with NDSolve[].> 2. Is is
possible to program a user friendly interface for my system>
simulation package with Mathematica or I have to use some
other> software?> Write a MathLink or J/Link frontend that
launch the kernel. But you> should keep> in mind that the
user interface is typical 80-90 % of your code.> If you just
whant to solve some ode's it is probably easyer to> include
one of the excelent ode-solvers from netlib in your C-code>
than to call Mathematica to do that. As long as you dont wish
to change> the ode's very often (than Mathematica is more
ßexible) you should> not use Mathematica.> 3. My
refrigeration system simulation package is likely to have>
approximately 60 First order Differential equations. Is is
possible to> solve these in Mathematica ?> Sure.> If yes
then can anybody here guide me about> this further.> Write
down the equations and call NDSolve[].> I am explaining
below in short about the objectives I want to fulfill> from>
coding out of my main input file> 1. Example from Main Input
File ( this will contain about 200-250> variables> which
will be entered by the user of this package)> This sounds
like a *very* userfiendly interface ;-)> Below is examples
of two variables entered into this file, which will be> used
in other analysis files for further evaluation.> 2. Example
from other analysis file ( there will be about 20-25 other>
such> component analysis files ) where the above mentioned
variables from main> input file will be used for further
evaluations:-> Below is one example from this file explaining
how the variables from> main> input file will be used in
other files.> I hope that this short information will be
useful for guiding me to> solve> the following problems
that I am facing. I am facing follwing problems> or>
objectives:-> 1. My 1st Objective:- The user of this package
must be able to change> only> the value of the variable in
the main input file but he must not be able> to> change the
name of the variable itself. For example he must be able to>
change the value of the variable   but he must not be able
to change> the> name of this variable itself.> Here our
problem is how to achieve or program it so that our
objective> will> be fullfilled.> Options with defaulf
values ? or something like>
{aParam,bParam}={ODEParameter1,ODEParameter2} /.> userRules
/.> {ODEParameter1->1,ODEParameter2->2}> 2. My 2nd
Objective:- How I can program the main input file so that
it> will> be user friendly in terms of its visuals and
satisfying the constraint> mentioned above in objective1.>
What is *userfiendly* in a file with 250 variables 
???> 3. My
3rd Objective:- How can I program the optional values for
each> variable in the main input file ? so that there will be
always a value> assigned to each variable listed in main
input file whenever the user> opens> up this file. 
If user
want to change the values of some variables then> he> can
change them and run the simulation otherwise the simulation
run will> be> done with optional values assigned to each
variable in the input file.> See above.> 4. My 4th
Objective:- How can I program the check for correctness of
the> input values supplied by the package user ?> And @@
(NumericQ /@ {aListOfAllYourNumericParameters})>
Jens>Reply-To: kuska@informatik.uni-leipzig.de
====
> My
name is Nagesh and pursuing research studies in
Refrigeration. At> present I am writing a Dynamic
Refrigeration System Simulation Package. I> am using
Mathematica as a programming language for the same since last
one> year. I don't have any programming experience before
this. I have following> querries:-> 1. Is any body here
have expertise or information about the capability of>
Mathematica as a system simulation tool?Since the most
system simulation tools are simply solving a system
ofordinary differntial equations it is simple to do this
with NDSolve[].> 2. Is is possible to program a user friendly
interface for my system> simulation package with Mathematica
or I have to use some other software?Write a MathLink or
J/Link frontend that launch the kernel. But youshould
keepin mind that the user interface is typical 80-90 % of
your code.If you just whant to solve some ode's it is
probably easyer toinclude one of the excelent ode-solvers
from netlib in your C-codethan to call Mathematica to do
that. As long as you dont wish to changethe ode's very
often (than Mathematica is more ßexible) you shouldnot use
Mathematica.> 3. My refrigeration system simulation package
is likely to have> approximately 60 First order Differential
equations. Is is possible to> solve these in Mathematica
?Sure. > If yes then can anybody here guide me about> this
further.Write down the equations and call NDSolve[].> I am
explaining below in short about the objectives I want to
fulfill from> coding out of my main input file> 1. 
Example
from Main Input File ( this will contain about 200-250
variables> which will be entered by the user of this
package)This sounds like a *very* userfiendly interface ;-)>
Below is examples of two variables entered into this file,
which will be> used in other analysis files for further
evaluation.> 2. Example from other analysis file ( there will
be about 20-25 other such> component analysis files ) where
the above mentioned variables from main> input file will be
used for further evaluations:-> Below is one example from
this file explaining how the variables from main> input 
file
will be used in other files.> I hope that this short
information will be useful for guiding me to solve> the
following problems that I am facing. I am facing follwing
problems or> objectives:-> 1. My 1st Objective:- The user of
this package must be able to change only> the value of the
variable in the main input file but he must not be able to>
change the name of the variable itself. For example he must
be able to> change the value of the variable   but he must
not be able to change the> name of this variable itself.>
Here our problem is how to achieve or program it so that our
objective will> be fullfilled.Options with defaulf values ?
or something
like{aParam,bParam}={ODEParameter1,ODEParameter2} /.
userRules /. {ODEParameter1->1,ODEParameter2->2}> 2. My 2nd
Objective:- How I can program the main input file so that it
will> be user friendly in terms of its visuals and
satisfying the constraint> mentioned above in
objective1.What is *userfiendly* in a file with 
250 variables
???> 3. My 3rd Objective:- How can I program the optional
values for each> variable in the main input file ? so that
there will be always a value> assigned to each variable
listed in main input file whenever the user opens> up this
file. If user want to change the values of some variables then
he> can change them and run the simulation otherwise the
simulation run will be> done with optional values assigned
to each variable in the input file.See above.> 4. My 4th
Objective:- How can I program the check for correctness of
the> input values supplied by the package user ?And @@
(NumericQ /@ {aListOfAllYourNumericParameters})>
Jens
====
You would need to make assumptions about y, and you
can't. The functionand the limits have to take care of that,
and when you try to do that,you end up with an expression
that has different antiderivatives ondifferent regions, for
different values of x. So, you have to break itup. The
assumption x > 0 implies that x is real, so Im[x]==0
isunnecessary.one = Integrate[Cosh[2 (y - x)] 2 y, {y,
Min[x, 1/2], 1/2}, Assumptions -> {x > 0}];two =
Integrate[Cosh[2 (x - y)] 2 y, {y, 0, Min[x, 1/2]},
Assumptions -> {x > 0}];one + two //
FullSimplifyPlot[{one, two, one + two}, {x, 0, 1/2}];
(1/4)*E^(-1 - 2*x)*(E + (-2 + E)*E^(4*x))Bobby
Treat-----Original Message----- there something implicit
that I am missing in the Assumptions ?MS.-- 12:02am up 5:06,
1 user, load average: 0.54, 0.22, 0.08Reply-To:
spammers-go-here@yahoo.com
====
On Friday 27 September 2002
04:18 am, DrBob, as drbob@bigfoot.com, held forth the
following in comp.soft-sys.math.mathematica
(<an14h9$ie2$1@smc.vnet.net>) :> You would need to make
assumptions about y, and you can't. The function> and the
limits have to take care of that, and when you try to do
that,> you end up with an expression that has different
antiderivatives on> different regions, for different values
of x. So, you have to break it> up. The assumption x > 0
implies that x is real, so Im[x]==0 is> unnecessary.> one =
Integrate[Cosh[2 (y - x)]> 2 y, {y, Min[x, 1/2], 1/2},
Assumptions -> {x > 0}];> two = Integrate[Cosh[2 (x - y)] 2>
y, {y, 0, Min[x, 1/2]}, Assumptions -> {x > 0}];> one + two
// FullSimplify> Plot[{one, two, one + two}, {x, 0, 1/2}];>
(1/4)*E^(-1 - 2*x)*(E + (-2 + E)*E^(4*x))> Bobby Treat>
-----Original Message-----> I have been trying to integrate
the following :> Integrate[Cosh[2 Abs[x-y]] 2 y, {y,0,1/2},>
Assumptions->{Im[x]==0,x>0}]> However, Mathematica chokes and
simply returns the integral as> it is. However, if I> split
up the integral into two portions, it quickly gives me an
answer> for the parts. Is> there something implicit that I
am missing in the Assumptions ?> MS.> ?Anyways, I will try
what you suggest.-- 12:53pm up 27 min, 1 user, load average:
0.19, 0.15, 0.09
====
Put this after the Plot statement, in
the same cell:SelectionMove[EvaluationNotebook[], All,
GeneratedCell]FrontEndTokenExecute[OpenCloseGroup]
FrontEndTokenExecute[SelectionAnimate]Bobby
Treat-----Original Message-----Do[Plot[Sin[t]*Sin[x], {x,
0, Pi}, PlotRange -> {{0, Pi}, {-1, 1}},ImageSize -> 400];
SelectionMove[EvaluationNotebook[], All,
GeneratedCell];FrontEndExecute[{FrontEnd`SelectionAnimate[
0.1]}]; FrontEndExecute[{FrontEndToken[Clear]}], {t, 0,
15, 0.1}]This works but the cell dividing line ßashes on and
off spoiling theanimation and if there is anything in the
cell below this jumps up anddown.Is there a proper way of
doing this?Hugh Goyder
====
>>So would it take about the same
amont of time for the completeprintoutof digits? Of course
it would take a few additional seconds to formatthe
output...I think it would take FAR more time for a complete
printout, and mightcrash the Front End. I was thinking
about the fact that I calculatedall those digits and then
threw them away. I could save them with Saveor DumpSave,
and read them in with Get the next time I wanted any
ofthem, although the file would be close to 70 MB (if not
more). I may dothat, in fact -- I have plenty of disk
space.The next step would be to somehow reuse the stored
digits if I wantedMORE digits. But how?The
Bailey-Borwein-Plouffe Pi algorithm is an avenue of attack,
since itcan calculate digits far from the decimal point,
without calculatingthose in between. Unfortunately, it
calculates hexadecimal digits inthat way, not decimal
digits. (That's true for the version I've 
seen,anyway.)
Still, I could take the stored digits, convert to
hexadecimal,add more hexadecimal digits with the B-B-P
algorithm, and then convertback to decimal. In both
conversions, I'd have to be very cognizant ofhow much
precision I end up with, but that shouldn't be too
difficult.It might go faster if I store hexadecimal digits,
as well as decimaldigits, to eliminate one of those
conversions at each increase in thenumber of digits.The
next step would be to set up an application that allowed
anyone toping for digits across the Internet, and would
return them if they'restored.Hasn't someone 
already done
that? It seems as if someone would have.Bobby
Treat-----Original Message-----> Could you tell me the CPU
you used and its speed etc...i am curious,to> other programs
out there.> I used one processor of a dual 1GH Mac and got the
same answer withthe> following speed:> 4.2 for Mac OS X
(June 4, 2002)> oldmax = $MaxPrecision> 6> 1. 10>
$MaxPrecision = Infinity> Infinity> With[{n = 
2^26}, Timing[>
pd = RealDigits[N[Pi, n + 1], 10, 20,> 19 - n]; ]]> {28794.1
Second, Null}> MaxMemoryUsed[]> 512055204> pd> {{3, 3, 8, 6,
3, 2, 2, 0, 8, 9, 6, 2, 2, 3,> 4, 0, 9, 8, 0, 3},
-67108844}> Tom Burton
====
I've been trying to get my
Mathematica 4.1 properly configured. I
set:#########################################################
##########/usr/local/mathematica/SystemFiles/FrontEnd/
TextResources/X/Specific.tr: @@resource maxForXListFonts
10000# xlsfonts | wc -l 5572/etc/X11/XF86Config: FontPath
/usr/X11R6/lib/X11/fonts/100dpi:unscaled FontPath
/usr/X11R6/lib/X11/fonts/75dpi:unscaled FontPath
/usr/X11R6/lib/X11/fonts/CID FontPath
/usr/X11R6/lib/X11/fonts/Speedo FontPath
/usr/X11R6/lib/X11/fonts/Type1 FontPath
/usr/X11R6/lib/X11/fonts/URW FontPath
/usr/X11R6/lib/X11/fonts/kwintv:unscaled FontPath
/usr/X11R6/lib/X11/fonts/latin2/Type1 FontPath
/usr/X11R6/lib/X11/fonts/local/mma/Type1 FontPath
/usr/X11R6/lib/X11/fonts/local/mma/X:unscaled FontPath
/usr/X11R6/lib/X11/fonts/misc:unscaled FontPath
/usr/X11R6/lib/X11/fonts/misc/sgi:unscaled FontPath
/usr/X11R6/lib/X11/fonts/truetype FontPath
/usr/X11R6/lib/X11/fonts/uni:unscaled# ls -R
/usr/X11R6/lib/X11/fonts/ | grep
//usr/X11R6/lib/X11/fonts/:/usr/X11R6/lib/X11/fonts/100dpi:
/usr/X11R6/lib/X11/fonts/75dpi:/usr/X11R6/lib/X11/fonts/CID:/
usr/X11R6/lib/X11/fonts/Speedo:/usr/X11R6/lib/X11/fonts/Type1
:/usr/X11R6/lib/X11/fonts/URW:/usr/X11R6/lib/X11/fonts/
encodings:/usr/X11R6/lib/X11/fonts/encodings/large:/usr/X11R6
/lib/X11/fonts/kwintv:/usr/X11R6/lib/X11/fonts/latin2:/usr/
X11R6/lib/X11/fonts/latin2/Type1:/usr/X11R6/lib/X11/fonts/
local:/usr/X11R6/lib/X11/fonts/local/mma:/usr/X11R6/lib/X11/
fonts/local/mma/Type1:/usr/X11R6/lib/X11/fonts/local/mma/X:/
usr/X11R6/lib/X11/fonts/misc:/usr/X11R6/lib/X11/fonts/misc/
sgi:/usr/X11R6/lib/X11/fonts/truetype:/usr/X11R6/lib/X11/
fonts/uni:/usr/X11R6/lib/X11/fonts/util:#####################
###################################When I open the
Mathematica Book Reference Guide in the Help Browser, I get
a beep and the message says:Unable to find font with
family Helvetica, weight Bold, slant Plain, and size 26.
Substituting Courier.Compared to the things which *were*
broken, this is a minor problem. I can live with the beep.
What I would now like to know is how to tell Mathematica 
what fonts to use by default. This seemingly simple question
seems to have no simple answer. Could someone please help
me.TIA,^LReply-To: kuska@informatik.uni-leipzig.de
====
I
expect that the Helvetica request comes from theFrontend
menues and this is lited I your X
resources$TopDirectory/SystemFiles/FontEnd/SystemResources/X
/XMathematicarequest the helvetica
font:!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
!!!!!!!!!!!!!!!!!!! GENERAL GUI STYLE / COLOR
SETTINGS!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
!!!!!!!!!!!!!!!!!!!!!!!! Choose your GUI widget background
color.! CDE: Comment this out.*background: #c0c0c0! Choose a
font for buttons/text fields, etc.! You can use xfontsel to
get a listing of fonts.!*Menu*fontList:
-*-helvetica-medium-r-*-*-*-100-*-*-*-*-*-*! Or just set all
the fonts at once.*fontList:
-*-helvetica-medium-r-*-*-*-100-*-*-*-*-*-* Jens> I've been
trying to get my Mathematica 4.1 properly configured.> I set:>
##############################################################
#####>
/usr/local/mathematica/SystemFiles/FrontEnd/TextResources/X/
Specific.tr:> @@resource maxForXListFonts> 10000> # xlsfonts
| wc -l> 5572> /etc/X11/XF86Config:> FontPath
/usr/X11R6/lib/X11/fonts/100dpi:unscaled> FontPath
/usr/X11R6/lib/X11/fonts/75dpi:unscaled> FontPath
/usr/X11R6/lib/X11/fonts/CID> FontPath
/usr/X11R6/lib/X11/fonts/Speedo> FontPath
/usr/X11R6/lib/X11/fonts/Type1> FontPath
/usr/X11R6/lib/X11/fonts/URW> FontPath
/usr/X11R6/lib/X11/fonts/kwintv:unscaled> FontPath
/usr/X11R6/lib/X11/fonts/latin2/Type1> FontPath
/usr/X11R6/lib/X11/fonts/local/mma/Type1> FontPath
/usr/X11R6/lib/X11/fonts/local/mma/X:unscaled> FontPath
/usr/X11R6/lib/X11/fonts/misc:unscaled> FontPath
/usr/X11R6/lib/X11/fonts/misc/sgi:unscaled> FontPath
/usr/X11R6/lib/X11/fonts/truetype> FontPath
/usr/X11R6/lib/X11/fonts/uni:unscaled> # ls -R
/usr/X11R6/lib/X11/fonts/ | grep />
/usr/X11R6/lib/X11/fonts/:> /usr/X11R6/lib/X11/fonts/100dpi:>
/usr/X11R6/lib/X11/fonts/75dpi:>
/usr/X11R6/lib/X11/fonts/CID:>
/usr/X11R6/lib/X11/fonts/Speedo:>
/usr/X11R6/lib/X11/fonts/Type1:>
/usr/X11R6/lib/X11/fonts/URW:>
/usr/X11R6/lib/X11/fonts/encodings:>
/usr/X11R6/lib/X11/fonts/encodings/large:>
/usr/X11R6/lib/X11/fonts/kwintv:>
/usr/X11R6/lib/X11/fonts/latin2:>
/usr/X11R6/lib/X11/fonts/latin2/Type1:>
/usr/X11R6/lib/X11/fonts/local:>
/usr/X11R6/lib/X11/fonts/local/mma:>
/usr/X11R6/lib/X11/fonts/local/mma/Type1:>
/usr/X11R6/lib/X11/fonts/local/mma/X:>
/usr/X11R6/lib/X11/fonts/misc:>
/usr/X11R6/lib/X11/fonts/misc/sgi:>
/usr/X11R6/lib/X11/fonts/truetype:>
/usr/X11R6/lib/X11/fonts/uni:>
/usr/X11R6/lib/X11/fonts/util:>
########################################################>
When I open the Mathematica Book Reference Guide in the Help
Browser, I get> a beep and the message says:> Unable to
find font with family Helvetica, weight Bold, slant Plain,
and> size 26. Substituting Courier.> Compared to the
things which *were* broken, this is a minor problem. I can>
live with the beep. What I would now like to know is how to
tell Mathematica what> fonts to use by default. This
seemingly simple question seems to have no> simple answer.>
Could someone please help me.> TIA,> ^L
====
> I expect that
the Helvetica request comes from the> Frontend menues and
this is lited I your X resources>
$TopDirectory/SystemFiles/FontEnd/SystemResources/X/
XMathematica$PreferencesDirectory? This is where I expected
to find this kind of thing in the first place. The 
idea of
chaning user preferences in a system file bothers me.>
request the helvetica font:>
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
!!!!!!!!!!!!> !> ! GENERAL GUI STYLE / COLOR SETTINGS> !>
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
!!!!!!!!!!!!!!> ! Choose your GUI widget background color.> !
CDE: Comment this out.> *background: #c0c0c0> ! Choose a font
for buttons/text fields, etc.> ! You can use xfontsel to get a
listing of fonts.> !*Menu*fontList:
-*-helvetica-medium-r-*-*-*-100-*-*-*-*-*-*> ! Or just set
all the fonts at once.> *fontList:
-*-helvetica-medium-r-*-*-*-100-*-*-*-*-*-*I still don't
know why Mathematica would ship with the font defaults
arranged in such a way as to cause it to give me errors
when viewing the help files. It makes me think something is
still amiss. Were they expecting my system to have these
already installed? Should I have other fonts installed?
Where would I get them, if I wanted them?> Jens>
STH
====
I have spent far more time attempting to get
Mathematica configured than I have using it. 
I've found the
the product to be exceptionally difficult to As an example,
I spent several hours trying to figure out how to tell
Mathematica to understand the delete key in the way most
contemporary systems understand it. I wanted to avoid
modifying system configuration files such
as:/usr/local/mathematica/SystemFiles/FrontEnd/TextResources/
KeyEventTranslations.trI expected to be able to change
something in my own ~/.Mathematic directory, but I could
not figure out an obvious way to affect this 
modification. I
want to adjust the font size used in the widgets, but again,
I see no ovbious means of modifying these attributes. I
suspect it can be accomplished by modifying the
~/.Mathematica/4.1/FrontEnd/init.m. Perhaps to an
experienced Mathematica user, the syntax and semantics of
this file are obvious. They aren't to me.I also 
find the
overall look & feel of the interface to be archaic. I
understand that Mathematica 4.1 was written years ago, and
much of the desktops for Unix which exist today did not
exist when it was written. I should be receiving Mathematica
4.2 in a few days, so perhaps these concerns will prove
undounded. Something tells me that not a lot has changed in
this respect.What have others observed?Reply-To:
kuska@informatik.uni-leipzig.de
====
I can't understand your
criticism. Open a terminal windowand type >maththe delete
key work perfect, due to your terminal settings anda command
line is the most modern interface I can imagine.It still
comes with Mathematica 4.2 and is perfect as before. Jens> I
have spent far more time attempting to get Mathematica
configured than I> have using it. I've found the 
the product
to be exceptionally difficult to> As an example, I spent
several hours trying to figure out how to tell Mathematica
to> understand the delete key in the way most contemporary
systems understand> it. I wanted to avoid modifying system
configuration files such as:>
/usr/local/mathematica/SystemFiles/FrontEnd/TextResources/
KeyEventTranslations.tr> I expected to be able to change
something in my own ~/.Mathematic directory,> but I could
not figure out an obvious way to affect this 
modification. I>
want to adjust the font size used in the widgets, but again,
I see no> ovbious means of modifying these attributes. I
suspect it can be> accomplished by modifying the
~/.Mathematica/4.1/FrontEnd/init.m. Perhaps> to an
experienced Mathematica user, the syntax and semantics of
this file are> obvious. They aren't to me.> I 
also find the
overall look & feel of the interface to be archaic. I>
understand that Mathematica 4.1 was written years ago, and
much of the desktops for> Unix which exist today did not
exist when it was written. I should be> receiving
Mathematica 4.2 in a few days, so perhaps these concerns will
prove> undounded. Something tells me that not a lot has
changed in this respect.> What have others observed?
====
> I
can't understand your criticism. Open a terminal window> and
type>>math> the delete key work perfect, due to your
terminal settings and> a command line is the most modern
interface I can imagine.> It still comes with Mathematica 4.2
and is perfect as before.> Jens> If I could use bash to edit
the command line, recall the command history, and other sorts
of things that bash is good at, that would make the command
line attractive. There *are* certain advantages to the gui
side of things. For example the M-k code completion. That
can really speed things up with languages such as Java and
C++. I don't believe there is any fundamental reason such a
thing could not exist with the command line. IMHO, a well
conceived XEmacs lisp package could provide a wonderful
frontend. I played around with mma.el for a few minutes,
but it wasn't acting like a normal XEmacs package, so I
switched back to cursing the Mod1<->Ctrl inversion of the
Mathematica frontend.gui front end, get in the way of what
really matters. The last thing I Mathematica. What I want to
do is improve the overal experience for all of
us.STH
====
Dear friendsI have the following problem with
Legend and LogPlot and LogPlotPlot:
Needs[Graphics`Graphics`]Needs[Graphics`Legend`]{q1[t
_],q2[t_],q3[t_]}={0.1 Exp[-0.02 t], 0.2 Exp[-0.025 t], 0.4
Exp[-0.028 t]};(*With Plot legend works
fine*)Plot[{q1[t],q2[t],q3[t]}, {t, 0,
100},PlotStyle[Rule]{ {AbsoluteThickness[0.5],
AbsoluteDashing[{4,4}]}, AbsoluteThickness[1.5],
{AbsoluteThickness[2], AbsoluteDashing[{1,8}]}},
AxesLabel[Rule]{Y, X}, PlotLabel[Rule]Title,
PlotLegend[Rule]{1,3,5}, LegendPosition[Rule]
{0.5,0}](*However with LogPlot or LogLogPlot the legend
desappear*)LogLogPlot[{q1[t],q2[t],q3[t]}, {t, 0,
100},PlotStyle[Rule]{ {AbsoluteThickness[0.5],
AbsoluteDashing[{4,4}]}, AbsoluteThickness[1.5],
{AbsoluteThickness[2], AbsoluteDashing[{1,8}]}},
AxesLabel[Rule]{Y, X}, PlotLabel[Rule]Title,
PlotLegend[Rule]{1,3,5}, LegendPosition[Rule]
{0.5,0}]I have shown a particular case, but I has this
problem always with Legend and LogPlot and LogPlotPlot. I
will appreciate any
help.GuillermoSanchez------------------------------------
---------This message was sent using Endymion
MailMan.
====
IIRC, there is a way to get a list of all the
symbols defined in the currently running session. I 
can't
seem to find the reference to that command. Could somone
point me in the direction of documentation which will tell
me how to get information about the current
session?TIA,Reply-To:
kuska@informatik.uni-leipzig.de
====
Mathematica's PostScript
does not need a 3d graphics cardand will not use it.Only
the RealTime3D` package make use of the 3d graphics and
it'smemory. Since it make an off screen rendering it will
be slow and since one must be mad to use it for 10^5
polygons it does notmatter how much graphics memory you
have. The tiny polygon countsof Plot3D[] & friends can be
handled wit 8 MByte :-)MathGL3d will run fine (not fast) with
8 MByte. For the mostMathGL3d applications 64 MByte
sufficient.With more than 10^6 polygons or many huge (1024^2)
textures you mayrun into problems. For the MathGL3d
development I have GeForce 3/4 cardswith 128 MByte RAM.BTW
the RAM of the 3d graphics cards is for textures and
Mathematicadoes not know what a texture is or how usefull
it is for scientificvisualization. Jens> We are about to
order new PCs for a university student lab in which>
Mathematica will be installed. Of course they will be using
2D and 3D> graphics -- plots of surfaces, e.g. Sooner or
later students will want> to rotate such plots, too.> An
unresolved issue is how much graphics RAM to get. On
existing> machines we typically have 64MB. But for the PCs
we are looking at,> manufacturer's limits on graphics RAM,
rather than cost, seems to limit> us to 32 MB.> Is 32 MB
adequate not just now, but likely to be adequate as well
for> the near future (say, a 3- to 5-year equipment
lifetime)?> --> Murray Eisenberg murray@math.umass.edu>
Mathematics & Statistics Dept.> Lederle Graduate Research
Tower phone 413 549-1020 (H)> University of Massachusetts 413
545-2859 (W)> 710 North Pleasant Street> Amherst, MA
01375Reply-To:
kuska@informatik.uni-leipzig.de
====
Names[Global`*]??
Jens> IIRC, there is a way to get a list of all the symbols
defined in the> currently running session. I 
can't seem to
find the reference to that> command. Could somone point me in
the direction of documentation which> will tell me how to get
information about the current session?> TIA,
====
To get a
complete list of all variables that have been defined for
thecurrent Mathematica session, please try:?`*which should
return a list of all variables that have been defined for
thecurrent Mathematica session.Steven Shippee
====
> IIRC,
there is a way to get a list of all the symbols defined in
the> currently running session. I can't seem to 
find the
reference to that> command. Could somone point me in the
direction of documentation which> will tell me how to get
information about the current session?> TIA,Let's begin with
a fresh sessionQuitMake some entries, notice that b, x and
y have no definitions - they aresimply created. a=3; b; p= 3x
+1; f[y_]:=y^2;We can find all the symbols we have created so
far. Names[`*] {a,b,f,p,x,y}Actually, they are the strings
of the symbols (otherwise, for example, awould immediately
evaluate to 3). InputForm[%] {a, b, f, p, x,
y}How can we take out the strings of the undefined
symbols?I make a function that test if the symbol has been
defined (or has anattribute assigned):
SetAttributes[definedQ, HoldFirst]; 
definedQ[x_String]:=
Or[DownValues@@#=!={}, UpValues@@#=!={},OwnValues@@#=!={},
SubValues@@#=!={},DefaultValues@@#=!={},NValues@@#=!={},
Attributes@@#=!={}]&[ ToExpression[x, InputForm, Hold]]
definedQ[x_]:=
definedQ[Evaluate[ToString[Unevaluated[x]]]]Using this we
get Select[Names[`*], definedQ] {a,definedQ,f,p}To 
get
information about the
symbolsInformation/@Select[Names[`*], definedQ]; a a = 3
definedQ Attributes[definedQ] = {HoldFirst} 
definedQ[x_String]
:= (DownValues @@ #1 =!= {} || UpValues @@ #1 =!= {} ||
OwnValues @@ #1 =!= {} || SubValues @@ #1 =!= {} ||
DefaultValues @@ #1 =!= {} || NValues @@ #1 =!= {} &
)[ToExpression[x, InputForm, Hold]] definedQ[x_] :=
definedQ[Evaluate[ ToString[Unevaluated[x]]]] f f[y_] := y^2
p p = 1 + 3*xI have assumed that you are interested only
in the symbols in the symbols inthe default context
Global`.Other context can be provided
for.--Allan---------------------Allan HayesMathematica
Training and ConsultingLeicester
UKwww.haystack.demon.co.ukhay@haystack.demon.co.ukVoice
: +44 (0)116 271 4198
====
There are several ways (note that
newly defined things are by default inGlobal` context) , one
would be ?@bye,Borut| IIRC, there is a way to get a list of
all the symbols defined in the| currently running session. I
can't seem to find the reference to that| 
command. Could
somone point me in the direction of documentation which|
will tell me how to get information about the current
session?||| TIA,|
====
Dear all,As far as I understand,
FindMinimum can be requested to use
theBerndt-Hall-Hall-Hausman method, if the option
Method->QuasiNewton is specified. I am pretty convinced that
there is no way to tellmathematica to produce as an output
the Jacobian vector and the Hessianmatrix at every step. In
addition, I have not found a way to specify
theDavidon-Fletcher-Powell algorithm to be used.Before I
start to program it myself, I was wondering if anybody has
writtensome [non-commercial] code that produces such an
output for BFGS and evenbetter for DFP
optimization.Kyriakos
_____+**+____+**+___+**+__+**+_Kyriakos
ChourdakisLecturer in Financial EconomicsURL:
http://www.qmw.ac.uk/~te9001tel: (++44) (+20) 7882 5086Dept
of EconomicsUniversity of London, QMLondon E1
4NSU.K.
====
Both Simplify and ImpliesQ need the assumption
that all variablesare real in order to use the CAD
algorithm. However, Simplifyuses many other methods trying
to simplify each subexpressionof the input, while ImpliesQ
only tries to prove that the wholeinput is implied by the
assumptions.Here is the series of simplifications that
Simplify does in thisexample.- An inequality simplification
heuristic using the inequality assumption y6 >= -1 is
applied to the whole system. We getsimplification:y4 >= -1
&& y6 >= -1 && y6 <= y5 <= 1 + y4 + y6 -> 1 + y4 >= 0 && y5
>= y6 && 1 + y4 + y6 >= y5- We go into simplification of
subexpressions, and get thesesimplifications using reduction
modulo a GroebnerBasis ofequation assumptions y4 == -1 and
y5 == y6 (Simplify keepsthe result of this transformation
even if it has the samecomplexity as the original
expression.)1 + y4 -> 0y5 -> y61 + y4 + y6 -> y6y5 -> y6-
As the result of subexpression simplifications the three
inequalities become then:1 + y4 >= 0 -> 0 >= 0y5 >= y6 ->
y6 >= y61 + y4 + y6 >= y5 -> y6 >= y6- 0 >= 0 evaluates to
True, y6 >= y6 is simplified to True using transformation x
>= y -> x - y >= 0.Adam StrzebonskiWolfram Research> That
makes everything clear, except for just one small mystery:>
In[1]:=> << Experimental`> In[2]:=> FullSimplify[y4 >= -1
&& y6 >= -1 &&> y6 <= y5 <= 1 + y4 + y6, y4 == -1 && y6 >= -1
&& y5 == y6]> Out[2]=> True> In[3]:=> ImpliesQ[y4 == -1 && y6
>= -1 && y5 == y6,> y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4
+ y6]> Out[3]=> False> I now understand why the last one
returns False, but why does the> second one return True?
Should not the same arument apply in both> cases? Or is it
because FullSimplify does not actually need the> assumption
that the variables are real while ImliesQ does?> Andrzej>
Actually, the reason why ImpliesQ (and FullSimplify) fail
to> prove the implication is not that the hypothesis is a
disjunction.> To use the cylindrical algebraic decomposition
algorithm they> need to know that the assumptions imply that
all variables are> real.> The assumptions mechanism infers
variable domains in a purely> syntactical way, i.e. v is
assumed to be real if there is> an Element[v, Reals]
statement or v appears in an inequality.> It does not attempt
to analyze assumptions further, to figure> out that, say y6 >=
-1 implies that y6 is real, and then if> we have y5 == y6
then y5 must be real too. Doing such an analysis> in general
would require solving the assumptions over complex> numbers,
and then finding out which variables need to be real.> This
would be in general too time consuming to do, but analyzing>
linear dependencies like the ones in your example is a
possible> future improvement.> ImpliesQ cannot prove the
implication here, because it knows only> that y6 is real.>
In[1]:= <<Experimental`> In[2]:= ImpliesQ[y4 == -1 && y6 >=
-1 && y5 == y6,> y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 +
y6]> Out[2]= False> If we add an explicit assumption that y4
and y5 are real, ImpliesQ> (and FullSimplify) can prove this
implication, and the full version> of your example.> In[3]:=
ImpliesQ[Element[y4|y5, Reals] && y4 == -1 && y6 >= -1 && y5
==> y6,> y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6]>
Out[3]= True> In[4]:= ImpliesQ[Element[y4|y5, Reals] && (y4
== -1 && y6 >= -1 &&> y5 == y6 || y4 > -1 && y6 >= -1 && y6
<= y5 <= 1 + y4 + y6),> y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1
+ y4 + y6]> Out[4]=> True> In[5]:= FullSimplify[y4 >= -1 &&
y6 >= -1 && y6 <= y5 <= 1 + y4 + y6,> Element[y4|y5, Reals]
&& (y4 == -1 && y6 >= -1 && y5 == y6 ||> y4 > -1 && y6 >= -1
&& y6 <= y5 <= 1 + y4 + y6)]> Out[5]=> True> Adam
Strzebonski> Wolfram Research> On second thoughts I
realized that there seems to be an inherent> ambiguity about
what one coudl mean by using alternatives (statements> joned
by Or) assumptions. In fact it now seems to me that the>
reasonable intertpretation for ImpliesQ and FullSimplify
ought to> perhaps be different. It seems to me that
ImpliesQ[Or[a,b],c] ought to> return True if aand only if
ImpliesQ[a,c] and ImpliesQ[b,c] both> return> True. If so
this could be acomplished by adding the rule>
ImpliesQ[Or[a,b],c] = And[ImpliesQ[a,c],ImpliesQ[b,c]]. That
could> then> be used in proving that the two answers to the
system of inequalities> that of Vincent's original posting
are equivalent. On the other hand> probably FullSimplify[a,
Or[p,q]] ought to return>
Or[FullSimplify[a,p],FullSimplify[a,q]] (or do nothing as it
doe> snow).> The first approach would seem to be consistent
with the way> FullSimplify works with domain specifications
but would however have> the strange effect of returning True
if just one of the alternatives> were true and the other
false. So perhaps after all it is best to> leave
FullSimplify as it is. However, it seems to me that
ImpliesQ> shoud be able to handle such cases (?)> Andrzej
Kozlowski> Toyama International University> JAPAN> The
modification to FullSimplify that I sent earlier works
correctly> only for assumptions of the form Or[a,b] (and
even then not is not> always what one would like). For what
it's worth here is a better> (but> slow) version:> In[1]:=>
Unprotect[FullSimplify];> In[2]:=> FullSimplify[expr_, x_ ||
y__] := FullSimplify[> FullSimplify[expr, x] ||
FullSimplify[expr, Or[y]]];> In[3]:=> Protect[FullSimplify];>
For example:> In[4]:=> FullSimplify[Sqrt[(x - 1)^2] + Sqrt[(x
- 2)^2] +> Sqrt[(x - 3)^2], x > 1 || x > 2 || x > 3]>
Out[4]=> -1 + x + Abs[-3 + x] + Abs[-2 + x] ||> -3 + 2*x +
Abs[-3 + x] || 3*(-2 + x)> Andrzej Kozlowski> Toyama
International University> JAPAN> On Thursday, September
26, 2002, at 11:14 AM, Andrzej Kozlowski> The reason why
InequalitySolve returns it's answer in what sometimes> turns
out to be unnecessarily complicated form is that the>
underlying> algorithm, Cylindrical Agebraic Decomposition
(CAD) returns its> answers in this form. Unfortunately it
seems to me unlikely that a> simplification of the kind you
need can be can be accomplished in> any> general way. To
see why observe the following. First of all:> In[1]:=>
FullSimplify[x > 0 || x == 0]> Out[1]=> x >= 0> This is fine.
However:> In[2]:=> FullSimplify[x > 0 && x < 2 || x == 0 && x
< 2]> Out[2]=> x == 0 || 0 < x < 2> Of course what you would
like is simply 0 <= x < 2. One reason why> you can't get it
is that while Mathematica can perform a> LogicalExpand,
as in:> In[3]:=> LogicalExpand[(x > 0 || x == 0) && x < 2]>
Out[3]=> x == 0 && x < 2 || x > 0 && x < 2> There i no
LogicalFactor or anything similar that would reverse>
what LogicalExpand does. if there was then you could perform
the> sort> of simplifications you need for:> In[4]:=>
FullSimplify[(x > 0 || x == 0) && x < 2]> Out[4]=> 0 <= x <
2> However, it does not seem to me very likely that such
logical> factoring can be performed by a general enough
algorithm (though I> am no expert in this field). In any
case, certainly Mathematica> can't> do this.> I also
noticed that Mathematica seems unable to show that the
answer> it returns to your problem is actually equivalent to
your simpler> one. In fact this looks like a possible bug in
Mathematica. Let's> first try the function 
ImpliesQ from the
Experimental context:> << Experimental`> Now Mathematica
correctly gives:> In[6]:=> ImpliesQ[y4 >= -1 && y6 >= -1 &&
y6 <= y5 <= 1 + y4 + y6,> y4 == -1 && y6 >= -1 && y5 == y6 ||
y4 > -1 && y6 >= -1 && y6 <=> y5> <= 1 + y4 + y6]> Out[6]=>
True> However:> In[7]:=> ImpliesQ[y4 == -1 && y6 >= -1 && y5
== y6 || y4 > -1 && y6 >= -1 &&> y6 <= y5 <= 1 + y4 + y6, y4
>= -1 && y6 >= -1 && y6 <= y5 <= 1 +> y4 + y6]> Out[7]=>
False> That simply means that ImpliesQ cannot show the
implication, not> that> it does not hold. ImpliesQ relies
on CAD, as does FullSimplify.> Switching to FullSimplify we
see that:> In[8]:=> FullSimplify[y4 == -1 && y6 >= -1 && y5
== y6 || y4 > -1 && y6 >= -1> &&> y6 <= y5 <= 1 + y4 + y6, y4
>= -1 && y6 >= -1 && y6 <= y5 <= 1 +> y4 + y6]> Out[8]=>
True> while> In[9]:=> FullSimplify[y4 >= -1 && y6 >= -1 &&
y6 <= y5 <= 1 + y4 + y6,> y4 == -1 && y6 >= -1 && y5 == y6 ||
y4 > -1 && y6 >= -1 && y6 <=> y5> <= 1 + y4 + y6]> Out[9]=> y4
>= -1 && y6 <= y5 <= 1 + y4 + y6> On the other hand, taking
just the individual summands of Or as> hypotheses;>
In[10]:=> FullSimplify[y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1
+ y4 + y6,> y4 > -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6]>
Out[10]=> True> In[11]:=> FullSimplify[y4 >= -1 && y6 >= -1
&& y6 <= y5 <= 1 + y4 + y6,> y4 == -1 && y6 >= -1 && y5 == y6
]> Out[11]=> True> In fact FullSimplify is unable to use Or
in assumptions, which can> be> demonstrated on an abstract
example:> In[12]:=> FullSimplify[C,(A||B)&&(C)]> Out[12]=>
True> In[13]:=> FullSimplify[C,LogicalExpand[(A||B)&&(C)]]>
Out[13]=> C> This could be fixed by modifying FullSimplify:>
In[14]:=> Unprotect[FullSimplify];> In[14]:=>
FullSimplify[expr_,Or[x_,y__]]:=Or[FullSimplify[expr,x],
FullSimplify> [e> xpr,y]];> In[15]:=>
Protect[FullSimplify];> Now at least we get as before:>
In[16]:=> FullSimplify[y4 == -1 && y6 >= -1 && y5 == y6 || y4
> -1 && y6 >= -1> &&> y6 <= y5 <= 1 + y4 + y6, y4 >= -1 && y6
>= -1 && y6 <= y5 <= 1 +> y4 + y6]> Out[16]=> True> but
also:> In[17]:=> FullSimplify[y4 >= -1 && y6 >= -1 && y6 <=
y5 <= 1 + y4 + y6,> y4 == -1 && y6 >= -1 && y5 == y6 || y4 >
-1 && y6 >= -1 && y6 <=> y5> <= 1 + y4 + y6]> Out[17]=>
True> This seems to me a possible worthwhile improvement in
FullSimplify,> though of course not really helpful for your
problem.> Andrzej Kozlowski> Toyama International
University> JAPAN> On Wednesday, September 25, 2002, at
02:51 PM, Vincent Bouchard> I have a set of inequalities
that I solve with InequalitySolve. But> then> it gives a
complete set of solutions, but not in the way I would> like
it> to be! :-) For example, the simple following calculation
will give:> In[1]:= ineq = {y4 >= -1, y5 >= -1, y6 + y4 >= y5
- 1, y5 >= y6, y6> = -1};> InequalitySolve[ineq,{y4,y6,y5}]>
Out[1]:= y4 == -1 && y6 >= -1 && y5 == y6 ||> y4 > -1 && y6
>= -1 && y6 <= y5 <= 1 + y4 + y6> the result is good, but I
would like it to be in the simpler but> equivalent form> y4
>= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6> How can I tell
InequalitySolve to do that? It is simple for this> example,>
but for a large set of simple inequalities InequalitySolve
gives> lines and> lines of results instead of a simple
result.> Vincent Bouchard> Andrzej Kozlowski> Toyama
International University> JAPAN
====
That makes everything
clear, except for just one small mystery:In[1]:=<<
Experimental`In[2]:=FullSimplify[y4 >= -1 && y6 >= -1 &&
y6 <= y5 <= 1 + y4 + y6, y4 == -1 && y6 >= -1 && y5 ==
y6]Out[2]=TrueIn[3]:=ImpliesQ[y4 == -1 && y6 >= -1 && y5
== y6, y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 +
y6]Out[3]=FalseI now understand why the last one returns
False, but why does the second one return True? Should not
the same arument apply in both cases? Or is it because
FullSimplify does not actually need the assumption that the
variables are real while ImliesQ does?Andrzej> Actually,
the reason why ImpliesQ (and FullSimplify) fail to> prove
the implication is not that the hypothesis is a disjunction.>
To use the cylindrical algebraic decomposition algorithm
they> need to know that the assumptions imply that all
variables are> real.> The assumptions mechanism infers
variable domains in a purely> syntactical way, i.e. v is
assumed to be real if there is> an Element[v, Reals]
statement or v appears in an inequality.> It does not attempt
to analyze assumptions further, to figure> out that, say y6 >=
-1 implies that y6 is real, and then if> we have y5 == y6
then y5 must be real too. Doing such an analysis> in general
would require solving the assumptions over complex> numbers,
and then finding out which variables need to be real.> This
would be in general too time consuming to do, but analyzing>
linear dependencies like the ones in your example is a
possible> future improvement.> ImpliesQ cannot prove the
implication here, because it knows only> that y6 is real.>
In[1]:= <<Experimental`> In[2]:= ImpliesQ[y4 == -1 && y6 >=
-1 && y5 == y6,> y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 +
y6]> Out[2]= False> If we add an explicit assumption that y4
and y5 are real, ImpliesQ> (and FullSimplify) can prove this
implication, and the full version> of your example.> In[3]:=
ImpliesQ[Element[y4|y5, Reals] && y4 == -1 && y6 >= -1 && y5
==> y6,> y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6]>
Out[3]= True> In[4]:= ImpliesQ[Element[y4|y5, Reals] && (y4
== -1 && y6 >= -1 &&> y5 == y6 || y4 > -1 && y6 >= -1 && y6
<= y5 <= 1 + y4 + y6),> y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1
+ y4 + y6]> Out[4]=> True> In[5]:= FullSimplify[y4 >= -1 &&
y6 >= -1 && y6 <= y5 <= 1 + y4 + y6,> Element[y4|y5, Reals]
&& (y4 == -1 && y6 >= -1 && y5 == y6 ||> y4 > -1 && y6 >= -1
&& y6 <= y5 <= 1 + y4 + y6)]> Out[5]=> True> Adam
Strzebonski> Wolfram Research On second thoughts I realized
that there seems to be an inherent> ambiguity about what one
coudl mean by using alternatives (statements> joned by Or)
assumptions. In fact it now seems to me that the> reasonable
intertpretation for ImpliesQ and FullSimplify ought to>
perhaps be different. It seems to me that ImpliesQ[Or[a,b],c]
ought to> return True if aand only if ImpliesQ[a,c] and
ImpliesQ[b,c] both > return> True. If so this could be
acomplished by adding the rule> ImpliesQ[Or[a,b],c] =
And[ImpliesQ[a,c],ImpliesQ[b,c]]. That could > then> be used
in proving that the two answers to the system of
inequalities> that of Vincent's original posting are
equivalent. On the other hand> probably FullSimplify[a,
Or[p,q]] ought to return>
Or[FullSimplify[a,p],FullSimplify[a,q]] (or do nothing as it
doe > snow).> The first approach would seem to be consistent
with the way> FullSimplify works with domain specifications
but would however have> the strange effect of returning True
if just one of the alternatives> were true and the other
false. So perhaps after all it is best to> leave
FullSimplify as it is. However, it seems to me that
ImpliesQ> shoud be able to handle such cases (?) Andrzej
Kozlowski> Toyama International University> JAPAN> The
modification to FullSimplify that I sent earlier works
correctly> only for assumptions of the form Or[a,b] (and
even then not is not> always what one would like). For what
it's worth here is a better > (but> slow) version: In[1]:=>
Unprotect[FullSimplify]; In[2]:=> FullSimplify[expr_, x_ ||
y__] := FullSimplify[> FullSimplify[expr, x] ||
FullSimplify[expr, Or[y]]]; In[3]:=> Protect[FullSimplify];
For example: In[4]:=> FullSimplify[Sqrt[(x - 1)^2] + Sqrt[(x
- 2)^2] +> Sqrt[(x - 3)^2], x > 1 || x > 2 || x > 3] Out[4]=>
-1 + x + Abs[-3 + x] + Abs[-2 + x] ||> -3 + 2*x + Abs[-3 + x]
|| 3*(-2 + x) Andrzej Kozlowski> Toyama International
University> JAPAN On Thursday, September 26, 2002, at 11:14
AM, Andrzej Kozlowski The reason why InequalitySolve returns
it's answer in what sometimes> turns out to be unnecessarily
complicated form is that the > underlying> algorithm,
Cylindrical Agebraic Decomposition (CAD) returns its>
answers in this form. Unfortunately it seems to me unlikely
that a> simplification of the kind you need can be can be
accomplished in > any> general way. To see why observe the
following. First of all: In[1]:=> FullSimplify[x > 0 || x ==
0] Out[1]=> x >= 0 This is fine. However: In[2]:=>
FullSimplify[x > 0 && x < 2 || x == 0 && x < 2] Out[2]=> x ==
0 || 0 < x < 2 Of course what you would like is simply 0 <= x
< 2. One reason why> you can't get it is that while
Mathematica can perform a> LogicalExpand, as in:>
In[3]:=> LogicalExpand[(x > 0 || x == 0) && x < 2] Out[3]=> x
== 0 && x < 2 || x > 0 && x < 2 There i no LogicalFactor or
anything similar that would reverse> what LogicalExpand does.
if there was then you could perform the > sort> of
simplifications you need for: In[4]:=> FullSimplify[(x > 0 ||
x == 0) && x < 2] Out[4]=> 0 <= x < 2 However, it does not
seem to me very likely that such logical> factoring can
be performed by a general enough algorithm (though I> am no
expert in this field). In any case, certainly Mathematica >
can't> do this. I also noticed that Mathematica seems unable
to show that the answer> it returns to your problem is
actually equivalent to your simpler> one. In fact this looks
like a possible bug in Mathematica. Let's> first 
try the
function ImpliesQ from the Experimental context: <<
Experimental` Now Mathematica correctly gives: In[6]:=>
ImpliesQ[y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6,> y4
== -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1 && y6 <=
> y5> <= 1 + y4 + y6] Out[6]=> True However: In[7]:=>
ImpliesQ[y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >=
-1 &&> y6 <= y5 <= 1 + y4 + y6, y4 >= -1 && y6 >= -1 && y6 <=
y5 <= 1 +> y4 + y6] Out[7]=> False That simply means that
ImpliesQ cannot show the implication, not > that> it does
not hold. ImpliesQ relies on CAD, as does FullSimplify.>
Switching to FullSimplify we see that: In[8]:=>
FullSimplify[y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 &&
y6 >= -1> &&> y6 <= y5 <= 1 + y4 + y6, y4 >= -1 && y6 >= -1
&& y6 <= y5 <= 1 +> y4 + y6] Out[8]=> True while In[9]:=>
FullSimplify[y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 +
y6,> y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1
&& y6 <= > y5> <= 1 + y4 + y6] Out[9]=> y4 >= -1 && y6 <= y5
<= 1 + y4 + y6 On the other hand, taking just the individual
summands of Or as> hypotheses;> In[10]:=> FullSimplify[y4 >=
-1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6,> y4 > -1 && y6 >=
-1 && y6 <= y5 <= 1 + y4 + y6] Out[10]=> True In[11]:=>
FullSimplify[y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 +
y6,> y4 == -1 && y6 >= -1 && y5 == y6 ] Out[11]=> True In
fact FullSimplify is unable to use Or in assumptions, which
can > be> demonstrated on an abstract example:> In[12]:=>
FullSimplify[C,(A||B)&&(C)] Out[12]=> True In[13]:=>
FullSimplify[C,LogicalExpand[(A||B)&&(C)]] Out[13]=> C This
could be fixed by modifying FullSimplify: In[14]:=>
Unprotect[FullSimplify]; In[14]:=>
FullSimplify[expr_,Or[x_,y__]]:=Or[FullSimplify[expr,x],
FullSimplify > [e> xpr,y]]; In[15]:=> Protect[FullSimplify];
Now at least we get as before: In[16]:=> FullSimplify[y4 == -1
&& y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1> &&> y6 <= y5
<= 1 + y4 + y6, y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 +> y4 +
y6] Out[16]=> True but also: In[17]:=> FullSimplify[y4 >= -1
&& y6 >= -1 && y6 <= y5 <= 1 + y4 + y6,> y4 == -1 && y6 >= -1
&& y5 == y6 || y4 > -1 && y6 >= -1 && y6 <= > y5> <= 1 + y4 +
y6] Out[17]=> True This seems to me a possible worthwhile
improvement in FullSimplify,> though of course not really
helpful for your problem.> Andrzej Kozlowski> Toyama
International University> JAPAN> On Wednesday, September
25, 2002, at 02:51 PM, Vincent Bouchard I have a set of
inequalities that I solve with InequalitySolve. But> then>
it gives a complete set of solutions, but not in the way I
would> like it> to be! :-) For example, the simple
following calculation will give: In[1]:= ineq = {y4 >= -1, y5
>= -1, y6 + y4 >= y5 - 1, y5 >= y6, y6> = -1};>
InequalitySolve[ineq,{y4,y6,y5}] Out[1]:= y4 == -1 && y6 >=
-1 && y5 == y6 ||> y4 > -1 && y6 >= -1 && y6 <= y5 <= 1 + y4
+ y6 the result is good, but I would like it to be in the
simpler but> equivalent form y4 >= -1 && y6 >= -1 && y6 <=
y5 <= 1 + y4 + y6 How can I tell InequalitySolve to do that?
It is simple for this> example,> but for a large set of
simple inequalities InequalitySolve gives> lines and> lines
of results instead of a simple result.> Vincent
BouchardAndrzej KozlowskiToyama International
UniversityJAPAN
====
inside a program I need to solve this
linear equation in terms of p1.However something odds
happens. Sometimes the solution is computed andsometimes
the result is empty [I mean no output...]. Is this a bug of
thesolve command or am I doing something wrong? The problem
is robust to:changing name to the variables and other
makeups..Davidps: Sorry for the stupid way in which I
copied the command...Solve[(x^2*((-0.9*x^7*(p^2*(-1 -
5.8*x^6 - 14.010000000000002*x^12 - 18.04*x^18 - 13.06*x^24 -
5.040000000000001*x^30 - 0.81*x^36) + x*(7.777777777777779 -
9.074074074074076*x + 30.333333333333336*x^6 -
21.51851851851852*x^7 - 16.333333333333336*x^8 +
44.33333333333334*x^12 + 3.188888888888883*x^13 -
65.68333333333332*x^14 + 28.777777777777786*x^18 +
47.937037037037044*x^19 - 100.10000000000002*x^20 + 7.*x^24 +
45.6037037037037*x^25 - 69.53333333333333*x^26 +
13.299999999999999*x^31 - 19.833333333333332*x^32 -
1.0499999999999996*x^38) + p*(-6 + 8.296296296296296*x -
28.799999999999997*x^6 + 32.785185185185185*x^7 +
9.333333333333336*x^8 - 55.260000000000005*x^12 +
49.04777777777776*x^13 + 38.38333333333334*x^14 -
52.980000000000004*x^18 + 34.20518518518518*x^19 +
60.20000000000001*x^20 - 25.380000000000003*x^24 +
11.736296296296294*x^25 + 43.63333333333334*x^26 - 4.86*x^30
+ 2.8999999999999986*x^31 + 13.533333333333333*x^32 +
0.81*x^37 + 1.0499999999999996*x^38)))/(x + 1.9*x^7 +
0.9*x^13)^2 - ((-1 + p - 7*x^6 + p*x^6 +
6*x^7)*(1.2962962962962965 - 3.111111111111112*x^6 +
9.333333333333336*x^7 - 10.111111111111114*x^12 + 22.05*x^13
- 5.703703703703705*x^18 + 17.15*x^19 +
5.483333333333331*x^25 + 1.0499999999999996*x^31 +
p1*x^5*(7.000000000000002 - 7.000000000000002*x +
14.000000000000004*x^6 - 14.000000000000004*x^7 +
7.000000000000002*x^12 - 6.999999999999998*x^13) -
1.166666666666667*p*x^4*x1 - 3.500000000000001*p*x^10*x1 -
1.0500000000000003*p*x^11*x1 - 3.500000000000001*p*x^16*x1 -
3.150000000000001*p*x^17*x1 - 1.166666666666667*p*x^22*x1 -
3.150000000000001*p*x^23*x1 -
1.0500000000000003*p*x^29*x1))/((1 + 0.9*x^6)^2*(1 +
x^6)^2)))/(p^2*(1 + x^6)^3) == 0, p1]
====
You're right; I
misunderstood your problem. First of all, never use Do--
forget it exists -- and don't put the SelectionMove, etc.
commandsinside a loop. That's your biggest problem.This
should help:Table[Plot[ Sin[t]*Sin[x], {x, 0, Pi}, PlotRange
-> {{0, Pi}, {-1, 1}}], {t, 0, 2Pi - Pi/32,
Pi/32}];SelectionMove[EvaluationNotebook[], All,
GeneratedCell]FrontEndTokenExecute[OpenCloseGroup]
FrontEndTokenExecute[SelectionAnimate]The first time
through the animation is a bit slow because the framesare
being generated, but then the cell group collapses and things
arebetter. I used an increment that's an exact divisor of
the period inorder to catch the max and min values of
Sin[t], as well as the zerovalue, on each swing. Use the
period MINUS the step-size as the upperlimit in order to
avoid having a last frame identical to the 
first.I'm a bit
annoyed at the tendency for the plot to jump near zero
andPi, but that's because the derivative of Sin is higher
there, and we'renot compensating by picking more points
there. If we do pick morepoints there, however, we won't
perceive t as time. If that's not aconsideration, you could
do it this way:f = Which[ -1 ? # ? 1, #, 1 < # ? 3, 2 - #,
True, f@Mod[#, 4, -1] ] &;Table[Plot[ f[t]*Sin[x], {x, 0,
Pi}, PlotRange -> {{0, Pi}, {-1, 1}}], { t, -1, 2.9,
0.1}];SelectionMove[EvaluationNotebook[], All,
GeneratedCell]FrontEndTokenExecute[OpenCloseGroup]
FrontEndTokenExecute[SelectionAnimate]Bobby-----Original
Message-----Put this after the Plot statement, in the same
cell:SelectionMove[EvaluationNotebook[], All,
GeneratedCell]FrontEndTokenExecute[OpenCloseGroup]
FrontEndTokenExecute[SelectionAnimate]Bobby
Treat-----Original Message-----Is it possible to tidy up
the generation of the graphic so that itbecomesthe
animation?I have tried the followingDo[Plot[Sin[t]*Sin[x],
{x, 0, Pi}, PlotRange -> {{0, Pi}, {-1, 1}},ImageSize ->
400]; SelectionMove[EvaluationNotebook[], All,
GeneratedCell];FrontEndExecute[{FrontEnd`SelectionAnimate[
0.1]}]; FrontEndExecute[{FrontEndToken[Clear]}], {t, 0,
15, 0.1}]This works but the cell dividing line ßashes on and
off spoiling theanimation and if there is anything in the
cell below this jumps up anddown.Is there a proper way of
doing this?Hugh Goyder
====
That's nice because it avoids
watching the frames being slowly created,and the group
doesn't always collapse as it should, the way 
I've
beendoing it.However, a couple of small changes give a
smoother animation with 64frames rather than 151, while
GENERATING only 33 frames.Block[{$DisplayFunction =
Identity, half, graphs, step = Pi/32}, half =
Table[GraphicCell[Plot[Sin[t]*Sin[x], {x, 0, Pi}, PlotRange
-> {{0, Pi}, {-1, 1}}, ImageSize -> 400]], {t, -Pi/2, Pi/2 -
step, step}]; graphs = Rest@Join[half, Rest@Reverse@half];
NotebookWrite[EvaluationNotebook[],
Cell[CellGroupData[graphs,Closed]]];
SelectionMove[EvaluationNotebook[], All, GeneratedCell];
FrontEndExecute[{FrontEndToken[EvaluationNotebook[],
SelectionAnimate]}]]Bobby-----Original
Message----->showing during generation is enough (but
uncontrolled).>Hugh GoyderThis creates a graphics cell from
a graphics expression.GraphicCell[graphics_] :=
Cell[GraphicsData[PostScript,
DisplayString[graphics]],Graphics]cellgroup.Block[{$
DisplayFunction=Identity, graphs}, graphs =
Table[GraphicCell[ Plot[Sin[t]*Sin[x], {x, 0, Pi}, PlotRange
-> {{0, Pi}, {-1,1}}, ImageSize -> 400]], {t,0,15,.1}];
NotebookWrite[EvaluationNotebook[],Cell[CellGroupData[graphs
,Closed]]]; SelectionMove[EvaluationNotebook[], All,
GeneratedCell];
FrontEndExecute[{FrontEndToken[EvaluationNotebook[],
SelectionAnimate]}]
]------------------------------------------------------------
--Omega ConsultingThe final answer to your Mathematica
needsSpend less time searching and more time
finding.http://www.wz.com/internet/Mathematica.html
====
I'm
woking on a kind of a Mathematica cheat-sheet. So I don't
have to repeatthe same learning process if I get pulled
away for another 6 months.I've attempted to get my domain
name to resolve to my IP address, but it seems Verisign and
I have different ideas about what 24 hours is.The site is
supposed to be www.globalsymmetry.com, but that will not
currently resolve. Here's the IP and
path:http://66.92.149.152/proprietary/com/wri/index.html
This is not a literary masterpiece. It's probably proof that
giving just anybody the power to publish is, perhaps, not a
guaranty that more quality publication will take place.If
anybody has answers to the questions I've come up with, or
comments about the answeres, etc. I'd be happy to
know.
====
>>I believe the complexity is O(n log n), so this
should be good enough.Umm ...good enough? I understand
the words individually, but thephrase makes no sense to
me.Bobby Treat-----Original Message-----> crash the Front
End. I was thinking about the fact that I calculated> all
those digits and then threw them away. I could save them
withSave> or DumpSave, and read them in with Get the next
time I wanted any of> them, although the file would be close
to 70 MB (if not more). I maydo> that, in fact -- I have
plenty of disk space.> The next step would be to somehow
reuse the stored digits if I wanted> MORE digits. But how?>
The Bailey-Borwein-Plouffe Pi algorithm is an avenue of
attack, sinceit> can calculate digits far from the decimal
point, without calculating> those in between. Unfortunately,
it calculates hexadecimal digits in> that way, not decimal
digits. (That's true for the version I've 
seen,> anyway.)
Still, I could take the stored digits, convert
tohexadecimal,> add more hexadecimal digits with the B-B-P
algorithm, and then convert> back to decimal. In both
conversions, I'd have to be very cognizantof> how much
precision I end up with, but that shouldn't be too 
difficult.>
It might go faster if I store hexadecimal digits, as well as
decimal> digits, to eliminate one of those conversions at
each increase in the> number of digits.> The next step would
be to set up an application that allowed anyone to> ping for
digits across the Internet, and would return them if
they're> stored.> Hasn't someone already done 
that? It seems
as if someone would have.> Bobby TreatIf you're interested
in decimal digits, I don't think the BBP algorithmis
theway to go. In order to get the nth decimal digit of Pi
you need tocompute theprevious n-1 digits, since base
conversion is global, not local. ThealgorithmMathematica
uses for computing Pi is quite fast - I believe
thecomplexity is O(nlog n), so this should be good
enough.David> -----Original Message-----calculation in>
So would it take about the same amont of time for the
completeprintout> of digits? Of course it would take a few
additional seconds to format> the output...> Or does
Mathematica take alot less time when it truncates the
output?<tburton@brahea.com>> Could you tell me the CPU you
used and its speed etc...i amcurious,performance> to>
other programs out there.> I used one processor of a dual 1GH
Mac and got the same answer with> the> following speed:> 4.2
for Mac OS X (June 4, 2002)> oldmax = $MaxPrecision> 6> 1.
10> $MaxPrecision = Infinity> Infinity> With[{n = 
2^26},
Timing[> pd = RealDigits[N[Pi, n + 1], 10, 20,> 19 - n]; ]]>
{28794.1 Second, Null}> MaxMemoryUsed[]> 512055204> pd> {{3,
3, 8, 6, 3, 2, 2, 0, 8, 9, 6, 2, 2, 3,> 4, 0, 9, 8, 0, 3},
-67108844}> Tom Burton
====
Edit ->Preferences -> Font
OptionsIn Preferences you will find everything you need to
configure yourMathematica environment. Also you may want to
look up Style Sheets in thebook or the on line help.Yas>
I've been trying to get my Mathematica 4.1 properly
configured.> I set:>
#############################################################
######>
/usr/local/mathematica/SystemFiles/FrontEnd/TextResources/X/
Specific.tr:> @@resource maxForXListFonts> 10000> # xlsfonts
| wc -l> 5572> /etc/X11/XF86Config:> FontPath
/usr/X11R6/lib/X11/fonts/100dpi:unscaled> FontPath
/usr/X11R6/lib/X11/fonts/75dpi:unscaled> FontPath
/usr/X11R6/lib/X11/fonts/CID> FontPath
/usr/X11R6/lib/X11/fonts/Speedo> FontPath
/usr/X11R6/lib/X11/fonts/Type1> FontPath
/usr/X11R6/lib/X11/fonts/URW> FontPath
/usr/X11R6/lib/X11/fonts/kwintv:unscaled> FontPath
/usr/X11R6/lib/X11/fonts/latin2/Type1> FontPath
/usr/X11R6/lib/X11/fonts/local/mma/Type1> FontPath
/usr/X11R6/lib/X11/fonts/local/mma/X:unscaled> FontPath
/usr/X11R6/lib/X11/fonts/misc:unscaled> FontPath
/usr/X11R6/lib/X11/fonts/misc/sgi:unscaled> FontPath
/usr/X11R6/lib/X11/fonts/truetype> FontPath
/usr/X11R6/lib/X11/fonts/uni:unscaled> # ls -R
/usr/X11R6/lib/X11/fonts/ | grep />
/usr/X11R6/lib/X11/fonts/:> /usr/X11R6/lib/X11/fonts/100dpi:>
/usr/X11R6/lib/X11/fonts/75dpi:>
/usr/X11R6/lib/X11/fonts/CID:>
/usr/X11R6/lib/X11/fonts/Speedo:>
/usr/X11R6/lib/X11/fonts/Type1:>
/usr/X11R6/lib/X11/fonts/URW:>
/usr/X11R6/lib/X11/fonts/encodings:>
/usr/X11R6/lib/X11/fonts/encodings/large:>
/usr/X11R6/lib/X11/fonts/kwintv:>
/usr/X11R6/lib/X11/fonts/latin2:>
/usr/X11R6/lib/X11/fonts/latin2/Type1:>
/usr/X11R6/lib/X11/fonts/local:>
/usr/X11R6/lib/X11/fonts/local/mma:>
/usr/X11R6/lib/X11/fonts/local/mma/Type1:>
/usr/X11R6/lib/X11/fonts/local/mma/X:>
/usr/X11R6/lib/X11/fonts/misc:>
/usr/X11R6/lib/X11/fonts/misc/sgi:>
/usr/X11R6/lib/X11/fonts/truetype:>
/usr/X11R6/lib/X11/fonts/uni:>
/usr/X11R6/lib/X11/fonts/util:>
########################################################>
When I open the Mathematica Book Reference Guide in the Help
Browser, I get> a beep and the message says:> Unable to
find font with family Helvetica, weight Bold, slant Plain,
and> size 26. Substituting Courier.> Compared to the
things which *were* broken, this is a minor problem. I can>
live with the beep. What I would now like to know is how to
tell Mathematica what> fonts to use by default. This
seemingly simple question seems to have no> simple answer.>
Could someone please help me.> TIA,> ^L
====
> Edit
->Preferences -> Font Options> In Preferences you will find
everything you need to configure your> Mathematica
environment. Also you may want to look up Style Sheets in
the> book or the on line help.> Yas> I went into the
preferences browser, and it was not clear to me what I was
modifying. At one point I clicked on a field 
filled with text.
I had inteded to edit it, and all the text vanished. It
didn't bother me as much as such things use to, because I
believe I know a backout strategy. It's been a while since I
looked at this stuff, and I have to admit it seems far more
tractible than it did a year ago. I'll look at the 
discussion
again, and see if it makes more sence to me now.STH
====
> As
an example, I spent several hours trying to figure out how to
tell> Mathematica to understand the delete key in the way
most contemporary> systems understand it. I wanted to avoid
modifying system> configuration files such as:>
/usr/local/mathematica/SystemFiles/FrontEnd/TextResources/
KeyEventTranslations.tr> I expected to be able to change
something in my own ~/.Mathematic> directory, but I could
not figure out an obvious way to affect this> 
modification.If
you could post a precise description of what you expect the
Delete keyto do when depressed, we could probably provide
you with a clear cutanswer of what needs to be done.> I
want to adjust the font size used in the widgets, but again,
I see> no ovbious means of modifying these attributes. I
suspect it can be> accomplished by modifying the
~/.Mathematica/4.1/FrontEnd/init.m.> Perhaps to an
experienced Mathematica user, the syntax and semantics> of
this file are obvious. They aren't to me.The 
size of fonts in
user interface elements is not specified through
theMathemtica init.m file. It is set through an X resource.
If you are notfamiliar with resources, you may want to
track down an introductory texton the X Window System.
Information on application-specific resourcesettings can be
found in the Mathematica Getting Started
Guide:http://documents.wolfram.com/v4/GettingStarted/
TroubleshootingUnixX.htmlThe setting that you would need to
adjust is XMathematica*fontList. Thevalue of the resource is
an X Logical Font Description field.> I also find 
the overall
look & feel of the interface to be archaic.That's because
Mathematica relies on the Motif library for user
interfaceelements.http://www.opengroup.org/desktop/
motif.htmlThe appearance of these elements, such as the
menu and scroll bars, wouldbe the same for any other Motif
application, such as the DDD debugger orreleases of
Netscape prior to verison 4.-- User Interface Programmer
paulh@wolfram.comWolfram Research, Inc.
====
hi,> I sholdn't
have to. If I start messing with X resource settings for my>
user environment, I am sure to break something else which is
configured> based on the current settings. There should
either be a GUI interface, or a> clearly documented, and
easily accessible configuration file to modify 
such>
properties as the size of the fonts in the GUI widgets. This
is> functionality which is rightfully expected of a modern
desktop UI.[snip]> And I'm sure there is some 
configuration
file in which I could place that,> and hope that what you
think will be read by my system *will* in fact be> read, and
not subsequently overridden during xsession startup. Things>
aren't the way they used to be back in the 1980s. The modern
Unix desktop> has moved beyond the paradigm of openlook and
motif. See for example> http://www.trolltech.com,
http://www.gnome.org, and http://www.kde.orgmoving the
frontend over to QT would have some neat side effects:
consistent look & feel with the modern linux gui,
themeability, source code truetype fonts as QT supports
Xrender and Xft (looks great - see KDE3). i think all of
those points are of value, but the most important might be
source compatibility. ONE frontend for MOST (or ALL)
platforms - sounds like a dream :-))gerald --
*************************************Gerald RothM@th
Desktop
Development*************************************
====
> If
you could post a precise description of what you expect the
Delete key> to do when depressed, we could probably provide
you with a clear cut> answer of what needs to be
done.Item[KeyEvent[Delete], DeleteNext]'Most' 
means
Ômore than half.' > I want to adjust the font 
size used in
the widgets, but again, I see> no ovbious means of modifying
these attributes. I suspect it can be> accomplished by
modifying the ~/.Mathematica/4.1/FrontEnd/init.m.> Perhaps to
an experienced Mathematica user, the syntax and semantics> of
this file are obvious. They aren't to me.> The 
size of fonts in
user interface elements is not specified through the>
Mathemtica init.m file. It is set through an X resource. If
you are not> familiar with resources, you may want to track
down an introductory text> on the X Window System. I
sholdn't have to. If I start messing with X resource 
settings
for my user environment, I am sure to break something else
which is configured based on the current settings. There
should either be a GUI interface, or a clearly documented,
and easily accessible configuration file to modify 
such
properties as the size of the fonts in the GUI widgets. This
is functionality which is rightfully expected of a modern
desktop UI.> Information on application-specific resource>
settings can be found in the Mathematica Getting Started
Guide:>
http://documents.wolfram.com/v4/GettingStarted/
TroubleshootingUnixX.htmlIt should be in a clear and easy
to access configuraton interface, or at least be redily
available through the help system in such a way that
reasonable queries will locate it. Changing fonts does not
belong in a section on trouble shooting, unless this is an
acknowledgement that the UI is broken.> The setting that you
would need to adjust is XMathematica*fontList. The> value of
the resource is an X Logical Font Description field.And 
I'm
sure there is some configuration file in which I 
could place
that, and hope that what you think will be read by my system
*will* in fact be read, and not subsequently overridden
during xsession startup. Things aren't the way they used to
be back in the 1980s. The modern Unix desktop has moved
beyond the paradigm of openlook and motif. See for example
http://www.trolltech.com, http://www.gnome.org, and
http://www.kde.org > I also find the overall look & feel of
the interface to be archaic.> That's because Mathematica
relies on the Motif library for user interface> elements.>
http://www.opengroup.org/desktop/motif.html> The appearance
of these elements, such as the menu and scroll bars, would>
be the same for any other Motif application, such as the DDD
debugger or> releases of Netscape prior to verison 4.My
point exactly.
====
Awk! Legends!Basically, the answer to
your question is that the PlotLegend option worksONLY for
the Plot command and does not work for other types of plots.
Forother types of plots you have to use ShowLegend. And
ShowLegend is not allthat easy to use, especially since WRI
does not give an example for multiplecurves in the
Help.Needs[Graphics`Graphics`]Needs[Graphics`Legend`]
{q1[t_], q2[t_], q3[t_]} = {0.1 Exp[-0.02 t], 0.2 Exp[-0.025
t], 0.4 Exp[-0.028 t]};Let's look at your first
plot.Plot[{q1[t], q2[t], q3[t]}, {t, 0, 100}, PlotStyle ->
{{AbsoluteThickness[0.5], AbsoluteDashing[{4, 4}]},
AbsoluteThickness[1.5], {AbsoluteThickness[2],
AbsoluteDashing[{1, 8}]}}, AxesLabel -> {Y, X},
PlotLabel -> Title, PlotLegend -> {1, 3, 5},
LegendPosition -> {0.5, 0}, ImageSize -> 500];The legend is
almost as big as the plot. It distracts from the
realinformation you are trying to convey. Furthermore, the
order of the curvesin the legend is the reverse of their
order in the plot.The following shows how to put the legend
in a LogLogPlot, or other types ofplots. I defined the plot
styles independently because they are used inseveral places.
I made the legend much smaller and put it in an empty areaof
the plot. I also reversed the order of the keys so they would
match theorder of the curves in the
plot.styles={{AbsoluteThickness[0.5],
AbsoluteDashing[{4,4}]},{AbsoluteThickness[1.5]},{
AbsoluteThickness[ 2],AbsoluteDashing[{1,8}]}};ShowLegend[
LogLogPlot[{q1[t], q2[t], q3[t]}, {t, 0, 100}, PlotStyle ->
styles, AxesLabel -> {Y, X}, PlotLabel -> Title,
ImageSize -> 500, DisplayFunction -> Identity],
{MapThread[{Graphics[{Sequence @@ #1, Line[{{0, 0}, {1,
0}}]}], #2} &, {styles, {1, 3, 5}}] // Reverse,
LegendPosition -> {-0.7, -0.4}, LegendSize -> {0.2, 0.3},
LegendShadow -> {0.02, -0.02}, LegendSpacing -> 0.5} ];But
why use a legend at all? After all, a legend is nothing but
another plotin which you have put labels on the curves.
Why not put the labels directlyon the curves in the real
plot in the first place?LogLogPlot[{q1[t], q2[t], q3[t]}, {t,
0, 100}, PlotStyle -> styles, AxesLabel -> {Y, X},
PlotLabel -> Title, ImageSize -> 500, Epilog -> MapThread[
Text[SequenceForm[Case , #1], {Log[10, 0.01], Log[10,
#2[0.01]]}, {0, -1}] &, {{1, 2, 3}, {q1, q2, q3}}]];In the
legend you have keyed the curves to numbers 1, 3 and 5.
(Perhaps youjust used these as examples and meant to use
something different in the realplots?) But these don't
seem to have any obvious relation to your functions.I
suppose the reader will have to look at another table or look
into thetext of your paper or notebook to find out what 1, 3
and 5 mean. So thereader has to go from the graph to the
legend then to the text and thenmentally transfer the
meaning of the curve back to the main plot. It is somuch
nicer to put the meaning right on the curve if you can.For
the most part, legends are just computer junk and not even
easy tonicely construct. When the legend urge comes over you
- try to resist.David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/
AbsoluteThickness[1.5], {AbsoluteThickness[2],
AbsoluteDashing[{1,8}]}}, AxesLabel[Rule]{Y, X},
PlotLabel[Rule]Title, PlotLegend[Rule]{1,3,5},
LegendPosition[Rule] {0.5,0}](*However with LogPlot or
LogLogPlot the legend
desappear*)LogLogPlot[{q1[t],q2[t],q3[t]}, {t, 0,
100},PlotStyle[Rule]{ {AbsoluteThickness[0.5],
AbsoluteDashing[{4,4}]}, AbsoluteThickness[1.5],
{AbsoluteThickness[2], AbsoluteDashing[{1,8}]}},
AxesLabel[Rule]{Y, X}, PlotLabel[Rule]Title,
PlotLegend[Rule]{1,3,5}, LegendPosition[Rule]
{0.5,0}]I have shown a particular case, but I has this
problem always with LegendandLogPlot and LogPlotPlot. I
will appreciate any
help.GuillermoSanchez------------------------------------
---------This message was sent using Endymion
MailMan.
====
Using the Front End as a interface with the
kernel I was running some calulations when suddenly pressing
Shift+Enter causes the contents of the cell being evaluated
to transform to the next text underlined with a red line:
NotebookObject[FrontEndObject[LinkObject[dd8,1,1]],8]
foollowed by the next messages:An unknown box name
(NotebookObject) was sent as the BoxForm for the expression.
Check the format rules for the expression.An unknown box name
(FrontEndObject) was sent as the BoxForm for the expression.
Check the format rules for the expression.An unknown box
name (LinkObject) was sent as the BoxForm for the expression.
Check the format rules for the expression.An invalid typeset
structure was generated: Missing BoxFormData.Any suggestions
will be very aprreciated.Cesar 
====
I have an odd problem.
I need to use and simplify functions that havebeen provided
by a piece of software that insists on outputing
thefunctional results of a data mining proceedure, using
e whenoutputing numbers in scientific 
notation.I'm
having difficultly using Replace, Hold, etc. to correctly
evaluatethese types of function formats. For example, y =
5e+5x1+2e-1x2,should be transcribed into 5 10^5 x1 + 0.2
x2.ChuckReply-To: kuska@informatik.uni-leipzig.de
====
str
= 5e+5x1+2e-1x2;StringJoin[Characters[str] /. e ->
*10^] // ToExpression??Work fine for me.But this type of
output is typical generated by a C/FORTRANProgram and you
should rewrite the formating rules inthe code that produce
this output. Jens> I have an odd problem. I need to use
and simplify functions that have> been provided by a piece
of software that insists on outputing the> functional
results of a data mining proceedure, using e when>
outputing numbers in scientific notation.> I'm 
having
difficultly using Replace, Hold, etc. to correctly evaluate>
these types of function formats. For example, y =
5e+5x1+2e-1x2,> should be transcribed into 5 10^5 x1 + 0.2
x2.> Chuck
====
RB> I am considering the following
integralRB> W[m_,n_]:=Integrate[BesselJ[m, x]*BesselJ[n,
x], {x, 0, Infinity}]RB> where m,n are reals >=0. With
Mathematica 4.1 I obtain:RB> If[Re[m+n]>-1,
-Cos[(m-n)Pi/2]/(2 Pi)*RB> (2 EulerGamma + Log[4] +RB>
PolyGamma[0, 1/2(1 + m - n)] +RB> PolyGamma[0, 1/2(1 - m +
n)] +RB> 2PolyGamma[0, 1/2(1 + m + n)])RB> Any explanation
about the analytical expression will beRB> gratefully
accepteed.The expression for W[m_,n_] returned by
Mathematica is wrong.To prove, just substitute m = n = 0
which is exactly what you had doneand observe that the
output you had hadW[0,0]=-(2 EulerGamma + Log[4] + 4
PolyGamma[0, 1/2])/(2 Pi)= 0.84564was incorrect. The correct
answer is 1/2.Mathematica can handle the numeric integration
successfullyIn[1] := NIntegrate[BesselJ[1, x]*BesselJ[0,
x], {x, 0, Infinity}, Method -> Oscillatory] (* The warnings
are skipped *)Out[1] = 0.5Using NIntegrate[BesselJ[0,
x]*BesselJ[0, x], {x, 0, Infinity}]without Method ->
Oscillatory is not the optimal choice asthe integrand
oscillates fairly rapidly over the integrationregion.RB> I
suspect that these integrals are divergent (*).In fact, not
exactly.Integrate[BesselJ[1, x]*BesselJ[0, x], {x, 0,
Infinity}]is equal to 1/2, and Mathematica 4.1 for Microsoft
Windows(November 2, 2000) does it correctly, while
Mathematica 4.2for Microsoft Windows (February 28, 2002)
concocts a strangemixture of a wrong divergence message and
the warning thatit cannot check the convergence [should I
trust to the secondwarning? or the first?]As a matter of
fact,Integrate[BesselJ[1, x]*BesselJ[0, x], {x, 0,
Infinity}]converges because the integrand is regular at x=0,
bounded overthe whole right semi-axis, and decays
as2*Cos[Pi/4 - x]*Cos[(3*Pi)/4 - x]/(Pi*x) + o(1/x)at x ->
Infinity .Say, calculateNormal[Series[BesselJ[1, x], {x,
Infinity, 1}]] Normal[Series[BesselJ[0, x], {x, 
Infinity, 1}]]
// InputForm->(2*(Cos[Pi/4 - x] - Sin[Pi/4 -
x]/(8*x))*(Cos[(3*Pi)/4 - x] +(3*Sin[(3*Pi)/4 -
x])/(8*x)))/(Pi*x)then Plot[%,{x,1,10}]and
Plot[BesselJ[1,x]*BesselJ[0,x],{x,1,10}]and you could hardly
see the difference.Generally, to get to the convergence
domain for W in terms ofm and n is easy via the asymtotics
of the Bessel functions(use something
likeExpand[Normal[Series[BesselJ[m, x], {x, Infinity,
1}]]Normal[Series[BesselJ[n, x], {x, Infinity, 1}]]]then
analyze the main term).Best wishes,Vladimir
BondarenkoMathematical DirectorSymbolic Testing
GroupWeb : http://www.CAS-testing.org/
http://maple.bug-list.org/VER2/ (under tuning)
http://maple.bug-list.org/VER3/ (under tuning)
http://maple.bug-list.org/VER1/ (under tuning)
http://www.beautyriot.com/ (teamwork) http://www.ohaha.com/
(teamwork) Voice: (380)-652-447325 Mon-Fri 9 a.m. - 6
p.m.Mail : 76 Zalesskaya Str, Simferopol, Crimea,
Ukraine
====
> inside a program I need to solve this linear
equation in terms of p1.> However something odds happens.
Sometimes the solution is computed and> sometimes the result
is empty [I mean no output...]. Is this a bug of the> solve
command or am I doing something wrong? The problem is robust
to:> changing name to the variables and other makeups..>
DavidThat's the weirdest bug I've seen in 
weeks. As it
happens, it's mine. Atleast the inconsistent behavior, that
is. I'll fix it, and maybe alsotry to address 
the issue of
how to handle approximate numbers in testingsubexpressions
for zero.I've excised your code and put in place a
substantially smaller examplethat I believe is responsible.
The table will tend to give erraticresults.zz =
(-1.*x^7*(-1. + p - 7.*x^6 + p*x^6 + 6.*x^7)*
(7.000000000000002 - 7.000000000000002*x +
14.000000000000004*x^6 - 14.000000000000004*x^7 +
7.000000000000002*x^12 -6.999999999999998*x^13))/ (p^2*(1. +
0.9*x^6)^2*(1. + x^6)^5);One workaround would be to use
exact input, say by preprocessing withRationalize.Daniel
LichtblauWolfram Research
====
>inside a program I need to
solve this linear equation in terms of p1.>However something
odds happens. Sometimes the solution is computed
and>sometimes the result is empty [I mean no output...]. Is
this a bug of the>solve command or am I doing something
wrong? The problem is robust to:>changing name to the
variables and other makeups..>David>ps: Sorry for the stupid
way in which I copied the
command...>Solve[(x^2*((-0.9*x^7*(p^2*(-1 - 5.8*x^6 -
14.010000000000002*x^12 -> 18.04*x^18 - 13.06*x^24 ->
5.040000000000001*x^30 - 0.81*x^36) +> x*(7.777777777777779 -
9.074074074074076*x +> 30.333333333333336*x^6 ->
21.51851851851852*x^7 -> 16.333333333333336*x^8 +>
44.33333333333334*x^12 +> 3.188888888888883*x^13 ->
65.68333333333332*x^14 +> 28.777777777777786*x^18 +>
47.937037037037044*x^19 -> 100.10000000000002*x^20 + 7.*x^24
+> 45.6037037037037*x^25 -> 69.53333333333333*x^26 +>
13.299999999999999*x^31 -> 19.833333333333332*x^32 ->
1.0499999999999996*x^38) +> p*(-6 + 8.296296296296296*x ->
28.799999999999997*x^6 +> 32.785185185185185*x^7 +>
9.333333333333336*x^8 -> 55.260000000000005*x^12 +>
49.04777777777776*x^13 +> 38.38333333333334*x^14 ->
52.980000000000004*x^18 +> 34.20518518518518*x^19 +>
60.20000000000001*x^20 -> 25.380000000000003*x^24 +>
11.736296296296294*x^25 +> 43.63333333333334*x^26 - 4.86*x^30
+> 2.8999999999999986*x^31 +> 13.533333333333333*x^32 +
0.81*x^37 +> 1.0499999999999996*x^38)))/(x + 1.9*x^7 +>
0.9*x^13)^2 - ((-1 + p - 7*x^6 + p*x^6 +>
6*x^7)*(1.2962962962962965 - 3.111111111111112*x^6>+>
9.333333333333336*x^7 - 10.111111111111114*x^12>+> 22.05*x^13
- 5.703703703703705*x^18 + 17.15*x^19>+>
5.483333333333331*x^25 + 1.0499999999999996*x^31>+>
p1*x^5*(7.000000000000002 - 7.000000000000002*x>+>
14.000000000000004*x^6 -> 14.000000000000004*x^7 +>
7.000000000000002*x^12 -> 6.999999999999998*x^13) ->
1.166666666666667*p*x^4*x1 -> 3.500000000000001*p*x^10*x1 ->
1.0500000000000003*p*x^11*x1 -> 3.500000000000001*p*x^16*x1
-> 3.150000000000001*p*x^17*x1 -> 1.166666666666667*p*x^22*x1
-> 3.150000000000001*p*x^23*x1 ->
1.0500000000000003*p*x^29*x1))/((1 + 0.9*x^6)^2*(1>+>
x^6)^2)))/(p^2*(1 + x^6)^3) == 0, p1]You might find it more
robust (and the results cleaner) if you Simplify the
equation prior to using Solve. Such asSolve[eqn //
Rationalize // Simplify, p1]However, if you are assigning
values to p or x prior to using Solve, there may not be a
solution. That is, for whenever the numerator of the
expression for p1 would be zero, e.g., p = (-6 x^7 + 7 x^6
+1)/(x^6 + 1).Bob Hanlon
====
I prefer to delete all output
and then Copy As>Notebook expression. ItBobby-----Original
Message-----andoutput --- I try to use one tab indent for
input and two tabs indent foroutput, plus some blank line
adjustment.I wonder if anyone has a way of automatically
achieving
thisreformating.--Allan---------------------Allan
HayesMathematica Training and ConsultingLeicester
UKwww.haystack.demon.co.ukhay@haystack.demon.co.ukVoice
: +44 (0)116 271 4198> Often posters to MathGroup copy and
paste in the complete cellexpression,> including the In and
Out numbers, when posting to MathGroup.> I wonder if this is
the best method because one can't then just copyoutall>
the statements and paste them into a Mathematica notebook.
All thestatement> numbers have to be edited out and if
there are many statementdefinitions> this is an extended
task for any responder. This, of course, decreasesthe>
chances for a response. A better method is for the poster to
just copyand> paste the CONTENTS of each cell. This is
more work for the poster, butit> may pay off in better
responses.> David Park> djmp@earthlink.net>
http://home.earthlink.net/~djmp/
====
Could someone explain
what is going on here, please?In[1]:= a = 77617.; b =
33096.; In[2]:=SetAccuracy[a, Infinity]; SetAccuracy[b,
Infinity]; SetPrecision[a, Infinity]; 
SetPrecision[b,
Infinity]; In[4]:=f := 333.75*b^6 + a^2*(11*a^2*b^2 - b^6 -
121*b^4 - 2) + 5.5*b^8 + a/(2*b)In[5]:=SetAccuracy[f,
Infinity]; SetPrecision[f, Infinity];
In[6]:=fOut[6]=-1.1805916207174113*^21In[7]:=a = 77617;
b = 33096; In[8]:=g := (33375/100)*b^6 + a^2*(11*a^2*b^2 -
b^6 - 121*b^4 - 2) + (55/10)*b^8 +
a/(2*b)In[9]:=gOut[9]=-(54767/66192)In[10]:=N[%]Out[
10]=-0.8273960599468214PK
====
Peter,I hope that the
following example will help - it is a matter or when
thingsevaluate.The a in SetAccuracy[a, Infinity], below,
evaluates before SetAccuracy acts,so we
getSetAccuracy[2.3, Infinity]. The value of a is not
changed. a = 2.3; aa = SetAccuracy[a, Infinity]
2589569785738035/1125899906842624But, a 2.3Whereas aa
2589569785738035/1125899906842624--Allan-------------------
--Allan HayesMathematica Training and
ConsultingLeicester
UKwww.haystack.demon.co.ukhay@haystack.demon.co.ukVoice
: +44 (0)116 271 4198> Could someone explain what is going on
here, please?> In[1]:=> a = 77617.; b = 33096.;> In[2]:=>
SetAccuracy[a, Infinity]; SetAccuracy[b, 
Infinity];>
SetPrecision[a, Infinity]; SetPrecision[b, 
Infinity];> In[4]:=>
f := 333.75*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) +
5.5*b^8 + a/(2*b)> In[5]:=> SetAccuracy[f, Infinity];
SetPrecision[f, Infinity];> In[6]:=> f> Out[6]=>
-1.1805916207174113*^21> In[7]:=> a = 77617; b = 33096;>
In[8]:=> g := (33375/100)*b^6 + a^2*(11*a^2*b^2 - b^6 -
121*b^4 - 2) + (55/10)*b^8+ a/(2*b)> In[9]:=> g> Out[9]=>
-(54767/66192)> In[10]:=> N[%]> Out[10]=>
-0.8273960599468214> PK
====
In receiving notebooks from many
different people I have noticed thatbeginners often do not
know how to use Text cells and write all of theircomments
as Input cells. I have even run across some extremely
advancedusers who did not know the easy method for entering
Text cells. A goodnotebook is usually a blend of Text cells,
Input/Output cells and graphicscells. Text cells are very
useful for documenting what you are doing andpassing
information to other people. Since many people do not know
how touse Text cells, I thought I would write a little
explanation for beginnerswho are followers of
MathGroup.The very easiest method for entering a Text cell
is to put the insertionpoint where you want the new cell to
be (at the end of the notebook orbetween two existing cells)
and then type Alt-7. Then just start typing andyou will have
a Text cell.Alternatively you can use
MenuFormatStyleText to start a new Text cell.Often,
it is useful to put the ToolBar at the top of the notebook.
UseMenuFormatShow ToolBar. The drop-down menu on the
ToolBar has the variouskinds of cells available for the
current style of the notebook. You canselect Text (or any
other style) from there.Some users may hesitate to use Text
cells because they want to include amathematical expression
in the comments. However, that is also very easy.Just use an
Inline cell within the text cell. At the point within the
textcell where you want to include a mathematical
expression, start an Inlinecell by typing Ctrl-(. A
selection placeholder will appear on a pinkbackground. You
can type a Mathematica expression there just as in an
Inputcell. Use Ctrl-) to complete the Inline cell, or
Shift-Space. You can evenselect an Inline cell and evaluate
it with Shift-Ctrl-Enter.Putting comments in Text cells is
far better than using Input cells (or cellgroup header
cells). Mathematica won't try to evaluate Text cells, the
textwill wrap properly and adjust better to the notebook
width if you change it.You can also check the spelling of
words by putting the cursor after a wordand using Ctrl-K.
(In an Input cell Mathematica doesn't use the dictionary,but
uses the table of symbols instead and hence it won't check
spelling.)David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/===
=This is because inline cells are not in StandardForm by
default, but TraditionalForm.Use the menu item Cell ->
Default Inline Format Type -> StandardForm.>David Park's
posting reminded me of a frequent annoyance when I am>trying
to include some Mathematica expressions within text cells --
a>Mathematica input expression in Standard Form that
involves use of a>Control-key combination to form a
superscript, square-root, or built-up>fraction:>For example,
suppose I want to include within a text cell a Standard>Form
expression for the square of x, with the exponent 2 raised.
If I>type the x first, even if I immediately highlight it and
change it to>Courier (to match the default font for Input
cells in Standard Form), as>soon as I press the Control-^
key combination, an Inline cell is created>beginning with
the x, and then when I type the exponent 2 everything
in>that Inline cell is now in Times, and the x is Italic. To
change both>characters to Courier is not so easy: it seems to
require separately>the entire Inline cell and selecting
Courier does not change the exponent!)>So to avoid this
annoyance I normally must first type the desired>expression
in a separate Input cell, then copy the contents of that
cell>to the desired point in the Text cell.>Any suggestions
on a more efficient method for handling this?> In receiving
notebooks from many different people I have noticed that>
beginners often do not know how to use Text cells ....> Some
users may hesitate to use Text cells because they want to
include a> mathematical expression in the comments....>
Just use an Inline cell within the text cell....>-->Murray
Eisenberg murray@math.umass.edu>Mathematics & Statistics
Dept.>Lederle Graduate Research Tower phone 413 549-1020
(H)>University of Massachusetts 413 545-2859 (W)>710 North
Pleasant Street>Amherst, MA
01375--------------------------------------------------------
------Omega ConsultingThe final answer to your Mathematica
needsSpend less time searching and more time
finding.http://www.wz.com/internet/Mathematica.htmlReply-To
: murray@math.umass.edu
====
David Park's posting reminded me
of a frequent annoyance when I am trying to include some
Mathematica expressions within text cells -- a Mathematica
input expression in Standard Form that involves use of a
Control-key combination to form a superscript, square-root,
or built-up fraction:For example, suppose I want to include
within a text cell a Standard Form expression for the square
of x, with the exponent 2 raised. If I type the x first, even
if I immediately highlight it and change it to Courier (to
match the default font for Input cells in Standard Form), as
soon as I press the Control-^ key combination, an Inline
cell is created beginning with the x, and then when I type
the exponent 2 everything in that Inline cell is now in
Times, and the x is Italic. To change both characters to
Courier is not so easy: it seems to require separately the
entire Inline cell and selecting Courier does not change the
exponent!)So to avoid this annoyance I normally must first
type the desired expression in a separate Input cell, then
copy the contents of that cell to the desired point in the
Text cell.Any suggestions on a more efficient method for
handling this?> In receiving notebooks from many different
people I have noticed that> beginners often do not know how
to use Text cells ....> Some users may hesitate to use Text
cells because they want to include a> mathematical
expression in the comments....> Just use an Inline cell
within the text cell....-- Murray Eisenberg
murray@math.umass.eduMathematics & Statistics Dept.Lederle
Graduate Research Tower phone 413 549-1020 (H)University of
Massachusetts 413 545-2859 (W)710 North Pleasant
StreetAmherst, MA 01375
====
correctly, all you need to do
is to make sure the default inline cell formatis
StandardForm. Go to the menu item Cell, select Default Inline
FormatType,and change it to StandardForm.Carl WollPhysics
DeptU of Washington----- Original Message -----> type the x
first, even if I immediately highlight it and change it to>
Courier (to match the default font for Input cells in
Standard Form), as> soon as I press the Control-^ key
combination, an Inline cell is created> beginning with the
x, and then when I type the exponent 2 everything in> that
Inline cell is now in Times, and the x is Italic. To change
both> characters to Courier is not so easy: it seems to
require separately> the entire Inline cell and selecting
Courier does not change theexponent!)> So to avoid this
annoyance I normally must first type the desired> expression
in a separate Input cell, then copy the contents of that
cell> to the desired point in the Text cell.> Any
suggestions on a more efficient method for handling this?> In
receiving notebooks from many different people I have noticed
that> beginners often do not know how to use Text cells ....>
Some users may hesitate to use Text cells because they want to
include a> mathematical expression in the comments....> Just
use an Inline cell within the text cell....> --> Murray
Eisenberg murray@math.umass.edu> Mathematics & Statistics
Dept.> Lederle Graduate Research Tower phone 413 549-1020
(H)> University of Massachusetts 413 545-2859 (W)> 710 North
Pleasant Street> Amherst, MA 01375Reply-To:
kuska@informatik.uni-leipzig.de
====
just one comment: the
meaning of the Alt-7 key depend on the style sheet that is
in use.The TMJ style use Alt-8 for text and one hasto
learn new key short-cuts for every style sheet ! Jens> In
receiving notebooks from many different people I have noticed
that> beginners often do not know how to use Text cells and
write all of their> comments as Input cells. I have even run
across some extremely advanced> users who did not know the
easy method for entering Text cells. A good> notebook is
usually a blend of Text cells, Input/Output cells and
graphics> cells. Text cells are very useful for documenting
what you are doing and> passing information to other people.
Since many people do not know how to> use Text cells, I
thought I would write a little explanation for beginners>
who are followers of MathGroup.> The very easiest method for
entering a Text cell is to put the insertion> point where
you want the new cell to be (at the end of the notebook or>
between two existing cells) and then type Alt-7. Then just
start typing and> you will have a Text cell.> Alternatively
you can use MenuFormatStyleText to start a new Text
cell.> Often, it is useful to put the ToolBar at the top of
the notebook. Use> MenuFormatShow ToolBar. The drop-down
menu on the ToolBar has the various> kinds of cells
available for the current style of the notebook. You can>
select Text (or any other style) from there.> Some users may
hesitate to use Text cells because they want to include a>
mathematical expression in the comments. However, that is
also very easy.> Just use an Inline cell within the text
cell. At the point within the text> cell where you want to
include a mathematical expression, start an Inline> cell by
typing Ctrl-(. A selection placeholder will appear on a
pink> background. You can type a Mathematica expression
there just as in an Input> cell. Use Ctrl-) to complete the
Inline cell, or Shift-Space. You can even> select an Inline
cell and evaluate it with Shift-Ctrl-Enter.> Putting comments
in Text cells is far better than using Input cells (or cell>
group header cells). Mathematica won't try to evaluate Text
cells, the text> will wrap properly and adjust better to
the notebook width if you change it.> You can also check the
spelling of words by putting the cursor after a word> and
using Ctrl-K. (In an Input cell Mathematica doesn't use the
dictionary,> but uses the table of symbols instead and hence
it won't check spelling.)> David Park> djmp@earthlink.net>
http://home.earthlink.net/~djmp/
====
Solve[youre equation,
p1, VerifySolutions->True] will return a solution. So
willSolve[Rationalize[your equation],p1].Andrzej
KozlowskiToyama International UniversityJAPAN> inside a
program I need to solve this linear equation in terms of p1.>
However something odds happens. Sometimes the solution is
computed and> sometimes the result is empty [I mean no
output...]. Is this a bug of > the> solve command or am I
doing something wrong? The problem is robust to:> changing
name to the variables and other makeups..> David> ps: Sorry
for the stupid way in which I copied the command...>
Solve[(x^2*((-0.9*x^7*(p^2*(-1 - 5.8*x^6 -
14.010000000000002*x^12 -> 18.04*x^18 - 13.06*x^24 ->
5.040000000000001*x^30 - 0.81*x^36) +> x*(7.777777777777779 -
9.074074074074076*x +> 30.333333333333336*x^6 ->
21.51851851851852*x^7 -> 16.333333333333336*x^8 +>
44.33333333333334*x^12 +> 3.188888888888883*x^13 ->
65.68333333333332*x^14 +> 28.777777777777786*x^18 +>
47.937037037037044*x^19 -> 100.10000000000002*x^20 + 7.*x^24
+> 45.6037037037037*x^25 -> 69.53333333333333*x^26 +>
13.299999999999999*x^31 -> 19.833333333333332*x^32 ->
1.0499999999999996*x^38) +> p*(-6 + 8.296296296296296*x ->
28.799999999999997*x^6 +> 32.785185185185185*x^7 +>
9.333333333333336*x^8 -> 55.260000000000005*x^12 +>
49.04777777777776*x^13 +> 38.38333333333334*x^14 ->
52.980000000000004*x^18 +> 34.20518518518518*x^19 +>
60.20000000000001*x^20 -> 25.380000000000003*x^24 +>
11.736296296296294*x^25 +> 43.63333333333334*x^26 - 4.86*x^30
+> 2.8999999999999986*x^31 +> 13.533333333333333*x^32 +
0.81*x^37 +> 1.0499999999999996*x^38)))/(x + 1.9*x^7 +>
0.9*x^13)^2 - ((-1 + p - 7*x^6 + p*x^6 +>
6*x^7)*(1.2962962962962965 - > 3.111111111111112*x^6 +>
9.333333333333336*x^7 - > 10.111111111111114*x^12 +>
22.05*x^13 - 5.703703703703705*x^18 + > 17.15*x^19 +>
5.483333333333331*x^25 + > 1.0499999999999996*x^31 +>
p1*x^5*(7.000000000000002 - > 7.000000000000002*x +>
14.000000000000004*x^6 -> 14.000000000000004*x^7 +>
7.000000000000002*x^12 -> 6.999999999999998*x^13) ->
1.166666666666667*p*x^4*x1 -> 3.500000000000001*p*x^10*x1 ->
1.0500000000000003*p*x^11*x1 -> 3.500000000000001*p*x^16*x1
-> 3.150000000000001*p*x^17*x1 -> 1.166666666666667*p*x^22*x1
-> 3.150000000000001*p*x^23*x1 ->
1.0500000000000003*p*x^29*x1))/((1 + > 0.9*x^6)^2*(1 +>
x^6)^2)))/(p^2*(1 + x^6)^3) == 0, p1]Reply-To:
murray@math.umass.edu
====
For all names, perhaps:
Names[*`*]For names you defined at a normal session
(without changing to some other context than the default
Global`): Names[Global`*]> IIRC, there is a way to get a
list of all the symbols defined in the > currently running
session. I can't seem to find the reference to 
that > command.
Could somone point me in the direction of documentation which
> will tell me how to get information about the current
session?> TIA,> -- Murray Eisenberg
murray@math.umass.eduMathematics & Statistics Dept.Lederle
Graduate Research Tower phone 413 549-1020 (H)University of
Massachusetts 413 545-2859 (W)710 North Pleasant
StreetAmherst, MA 01375
====
> For all names, perhaps:>
Names[*`*]> For names you defined at a normal session
(without changing to some> other context than the default
Global`):> Names[Global`*]> IIRC, there is a way to get
a list of all the symbols defined in the> currently running
session. I can't seem to find the reference to 
that> command.
Could somone point me in the direction of documentation
which> will tell me how to get information about the current
session?> TIA,> I think I asked the wrong question. I'll
have to look at things some more. What you gave me resulted
in far more than I was looking
for.http://public.globalsymmetry.com/proprietary/com/wri/
system-symbols.htmlhttp://66.92.149.152/proprietary/com/wri
/system-symbols.htmlI think I really hosed the code for
generating that table. I used 5 lines. I probably didn't
need more than two, but I'm too tired right now to think
about it. Mathematica is totaly awesome when it comes to
what it was intended for. They really need to rent a Troll
for a few months and fix this interface. Qt will run on just
about anything. Heck, my Win-XP partition runs XFree86, with
the KDE, or it did when I booted into XP a month
ago.STH
====
?*does the trick. You can limit it to Global
variables with?Global`*Bobby-----Original
Message-----Sender: steve@smc.vnet.netApproved: Steven M.
Christensen <steve@smc.vnet.net>, Moderator
====
Well, first
of of all, your using SetAccuracy and SetPrecision does
nothing at all here, since they do not change the value of a
or b. You should use a = SetAccuracy[a, Infinity] etc. But
even then you won't get the same answer as when you use
exact numbers because of the way you evaluate f. Here is the
order of evaluation that will give you the same answer, and
should explain what is going on:f = SetAccuracy[333.75*b^6 +
a^2*(11*a^2*b^2 - b^6 - 121* b^4 - 2) + 5.5*b^8 + a/(2*b),
Infinity];a = 77617.; b = 33096.;a = SetAccuracy[a,
Infinity]; b = SetAccuracy[b, Infinity];f 
54767-(-----)
66192Andrzej KozlowskiToyama International
UniversityJAPAN> Could someone explain what is going on
here, please?> In[1]:=> a = 77617.; b = 33096.;> In[2]:=>
SetAccuracy[a, Infinity]; SetAccuracy[b, 
Infinity];>
SetPrecision[a, Infinity]; SetPrecision[b, 
Infinity];> In[4]:=>
f := 333.75*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) +
5.5*b^8 + > a/(2*b)> In[5]:=> SetAccuracy[f, Infinity];
SetPrecision[f, Infinity];> In[6]:=> f> Out[6]=>
-1.1805916207174113*^21> In[7]:=> a = 77617; b = 33096;>
In[8]:=> g := (33375/100)*b^6 + a^2*(11*a^2*b^2 - b^6 -
121*b^4 - 2) + > (55/10)*b^8 + a/(2*b)> In[9]:=> g> Out[9]=>
-(54767/66192)> In[10]:=> N[%]> Out[10]=> -0.8273960599468214>
PK
====
 ====By taking sums and differences of the
counterweights 1, 3, ... 3^(k-1)you can precisely express
all integers up to the integer whose base 3notation is
composed of k ones.By doubling those counterweights, you can
precisely express all the EVENintegers up to TWICE that
limit, or -1 + 3^k. If you know that unknownsare limited to
the integers from 1 to 3^k, this allows you to
preciselyweight the even numbers and bracket the odd
numbers. Hencecounterweights 2, 6, ... 2*3^(k-1) are
sufficient for unknowns up to3^k.Notice that this is an
efficient coding scheme: Each counterweight ismultiplied by
-1, 0, or 1, so there can be at most 3^k
distinctcounterweight sums. More than half are negative or
zero, however; thepositive sums and differences number at
most (3^k - 1)/2. Since we needto enumerate (and CAN
enumerate with even weights) only the even numbersand
because we get the maximum unknown 3^k by elimination, we
cover upto twice as many as we can enumerate, plus one.
Twice (3^k - 1)/2 plusone is 3^k, so we're getting the
maximum coverage possible for a givennumber of
counterweights.Bobby Treat-----Original Message-----For
these 40 positive integers to match 1 to 40 correspondingly,
thebiggestshould be 40 (also, the smallest should be 1).
So we have:a+b+c+d=40 or 1+b+c+d=40 (we state that
a<b<c<d)This could be a particular case (for a=3 and n=4)
of:In[1]:= Sum[a^i , {i, 0, n - 1}] == (a^n-1)/(a-1)Out[1]=
TrueWith 5 counterweights, we would be able to determine the
weight of 121elementsIn[]:= n = 5; a = 3; (a^n - 1)/(a -
1)Out[]= 121And the 5 counterweights would be the 5
elements of the correspondingsum.In[]:=Table[a^h, {h, 0,
n - 1}]Out[]={1, 3, 9, 27, 81}I did not understand what
Bryan said about 2,4,10,28 being a solution.At first, I
thought b could be either 2 or 3. b, the second of the
40positive combinations, could beb=2 or b-a=2, meaining
b=3But, if b=2, we would have 2 of the 40 positive
combinations that wouldequal1:a=1 and b-a=1This means we
would not be able to get all 40 values with the
40positivecombinations. so b must be 3.Maybe the same
reasoning would lead to 9 and 27 being also unique.{1,3,9,27}
could be the most efficient solution (of a+b+c+d=40),
inthesense that it would not generate any repetition. And
the most efficientsolution would be the only solution, since
the repetitions are notacceptable.I did not understand,
either, what Bob said about 34 being the
largestpossibleweight.I am very curious about these two
issues (a second solution, and 34), soIwould very much
appreciate if you could elaborate on them.Julio-----
Original Message -----> 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 anature> 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 tosolve> 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 thismath> 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}}> Bob Hanlon> Chantilly, VA
USA>
====
Plot[Evaluate[Sin[x]/Sin[x]], {x, -2.5,
-1.5}]Gives the expected graph.Bobby Treat-----Original
Message-----substituting a ßoating-point value for x results
in composite divisionbeing performed numerically, with
consequent inaccuracy. This can beverified by the
following:In[1]:=
FullForm[HoldForm[Sin[x]/Sin[x]]Out[1]//FullForm=
HoldForm[Times[Sin[x], Power[Sin[x], -1]]]By default,
Mathematica scales the y axis such that the
smalldiscrepanciesare visible. The PlotRange option may
be given to explicitly specify therange for the y axis (for
example PlotRange->{0, 2}).Ian McInnes.> I would expect that
Plot[Sin[x]/Sin[x], {x, -2.5, -1.5}] would produce a
horizontal line at y=1. However, on my Windows XP computer it
produces a graph where the yvalueis> less than 1 at
several points. Most notibly between x=-1.6 and x=-1.8 Is
this just an isolated case? Or does it happen to others? If
so -why? Ken.>
====
>>But my actual output reproduces the
input form, i.e., it is
simplyf[...listA,newEntry,...].That means your arguments
don't fit a pattern for which the function 
isdefined. That
is, it has nothing to do with what's on the right side
of:=. It's about making sure there are the right number
and type ofarguments passed to the function. You haven't
told us what the patternis or what the arguments are, so we
can't help you.Bobby-----Original Message----- I 
define a
function of the form: f[...,listA_, newEntry_,...]:=
Which[...]Here Which has the form of a set of logically
exclusive and exhausivetests, each test with its
ownaction that modifies the concrete list subsituted for
listA_. For instance, if there were only two tests the rhs
above wouldbeone that puts the newEntryat the start of
listA and the other that puts it at the end: Which[ test1,
listA = Insert[listA, newEntry, 1], test2, listA =
Insert[listA, newEntry, -1] ]Then I try to use the function,
substituting values for the arguments:f[..., listA, newEntry,
...]. Theoutput should be the appropriately modified list.
But my actual outputreproduces the input form,i.e., it is
simply f[...listA,newEntry,...]. Yet, the program does do
something, as the use of evaluate inplaceshows: Suppose
that test1 is satisfied by the particular values of
thearguments. Then whenI apply evaluate in place to
the lhs of the action that is supposed tooccur when test1
is true,in fact the value of Insert[listA,newEntry,1] is
correct, but the rhsgivesonly the initial value oflistA
and not the modified value that is on the rhs. Concretely,
suppose that in the function argument listA is
{121}andnewEntry is 200.Then after evaluation of the
function (input to the kernel), that linereads, according
toevaluate in place: true, {121} = {200, 121}So it seems
that only part of the correct action was taken: 200 was
putatthe start of listA. Butthe assignment of this
value -- the extended list -- to be the new valueoflistA
did not take place. I then tried using a new name for the
modified list, substituting thisprogram line for test1:
test1, newListA = Insert[listA,newEntry,1]What I get here,
applying evaluate in place after evaluating
thefunctionwith concrete values as aboveis: true,
newListA = {200, 121} Trace doesn't help because all I get
back is the function namewithits concrete arguments. Any
thoughts? Tom
====
The problem is that I do not think there
is any bounded 3d body described by your conditions. Anyway,
this is how you can use Mathematica (in general) to solve
this sort of problem.You need two
packages:In[1]:=<<Graphics`InequalityGraphics`In[2]:=<<
Calculus`Integration`To see your object
use:InequalityPlot3D[y > 3x && y < 4 - x^2 && z < x^2 + 4,
{x}, {y}, {z}]However, you will just get some error messages
and a picture that looks two dimensional. If you wanted the
volume,
evaluate:In[3]:=Integrate[Boole[y>3x&&y<4-x^2&&z<x^2+4],{x}
,{y},{z}]Out[3]=-InfinityIf you impose some limits on z you
will get a finite positive answer, but one that clearly is
unbounded:In[20]:=NIntegrate[Boole[y>3x&&y<4-x^2&&z<x^2+4],
{x},{y},{z,-100,100}]Out[20]=2239.58In[21]:=NIntegrate[
Boole[y>3x&&y<4-x^2&&z<x^2+4],{x},{y},{z,-1000,1000}]Out[21]
=20989.6Andrzej Sorry, I must have something missing in my
previous description. I need to find out the volumn of a 3D
object which form by the > equation :> z=x^2 +4 (as bottom
surface)> and on the xy plane which bounded by a parabola
y=4-x^2 and y=3x line. How would I use Mathematica to plot
out this 3D object or find out its > volumn with only the
equation given? Shz Shz> Andrzej Kozlowski
<akoz@mimuw.edu.pl> 3/September/2002 04:33pm Mathematica
could do this sort of thing if there were a three>
dimensional object described by your equations (as
boundaries) but> there isn't one. More precisely, the pair
of equations {z=x^2 +4,> y=4-x^2} describes a parabola in
three space which you can plot with: g1=ParametricPlot3D[{x,
4 - x^2, x^2 + 4}, {x, -5, 5}] The equation y=3x describes
the plane: g2 = ParametricPlot3D[{x, 3x, z}, {x, -5, 5}, {z,
-25, 25}]> You can see the two together in> <<RealTime3D`>
Show[{g1,g2}] There are clearly two points of intersection.
They can be found with:> Solve[{z == x^2 + 4, y == 4 - x^2, y
== 3*x}, {x, y, z}]> {{z -> 5, y -> 3, x -> 1}, {z -> 20, y ->
-12, x -> -4}} So where is the 3D object whose volume you want
to find? Andrzej Kozlowski> Toyama International University>
JAPAN Can I use Mathematica to find out the volumn of this 3
dimensional> object from> the equations : z=x^2 +4,
y=4-x^2, y=3x> Shz Shz <DISCLAIM.TXT>
====
 How can I run
mathematica program in background? Is there any waysimilar
to fortran or c where one can runs his program in
thebackground? Please suggest. RajReply-To:
kuska@informatik.uni-leipzig.de
====
math <<somecommands.m >
someresults.m &???BTW this is a problem of your operating
systemand not of Mathematica. Jens> How can I run
mathematica program in background? Is there any way> similar
to fortran or c where one can runs his program in the>
background?> Please suggest.> Raj
====
All works fine but now
I wanted to generate an animated gif with help of Now the
same file which worked earlier produces
errors:In[4]:=Export[schwing.gif, schwing,
GIF]LinkConnect::linkc: !([LeftSkeleton] 1
[RightSkeleton]) is dead; attempt to connect
failed.Out[4]=schwing.gifAlso a dialogue box appears
withsuch file or directorysuch file or 
directoryThe files
are at the place and are executable etc. But if one tries to
execute them always this No such file or directory error
message appears.Does anyone know what to do?-- Hendrik van
Hees Fakult.8at f.9fr Physik http://theory.gsi.de/~vanhees/
D-33615 Bielefeld 
====
First define the functionsz[x_] := x^2
+ 4;y1[x_] := 4 - x^2;y2[x_] := 3x;Figure out where the y
limits coincide:Solve[y1[x] == y2[x], x]{{x -> -4}, {x ->
1}}Which is bigger in between?y1[0] - y2[0]4y1 is. Confirm
that with a plot. (y2 is linear):Plot[{y1[x], y2[x]}, {x,
-5, 5}]Next figure out where z is zero:Solve[z[x] == 0,
x]{{x -> -2*I}, {x -> 2*I}}It's never zero for real x, but
is it positive, or negative?z[0]4Look at the plot, just for
fun:Plot[{z[x]}, {x, -4, 1}]z is positive for all x (but
only the interval [-4,1] MATTERS). Thevolume you want,
therefore, isIntegrate[z[x]*(y1[x] - y2[x]), {x, -4,
1}]625/4or:g[x_] = Integrate[z[x]*(y1[x] - y2[x]), x]g[1]
- g[-4]16*x - 6*x^2 - (3*x^4)/4 - x^5/5625/4Bobby
Treat-----Original Message-----Sender:
steve@smc.vnet.netApproved: Steven M. Christensen
<steve@smc.vnet.net>, Moderator
====
>-----Original
Message----->Sent: Wednesday, September 04, 2002 8:56
AM>previous output?>>What is a efficient way to use output
expression to define new >function. 
PeterPeter,I'd say
none,... Ôcause look at In[1]:= y = x^2Out[1]= x^2In[2]:=
f1[x_] = Out[1]Out[2]= x^2In[3]:= f2[x_] := Out[1]In[4]:= x
= 5Out[4]= 5In[5]:= f3[x_] = Out[1]Out[5]= 25In[6]:=
Block[{x}, f4[x_] = Out[1]]Out[6]= 25In[7]:= Block[{x},
f5[x_] := Out[1]]In[9]:= f6[x_] = yOut[9]= 25In[10]:=
Block[{x}, f7[x_] = y]Out[10]= 25In[11]:= f8[x_] :=
yIn[12]:= Through[{f1, f2, f3, f4, f5, f6, f7, f8}[#]] &
/@ {0, 1}Out[12]= {{0, 25, 25, 0, 25, 25, 0, 25}, {1, 25,
25, 1, 25, 25, 1, 25}}In[13]:= Block[{x}, f9[x_] =
Out[1]]Out[13]= %1In[14]:= Through[{f1, f2, f3, f4, f5,
f6, f7, f8, f9}[#]] & /@ {0, 1}Out[14]= {{0, %1, 25, 0, %1,
25, 0, 25, %1}, {1, %1, 25, 1, %1, 25, 1, 25, %1}}f1, f4 and
f7 come out right, but f1 is based on the condition that
thepattern variables of the expression have not been
redefined, and f4 is onlycorrect if the history is
remembered well. So what rests is f7, but thatwasn't you
question.--Hartmut Wolf
====
If we are in the business of
giving good links to Doppler demonstrations,here is another
one
(freeware):http://www.phy.ntnu.edu.tw/java/Doppler/
Doppler.htmlFor anyone interested in good Java
implementations of physics phenomena,NTNU Virtual Physics
Laboratory athttp://www.phy.ntnu.edu.tw/java/index.html is
a treasure chest.This is not Mathematica stuff (as far as I
know), but might be of interestto several Mathgroup
readers.Ingolf DahlChalmers
UniversitySwedenf9aid@fy.chalmers.se-----Original
Message-----Sender: steve@smc.vnet.netApproved: Steven M.
Christensen <steve@smc.vnet.net>, Moderator
====
My main
application for this is in drawing a piece of a plane at a
pointnormal to a vector. For me, efficiency is not an
important criterion. But Iwant orthogonal vectors so the
plane will be composed of squares and notdiamonds. I want
unit vectors so that I can easily specify the size of
thepiece I am drawing.I posted a summary last evening and
it doesn't seem to have appeared today.In any case, I have
realized that none of the routines really require
theLinearAlgebra`Orthogonalization package, and run faster
without it. I havemodified the original routines to use a
pure function instead of Normalizeor GramSchmidt.Here is a
summary of the responses so far. The timings were done with
thefollowing exact vector. They are faster with a machine
precision vector.vtest = {1, 2,
1};__________________________________________________________
Hugh Goyder...The original routine
wasOrthogonalUnitVectors[v : {_, _, _}] :=
GramSchmidt[NullSpace[{v}]]but I changed it
toOrthogonalUnitVectors[v : {_, _, _}] := #/Sqrt[#.#] & /@
NullSpace[{v}]OrthogonalUnitVectors[vtest]{{-(1/Sqrt[2]), 0,
1/Sqrt[2]}, {-(1/Sqrt[3]), 1/Sqrt[3],
-(1/Sqrt[3])}}Do[OrthogonalUnitVectors[vtest], {10000}] //
Timing{2.53 Second, Null}This appears to be the shortest
and fastest routine. I would give it theprize for elegance.
It is certainly easy to
remember.____________________________________________________
____Ingolf Dahl...OrthogonalUnitVectors[v:{_, _, _}] :=
(#1/Sqrt[#1 . #1] & ) /@ NullSpace[{v, {0, 0, 0}, {0, 0,
0}}]OrthogonalUnitVectors[vtest]{{-(1/Sqrt[2]), 0,
1/Sqrt[2]}, {-(2/Sqrt[5]), 1/Sqrt[5],
0}}Do[OrthogonalUnitVectors[vtest], {10000}] // Timing{2.59
Second, Null}This is really the same as Hugh Goyder's
routine. It isn't necessary to addthe zero rows to the
matrix. NullSpace automatically does
it.________________________________________________________
Selwyn Hollis...OrthogonalUnitVectors[v : {_, _, _}] :=
Module[{u, w}, u = Which[(w = {0, v[[3]], -v[[2]]}).w != 0,
w, (w = {v[[3]], 0, -v[[1]]}).w != 0, w, (w = {v[[2]],
-v[[1]], 0}).w != 0, w, True, Return[$Failed]]; #/Sqrt[#.#] &
/@ {u, Cross[u, v]} ]OrthogonalUnitVectors[vtest]{{0,
1/Sqrt[5], -(2/Sqrt[5])}, {Sqrt[5/6], -Sqrt[2/15],
-(1/Sqrt[30])}}Do[OrthogonalUnitVectors[vtest], {10000}] //
Timing{11.92 Second, Null}I wish I understood the reason
for using rows of the v Cross matrix as testvectors. It
does have the advantage that a dot product is used to give
ascalar test for zero. Why not...OrthogonalUnitVectors[v :
{_, _, _}] := Module[{u, w}, u = Which[(w = {1, 0,
0})[Cross]v != {0, 0, 0}, w, (w = {0, 1, 0})[Cross]v !=
{0, 0, 0}, w, True, Return[$Failed]]; #/Sqrt[#.#] & /@ {u,
Cross[u, v]} ]OrthogonalUnitVectors[vtest]{{1, 0, 0}, {0,
-(1/Sqrt[5]), 2/Sqrt[5]}}Do[OrthogonalUnitVectors[vtest],
{10000}] // Timing{10.33 Second,
Null}________________________________________________________
____John Browne...OrthogonalUnitVectors[v : {_, _, _}] :=
Module[{r, v1, v2}, r = {Random[], Random[], Random[]}; v1 =
Cross[v, r]; v2 = Cross[v1, v]; {v1/Sqrt[Dot[v1, v1]],
v2/Sqrt[Dot[v2, v2]]}]OrthogonalUnitVectors[vtest]{{0.26601,
-0.534209, 0.802408}, {-0.873254, 0.218984,
0.435286}}Do[OrthogonalUnitVectors[vtest], {10000}] //
Timing{9.17 Second, Null}This has the problem that the r
vector might be parallel to v. But in 100000test cases it
never
happened.____________________________________________________
________Daniel Lichblau...(Daniel left out the
normalization, which I added.)OrthogonalUnitVectors[v:{_, _,
_}] := With[{vecs = NullSpace[{v}]}, (#1/Sqrt[#1 . #1] & ) /@
{vecs[[1]], vecs[[2]] - vecs[[2]] .
vecs[[1]]*vecs[[1]]}]OrthogonalUnitVectors[vtest]{{-(1/Sqrt[
2]), 0, 1/Sqrt[2]}, {0, 1/Sqrt[5],
-(2/Sqrt[5])}}Do[OrthogonalUnitVectors[vtest], {10000}] //
Timing{3.35 Second,
Null}________________________________________________________
______Garry HelzerOrthogonalVectors[v : {a1_, a2_, a3_}]
:= With[{w = First[Sort[{{a2, -a1, 0}, {a3, 0, -a1}, {0, a3,
-a2}}, OrderedQ[{Plus @@ Abs[#2], Plus @@ Abs[#1]}] &]]}, {w,
Cross[v, w]} ]OrthogonalUnitVectors[vtest]{{-(1/Sqrt[2]), 0,
1/Sqrt[2]}, {-(2/Sqrt[5]), 1/Sqrt[5],
0}}Do[OrthogonalUnitVectors[vtest], {10000}] // Timing{2.64
Second,
Null}________________________________________________________
__________Tom BurtonA nice feature of the following
simple scheme is that you know that the twopoints of
discontinuity are where r and -r cross the unit
sphere.triad[u_, r_:{1, 0, 0}] := Module[{v = Cross[u, r]},
#/Sqrt[# . #] & /@{u, v, Cross[u,
v]}]triad[vtest]{{1/Sqrt[6], Sqrt[2/3], 1/Sqrt[6]}, {0,
1/Sqrt[5], -(2/Sqrt[5])}, {-Sqrt[5/6], Sqrt[2/15],
1/Sqrt[30]}}Do[triad[vtest], {10000}] // Timing{17.02
Second, Null}David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/
djmp@earthlink.nethttp://home.earthlink.net/~djmp/
====

CeZaR,Sort of confused about what you really wanted. If you
have forgotten toassign x = ? then you can do the
following,Clear[x, eq, m, l]eq = m x^2 + 2(m + 1)x + m +
2l = {};x = 1;For[i = -30, i < 30, i++, AppendTo[l, eq /.
m -> i]]ListPlot[l]But if you really intend to make a list
of quadratics and then plot theresults for each of the
quadratics then you can try the followingClear[x, eq, m,
l]eq = m x^2 + 2(m + 1)x + m + 2;l = {};For[i = -30, i <
30, i++, AppendTo[l, eq /. m -> i]];test = Table[l[[i]], {i,
1, Length[l], 1}, {x, -50, 50}];MultipleListPlot[test[[5]],
test[[10]], test[[15]], test[[20]],test[[25]], test[[30]],
test[[35]], test[[40]], test[[45]], test[[50]], test[[55]],
test[[60]], PlotJoined -> True, SymbolShape -> None,
ImageSize -> 500 ]Essentially your For loop generates a list
of quadratics if x is left out.Yas I want to do these
operations: eq = m x^2 + 2(m + 1)x + m + 2> l = {}> For[i =
-30, i < 30, i++, AppendTo[l, eq /. m -> i]]> Plot[l, {x,
-50, 50}] but i get this error:> l is not a machine-size real
number at x = -49.99999583333334 Can someone exaplain to me
what i am doing wrong here??> CeZaR
====
Plot[l // Evaluate,
{x, -30, 30}, Frame -> True];CeZaR
====
> I want to get some
F and R such that:> F[n,p] + R[n,p] = Sum[Binomial[n,k]
p^(n-k) (1-p)^k, {k, 0, Floor[n/2] - 1}],> when F[n,p] is an
approximation to the sum and the R is the remaining error.>
Constantine.>>I'm looking for a way of finding 
the
approximation for partitial binomial>>sum.>>I'll be
pleasant for any hint..> [...]> Office: Taub 411You can get a
closed form in terms of special functions if you splitinto
two cases depending on whether n is even or odd.In[39]:= n =
2*m;In[40]:= InputForm[Sum[Binomial[n,k]*p^(n-k)*(1-p)^k,
{k,0,m-1}]]Out[40]//InputForm= p^(2*m)*((p^(-1))^(2*m) -
((-4 + 4/p)^m*Gamma[1/2 + m]* Hypergeometric2F1[1, -m, 1 + m,
(-1 + p)/p])/(Sqrt[Pi]*Gamma[1 +m]))In[41]:= n =
2*m+1;In[42]:= InputForm[Sum[Binomial[n,k]*p^(n-k)*(1-p)^k,
{k,0,m-1}]]Out[42]//InputForm= p^(1 + 2*m)*((p^(-1))^(1 +
2*m) - (2^(1 + 2*m)*(-1 + p^(-1))^m*Gamma[3/2+ m]*
Hypergeometric2F1[1, -1 - m, 1 + m, (-1 +
p)/p])/(Sqrt[Pi]*Gamma[2 +m]))Daniel LichtblauWolfram
Research
====
I want to get some F and R such that:F[n,p] +
R[n,p] = Sum[Binomial[n,k] p^(n-k) (1-p)^k, {k, 0, Floor[n/2]
- 1}],when F[n,p] is an approximation to the sum and the R
is the remaining error.Constantine.>>I'm looking for a way
of finding the approximation for partitial binomial
>>sum.>>I'll be pleasant for any hint..>Use the standard
add-on package Statistics`NonlinearFit` to do a >NonlinearFit
to whatever model you want to use for the approximation.>Bob
Hanlon>Chantilly, VA USAConstantine ElsterComputer
Science Dept.Technion I.I.T.Office: Taub
411
====
Constantine,If you break your problem up into two
cases, even n and odd n, thenMathematica can sum up your
problem and get results, albeit withhypergeometric
functions. Consider the following (make sure you look at
thiswith a fixed font):In[21]:=evenans =
Sum[Binomial[2*n, k]*p^(2*n - k)*(1 - p)^k, {k, 0, n -
1}];In[22]:=PowerExpand[FunctionExpand[FullSimplify[evenans
, n [Element] Integers]]]Out[22]= 2 n n n 1 p - 1 2 (1 -
p) p Gamma[n + -] Hypergeometric2F1[1, -n, n + 1, -----] 2
p1 -
-------------------------------------------------------------
------- Sqrt[Pi] Gamma[n + 1]In[23]:=oddans =
Sum[Binomial[2*n + 1, k]*p^(2*n + 1 - k)*(1 - p)^k, {k, 0, n
-
1}];In[24]:=PowerExpand[FunctionExpand[FullSimplify[oddans
, n [Element] Integers]]]Out[24]= 2 n + 1 n n + 1 3p - 1
2 (1 - p) p Gamma[n + -] Hypergeometric2F1[1, -n - 1, n +1,
-----] 2p1 -
------------------------------------------------------------
-------------------- Sqrt[Pi] Gamma[n + 2]Is this what you
were looking for?Carl WollPhysics DeptU of Washington>
I want to get some F and R such that: F[n,p] + R[n,p] =
Sum[Binomial[n,k] p^(n-k) (1-p)^k, {k, 0, Floor[n/2] -1}],>
when F[n,p] is an approximation to the sum and the R is the
remainingerror. Constantine.>>I'm looking for a way of
finding the approximation for 
partitialbinomial>>sum.>>I'll
be pleasant for any hint..>Use the standard add-on package
Statistics`NonlinearFit` to do a>NonlinearFit to whatever
model you want to use for the approximation.>>Bob
Hanlon>Chantilly, VA USA Constantine Elster> Computer
Science Dept.> Technion I.I.T.> Office: Taub 411
====
>I want to
do these operations:>>eq = m x^2 + 2(m + 1)x + m + 2>l =
{}>For[i = -30, i < 30, i++, AppendTo[l, eq /. m ->
i]]>Plot[l, {x, -50, 50}]>>but i get this error:>l is not a
machine-size real number at x = -49.99999583333334>>Can
someone exaplain to me what i am doing wrong here??>eq = m
x^2 + 2(m + 1)x + m + 2;l = {};For[i = -30, i < 30, i++,
AppendTo[l, eq /. m -> i]];It would be more straightforward
and efficient to define the list using Tablel == 
Table[eq,
{m, -30, 29}]TrueTo Plot, use EvaluatePlot[Evaluate[l],
{x, -50, 50}];Bob Hanlon
====
>What is a efficient way to use
output expression to define new function.Use the menu command
Input/Copy Output from Aboveor use % (Out)Bob
Hanlon
====
>-----Original Message----->Sent: Wednesday,
September 04, 2002 8:57 AM>The problem is that I do not
think there is any bounded 3d body >described by your
conditions. Anyway, this is how you can use >Mathematica (in
general) to solve this sort of problem.>You need two
packages:>>In[1]:=><<Graphics`InequalityGraphics`>>In[2]:=><<
Calculus`Integration`>To see your object
use:>InequalityPlot3D[y > 3x && y < 4 - x^2 && z < x^2 + 4,
{x}, {y}, {z}]>>However, you will just get some error
messages and a picture >that looks >two dimensional. If you
wanted the volume,
evaluate:>>In[3]:=>Integrate[Boole[y>3x&&y<4-x^2&&z<x^2+4],{x
},{y},{z}]>>Out[3]=>-Infinity>>If you impose some limits on z
you will get a finite positive answer, >but one that clearly 
is
unbounded:>>In[20]:=>NIntegrate[Boole[y>3x&&y<4-x^2&&z<x^2+4],
{x},{y},{z,-100,100}]>>Out[20]=>2239.58>>In[21]:=>NIntegrate[
Boole[y>3x&&y<4-x^2&&z<x^2+4],{x},{y},{z,-1000,1000}]>>Out[21
]=>20989.6>>Andrzej Sorry, I must have something missing in
my previous description. I need to find out the volumn of a 3D
object which form by the > equation :> z=x^2 +4 (as bottom
surface)> and on the xy plane which bounded by a parabola
y=4-x^2 and >y=3x line. How would I use Mathematica to plot
out this 3D object or >find out its > volumn with only the
equation given? Shz Shz> Andrzej Kozlowski
<akoz@mimuw.edu.pl> 3/September/2002 04:33pm Mathematica
could do this sort of thing if there were a three>
dimensional object described by your equations (as
boundaries) but> there isn't one. More precisely, the pair
of equations {z=x^2 +4,> y=4-x^2} describes a parabola in
three space which you can >plot with: g1=ParametricPlot3D[{x,
4 - x^2, x^2 + 4}, {x, -5, 5}] The equation y=3x describes the
plane: g2 = ParametricPlot3D[{x, 3x, z}, {x, -5, 5}, {z, -25,
25}]> You can see the two together in> <<RealTime3D`>
Show[{g1,g2}] There are clearly two points of intersection.
They can be found with:> Solve[{z == x^2 + 4, y == 4 - x^2, y
== 3*x}, {x, y, z}]> {{z -> 5, y -> 3, x -> 1}, {z -> 20, y ->
-12, x -> -4}} So where is the 3D object whose volume you want
to find? Andrzej Kozlowski> Toyama International University>
JAPAN Can I use Mathematica to find out the volumn of this 3
dimensional> object from> the equations : z=x^2 +4,
y=4-x^2, y=3x> Shz Shz>> <DISCLAIM.TXT>>Shz Shz,as
Andrzej having said everything about the calculation of the
volume, andas still your specification is incomplete, 
I'm
going to show how you canplot at least all you have
(communicated).As you said something obout the bottom --
which direction is up? --assuming positive z-direction
the I come to the inequalities: y >= 3 x, y <= 4 -x^2, z >= 4
+ x^2and arbitrarily, to make the example complete I'll show
only parts with z < 20but do not complete the shape.To
begin building the graphics we draw the boundary surfaces:[1]
the bottom{xmin, xmax} = x /. NSolve[4 - x^2 == 3x, {x}]{-4.,
1.}Plot[{4 - x^2, 3x}, {x, xmin, xmax}]{ymin, ymax} = {3
xmin, 4}g3 = Graphics3D[Plot3D[x^2 + 4, {x, xmin, xmax}, {y,
ymin, ymax}]][2] the wallsg = ParametricPlot3D[{{x, 4 - x^2,
z}, {x, 3x, z}}, {x, xmin, xmax}, {z, 0,  20}]One idea now
would be to exploit the Mathematica rendering algorithm, to
cutoff the undesired parts of the boundary surfaces:gg =
Show[g3, g, PolygonIntersections -> False][Epsilon] = 1.
10^-4g4 = gg /. Line[_] -> {} /. p : Polygon[pts_] :> If[Or
@@ (#2 > 4 - #1^2 + [Epsilon] || #2 < 3 #1 - [Epsilon] ||
#3 < #1^2 + 4 - [Epsilon] &) @@@  pts, {}, p]Show[g4 /.
Line[_] -> {}, BoxRatios -> {1, 1, .5}, ViewPoint -> {1.3,
-2.4, 5}]But alas! The edges are rather gnawed off.
Increasing epsilon doesn't help,unwanted parts will show up
and still some polygons are nibbled away.So we have to do it
ourselves:We definetag[inequality_][p_] := Block[{x, y, z},
{x, y, z} = p; If[inequality, {p, inside}, {p,
outside}]]...tags points whether inside ore
outside.sol[equality_, {p1_, _}, {p2_, _}] := Block[{p, d,
x, y, z}, {x, y, z} = p = p1 (1 - d) + p2 d;
First[Cases[Solve[equality, d], s_ /; NonNegative[d /. s] :>
(p /. s)]] ]...computes the cut on the line between to points
of opposite sides.sep[equality_][pp1_, pp2_] := If[pp1[[2]]
=== pp2[[2]], {pp1}, ppx = sol[equality, pp1, pp2]; {pp1,
{ppx, pp1[[2]]}, {ppx, pp2[[2]]}}]...effectively cuts a
segment crossing the border into two
pieces.cutGraphics3D[g_, inequality_] := Module[{equality =
Equal @@ inequality}, g /. Polygon[pts_] :> Block[{tagpts =
tag[inequality] /@ pts}, With[{t = sep[equality] @@@
Transpose[{tagpts, RotateLeft[tagpts]}]},  With[{newpts =
Cases[t, {p_, label_} /; label == inside :> p,2]},
If[Length[newpts] > 2, Polygon[newpts], {}]] ]]]...Polygons
crossing a border are cut, only those inside are kept.We
first cut the bottom...g3new = Fold[cutGraphics3D, g3, {y >=
3x, y <= 4 - x^2}]...next the sides...gnew =
cutGraphics3D[g, z >= 4 + x^2]...and display:Show[gnew,
g3new, BoxRatios -> {1, 1, .5}, ViewPoint -> {1.3, -2.4,
5}]Show[gnew, g3new, BoxRatios -> {1, 1, .5}, ViewPoint ->
{1.3, -2.4, 15}]Show[gnew, g3new, BoxRatios -> {1, 1, .5},
ViewPoint -> {1.3, -2.4, -2}]Show[gnew, g3new, BoxRatios ->
{1, 1, .5}, ViewPoint -> {1.3, 2.5, 2}]Here I have brought a
method, I had already posted twice, to a more handyform.To
make that reminiscence complete, we could also define a
coloring functioncolorGraphics3D[g_, inequality_,
colorinside_, coloroutside_] := Module[{equality = Equal @@
inequality}, g /. Polygon[pts_] :> Block[{tagpts =
tag[inequality] /@ pts}, With[{t = sep[equality] @@@
Transpose[{tagpts, RotateLeft[tagpts]}]}, {With[{newpts = 
Cases[t, {p_, label_} /; label == outside :> p, 2]},
If[Length[newpts] > 2,{FaceForm[SurfaceColor[coloroutside]],
Polygon[newpts]}, {}]], With[{newpts = Cases[t, {p_, label_}
/; label == inside :> p, 2]}, If[Length[newpts] > 2,
{FaceForm[SurfaceColor[colorinside]], Polygon[newpts]},
{}]]} ]]]and with...g3D = Graphics3D[Plot3D[E^(-(x^2/2) -
y^2/2)/(2*Pi) , {x, -2, 2}, {y,
-2,2}]]Show[colorGraphics3D[g3D, With[{e = Take[{x, y, z},
2] - {0.8, -0.5}}, e.e] <= 1, Hue[0, 0.3, 1],  Hue[0.6, 0.3,
1]]]--Hartmut Wolf
====
This result seems to me to be wrong
(it's right?):In[1]:=Integrate[Sqrt[a^2*Cos[t]^2 +
b^2*Sin[t]^2], {t, 0, 2*Pi}, Assumptions->{Im[a]==0,
Im[b]==0}]Out[1]=0....but written in another way, it's
right:In[2]:=Integrate[Sqrt[a^2 + (b^2 - a^2)*Sin[t]^2],
{t, 0, 2*Pi}, Assumptions -> {Im[a] == 0, Im[b] ==
0}]Out[2]=If[a^2/(-a^2 + b^2) >= 0 || b^2/(-a^2 + b^2) <= 0
|| Im[a^2/(-a^2 + b^2)] != 0, (4*a^2*Sqrt[-1 +
b^2/a^2]*EllipticE[ 1 - b^2/a^2])/Sqrt[-a^2 + b^2],
Integrate[Sqrt[a^2 + (-a^2 + b^2)*Sin[t]^2], {t, 0,
2*Pi}]]Someone can explane me this
difference?Raf.
====
involved Bessel
equations(electro-magnetic field in coaxial) but when I
initialize and calculate that2 eqs for the 2nd or 3rd time,
Ôcause I changed some parameters, I see astrange
Removed[n] instead of the value of n (=0).I use the
line Remove[Global`*];at the beginning to clear any
variable and I think it could be the problem.So, have I
always to quit and restart the program in order to
recalculatethat equations?Filippo Sola
====
Can someone
provide me more information on how ListIntegrate worksthan
what is contained in the version 4 manual ? Is there a
websiteperhaps or tutorial ?I would like to know how the
beginning and end of a list of {x,y}pairs that is being
integrated is dealt with by ListIntegrate.I know a series of
polynomials is somehow used but how ? Do theseoverlap ? Are
they piecewise continuous ?Are these polynomials available
for inspection ? How do they change asa function of k
?optimum k value for a given list ? For any given list,
will accuracymonotonically increase with increasing values
of k ? Is thereanything in the Option Inspector that
could cause unexpected behaivorwith ListIntegrate ?Are
there any good rules of thumb or procedures that will help
ensurea reasonable answer is produced ?Is there a way of
estimating the error or accuracy of integrationperformed by
ListIntegrate ?
====
Be sure to note the following and what
comes after it near the end of the Help Browser info on
ListIntegrate:``This package has been included for
compatibility with previous versions of Mathematica. The
functionality of this package has been superseded by
improvements made to InterpolatingFunction.In other words,
ListIntegrate is a dinosaur that you don't need at all!To
integrate a list of data with Mathematica, one can proceed in
either of two ways: (1) Construct an interpolating function
and use NIntegrate (or, better,
NIntegrateInterpolatingFunction) on that (which is what
ListIntegrate apparently does); or (2) apply a simple
routine that implements the trapezoidal rule, Simpson's
rule, or maybe some higher order method.Assuming you've
chosen to take path #1, you need to realize the following:(a)
NIntegrate[Interpolation[data, InterpolationOrder->1][x],
{x,a,b}] is equivalent to the trapezoidal rule;(b)
NIntegrate[Interpolation[data, InterpolationOrder->2][x],
{x,a,b}] is *not* equivalent to Simpson's rule (because of
the peculiar way that Interpolation works);(c)
Interpolation[data, InterpolationOrder->k] generally does not
return a smooth function unless you set
InterpolationOrder->n-1, where n is the number of data
points, which is the case where a single polynomial of
degree n-1 fits the data points.(d) If you want to integrate
a smooth interpolant, you can do this:
<<NumericalMath`SplineFit` NIntegrate[SplineFit[data,
Cubic][x][[2]], {x,a,b}]To understand better the way
Interpolation works, look closely at the plots created by
the following: data = Table[{i, Random[]}, {i, 0, 5}];
Do[Plot[Interpolation[data, InterpolationOrder->k][x], {x, 0,
5}, PlotRange -> All], {k, 1, 6}](The last of those plots will
give a warning message.)Having said all that, you really
should consider path #2 instead. Here are a couple of links
to a MathGroup discussion of last July (somehow the thread
got split
up):http://library.wolfram.com/mathgroup/archive/2002/Jul/
msg00490.htmlhttp://library.wolfram.com/mathgroup/archive/
2002/Jul/msg00519.htmlI hope all this helps
some.----Selwyn Hollis > Can someone provide me more
information on how ListIntegrate works > than what is
contained in the version 4 manual ? Is there a website >
perhaps or tutorial ? > I would like to know how the
beginning and end of a list of {x,y} > pairs that is being
integrated is dealt with by ListIntegrate. > I know a
series of polynomials is somehow used but how ? Do these >
overlap ? Are they piecewise continuous ? > Are these
polynomials available for inspection ? How do they change as
> a function of k ? > optimum k value for a given list
? For any given list, will accuracy > monotonically increase
with increasing values of k ? Is there > anything in the
Option Inspector that could cause unexpected behaivor > with
ListIntegrate ? > Are there any good rules of thumb or
procedures that will help ensure > a reasonable answer is
produced ? > Is there a way of estimating the error or
accuracy of integration > performed by ListIntegrate ? >
>
====
I timed Daniel's three solutions and Gary's 
one (plus a
couple of my owna little later):perps1[v_] := If[v[[1]] ==
v[[2]] == 0, {{1, 0, 0}, {0, 1, 0}}, {{v[[2]], -v[[ 1]], 0},
Cross[v, {v[[2]], -v[[1]], 0}]}]perps2[v_] := With[{vecs =
NullSpace[{v}]}, {vecs[[ 1]], vecs[[2]] -
(vecs[[2]].vecs[[1]])*vecs[[1]]}]perps2C = Compile[{{v,
_Real, 1}}, Module[{vecs = NullSpace[{v}]}, { vecs[[1]],
vecs[[2]] - (vecs[[2]].vecs[[1]])*vecs[[1]]}]]helzer[v :
{a1_, a2_, a3_}] := With[{w = First[Sort[{{a2, -a1, 0}, {a3,
0, -a1}, {0, a3, -a2}}, OrderedQ[{Plus @@ Abs[#2], Plus @@
Abs[#1]}] &]]}, {w, Cross[v, w]}]vecs = Table[Random[],
{10000}, {3}]; Timing[perps1 /@ vecs; ]Timing[perps2 /@
vecs; ]Timing[perps2c /@ vecs; ]Timing[helzer /@ vecs;
]{1.7350000000000012*Second, Null}{0.5619999999999994*Second,
Null}{0.219 Second, Null}{2.7349999999999994*Second, Null}I
made a small change to Daniel's perps1, and got a solution 
as
fast asperps2c, WITHOUT compiling. Compiling tripled the
speed again, sotreatC is the fastest solution I've seen
so far.treat[{a_, b_, c_}] := If[a == b == 0, {{1, 0, 0},
{0, 1, 0}}, {{b, -a, 0}, {a*c, b*c, -a^2 -b^2}}]treatC =
Compile[{{v, _Real, 1}}, If[v[[1]] == v[[2]] == 0, {{1, 0,
0}, {0, 1, 0}}, {{v[[2]], -v[[1]], 0}, {v[[1]]*v[[3]],
v[[2]]*v[[3]], -v[[1]]^2 - v[[2]]^2}}]]vecs =
Table[Random[], {10000}, {3}]; Timing[perps2c /@ vecs;
]Timing[helzer /@ vecs; ]Timing[treat /@ vecs;
]Timing[treatC /@ vecs;]{0.2190000000000083*Second,
Null}{2.7339999999999947*Second, Null}{0.25*Second,
Null}{0.07800000000000296*Second, Null}None of these
solutions reliably return normalized vectors.Bobby
Treat-----Original Message-----> pose the problem to
MathGroup. Who has the most elegant Mathematica> routine...>
OrthogonalUnitVectors::usage =
OrthogonalUnitVectors[v:{_,_,_}] willreturn> two unit
vectors orthogonal to each other and to v.> You can assume
that v is nonzero.> David Park> djmp@earthlink.net>
http://home.earthlink.net/~djmp/Some
possibilities:perps1[v_] := If [v[[1]]==v[[2]]==0,
{{1,0,0},{0,1,0}}, {{v[[2]],-v[[1]],0},
Cross[v,{v[[2]],-v[[1]],0}]} ]perps2[v_] :=
With[{vecs=NullSpace[{v}]}, {vecs[[1]], vecs[[2]] -
(vecs[[2]].vecs[[1]])*vecs[[1]]} ]This appears to be 2-3
times faster than perps1 for vectors of machinereals. I get
another factor of 2 using Compile, which is appropriate
fore.g. graphics use.perps2C = Compile[{{v,_Real,1}},
Module[{vecs=NullSpace[{v}]}, {vecs[[1]], vecs[[2]] -
(vecs[[2]].vecs[[1]])*vecs[[1]]} ]]In[61]:= vecs =
Table[Random[], {10000}, {3}];In[62]:= Timing[p2 =
Map[perps1C,vecs];]Out[62]= {0.49 Second, Null}This is on a
1.5 GHz processor.Daniel LichtblauWolfram Research
====
>I
was wondering if there is a method for resizing Raster
graphics>(resizing the actual matrix of pixels, not just the
display size). I>am processing a large number of JPEG images,
and we sometimes need to>reduce the image size to allow data
processing algorithms to function>without running out of
memory. In the past we simply used a program>such as
Photoshop to resize them before importing them
into>Mathematica. Due to the number of images we are
processing now this>is very inconvenient and it would be
very useful if there was a method>for accomplishing it in
Mathematica, but I can't find one. I 
also>thought about doing
something simple like sampling every few pixels or>averaging,
but I thought there might be a method with more efficacy>than
this. I also tried exporting the graphics with the Export
command>as new JPEGs and manipulating the ImageResolution
and ImageSize>options but this seemed to have no effect. Any
help would be much>appreciated.>Aaron UrbasIn 4.2, the
following should work (as long as you define newsize).in =
Import[file.jpg];Export[newfile.jpg, in,
ImageSize->newsize]In 4.1 and earlier, this will not work
because an optimization interferes with the ImageSize and
rasters are written out with the raster size, not the
ImageSize. There is a ConversionOption to force the behavior
you want.in = 
Import[file.jpg];Export[newfile.jpg,
Show[in, ImageSize->newsize],
ConversionOptions->{RasterExport->Graphics}]-Dale
====

Borrowing liberally from Daniel, I like the
following:ClearAll[sumBin, sumBinOdd, sumBinEven,
index]sumBinOdd = Sum[Binomial[2index + 1, k]*p^(2index + 1
- k)*(1 - p)^k, { k, 0, index - 1}];sumBinEven =
Sum[Binomial[2index, k]* p^(2index - k)*(1 - p)^k, {k, 0,
index - 1}];sumBin[n_, Odd] = sumBinOdd /. {index -> (n -
1)/2};sumBin[n_, Even] = sumBinEven /. {index ->
n/2};sumBin[n_?EvenQ] = sumBin[n, Even];sumBin[n_?OddQ] =
sumBin[n, Odd];It allows you to see the solution
symbolically for both odd and even n,and also to calculate
it when n is a known integer. We also have theopportunity,
for instance, to assume that 3x is even and
calculatesumBin[3x, Even]p^(3*x)*((1/p)^(3*x) - ((-4 +
4/p)^((3*x)/2)* Gamma[1/2 + (3*x)/2]*Hypergeometric2F1[1,
-((3*x)/2), 1 + (3*x)/2, (-1 + p)/p])/(Sqrt[Pi]*Gamma[1
+(3*x)/2]))or assume 3x is odd and calculatesumBin[3*x,
Odd]p^(3*x)*((1/p)^(3*x) - (2^(3*x)*(-1 + 1/p)^((1/2)*(-1 +
3*x))* Gamma[3/2 + (1/2)*(-1 + 3*x)]*Hypergeometric2F1[1, -1
+ (1/2)*(1 - 3*x), 1 + (1/2)*(-1 + 3*x), (-1
+p)/p])/(Sqrt[Pi]* Gamma[2 + (1/2)*(-1 + 3*x)]))Bobby
Treat-----Original Message-----> when F[n,p] is an
approximation to the sum and the R is the remainingerror.>
Constantine.>In a message dated 8/28/02 4:44:13 AM,
celster@cs.technion.ac.il>>I'm looking for a way of 
finding
the approximation for partitialbinomial>>sum.>>I'll be
pleasant for any hint..> [...]> Office: Taub 411You can get a
closed form in terms of special functions if you splitinto
two cases depending on whether n is even or odd.In[39]:= n =
2*m;In[40]:= InputForm[Sum[Binomial[n,k]*p^(n-k)*(1-p)^k,
{k,0,m-1}]]Out[40]//InputForm= p^(2*m)*((p^(-1))^(2*m) -
((-4 + 4/p)^m*Gamma[1/2 + m]* Hypergeometric2F1[1, -m, 1 + m,
(-1 + p)/p])/(Sqrt[Pi]*Gamma[1 +m]))In[41]:= n =
2*m+1;In[42]:= InputForm[Sum[Binomial[n,k]*p^(n-k)*(1-p)^k,
{k,0,m-1}]]Out[42]//InputForm= p^(1 + 2*m)*((p^(-1))^(1 +
2*m) - (2^(1 + 2*m)*(-1 + p^(-1))^m*Gamma[3/2+ m]*
Hypergeometric2F1[1, -1 - m, 1 + m, (-1 +
p)/p])/(Sqrt[Pi]*Gamma[2 +m]))Daniel LichtblauWolfram
Research
====
I've found that Mathematica 4.2 takes too long
to finish to generate firsttime help browser. Is 
this normal?
How many time should it take?Calimero
====
Please excuse me
if this has been discussed before; I haven't been monitoring
the group very much.A long time ago I posted to this group as
well as to Wolfram about the errors in many or all functions
involving Fourier transforms when values of
FourierParameters other than the default values are used. The
group responded that indeed there was a problem so I'm
wondering if it was ever fixed. My version (then and now) is
4.0.1 for Macintosh.Jerry
====
Could you help me?I'm to
solve heat conductivity equation (Laplas equation) - partial
differential equation. Are there any ready-to-use packages
that will help me do the job?Standard function DSolve
cannot! And so do functions from package
Calculus`DSolveIntegrals`.More info: the space where the
equation is to be solved is cylindre (not infinite).
Andrew.
====
 When I multiply an expression with 0 it is giving
0.expression whichis creating problem in the further
calculation.eg.In[1]:=func1[r_]:=Exp[-r/2]In[2]:coeff[[1,1
]]=0.0;;;In[34]:=func1[r]coeff[[1,1]]Out[34]:=0.Exp[-r/2]I
want anything to be multiply by zero must be zero. How can I
do that? Your suggestion will be highly appreciated.
Raj
====
> When I multiply an expression with 0 it is giving
0.expression which> is creating problem in the further
calculation.No. Your trouble comes when you multiply an
expression by theßoating-point number 0.0, rather than by
the precise symbolic 0 (having nodecimal point). Merely
make the coefficient 0 (rather than 0.0) andeverything
should work as you wish.David> eg.
In[1]:=func1[r_]:=Exp[-r/2]> In[2]:coeff[[1,1]]=0.0;> ;> ;>
In[34]:=func1[r]coeff[[1,1]] Out[34]:=0.Exp[-r/2] I want
anything to be multiply by zero must be zero. How can I do
that?-- -------------------- http://NewsReader.Com/
-------------------- Usenet Newsgroup ServiceReply-To:
kuska@informatik.uni-leipzig.de
====
Unprotect[Times]Times[0
., __] := 0Protect[Times] Jens> When I multiply an
expression with 0 it is giving 0.expression which> is
creating problem in the further calculation.> eg.>
In[1]:=func1[r_]:=Exp[-r/2]> In[2]:coeff[[1,1]]=0.0;> ;> ;>
In[34]:=func1[r]coeff[[1,1]]> Out[34]:=0.Exp[-r/2]> I want
anything to be multiply by zero must be zero.> How can I do
that?> Your suggestion will be highly appreciated.>
Raj
====
Now I'm trying to calculate this formula:Delta[eq_,
x_]:=Coefficient[eq, x]^2 - 4 Coefficient[eq, 
x^2]
Coefficient[eq, x^0]eq has this form a x^2 + b x + cBut
there is a problem with the x^0 coefficient!How can I
overcome that?CeZaRReply-To:
kuska@informatik.uni-leipzig.de
====
andDelta[eq_,
x_]:=Coefficient[eq, x]^2 - 4 Coefficient[eq,
x,2]Coefficient[eq, x,0]does what you want. Because the
Help-Browser say:Coefficient[expr, form, 0] picks out terms
that are not proportional toform. Jens> Now I'm trying
to calculate this formula:> Delta[eq_, x_]:=Coefficient[eq,
x]^2 - 4 Coefficient[eq, x^2] Coefficient[eq, 
x^0]> eq has
this form a x^2 + b x + c> But there is a problem with the
x^0 coefficient!> How can I overcome that?> CeZaR
====
I
need to fill the space between two contour lines, C1 and C2,
withred color, and leave the other place white. What trick
I have to use?Any suggestion and advice will be
appreciated.Jun LinReply-To:
kuska@informatik.uni-leipzig.de
====
gr = ContourPlot[x^2 +
y^2, {x, -1, 1}, {y, -1, 1}, Contours -> {0.2, 0.4},
ColorFunction -> (If[# >= 0.2 && # <= 0.4, RGBColor[1, 0, 0],
RGBColor[1, 1, 1]] &), ColorFunctionScaling -> False ] Jens>
I need to fill the space between two contour lines, C1 and C2,
with> red color, and leave the other place white. What trick
I have to use?> Any suggestion and advice will be
appreciated.> Jun Lin
====
I am trying to find an example that
will demonstrate the difference between$PrePrint and $Post. I
found an old thread in this news group where auser wanted to
display all matrices using MatrixForm. Some users
suggestedthe following: In[1]:=
$Post=(#/.mtrx_?MatrixQ:>MatrixForm[mtrx]&);Then Dave
Withoff said it's better to assign this to $PrePrint since
theobjective here is to adjust the display rather than the
result of thecalculation. With the assignment to $Post you
could, for example, getunexpected results from calculations
using %, since matrices will be wrappedin
MatrixForm.--------However, if we use $Post above, the next
input will compute the inversethe matrix. I did verify that
Inverse can't take a matrix wrapped inMatrixForm. Can
somebody give an example where doing this with
$PrePrintinstead of $Post gives a different result. In[2]:=
m={{2,3},{0,1}}; Inverse[%] Out[3]= (* Inverse of (m) in
MatrixForm, not shown. *)------ Ted Ersek Get Mathematica
tips, tricks from
http://www.verbeia.com/mathematica/tips/Tricks.htmlReply-To
: kuska@informatik.uni-leipzig.de
====
yo can just
tryIn[]:=$Post = (# /. mtrx_?MatrixQ :> AnyHead[mtrx]
&);In[]:=m = {{2, 3}, {0,
1}}In[]:=q=%;In[]:=Head[q]andIn[]:=$PrePost = (# /.
mtrx_?MatrixQ :> AnyHead[mtrx] &);In[]:=m = {{2, 3}, {0,
1}}In[]:=q=%;In[]:=Head[q]But you are right -- the
behaviour of MatrixForm[] in your exampleis strange. Jens>
I am trying to find an example that will demonstrate the
difference between> $PrePrint and $Post. I found an old
thread in this news group where a> user wanted to display
all matrices using MatrixForm. Some users suggested> the
following:> In[1]:=
$Post=(#/.mtrx_?MatrixQ:>MatrixForm[mtrx]&);> Then Dave
Withoff said it's better to assign this to $PrePrint since
the> objective here is to adjust the display rather than
the result of the> calculation. With the assignment to $Post
you could, for example, get> unexpected results from
calculations using %, since matrices will be wrapped> in
MatrixForm.> --------> However, if we use $Post above, the
next input will compute the inverse> the matrix. I did
verify that Inverse can't take a matrix wrapped in>
MatrixForm. Can somebody give an example where doing this
with $PrePrint> instead of $Post gives a different result.>
In[2]:= m={{2,3},{0,1}};> Inverse[%]> Out[3]= (* Inverse of
(m) in MatrixForm, not shown. *)> ------> Ted Ersek> Get
Mathematica tips, tricks from>
http://www.verbeia.com/mathematica/tips/Tricks.html
====

myArray is a list of triplets which look like {n,k,p}, where
n and kare integers and p is real number. repetitions in n
and k are allowedbut no two triplets ar the same.I define a
function, myfunc[n_,pthreshold_] as
follows.myfunc[n_,pthreshold_] :=
Module[{t},t=Max[Cases[myArray, {n, k_, p_} /; (p >=
pthreshold) -> p, 2]];If[t>=0,t,-1]]It prints the largest k
for which p is greater than pthreshold for agiven n and-1
if there is no such k. The function does its job fine. My
problem is as follows. I would like to plot a collection of
plots,myfunc[n,p], for 1<=n<=21 with respect to p.
unfortunately thefollowing does not work as
desired.Plot[Table[myfunc[n,p],{n,1,21}],{p,0.0,1.0}]
because the Table gets evaluated to numeric value (-1)
before Plot isinvoked.
====
I made a mistake in the
definition of the question. The function is correctly 
defined
as myfunc[n_,pthreshold_] := Module[{t}, t=Max[Cases[myArray,
{n, k_, p_} /; (p >= pthreshold) -> k, 2]];
If[t>=0,t,-1]]Please note that inserting Evaluate before
Table does not workcorrectly either, because evaluating
myfunc[n,p] when p is not anumber, gives -1.>
Plot[Table[myfunc[n,p],{n,1,21}],{p,0.0,1.0}] > because the
Table gets evaluated to numeric value (-1) before Plot is>
invoked.Reply-To: kuska@informatik.uni-leipzig.de
====
could
you supply a complete example next time ?Try myfunc[] with a
more restrictive pattern and it works:myArray =
Flatten[Table[{n, 1, Random[]}, {15}, {n, 1, 21}],
1];myfunc[n_, pthreshold_?NumericQ] := Module[{t}, t =
Max[Cases[myArray, {n, k_, p_} /; (p >= pthreshold) -> p,
2]]; If[t >= 0, t, -1]]Plot[Evaluate[Table[myfunc[n, p], {n,
1, 21}]], {p, 0.0, 1.0}] Jens> myArray is a list of triplets
which look like {n,k,p}, where n and k> are integers and p
is real number. repetitions in n and k are allowed> but no
two triplets ar the same.> I define a function,
myfunc[n_,pthreshold_] as follows.> myfunc[n_,pthreshold_] :=
Module[{t},> t=Max[Cases[myArray, {n, k_, p_} /; (p >=
pthreshold) -> p, 2]];> If[t>=0,t,-1]]> It prints the largest
k for which p is greater than pthreshold for a> given n and>
-1 if there is no such k. The function does its job fine.> My
problem is as follows. I would like to plot a collection of
plots,> myfunc[n,p], for 1<=n<=21 with respect to p.
unfortunately the> following does not work as desired.>
Plot[Table[myfunc[n,p],{n,1,21}],{p,0.0,1.0}]> because the
Table gets evaluated to numeric value (-1) before Plot is>
invoked.
====
For my PDE class we have been calculating
Fourier transforms. The instructorarrived today with a
printout of two plots of a certain Fourier transform,done
with a different CAS. The first plot was to 30 terms, the
second was to120 terms. Curious, I translated the functions
into Mathematica (4.0 on Windows2000on a PIII 700) to see
how much time this required to process. I wasStaggered at
how much time it took. Here's the code:L = 2;f[x_] :=
UnitStep[x - 1];b[n_] := (2/L)*Integrate[f[x]*Sin[n*Pi*x/L],
{x, 0, L}];FS[N_, x_] := Sum[b[n]*Sin[n*Pi*x/L], {n, 1,
N}];Timing[Plot[FS[30, x], {x, 0, 2}]]Out[23]={419.713
Second,
[SkeletonIndicator]Graphics[SkeletonIndicator]}In this
case the number of terms is 30.The time required per number
of terms seems to fit the following polynomial:y = 0.3926x^2
+ 2.2379xThis is a large amount of time. I understand that
the code is not optimized,and was more or less copied from
the code in the other CAS, but is this areasonable amount
of time, or is something going wrong? I don't use
Mathematicabecause of the speed, but should it be this
slow?Just curious,Steve Story
====
> I don't use
Mathematica> because of the speed, but should it be this
slow?Mathematica has become more competitive in the speed
department in recentyears. See for example the attached
comparison (not sent to newsgroup) byStephan Steinhaus
(steinhaus-net.de). So when Mathematica takes a very
longtime, you should investigate. In this case inserting
Evaluate[] in twoplacesIn[91]:=b[n_] :=
Evaluate[(2/L)*Integrate[f[x]*Sin[n*Pi*x/L], {x, 0,
L}]];....In[104]:=Timing[Plot[Evaluate[FS[120, x]], {x, 0,
2}]]Out[104]={0.18
Second,[SkeletonIndicator]Graphics[SkeletonIndicator]}
speeds the process enormously (18 milliseconds to plot 120
terms on myfeeble old 500MHz PowerBook).Why was it so slow
before? When I switch from an ordinary numerical languageto
Mathematica, I enter into an implicit bargain
withMathematica: the software will go the extra mile to get
me a good answer,including (1) using exra precision
(sometimes without being asked) and (2)carrying around
unevaluated mathematical expressions (usually without
beingasked) that could possibly be evaluated more
appropriately at a later time.Most tools cannot do either of
these things, so I don't have to worry aboutit, except for
the bad answers that result now and then. But I need to
takecare that Mathematica does not burden itself
unnecessarily. That's my sideof the bargain.Number (2) is
the issue here. Your definition of b[n] is written so
thatMathematica analytically evaluates b separately for
each n. But you know inthis case that the integration can
be done safely once for all n. So do it!The huge difference,
though, comes from pre-evaluating the argument to Plot.Read
the on-line help! You should pre-evaluate where possible. In
somecases, the most common of which involve branching
within the definition offunction to plot, you cannot
pre-evaluate so, in keeping with the bargain,Mathematica
goes the extra mile and holds back just in case. You need
tosteer it into the shortcut when it's OK.Hope this
helps,Tom Burton-- Reply-To:
kuska@informatik.uni-leipzig.de
====
with your code you
compute the expansion coefficents everytime when FS[] is
evaluated. Store the values for b[n] withL = 2;f[x_] :=
UnitStep[x - 1];b[n_] := b[n] =
(2/L)*Integrate[f[x]*Sin[n*Pi*x/L], {x, 0, L}];FS[N_, x_] :=
Sum[b[n]*Sin[n*Pi*x/L], {n, 1, N}];Timing[Plot[FS[30, x], {x,
0, 2}]]and you need only a 1-3 seconds (depending on your
machine) Jens> For my PDE class we have been calculating
Fourier transforms. The instructor> arrived today with a
printout of two plots of a certain Fourier transform,> done
with a different CAS. The first plot was to 30 terms, the
second was to> 120 terms.> Curious, I translated the
functions into Mathematica (4.0 on Windows2000> on a PIII
700) to see how much time this required to process. I was>
Staggered at how much time it took. Here's the code:> L = 
2;>
f[x_] := UnitStep[x - 1];> b[n_] :=
(2/L)*Integrate[f[x]*Sin[n*Pi*x/L], {x, 0, L}];> FS[N_, x_]
:= Sum[b[n]*Sin[n*Pi*x/L], {n, 1, N}];> Timing[Plot[FS[30,
x], {x, 0, 2}]]> Out[23]=> {419.713 Second,
[SkeletonIndicator]Graphics[SkeletonIndicator]}> In this
case the number of terms is 30.> The time required per number
of terms seems to fit the following polynomial:> y = 0.3926x^2
+ 2.2379x> This is a large amount of time. I understand that
the code is not optimized,> and was more or less copied from
the code in the other CAS, but is this a> reasonable amount
of time, or is something going wrong? I don't use
Mathematica> because of the speed, but should it be this
slow?> Just curious,> Steve Story
====
I would like to use
mathematica type papers for my math courses, butI'm having
trouble formatting documents. Despite searching, I've
beenunable to find a complete guide to word processing with
mathematica.Does anyone know where such a document could be
found?
====
Look at in the Help browser or in the Mathematica
Book Style Sheet Also, you have the documentation
included for package Author Tools(only in Mathematica
4.2).Guillermo Sanchez > I would like to use mathematica
type papers for my math courses, but> I'm having trouble
formatting documents. Despite searching, I've been> unable
to find a complete guide to word processing with mathematica.>
Does anyone know where such a document could be
found?
====
Kenny,Sympathy but no solution.I too have been
trying to use Mathematica (v4.2 most recently) to typemaths
papers and the like but I'm not ready to ditch LaTeX yet.
Thereare just too many cases where I cannot figure out how
to achieve what Iwant in Mathematica, things like:- left
brackets spanning multiple lines for defining hybrid
functions;- vertical alignment of equals signs in multi-line
equations orderivations;- setting typefaces in tables of
material.I figure most of this is do-able, but I 
don't have
the time, orpatience, to spend too much time on it. So,
I'll be the first customerwhen you write the 
guide to math
publishing in Mathematica - I just hopeyou won't have to
use LaTeX to write it.Mark Westwood> I would like to use
mathematica type papers for my math courses, but> I'm having
trouble formatting documents. Despite searching, I've been>
unable to find a complete guide to word processing with
mathematica.> Does anyone know where such a document could be
found?
====
OrthogonalUnitVectors that I sent a few minutes ago
doesn't do what Ithought it would.After making either
definition below I get the following:In[2]:=
s1=OrthogonalUnitVectors[{1,0,1/2,1,0},{0,1,-1,1/2,2}]Out[2]
= {{0, -2/Sqrt[5], 0, 0, 1/Sqrt[5]}, {-2/3, -1/3, 0, 2/3, 0},
{-1/3, 2/3,2/3, 0, 0}}The dot products below aren't zero, so
the vectors aren't orthogonal. Whatwent wrong?In[3]:=
Part[s1,1].Part[s1,2]Out[3]= 2/(3*Sqrt[5])In[4]:=
Part[s1,1].Part[s1,3]Out[3]= -4/(3*Sqrt[5])-------------->
Hugh Goyder and David Park gave a most elegant function to
find two> vectors that are orthogonal to one vector in 3D.
The key to coming up> with the elegant solution is an
understanding of Mathematica's NullSpace> function. We can
easily make the version from Hugh and David much more>
general with the version below.> ------------->
OrthogonalUnitVectors[vect__?VectorQ]:= >
#/Sqrt[#.#]&/@NullSpace[{vect}]> -------------> The version
above will give a set of unit orthogonal vectors if given
any> number of vectors in any dimension. > So besides giving
it a 3D vector we can give it the following:>
OrthogonalUnitVectors[{2,1,0,-1,1}]> or>
OrthogonalUnitVectors[{0,1,0,1/2,1},{1,0,-1,1/2}]>
------------> But the short version above isn't very 
robust.>
(1) Clear[x,y,z];NullSpace[{{x,y,z}}]> returns two vectors
orthogonal to {x,y,z}, but the two vectors> NullSpace
returns aren't orthogonal to each other. > So
(OrthogonalUnitVectors) should only work with numeric
vectors.> (2) We should ensure all the vectors have the same
dimension and length>1.> I give a less concise version below
that corrects these problems.> ------------>
OrthogonalUnitVectors[vect__?(VectorQ[#,NumericQ]&)]/;>
(SameQ@@Length/@{vect})&&(Length[First[{vect}]]>1):= >
#/Sqrt[#.#]&/@NullSpace[{vect}]> --------------> Ted Ersek>
Get Mathematica tips, tricks from>
http://www.verbeia.com/mathematica/tips/Tricks.html>

====
Daniel Lichtblau has pointed out that NullSpace does not
generally giveorthogonal vectors. Therefore the routines
that depended upon that were inerror. He says that it does
give orthogonal vectors when the input vectorcontains
approximate numbers. For graphical purposes this will be
goodenough for me. Therefore I modify Ted's routine
toOrthogonalUnitVectors[vect__?(VectorQ[#, NumericQ] &)] /;
(SameQ @@ Length /@ {vect}) && (Length[First[{vect}]] > 1) :=
#/Sqrt[#.#] & /@ NullSpace[{vect}// N]and the short version
for 3D vectorsOrthogonalUnitVectors[v : {_, _, _}] :=
#/Sqrt[#.#] & /@ NullSpace[{v//N}]For exact vectors I might
use for 3DOrthogonalUnitVectors[v : {_, _, _}] :=
#/Sqrt[#.#] & /@ {temp = First[NullSpace[{v}]],
v[Cross]temp}I'm still looking for something that is easy
to remember.David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/
The version above will give a set of unit orthogonal vectors
if given anynumber of vectors in any dimension.So besides
giving it a 3D vector we can give it the following:
OrthogonalUnitVectors[{2,1,0,-1,1}] or
OrthogonalUnitVectors[{0,1,0,1/2,1},{1,0,-1,1/2}]------------
But the short version above isn't very robust.(1)
Clear[x,y,z];NullSpace[{{x,y,z}}] returns two vectors
orthogonal to {x,y,z}, but the two vectorsNullSpace returns
aren't orthogonal to each other. So (OrthogonalUnitVectors)
should only work with numeric vectors.(2) We should ensure
all the vectors have the same dimension and length >1.I
give a less concise version below that corrects these
problems.------------OrthogonalUnitVectors[vect__?(VectorQ[#
,NumericQ]&)]/;
(SameQ@@Length/@{vect})&&(Length[First[{vect}]]>1):=
#/Sqrt[#.#]&/@NullSpace[{vect}]-------------- Ted Ersek Get
Mathematica tips, tricks from
http://www.verbeia.com/mathematica/tips/Tricks.html
====
 I am
unsure if my messages are making it through to the usegroup so
I will try again. I have several differential equations I
would like to plot in Mathematica. I would like to plot the
Slope Fields of them though. Can anyone lead me in the right
direction? I can solve the equations trivially but I want to
display the slope fields. An example follows :y' 
+ 2y =
3-John
====
John,You can do it from scratch with
PlotVectorField from the Graphics`PlotField` package, but
you make make your life easier (and get prettier plots) by
using my DEGraphics package, which can be found at
MathSource (it's part of the DiffEqs suite of packages)
or
here:http://www.math.armstrong.edu/faculty/hollis/mmade/
DiffEqs---Selwyn Hollis> I am unsure if my messages are
making it through to the usegroup so I > will try again. I
have several differential equations I would like to > plot in
Mathematica. I would like to plot the Slope Fields of them >
though. Can anyone lead me in the right direction? I can
solve the > equations trivially but I want to display the
slope fields. An example > follows :> y' + 2y = 
3> -John>
Reply-To:
kuska@informatik.uni-leipzig.de
====
Needs[Graphics`
PlotField`]PlotVectorField[{x, 3 - 2 y}, {x, 0, 4}, {y, -1,
4}, Axes -> True]??? Jens> I am unsure if my messages are
making it through to the usegroup so I> will try again. I
have several differential equations I would like to> plot in
Mathematica. I would like to plot the Slope Fields of them>
though. Can anyone lead me in the right direction? I can
solve the> equations trivially but I want to display the
slope fields. An example> follows :> y' + 2y = 
3>
-John
====
You can proceed like this :delta[poly2_, var_] :=
Coefficient[poly2, var, 1]^2 - 
4*Coefficient[poly2,var,
0]*Coefficient[poly2, var, 2]Meilleures salutationsFlorian
Jaccard-----Message d'origine-----Envoy.8e : ven., 6.
septembre 2002 09:17è : mathgroup@smc.vnet.netObjet :
Coefficient problemNow I'm trying to calculate 
this
formula:Delta[eq_, x_]:=Coefficient[eq, x]^2 - 4
Coefficient[eq, x^2]Coefficient[eq, x^0]eq has 
this form a
x^2 + b x + cBut there is a problem with the x^0
coefficient!How can I overcome that?CeZaR
====
This
works:eq = a x^2 + b x + cdiscriminant[eq_, x_] :=
Coefficient[eq, x]^2 - 4 Coefficient[eq, x, 2] 
Coefficient[eq,
x, 0]discriminant[eq, x]x^0 is reduced to 1 and the
Coefficient of 1 doesn't make sense 
toMathematica, because
it depends on what the variable is (a, b, c, orx?). So, the
other form of the Coefficient call is needed. I used itfor
the second power too, but that wasn't necessary. I think 
that
formis best, though, since it allows no ambiguity.I renamed
the function because that's what the quantity is often
called,for a quadratic.Bobby Treat-----Original
Message-----CeZaR
====
>-----Original Message----->Sent:
Friday, September 06, 2002 9:17 AM>myArray is a list of
triplets which look like {n,k,p}, where n and k>are integers
and p is real number. repetitions in n and k are allowed>but
no two triplets ar the same.>>I define a function,
myfunc[n_,pthreshold_] as follows.>>myfunc[n_,pthreshold_] :=
Module[{t},>t=Max[Cases[myArray, {n, k_, p_} /; (p >=
pthreshold) -> p, 2]];>If[t>=0,t,-1]]>>It prints the largest
k for which p is greater than pthreshold for a>given n
and>-1 if there is no such k. The function does its job fine.
>>My problem is as follows. I would like to plot a collection
of plots,>myfunc[n,p], for 1<=n<=21 with respect to p.
unfortunately the>following does not work as
desired.>>Plot[Table[myfunc[n,p],{n,1,21}],{p,0.0,1.0}]
>>because the Table gets evaluated to numeric value (-1)
before Plot is>invoked.>The problem with your calculation
indented is (1) to evaluate Table (withinPlot), which gives
{myfunc[1, p], ..., myfunc[21, p]}but (2) *then* to prevent
further evaluation of myfunc[.., p] to -1. Thereare several
tricks to do so, e.g. ...Hold[Plot[toPlot, {p, 0, 1}]] /.
toPlot -> Table[headPlaceholder[n, p], {n, 3}] /.
headPlaceholder -> myfunc // ReleaseHold...but in your case
just simply prevent evaluation of myfunc[.., p] with pbeing
a Symbol by defining:myfunc2[n_, pthreshold_?NumericQ] :=
...now...Plot[Evaluate[Table[myfunc2[n, p], {n, 21}]], {p,
0, 1}]...works as expected.BTW, what do you want to read off
the plot, that you didn't know fromTable[myfunc[n, 0], {n,
21}] ?--Hartmut Wolf
====
>-----Original Message----->Sent:
Friday, September 06, 2002 9:17 AM When I multiply an
expression with 0 it is giving 0.expression which>is
creating problem in the further
calculation.>>eg.>>In[1]:=func1[r_]:=Exp[-r/2]>In[2]:coeff[[
1,1]]=0.0;>;>;>In[34]:=func1[r]coeff[[1,1]]>>Out[34]:=0.Exp[-
r/2]>>I want anything to be multiply by zero must be zero.
How can I do that? >Your suggestion will be highly
appreciated.> Raj>Raj,this simply is, because 0. Exp[-r/2]
is not always zero! func1[-Infinity]coeff[[1,
1]]Infinity::indet: Indeterminate expression 0. 
Infinity
encountered.Out[90]= Indeterminate--Hartmut Wolf
====
I
have a mathematica notebook showing using the alternating
direction implicit method for solving heat conduction in 2D
via finite difference methods. Moving from 2D to 3D is pretty
simple although I can't find my notebook on this 
anymore :( .
The 3D method is sometimes called Brian's method.
http://mid-ohio.mse.berkeley.edu/scott/projects/index.html
Scott> Could you help me?> I'm to solve heat conductivity
equation (Laplas > equation) - partial differential equation.
Are there > any ready-to-use packages that will help me do the
job?> Standard function DSolve cannot! And so do functions
> from package Calculus`DSolveIntegrals`.> More info: the
space where the equation is to be > solved is cylindre (not
infinite).> Andrew.> 
====
 I am working with Mathematica 4.0,
and I would like pass my data todo one graphic in Origin
5.0. How can I do this in more simple manner?
====
> I need to
fill the space between two contour lines, C1 and C2, with>
red color, and leave the other place white. What trick I have
to use?> Any suggestion and advice will be appreciated.Over
the years I have alighted upon the following scheme. Suppose
I want tocolor only between contour levels 1 and
2:In[84]:=Needs[Graphics`Colors`]In[88]:=ContourPlot[x,
{x, -1, 3}, {y, 0, 1}, Contours -> {0, 1, 2, 3, 4},
ColorFunction -> (If[1 < # < 2, Red, White] & ),
ColorFunctionScaling -> False];1. Specify the specific contour
levels instead of specifying only the count.2. Disable
ColorFunctionScaling so the argument to ColorFunction
correspondsto the contour levels.Hope this helps,Tom
Burton
====
David Park replied with----------------Daniel
Lichtblau has pointed out that NullSpace does not generally
giveorthogonal vectors. Therefore the routines that
depended upon that were inerror. He says that it does give
orthogonal vectors when the input vectorcontains
approximate numbers. For graphical purposes this will be
goodenough for me. Therefore I modify Ted's routine
toOrthogonalUnitVectors[vect__?(VectorQ[#, NumericQ] &)] /;
(SameQ @@ Length /@ {vect}) && (Length[First[{vect}]] > 1) :=
#/Sqrt[#.#] & /@ NullSpace[{vect}// N]----------------Lets
see what NullSpace does with approximate complex
vectors.In[1]:= v1 = {1.0 I, 0.0, 0.5 I, 0.0, 1.0}; v2 =
{0.0, 2.0, 1.0 I, 2.0, 0.5}; {v3,v4,v5} =
NullSpace[{v1,v2}]Out[3]= {{-0.730153 + 0.*I, 0. -
0.138254*I, 0.250585 + 0.*I, 0. - 0.138254*I, 0.+
0.60486*I}, {0. + 0.*I, -0.515861 + 0.*I, 0. + 0.457321*I,
0.687357 + 0.*I, 0.22866+ 0.*I}, {0. + 0.*I, 0.510406 +
0.*I, 0. + 0.740442*I, -0.23274 + 0.*I, 0.370221+
0.*I}}--------In the next line we see NullSpace returned
vectors that are orthogonal tothe vectors we gave
NullSpace.In[4]:= {v1.v3, v1.v4, v1.v5, v2.v3, v2.v4,
v2.v5}//ChopOut[4]= {0, 0, 0, 0, 0, 0}----------However,
the vectors returned aren't orthogonal to each other.In[5]:=
{v3.v4, v3.v5, v4.v5}//ChopOut[5]= {0.229195*I, 0.371087*I,
-0.677239}---------I suppose an OrthogonalUnitVectors
function that uses NullSpace should (1) Only accept real
valued vectors. (2) Ensure NullSpace is given approximate
vectors.------ Ted
Ersek
====
>OrthogonalUnitVectors[vect__?(VectorQ[#, NumericQ]
&)] /;> (SameQ @@ Length /@ {vect}) && (Length[First[{vect}]]
> 1) :=> #/Sqrt[#.#] & /@ NullSpace[{vect}//
N]---------------->>Lets see what NullSpace does with
approximate complex vectors.>>In[1]:=> v1 = {1.0 I, 0.0, 0.5
I, 0.0, 1.0};> v2 = {0.0, 2.0, 1.0 I, 2.0, 0.5};> {v3,v4,v5}
= NullSpace[{v1,v2}]>>Out[3]=> {{-0.730153 + 0.*I, 0. -
0.138254*I, 0.250585 + 0.*I, 0. - 0.138254*I,>0.>+
0.60486*I}, > {0. + 0.*I, -0.515861 + 0.*I, 0. + 0.457321*I,
0.687357 + 0.*I, 0.22866>+ 0.*I}, > {0. + 0.*I, 0.510406 +
0.*I, 0. + 0.740442*I, -0.23274 + 0.*I, 0.370221>+
0.*I}}-------->In the next line we see NullSpace returned
vectors that are orthogonal to>the vectors we gave
NullSpace.>>In[4]:=> {v1.v3, v1.v4, v1.v5, v2.v3, v2.v4,
v2.v5}//Chop>>Out[4]=> {0, 0, 0, 0, 0, 0}---------->However,
the vectors returned aren't orthogonal to each
other.>>In[5]:=> {v3.v4, v3.v5, v4.v5}//Chop>>Out[5]=>
{0.229195*I, 0.371087*I, -0.677239}--------->I suppose an
OrthogonalUnitVectors function that uses NullSpace should >
(1) Only accept real valued vectors. > (2) Ensure NullSpace
is given approximate vectors.------> Ted ErsekI think you
will find that the output vectors are orthogonal if you use
thecomplex conjugate. for example v4.Conjugate[v5] is
zero.Dennis Wangsness
====
>I need to fill the space between
two contour lines, C1 and C2, with>red color, and leave the
other place white. What trick I have to use?>Any suggestion
and advice will be
appreciated.Needs[Graphics`FilledPlot`];Needs[Graphics
`Colors`];FilledPlot[{4 - x^2, 3x}, {x, -4, 1}, Fills ->
Red];Bob Hanlon
====
>For my PDE class we have been
calculating Fourier transforms. The instructor>arrived
today with a printout of two plots of a certain Fourier
transform,>done with a different CAS. The first plot was to 30
terms, the second was>to>120 terms.> Curious, I translated
the functions into Mathematica (4.0 on Windows2000>on a PIII
700) to see how much time this required to process. I
was>Staggered at how much time it took. Here's the code:>>L
= 2;>f[x_] := UnitStep[x - 1];>b[n_] :=
(2/L)*Integrate[f[x]*Sin[n*Pi*x/L], {x, 0, L}];>>FS[N_, x_]
:= Sum[b[n]*Sin[n*Pi*x/L], {n, 1, N}];>>Timing[Plot[FS[30,
x], {x, 0, 2}]]>>Out[23]=>{419.713 Second,
[SkeletonIndicator]Graphics[SkeletonIndicator]}>>In this
case the number of terms is 30.>>The time required per number
of terms seems to fit the following polynomial:>>y = 0.3926x^2
+ 2.2379x>>This is a large amount of time. I understand that
the code is not optimized,>and was more or less copied from
the code in the other CAS, but is this>a>reasonable amount
of time, or is something going wrong? I don't use
Mathematica>because of the speed, but should it be this
slow?>>Just curious,Your definition of b recalculates the
integral for every call. To evaluate the integral
onceinclude Evaluate.Clear[f, b, FS];L = 2;f[x_] :=
UnitStep[x - 1];b[n_] := Evaluate[
(2/L)*Integrate[f[x]*Sin[n*Pi*x/L], {x, 0, L}]];In the case
of FS it is best to wait for an integer value of N prior to
performing the Sum.FS[N_Integer, x_] :=
Sum[b[n]*Sin[n*Pi*x/L], {n, 1, N}];Now Evaluate the argument
of the Plot to cause the Sum to be done
once.Timing[Plot[Evaluate[FS[120, x]], {x, 0,
2}]][[1]]0.366667 SecondWhile my computer may be faster
than yours, this result for N=120 is 1000 times faster than
your result for N=30.Bob Hanlon
====
> Now I'm trying to
calculate this formula:> Delta[eq_, x_]:=Coefficient[eq, x]^2
- 4 Coefficient[eq, x^2] Coefficient[eq, x^0]> eq 
has this
form a x^2 + b x + c> But there is a problem with the x^0
coefficient!> How can I overcome that?> 
CeZaRCoefficient
cannot figure out who is and is not a variable when
avariable of 1 is specified. To work around this you might
instead dodelta1[poly_, x_] := Coefficient[poly,x]^2 -
4*Coefficient[poly,x,2]*Coefficient[poly,x,0]I 
prefer instead
to use CoefficientList:delta2[poly_, x_] := (#[[2]]^2 -
4*#[[1]]*#[[3]])&[CoefficientList[poly,x]]In[25]:= poly =
a*x^2+b*x+c;In[26]:= delta1[poly,x] ===
delta2[poly,x]Out[26]= TrueDaniel LichtblauWolfram
Research
====
>>Delta[eq_, x_]:=Coefficient[eq, x]^2 - 4
Coefficient[eq, x^2] Coefficient[eq,>x^0]>>eq has 
this form a
x^2 + b x + c>>But there is a problem with the x^0
coefficient!>How can I overcome that?see on-line help for
Coefficientdelta[eq_, x_] := Coefficient[eq, x, 
1]^2 - 4
*Coefficient[eq, x, 2] *Coefficient[eq, x, 0];eq = 
a*x^2 +
b*x + c;delta[eq, x]b^2 - 4*a*cBob Hanlon
====
Would
someone with a very fast machine and lots of memory be
willing to try this Solve for me?It is the inverse of the
20 node quadratic hexahedral mapping used in finite element
analysis.None of my computers can handle this - they run out
of memory (using (*Hex20 Node definition in global coordinates
*)Clear[x1, y1, z1,x2, y2, z2,x3, y3, z3,x4, y4, z4,x5,
y5, z5,x6, y6, z6,x7, y7, z7,x8, y8, z8,x9, y9, z9,x10,
y10, z10,x11, y11, z11,x12, y12, z12,x13, y13, z13,x14,
y14, z14,x15, y15, z15,x16, y16, z16,x17, y17, z17,x18,
y18, z18,x19, y19, z19,x20, y20, z20];(* local coordinates
*)Clear[u, v, w];(* Global co-ordinates *)Clear[x, y, z];(*
corner nodes *)N1= (1-u)*(1-v)*(1-w)*(-2-u-v-w)/8;N3=
(1+u)*(1-v)*(1-w)*(-2+u-v-w)/8;N5=
(1+u)*(1+v)*(1-w)*(-2+u+v-w)/8;N7=
(1-u)*(1+v)*(1-w)*(-2-u+v-w)/8;N13=(1-u)*(1-v)*(1+w)*(-2-u-v
+w)/8;N15=(1+u)*(1-v)*(1+w)*(-2+u-v+w)/8;N17=(1+u)*(1+v)*(1
+w)*(-2+u+v+w)/8;N19=(1-u)*(1+v)*(1+w)*(-2-u+v+w)/8;(* to u
nodes *)N2= (1-u^2)*(1-v)*(1-w)/4;N6=
(1-u^2)*(1+v)*(1-w)/4;N14=(1-u^2)*(1-v)*(1+w)/4;N18=(1-u^2)
*(1+v)*(1+w)/4;(* to v nodes *)N4=
(1+u)*(1-v^2)*(1-w)/4;N8=
(1-u)*(1-v^2)*(1-w)/4;N16=(1+u)*(1-v^2)*(1+w)/4;N20=(1-u)*(
1-v^2)*(1+w)/4;(* to w nodes *)N9=
(1-u)*(1-v)*(1-w^2)/4;N10=(1+u)*(1-v)*(1-w^2)/4;N11=(1+u)*(
1+v)*(1-w^2)/4;N12=(1-u)*(1-v)*(1-w^2)/4;(* solve the
inverse transform
*)Solve[{x1*N1+x2*N2+x3*N3+x4*N4+x5*N5+x6*N6+x7*N7+x8*N8+x9
*N9+x10*N10+x11*N11+x12*N12+x13*N13+x14*N14+x15*N15+x16*N16+
x17*N17+x18*N18+x19*N19+x20*N20-x==0,y1*N1+y2*N2+y3*N3+y4*
N4+y5*N5+y6*N6+y7*N7+y8*N8+y9*N9+y10*N10+y11*N11+y12*N12+y13
*N13+y14*N14+y15*N15+y16*N16+y17*N17+y18*N18+y19*N19+y20*N20
-y==0,z1*N1+z2*N2+z3*N3+z4*N4+z5*N5+z6*N6+z7*N7+z8*N8+z9*N9+
z10*N10+z11*N11+z12*N12+z13*N13+z14*N14+z15*N15+z16*N16+z17*
N17+z18*N18+z19*N19+z20*N20-z==0},{u,v,w}]Christopher J.
PurcellDefence R&D Canada .9a Atlantic9 Grove St., PO Box
1012Dartmouth NS Canada B2Y
3Z7
====
Needs[Graphics`Colors`];ContourPlot[Sin[x y], {x,
-5, 5}, {y, -5, 5}, ColorFunction -> (If[0.5 < # < 0.7, Red,
White] &)];Bob Hanlon>ContourPlot?>Jun Lin > In a message
dated 9/6/02 3:53:58 AM,>I need to fill the space between two
contour lines,> C1 and C2, with>red color, and leave the
other place white. What> trick I have to use?>Any suggestion
and advice will be appreciated.>
Needs[Graphics`FilledPlot`];> Needs[Graphics`Colors`];>
FilledPlot[{4 - x^2, 3x}, {x, -4, 1}, Fills -> Red];
====
Jun
Lin,Here is an example.Needs[Graphics`Colors`]Let's
make a contour plot of this function.f[x_, y_] :=
Sin[x]Sin[2y]Let's specify the exact contours to use. I got
rid of the 0. contour becauseit is difficult to obtain in
this plot.contourvalues = Complement[Range[-1, 1, 0.2],
{0.}]{-1, -0.8, -0.6, -0.4, -0.2, 0.2, 0.4, 0.6, 0.8, 1.}Now
we define a ColorFunction for the plot. I actually colored two
differentbands to show how you can make a general color
function to give each band adesired color.cfun[z_] :=
Which[ -0.6 < z < -0.42, RoyalBlue, 0.4 < z < 0.6, Red, True,
White]ContourPlot[f[x, y], {x, 0, Pi}, {y, 0, Pi}, PlotPoints
-> 30, ColorFunctionScaling -> False, ColorFunction -> cfun,
Contours -> contourvalues];Using the option
ColorFunctionScaling -> False says that the z value will
bethe actual value of f[x,y]. Otherwise, in general, it
will be scaled between0 and 1. It is easier to write a
color function when z is the actual valueof the
function.David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/
Sender: steve@smc.vnet.netApproved: Steven M. Christensen
<steve@smc.vnet.net>, ModeratorReply-To:
<drbob@bigfoot.com>
====
I checked again the six solutions I
had previously timed, and they DOgive orthogonal results.
(None of them depend on NullSpace for that.)By the way, I
reused my combinations function (from a recent problemon
adding fractions to get 1) to check for orthogonality:
ClearAll[orthogonalQ] orthogonalQ[v : {__?VectorQ}] := And @@
(Chop@(Dot @@ #) == 0 & /@ combinations[v, {2}]) <<
DiscreteMath`Combinatorica`; ClearAll[combinations]; r =
Range[1, 9]; combinations::usage =
combinations[list,n:{__Integer}] lists the combinations of
list taken n at a time; combinations[r_List, n_Integer, {}]
:= If[n > Length@r, {},
DiscreteMath`Combinatorica`KSubsets[r, n]];
combinations[r_List, n_Integer, e_?VectorQ] := Join[e, #] &
/@ DiscreteMath`Combinatorica`KSubsets[Complement[r, e],
n]; combinations[r_List, n_Integer, e : {__?VectorQ}] :=
Flatten[ combinations[r, n, #] & /@ e, 1];
combinations[r_List, n : {__Integer}] := Which[Plus @@ n
==Length@r, Join[#, Complement[r, #]] & /@
combinations[r, Drop[n, -1]], Plus @@ n > Length@r, {}, True,
Fold[ combinations[r, #2, #1] &, {}, n]]Bobby-----Original
Message----- (SameQ @@ Length /@ {vect}) &&
(Length[First[{vect}]] > 1) := #/Sqrt[#.#] & /@
NullSpace[{vect}// N]and the short version for 3D
vectorsOrthogonalUnitVectors[v : {_, _, _}] := #/Sqrt[#.#]
& /@NullSpace[{v//N}]For exact vectors I might use for
3DOrthogonalUnitVectors[v : {_, _, _}] := #/Sqrt[#.#] & /@
{temp = First[NullSpace[{v}]], v[Cross]temp}I'm still
looking for something that is easy to remember.David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/
OrthogonalUnitVectors[vect__?VectorQ]:=
#/Sqrt[#.#]&/@NullSpace[{vect}]-------------The version
above will give a set of unit orthogonal vectors if
givenanynumber of vectors in any dimension.So besides
giving it a 3D vector we can give it the following:
OrthogonalUnitVectors[{2,1,0,-1,1}] or
OrthogonalUnitVectors[{0,1,0,1/2,1},{1,0,-1,1/2}]------------
But the short version above isn't very robust.(1)
Clear[x,y,z];NullSpace[{{x,y,z}}] returns two vectors
orthogonal to {x,y,z}, but the two vectorsNullSpace returns
aren't orthogonal to each other. So (OrthogonalUnitVectors)
should only work with numeric vectors.(2) We should ensure
all the vectors have the same dimension and length>1.I give
a less concise version below that corrects these
problems.------------OrthogonalUnitVectors[vect__?(VectorQ[#
,NumericQ]&)]/;
(SameQ@@Length/@{vect})&&(Length[First[{vect}]]>1):=
#/Sqrt[#.#]&/@NullSpace[{vect}]-------------- Ted Ersek Get
Mathematica tips, tricks from
http://www.verbeia.com/mathematica/tips/Tricks.html
====
I
have substituted the letters SS, CC, X, Y and Z1 and Z2 for
some complicated expressions just to illustrate the form of
the function. This function works and all the conditions are
necessary but I am sure a more elegant programming solution
perhaps using While could be found. Any f[{Sa_, Ca_, Aa_,
Sb_, Cb_, Ab_, a_, b_}] := {If[Aa == Ab, a Sa + b Sb, If[Ca
== 0 || Cb == 0, a Sa + b Sb, SS]], If[Aa == Ab, a Ca + b Cb,
If[Ca == 0 || Cb == 0, a Ca + b Cb, CC]], If[Ca == 0 && Cb ==
0, Nil, If[Z1 == Z2, If[Ca < 0, Aa, If[Cb < 0, Ab, If[Ca
> 0, If[Aa > 90, Aa - 90, Aa + 90], If[Cb > 0, If[Ab > 90, Ab
- 90, Ab + 90]]]]], If[Aa == Ab, Aa, If[Ca == 0 && Cb == 0,
Nil, If[Ca == 0, Ab, If[Cb == 0, Aa, If[Y == 0, Nil,
If[X > 180, X - 180, If[X < 0, X + 180, X]]]]]]]]]}
====
Check
out the Help for Coefficient.Coefficient[expr, 
form, 0] picks
out terms that are not proportional toform.Delta[eq_, x_]
:= Coefficient[eq, x]^2 - 4 Coefficient[eq, x^2] 
Coefficient[eq,
x, 0]Delta[a x^2 + b x + c, x]b^2 - 4*a*cDavid
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/===
=Dear all,1.) How to find the General Solution for 
below's
partial differential equation? (y + u) du/dx + y (du/dy) = x
- y** I use d to represent the partial differential
symbol.Can it be solved by function NDSolve in mathematica
4.1? How?2.) I manage to get the roots of complex equation
z^5 = i fromSolve[z^5 == i, z]. It gave me straight the 5
roots in the output.Is there any way to view the steps in
mathematica?Shz
Shz
====
=====================================================

====
=====================of the individual or entity to
which they are addressed. Any disclosure,
copying,distribution and diversion contrary to the
applicable export control laws andregulations including US
Export Administration Regulations is strictly prohibited.and
do not disclose it to others. Please notify the
postmaster@hitachi.com.myof the delivery error by
replying to this message and then delete it from
your
====
===================================================

====
===========================I have been trying to code
Sethian's Fast Marching Method in 2D but Mathematicahas
been very slow (taking something like 1-2 hours for something
thatshould take much less than a second in C++). I am sure
part of the problemmy time.I looked at the list archives
and there was mention of an profilingpackage for Mathematica
but a)I can't find it & b)It may not work with 
Mathematica
4.*.My questions:1. Any general suggestions on how to figure
out which functions are takingmost of the time? I guess I
could manually have each function I aminterested in
monitoring keep a variable that counts the amount of
CPUtime that has been spent on it by doing something
like:function[args_]:=Module[{},functionTimer+=Timing[ ....
My actual function..... ][[1]]]but it would very cumbersome
to do this to all of the functions in myprogram and I am not
sure I will get accurate results anyway.2. Compiling functions
is not always that easy. I did read the on-linedocs and the
archives and it does take some work to make a
functioncompile usefully. Is there an FAQ or a tutorial
somewhere?3. Am I the only one who finds the lack of a 
profiler
really reallyannoying ? Mathematica is powerful and it is
usually easy to ask it to do what uwant. The challenge a
lot of times is doing so without taking too long.HusainPS:
One more quick one: Why does the front end act funny when I
have the
====
Caution: the definitionb[n_] :=
Evaluate[(2/L)*Integrate[f[x]*Sin[n*Pi*x/L], {x, 0,
L}]];doesn't immediately compute the integral unless f is
already defined atthis point, and in that case you may as
well writeb[n_] = (2/L)*Integrate[f[x]*Sin[n*Pi*x/L], {x,
0, L}];instead. If b will be computed more than once for the
same n, its evenbetter to do it THIS way (if f and L do not
change):f[x_] = Cos[x] (* for example
*)Simplify[Integrate[f[x]*Sin[n*Pi*(x/L)], {x, 0,
L}]](L*((-n)*Pi + n*Pi*Cos[L]*Cos[n*Pi] +
L*Sin[L]*Sin[n*Pi]))/ ((L - n*Pi)*(L + n*Pi))b[n_] := b[n] =
(L*((-n)*Pi + n*Pi*Cos[L]*Cos[n*Pi] + L*Sin[L]*Sin[n*Pi]))/
((L - n*Pi)*(L + n*Pi))Bobby-----Original
Message-----Stephan Steinhaus (steinhaus-net.de). So when
Mathematica takes a verylongtime, you should investigate.
In this case inserting Evaluate[] in
twoplacesIn[91]:=b[n_] :=
Evaluate[(2/L)*Integrate[f[x]*Sin[n*Pi*x/L], {x,
0,L}]];....In[104]:=Timing[Plot[Evaluate[FS[120, x]], {x,
0, 2}]]Out[104]={0.18
Second,[SkeletonIndicator]Graphics[SkeletonIndicator]}
speeds the process enormously (18 milliseconds to plot 120
terms on myfeeble old 500MHz PowerBook).Why was it so slow
before? When I switch from an ordinary numericallanguageto
Mathematica, I enter into an implicit bargain
withMathematica: the software will go the extra mile to get
me a goodanswer,including (1) using exra precision
(sometimes without being asked) and(2)carrying around
unevaluated mathematical expressions (usually
withoutbeingasked) that could possibly be evaluated more
appropriately at a latertime.Most tools cannot do either
of these things, so I don't have to worryaboutit, except
for the bad answers that result now and then. But I need
totakecare that Mathematica does not burden itself
unnecessarily. That's mysideof the bargain.Number (2) is
the issue here. Your definition of b[n] is written so
thatMathematica analytically evaluates b separately for
each n. But you knowinthis case that the integration can
be done safely once for all n. So doit!The huge
difference, though, comes from pre-evaluating the argument
toPlot.Read the on-line help! You should pre-evaluate
where possible. In somecases, the most common of which
involve branching within the definitionoffunction to plot,
you cannot pre-evaluate so, in keeping with
thebargain,Mathematica goes the extra mile and holds back
just in case. You need tosteer it into the shortcut when
it's OK.Hope this helps,Tom Burton-- 
====
With the second
method below (mine), the times add up to 40% less than with
the first method (Jens-Peer's), for what SEEMS 
to be exactly
the same work. Go figure!Can anybody explain that?(*
Jens-Peer *)ClearAll[f, b, FS]L = 2;f[x_] := UnitStep[x -
1];b[n_] := b[n] = (2/L)*Integrate[f[x]*Sin[n*Pi*x/L], {x,
0, L}];FS[N_, x_] := Sum[b[n]*Sin[n*Pi*x/L], {n, 1,
N}];Timing[Plot[FS[30, x], {x, 0, 2}]]{0.547 Second,
.89»¡Graphics.89»[DownExcl
amation]}(* Treat #1
*)Timing[(2/L)*Integrate[f[x]*Sin[n*Pi*(x/L)], {x, 0,
L}]]{0.016000000000000014*Second, (2*(Cos[(n*Pi)/2] -
Cos[n*Pi]))/(n*Pi)}ClearAll[f, b, FS]L = 2;f[x_] :=
UnitStep[x - 1];b[n_] := b[n] = (2*(Cos[(n*Pi)/2] -
Cos[n*Pi]))/(n*Pi);FS[N_, x_] := Sum[b[n]*Sin[n*Pi*(x/L)],
{n, 1, N}];Timing[Plot[FS[30, x], {x, 0, 2}]]{0.297 Second,
.89»¡Graphics.89»[DownExcl
amation]}Even stranger, it SLOWS the Plot if we
precompute b before Timing starts:(* Treat #2 *)ClearAll[f,
b, FS]L = 2; f[x_] := UnitStep[x - 1]; b[n_] := b[n] =
(2*(Cos[(n*Pi)/2] - (-1)^n))/(n*Pi); b /@ Range[30]; FS[N_,
x_] := Sum[b[n]*Sin[n*Pi*(x/L)], {n, 1, N}];
Timing[Plot[FS[30, x], {x, 0, 2}]]{0.328 Second,
.89»¡Graphics.89»[DownExcl
amation]}But here's a winner:(* Treat #3
*)ClearAll[f, b, FS]L = 2; f[x_] := UnitStep[x - 1];
b[n_] := b[n] = N[(2*(Cos[(n*Pi)/2] - (-1)^n))/ (n*Pi)];
FS[N_, x_] := Sum[b[n]*Sin[n*Pi*(x/L)], {n, 1, N}];
Timing[Plot[FS[30, x], {x, 0, 2}]]{0.204 Second,
.89»¡Graphics.89»[DownExcl
amation]}Apparently, computing b within the Plot
causes machine-precision arithmetic to be used, and that
saves time. Precomputing b and then converting exact
expressions to approximate ones within the Plot seems to
take longer. For that to make sense, I think it must be that
n is approximate (not Integer) when it is passed to b within
Plot.Bobby Treat-----Original Message-----and you need
only a 1-3 seconds (depending on your machine) Jens For my
PDE class we have been calculating Fourier transforms. The
instructor> arrived today with a printout of two plots of a
certain Fourier transform,> done with a different CAS. The
first plot was to 30 terms, the second was to> 120 terms.>
Curious, I translated the functions into Mathematica (4.0 on
Windows2000> on a PIII 700) to see how much time this
required to process. I was> Staggered at how much time it
took. Here's the code: L = 2;> f[x_] := UnitStep[x - 1];>
b[n_] := (2/L)*Integrate[f[x]*Sin[n*Pi*x/L], {x, 0, L}];
FS[N_, x_] := Sum[b[n]*Sin[n*Pi*x/L], {n, 1, N}];
Timing[Plot[FS[30, x], {x, 0, 2}]] Out[23]=> {419.713 Second,
[SkeletonIndicator]Graphics[SkeletonIndicator]} In this
case the number of terms is 30. The time required per number
of terms seems to fit the following polynomial: y = 0.3926x^2
+ 2.2379x This is a large amount of time. I understand that
the code is not optimized,> and was more or less copied from
the code in the other CAS, but is this a> reasonable amount
of time, or is something going wrong? I don't use
Mathematica> because of the speed, but should it be this
slow? Just curious, Steve Story 
====
Even better -- MUCH
better -- add Bob Hanlon's Evaluate to the other
tricks:ClearAll[f, b, FS]L = 2; f[x_] := UnitStep[x - 1];
b[n_] := b[n] = N[(2*(Cos[(n*Pi)/2] - (-1)^n))/(n*Pi)];
FS[N_, x_] := Sum[b[n]*Sin[n*Pi*(x/L)], {n, 1, N}];
Timing[Plot[Evaluate[FS[30, x]], {x, 0, 2}]]{0.015 Second,
.89»¡Graphics.89»[DownExcl
amation]}ClearAll[f, b, FS]L = 2; f[x_] :=
UnitStep[x - 1]; b[n_] := b[n] = N[(2*(Cos[(n*Pi)/2] -
(-1)^n))/(n*Pi)]; FS[N_, x_] := Sum[b[n]*Sin[n*Pi*(x/L)],
{n, 1, N}]; Timing[Plot[Evaluate[FS[120, x]], {x, 0,
2}]]{0.063 Second, 
.89»¡Graphics.89»[DownExcla
mation]}Bobby
Treat-----Original Message-----Timing[Plot[FS[30, x], {x,
0, 2}]]{0.547 Second, 
.89»¡Graphics.89»[DownExcla
mation]}(* Treat #1
*)Timing[(2/L)*Integrate[f[x]*Sin[n*Pi*(x/L)], {x, 0,
L}]]{0.016000000000000014*Second, (2*(Cos[(n*Pi)/2] -
Cos[n*Pi]))/(n*Pi)}ClearAll[f, b, FS]L = 2;f[x_] :=
UnitStep[x - 1];b[n_] := b[n] = (2*(Cos[(n*Pi)/2] -
Cos[n*Pi]))/(n*Pi);FS[N_, x_] := Sum[b[n]*Sin[n*Pi*(x/L)],
{n, 1, N}];Timing[Plot[FS[30, x], {x, 0, 2}]]{0.297 Second,
.89»¡Graphics.89»[DownExcl
amation]}Even stranger, it SLOWS the Plot if we
precompute b before Timing starts:(* Treat #2 *)ClearAll[f,
b, FS]L = 2; f[x_] := UnitStep[x - 1]; b[n_] := b[n] =
(2*(Cos[(n*Pi)/2] - (-1)^n))/(n*Pi); b /@ Range[30]; FS[N_,
x_] := Sum[b[n]*Sin[n*Pi*(x/L)], {n, 1, N}];
Timing[Plot[FS[30, x], {x, 0, 2}]]{0.328 Second,
.89»¡Graphics.89»[DownExcl
amation]}But here's a winner:(* Treat #3
*)ClearAll[f, b, FS]L = 2; f[x_] := UnitStep[x - 1];
b[n_] := b[n] = N[(2*(Cos[(n*Pi)/2] - (-1)^n))/ (n*Pi)];
FS[N_, x_] := Sum[b[n]*Sin[n*Pi*(x/L)], {n, 1, N}];
Timing[Plot[FS[30, x], {x, 0, 2}]]{0.204 Second,
.89»¡Graphics.89»[DownExcl
amation]}Apparently, computing b within the Plot
causes machine-precision arithmetic to be used, and that
saves time. Precomputing b and then converting exact
expressions to approximate ones within the Plot seems to
take longer. For that to make sense, I think it must be that
n is approximate (not Integer) when it is passed to b within
Plot.Bobby Treat-----Original Message-----and you need
only a 1-3 seconds (depending on your machine) Jens For my
PDE class we have been calculating Fourier transforms. The
instructor> arrived today with a printout of two plots of a
certain Fourier transform,> done with a different CAS. The
first plot was to 30 terms, the second was to> 120 terms.>
Curious, I translated the functions into Mathematica (4.0 on
Windows2000> on a PIII 700) to see how much time this
required to process. I was> Staggered at how much time it
took. Here's the code: L = 2;> f[x_] := UnitStep[x - 1];>
b[n_] := (2/L)*Integrate[f[x]*Sin[n*Pi*x/L], {x, 0, L}];
FS[N_, x_] := Sum[b[n]*Sin[n*Pi*x/L], {n, 1, N}];
Timing[Plot[FS[30, x], {x, 0, 2}]] Out[23]=> {419.713 Second,
[SkeletonIndicator]Graphics[SkeletonIndicator]} In this
case the number of terms is 30. The time required per number
of terms seems to fit the following polynomial: y = 0.3926x^2
+ 2.2379x This is a large amount of time. I understand that
the code is not optimized,> and was more or less copied from
the code in the other CAS, but is this a> reasonable amount
of time, or is something going wrong? I don't use
Mathematica> because of the speed, but should it be this
slow? Just curious, Steve Story 
====
How can I plot with
Mathematica two function in the same graphic?I explain
better:compare a and b).
====
(1) Is there a way in
Mathematica 4.2 to put two separate gifs into asingle, cell
side by side (with some intervening space), withouthaving to
combine them in some graphics program first?In particular, 
I'd
like to do that within a text cell.Even in a new Input cell,
if I first create a GridBox (via
Inut>CreateTable/Matrix/Palette) and then try to insert the
first gif (viaEdit>Insert Object>Create from File ....),
Mathematica promptlycrashes. (Mathematica 4.2 under Windows
2000.)I just don't see how to get anything other than a
single gif into acell.(2) Is there a way to cause a gif
imported into a Mathematica 4.2notebook to become a
hyperlink -- so that when the user clicks on thegif the
hyperlink's target is summoned?(My aim in all this is to use
Mathematica to create web pages withhigh mathematical
content -- saving the notebook as HTML+MathML --without
having to do any extensive editing of the resulting .xml
andrelated files. That way as the source Mathematica
notebook changes, Iwould need only to re-export without
further tinkering with the .xmlfile, etc.) -- Murray
Eisenberg Internet: murray@math.umass.edu Mathematics &
Statistics Dept. Voice: 413-545-2859 (W) University of
Massachusetts 413-549-1020 (H)
====
Your method 1), below,
does not produce correct results if one thenuses menu>Save
As Special>HTML+MathML.I view the resulting .xml file in a
MathML-enabled browser (e.g.,Mozilla with appropriate
TrueType fonts installed -- the four BaKoMacm fonts, the
old Mathematica Math1, etc., fonts, and the two
MTfonts. Then the everything in the originally multi-line
inline cellappears on one line separate by a ?, as do the
alignment markers. Inthe page source, these unrendered
symbols have codes #8289 and #63328,respectively.Is there
an issue of the encoding used in the browser here? I did
tryUnicode-7, Unicode-8, and several of the Western
encodings availableunder Mozilla's Default Character
Encoding.Or is it something else? Note that I do also have
the newMathematica TrueType fonts (Mathematica1,
etc.).P.S. I don't think Mathematica documents this, but if
one wants a fileexported from Mathematica as HTML+MathML to
be rendered correctly by abrowser, it seems to require an
extension of .xml rather than .html. >>When writing a
series of equations in mathematica (in a text cell) >>is
there any way to align the equations at the = similar to
>>what can be done in the equation editor made by math type
(used >>in word etc.)? > Mike, > Two ways > 1) Start an
inline cell in your text cell
(menu>Edit>ExpressionInput>Start > Inline Cell) > Type in
your equations with the alignment marker (Esc am Esc)
inafter each > equal sign. > Close the inline cell
(menu>Edit>Expression Input>End Inline Cell). > Select the
text cell (or the inline cell, though this is a delicate >
operation, do it by repeatedly double clicking in it}. > Use
menu> Format>Text Alignment>On Alignment Marker. > -- you
will find that all single letters are now italic - you can
select > individual letters and change this - alternatively
you can select theinline > cell and use the option
inspector (menu>Format>Option Inspector) to set >
SingleLetterItalics->False) > -- you can put existing text
in an inline cell by selcting it and using >
menu>Edit>Expression Input>Start Inline Cell, but you may
have to adjust > the line breaks after this.... -- Murray
Eisenberg Internet: murray@math.umass.edu Mathematics &
Statistics Dept. Voice: 413-545-2859 (W) University of
Massachusetts 413-549-1020 (H)
====
Group,Can anyone help me
with the following? I guess this is something trivial for
most of you, but I have being struggling with it for
weeks.The problem is to construct a taxonomic hierarchy
whose structure is hereby described by example:<<
DiscreteMath`Combinatorica`n[1]:= m = 10;(*with the
following I get the 1th level (1--element)
partition*)in[2]:= ks1 = Partition[Range[m], 1](*Now the
following gives all the 2th-level (?) different partitions
that can be set up by joining two elements in the 1-element
partition and leaving the rest as they are?*)(*First, these
are all possible Ôjoinings' to generate the two 
elements
Ôpieces' from 1th level (1--element) 
partition*)in[3]:= ks2
= KSubsets[Range[m], 2](*And this generates all the 2th-level
(?) different partitions that can be set up by joining two
elements in the 1-element partition and leaving the rest as
they are.*)in[4]:= p2 = MapThread[Complement[Append[#1, #2],
(Sequence @@ {#} & /@ Partition[#2, 1])] &, {Array[ks1&,
Length[ks2]], ks2}];Length[p2]in[5]:= p2[[1]]out[5]:=
{{3}, {4}, {5}, {6}, {7}, {8}, {9}, {10}, {1, 2}}in[6]:=
p2[[Length[p2]]]out[6]:= {{1}, {2}, {3}, {4}, {5}, {6}, {7},
{8}, {9, 10}}(*compute in[4] t and see the full output*)(*Now
I need to solve the problem of given the ith-level (?)
different partitions to set up all the i+1th different
partitions by joining two elements in the ith level
partition and leaving the rest as they are? *)(*for example
given the following 2th-level partition*){3}, {4}, {5}, {6},
{7}, {8}, {9}, {10}, {1, 2}(*we should get*){{{4}, {5}, {6},
{7}, {8}, {9}, {10}, {1, 2, 3}},{{3}, {5}, {6}, {7}, {8},
{9}, {10}, {1, 2, 4}},{{3}, {4}, {6}, {7}, {8}, {9}, {10},
{1, 2, 5}},{{3}, {4}, {5}, {7}, {8}, {9}, {10}, {1, 2,
6}},{{3}, {4}, {5}, {6}, {8}, {9}, {10}, {1, 2, 7}},{{3},
{4}, {5}, {6}, {7}, {9}, {10}, {1, 2, 8}},{{3}, {4}, {5},
{6}, {7}, {8}, {10}, {1, 2, 9}},{{3}, {4}, {5}, {6}, {7},
{8}, {9}, {1, 2, 10}}}(*The process would end up to
*){1,{2,3,4,5,6,7,8,9,10}}{2,{1,3,4,5,6,7,8,9,10}}{3,{
1,2,4,5,6,7,8,9,10}}{4,{1,2,3,5,6,7,8,9,10}}{5,{
1,2,3,4,6,7,8,9,10}}{6,{1,2,3,4,5,7,8,9,10}}{7,{
1,2,3,4,5,6,8,9,10}}{8,{1,2,3,4,5,6,7,9,10}}{9,{
1,2,3,4,5,6,7,8,10}}{10,{1,2,3,4,5,6,7,8,9}}(*
and*){1,2,3,4,5,6,7,8,9,10}Emilio Martin-Serrano
====
> I too
have been trying to use Mathematica (v4.2 most recently) to
type> maths papers and the like but I'm not ready to ditch
LaTeX yet. There> are just too many cases where I cannot
figure out how to achieve what I> want in Mathematica, things
like:> - left brackets spanning multiple lines for defining
hybrid functions;You can accomplish this by doing the
following:1) Put your function braches in the rows of a grid
box structure.2) Add the following options to your cell:
ShowAutoStyles -> False SpanMaxSize -> InfinityThe
following cell snippet demonstrates how this inßuences the
result. To view it, paste the Cell[] expression into a
notebook and then click on Yes when you are prompted on
whether the front end should interpret the
result.Cell[BoxData[ FormBox[ RowBox[{ RowBox[{f, (,
x, )}], =, RowBox[{{, GridBox[{ {x, RowBox[{x,
 , <, 0}]}, { SuperscriptBox[x, 2], RowBox[{0,
[LessEqual], x, <, 1}]}, { RowBox[{sin, (,
x, )}], RowBox[{1, [LessEqual], x, <, 2}]},
{ RowBox[{[CapitalGamma], (, x, )}], RowBox[{x,
[GreaterEqual], 2}]} }]}]}], TraditionalForm]],
DisplayFormula, ShowAutoStyles->False,
SpanMaxSize->Infinity]> - vertical alignment of equals signs
in multi-line equations or> derivations;Put your equations
in a GridBox and set the ColumnAlignments option to a string
containing the equal sign.Cell[BoxData[ FormBox[GridBox[{ {
RowBox[{ RowBox[{ RowBox[{3, x}],  , +,  ,
RowBox[{4,  , y}]}],  , =,  , 9}]}, {
RowBox[{ RowBox[{ RowBox[{2, x}],  , -,  ,
RowBox[{7,  , y}]}], =, RowBox[{32,  , -,  ,
RowBox[{sin, (, x, )}]}]}]} }], TraditionalForm]],
DisplayFormula,
GridBoxOptions->{ColumnAlignments->{=}}]> - setting
typefaces in tables of material.I think the Author Tools
material that comes with Mathematica 4.2 might be able to
help you do this.-- User Interface Programmer
paulh@wolfram.comWolfram Research, Inc.
====
You have three
nonlinear (fourth-order) equations and 23 unknowns. Afaster
computer won't help any.Bobby-----Original
Message-----(*Hex20 Node definition in global coordinates
*)Clear[x1, y1, z1,x2, y2, z2,x3, y3, z3,x4, y4, z4,x5,
y5, z5,x6, y6, z6,x7, y7, z7,x8, y8, z8,x9, y9, z9,x10,
y10, z10,x11, y11, z11,x12, y12, z12,x13, y13, z13,x14,
y14, z14,x15, y15, z15,x16, y16, z16,x17, y17, z17,x18,
y18, z18,x19, y19, z19,x20, y20, z20];(* local coordinates
*)Clear[u, v, w];(* Global co-ordinates *)Clear[x, y, z];(*
corner nodes *)N1= (1-u)*(1-v)*(1-w)*(-2-u-v-w)/8;N3=
(1+u)*(1-v)*(1-w)*(-2+u-v-w)/8;N5=
(1+u)*(1+v)*(1-w)*(-2+u+v-w)/8;N7=
(1-u)*(1+v)*(1-w)*(-2-u+v-w)/8;N13=(1-u)*(1-v)*(1+w)*(-2-u-v
+w)/8;N15=(1+u)*(1-v)*(1+w)*(-2+u-v+w)/8;N17=(1+u)*(1+v)*(1
+w)*(-2+u+v+w)/8;N19=(1-u)*(1+v)*(1+w)*(-2-u+v+w)/8;(* to u
nodes *)N2= (1-u^2)*(1-v)*(1-w)/4;N6=
(1-u^2)*(1+v)*(1-w)/4;N14=(1-u^2)*(1-v)*(1+w)/4;N18=(1-u^2)
*(1+v)*(1+w)/4;(* to v nodes *)N4=
(1+u)*(1-v^2)*(1-w)/4;N8=
(1-u)*(1-v^2)*(1-w)/4;N16=(1+u)*(1-v^2)*(1+w)/4;N20=(1-u)*(
1-v^2)*(1+w)/4;(* to w nodes *)N9=
(1-u)*(1-v)*(1-w^2)/4;N10=(1+u)*(1-v)*(1-w^2)/4;N11=(1+u)*(
1+v)*(1-w^2)/4;N12=(1-u)*(1-v)*(1-w^2)/4;(* solve the
inverse transform
*)Solve[{x1*N1+x2*N2+x3*N3+x4*N4+x5*N5+x6*N6+x7*N7+x8*N8+x9
*N9+x10*N10+x11*N11+x12*N12+x13*N13+x14*N14+x15*N15+x16*N16+
x17*N17+x18*N18+x19*N19+x20*N20-x==0,y1*N1+y2*N2+y3*N3+y4*
N4+y5*N5+y6*N6+y7*N7+y8*N8+y9*N9+y10*N10+y11*N11+y12*N12+y13
*N13+y14*N14+y15*N15+y16*N16+y17*N17+y18*N18+y19*N19+y20*N20
-y==0,z1*N1+z2*N2+z3*N3+z4*N4+z5*N5+z6*N6+z7*N7+z8*N8+z9*N9+
z10*N10+z11*N11+z12*N12+z13*N13+z14*N14+z15*N15+z16*N16+z17*
N17+z18*N18+z19*N19+z20*N20-z==0},{u,v,w}]Christopher J.
PurcellDefence R&D Canada - Atlantic9 Grove St., PO Box
1012Dartmouth NS Canada B2Y 3Z7
====
If I haven't missed a
step, the following should work nicely -- bestdone before
defining any of the symbols that appear:Attributes[dummyIf] =
HoldAll;Attributes[dummyWhich] = HoldAll;expr = {If[Aa ==
Ab, a Sa + b Sb, If[Ca == 0 || Cb == 0, a Sa + b Sb,SS]],
If[Aa == Ab, a Ca + b Cb, If[Ca == 0 || Cb == 0, a Ca + b
Cb, CC]], If[Ca == 0 && Cb == 0, Nil, If[Z1 == Z2, If[Ca <
0, Aa, If[Cb < 0, Ab, If[Ca > 0, If[Aa > 90, Aa - 90, Aa +
90], If[Cb > 0, If[Ab > 90, Ab - 90, Ab + 90]]]]], If[Aa ==
Ab, Aa, If[Ca == 0 && Cb == 0, Nil, If[ Ca == 0, Ab, If[Cb
== 0, Aa, If[Y == 0, Nil, If[X > 180, X - 180, If[X < 0, X
+ 180, X]]]]]]]]]} /. {If -> dummyIf};rule1 = dummyIf[a_,
b_, dummyIf[c_, d_, e_]] -> dummyWhich[a, b, c, d, True,
e];rule2 = dummyIf[a_, dummyIf[ b_, c_, d_], e_] ->
dummyWhich[a && b, c, a, d, True,e];rule3 = dummyIf[a_,
dummyIf[b_, c_, d_]] -> dummyWhich[a && b, c, a, d];rule4 =
dummyWhich[a__, b_, dummyIf[c_, d_, e_], f___] ->
dummyWhich[a, b && c, d, b, e, f];rule5 = dummyWhich[a__,
b_, dummyWhich[c_, d_, e___], f___] -> dummyWhich[a, b && c,
d, b, dummyWhich[e], f];rule6 = dummyWhich[] :> Null;rule7
= HoldPattern[True && a_] :> a;expr //. {rule1, rule2,
rule3, rule4, rule5, rule6, rule7} /. {dummyWhich ->
WhichThe result of that last line is:{Which[Aa == Ab, a Sa
+ b Sb, Ca == 0 || Cb == 0, a Sa + b Sb, True, SS], Which[Aa
== Ab, a Ca + b Cb, Ca == 0 || Cb == 0, a Ca + b Cb, True,
CC], Which[Ca == 0 && Cb == 0, Nil, Z1 == Z2 && Ca < 0,
Aa, Z1 == Z2 && Cb < 0, Ab, (Z1 == Z2 && Ca > 0) && Aa > 90,
Aa - 90, Z1 == Z2 && Ca > 0, Aa + 90, Z1 == Z2 && (Cb > 0 &&
Ab > 90), Ab - 90, Z1 == Z2 && Cb > 0, Ab + 90, Z1 == Z2,
Null, Aa == Ab, Aa, Ca == 0 && Cb == 0, Nil, Ca == 0, Ab,
Cb == 0, Aa, Y == 0, Nil, X > 180, X - 180, X < 0, X +
180, True, X]}I've used Null where that would be the result
of your original logic --maybe you want it to be Nil. Or
maybe you want to use Null instead ofNil. Your choice.
Wherever you see Null, there's possibly a case 
youhaven't
covered.Bobby Treat-----Original Message----- If[Aa == Ab,
a Ca + b Cb, If[Ca == 0 || Cb == 0, a Ca + b Cb, CC]], If[Ca
== 0 && Cb == 0, Nil, If[Z1 == Z2, If[Ca < 0, Aa, If[Cb <
0,Ab, If[Ca > 0, If[Aa > 90, Aa - 90, Aa + 90], If[Cb > 0,
If[Ab > 90, Ab - 90, Ab + 90]]]]], If[Aa == Ab, Aa, If[Ca ==
0 && Cb == 0, Nil, If[Ca == 0, Ab, If[Cb == 0, Aa, If[Y ==
0, Nil, If[X > 180, X - 180, If[X < 0, X + 180,
X]]]]]]]]]}
====
Dear Group,I have a program to take the
local polynomial non parametric regressionof two variables.
It uses Compile, unfortunately, and regularly,
butinconsistently, causes Mathematica to crash (in Win2K,
with I forgetwhat error, and in Win98/Mathematica4.0 with
an invalid memory access fromMathDLL.dll). I am running
Mathematica 4.1, and have this problem consistentlyon 4
different machines.Here is the code, if it doesn't crash the
first time, it will the secondor third:(*w = Compile[{{xj,
_Real, 0}, {XX, _Real, 1}, {YY, _Real, 1}, {h, _Real,0}, {nn,
_Integer, 0}, {ord, _Integer,
0}},First[Inverse[Sum[Outer[Times, Table[If[Positive[q],
(XX[[i]] - xj)^q,1], {q, 0, ord}],Table[If[Positive[q],
(XX[[i]] - xj)^q, 1.], {q, 0,ord}]*E^(-0.5*((XX[[i]] -
xj)/h)^2)], {i, nn}]] .Sum[(Table[If[Positive[q], (XX[[i]] -
xj)^q, 1.], {q, 0, ord}]*YY[[i]])*E^(-0.5*((XX[[i]] -
xj)/h)^2), {i, nn}]]];*)(*!(tt = MemoryInUse[];
ListPlot[Table[{((i + 1))^2, (Timing[ nn = ((i +
1))^2; [Epsilon] = Table[Random[]*3, {i,nn}]; testX =
Table[i* .3 - Random[]*2, {i, nn}]; testY = Table[Sin[i/3] -
2, {i, nn}] + [Epsilon]; Map[w[#, testX, testY, .1 +
Random[], nn, 0] &, testX];])[([1])]/Second}, {i, 15}],
PlotJoined ->True, PlotLabel -> <Calculation Times as a
function of n>]; MemoryInUse[] - tt)*)I have followed
Ted Ersek's tips and
tricks(http://www.verbeia.com/mathematica/tips/tip_
index.html) about Compile,and have tried changing all
variable names (works sometimes), removingany hidden
spaces, restructuring the formulas, changing all 0's to 0.
Ôsand all 1's to 1. Ôs, etcetera 
etcetera, but I still
can't comprehendwhat the problem might be. Doing w[[-2]]
shows a list of op-codenumbers, and one function name,
Inverse[#1]&, so it seems to me thatthere are no problems
with the use of Compile here. Would anyone haveany
thoughts???Bernard
Gressburnthebiscuit@netscape.net
====
Expanding on my
earlier simplification of the Moran Research code, Istudied
the problem a little more and found another useful Rule and
aRule that should help but doesn't. Maybe somebody can
explain what I'mdoing wrong on that one.My earlier result
(after applying rule1, ... rule7) had the following asthird
element of a List in the definition of f:Which[ Ca == 0 && Cb
== 0, Nil, Z1 == Z2 && Ca < 0, Aa, Z1 == Z2 && Cb < 0, Ab,
(Z1 == Z2 && Ca > 0) && Aa > 90, Aa - 90, Z1 == Z2 && Ca > 0,
Aa + 90, Z1 == Z2 && (Cb > 0 && Ab > 90), Ab - 90, Z1 == Z2
&& Cb > 0, Ab + 90, Z1 == Z2, Null, Aa == Ab, Aa, Ca == 0 &&
Cb == 0, Nil, Ca == 0, Ab, Cb == 0, Aa, Y == 0, Nil, X
> 180, X - 180, X < 0, X + 180, True, X]Several things
occurred to me in terms of simplifying this. The firstwas
that the last six arguments of Which could be replaced with
two, asfollows:Which[ Ca == 0 && Cb == 0, Nil, Z1 == Z2
&& Ca < 0, Aa, Z1 == Z2 && Cb < 0, Ab, (Z1 == Z2 && Ca > 0) &&
Aa > 90, Aa - 90, Z1 == Z2 && Ca > 0, Aa + 90, Z1 == Z2 && (Cb
> 0 && Ab > 90), Ab - 90, Z1 == Z2 && Cb > 0, Ab + 90, Z1 ==
Z2, Null, Aa == Ab, Aa, Ca == 0 && Cb == 0, Nil, Ca == 0,
Ab, Cb == 0, Aa, Y == 0, Nil, True,Mod[X,180]]This
probably has no advantage other than clarity, but that's
worththe others:rule8 = dummyIf[a_ > b_, a_ - b_,
dummyIf[a_ < 0, a_ + b_, a_]] :>Mod[a, b];but it had no
effect on the expression. Apparently there was no match,and
I can't see why.The second thing I noticed was that the
conditionCa == 0 && Cb == 0occurs twice in the Which
statement. The following rule fixes that kindof
situation:rule9 = dummyWhich[a___, b_, c_, d__, b_, e_, f__]
/; EvenQ[Length[List[a]]] &&EvenQ[Length[List[d]]] &&
EvenQ[Length[List[f]]] :> dummyWhich[a, b, c, d, f];expr //.
{rule1, rule2, rule3, rule4, rule5, rule6, rule7, rule9}The
third thing I noticed was the treatment of Aa and Ab.
TheexpressionsIf[Aa > 90, Aa - 90, Aa + 90]If[Ab > 90,
Ab - 90, Ab + 90]in the original expression may be
equivalent (depending on what's knownabout Aa and Ab a
priori) to:Mod[Aa+90,180]Mod[Ab+90,180]If that is a valid
assumption in the situation (no way for me to know),the
expression is nowWhich[ Ca == 0 && Cb == 0, Nil, Z1 ==
Z2 && Ca < 0, Aa, Z1 == Z2 && Cb < 0, Ab, Z1 == Z2 && Ca > 0,
Mod[Aa + 90, 180], Z1 == Z2 && Cb > 0, Mod[Ab + 90, 180], Z1
== Z2, Null, Aa == Ab, Aa, Ca == 0, Ab, Cb == 0, Aa, Y == 0,
Nil, True, Mod[X, 180]]The next thing I notice is that
the sixth condition, Z1==Z2, whichresults in Null if found
True, isn't needed. If Z1==Z2 then one of 
thefirst five
conditions is also true, so execution wouldn't get that
far.Hence we can delete that condition-response
pair.Finally, I'm stopping with this:Which[ Ca == 0 == Cb,
Nil, Z1 == Z2 && Ca < 0, Aa, Z1 == Z2 && Cb < 0, Ab, Z1 ==
Z2 && Ca > 0, Mod[Aa + 90, 180], Z1 == Z2 && Cb > 0, Mod[Ab +
90, 180], Aa == Ab, Aa, Ca == 0, Ab, Cb == 0, Aa, Y == 0,
Nil, True, Mod[X, 180]]I have no idea what you're
actually doing with this code, but it looksweird even AFTER
I've simplified it.Bobby Treat-----Original 
Message----- 0,
Aa, If[Cb < 0, Ab, If[Ca > 0, If[Aa > 90, Aa - 90, Aa + 90],
If[Cb > 0, If[Ab > 90, Ab - 90, Ab + 90]]]]], If[Aa == Ab,
Aa, If[Ca == 0 && Cb == 0, Nil, If[ Ca == 0, Ab, If[Cb ==
0, Aa, If[Y == 0, Nil, If[X > 180, X - 180, If[X < 0, X +
180, X]]]]]]]]]} /. {If -> dummyIf};rule1 = dummyIf[a_, b_,
dummyIf[c_, d_, e_]] -> dummyWhich[a, b, c, d, True,
e];rule2 = dummyIf[a_, dummyIf[ b_, c_, d_], e_] ->
dummyWhich[a && b, c, a, d, True,e];rule3 = dummyIf[a_,
dummyIf[b_, c_, d_]] -> dummyWhich[a && b, c, a, d];rule4 =
dummyWhich[a__, b_, dummyIf[c_, d_, e_], f___] ->
dummyWhich[a, b && c, d, b, e, f];rule5 = dummyWhich[a__,
b_, dummyWhich[c_, d_, e___], f___] -> dummyWhich[a, b && c,
d, b, dummyWhich[e], f];rule6 = dummyWhich[] :> Null;rule7
= HoldPattern[True && a_] :> a;expr //. {rule1, rule2,
rule3, rule4, rule5, rule6, rule7} /. {dummyWhich ->
WhichThe result of that last line is:{Which[Aa == Ab, a Sa
+ b Sb, Ca == 0 || Cb == 0, a Sa + b Sb, True, SS], Which[Aa
== Ab, a Ca + b Cb, Ca == 0 || Cb == 0, a Ca + b Cb, True,
CC], Which[Ca == 0 && Cb == 0, Nil, Z1 == Z2 && Ca < 0,
Aa, Z1 == Z2 && Cb < 0, Ab, (Z1 == Z2 && Ca > 0) && Aa > 90,
Aa - 90, Z1 == Z2 && Ca > 0, Aa + 90, Z1 == Z2 && (Cb > 0 &&
Ab > 90), Ab - 90, Z1 == Z2 && Cb > 0, Ab + 90, Z1 == Z2,
Null, Aa == Ab, Aa, Ca == 0 && Cb == 0, Nil, Ca == 0, Ab,
Cb == 0, Aa, Y == 0, Nil, X > 180, X - 180, X < 0, X +
180, True, X]}I've used Null where that would be the result
of your original logic --maybe you want it to be Nil. Or
maybe you want to use Null instead ofNil. Your choice.
Wherever you see Null, there's possibly a case 
youhaven't
covered.Bobby Treat-----Original Message----- If[Aa == Ab,
a Ca + b Cb, If[Ca == 0 || Cb == 0, a Ca + b Cb, CC]], If[Ca
== 0 && Cb == 0, Nil, If[Z1 == Z2, If[Ca < 0, Aa, If[Cb <
0,Ab, If[Ca > 0, If[Aa > 90, Aa - 90, Aa + 90], If[Cb > 0,
If[Ab > 90, Ab - 90, Ab + 90]]]]], If[Aa == Ab, Aa, If[Ca ==
0 && Cb == 0, Nil, If[Ca == 0, Ab, If[Cb == 0, Aa, If[Y ==
0, Nil, If[X > 180, X - 180, If[X < 0, X + 180,
X]]]]]]]]]}
====
Sorry... ONE MORE simplification:Which[ Ca ==
0 == Cb, Nil, Z1 == Z2, Which[Ca < 0, Aa, Cb < 0, Ab, Ca >
0, Mod[Aa + 90, 180], Cb > 0, Mod[Ab + 90, 180]], Aa == Ab,
Aa, Ca == 0, Ab, Cb == 0, Aa, Y == 0, Nil, True, Mod[X,
180]]I didn't like repeating the Z1==Z2 test.Upon noticing
this, I thought at first that my rule5 caused thissituation
to occur, but actually I have to delete both rule 4 AND
rule5to eliminate it, and the resulting expression wouldn't
be as easy tounderstand.Bobby Treat-----Original
Message----- Z1 == Z2 && Ca < 0, Aa, Z1 == Z2 && Cb < 0, Ab,
(Z1 == Z2 && Ca > 0) && Aa > 90, Aa - 90, Z1 == Z2 && Ca > 0,
Aa + 90, Z1 == Z2 && (Cb > 0 && Ab > 90), Ab - 90, Z1 == Z2 &&
Cb > 0, Ab + 90, Z1 == Z2, Null, Aa == Ab, Aa, Ca == 0 && Cb
== 0, Nil, Ca == 0, Ab, Cb == 0, Aa, Y == 0, Nil, X >
180, X - 180, X < 0, X + 180, True, X]Several things
occurred to me in terms of simplifying this. The firstwas
that the last six arguments of Which could be replaced with
two, asfollows:Which[ Ca == 0 && Cb == 0, Nil, Z1 == Z2
&& Ca < 0, Aa, Z1 == Z2 && Cb < 0, Ab, (Z1 == Z2 && Ca > 0) &&
Aa > 90, Aa - 90, Z1 == Z2 && Ca > 0, Aa + 90, Z1 == Z2 && (Cb
> 0 && Ab > 90), Ab - 90, Z1 == Z2 && Cb > 0, Ab + 90, Z1 ==
Z2, Null, Aa == Ab, Aa, Ca == 0 && Cb == 0, Nil, Ca == 0,
Ab, Cb == 0, Aa, Y == 0, Nil, True,Mod[X,180]]This
probably has no advantage other than clarity, but that's
worththe others:rule8 = dummyIf[a_ > b_, a_ - b_,
dummyIf[a_ < 0, a_ + b_, a_]] :>Mod[a, b];but it had no
effect on the expression. Apparently there was no match,and
I can't see why.The second thing I noticed was that the
conditionCa == 0 && Cb == 0occurs twice in the Which
statement. The following rule fixes that kindof
situation:rule9 = dummyWhich[a___, b_, c_, d__, b_, e_, f__]
/; EvenQ[Length[List[a]]] &&EvenQ[Length[List[d]]] &&
EvenQ[Length[List[f]]] :> dummyWhich[a, b, c, d, f];expr //.
{rule1, rule2, rule3, rule4, rule5, rule6, rule7, rule9}The
third thing I noticed was the treatment of Aa and Ab.
TheexpressionsIf[Aa > 90, Aa - 90, Aa + 90]If[Ab > 90,
Ab - 90, Ab + 90]in the original expression may be
equivalent (depending on what's knownabout Aa and Ab a
priori) to:Mod[Aa+90,180]Mod[Ab+90,180]If that is a valid
assumption in the situation (no way for me to know),the
expression is nowWhich[ Ca == 0 && Cb == 0, Nil, Z1 ==
Z2 && Ca < 0, Aa, Z1 == Z2 && Cb < 0, Ab, Z1 == Z2 && Ca > 0,
Mod[Aa + 90, 180], Z1 == Z2 && Cb > 0, Mod[Ab + 90, 180], Z1
== Z2, Null, Aa == Ab, Aa, Ca == 0, Ab, Cb == 0, Aa, Y == 0,
Nil, True, Mod[X, 180]]The next thing I notice is that
the sixth condition, Z1==Z2, whichresults in Null if found
True, isn't needed. If Z1==Z2 then one of 
thefirst five
conditions is also true, so execution wouldn't get that
far.Hence we can delete that condition-response
pair.Finally, I'm stopping with this:Which[ Ca == 0 == Cb,
Nil, Z1 == Z2 && Ca < 0, Aa, Z1 == Z2 && Cb < 0, Ab, Z1 ==
Z2 && Ca > 0, Mod[Aa + 90, 180], Z1 == Z2 && Cb > 0, Mod[Ab +
90, 180], Aa == Ab, Aa, Ca == 0, Ab, Cb == 0, Aa, Y == 0,
Nil, True, Mod[X, 180]]I have no idea what you're
actually doing with this code, but it looksweird even AFTER
I've simplified it.Bobby Treat-----Original 
Message----- 0,
Aa, If[Cb < 0, Ab, If[Ca > 0, If[Aa > 90, Aa - 90, Aa + 90],
If[Cb > 0, If[Ab > 90, Ab - 90, Ab + 90]]]]], If[Aa == Ab,
Aa, If[Ca == 0 && Cb == 0, Nil, If[ Ca == 0, Ab, If[Cb ==
0, Aa, If[Y == 0, Nil, If[X > 180, X - 180, If[X < 0, X +
180, X]]]]]]]]]} /. {If -> dummyIf};rule1 = dummyIf[a_, b_,
dummyIf[c_, d_, e_]] -> dummyWhich[a, b, c, d, True,
e];rule2 = dummyIf[a_, dummyIf[ b_, c_, d_], e_] ->
dummyWhich[a && b, c, a, d, True,e];rule3 = dummyIf[a_,
dummyIf[b_, c_, d_]] -> dummyWhich[a && b, c, a, d];rule4 =
dummyWhich[a__, b_, dummyIf[c_, d_, e_], f___] ->
dummyWhich[a, b && c, d, b, e, f];rule5 = dummyWhich[a__,
b_, dummyWhich[c_, d_, e___], f___] -> dummyWhich[a, b && c,
d, b, dummyWhich[e], f];rule6 = dummyWhich[] :> Null;rule7
= HoldPattern[True && a_] :> a;expr //. {rule1, rule2,
rule3, rule4, rule5, rule6, rule7} /. {dummyWhich ->
WhichThe result of that last line is:{Which[Aa == Ab, a Sa
+ b Sb, Ca == 0 || Cb == 0, a Sa + b Sb, True, SS], Which[Aa
== Ab, a Ca + b Cb, Ca == 0 || Cb == 0, a Ca + b Cb, True,
CC], Which[Ca == 0 && Cb == 0, Nil, Z1 == Z2 && Ca < 0,
Aa, Z1 == Z2 && Cb < 0, Ab, (Z1 == Z2 && Ca > 0) && Aa > 90,
Aa - 90, Z1 == Z2 && Ca > 0, Aa + 90, Z1 == Z2 && (Cb > 0 &&
Ab > 90), Ab - 90, Z1 == Z2 && Cb > 0, Ab + 90, Z1 == Z2,
Null, Aa == Ab, Aa, Ca == 0 && Cb == 0, Nil, Ca == 0, Ab,
Cb == 0, Aa, Y == 0, Nil, X > 180, X - 180, X < 0, X +
180, True, X]}I've used Null where that would be the result
of your original logic --maybe you want it to be Nil. Or
maybe you want to use Null instead ofNil. Your choice.
Wherever you see Null, there's possibly a case 
youhaven't
covered.Bobby Treat-----Original Message----- If[Aa == Ab,
a Ca + b Cb, If[Ca == 0 || Cb == 0, a Ca + b Cb, CC]], If[Ca
== 0 && Cb == 0, Nil, If[Z1 == Z2, If[Ca < 0, Aa, If[Cb <
0,Ab, If[Ca > 0, If[Aa > 90, Aa - 90, Aa + 90], If[Cb > 0,
If[Ab > 90, Ab - 90, Ab + 90]]]]], If[Aa == Ab, Aa, If[Ca ==
0 && Cb == 0, Nil, If[Ca == 0, Ab, If[Cb == 0, Aa, If[Y ==
0, Nil, If[X > 180, X - 180, If[X < 0, X + 180,
X]]]]]]]]]}
====
If x1 through x20 aren't unknowns, you should
have told me their values.If you can't, they are
UNKNOWN.True, you're not solving for them... but they affect
the dimensionalityof the problem just as if you WERE solving
for them. You have a20-dimensional space full of
contingencies -- solution forms that dependon the values of
x1 through x20. The Solve function won't deal
withcontingencies even if it could, and if it tried, there
would be toomany.Bobby-----Original
Message----->Bobby-----Original Message----->Sent: Saturday,
September 07, 2002 1:54 AM>resend>Would someone with a very
fast machine and lots of memory be willing to>try>this
Solve for me?>It is the inverse of the 20 node quadratic
hexahedral mapping used in>finite element analysis.>None of
my computers can handle this - they run out of memory
(using(*Hex20 Node definition in global coordinates
*)>Clear[>x1, y1, z1,>x2, y2, z2,>x3, y3, z3,>x4, y4, z4,>x5,
y5, z5,>x6, y6, z6,>x7, y7, z7,>x8, y8, z8,>x9, y9, z9,>x10,
y10, z10,>x11, y11, z11,>x12, y12, z12,>x13, y13, z13,>x14,
y14, z14,>x15, y15, z15,>x16, y16, z16,>x17, y17, z17,>x18,
y18, z18,>x19, y19, z19,>x20, y20, z20];(* local coordinates
*)>Clear[u, v, w];(* Global co-ordinates *)>Clear[x, y, z];(*
corner nodes *)>N1= (1-u)*(1-v)*(1-w)*(-2-u-v-w)/8;>N3=
(1+u)*(1-v)*(1-w)*(-2+u-v-w)/8;>N5=
(1+u)*(1+v)*(1-w)*(-2+u+v-w)/8;>N7=
(1-u)*(1+v)*(1-w)*(-2-u+v-w)/8;>N13=(1-u)*(1-v)*(1+w)*(-2-u-v
+w)/8;>N15=(1+u)*(1-v)*(1+w)*(-2+u-v+w)/8;>N17=(1+u)*(1+v)*(1
+w)*(-2+u+v+w)/8;>N19=(1-u)*(1+v)*(1+w)*(-2-u+v+w)/8;>(* to u
nodes *)>N2= (1-u^2)*(1-v)*(1-w)/4;>N6=
(1-u^2)*(1+v)*(1-w)/4;>N14=(1-u^2)*(1-v)*(1+w)/4;>N18=(1-u^2)
*(1+v)*(1+w)/4;>(* to v nodes *)>N4=
(1+u)*(1-v^2)*(1-w)/4;>N8=
(1-u)*(1-v^2)*(1-w)/4;>N16=(1+u)*(1-v^2)*(1+w)/4;>N20=(1-u)*(
1-v^2)*(1+w)/4;>(* to w nodes *)>N9=
(1-u)*(1-v)*(1-w^2)/4;>N10=(1+u)*(1-v)*(1-w^2)/4;>N11=(1+u)*(
1+v)*(1-w^2)/4;>N12=(1-u)*(1-v)*(1-w^2)/4;(* solve the
inverse transform
*)>Solve[{>x1*N1+x2*N2+x3*N3+x4*N4+x5*N5+x6*N6+x7*N7+x8*N8+x9
*N9+x10*N10+>x11*N11+x12*N12+x13*N13+x14*N14+x15*N15+x16*N16+
x17*N17+x18*N18+x19*N19+>x20*N20-x==0,>y1*N1+y2*N2+y3*N3+y4*
N4+y5*N5+y6*N6+y7*N7+y8*N8+y9*N9+y10*N10+>y11*N11+y12*N12+y13
*N13+y14*N14+y15*N15+y16*N16+y17*N17+y18*N18+y19*N19+>y20*N20
-y==0,>z1*N1+z2*N2+z3*N3+z4*N4+z5*N5+z6*N6+z7*N7+z8*N8+z9*N9+
z10*N10+>z11*N11+z12*N12+z13*N13+z14*N14+z15*N15+z16*N16+z17*
N17+z18*N18+z19*N19+>z20*N20-z==0},>{u,v,w}]>Christopher J.
Purcell>Defence R&D Canada - Atlantic>9 Grove St., PO Box
1012>Dartmouth NS Canada B2Y 3Z7Christopher J.
PurcellDefence R&D Canada - Atlantic9 Grove St., PO Box
1012Dartmouth NS Canada B2Y 3Z7
====
If we do think of
x1...x20 as constants, it's still fourth-order, withthree
simultaneous equations. Evaluate this:Solve[a x^4 + b x^3 +
c x^2 + d x + e == 0, x]and look at the four solutions found
-- for a single variable and oneequation!For your problem,
I'm thinking there'd be at least 4^3 
solutions, eachof them
even more complicated than these, and an unknown number
ofcontingencies to deal with, depending on values of the
xi. Check eachradical for a negative argument, and the
special cases multiply quickly.For some values of the xi,
there'd be no solutions; for other values,many solutions.If
you do manage to solve this, I'd love to see the
method!Bobby-----Original
Message-----many.Bobby-----Original
Message----->Bobby-----Original Message----->Sent: Saturday,
September 07, 2002 1:54 AM>resend>Would someone with a very
fast machine and lots of memory be willing to>try>this
Solve for me?>It is the inverse of the 20 node quadratic
hexahedral mapping used in>finite element analysis.>None of
my computers can handle this - they run out of memory
(using(*Hex20 Node definition in global coordinates
*)>Clear[>x1, y1, z1,>x2, y2, z2,>x3, y3, z3,>x4, y4, z4,>x5,
y5, z5,>x6, y6, z6,>x7, y7, z7,>x8, y8, z8,>x9, y9, z9,>x10,
y10, z10,>x11, y11, z11,>x12, y12, z12,>x13, y13, z13,>x14,
y14, z14,>x15, y15, z15,>x16, y16, z16,>x17, y17, z17,>x18,
y18, z18,>x19, y19, z19,>x20, y20, z20];(* local coordinates
*)>Clear[u, v, w];(* Global co-ordinates *)>Clear[x, y, z];(*
corner nodes *)>N1= (1-u)*(1-v)*(1-w)*(-2-u-v-w)/8;>N3=
(1+u)*(1-v)*(1-w)*(-2+u-v-w)/8;>N5=
(1+u)*(1+v)*(1-w)*(-2+u+v-w)/8;>N7=
(1-u)*(1+v)*(1-w)*(-2-u+v-w)/8;>N13=(1-u)*(1-v)*(1+w)*(-2-u-v
+w)/8;>N15=(1+u)*(1-v)*(1+w)*(-2+u-v+w)/8;>N17=(1+u)*(1+v)*(1
+w)*(-2+u+v+w)/8;>N19=(1-u)*(1+v)*(1+w)*(-2-u+v+w)/8;>(* to u
nodes *)>N2= (1-u^2)*(1-v)*(1-w)/4;>N6=
(1-u^2)*(1+v)*(1-w)/4;>N14=(1-u^2)*(1-v)*(1+w)/4;>N18=(1-u^2)
*(1+v)*(1+w)/4;>(* to v nodes *)>N4=
(1+u)*(1-v^2)*(1-w)/4;>N8=
(1-u)*(1-v^2)*(1-w)/4;>N16=(1+u)*(1-v^2)*(1+w)/4;>N20=(1-u)*(
1-v^2)*(1+w)/4;>(* to w nodes *)>N9=
(1-u)*(1-v)*(1-w^2)/4;>N10=(1+u)*(1-v)*(1-w^2)/4;>N11=(1+u)*(
1+v)*(1-w^2)/4;>N12=(1-u)*(1-v)*(1-w^2)/4;(* solve the
inverse transform
*)>Solve[{>x1*N1+x2*N2+x3*N3+x4*N4+x5*N5+x6*N6+x7*N7+x8*N8+x9
*N9+x10*N10+>x11*N11+x12*N12+x13*N13+x14*N14+x15*N15+x16*N16+
x17*N17+x18*N18+x19*N19+>x20*N20-x==0,>y1*N1+y2*N2+y3*N3+y4*
N4+y5*N5+y6*N6+y7*N7+y8*N8+y9*N9+y10*N10+>y11*N11+y12*N12+y13
*N13+y14*N14+y15*N15+y16*N16+y17*N17+y18*N18+y19*N19+>y20*N20
-y==0,>z1*N1+z2*N2+z3*N3+z4*N4+z5*N5+z6*N6+z7*N7+z8*N8+z9*N9+
z10*N10+>z11*N11+z12*N12+z13*N13+z14*N14+z15*N15+z16*N16+z17*
N17+z18*N18+z19*N19+>z20*N20-z==0},>{u,v,w}]>Christopher J.
Purcell>Defence R&D Canada - Atlantic>9 Grove St., PO Box
1012>Dartmouth NS Canada B2Y 3Z7Christopher J.
PurcellDefence R&D Canada - Atlantic9 Grove St., PO Box
1012Dartmouth NS Canada B2Y 3Z7
====
The documentation you're
talking about doesn't even deserve to be
calleddocumentation. A few hints; that's it.Bobby
Treat-----Original Message-----> unable to find a complete
guide to word processing with mathematica.> Does anyone know
where such a document could be found?
====
This is a question
that's been asked and answered a number of times. One
answer, by me, can be found by going to www.wolfram.com and
selecting first Resource Library and then MathGroup on the
Resource Library page; then search for direction field.
You'll find, among many others, the URL:
http://library.wolfram.com/mathgroup/archive/2002/Jul/
msg00163.html(This is not to chastise you for not first
looking there, but to suggest a way for you to find answers
more efficiently than waiting for the ou need to load the
package, then manufacture a vector field whose plotis the
slope field (what you call vector ßow diagram). To do
thelatter, let me separately define the function giving the
right-hand sideof the ODE; you don't really have to do
that, but could instead directlyuse as the first argument to
PlotVectorField the right-hand side of mydefinition of f. <<
Graphics`PlotField` f[t_, y_] := {1, 1/y} field =
PlotVectorField[f[t, y], {t, 0, 3}, {y, 0.05, 3}];I gave a
name to the graphics result just in order to re-display
thefield below along with the graph of one solution. First,
find a generalsolution: soln = First @ 
DSolve[{y'[t] ==
1/y[t], y[0] == 1}, y[t], t]{y[t] -> Sqrt[2]*Sqrt[1/2 + t]}
grf = Plot[y[t] /. soln, {t, 0, 3}];Now combine the plot
with the field: Show[grf, field];Note that 
PlotVectorField
does not always give quite the result youmight want, since
the vector lengths are scaled with respect to themagnitude
of the actual vectors. You might want just a
directionfield in which all arrows have the same length.
You can produce such agraphic by using the ScaleFunction
and ScaleFactor options toPlotVectorField. For example:
field = PlotVectorField[f[t, y], {t, 0, 3}, {y, 0.05, 3},
ScaleFunction -> (1 &), ScaleFactor -> 0.175];... I have
several differential equations I would like toplot in
Mathematica. I would like to plot the Slope Fields of them
though. Can anyone lead me in the right direction? I can
solve the equations trivially but I want to display the
slope fields. An example follows :y' + 2y = 
3...-- Murray
Eisenberg murray@math.umass.eduMathematics & Statistics
Dept.Lederle Graduate Research Tower phone 413 549-1020
(H)University of Massachusetts 413 545-2859 (W)710 North
Pleasant StreetAmherst, MA 01375
====
I am trying to
implement a very simple sorted tree to quickly store
somereal numbers I need. I have written an add, delete,
minimum, and pop(delete the lowest value) function and they
seem to work ok but are veryslow. Let's just look @ my
implementation of the add part:nums=Null;(*my initial blank
Tree)In[326]:=Clear[add]In[327]:=add[Null,x_Real]:=node[x
,Null,Null]add[Null,node[x_Real,lower_,higher_]]=node[x,
lower,higher]In[328]:=add[node[x_Real,lower_,higher_],y_
Real]:=
If[x>y,node[x,add[lower,y],higher],node[x,lower,add[higher,y]
]]In[288]:=add[node[x_Real,lowerx_,higherx_],node[y_Real,
lowery_,highery_]]:=If[x>y,
node[x,add[lowerx,node[y,lowery,highery]],higherx],
node[x,lowerx,add[higherx,node[y,lowery,highery]]] ]Now this
is my attempt to test how fast my add
works:SeedRandom[5];Do[nums=add[nums,Random[]],{5000}];//
TimingOut[333]={13.279 Second,Null}running on an 1.4GHz
Athlon with 1GB of ram).Questions:1. Is this as fast as I
can get my code to run?2. Am I doing something obviously
stupid?3. would Compiling things help?Husain
====
The
vectors that NullSpace gives do not need to be orthogonal
toeach other. What about the following variant, that adds
one vectorat a time? It may not be fast enough for serious
number crunching.OrthogonalComplement[v__ /; MatrixQ[{v}]]
:= With[{n = Length[NullSpace[{v}]]},
Take[Nest[Join[{First[NullSpace[#]]}, #] &, {v}, n],
n]];OrthonormalComplement[v__ /; MatrixQ[{v}]] :=
Map[#/Sqrt[#.#] &, OrthogonalComplement[v]];When the input
of OrthogonalComplement is integer, I expect the outputto
be integer too:OrthogonalComplement[{1, 2, 3}]{{1, -5, 3},
{-3, 0, 1}} Gianluca Gorni> OrthogonalUnitVectors that I
sent a few minutes ago doesn't do what I> thought it would.>
After making either definition below I get the following:>
In[2]:=>
s1=OrthogonalUnitVectors[{1,0,1/2,1,0},{0,1,-1,1/2,2}]>
Out[2]=> {{0, -2/Sqrt[5], 0, 0, 1/Sqrt[5]}, {-2/3, -1/3, 0,
2/3, 0}, {-1/3, 2/3,> 2/3, 0, 0}}> The dot products below
aren't zero, so the vectors aren't orthogonal. 
What> went
wrong?> In[3]:= > Part[s1,1].Part[s1,2]> Out[3]=>
2/(3*Sqrt[5])> In[4]:= > Part[s1,1].Part[s1,3]> Out[3]=>
-4/(3*Sqrt[5])> --------------> Hugh Goyder and David Park
gave a most elegant function to find two> vectors that are
orthogonal to one vector in 3D. The key to coming up> with
the elegant solution is an understanding of Mathematica's
NullSpace> function. We can easily make the version from
Hugh and David much more> general with the version below.>
-------------> OrthogonalUnitVectors[vect__?VectorQ]:=>
#/Sqrt[#.#]&/@NullSpace[{vect}]> -------------> The version
above will give a set of unit orthogonal vectors if given
any> number of vectors in any dimension.> So besides giving
it a 3D vector we can give it the following:>
OrthogonalUnitVectors[{2,1,0,-1,1}]> or>
OrthogonalUnitVectors[{0,1,0,1/2,1},{1,0,-1,1/2}]>
------------> But the short version above isn't very 
robust.>
(1) Clear[x,y,z];NullSpace[{{x,y,z}}]> returns two vectors
orthogonal to {x,y,z}, but the two vectors> NullSpace
returns aren't orthogonal to each other.> So
(OrthogonalUnitVectors) should only work with numeric
vectors.> (2) We should ensure all the vectors have the same
dimension and length> 1.> I give a less concise version
below that corrects these problems.> ------------>
OrthogonalUnitVectors[vect__?(VectorQ[#,NumericQ]&)]/;>
(SameQ@@Length/@{vect})&&(Length[First[{vect}]]>1):=>
#/Sqrt[#.#]&/@NullSpace[{vect}]> --------------> Ted Ersek>
Get Mathematica tips, tricks from>
http://www.verbeia.com/mathematica/tips/Tricks.html> 
====
I
am using MultipleListPlot to plot a range of 2D graphs and
have found thatthe last graph, which has a much greater
(by a factor of 4) x range than theothers does not plot
completely but is truncated in the x direction. Byremoving
the Frame, I can see the points where these lie outside the
Framebut these are not joined etc. How do I get over
this?-Malcolm Woodruff
====
Group,Sorry for the previous
posting. The following solves (to some extent my
problem)<<DiscreteMath`Combinatorica`lPartition[ksN_,r_]:=
With[{ks=KSubsets[Range[r],ksN],ks1=Partition[Range[r],1]},
MapThread[Complement[Append[#1,#2],(Sequence@@{#}&/@Partition
[#2,1])]&,{Array[ks1&,Length[ks]],ks}]][ksN<(r=+1),s=fTaxon[
ksN,r];ksN=ksN+1;Print[s]]]However, I do not like this
solution since to my taste is to complex and rather
inelegant. Nest or Fold would do better. Any Idea?Just an
additional remark.Besides, I do not need the whole taxonomy,
but a subset satisfying a minimizing condition on
connectivity implied by some relationships among the
elements. This condition is satisfied by one and only one
element, or class of equivalence in each level of the
hierarchy. So it is that very class which matter and
consequently the procedure should give control on each class
as the total structure is being generated. Once it is know
which class satisfy the condition it is possible to avoid
generating the whole structure and the subsequent
combinatorial explosion. With just Emilio
Martin-Serrano
====
> (1) Is there a way in Mathematica 4.2 to
put two separate gifs into a> single, cell side by side (with
some intervening space), without> having to combine them in
some graphics program first? In particular, I'd 
like to do
that within a text cell. Even in a new Input cell, if I first
create a GridBox (via Inut>Create> Table/Matrix/Palette) and
then try to insert the first gif (via> Edit>Insert
Object>Create from File ....), Mathematica promptly>
crashes. (Mathematica 4.2 under Windows 2000.)You could do
something like this:grlist = Map[Import[#, GIF]&, {<first
file>, <second file>}]celist =
Map[Cell[GraphicsData[PostScript, DisplayString[#]]]&,
grlist]CellPrint[Cell[BoxData[GridBox[{celist}]],
Text]]This embeds the PostScript versions of these
graphics each as inline cellsin a GridBox[] structure.> (2)
Is there a way to cause a gif imported into a Mathematica 4.2>
notebook to become a hyperlink -- so that when the user clicks
on the> gif the hyperlink's target is summoned?Again, we
embed the PostScript graphic within an inline cell. Here is
afunction that does what you
describe:createGIFHyperlink[gifImage_String, url_String] :=
NotebookWrite[EvaluationNotebook[], Cell[BoxData[ ButtonBox[
Cell[GraphicsData[PostScript,
DisplayString[Import[gifImage, GIF]]]], ButtonStyle ->
Hyperlink, ButtonData :> {URL[url], None}]], Text]]--
User Interface Programmer paulh@wolfram.comWolfram
Research, Inc.
====
> 1.) How to find the General Solution for
below's partial differential> equation?> (y + u) du/dx + y
(du/dy) = x - y> ** I use d to represent the partial
differential symbol.> Can it be solved by function NDSolve in
mathematica 4.1? How?It's been a while, but I suspect that
the presence of the nonlinearconvective term u du/dx alone
makes a general solution unlikely. It's easyto get a
numerical solution for a particular set of boundary
conditions, butyou are not guaranteed a solution. For
instance, the following choice of aand b are skirting
failure, as the diagnostic contour plot shows.Tom
BurtonNeeds[Graphics`Colors`]!((soln = With[{a =
(-1), b = 1.27, e = 0.001}, solutions = NDSolve[{((y +
u[x, y])) [PartialD]_x u[x, y] + y [PartialD]_y
u[x, y] == x - y, u[x, b] == 0, u[a, y] == 0}, u, {x, a, 10},
{y, b, 10}]; [IndentingNewLine]Plot3D[ Evaluate[u[x, y] /.
[InvisibleSpace]First[solutions]], {x, a, 10}, {y, b, 10},
PlotPoints [Rule] 50]; [IndentingNewLine]ContourPlot[
Evaluate[y + u[x, y] /. [InvisibleSpace]First[solutions]],
{x, a, 10}, {y, b, 10}, PlotPoints [Rule] 50,
ColorFunction [Rule] ((If[#1 > e, Green, Red]
&))][IndentingNewLine]];))
====
I'd be happier if I
could just control pagination, or if automaticpagination
were done sensibly.Yes, I can make a page break where I want
one, but I can't eliminatepage-breaks that leave tons of
white-space at the end of some pages. Insome of my
notebooks, Mathematica puts the first cell on a page
byitself, using less than an inch --- and NOT because the
next cell takesa whole page. If I force the first two cells
to stay together, theproblem just moves to the next
page.It's enough to make Mathematica useless as a word
processor all byitself.Bobby Treat-----Original
Message-----derivations;- setting typefaces in tables of
material.I figure most of this is do-able, but I 
don't have
the time, orpatience, to spend too much time on it. So,
I'll be the first customerwhen you write the 
guide to math
publishing in Mathematica - I just hopeyou won't have to
use LaTeX to write it.Mark Westwood> I would like to use
mathematica type papers for my math courses, but> I'm having
trouble formatting documents. Despite searching, I've been>
unable to find a complete guide to word processing with
mathematica.> Does anyone know where such a document could be
found?
====
> Wouldn't it be nice if NullSpace's behavior 
were
DOCUMENTED? Otherwise,> it's futile to give it approximate
numbers expecting any particular> behavior. Even if it
always works, it may not work in the next version> of
Mathematica.> BobbyThe expected, and documented, behavior
is that the output should be abasis for the null space,
that is, solutions of the homogeneous matrixequation
A.x==0. If this were to stop working then that would be
aserious bug. Is this the behavior you mean?The
implementation notes of the manual mention that
approximateNullSpace is based on a singular values
decomposition. This in factgives resulting vectors that are
orthonormal by the usualconjugate-symmetric inner product on
C (though these are now notnormal to the original vector
in this same inner product, unless theyare real-valued).
But this basis-orthogonality is not part of themission of
NullSpace and moreover should not become part of it.
Hencethat particular (and implementation dependent) aspect
of NullSpaceshould not become documented.Daniel
LichtblauWolfram Research
====
Wouldn't it be nice if
NullSpace's behavior were DOCUMENTED? 
Otherwise,it's futile
to give it approximate numbers expecting any
particularbehavior. Even if it always works, it may not
work in the next versionof Mathematica.Bobby-----Original
Message-----enough for me. Therefore I modify Ted's routine
toOrthogonalUnitVectors[vect__?(VectorQ[#, NumericQ] &)] /;
(SameQ @@ Length /@ {vect}) && (Length[First[{vect}]] > 1) :=
#/Sqrt[#.#] & /@ NullSpace[{vect}// N]----------------Lets
see what NullSpace does with approximate complex
vectors.In[1]:= v1 = {1.0 I, 0.0, 0.5 I, 0.0, 1.0}; v2 =
{0.0, 2.0, 1.0 I, 2.0, 0.5}; {v3,v4,v5} =
NullSpace[{v1,v2}]Out[3]= {{-0.730153 + 0.*I, 0. -
0.138254*I, 0.250585 + 0.*I, 0. -0.138254*I, 0.+ 0.60486*I},
{0. + 0.*I, -0.515861 + 0.*I, 0. + 0.457321*I, 0.687357 +
0.*I,0.22866+ 0.*I}, {0. + 0.*I, 0.510406 + 0.*I, 0. +
0.740442*I, -0.23274 + 0.*I,0.370221+ 0.*I}}--------In the
next line we see NullSpace returned vectors that are
orthogonaltothe vectors we gave NullSpace.In[4]:=
{v1.v3, v1.v4, v1.v5, v2.v3, v2.v4, v2.v5}//ChopOut[4]= {0,
0, 0, 0, 0, 0}----------However, the vectors returned aren't
orthogonal to each other.In[5]:= {v3.v4, v3.v5,
v4.v5}//ChopOut[5]= {0.229195*I, 0.371087*I,
-0.677239}---------I suppose an OrthogonalUnitVectors
function that uses NullSpace should (1) Only accept real
valued vectors. (2) Ensure NullSpace is given approximate
vectors.------ Ted Ersek
====
Plot[{x,x^2},{x,0,10}]Tomas
GarzaMexico City----- Original Message
-----
====
Mario,Here is a fancy version of your plot. I used
Text statements within anEpilog option to label the two
curves. The regular plot statement allows youto plot a
series of functions inclosed in a
list.Needs[Graphics`Colors`]Plot[{x, x^2}, {x, 0, 1},
PlotStyle -> {Black, Blue}, Frame -> True, FrameLabel -> {x,
y}, PlotLabel -> Comparison of Two Functions, Epilog ->
{Text[x, {0.5, 0.55}], Blue, Text[x^2, {0.7, 0.4}]},
Background -> Linen, ImageSize -> 500];David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/
Sender: steve@smc.vnet.netApproved: Steven M. Christensen
<steve@smc.vnet.net>, Moderator
====
>How can I plot with
Mathematica two function in the same graphic? I>explain
better: I'd like to draw a) y=x and b) y=x^2 in the same
axesPlot[{x, x^2},{x,-2,2}]
====
This might be a convenient
way of defining color
functions:ClearAll[cfun]cfun[colors_List, brkPts_List] /;
Length@colors == Length@brkPts := Function[z, Which @@
Sequence@Flatten@Transpose[{Less[z, #] & /@
brkPts,colors}]]colors = {White, RoyalBlue, White, Red,
White};brkPts = {-0.6, -0.42, 0.4, 0.6,
Infinity};ContourPlot[f[x, y], { x, 0, Pi}, {y, 0, Pi},
PlotPoints -> 30, ColorFunctionScaling ->False,
ColorFunction -> cfun[colors, brkPts], Contours ->
contourvalues];colors = {Yellow, Peru, Salmon, Apricot,
HotPink, Linen};brkPts = {-0.6, -0.42, 0.2, 0.4, 0.6,
Infinity};Timing[ContourPlot[f[x, y], {x, 0, Pi}, {y, 0, Pi},
PlotPoints -> 30, ColorFunctionScaling -> False, ColorFunction
-> cfun[colors,brkPts], Contours -> contourvalues];]Bobby
Treat-----Original Message-----it is difficult to obtain in
this plot.contourvalues = Complement[Range[-1, 1, 0.2],
{0.}]{-1, -0.8, -0.6, -0.4, -0.2, 0.2, 0.4, 0.6, 0.8, 1.}Now
we define a ColorFunction for the plot. I actually colored
twodifferentbands to show how you can make a general
color function to give eachband adesired color.cfun[z_]
:= Which[ -0.6 < z < -0.42, RoyalBlue, 0.4 < z < 0.6, Red,
True, White]ContourPlot[f[x, y], {x, 0, Pi}, {y, 0, Pi},
PlotPoints -> 30, ColorFunctionScaling -> False,
ColorFunction -> cfun, Contours -> contourvalues];Using the
option ColorFunctionScaling -> False says that the z
valuewill bethe actual value of f[x,y]. Otherwise, in
general, it will be scaledbetween0 and 1. It is easier to
write a color function when z is the actualvalueof the
function.David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/
Sender: steve@smc.vnet.netApproved: Steven M. Christensen
<steve@smc.vnet.net>, Moderator
====
For what it's worth,
here's a version of Simpson's rule that works 
with both
equally spaced and unequally spaced points.simp =
Compile[{x1,y1,x2,y2,x3,y3},With[{h13=x1-x3, h12=x1-x2,
h23=x2-x3},(-y1*h23*(2x1-3x2+x3) - y2*h13^2 +
y3*h12*(x1-3x2+2x3))*h13/(6h12*h23)]];
ListSimpson[vals_?VectorQ, dx_Real]:=
With[{n=Length[vals]},If[OddQ[n], (dx/3)*(First[vals] +
Last[vals] + 4*Plus@@vals[[Range[2,n-1,2]]] +
2*Plus@@vals[[Range[3,n-2,2]]]),
$Failed]];ListSimpson[datapts_?MatrixQ] :=
If[OddQ[Length[datapts]], Plus@@(simp@@Flatten[#]&/@
Partition[datapts,3,2]), $Failed] (* equally spaced
points*)data = Table[100*Sin[x], {x, 0, 100,
0.001}];ListSimpson[data, .001] // Timing {0.21 Second,
13.7681}(* randomly spaced points *)Table[{data =
Sort[Table[With[{x =
Random[Real,{0,100}]},{x,100*Sin[x]}],{100001}]];
ListSimpson[data] // Timing,With[{xvals =
Transpose[data][[1]]},gaps = Drop[RotateLeft[xvals] - xvals,
-1];{Min[gaps], Max[gaps]}]},{4}] {{{1.62 Second, 13.8348},
{1.32223*10^-8, 0.0114454}}, {{1.59 Second, 13.8261},
{2.42163*10^-8, 0.0127803}}, {{1.62 Second, 13.8455},
{1.08923*10^-9, 0.0119059}}, {{1.54 Second, 13.7987},
{9.66814*10^-9, 0.0106988}}}---Selwyn Hollis> Mathew, Some
possibilities <<NumericalMath`ListIntegrate`
ListIntegrate[data]//Timing {6.59 Second,13.7681} The
following is suggested in the Help Browser entry for the
package> Integrate[> Interpolation[data,
InterpolationOrder[Rule]1][x],> {x,0,100}]//Timing {4.56
Second,13.768} Trapezium rule with equal steps:
#[[1]]+#[[-1]]+ 2 Tr[Take[#,{2,-2}]]&[data[[All,2]]]
0.01/2//Timing {0.22 Second,13.768} Trapezium rule with
possibly unequal steps (Drop[#1,1] -
Drop[#1,-1]).(Drop[#2,-1] + Drop[#2,1])&[> data[[All,1]],
data[[All,2]]]/2//Timing {0.83 Second,13.768} --> Allan
---------------------> Allan Hayes> Mathematica Training and
Consulting> Leicester UK> www.haystack.demon.co.uk>
hay@haystack.demon.co.uk> Voice: +44 (0)116 271 4198> I've
tracked down the slow operation of my Mathematica simulation
> code to> lie in the ListIntegrate command:> G[n_] :=
ListIntegrate[xsec Phi[n], 1]> where both xsec and Phi[n] are
400 values long.> Is there a way to speed up ListIntegrate via
Compile or a similar > technique?> Matt> ---> Matthew
Rosen> Harvard-Smithsonian Center for Astrophysics> Mail
Stop 59> 60 Garden Street> Cambridge, MA 02138> e:
mrosen@cfa.harvard.edu> o: (617) 496-7614>
====
Daniel,I
don't mean to be overly critical of WRI's 
documentation --
it's verygood, as such things go. Nor do I like to overlook
chances to make itbetter. Do you?Your own comments below
point out that we should expect vectorsresulting from
NullSpace to be orthonormal by the
usualconjugate-symmetric inner product on C. But that's
not spelled out(documented) for dummies like myself who
don't know that's a naturalresult of singular 
values
decomposition.The mission of NullSpace (or any function)
is to adhere todocumentation, so reasonable persons may
differ on whether orthogonalityis a feature we should
depend on.The mission of documentation is to tell us what to
expect. When itdoesn't, the result is that we spend all this
time discussing issuesonline, trying to figure things out. A
simple don't depend onorthogonal results would be nice,
if that's the intent.In any case, I just spent ten minutes
LOOKING for implementation notesfor NullSpace, and have not
found any. Searching for implementationnotes doesn't help
and there's no link from NullSpace. What use isdocumentation
I can't find?In general, I don't 
like Mathematica's quirky
Help Browser, in which Icannot search for anything that's
not indexed. Every other help engineon my computer (and
there are hundreds) allows me to search for words,and that's
exactly what I need in order to find all mentions
ofNullSpace.Bobby Treat-----Original Message-----> of
Mathematica.> BobbyThe expected, and documented, behavior
is that the output should be abasis for the null space,
that is, solutions of the homogeneous matrixequation
A.x==0. If this were to stop working then that would be
aserious bug. Is this the behavior you mean?The
implementation notes of the manual mention that
approximateNullSpace is based on a singular values
decomposition. This in factgives resulting vectors that are
orthonormal by the usualconjugate-symmetric inner product on
C (though these are now notnormal to the original vector
in this same inner product, unless theyare real-valued).
But this basis-orthogonality is not part of themission of
NullSpace and moreover should not become part of it.
Hencethat particular (and implementation dependent) aspect
of NullSpaceshould not become documented.Daniel
LichtblauWolfram Research
====
P.J.,I'd be very interested
in how you created these fancy cells. Do youopen a cell,
push Ctrl-Shift-e, type in all that text, then
pushCtrl-Shift-e again?Bobby Treat-----Original
Message-----1) Put your function braches in the rows of a
grid box structure.2) Add the following options to your cell:
ShowAutoStyles -> False SpanMaxSize -> InfinityThe following
cell snippet demonstrates how this inßuences the result.To
view it, paste the Cell[] expression into a notebook and then
clickon Yes when you are prompted on whether the front end
should interpret the result.Cell[BoxData[ FormBox[ RowBox[{
RowBox[{f, (, x, )}], =, RowBox[{{, GridBox[{
{x, RowBox[{x,  , <, 0}]}, { SuperscriptBox[x,
2], RowBox[{0, [LessEqual], x, <, 1}]}, {
RowBox[{sin, (, x, )}], RowBox[{1,
[LessEqual], x, <, 2}]}, {
RowBox[{[CapitalGamma], (, x, )}], RowBox[{x,
[GreaterEqual], 2}]} }]}]}], TraditionalForm]],
DisplayFormula, ShowAutoStyles->False,
SpanMaxSize->Infinity]> - vertical alignment of equals signs
in multi-line equations or> derivations;Put your equations
in a GridBox and set the ColumnAlignments option to astring
containing the equal sign.Cell[BoxData[ FormBox[GridBox[{ {
RowBox[{ RowBox[{ RowBox[{3, x}],  , +,  ,
RowBox[{4,  , y}]}],  , =,  , 9}]}, {
RowBox[{ RowBox[{ RowBox[{2, x}],  , -,  ,
RowBox[{7,  , y}]}], =, RowBox[{32,  , -,  ,
RowBox[{sin, (, x, )}]}]}]} }], TraditionalForm]],
DisplayFormula,
GridBoxOptions->{ColumnAlignments->{=}}]> - setting
typefaces in tables of material.I think the Author Tools
material that comes with Mathematica 4.2 mightbe able to
help you do this.-- User Interface Programmer
paulh@wolfram.comWolfram Research, Inc.
====
> I'd be very
interested in how you created these fancy cells. Do you>
open a cell, push Ctrl-Shift-e, type in all that text, then
push> Ctrl-Shift-e again?Surely you have grossly
overestimated my expertise with Mathematicatypesetting
language syntax. :-)Here is the walkthrough for creating
each sample cell.Example 1:1) Set down a cell insertion
point in your notebook.3) Click on the cell's bracket.4)
Click on the front end menu command sequence:Format -> Cell
Style -> DisplayFormula5) Click on the front end menu
command sequence:Cell -> Display As -> TraditionalForm6)
Click on the front end menu command:Format -> Option
Inspector...7) Make sure that the Option Inspector scope
indicator is set toselection.8) Enter the name of the
option ShowAutoStyles into the search field andpress the
Lookup button.9) Change the value of the option
ShowAutoStyles from True to False. Thisprevents the front
end from showing the left brace with the umatchedsyntax
coloring.10) Look up the option SpanMaxSize as was done in
step (8).11) Change the value of this option to Infinity.12)
Close the Option Inspector dialog.13) Click within the cell
you just created to create an editor caret.14) Type in the
text f(x) = {4 x 2.17) Enter in the function branch
definitions in the left column, and thedomains of 
definition
in the right column. You can use the Tab key tonavigate
between grid elements.Example 2:1) Repeat Steps (1) - (11)
in Example 1.2) With the cell bracket still selected, look up
the option namedColumnAlignments. Change it from {Center} to
{=}.3) Close the Option Inspector dialog.4) Click within the
cell you just created to create an editor caret.6) Enter in
the equations, with one equation per grid element.I used the
front end menu command Edit -> Copy As -> Cell Expression
toget the underlying expressions.Hope that demystifies the
procedure for you.-- User Interface Programmer
paulh@wolfram.comWolfram Research, Inc.
====
You're not
doing anything dumb as far as I can see, but you're using
farmore memory than necessary. Here are statistics for your
algorithm onmy machine:SeedRandom[5];nums = NullDo[nums
= add[nums, Random[]], {5000}]; // Timingnums //
LeafCountnums // ByteCountCount[nums, Null,
Infinity]Count[nums, node[_, Null, _], 
Infinity]Count[nums,
node[_, _, Null], Infinity]Count[nums, node[_, Null, Null],
Infinity]Count[nums, node[___], 
Infinity]Count[nums, node[_,
_node | _?NumericQ, _node | _?NumericQ],
Infinity]Depth[nums]{4.406000000000001*Second,
Null}150012400005001 (* Null values in the tree *)2458 (*
nodes without left offspring *)2543 (* nodes without right
offspring *)1669 (* leaf nodes *)4999 (* total nodes *)1667
(* nodes with both left and right offspring *)28 (* minus
one, makes 27 levels in the tree *)It's clearly not ideal to
store 5001 Nulls for 5000 actual values!Here's a method that
doesn't change the algorithm much, but changes thestorage
method a great deal:First of all, I'm lazy, so 
I'll redefine
<,>,<=,>=, etc. to make thecode simpler:Unprotect[Less,
Greater, LessEqual, GreaterEqual];Less[a : _node ..] := Less
@@ (First /@ {a})LessEqual[a : _node ..] := LessEqual @@
(First /@ {a})Greater[a : _node ..] := Greater @@ (First /@
{a})GreaterEqual[a : _node ..] := GreaterEqual @@ (First /@
{a})Protect[Less, Greater, LessEqual, GreaterEqual];Again
because I'm lazy, I'll define 
node so that I don't have
toconsciously avoid leaf nodes and unnecessary
nesting:ClearAll[node]node[a___, node[b_], c___] = node[a,
b, c];node[node[a__]] = node[a];Here's my add
function:Clear[add]add[Null, x_] = node[x];add[x_,
(y_)?NumericQ] := add[node[x], node[y]]add[(x_)?NumericQ,
y_] := add[node[x], node[y]]add[node[x_], node[y_]] := If[x
> y, node[x, y], node[y, x]] (* <--THERE! *)add[node[x_,
lower_], y_node] := Which[y >= node[x], node[x, lower, y], y
<= node[lower], node[lower, y, x], True, node[y, lower,
x]]add[node[x_, lower_, higher_], y_node] /; node[y] <=
node[x] := node[x, add[lower, y], higher]add[node[x_,
lower_, higher_], y_node] := node[x, lower,
add[higher,y]]Where I have a pointer is the code that
prevents adding right-offspringto a leaf node. That avoids
nodes like your node[a,Null,c]. When youwould have
node[a,b,Null], I have node[a,b]. When you would
havenode[a,Null,Null], I use a itself.Timing and space
requirements are much better now:SeedRandom[5];nums =
NullTiming[Do[nums = add[nums, Random[]], {5000}];
]{2.109*Second,
Null}LeafCount[nums]ByteCount[nums]Count[nums, Null,
Infinity]Count[nums, node[_], Infinity]Count[nums, 
node[_,
_], Infinity]Count[nums, node[_, _, _], 
Infinity]Count[nums,
node[___], Infinity]Depth[nums]7852 (* 48% fewer leaf
expressions, mostly due to Nulls and right-offspring
eliminated *)165624 (* 31% less storage *)0 (* no Nulls,
versus 5001 *)0 (* no leaf nodes, versus 1669 *) (* no nodes
without left-offspring, versus 2458 *)705 (* nodes without
right offspring, versus 2543 *)2146 (* nodes with both left
and right offspring, versus 1667 *)2851 (* total nodes *)21
(* 21 tree levels versus 27 *)The logical structure is
identical to yours except that there are nonodes with only
right-offspring. This incidentally tends to reduce
treedepth.The storage format is far different: there are
no trivial (childless)nodes, and nodes with left-offspring
but no right-offspring are storedwithout a Null on the
right.I could make this quite a bit faster, I think, if I
spent time on thecode to eliminate changes to Less,
Greater, etc. and the frequentnesting followed by unnesting
that goes on in the algorithm. It mighttake twice as much
code, however, so I like it as is.You're probably better off
storing these things in a heap, of course.Or -- better yet --
use the built-in Sort and Ordering functions.Bobby
Treat-----Original
Message-----In[327]:=add[Null,x_Real]:=node[x,Null,Null]
add[Null,node[x_Real,lower_,higher_]]=node[x,lower,higher]In
[328]:=add[node[x_Real,lower_,higher_],y_Real]:=
If[x>y,node[x,add[lower,y],higher],node[x,lower,add[higher,y]
]]In[288]:=add[node[x_Real,lowerx_,higherx_],node[y_Real,
lowery_,highery_]]:=If[x>y,
node[x,add[lowerx,node[y,lowery,highery]],higherx],
node[x,lowerx,add[higherx,node[y,lowery,highery]]] ]Now this
is my attempt to test how fast my add
works:SeedRandom[5];Do[nums=add[nums,Random[]],{5000}];//
TimingOut[333]={13.279 Second,Null}RH7.3running on an
1.4GHz Athlon with 1GB of ram).Questions:1. Is this as fast
as I can get my code to run?2. Am I doing something obviously
stupid?3. would Compiling things help?Husain