A55 
==
NDSolve seems to have difficulties with solving
integral equation.> n = 5; NDSolve[{D[[Sigma]norm[z, t],
t] == 3*z*Integrate[[Sigma]norm[z,> t]^n*z, {z, 0, 1}] -
[Sigma]norm[z, t]^n,> [Sigma]norm[z, 0] == 1.5*z,
[Sigma]norm[0, t] == 0}*[Sigma]norm[z,> t], {z, 0.01,
1}, {t, 0.01, 2}]> Mathematica returns a message>
NDSolve::deql: The first argument must have both an
equation and an > initial condition.> which I cannot
understand.> Can anybody tell what's wrong with my attempt?>
> -Toshi> I am not really sure that your question isn't a
joke.NDSolve solves differential equations, not integral
equations.Although there are relations between these topics,
they are certainlynot the same.Alois-- Vienna University
of Technology,
====
evaluating the following gives you a
sample x=x+1 button:NotebookPut@Notebook[{Cell[BoxData[
ButtonBox[(x = (x + 1)), RuleDelayed[ButtonFunction,
CompoundExpression[If[Not[ ValueQ[x]], Set[x, 0]], Set[x,
Plus[x, 1]]]], Rule[ButtonEvaluator, Automatic]] ],
NotebookDefault, PageBreakAbove -> True, CellTags ->
GeneratedButtonBoxx=x+1]}, ClosingAutoSave -> True,
Editable -> False, WindowToolbars -> {}, PageWidth -> 299.5,
WindowSize -> {89., 29.}, WindowMargins -> {{92., Automatic},
{Automatic, 56.}}, WindowFrame -> Palette, WindowElements
-> {}, WindowFrameElements -> CloseBox, WindowClickSelect
-> False, ScrollingOptions -> {PagewiseScrolling -> True},
ShowCellBracket -> False, CellMargins -> {{0., 0.},
{Inherited, 0.}}, Active -> True, CellOpen -> True,
ShowCellLabel -> False, ShowCellTags -> False, ImageMargins
-> {{0., Inherited}, {Inherited, 0.}}, Magnification -> 1.5](*
********************************* *)Now, how do you create
such a button in less than a minute? ... :One way is to just
create a section cell and the underlying ButtonFunction
code as input cells, i.e., type interactively such that you
get something like:NotebookPut[Notebook[
{Cell[CellGroupData[ {Cell[x=x+1, Section],
Cell[If[!ValueQ[x], x=0], Input], Cell[x=x+1,
Input]}, Open]]}]]Then hit the F2B (function to
Button) button in ButtonTools.nb ( my freewarebutton tools
from http://www.mertig.com/mathdepot ) and you get the
button.With the HP and VP you can easily and quickly
generate (horizontally orverically) palettes. Check out
the Help button, or also the source code.It basically is
all straightforward and there is actually documentation about
all those ButtonFunction features somewhere. I agree that the
whole Button-design could have been made better, but up to a
point is quite useful. Of course the world is used to
better GUI's these days but if you really need nice 
GUI's
and buttons, use Java and JLink ( and there are also simple
examples in the JLink manual of how to do this ).If you
don't like Java, go with VBA and use the nice Mahematica
for Active Xproduct from http://www.episoft.comRolf
MertigMertig
Consultinghttp://www.mertig.com
====
tryIn[1]:=Clear[a,b,
c,d,x,y]x= {{0,0,0,1},{1,0,0,1},{0,1,0,1},{1,1,1,1}};y =
{a,b,c,d}; LinearSolve[x,y]s01=
LinearSolve[x,y][[1]]Out[4]={-a+b,-a+c,a-b-c+d,a}Out[5]=-a+
bIn[6]:=g[a_,b_]
=s01Out[6]=-a+bIn[7]:=g[1,3]Out[7]=2*NEVER* use capital
letters at the beginning of a variable's name!
Never!Matthias BodeSal. Oppenheim jr. & Cie.
KGaAKoenigsberger Strasse 29D-60487 Frankfurt am
MainGERMANYMobile: +49(0)172 6 74 95 77Internet:
http://www.oppenheim.de-----Ursprí.b9ngliche
Nachricht-----Gesendet: Freitag, 23. August 2002 06:25An:
mathgroup@smc.vnet.netBetreff: := Does not assign
variables properly.Why?Here's a piece of a conversion I had
with Mathematica.Why is a[A_,B_] := LinearSolve[X,Y][[1]]
not givingme the function I expect?In[261]:= X =
{{0,0,0,1},{1,0,0,1},{0,1,0,1},{1,1,1,1}};In[262]:= Y =
{A,B,C,D};In[263]:= LinearSolve[X,Y]Out[263]=
{-A+B,-A+C,A-B-C+D,A}In[264]:=
LinearSolve[X,Y][[1]]Out[264]= -A+BIn[265]:= a[A_,B_] :=
LinearSolve[X,Y][[1]]In[266]:= a[1,3]Out[266]= -A+BThe
output above is not what I want. I want 2. Here'swhat I
expect:In[267]:= a[A_,B_] := -A+B;In[268]:=
a[1,3]Out[268]= 2This output is what I expect. What is the
difference betweenthe two?
====
First of all, just look at
your own posting below. You clearly have a *(Times) where a ,
(comma) should be. Presumably your input ought to be:n = 5;
NDSolve[{D[[Sigma]norm[z, t], t] ==
3*z*Integrate[[Sigma]norm[z,t]^n*z, {z, 0, 1}] -
[Sigma]norm[z, t]^n, [Sigma]norm[z, 0] == 1.5*z,
[Sigma]norm[0, t] == 0},[Sigma]norm[z,t], {z, 0.01, 1},
{t, 0.01, 2}]However, even in the corrected version the
equation can't be solved. First of all you will get the
complaint:NDSolve::bcedge: Boundary conditions must be
specified at the edge of the spatial domain.In other
words Mathematica wants a boundary condition for
[Sigma]norm[z, 0.1] or alternatively you should use {z,0,1}
in NDSolve. But actually I do not think this equation is
solvable by any numerical scheme even if you could provide
the initial conditions at the edge of the boundary that
Mathematica requests. To evaluate the integral in your
equation NDSolve needs to know the values of
[Sigma]norm[z, t] for all z between 0 and 1 and a given t,
but this knowledge is not available at any stage of the
evaluation. I am not really an expert, but this seems to me
a clear example of an equation that is not solvable by any
numerical means.By the way, the fact that you know a
solution to a differential equation, and even the fact that
the solution is very simple does not imply that the equation
can be solved by any known method, except of course guessing,
which computer programs generally do not use. Andrzej
KozlowskiToyama International UniversityJAPANOn Friday,
August 23, 2002, at 05:25 AM, Toshiyuki ((Toshi)) Meshii >>
NDSolve seems to have difficulties with solving integral
equation.>> n = 5; NDSolve[{D[[Sigma]norm[z, t], t] == >
3*z*Integrate[[Sigma]norm[z,> t]^n*z, {z, 0, 1}] -
[Sigma]norm[z, t]^n,> [Sigma]norm[z, 0] == 1.5*z,
[Sigma]norm[0, t] == > 0}*[Sigma]norm[z,> t], {z, 0.01,
1}, {t, 0.01, 2}]>> Mathematica returns a message>>
NDSolve::deql: The first argument must have both an
equation and an > initial condition.>> which I cannot
understand.> Can anybody tell what's wrong with my 
attempt?>>
-Toshi>
====
>Here's a piece of a conversion I had with
Mathematica.>Why is a[A_,B_] := LinearSolve[X,Y][[1]] not
giving>me the function I expect?>>In[261]:= X =
{{0,0,0,1},{1,0,0,1},{0,1,0,1},{1,1,1,1}};>In[262]:= Y =
{A,B,C,D};>In[263]:= LinearSolve[X,Y]>Out[263]=
{-A+B,-A+C,A-B-C+D,A}>In[264]:=
LinearSolve[X,Y][[1]]>Out[264]= -A+B>In[265]:= a[A_,B_] :=
LinearSolve[X,Y][[1]]>In[266]:= a[1,3]>Out[266]= -A+B>>The
output above is not what I want. I want 2. Here's>what I
expect:>>In[267]:= a[A_,B_] := -A+B;>In[268]:=
a[1,3]>Out[268]= 2>>This output is what I expect. What is the
difference between>the two?X={{0,0,0,1},{1,0,0,1},
{0,1,0,1},{1,1,1,1}};Y={A,B,C,D};a[A_,B_]:=
LinearSolve[X,Y][[1]];?aGlobal`aa[A_, B_] :=
LinearSolve[X,Y][[1]]Note that the RHS of the stored
definition is not a function of the arguments. Now add
Evaluate to RHSa[A_,B_]:=
Evaluate[LinearSolve[X,Y][[1]]];?aGlobal`aa[A_, B_] := -A
+ Ba[1,3]2Bob HanlonChantilly, VA USA
====
All of this
looks like a mistake to me because it seems far too easy.
But anyway, here is the solution that makes almost no use of
Mathematica. First of all, your equation is not a
differential equation so there is no point using
DSolve.Secondly the use of z in
Integrate[[Sigma]norm[z]^n*z, {z, 0, d}] is deceptive,
since you are integrating over z, so let's replace it by
something else, say s. So your equation
is:(3*z)/d^3)*Integrate[[Sigma]norm[s]^n*s, {s, 0, d}]
==[Sigma]norm[z]^nwhich is supposed to hold true for
every z>0. Re-write it asIntegrate[[Sigma]norm[s]^n*s,
{s, 0, d}]/d^3 =[Sigma]norm[z]^n/3zfor all z. However,
the left hand side is a function of d, independent of z, so
we can write:[Sigma]norm[z_]:=(3z*g[d])^(1/n)Let's take
this as a definition and substitute in the original
equationIn[2]:=Simplify[((3*z)*Integrate[[Sigma]norm[s]^
n*s, {s, 0, d}])/d^3 == [Sigma]norm[z]^n, {d > 0, n > 0, z
> 0}]Out[2]=TrueThat means you can take g to be an
arbitrary function of d.Andrzej KozlowskiToyama
International UniversityJAPANOn Friday, August 23, 2002,
at 05:25 AM, Toshiyuki ((Toshi)) Meshii > How can I solve
the following integral equation?> Mathematica seems not to
work.> Is there any way?>>
DSolve[((3*z)/d^3)*Integrate[[Sigma]norm[z]^n*z, {z, 0, d}]
==> [Sigma]norm[z]^n, [Sigma]norm[z], z]>> note: z>0 &
n>1>> I know that the answer is simple and> $B&R(Bnorm[z_]
= (1 + 1/(2*n))*(z/d)^(1/n)>> -Toshi>
====
But suddenly
the following Errormessage
appears:/usr/bin/local/mathematica: file or directory not
foundMandrake-updates ... but so far everything exceptof
Mathematica seems to work fine.)Of course I checked for the
File (it is indeed there) and (a hint from a Unix-usegroup)
the needed libs libc.so.5, libm.so.5 are in /lib/ too.Had
anyone had, ore better solved, a similar problem?greetings
Detlef
====
How can I create a Mathematica thing (function?
program?) that wouldautomatically open a browser page, and
give my username and pwd to log meinto a https:....
site? The server's login page has a script 
resettingfields:
<script language=javascript> function ResetFormFields() {
document.Login.username.value = ;
document.SearchUserForm.search.value = ;
document.FindCourseForm.searchText.value = ; } </script>
<body . . . ONLOAD=ResetFormFields();
document.Login.username.select();
document.Login.username.focus();>and the relevant input
cells are<input type=text name=username . . . ><input
type=password name=password . . . ><input
type=submit name=login
value=Login>Nicholas
====
Howdy,I'm trying to figure out
the correct syntax to do the following. I havesome function
with three arguments, and I want to syntactically describe
thesingle-argument function that holds two of those
arguments constant (i.e.without creating that
single-argument function).More specifically, I have 
defined
Machine[radix_,multiplier_,state_] := Module [{c,s}, c =
Floor[state/base]; s = Mod[state,base]; multiplier*s + c
]where I have a generalize 'machine', 
defined by the radix
and multiplier,which converts one state into another state.
So I'd like to be able to dosomething like this:
NestList[Machine[10,7,#], 3, 22]to get the series of states
that the radix-10 multiplier-7 machine runsthrough
(starting with state 3). However, this syntax doesn't seem 
to
dowhat I want.I hope that description makes sense. It seems
like there must be a syntaxto describe the function
Machine[10,7,#].Anyone have any ideas?Bob H
====
>>There is
a following problem with Mathematica 4.2:>when i try to load
Help Browser, i get a message:>building help browser index
(first time only)>scanning index file>and 
Mathematica stops
responding: you have to>kill process. There was no such a
problem with>version 4.1.>>Is there is a solution to
that?I'd start with the following
FAQ.http://support.wolfram.com/mathematica/interface/
helpbrowser/howrebuildindex.html-Dale
====
> There is a
following problem with Mathematica 4.2:> when i try to load
Help Browser, i get a message:> building help browser index
(first time only)> scanning index file> and 
Mathematica stops
responding: you have to> kill process. There was no such a
problem with > version 4.1.> Is there is a solution to
that?> I have the exact same problem with 4.2 (on Win2K)
:(
====
>I'd start with the following
FAQ.>>http://support.wolfram.com/mathematica/interface/
helpbrowser/howrebuildindex.html>>-DaleSorry that i
failed say it immediately in a first place, but of coursei
did try it FAQ at first, and both tried to delete cache and
rebuildindex, but results where the same - whenever i try
to invoke help browser (or rebuild index, for that matter),
mathematica stops responding (i did read your answer to the
same question asked a week ago before - actually that is why
i turned to the FAQ).
====
People encounter this all the time.
It is because SelectionEvaluate does not do what you think.
It does not work like ToExpression, which causes immediate
kernel evaluation. Instead it works like when you press
Shift-Enter, which selects a cell for evaluation after all
current evaluations have finished.See
http://support.wolfram.com/mathematica/kernel/interface/
selectionevaluate.html-Dale>>Trying to manipulate
notebooks from the kernel I found>an unexpected bahavior
with Mathematica. First I tried>>the following commands one
by one (they are not int>the same cell, and they are not
selected at same time>for evaluation)>>nb =
NotebookCreate[];>i = 0;>(* Purpose is a Do loop here
*)>NotebookDelete[nb] (*1*)>NotebookWrite[nb, ++i, All]
(*2*)>SelectionEvaluate[nb] (*3*)>SelectionMove[nb, All,
Cell] (*4*)>>If I repeat evaluating coomands (*1,2,3,4*) one
by one>what is shown in the created notebook is an
animation>of the index i in the same cell.>Naturally a loop
must do the job. However when I>intent to collect
(*1,2,3,4*) in the same cell the>result is not the same even
whitout the Do loop (I>mean just evaluating this cell several
times)>>I would be grateful if some can explain what's
going>on here or if there is something wrong with
my>machine.>>Cesar>________________________________
__________________>Do You Yahoo!?>Yahoo! Finance - Get
real-time stock quotes>http://finance.yahoo.com
====
>Here's a
piece of a conversion I had with Mathematica.>Why is
a[A_,B_] := LinearSolve[X,Y][[1]] not giving>me the
function I expect?>>In[261]:= X =
{{0,0,0,1},{1,0,0,1},{0,1,0,1},{1,1,1,1}};>In[262]:= Y =
{A,B,C,D};>In[263]:= LinearSolve[X,Y]>Out[263]=
{-A+B,-A+C,A-B-C+D,A}>In[264]:=
LinearSolve[X,Y][[1]]>Out[264]= -A+B>In[265]:= a[A_,B_] :=
LinearSolve[X,Y][[1]]>In[266]:= a[1,3]>Out[266]= -A+B>>The
output above is not what I want. I want 2. Here's>what I
expect:>>In[267]:= a[A_,B_] := -A+B;>In[268]:=
a[1,3]>Out[268]= 2>>This output is what I expect. What is the
difference between>the two?This is a common misconception
about what := does. What it does is set up a variable
replacement for the unevaluated expression, not the evaluated
expression. Soa[A_,B_] := LinearSolve[X,Y][[1]]a[1,3]is
similar to doingReleaseHold[ Hold[ LinearSolve[X,Y][[1]] ]
/. {A->1, B->3}]which means that only explicit instances of
A and B are replaced. What you are attempting is done with
=. This assigns the function to the evaluated
expression.In[5]:=a[A_,B_] =
LinearSolve[X,Y][[1]]Out[5]=-A+BIn[6]:=a[1,3]Out[6]=2
There are certain situations where you get a different result
if you evaluate with the values or replace the values in the
symbolic evaluation. (This example is not one of them.) Using
= does the later. If you want to do the former, you should
use := and have the right hand side use A and B explicitly
or you could use a Block.In[7]:=a[b_,c_] := Block[{A=b,
B=c}, LinearSolve[X,Y][[1]]
]In[8]:=a[1,3]Out[8]=2-------------------------------------
-------------------------Omega ConsultingThe final answer
to your Mathematica needsSpend less time searching and more
time
finding.http://www.wz.com/internet/Mathematica.html
====
>
Normally I don't feel so stupid, but I'm 
trying to create an
interactive>Mathematica notebook, and I'm stuck at square
one.>>Specifically, how do I do something like create a button
to add a number (or>perform other mathematical functions)
and then display or otherwise>manipulate the result
(eventually I want to be able to click a button
that>increments/decrements an angle and animate the
resulting transformation of a>vector, with an eye to finally
simulating a simple robotic arm complete with>simple
controls to manipulate the arm).>>I want to have a variable
(or matrix or whatever) defined as a global>variable x, and
then perform x = x+1 when a button is clicked. I've
been>using and programming computers for almost 25 years and
I can't follow>Wolfram's documentation. Is it 
just me, or is
he always this obtuse when>explaining things? I mean, given
the amazing power of the Mathematica>system, a sample list
of buttons in a notebook that you could select and>examine
how they were implemented, would have been nice. I can't 
find
such a>list, and this seems to be par for the course for
the rest of the>documentation as well.Button programming
can be very confusing at first. It has a whole series of
quirks that make it in many ways unique to Mathematica and
programming in general. This causes a lot of
head-scratching, but once you understand what's going on
things get easier.The key is to create a button that uses
the kernel (by default it only uses the front end). Here's
a simple example to help you get started.In[1]:=
x=1;In[2]:=ButtonBox[Increment x,Active->True,
ButtonEvaluator->Automatic, ButtonFunction:>Print[x =
,++x] ]//DisplayFormSome random things to note:-
ButtonEvaluator->Automatic. This says use the kernel to
implement the ButtonFunction.- Buttons only create
side-effects. They generate no output. What you see when you
press the button is the result of 2 side effects. One from ++,
which changes the value of x. The other from Print, which
creates a cell.- Print is, in general, a poor side-effect to
use in a button. It's difficult to control where 
the Print
cell is placed. It is worth your while to learn how to use
other front end side-effect functions (such as NotebookRead
and NotebookWrite) when using buttons.Here are some further
resources you might find
helpful:http://support.wolfram.com/mathematica/interface/
buttons/http://library.wolfram.com/conferences/devconf99/
hinton/Buttons19991022.nbhttp://library.wolfram.com/
conferences/devconf2001/horton2/horton2.nb------------------
--------------------------------------------Omega
ConsultingThe final answer to your Mathematica needsSpend
less time searching and more time
finding.http://www.wz.com/internet/Mathematica.html
====
How
do I change format used for tick mark labels in a 2D plot? I
wouldlike to use DigitBlock->3 option of NumberForm to
format large numbersappearing as tick mark labels on a
histogram.Alexander
====
As Daniel Lichtblau pointed out,
the statement below about vertices is nonsense. Consider two
overlapping rectangles arranged as a cross. You need to
compute intersections and test them instead of
vertices.Begin forwarded message:>> Begin forwarded
message:>> Dear colleagues,>> any hints on how to implement a
very fast routine in Mathematica for> testing if two
rectangles have an intersection area?> Frank Brand> Here
is one approach.>> Given three points
{x1,y1},{x2,y2},{x3,y3}, switch to homogenous> coordinates
a={1,x1,y1}, b={1,x2,y2}, c={1,x3,y3}. Then>
Sign[Det[{a,b,c}]] is +1 if and only if the point a lies on
your left as> you walk along the line though b and c in the
direction from b to c.> ( If the result is zero, then a lies
on the line.)>> The value of the determinant is
x2y3-x3y2-x1y3+x3y1+x1y2-x2y1, and the> speed of the
algorithm depends essentially on how fast this quantity can>
be computed. Suppose we write a function LeftSide[a,{b,c}]
that computes> the sign of the determinant.>> Now let
{p1,p2, . . ., pn} be a list of vertices (pi={xi,yi}) of a>
convex polygon traced counterclockwise. Then a lies within or
on the> boundary of the polygon if and only if none of the
numbers> LeftSide[a,{pi,p(i+1)}] are -1. That is, if -1 does
not appear in the> list LeftSide[a,#]&/@Partition[{p1,p2,. .
.,pn,p1},2,1].>> Now use the fact that if two convex
polynomials overlap, then some> vertex of one of them must
lie inside or on the boundary of the other.>> If an overlap
of positive area is required, then the check is that only>
+1 appears--not that -1 does not appear.>> For two rectangles
( or parallelograms) this approach requires the> evaluation
of 16 determinants, so it may be a bit expensive. If the>
points have rational coordinates, then (positive)
denominators may be> cleared in the homogeneous coordinates
and the computations can be done> in integer arithmetic, at
the cost of at least three more> multiplications per
determinant.>>Garry HelzerDepartment of
MathematicsUniversity of MarylandCollege Park, MD
20742301-405-5176gah@math.umd.edu>>
====
While playing
arounf with patterns and substitutions, I came across
thefollowing behavior which I didn't expect:z :=
SomeHead[{{1, 2}, {3, 4}}]z /. {SomeHead[q_] ->
Flatten[q]}While this _does_ yield the desired result
{1,2,3,4}, Mathematica complains:Flatten::normal:
Nonatomic expression expected at position 1
inFlatten[q].........as if it is trying to evaluate
Flatten[q], with q not bound to{{1,2},{3,4}}.Could anybody
explain why this happens? Sidney Cadot
====
> While
playing arounf with patterns and substitutions, I came across
the> following behavior which I didn't expect:> z :=
SomeHead[{{1, 2}, {3, 4}}]> z /. {SomeHead[q_] ->
Flatten[q]}> While this _does_ yield the desired result
{1,2,3,4}, Mathematica complains:> Flatten::normal:
Nonatomic expression expected at position 1 in> Flatten[q].>
> ........as if it is trying to evaluate Flatten[q], with q
not bound to> {{1,2},{3,4}}.> Could anybody explain why
this happens?> Sidney Cadotduring the Replaceall (/.)
Mathematica first evaluate the ruleSomeHead[q_] ->
Flatten[q] . q is stilla symbol then and will bereplaces
when /. is used.You can avoid this Problem by using a
delayed rule SomeHead[q_] :>Flatten[q] or using thr Rule
Somehead->Flatten.Yours, Alexander-- / Alexander Dreyer,
Dipl.-Math. - Abteilung Adaptive Systeme / Fraunhofer
Institut fuer Techno- und Wirtschaftsmathematik (ITWM) 
Gottlieb-Daimler-Strasse, Geb. 7^2=49/313 D-67663
Kaiserslautern /
====
> ............... I came across the>
following behavior which I didn't expect:>> z :=
SomeHead[{{1, 2}, {3, 4}}]> z /. {SomeHead[q_] ->
Flatten[q]}>> While this _does_ yield the desired result
{1,2,3,4}, Mathematicacomplains:>> Flatten::normal:
Nonatomic expression expected at position 1 in>
Flatten[q].>............> Could anybody explain why this
happens?Sidney,The clue is that, before any replacing is
done, z is evaluated to SomeHead[{{1, 2}, {3, 4}}];
SomeHead[q_] is evaluated to SomeHead[q_] Flatten[q] is
evaluated to Flatten[q] and the message is generatedsaying
essentially nothing to ßatten.*Then* the replacement
SomeHead[{{1, 2}, {3, 4}}] /. {SomeHead[q_] -> Flatten[q]}is
performed.It would be better here to use RuleDelayed, :>,
instead of Rule, ->, sinceFlatten[q] would not then be
evaluated. z:=SomeHead[{{1,2},{3,4}}]
z/.{SomeHead[q_]:>Flatten[q]}
{1,2,3,4}--Allan---------------------Allan
HayesMathematica Training and ConsultingLeicester
UKwww.haystack.demon.co.ukhay@haystack.demon.co.ukVoice
: +44 (0)116 271 4198
====
Never mind my question, I should
have used a delayed rule there (:> insteadof ->).Bit silly
of me.
====
Suppose f is a function of n real variables, and
returns avector of n real variables. What is the correct
syntax to find a rootof f using FindRoot?For instance, the
following works with n=7, but I would need to changethe
code if I changed the value of n:f[p_] := Table[fk[p, k],
{k,n}];pn = {p1,p2,p3,p4,p5,p6,p7};f1 := f[pn] [[1]];f2 :=
f[pn] [[2]];f3 := f[pn] [[3]];f4 := f[pn] [[4]];f5 := f[pn]
[[5]];f6 := f[pn] [[6]];f7 := f[pn] [[7]];theRoot =
FindRoot[{f1==0,f2==0,f3==0,f4==0,f5==0,f6==0,f7==0}, {p1,
1/n},{p2, 1/n},{p3, 1/n},{p4, 1/n},{p5, 1/n}, {p6, 1/n},{p7,
1/n}];-- John MacCormick Systems Research Center, HP Labs,
1501 Page Mill Road,
====
>I'm trying to figure out the correct
syntax to do the following. I have>some function with three
arguments, and I want to syntactically
describe>the>single-argument function that holds two of
those arguments constant (i.e.>without creating that
single-argument function).>>More specifically, I have
defined>> Machine[radix_,multiplier_,state_] := Module
[{c,s},> c = Floor[state/base]; s = Mod[state,base];>
multiplier*s + c> ]>>where I have a generalize 
'machine',
defined by the radix and multiplier,>which converts one state
into another state. So I'd like to be able to>do>something
like this:>> NestList[Machine[10,7,#], 3, 22]>>to get the
series of states that the radix-10 multiplier-7 machine
runs>through (starting with state 3). However, this syntax
doesn't seem to>do>what I want.>>I hope that description
makes sense. It seems like there must be a syntax>to
describe the function Machine[10,7,#].>I assume that you
want base rather than radix in the definition (or vice
versa).Machine[base_, multiplier_, state_]:= Module[{c, s},
c=Floor[state/base]; s=Mod[state, base];
multiplier*s+c];NestList[Function[Machine[10,7,#]],3,22]{3,
21, 9, 63, 27, 51, 12, 15, 36, 45, 39, 66, 48, 60, 6, 42, 18,
57, 54, 33, 24, 30, 3}The abbreviation for Function[body] is
body&%==NestList[Machine[10, 7, #]&, 3, 22]TrueBob
HanlonChantilly, VA USA
====
>While playing arounf with
patterns and substitutions, I came across the>following
behavior which I didn't expect:>>z := SomeHead[{{1, 2}, {3,
4}}]>z /. {SomeHead[q_] -> Flatten[q]}>>While this _does_
yield the desired result {1,2,3,4}, Mathematica
complains:>>Flatten::normal: Nonatomic expression
expected at position 1 in>Flatten[q].>>........as if it is
trying to evaluate Flatten[q], with q not bound
to>{{1,2},{3,4}}.>>Could anybody explain why this
happens?z:=SomeHead[{{1,2},{3,4}}]The error is associated
with defining the Rule not executing it. Note that you get
the error with{SomeHead[q_]->Flatten[q]};Use
RuleDelayedz/.{SomeHead[q_]:>Flatten[q]}{1, 2, 3, 4}Bob
HanlonChantilly, VA USA
====
Can anyone help me with this
problem. If I have an n-element list, (say where each
element is itself alist), such as {{a,b}, {a,b}, {a,b}}is
there a way to strip off the outermost nesting of the list
toobtain just a sequence of of these n elements, that
is{a,b},{a,b},{a,b} so that I can use this for input for
some function.I would like to do something
likeOuter[SomeFunction, Table[{a,b},{N} ]] where I can
enter Ndynamically.The problem, of course, is that the
output of the Table command is onebig listand Outer is
expecting a sequence of N separate lists afterSomeFunction.

====
Bob, Apply[Sequence,{{a,b},{c,d}}] Sequence[{a,b},{c,d}]
Outer[F,Apply[Sequence,Table[{a,b},{3}]]]
{{{F[a,a,a],F[a,a,b]},{F[a,b,a],F[a,b,b]}},{{F[b,a,a],F[b,a,b
]},{F
[b,b,a],F[b,b,b]}}}---Allan---------------------Allan
HayesMathematica Training and ConsultingLeicester
UKwww.haystack.demon.co.ukhay@haystack.demon.co.ukVoice
: +44 (0)116 271 4198> Can anyone help me with this
problem.>> If I have an n-element list, (say where each
element is itself a> list), such as {{a,b}, {a,b}, {a,b}}>
is there a way to strip off the outermost nesting of the list
to> obtain just a sequence of of these n elements, that is>
{a,b},{a,b},{a,b} so that I can use this for input for some
function.>> I would like to do something like>
Outer[SomeFunction, Table[{a,b},{N} ]] where I can enter N>
dynamically.> The problem, of course, is that the output of
the Table command is one> big list> and Outer is expecting
a sequence of N separate lists after> SomeFunction.>>
====
>
Can anyone help me with this problem.> If I have an
n-element list, (say where each element is itself a> list),
such as {{a,b}, {a,b}, {a,b}}> is there a way to strip off
the outermost nesting of the list to> obtain just a sequence
of of these n elements, that is> {a,b},{a,b},{a,b} so that I
can use this for input for some function.> I would like to
do something like> Outer[SomeFunction, Table[{a,b},{N} ]]
where I can enter N> dynamically.> The problem, of course,
is that the output of the Table command is one> big list>
and Outer is expecting a sequence of N separate lists after>
SomeFunction.> Sequence@@{{a,b}, {a,b}, {a,b}}resp.
Apply[Sequence, {{a,b}, {a,b}, {a,b}}]will do it.CU,
Alexander-- / Alexander Dreyer, Dipl.-Math. - Abteilung
Adaptive Systeme / Fraunhofer Institut fuer Techno- und
Wirtschaftsmathematik (ITWM)  Gottlieb-Daimler-Strasse,
Geb. 7^2=49/313 D-67663 Kaiserslautern /
====
some function
with three arguments, and I want to syntactically describe
thesingle-argument function that holds two of those
arguments constant (i.e.without creating that
single-argument function).More specifically, I have 
defined
Machine[radix_,multiplier_,state_] := Module [{c,s}, c =
Floor[state/base]; s = Mod[state,base]; multiplier*s + c
]where I have a generalize 'machine', 
defined by the radix
and multiplier,which converts one state into another state.
So I'd like to be able to dosomething like this:
NestList[Machine[10,7,#], 3, 22]to get the series of states
that the radix-10 multiplier-7 machine runsthrough
(starting with state 3). However, this syntax doesn't seem 
to
dowhat I want.I hope that description makes sense. It seems
like there must be a syntaxto describe the function
Machine[10,7,#].Anyone have any ideas?Bob
H
====
Sidney,With a direct rule Mathematica tries to
Flatten the symbol q immediately.You want a delayed rule to
avoid the error message.z /. SomeHead[q_] :>
Flatten[q]David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/{{
1,2},{3,4}}.Could anybody explain why this happens? Sidney
Cadot
====
You are quite right, Mathematica does evaluate
Flatten[q] before substituting {{1, 2}, {3, 4}}. It then
issues the error message and returns Flatten[q]. Only now
{{1, 2}, {3, 4}} is substituted for q, Flatten[{{1, 2}, {3,
4}}] is evaluated and you get the right answer. To avoid all
this just use RuleDelayed instead of Rule:In[1]:=z :=
SomeHead[{{1, 2}, {3, 4}}];In[2]:=z /. {SomeHead[q_] :>
Flatten[q]}Out[2]={1,2,3,4} Andrzej KozlowskiToyama
International UniversityJAPAN>> While playing arounf with
patterns and substitutions, I came across the> following
behavior which I didn't expect:>> z := SomeHead[{{1, 2}, {3,
4}}]> z /. {SomeHead[q_] -> Flatten[q]}>> While this _does_
yield the desired result {1,2,3,4}, Mathematica >
complains:>> Flatten::normal: Nonatomic expression
expected at position 1 in> Flatten[q].>> ........as if it is
trying to evaluate Flatten[q], with q not bound to>
{{1,2},{3,4}}.>> Could anybody explain why this happens?>
Sidney Cadot>>
====
> [...] Y = {A,B,C,D}; [...]>
In[265]:= a[A_,B_] := LinearSolve[X,Y][[1]]> In[266]:=
a[1,3]> Out[266]= -A+B> The output above is not what I
want. I want 2.> [...]You are mixing the (global (*))
symbols A and B with the (local (*)) patterns A_ and B_. If
you want to replace the (global) solution with local values
you should writeIn[4]:= a[AA_, BB_] := LinearSolve[X,
Y][[1]] /. {A -> AA, B -> BB}; a[1, 3]Out[5]=
2________________(*) as far we can say that in Mathematica--
Rainer Gruber
====
> NestList[Machine[10,7,#], 3, 22]You
almost have it:NestList[Machine[10,7,#]&, 3, 22]More
elegantly, you can defined Machine this
way:Machine[radix_,multiplier_][state_] := ...Then you can
writeNestList[Machine[10,7], 3, 22]Tom Burton
====
> z :=
SomeHead[{{1, 2}, {3, 4}}]> z /. {SomeHead[q_] ->
Flatten[q]}> While this _does_ yield the desired result
{1,2,3,4}, Mathematica complains:> Flatten::normal:
Nonatomic expression expected at position 1 in>
Flatten[q].The -> operator is prompt: It permits evaluation
of the RHS Flatten[q]immediately, before it is used in the
substitution. Evidently, q wasundefined before you tried
this, so Flatten[q]'s evaluation yielded itself.Later, in
the process of substitution, with q set as you intended,
Flattenevaluated again, this time yielding the desired
result.Try it again after setting q to something. Then you
get not only thecomplaint but also the wrong answer.To
avoid these issues, use the delayed operator :> instead of
the promptoperator ->. A rule of thumb is: Use delayed
operators := and :> when theLHS (SomeHead[q_] in your case)
contains a blank (_).Tom Burton
====
 Mark:In: ?KOut: K is
a default generic name for a summation index in a symbolic
sum.Turns out that there are seven single-letter
symbols.In: ClearAll[Global`*]; Select[Names[*],
(StringLength[#] == 1) &]Out: {C, D, E, I, K, N, O}K is a
little weird, because it's not Protected.In: Attributes[{C,
D, E, I, K, N, O}]Out: {{NHoldAll, Protected}, {Protected,
ReadProtected}, {Constant, Protected, ReadProtected},
{Locked, Protected, ReadProtected}, {}, {Protected},
{Protected, ReadProtected}}----Selwyn Hollis> I just
learned that K is a System` Symbol:> Information[K]>
Context[K]> I learned this due to an error message:>
Block[{K = 1}, Sum[j, {j, i}]]> The same message can be
generated by the following:> K := 1> Sum[j, {j, i}]> The
message can be eliminated by Removing K:> Remove[K]>
Block[{K = 1}, Sum[j, {j, i}]]> K := 1> Sum[j, {j, i}]>
Surely, this is not intentional.> --Mark.> 
====
Dear
MathgroupI have convert pde to ode like thisdu_i/dt = 1-4
u_i + .02 (u_i-1 .9a 2 u_i + u i+1)/(delta (x))^3 + (u_i)^3
v_idv_i /dt = 3 u_i + .02 (v_i-1 .9a 2 v_i + u i+1)/(delta
(x))^3 - (u_I)^3 v_idelta(x)=(i)/(N+1)x_i=
(i)/(N+1)Boundary conditionu_0 (t) = 1 = u_N+1(t)v_0(t) =
3 = v_N+1(t),Initial conditiou_i(0) = 1+sin (2 pi x_i )v_i
= 3For i = 1,á., NTime Interval =[t_o, t_end] =
[0,10]Could i get code in Mathematica (by using Euler of 4
Runge - Kutta..)to solve this ordinary differential
equation.I am very happy if you give me
helpKhaled________________________________________________
_________________MSN Photos is the easiest way to share and
print your photos:
http://photos.msn.com/support/worldwide.aspx
====
> Dear
Mathgroup> I have convert pde to ode like this> du_i/dt =
1-4 u_i + .02 (u_i-1 ? 2 u_i + u i+1)/(delta (x))^3 + (u_i)^3
> v_i> dv_i /dt = 3 u_i + .02 (v_i-1 ? 2 v_i + u
i+1)/(delta (x))^3 - (u_I)^3 > v_i> delta(x)=(i)/(N+1)>
x_i= (i)/(N+1)> Boundary condition> u_0 (t) = 1 = u_N+1(t)>
v_0(t) = 3 = v_N+1(t),> Initial conditio> u_i(0) = 1+sin (2
pi x_i )> v_i = 3> For i = 1,?., N> Time Interval =[t_o,
t_end] = [0,10]Since this is 1+1 dimensional initial value
problem, Mathematica can do the discretization for you
automatically. Just useNDSolve[{ D[u[t, x], t] == 1 - 4
u[t, x] + 0.02 D[u[t, x], x, x] + u[t, x]^3 v[t, x], D[v[t,
x], t] == 3u[t, x] + 0.02 D[v[t, x], x, x] + u[t, x] ^3v[t,
x], u[t, 0] == u[t, 1] == 1, u[0, x] == 1 + Sin[2 Pi x], v[t,
0] == v[t, 1] == 3, v[0, x] == 3}, {u, v}, {t, 0, 10}, {x, 0,
1}]Mathematica by default uses fourth order differences
instead of the second order you specified above. If you
really want second order spatial differences with the exact
spacing you defined, you can useNDSolve[{D[u[t, x], t] == 1 -
4 u[t, x] + 0.02 D[u[t, x], x, x] + u[t, x]^3 v[t, x], D[v[t,
x], t] == 3u[t, x] + 0.02 D[v[t, x], x, x] + u[t, x] ^3v[t,
x], u[t, 0] == u[t, 1] == 1, u[0, x] == 1 + Sin[2 Pi x], v[t,
0] == v[t, 1] == 3, v[0, x] == 3}, {u, v}, {t, 0, 10}, {x, 0,
1}, StartingStepSize -> 1./206, MaxSteps -> {1000, 300},
DifferenceOrder -> 2]Interestingly enough, either way you do
it, Mathematica is only able to carry out the solution out
to about t == 0.035 or so. This appears to be because the
nonlinearity is causing the solution to form a
nonsingularity which appears as a cusp with rapidly rising
height.is because of the (delta (x))^3 in .02 (u_i-1 ? 2 u_i
+ u i+1)/(delta (x))^3 Any finite difference formula for the
second derivative on a uniform grid will involve (delta
(x))^2 -- i.e. squared, not cubed. At a fixed grid space, the
extra power will have the effect of making the diffusion
coefficient larger by a factor of n (which for your range of
interest effectively removes to formation of the
discontinuity). Presumably the (delta (x))^3 was a typo as
was the u_i+1 instead of v_i+1 in .02 (v_i-1 ? 2 v_i + u
i+1)However, if you really wanted the cubed power, here is
how you could manually do the discretization with
Mathematica:n = 205;X = N[Range[1, n]/(n + 1)];U[t_] =
Map[u[#][t] &, Range[1, n]];V[t_] = Map[v[#][t] &, Range[1,
n]];eqns = Join[ Thread[D[U[t], t] == 1 - 4 U[t] + 0.02
ListCorrelate[N[{1, -2, 1} n^3], U[t], {2, 2}, 1] + U[t]^3
V[t]], Thread[ D[V[t], t] == 3 U[t] + 0.02
ListCorrelate[N[{1, -2, 1} n^3], V[t], {2, 2}, 3] + U[t]^3
V[t]], Thread[U[0] == 1 + Sin[X]], Thread[V[0] == 3 + 0.
X]];NDSolve[eqns, Join[U[t], V[t]], {t, 0, 10}]> Could i
get code in Mathematica (by using Euler of 4 Runge -
Kutta..)to > solve this ordinary differential equation.I do
not recommend solving this with either an Euler method or a
RungeKutta method for the reason that the potential
formation of discontinuities could make the ODEs stiff. The
default method Mathematica uses automatically switches to
methods appropriate for stiff ODEs when needed. If you want
to use a RungeKutta method, you can useNDSolve[eqns,
Join[U[t], V[t]], {t, 0, 10}, Method->RungeKutta]This uses
the Runge-Kutta-Fehlberg 4(5) method.> I am very happy if you
give me help> Khaled>
_____________________________________________________________
____> MSN Photos is the easiest way to share and print your
photos: > http://photos.msn.com/support/worldwide.aspx>

====
 I am a newbie to mathematica. I have a 14 functions
which are thefunction of r,theta and phi. I want to do some
mathematical operationover them. How can I do? Can it be
possible to call them in Do or Forloop with some index?
Please suggest. Raj
====
Is there an easy (elegant?) way to
generate the set of all k-tuplestaking values from some set
(list) S? I want the arguments of thefunction to be k (the
length of the tuples) and the set S. That
is,KTuples[3,{a,b}] should
produce{{a,a,a},{a,a,b},{a,b,a},{a,b,b},{b,a,a},{b,a,b},{b,b
,a},{b,b,b}}.
====
Bob, KTuples[n_,lst_]:=
Distribute[Table[{a,b},{n}],List] KTuples[3,{a,b}]
{{a,a,a},{a,a,b},{a,b,a},{a,b,b},{b,a,a},{b,a,b},{b,b,a},{b,b
,b}}--Allan---------------------Allan HayesMathematica
Training and ConsultingLeicester
UKwww.haystack.demon.co.ukhay@haystack.demon.co.ukVoice
: +44 (0)116 271 4198> Is there an easy (elegant?) way to
generate the set of all k-tuples> taking values from some
set (list) S? I want the arguments of the> function to be k
(the length of the tuples) and the set S. That is,>
KTuples[3,{a,b}] should produce>
{{a,a,a},{a,a,b},{a,b,a},{a,b,b},{b,a,a},{b,a,b},{b,b,a},{b,b
,b}}.>
====
Here's my
contestant:<<DiscreteMath`Combinatorica`KTuples[k_Integer,
vals_List] := Union[KSubsets[PadRight[vals, k*Length[vals],
vals], k]]---Selwyn Hollis> Is there an easy (elegant?) way
to generate the set of all k-tuples> taking values from some
set (list) S? I want the arguments of the> function to be k
(the length of the tuples) and the set S. That is,>
KTuples[3,{a,b}] should produce>
{{a,a,a},{a,a,b},{a,b,a},{a,b,b},{b,a,a},{b,a,b},{b,b,a},{b,b
,b}}.> 
====
> Is there an easy (elegant?) way to generate the
set of all k-tuples> taking values from some set (list) S? I
want the arguments of the> function to be k (the length of
the tuples) and the set S. That is,> KTuples[3,{a,b}] should
produce>
{{a,a,a},{a,a,b},{a,b,a},{a,b,b},{b,a,a},{b,a,b},{b,b,a},{b,b
,b}}.Here's a first implementation:ktuples[k_, 
set_List] :=
Map[ set[[#]] &, Flatten[ Array[ List, Table[ Length[set],
{k}]], k - 1]]In[31]:=
ktuples[3,{a,b}]Out[31]={{a,a,a},{a,a,b},{a,b,a},{a,b,b},{b,
a,a},{b,a,b},{b,b,a},{b,b,b}}-Nevin
====
Does anyone know if
there are any publications that describe theintegration of
above?In particular I am working with a 4 gaussian process
with joint pdfp(x1,y1,x2,y2) transformed to->
p(r1,r2,theta1,theta2), the processeshave non-zero mean and
I would like to integrate the joint pdf twice(over r1 and r2)
and obtain an expression for the pdf of the phaseangles
between p(theta1,theta2). I could assume that the means are
thesame (to make life easier). Has anyone come accross this
- it doesn'tseem that unusual to me. I am asking because the
integrals arebecoming 'interesting'.thankyou for 
any
tips.
====
Start Excel and go to Tools -> Macro ->
Security..., then on the SecurityLevel tab click on
Medium and then click OK.----- Original Message ----->>
> My original version of MathLink for Excel (MLX.xla) was for
Excel 97,and> It worked fine. Then I upgraded the package
to Excel 2000 by unloadingthe> macro from the Wolfram
site. It continued working perfectly on my old>
environment (Windows 98 + Mathematica 4.1).>> But I
recently upgraded to Mathematica 4.2 under Windows 2000; and
when> trying to load the macro (MLX.xla) for Excel2000, I
got (in Excel 2000)> the> following message: This
workbook contains a type of macro (Microsoft> Excel>
version 4.0 macro) that cannot be disable nor signed.
Therefore, this> workbook cannot be opened under high
security level.>> Does any one have any experience on
this?> Emilio Martin-Serrano>
====
Download the
newest patch from wolfram research - this should
theoreticallyhelp.> Group,>> My original version of
MathLink for Excel (MLX.xla) was for Excel 97, and> It
worked fine. Then I upgraded the package to Excel 2000 by
unloading the> macro from the Wolfram site. It continued
working perfectly on my old> environment (Windows 98 +
Mathematica 4.1).>> But I recently upgraded to Mathematica
4.2 under Windows 2000; and when> trying to load the macro
(MLX.xla) for Excel2000, I got (in Excel 2000)the>
following message: This workbook contains a type of macro
(MicrosoftExcel> version 4.0 macro) that cannot be disable
nor signed. Therefore, this> workbook cannot be opened under
high security level.>> Does any one have any experience on
this?> Emilio Martin-Serrano>
====
Find the link and
instructions on this
page:http://support.wolfram.com/applicationpacks/excel_link/
excelxp.html> Download the newest patch from wolfram
research - this shouldtheoretically> help.>> Group,>>
> My original version of MathLink for Excel (MLX.xla) was for
Excel 97,and> It worked fine. Then I upgraded the package
to Excel 2000 by unloadingthe> macro from the Wolfram
site. It continued working perfectly on my old>
environment (Windows 98 + Mathematica 4.1).>> But I
recently upgraded to Mathematica 4.2 under Windows 2000; and
when> trying to load the macro (MLX.xla) for Excel2000, I
got (in Excel 2000)> the> following message: This
workbook contains a type of macro (Microsoft> Excel>
version 4.0 macro) that cannot be disable nor signed.
Therefore, this> workbook cannot be opened under high
security level.>> Does any one have any experience on
this?> Emilio Martin-Serrano>
====
>I'm sorry for
that my question is not clear,I have correct below.> I have
a very interesting math problem:If I have a scales,and I>>
have 40 things that their mass range from 1~40 which each is
a nature>> number,and now I can only make 4 counterweights
to measure out each>> mass of those things.Question:What
mass should the counterweights>> be???>> The answer is that
1,3,9,27 and I wnat to use mathematica to solve>> this
problem.>> In fact,I think that this physical problem has
various>> answer,ex.2,4,10,28>> this way also work,because
if I have a thing which weight 3 , and I>> can measure out
by comparing 2<3<4 . But,If I want to solve this math>>
problem:>> {x|x=k1*a+k2*b+k3*c+k4*d}={1,2,3,4,,,,,,40} where
a,b,c,d is nature numbers.>> and {k1,k2,k3,k4}={1,0,-1}>>
How to solve it ?? >> mathematica solving method. appreciate
any idea sharing>> sincerely >> bryan>Just use brute
force.Needs[DiscreteMath`Combinatorica`];var = {a, b, c,
d}; n = Length[var];s = Outer[Times, var, {-1, 0, 1} ];f =
Flatten[Outer[Plus, Sequence@@s]];Since the length of f is
just 3^n then the range of numbersto be covered should be
{-(3^n-1)/2, (3^n-1)/2}.Consequently, the largest of the
weights can not exceed(3^n-1)/2 - (1+2+...+(n-1)) or
((3^n-1) - n(n-1))/234Thread[var->#]& /@ (First /@
Select[{var,f} /. Thread[var->#]& /@ KSubsets[Range[((3^n-1)
- n(n-1))/2], n], Sort[#[[2]]] ==
Range[-(3^n-1)/2,(3^n-1)/2]&]){{a -> 1, b -> 3, c -> 9, d ->
27}}Bob HanlonChantilly, VA USA
====
Can we please get a
response from WRI?i.e. regarding:In[1]:= Limit[
(Log[x]^Log[Log[x]])/ x , x->Infinity]Out[2]:= 
InfinityIt
should be 0Jonathan Rockmannmtheory@msn.com
====
I'm
finding that the ImageSize option in Export has no effect when
exporting Cell or Notebook objects. For instance, the
following two commands produce precisely the same graphic:
Export[image1.jpg, Cell[Some cell contents, Text,
FontSize -> 100]] Export[image2.jpg, Cell[Some cell
contents, Text, FontSize -> 100], ImageSize -> {576,
288}]Has anyone encountered this problem before?(This is
with Mathematica 4.1.5 and Mac OS X.)----Selwyn
Hollisslhollis@mac.com
====
Does anyone know how to get the
JavaPlot window (or any windows of this type)which can be
seen at
http://www.wolfram.com/products/mathematica/newin42/
java.htmlthat WRI advertises comes with
4.2?Jonathanmtheory@msn.com
====
How can I get mathematica
to display the inverse of functions like:f(x) = x^2 - 7*x +
10orf(x) = cos(3*x + 1/2*pi)orf(x) = (x - 3) / (x +
2)I'm having trouble getting the syntax
right.
====
G'day,Looks like you forgot to define base in 
your
Machine function and whenusing pure functions, don't forget
the ampersand.ThusNestList[Machine[10,7,#]&, 3,
22]returns{3, 8, 4, 2, 1, 7, 10, 5, 9, 11, 12, 6, 3, 8, 4,
2, 1, 7, 10, 5, 9, 11,12}with base = 2.Yas> Howdy,>> I'm
trying to figure out the correct syntax to do the following. I
have> some function with three arguments, and I want to
syntactically describe the> single-argument function that
holds two of those arguments constant (i.e.> without creating
that single-argument function).>> More specifically, I have
defined>> Machine[radix_,multiplier_,state_] := Module
[{c,s},> c = Floor[state/base]; s = Mod[state,base];>
multiplier*s + c> ]>> where I have a generalize 
'machine',
defined by the radix and multiplier,> which converts one state
into another state. So I'd like to be able to do> something
like this:>> NestList[Machine[10,7,#], 3, 22]>> to get the
series of states that the radix-10 multiplier-7 machine
runs> through (starting with state 3). However, this syntax
doesn't seem to do> what I want.>> I hope that description
makes sense. It seems like there must be a syntax> to
describe the function Machine[10,7,#].>> Anyone have any
ideas?>> Bob H>
====
Some functions with vector arguments
work as expected and some giveerrors. I would appreciate if
someone can clarify why.First an example of a function
definition that works fine :f[u_,v_] := u.vCalling 
it with
f[{x1,y1},{x2,y2}] gives the expected result x1 x2 +y1
y2.Now an example that does not work :g[u_,v_] :=
u-vCalling it with g[{x1,y1},{x2,y2}] one would expect the
answer {x1-x2,y1-y2} but instead one gets error
messages.Why ? And how do I fix g (i.e write a function that
outputs thedifference of 2 vectors).
====
I do not know why
you are getting a message, using your own code I get
theresult you expect.> Some functions with vector arguments
work as expected and some give> errors. I would appreciate if
someone can clarify why.>> First an example of a function
definition that works fine :> f[u_,v_] := u.v> 
Calling it with
f[{x1,y1},{x2,y2}] gives the expected result x1 x2 +> y1 y2.>>
Now an example that does not work :> g[u_,v_] := u-v> Calling
it with g[{x1,y1},{x2,y2}] one would expect the answer
{x1-x2,> y1-y2} but instead one gets error messages.>> Why ?
And how do I fix g (i.e write a function that outputs the>
difference of 2 vectors).>>
====
> Now an example that does not
work :> g[u_,v_] := u-v> Calling it with g[{x1,y1},{x2,y2}]
one would expect the answer {x1-x2,> y1-y2} but instead one
gets error messagesWorks fine for me.Please check that you
have no existing definitions for the symbols used.
Clear[g,x1,x2,y1,y2]; g[u_,v_]:=u-v g[{x1,y1},{x2,y2}]
{x1-x2,y1-y2}--Allan---------------------Allan
HayesMathematica Training and ConsultingLeicester
UKwww.haystack.demon.co.ukhay@haystack.demon.co.ukVoice
: +44 (0)116 271 4198> Some functions with vector arguments
work as expected and some give> errors. I would appreciate
if someone can clarify why.>> First an example of a function
definition that works fine :> f[u_,v_] := u.v> 
Calling it with
f[{x1,y1},{x2,y2}] gives the expected result x1 x2 +> y1 y2.>>
Now an example that does not work :> g[u_,v_] := u-v> Calling
it with g[{x1,y1},{x2,y2}] one would expect the answer
{x1-x2,> y1-y2} but instead one gets error messages.>> Why ?
And how do I fix g (i.e write a function that outputs the>
difference of 2 vectors).>>
====
> Some functions with vector
arguments work as expected and some give> errors. I would
appreciate if someone can clarify why.> First an example of
a function definition that works fine :> f[u_,v_] 
:= u.v>
Calling it with f[{x1,y1},{x2,y2}] gives the expected result
x1 x2 +> y1 y2.> Now an example that does not work :>
g[u_,v_] := u-v> Calling it with g[{x1,y1},{x2,y2}] one
would expect the answer {x1-x2,> y1-y2} but instead one gets
error messages.> Why ? And how do I fix g (i.e write a
function that outputs the> difference of 2 vectors).> in
fact this call works fine on my Mathematicas (3.0 up to 4.1).
Did Youreally use the minus character? Maybe some erroneous
code was assignedto g before. Restart Mathematica or use
Remove[g] to really start upfrom the beginning.Sincerly
Alexander-- / Alexander Dreyer, Dipl.-Math. - Abteilung
Adaptive Systeme / Fraunhofer Institut fuer Techno- und
Wirtschaftsmathematik (ITWM)  Gottlieb-Daimler-Strasse,
Geb. 7^2=49/313 D-67663 Kaiserslautern /
====
Garry, Here's a
solution using your LeftSide concept; it works perfectly
but takes twice as much time as my solution. Both solutions
look at every vertex of both rectangles, but mine uses two
sides from each and yours requires looking at all four sides
of each rectangle. I'd think yours should be a triße 
faster
than this, though. There may be efficiencies I'm 
missing (in
both solutions). ClearAll[cis, rect, pickRect, extent,
cannotIntersect, intersects, daveRect] cis[t_] := {Cos@t,
Sin@t} rect[{pt : {_, _}, angle_, {len1_, len2_}}] :=
Module[{pt2}, {pt, pt2 = pt + len1 cis[angle], pt2 - len2
cis[angle - Pi/2], pt - len2 cis[angle - Pi/2]}] daveRect :=
{{Random[], Random[]}, Random[] + Pi/2, {Random[], Random[]}}
pickRect := rect@daveRect extent[r1_, r2_] := {Min@#, Max@#}
& /@ ((Take[r1, 2] - r1[[{2, 3}]]).Transpose@r2)
cannotIntersect[{{min1_, max1_}, {min2_, max2_}}] := max2 <
min1 || min2 > max1 intersects[r1_, r2_] := Catch[
If[cannotIntersect[#], Throw[False]] & /@
Flatten[Transpose[Outer[extent,  {r1}, {r1, r2},
1]~Join~Outer[extent, {r2}, {r2, r1}, 1], {1, 3, 2}], 1];
Throw[True]] ClearAll[leftSide,leftIntersects,sides]
sides[a_List]:=Partition[Join[a,{First@a}],2,1]
leftSide[{a_,b_},{{c_,d_},{e_,f_}}]:=-b c+a d+b e-d e-a f+c
f>0 leftSide[a:{{_,_}..},b:{{_,_},{_,_}}]:=leftSide[#,b]&/@a
leftSide[a_List,b:{{{_,_},{_,_}}..}]:=leftSide[a,#]&/@b
leftIntersects[a_,b_]:=!Or@@(And@@#&/@leftSide[a,sides@b])&&
! Or@@(And@@#&/@leftSide[b,sides@a])
davePairs={daveRect,daveRect}&/@Range[10000];
rectanglePairs=Map[Reverse@rect[#]&,davePairs,{2}];
Timing[right=intersects[Sequence@@#]&/@rectanglePairs;]
Timing[test=leftIntersects[Sequence@@#]&/@rectanglePairs;]
right[Equal]test {3.187999999999999*Second, Null}
{6.765000000000001*Second, Null} True Bobby Treat
-----Original Message----- intersection > Begin forwarded
message: > Dear colleagues, > any hints on how to
implement a very fast routine in Mathematica for > testing if
two rectangles have an intersection area? > Frank Brand >
Here is one approach. > Given three points
{x1,y1},{x2,y2},{x3,y3}, switch to homogenous > coordinates
a={1,x1,y1}, b={1,x2,y2}, c={1,x3,y3}. Then >
Sign[Det[{a,b,c}]] is +1 if and only if the point a lies on
your left as > you walk along the line though b and c in the
direction from b to c. > ( If the result is zero, then a lies
on the line.) > The value of the determinant is
x2y3-x3y2-x1y3+x3y1+x1y2-x2y1, and the > speed of the
algorithm depends essentially on how fast this quantity can
> be computed. Suppose we write a function LeftSide[a,{b,c}]
that computes > the sign of the determinant. > Now let
{p1,p2, . . ., pn} be a list of vertices (pi={xi,yi}) of a >
convex polygon traced counterclockwise. Then a lies within or
on the > boundary of the polygon if and only if none of the
numbers > LeftSide[a,{pi,p(i+1)}] are -1. That is, if -1 does
not appear in the > list LeftSide[a,#]&/@Partition[{p1,p2,. .
.,pn,p1},2,1]. > Now use the fact that if two convex
polynomials overlap, then some > vertex of one of them must
lie inside or on the boundary of the other. > If an overlap
of positive area is required, then the check is that only >
+1 appears--not that -1 does not appear. > For two
rectangles ( or parallelograms) this approach requires the >
evaluation of 16 determinants, so it may be a bit expensive.
If the > points have rational coordinates, then (positive)
denominators may be > cleared in the homogeneous coordinates
and the computations can be done > in integer arithmetic, at
the cost of at least three more > multiplications per
determinant. > Garry Helzer Department of Mathematics
University of Maryland College Park, MD 20742 301-405-5176
gah@math.umd.edu 
====
I would like to have a list of all the
directives that can be used forcommand in the Windows Start
Menu). I am especially interested in adirective with which
you can appoint a file which has to be evaluated by
thekernel after it has been launched. I imagine that such
a directive lookslike -f filename, but I can not 
find a list
of the correct directives.The Mathematica version I have is
4.1 on a Windows NT system.Rene Klaver
====
> 1) Get data
from excel into a coordinate > list {x,y,z}> e.g. Node 1,
{x1,y1,z1}> Node 2, {x2,y2,z2}> etc...Why not just use data
= ReadList[filename, Table[Number, {4}]];Alternatively,
data = ReadList[filename, Number, RecordLists->True]; data
= Partition[data, 4]; cord = {#[[1]], #[[2]], #[[3]]}& /@
data;> 2) Convert from Rectangular to > Cylindrical
(maybe)Coordinate tranforms live in
Calculus`VectorAnalysis`. It's straightforwardto write a
function that uses > 3) Plot3D the dataThis will be tricky
-- ListSurfacePlot3D plots f[x, y]. ListContourPlot3Dwhich
plots f[x, y, z] can be quite slow.> 4) Generate a harmonic
bessel function for that > plot3D/graph> 5) Find the
equation(s) that spits out > these harmonic > bessel
functions (which I think might be in the general form > of
Hankel > Function solutions to the Helmholtz equation which
shows > cylinder harmonics > of order v)Maybe I
misunderstand ... isn't this the same as fitting 
a bessel
functionto your data? For this, you can use
Statistics`NonlinearFit`. > I can figure out step 5 if I can
get steps 1 through 4 > figured out. If > anyone can write a
recipe for me to follow that would be > great, or even some >
tips and clues...Anything!!!Dave.
====
You've gotten some very
helpful responses so I'm just going to add a coupleof
comments. The following is an example of yet another type of
button that you can useas a model. Copy and paste the
following code into a notebook. When youevaluate it, it
will create a button within the same notebook which
whenpressed evaluates the entire notebook. Sort of like
selecting from the menuis something in the notebook
already to evaluate. So input 1+2 (and don'thit enter)
somewhere in the notebook so that you can see whatthe
effect of the button
is.DisplayForm[Button[Evaluate*Notebook,ButtonFunction:>
FrontEndExecute[FrontEndToken[EvaluateNotebook]],
ButtonEvaluator->None,Active->True]]It has the nice feature
that it doesn't get stuck in a loop creating the button over
and over again. However, with some enthusiasm you can make
thishappen too.Personally, I am fascinated by the Java
integration in version 4.2. But I wouldn't recommend it to
anyone who is just starting to program in
Mathematica.Further, there is hardly enough material
contained in the Help Browser to make sense of much of
it.It is always a good idea to have several basic constructs
of efficient, usefulprograms that can serve as models when
programming. If you look atsome of the MathGroup archived
threads you will find similar complaints that the Mathematica
documentation is confusing.Jonathan
Rockmannmtheory@msn.com----- Original Message -----I want
to have a variable (or matrix or whatever) defined as a
globalvariable x, and then perform x = x+1 when a button is
clicked. I've beenusing and programming computers for
almost 25 years and I can't followWolfram's 
documentation.
Is it just me, or is he always this obtuse whenexplaining
things? I mean, given the amazing power of the
Mathematicasystem, a sample list of buttons in a notebook
that you could select andexamine how they were implemented,
would have been nice. I can't find such alist, 
and this
seems to be par for the course for the rest of
thedocumentation as well.
====
This was REALLY interesting.
Here's a solution that looks only at the7,560 relevant
combinations. It first chooses three numerators. Thenit
chooses two denominators for the first fraction. Then
twodenominators for the second fraction. The last fraction
is determinedat that point.<<
DiscreteMath`Combinatorica`ClearAll[f, g, h, j]r = Range[1,
9];f = KSubsets[#1, #2] &;g[r_List, n_Integer, {}] := f[r,
n]g[r_List, n_Integer, e_?VectorQ] := Join[e, #] & /@
f[Complement[r, e],n]g[r_List, n_Integer, e : {__?VectorQ}]
:= Flatten[g[r, n, #] & /@ e, 1]h[r_List, e : {__?VectorQ}]
:= Join[#, Complement[r, #]] & /@ ej[{a_, b_, c_, d_, e_,
f_, g_, h_, i_}] := a/(d e) + b/(f g) + c/(h
i)Timing[Select[h[r, Fold[g[r, #2, #1] &, {}, {3, 2, 2}]],
j@# == 1 &]]{0.532 Second, {{1, 5, 7, 3, 6, 8, 9, 2,
4}}}Hence the only solution is1/(3*6)+5/(8*9)+7/(2*4)Bobby
Treat-----Original Message-----University Professor of
Philanthropy and the LawDirector, National Center on
Philanthropy and the LawNew York University School of
LawRoom 206A110 West 3rd StreetNew York, N.Y.
10012-1074-----Original
Message-----byran__________________________________________
______________________________service. For more information
on a proactive anti-virus service workingaround the clock,
around the globe, visit
http://www.messagelabs.com__________________________________
______________________________________
====
Garry,Here's a
solution using your LeftSide concept; it works perfectly
buttakes twice as much time as my solution. Both solutions
look at everyvertex of both rectangles, but mine uses two
sides from each and yoursrequires looking at all four sides
of each rectangle. I'd think yoursshould be a triße 
faster
than this, though. There may be efficienciesI'm 
missing (in
both solutions).ClearAll[cis, rect, pickRect, extent,
cannotIntersect, intersects,daveRect]cis[t_] := {Cos@t,
Sin@t}rect[{pt : {_, _}, angle_, {len1_, len2_}}] :=
Module[{pt2}, {pt, pt2 = pt + len1 cis[angle], pt2 - len2
cis[angle - Pi/2], pt - len2 cis[angle - Pi/2]}]daveRect :=
{{Random[], Random[]}, Random[] + Pi/2,
{Random[],Random[]}}pickRect := rect@daveRectextent[r1_,
r2_] := {Min@#, Max@#} & /@ ((Take[r1, 2] -
r1[[{2,3}]]).Transpose@r2)cannotIntersect[{{min1_, max1_},
{min2_, max2_}}] := max2 < min1 || min2 >
max1intersects[r1_, r2_] := Catch[ If[cannotIntersect[#],
Throw[False]] & /@Flatten[Transpose[Outer[extent, {r1},
{r1, r2}, 1]~Join~Outer[extent, {r2}, {r2, r1}, 1], {1, 3,
2}],1];
Throw[True]]ClearAll[leftSide,leftIntersects,sides]sides[a_
List]:=Partition[Join[a,{First@a}],2,1]leftSide[{a_,b_},{{c_
,d_},{e_,f_}}]:=-b c+a d+b e-d e-a f+c
f>0leftSide[a:{{_,_}..},b:{{_,_},{_,_}}]:=leftSide[#,b]&/@a
leftSide[a_List,b:{{{_,_},{_,_}}..}]:=leftSide[a,#]&/@b
leftIntersects[a_,b_]:=!Or@@(And@@#&/@leftSide[a,sides@b])&&!
Or@@(And@@#&/@leftSide[b,sides@a])davePairs={daveRect,
daveRect}&/@Range[10000];rectanglePairs=Map[Reverse@rect[#]&
,davePairs,{2}];Timing[right=intersects[Sequence@@#]&/@
rectanglePairs;]Timing[test=leftIntersects[Sequence@@#]&/@
rectanglePairs;]right[Equal]test{3.187999999999999*Second
, Null}{6.765000000000001*Second, Null}TrueBobby
Treat-----Original Message-----intersection>> Begin
forwarded message:>> Dear colleagues,>> any hints on how to
implement a very fast routine in Mathematica for> testing if
two rectangles have an intersection area?> Frank Brand>
Here is one approach.>> Given three points
{x1,y1},{x2,y2},{x3,y3}, switch to homogenous> coordinates
a={1,x1,y1}, b={1,x2,y2}, c={1,x3,y3}. Then>
Sign[Det[{a,b,c}]] is +1 if and only if the point a lies on
your leftas> you walk along the line though b and c in the
direction from b to c.> ( If the result is zero, then a lies
on the line.)>> The value of the determinant is
x2y3-x3y2-x1y3+x3y1+x1y2-x2y1, and the> speed of the
algorithm depends essentially on how fast this
quantitycan> be computed. Suppose we write a function
LeftSide[a,{b,c}] thatcomputes> the sign of the
determinant.>> Now let {p1,p2, . . ., pn} be a list of
vertices (pi={xi,yi}) of a> convex polygon traced
counterclockwise. Then a lies within or on the> boundary of
the polygon if and only if none of the numbers>
LeftSide[a,{pi,p(i+1)}] are -1. That is, if -1 does not
appear in the> list LeftSide[a,#]&/@Partition[{p1,p2,. .
.,pn,p1},2,1].>> Now use the fact that if two convex
polynomials overlap, then some> vertex of one of them must
lie inside or on the boundary of the other.>> If an overlap
of positive area is required, then the check is thatonly>
+1 appears--not that -1 does not appear.>> For two rectangles
( or parallelograms) this approach requires the> evaluation
of 16 determinants, so it may be a bit expensive. If the>
points have rational coordinates, then (positive)
denominators may be> cleared in the homogeneous coordinates
and the computations can bedone> in integer arithmetic, at
the cost of at least three more> multiplications per
determinant.>>Garry HelzerDepartment of
MathematicsUniversity of MarylandCollege Park, MD
20742301-405-5176gah@math.umd.edu>>
====
Garry,Also note
your solution requires rectangle points to be in
clockwiseorder (mine doesn't), but yours works for
arbitrary convex polygons aswritten.Bobby-----Original
Message-----ClearAll[cis, rect, pickRect, extent,
cannotIntersect, intersects,daveRect]cis[t_] := {Cos@t,
Sin@t}rect[{pt : {_, _}, angle_, {len1_, len2_}}] :=
Module[{pt2}, {pt, pt2 = pt + len1 cis[angle], pt2 - len2
cis[angle - Pi/2], pt - len2 cis[angle - Pi/2]}]daveRect :=
{{Random[], Random[]}, Random[] + Pi/2,
{Random[],Random[]}}pickRect := rect@daveRectextent[r1_,
r2_] := {Min@#, Max@#} & /@ ((Take[r1, 2] -
r1[[{2,3}]]).Transpose@r2)cannotIntersect[{{min1_, max1_},
{min2_, max2_}}] := max2 < min1 || min2 >
max1intersects[r1_, r2_] := Catch[ If[cannotIntersect[#],
Throw[False]] & /@Flatten[Transpose[Outer[extent, {r1},
{r1, r2}, 1]~Join~Outer[extent, {r2}, {r2, r1}, 1], {1, 3,
2}],1];
Throw[True]]ClearAll[leftSide,leftIntersects,sides]sides[a_
List]:=Partition[Join[a,{First@a}],2,1]leftSide[{a_,b_},{{c_
,d_},{e_,f_}}]:=-b c+a d+b e-d e-a f+c
f>0leftSide[a:{{_,_}..},b:{{_,_},{_,_}}]:=leftSide[#,b]&/@a
leftSide[a_List,b:{{{_,_},{_,_}}..}]:=leftSide[a,#]&/@b
leftIntersects[a_,b_]:=!Or@@(And@@#&/@leftSide[a,sides@b])&&!
Or@@(And@@#&/@leftSide[b,sides@a])davePairs={daveRect,
daveRect}&/@Range[10000];rectanglePairs=Map[Reverse@rect[#]&
,davePairs,{2}];Timing[right=intersects[Sequence@@#]&/@
rectanglePairs;]Timing[test=leftIntersects[Sequence@@#]&/@
rectanglePairs;]right[Equal]test{3.187999999999999*Second
, Null}{6.765000000000001*Second, Null}TrueBobby
Treat-----Original Message-----intersection>> Begin
forwarded message:>> Dear colleagues,>> any hints on how to
implement a very fast routine in Mathematica for> testing if
two rectangles have an intersection area?> Frank Brand>
Here is one approach.>> Given three points
{x1,y1},{x2,y2},{x3,y3}, switch to homogenous> coordinates
a={1,x1,y1}, b={1,x2,y2}, c={1,x3,y3}. Then>
Sign[Det[{a,b,c}]] is +1 if and only if the point a lies on
your leftas> you walk along the line though b and c in the
direction from b to c.> ( If the result is zero, then a lies
on the line.)>> The value of the determinant is
x2y3-x3y2-x1y3+x3y1+x1y2-x2y1, and the> speed of the
algorithm depends essentially on how fast this
quantitycan> be computed. Suppose we write a function
LeftSide[a,{b,c}] thatcomputes> the sign of the
determinant.>> Now let {p1,p2, . . ., pn} be a list of
vertices (pi={xi,yi}) of a> convex polygon traced
counterclockwise. Then a lies within or on the> boundary of
the polygon if and only if none of the numbers>
LeftSide[a,{pi,p(i+1)}] are -1. That is, if -1 does not
appear in the> list LeftSide[a,#]&/@Partition[{p1,p2,. .
.,pn,p1},2,1].>> Now use the fact that if two convex
polynomials overlap, then some> vertex of one of them must
lie inside or on the boundary of the other.>> If an overlap
of positive area is required, then the check is thatonly>
+1 appears--not that -1 does not appear.>> For two rectangles
( or parallelograms) this approach requires the> evaluation
of 16 determinants, so it may be a bit expensive. If the>
points have rational coordinates, then (positive)
denominators may be> cleared in the homogeneous coordinates
and the computations can bedone> in integer arithmetic, at
the cost of at least three more> multiplications per
determinant.>>Garry HelzerDepartment of
MathematicsUniversity of MarylandCollege Park, MD
20742301-405-5176gah@math.umd.edu>>
====
Try
this:Off[Remove::rmnsm]Remove[Global`p@,
Global`p@@]n = 7;pn = Unique[p] & /@ Range[10]f[p_]
= Array[fk[p, #] &, n]f[p]fEq[p_] = MapThread[Equal, {f[p],
Array[0 &, n]}]Bobby Treat-----Original Message-----f2 :=
f[pn] [[2]];f3 := f[pn] [[3]];f4 := f[pn] [[4]];f5 :=
f[pn] [[5]];f6 := f[pn] [[6]];f7 := f[pn] [[7]];theRoot =
FindRoot[{f1==0,f2==0,f3==0,f4==0,f5==0,f6==0,f7==0}, {p1,
1/n},{p2, 1/n},{p3, 1/n},{p4, 1/n},{p5, 1/n}, {p6, 1/n},{p7,
1/n}];-- John MacCormick Systems Research Center, HP Labs,
1501 Page Mill Road,
====
the problem, but I duplicated it
just now, so I'm puzzled. I'm usingWindows XP 
Home, and
System`Private`$BuildNumber is 168634.4.2 for Microsoft
Windows (June 5, 2002)f[n_] :=
Log[n]^Log[Log[n]]Limit[f[n]/n, n -> 
Infinity]InfinityBobby
Treat -----Original Message----- By the way, Mathematica
gets the following limit wrong:f[n_] :=
Log[n]^Log[Log[n]]Limit[f[n]/n, n -> 
Infinity]InfinityThat
limit is zero. For the Sieve to be polynomial, we only need
thesequence to be BOUNDED (for some power of n in the
denominator).Bobby Treat
====
Garry,No, you don't have to
compute intersections, and yes, you can testvertices only.
I haven't coded it yet, but the LeftSide idea seemslike
a good one.It is sufficient to test whether all vertices of
one convex polygon areon the left (out) side of some side
of the second polygon (both polygonsin clockwise order). If
that happens for any side of either polygon,the polygons
don't intersect. In the cross example, some vertices
areto the right for every side you try.Bobby
Treat-----Original Message-----intersection>> Begin
forwarded message:>> Dear colleagues,>> any hints on how to
implement a very fast routine in Mathematica for> testing if
two rectangles have an intersection area?> Frank Brand>
Here is one approach.>> Given three points
{x1,y1},{x2,y2},{x3,y3}, switch to homogenous> coordinates
a={1,x1,y1}, b={1,x2,y2}, c={1,x3,y3}. Then>
Sign[Det[{a,b,c}]] is +1 if and only if the point a lies on
your leftas> you walk along the line though b and c in the
direction from b to c.> ( If the result is zero, then a lies
on the line.)>> The value of the determinant is
x2y3-x3y2-x1y3+x3y1+x1y2-x2y1, and the> speed of the
algorithm depends essentially on how fast this
quantitycan> be computed. Suppose we write a function
LeftSide[a,{b,c}] thatcomputes> the sign of the
determinant.>> Now let {p1,p2, . . ., pn} be a list of
vertices (pi={xi,yi}) of a> convex polygon traced
counterclockwise. Then a lies within or on the> boundary of
the polygon if and only if none of the numbers>
LeftSide[a,{pi,p(i+1)}] are -1. That is, if -1 does not
appear in the> list LeftSide[a,#]&/@Partition[{p1,p2,. .
.,pn,p1},2,1].>> Now use the fact that if two convex
polynomials overlap, then some> vertex of one of them must
lie inside or on the boundary of the other.>> If an overlap
of positive area is required, then the check is thatonly>
+1 appears--not that -1 does not appear.>> For two rectangles
( or parallelograms) this approach requires the> evaluation
of 16 determinants, so it may be a bit expensive. If the>
points have rational coordinates, then (positive)
denominators may be> cleared in the homogeneous coordinates
and the computations can bedone> in integer arithmetic, at
the cost of at least three more> multiplications per
determinant.>>Garry HelzerDepartment of
MathematicsUniversity of MarylandCollege Park, MD
20742301-405-5176gah@math.umd.edu>>
====
> Dear All:> I want
to make an animation that simulate the Doppler Effect, just
2D> circles travel out one by one, and at the same time,the
origin of the> wave also moves toward one direction. I have
no idea to make the speed> of the wave origin and the speed
of traveling wave independent.Is> sincerely bryancircles,
but the second one includes two lines which approximate
theshock waves, I believe. The arguments of the functions
are vx & vy(velocities in x & y directions) and tstart,
tend, tstep whichdetermine how many circles to draw and the
spacing of the circles. Ijust used the Do loop to make the
pictures, & then you can select thecells & pick
Animate.dopp[vx_,vy_,tstart_,tend_,tstep_]:=Show[Graphics
[Table[{Hue[t/tend],Circle[{vx*t,vy*t},tend-t]},{t,tstart,
tend,tstep}]],AspectRatio->Automatic,Axes->True,PlotLabel->
Mach
Sqrt(vx^2+vy^2)];doppcone[vx_,vy_,tend_,tstep_]:=Show[
Graphics[Table[{Hue[t/tend],Circle[{vx*t,vy*t},tend-t]},{t,0
,tend,tstep}]],Graphics[Line[{{-tend*Sin[If[vx==0,0,ArcTan[
vy/vx]]],tend*Cos[If[vx==0,0,ArcTan[vy/vx]]]},{Sign[vx]*Cos[
If[vx==0,0,ArcTan[vy/vx]]]*(tend-tstep)*sqrt[vx^2+vy^2]-
tstep*Sin[If[vx==0,0,ArcTan[vy/vx]]],Sin[If[vx==0,0,ArcTan[
vy/vx]]]*Sign[vx]*(tend-tstep)*sqrt[vx^2+vy^2]+tstep*Cos[If[
vx==0,0,ArcTan[vy/vx]]]}}]],Graphics[Line[{{tend*Sin[If[vx==
0,0,ArcTan[vy/vx]]],-tend*Cos[If[vx==0,0,ArcTan[vy/vx]]]},{
Sign[vx]*Cos[If[vx==0,0,ArcTan[vy/vx]]]*(tend-tstep)*sqrt[vx^
2+vy^2]+tstep*Sin[If[vx==0,0,ArcTan[vy/vx]]],Sign[vx]*Sin[If
[vx==0,0,ArcTan[vy/vx]]]*(tend-tstep)*srt[vx^2+vy^2]-tstep
*Cos[If[vx==0,0,ArcTan[vy/vx]]]}}]],AspectRatio->Automatic,
Axes->True];Do[dopp[i,0,0,10,1],{i,0,1.4,0.2}];Hope I did
that without typos - I can't see how to post the
notebook
====
Neat!The animation was jagged, though, so I
fixed the text position and the imagesize; it 
won't work
for large vertical
velocities:margin=.8;dopp[vx_,vy_,tend_,tstep_]:=Show[
Graphics[ {Table[{Hue[t/tend],Circle[{vx*t,vy*t},tend-t]},
{t,0,tend,tstep}], Text[Mach <>ToString@Sqrt[vx^2+vy^2]
,{0,.5+tend},{-1,0}] }],AspectRatio->Automatic,Axes->True,
PlotRange-> {{-tend-margin,margin+Max[tend,(vx tend)+.5]},
{-tend-margin,2margin+Max[tend,(vy tend)]}},
Ticks->{{-10,-5,0,5,10},Automatic}, ImageSize->{10 2
margin+10 (tend+Max[tend,(vx tend)+.5]), 10 3 margin+10(tend
+Max[tend,(vy tend)])}];Do[dopp[i,0,10,1],{i,0,1.4,0.2}];By
the way to avoid the slashes, CopyAs/TextOnly.> Dear All:>
I want to make an animation that simulate the Doppler Effect,
just 2D> circles travel out one by one, and at the same
time,the origin of the> wave also moves toward one
direction. I have no idea to make the speed> of the wave
origin and the speed of traveling wave independent.Is>
sincerely bryan>> circles, but the second one includes two
lines which approximate the> shock waves, I believe. The
arguments of the functions are vx & vy> (velocities in x & y
directions) and tstart, tend, tstep which> determine how many
circles to draw and the spacing of the circles. I> just used
the Do loop to make the pictures, & then you can select the>
cells & pick
Animate.>>dopp[vx_,vy_,tstart_,tend_,tstep_]:=Show[
Graphics[Table[{Hue[t/tend],Circle[{vx*t,vy*t},tend-t]},{t,
tstart,tend,tstep}]],AspectRatio->Automatic,Axes->True,
PlotLabel->Mach>  Sqrt(vx^2+vy^2)];>>
doppcone[vx_,vy_,tend_,tstep_]:=Show[Graphics[Table[{Hue[t/
tend],Circle>[{vx*t,vy*t},tend-t]},{t,0,tend,tstep}]],
Graphics[Line[{{-tend*Sin[If[vx==0,0,ArcTan[vy/vx]]],tend*
Cos[If[vx==0,0,ArcTan[vy/vx]]]},{Sign[vx]*Cos[If[vx==0,0,
ArcTan[vy/vx]]]*(tend-tstep)*sqrt[vx^2+vy^2]>
-tstep*Sin[If[vx==0,0,ArcTan[vy/vx]]],Sin[If[vx==0,0,ArcTan[
vy/vx]]]*Sign[vx]*(tend-tstep)*sqrt[vx^2+vy^2]+tstep*Cos[If[
vx==0,0,ArcTan[vy/vx]]]}}]],Graphics[Line[{{tend*Sin[If[vx=
=0,0,ArcTan[vy/vx]]],-tend*Cos[If[vx==0,0,ArcTan[vy/vx]]]},
{Sign[vx]*Cos[If[vx==0,0,ArcTan[vy/vx]]]*(tend-tstep)*sqrt[vx
^2+vy^2]+tstep*Sin[If[vx==0,0,ArcTan[vy/vx]]],Sign[vx]*Sin[
If[vx==0,0,ArcTan[vy/vx]]]*(tend-tstep)*s>rt[vx^2+vy^2]-
tstep*Cos[If[vx==0,0,ArcTan[vy/vx]]]}}]],AspectRatio->
Automatic,Axes->True];>
Do[dopp[i,0,0,10,1],{i,0,1.4,0.2}];>> Hope I did that
without typos - I can't see how to post the notebook>
====
It
would take me a lot of time to understand howfunctions like
SelectionEvaluate works...Cesar.> People encounter this all
the time. It is because> SelectionEvaluate does > not do
what you think. It does not work like> ToExpression, which
causes > immediate kernel evaluation. Instead it works like>
when you press > Shift-Enter, which selects a cell for
evaluation> after all current > evaluations have finished.>
See
>http://support.wolfram.com/mathematica/kernel/interface/
selectionevaluate.html> -Dale>
__________________________________________________Do You
Yahoo!?Yahoo! Finance - Get real-time stock
quoteshttp://finance.yahoo.com
====
> How can I get
mathematica to display the inverse of functions like:> f(x)
= x^2 - 7*x + 10> or> f(x) = cos(3*x + 1/2*pi)> or>
f(x) = (x - 3) / (x + 2)> I'm having trouble getting the
syntax right.> Solve[y==x^2-7*x+10, x]{{x -> (1/2)*(7 -
Sqrt[9 + 4*y])}, {x -> (1/2)*(7 + Sqrt[9 + 4*y])}}x^2-7*x+10
/. % // ExpandAll{y, y}Solve[y == Cos[3*x+1/2*Pi], x]{{x ->
-(ArcSin[y]/3)}}Cos[3*x+1/2*Pi] /. %{y}Solve[y ==
(x-3)/(x+2), x]{{x -> (-3 - 2*y)/(-1 + y)}}(x-3)/(x+2) /. %
// Simplify{y}Bob HanlonChantilly, VA USA
====
> I am a
newbie to mathematica. I have a 14 functions which are the>
function of r,theta and phi. I want to do some mathematical
operation> over them. How can I do? Can it be possible to
call them in Do or For> loop with some index?> As a general
rule avoid Do and For loops and just operate on Lists or Map
(/@) onto lists.g /@ {f1[r,theta,phi], f2[r,theta,phi],
f3[r,theta,phi]}{g[f1[r, theta, phi]], g[f2[r, theta, phi]],
g[f3[r, theta, phi]]}This can be written more compactly
asg /@ (#[r,theta,phi]& /@ {f1,f2,f3}){g[f1[r, theta,
phi]], g[f2[r, theta, phi]], g[f3[r, theta, phi]]}Bob
HanlonChantilly, VA USA
====
> Is there an easy (elegant?)
way to generate the set of all k-tuples> taking values from
some set (list) S? I want the arguments of the> function to
be k (the length of the tuples) and the set S. That is,>
KTuples[3,{a,b}] should produce>
{{a,a,a},{a,a,b},{a,b,a},{a,b,b},{b,a,a},{b,a,b},{b,b,a},{b,b
,b}}.> kTuples[n_Integer?Positive, s_List] :=
Flatten[Outer[List, Sequence@@Table[s,{n}]],n-1];s =
{a,b};kTuples[3, {a,b}]{{a, a, a}, {a, a, b}, {a, b, a}, {a,
b, b}, {b, a, a}, {b, a, b}, {b, b, a}, {b, b,
b}}Length[kTuples[10, {a,b}]]1024Bob HanlonChantilly, VA
USA
====
> If I have an n-element list, (say where each
element is itself a> list), such as {{a,b}, {a,b}, {a,b}}>
is there a way to strip off the outermost nesting of the list
to> obtain just a sequence of of these n elements, that is>
{a,b},{a,b},{a,b} so that I can use this for input for some
function.> I would like to do something like>
Outer[SomeFunction, Table[{a,b},{N} ]] where I can enter N>
dynamically.> The problem, of course, is that the output of
the Table command is one> big list> and Outer is expecting
a sequence of N separate lists after> SomeFunction.> Use
SequencekTuples[n_Integer?Positive,s_List]:=
Flatten[Outer[List,Sequence@@Table[s,{n}]],n-1];s =
{a,b,c,d,e}; n =3;Length[kTuples[n,s]] ==
Length[s]^nTrueBob HanlonChantilly, VA
USA
====
f[Sequence@@{{a,b}, {a,b}, {a,b}}]f[{a, b}, {a, b},
{a, b}]Bobby Treat-----Original
Message-----dynamically.The problem, of course, is that the
output of the Table command is onebig listand Outer is
expecting a sequence of N separate lists afterSomeFunction.

====
f[x_] = x^2 - 7*x + 10g[x_] = Cos[3*x + 1/2*Pi]h[x_] =
(x - 3)/(x + 2)Off[Solve::ifun]Solve[f[x] == y,
x]Solve[g[x] == y, x]Solve[h[x] == y, x]Bobby
Treat-----Original Message-----
====
f[x_] = x^2 - 7*x +
10;g[x_] = Cos[3*x + 1/2*Pi];h[x_] = (x - 3)/(x + 2);#[x] &
/@ {f, g, h}{10 - 7*x + x^2, -Sin[3*x], (-3 + x)/(2 +
x)}Bobby Treat-----Original Message-----
====
I get the
expected results. I suspect g was already defined at
thetime. For instance you might have set g={a,b}
previously. ClearAll[g]goes before defining the function
g.Bobby Treat-----Original Message-----Calling it with
g[{x1,y1},{x2,y2}] one would expect the answer {x1-x2,y1-y2}
but instead one gets error messages.Why ? And how do I fix g
(i.e write a function that outputs thedifference of 2
vectors).
====
In[1]:=ktuples[n_Integer, d_List] :=
Flatten[Outer[List, Sequence @@ Table[d, {n}]],
2]In[2]:=ktuples[3, {a, b}]Out[2]={{a, a, a}, {a, a, b},
{a, b, a}, {a, b, b}, {b, a, a}, {b, a, b}, {b, b, a}, {b, b,
b}}Tomas GarzaMexico City----- Original Message -----

====
kTuples[k_Integer?Positive, s_List] :=
Partition[Flatten@Outer[List,Sequence @@ (s & /@ Range[k])],
k]kTuples[3, {a, b}]Bobby Treat-----Original
Message-----Sender: steve@smc.vnet.netApproved: Steven M.
Christensen <steve@smc.vnet.net>, Moderator
====
The trick is
to use the Sequence Function. Try the following code and
youwill see how it works.abcd = CharacterRange[a,
d]abcd1 = Flatten[Outer[{#1, #2} &, abcd, abcd], 1]abcd2
= Outer[First[#]*Last[#] &, Sequence[abcd1], 1]abcd2 uses the
list of lists you have created to load as a sequence
ofarguments to outer. You need the third argument to
operate on the parts ofthe list if this is your goal.Read
the description of Sequence. It also works in contexts other
thanOuter.Richard Palmer> Can anyone help me with this
problem.> If I have an n-element list, (say where each
element is itself a> list), such as {{a,b}, {a,b}, {a,b}}>
is there a way to strip off the outermost nesting of the list
to> obtain just a sequence of of these n elements, that is>
{a,b},{a,b},{a,b} so that I can use this for input for some
function.> I would like to do something like>
Outer[SomeFunction, Table[{a,b},{N} ]] where I can enter N>
dynamically.> The problem, of course, is that the output of
the Table command is one> big list> and Outer is expecting
a sequence of N separate lists after> SomeFunction.>

====
Dear GroupI would like to know, if is possible to
solveIntegrate[(exp[-a/x])/(x^2-b^2),{x,0,infinity}] witha
and b constant ( Real )using Mathematica.Valdeci
Mariano*****************************************************
****************Valdeci Mariano de SouzaMasterÇs Degree of
Applied Physics - Unesp/Rio Claro - State of S.8boPaulo -
BrazilLaboratory of Electrical Measurementsphone : (
0XX19 ) 526 -
2237*********************************************************
************
====
> Is there an easy (elegant?) way to generate
the set of all k-tuples> taking values from some set (list)
S? I want the arguments of the> function to be k (the length
of the tuples) and the set S. That is,> KTuples[3,{a,b}]
should produce>
{{a,a,a},{a,a,b},{a,b,a},{a,b,b},{b,a,a},{b,a,b},{b,b,a},{b,b
,b}}.Method 1:Distribute[Table[{a,b},{3}],List]Method
2:Needs[DiscreteMath`Combinatorica`];Strings[{a,b},3]
Rob PrattDepartment of Operations
Researchhttp://www.unc.edu/~rpratt/
====
> Does anyone know
how to get the JavaPlot window (or any windows of this
type)> which can be seen at >
http://www.wolfram.com/products/mathematica/newin42/java.html
> that WRI advertises comes with 4.2?Well, I don't know if
any tools (like that Window/Palette) written in Java are in
the M_4.2 box, but if you know a little Java Programming,
you can easily write something like that yourself; eg. using
the MathGraphicsJPanel which allows you to display
Mathematica GraphicsExpressions in a Java Component, simply
by calling its setCommand()Method (I hope that's correct, I
am writing that from memory); eg. a simple JFrame with a
MathGraphicsJPanel would be programmed likethis: (* This
code is necessary for setting up J/Link
*)<<JLink`InstallJava[ ](* now the GUI stuff *)theJFrame =
JavaNew[javax.swing.JFrame, My Java Plot
JFrame];theGraphicsPanel =
JavaNew[com.wolfram.jlink.MathGraphicsJPanel];cp@add[
theGraphicsPanel];theJFrame@pack[];theJFrame@setVisible[
True];(* now set the Graphics Expression, using eg. a Plot
commandpassed in as a
string*)theGraphicsPanel@setCommand[Plot[Sin[x], {x, -1,
1}]];This should give you a Java Swing JFrame showing the
result the Plot; (I hope this code works, I haven't tested
it; for further informationlook at the J/Link Userguide,
which should cover just about every Question about J/Link);
murphee
====
I'd like to import data from a file. The 
file is
a standard ASCII text file which contains a matrix of
numbers. Matrix columns are tab separated, matrix rows
are EOL seperated. Numbers use coma , as a decimal
point specifier.So the file looks like this:1,234 
1,567
1,892,234 2,567 2,893,234 3,567 3,89I know how to do this
using Import:Import[MyFile.txt, Table,
ConversionOptions -> {NumberPoint -> ,}] ;However, Import
is not the function I'd like to use, because it is
realyslow over large files.I wonder if someone knows a way
to use something faster than Import (I namely think of
ReadList) with coma-numbers?
====
I'm a poor physicist trying
to figure out how to sort out thephysical from the
non-physical solutions to a problem. To dothat, I need
to be able to look at an expression and pick out
asubexpression, the part under the radical.For example,
say I've got the expression a b x^2 + 5 x^3 + 5 Sqrt[4 -
x^2]I'd like to pick out 4 - x^2, which would then tell me
that x isbetween +/- 2. I know there has got to be an easy
way to do it, but Ican't find it. Any help would 
be
appreciated.Steve
Beachasb4@psu.eduhttp://www.thebeachfamily.org
====
I had
looked at using Cases, but had gotten tripped up
byforgettting to use Infinity to tell it to look at all
subexpressions. The tip from Bob Hanlon about how to use
Algebra`InequalitySolve`package helped too.> 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.>
For example, say I've got the expression > a b x^2 + 5 x^3
+ 5 Sqrt[4 - x^2]> I'd like to pick out 4 - x^2, which
would then tell me that x is> between +/- 2. I know there
has got to be an easy way to do it, but I> can't 
find it. Any
help would be appreciated.> Steve Beach> asb4@psu.edu>
http://www.thebeachfamily.org
====
 expr=a b x^2+5 x^3+5
Sqrt[4-x^2];First note that Mathematica interprets Sqrt[u]
as u^(1/2) FullForm[Sqrt[u]] Power[u,Rational[1,2]]Now we
can find the list of all u from subexpressions the form
Sqrt[u_]: Cases[expr, u_^(1/2) ->u,{0, Infinity}] {4 -
x^2}Here {0, Infinity} causes the search to be over
subexpressions at alllevels including level 0, which is
the whole expression. The search willnot go inside
heads unless we specify this: Cases[expr[y], u_^(1/2) ->
u, {0, Infinity}] {} Cases[expr[y], u_^(1/2) -> u, {0,
Infinity}, Heads -> True] {4 -
x^2}--Allan---------------------Allan HayesMathematica
Training and ConsultingLeicester
UKwww.haystack.demon.co.ukhay@haystack.demon.co.ukVoice
: +44 (0)116 271 4198> 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.>> For example, say
I've got the expression>> a b x^2 + 5 x^3 + 5 Sqrt[4 - 
x^2]>>
I'd like to pick out 4 - x^2, which would then tell me that
x is> between +/- 2. I know there has got to be an easy way
to do it, but I> can't find it. Any help would 
be
appreciated.> Steve Beach> asb4@psu.edu>
http://www.thebeachfamily.org>
====
Bob,Mathematica has
commands to do exactly what you wish and their use is
fairlycommon. The first command is Apply (@@ in 
prefix form)
and the secondcommand is Sequence.If you have a list of
arguments such as...arglist = {{a, b}, c, {d, e}, 3, Report
-> True};you can insert them into a function, f, simply by
applying f to the list.f @@ arglistf[{a, b}, c, {d, e}, 3,
Report -> True]If you want to insert them into another
function, h, that also has otherarguments, then you can use
Sequence and Apply.h[firstarg, Sequence @@ arglist, Compile ->
False]h[firstarg, {a, b}, c, {d, e}, 3, Report -> True,
Compile -> False]David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/
The problem, of course, is that the output of the Table
command is onebig listand Outer is expecting a sequence
of N separate lists
afterSomeFunction.
====
Bob,KTuples[n_Integer?Positive,
elements_List] := Flatten[Outer[List, Sequence @@
Table[elements, {n}]], n - 1]KTuples[3, {a, b}]{{a, a, a},
{a, a, b}, {a, b, a}, {a, b, b}, {b, a, a}, {b, a, b}, {b, b,
a}, {b, b, b}}Outer produces an array of all the elements but
we have to ßatten it to getit down to a two-level
array.David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/
Sender: steve@smc.vnet.netApproved: Steven M. Christensen
<steve@smc.vnet.net>, Moderator
====
Raj,You could include
an index for the function in the argument list, or make
aclear separation between parameters and variables as
follows (I just usetwo functions).f[1][r_, t_, p_] := 2r
Sin[t]f[2][r_, t_, p_] := r Cos[t]Sin[p]You could then sum
them, for example, bySum[f[i][r, t, p], {i, 1, 2}]r Cos[t]
Sin[p] + 2 r Sin[t]Or perhaps you wish to sum the derivatives
of the functions with respect tothe second argument,
t.Sum[Derivative[0, 1, 0][f[i]][r, t, p], {i, 1, 2}]2 r
Cos[t] - r Sin[p] Sin[t]Perhaps you can define all your
functions in terms of integer parameters.g[n_, m_][r_, t_,
p_] := r^(m - n)Sin[m t]Cos[n p]You could then sum as
follows.Sum[g[n, m][r, t, p], {m, 1, 3}, {n, 1,
3}]Cos[p]*Sin[t] + (Cos[2*p]*Sin[t])/r +
(Cos[3*p]*Sin[t])/r^2 + r*Cos[p]*Sin[2*t] + Cos[2*p]*Sin[2*t]
+ (Cos[3*p]*Sin[2*t])/r + r^2*Cos[p]*Sin[3*t] +
r*Cos[2*p]*Sin[3*t] + Cos[3*p]*Sin[3*t]Or perhaps you want
to take the square of the derivatives of the functionswith
respect to the first argument, r...Sum[(Derivative[1, 0,
0][g[n, m]][r, t, p])^2, {m, 1, 3}, {n, 1,
3}](Cos[2*p]^2*Sin[t]^2)/r^4 + (4*Cos[3*p]^2*Sin[t]^2)/ r^6 +
Cos[p]^2*Sin[2*t]^2 + (Cos[3*p]^2*Sin[2*t]^2)/ r^4 +
4*r^2*Cos[p]^2*Sin[3*t]^2 + Cos[2*p]^2*Sin[3*t]^2If
possible, try to steer away from For and Do loops and use
functionalprogramming as much as you can. At first it may
seem strange, but it is muchmore powerful and easier once
you get used to it. Ask further questions toMathGroup with
SPECIFIC examples and you will get a lot of help on how
touse functional programming.David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/
Sender: steve@smc.vnet.netApproved: Steven M. Christensen
<steve@smc.vnet.net>, Moderator
====
David,Except in a few
cases, Mathematica will not automatically generate
theinverse function for you. You will have to Solve the
equations and constructthe inverse function yourself. In
two of your three cases there are actuallymultiple inverse
functions.Taking the easy one first.Clear[x];Solve[f[x] ==
f, x][[1, 1]] // Simplifyx[f_] = x /. %x -> (3 + 2*f)/(1 -
f)(3 + 2*f)/(1 - f)So we now have the inverse
function.x[f](3 + 2*f)/(1 - f)For the quadratic equation
there are two solution.Clear[x]sols = Solve[f[x] == f,
x]{{x -> (1/2)*(7 - Sqrt[9 + 4*f])}, {x -> (1/2)*(7 + Sqrt[9
+ 4*f])}}We define the two solutions and identify them by an
index.x[1][f_] = x /. sols[[1, 1]]x[2][f_] = x /. sols[[2,
1]](1/2)*(7 - Sqrt[9 + 4*f])(1/2)*(7 + Sqrt[9 + 4*f])For the
third example there is a double infinity of
solutions.Clear[x]sols = Solve[f[x] == f,
x]Solve::ifun: Inverse functions are being used by
!(Solve), so some solutions may not be found.{{x
-> -(ArcSin[f]/3)}}Using some trigonometry we can define the
solutions as (I hope I got thisright)Clear[x];x[1,
n_][f_] = (-ArcSin[f] + 2 Pi n)/3x[2, n_][f_] = (-Pi +
ArcSin[f] + 2Pi n)/3Here are some of the solutions for f =
1/2.Table[x[1, n][1/2], {n, -5, 5}]~Join~Table[x[1, n][1/2],
{n, -5, 5}] // Sortf /@ %{-((61*Pi)/18), -((61*Pi)/18),
-((49*Pi)/18), -((49*Pi)/18), -((37*Pi)/18), -((37*Pi)/18),
-((25*Pi)/18), -((25*Pi)/18), -((13*Pi)/18), -((13*Pi)/18),
-(Pi/18), -(Pi/18), (11*Pi)/18, (11*Pi)/18, (23*Pi)/18,
(23*Pi)/18, (35*Pi)/18, (35*Pi)/18, (47*Pi)/18, (47*Pi)/18,
(59*Pi)/18, (59*Pi)/18}{1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2,
1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2,
1/2, 1/2, 1/2}Generally, when you want inverse functions you
are going have to do somework.David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/
Sender: steve@smc.vnet.netApproved: Steven M. Christensen
<steve@smc.vnet.net>, Moderator
====
g[u_, v_] := u -
vg[{x1, y1}, {x2, y2}]{x1 - x2, y1 - y2}David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/
y1-y2} but instead one gets error messages.Why ? And how do
I fix g (i.e write a function that outputs thedifference of 2
vectors).
====
>If I have an n-element list, (say where each
element is itself a>list), such as {{a,b}, {a,b}, {a,b}} is
there a way to strip off the>outermost nesting of the list
to obtain just a sequence of of these n>elements, that is
{a,b},{a,b},{a,b} so that I can use this for input>for some
function.Try Sequence@@ as in the
following:In[1]:=x={{a,b},{a,b},{a,b}};In[2]:=g[x_,y_,z_]
:=x(y.z)In[3]:=g[Sequence@@x]Out[3]=!({a ((a^2 +
b^2)), b ((a^2 + b^2))})
====
Why is it that
when I type, say, Table[ Plot[ Sin[n x],{x,0,Pi}], {n,
1,5}]my output is marked in the righthand column with 3 cell
``markers,withthe middle marker encompassing the output
(and *only* the output).However, when I try something like
Table[Show[Graphics[ etc., etc]], {n, 1,5}]the middle cell
marker on the righthand side of the notebook alsoincludes
the input command?The reason that this is important for me
is that I am doing ananimation; and after the animation I
would like to delete all thegraphic images; but when I use
Show[Graphics[ this would mean deletingthe input command as
well.Could someone explain this to me and also make a
suggestion so how Icould use Show[Graphics[ ] and have the
middle cell marker *only*encompass the
output.
====
Bobby,here my solution(s) as promised. As you
stated, my published solution wasonly for rectangles
oriented parallel to the X and Y axis, as this isperhaps
the predominant application.The idea was, to decide whether
all vertices of a rectangle lie to one sideof the other.
This idea expands to the general case. And effectively it
isthe same idea as of your suggestion, as well as of the
improved versionafter Garry Helzer, i.e. the
constructive solutions, as opposed to thoseinvolving
more abstract mathematical reasoning as of David Park and
AndrzejKozlowski. And these, esp. Andrzej's, are very
interesting, as they show howto derive a solution just
from this kind of reasoning.First some preliminaries: random
rectanglesI had chosen to describe the rectangles as a list
of counterclockwisevertices. To generate them, I had used a
similar description as David Park:toRectangle[corner_,
[Theta]_, base_, side_] := Module[{vb = {Cos[[Theta]],
Sin[[Theta]]}* base, vs = {-Sin[[Theta]],
Cos[[Theta]]}* side}, {corner, corner + vb, corner + vb +
vs, corner + vs}]makeRect[] := toRectangle[{Random[],
Random[]}, [Pi]/2 Random[], Random[],
Random[]]rectLine[{pFirst_, pRest__}] := Line[{pFirst,
pRest, pFirst}]ShowRects[r1_, r2_] :=
Show[Graphics[{PointSize[0.05], Hue[.5], rectLine[r1],
MapIndexed[{Point[#1], GrayLevel[0], Text[First[#2], #1]} &,
r1], Hue[0], rectLine[r2], MapIndexed[{Point[#1],
GrayLevel[0], Text[First[#2], #1]} &, r2]}], Background ->
GrayLevel[.8], AspectRatio -> Automatic, PlotRange ->
All]test cases:rect[1] = toRectangle[{0, 0}, 0, 1, 1]; (*
from David *)rect[2] = toRectangle[{1, 1}*0.9, [Pi]/4, 2,
1];ShowRects[rect[1], rect[2]]rx = rect[2] /. {x_?NumericQ,
y_} :> {x, y} + {-0.1, 0.15};ShowRects[rect[1], rx]rxx =
{{-.5, .2}, {1.5, .2}, {1.5, .8}, {-.5,
.8}};ShowRects[rect[1], rxx]test data:recs1000 =
Table[makeRect[],{1000}];Now here my solution. It is written
such as to communicate the idea, call
itelegant:offSide[r2_][{p1_, p2_}] := And @@ ((p2 - p1).(#
- p1) <= 0 &) /@ r2rectOverlap[r1_, r2_] := =AC (Or @@
offSide[r2] /@ Partition[r1, 2, 1, {1, 1}] [Or] Or @@
offSide[r1] /@ Partition[r2, 2, 1, {1, 1}])I came to this,
when I tried to expand my solution for evenly
orientedrectangles to the general case. I first tried
oblique coordinates, sightingalong the sides of the
rectangles, then saw that giving them in the dualbase is
computationally simpler, to recognize i.e. just the distance
of apoint from the side's (straight
line).Correctness:rectOverlap[rect[1],
rect[2]]TruerectOverlap[rect[1],
rx]FalserectOverlap[rect[1],
rxx]TruePerformance:(testr1 = MapThread[ rectOverlap,
{recs1000, RotateLeft[recs1000, 1]}]); // Timing{3.635
Second, Null}The costs for RotateLeft are negligable in
Timing.Garry Helzer had proposed a solution which wasn't
correct (e.g. for rect[1},rxx) as he noted himself. Bobby
Treat however noticed that it can be fixed.Here my version
thereof. (You can see how similar it is, starting
fromdifferent reasoning.)rightSide[a_, {b_, c_}] :=
Det[Prepend[#, 1] & /@ {a, b, c}] < 0vertexExcluded[r1_,
r2_] := =AC (Or @@ And @@@ Outer[rightSide[#2, #1] &,
Partition[r2, 2, 1, {1, 1}], r1, 1] [Or] Or @@ And @@@
Outer[rightSide[#2, #1] &, Partition[r1, 2, 1, {1, 1}], r2,
1])(testr4 = MapThread[ vertexExcluded, {recs1000,
RotateLeft[recs1000, 1]}]); // Timing{3.646 Second,
Null}testr1 === testr4Trueof quite similar
performance.Bobby Treat gave a different implementation of
this (as seen below) which ismarginally faster.Now here
Bobby's solution, rewritten in my style (which makes it
slightlyfaster):extent[r1_, r2_] := {Min[#], Max[#]} & /@
((r1[[{1, 2}]] - r1[[{2,
3}]]).Transpose[r2])cannotIntersect[{{min1_, max1_}, {min2_,
max2_}}] := max2 < min1 || max1 < min2intersects[r1_, r2_] :=
Catch[If[cannotIntersect[#], Throw[False]] & /@
Flatten[Transpose[ Join[Outer[extent, {r1}, {r1, r2}, 1],
Outer[extent, {r2}, {r2, r1}, 1]], {1, 3, 2}], 1];
True](testr5 = MapThread[intersects, {recs1000,
RotateLeft[recs1000, 1]}]); //Timing{1.502 Second,
Null}testr5 === testr1Truethe tests for all sides. Such
we come to use of the non-strict evaluation ofAnd[ ] and
Or[ ]. This effectively corresponds to a Catch and
Throw.rectOverlap2[r1 : {p1_, p2_, p3_, p4_}, r2 : {q1_,
q2_, q3_, q4_}] :=! Or[ ((p2 - p1).(# - p1) <= 0 &) /@ And @@
r2, ((p3 - p2).(# - p2) <= 0 &) /@ And @@ r2, ((p4 - p3).(# -
p3) <= 0 &) /@ And @@ r2, ((p1 - p4).(# - p4) <= 0 &) /@ And
@@ r2, ((q2 - q1).(# - q1) <= 0 &) /@ And @@ r1, ((q3 -
q2).(# - q2) <= 0 &) /@ And @@ r1, ((q4 - q3).(# - q3) <= 0
&) /@ And @@ r1, ((q1 - q4).(# - q4) <= 0 &) /@ And @@
r1](testr2 = MapThread[ rectOverlap2, {recs1000,
RotateLeft[recs1000, 1]}]); // Timing{1.502 Second,
Null}testr2 === testr1TrueEqual in performance to Bobby's
solution. The use of And with ((p2 - p1).(#- p1) <= 0 &) /@
And @@ r2, etc. is a bit tricky: it effectively preventsthe
evaluation of (p2 - p1).(q1 - p1) <= 0 etc., such that this is
evaluatedwithin And (non-standard evaluation) after
mapping. And @@ ((p2 - p1).(# -p1) <= 0 &) /@ r2 to the
contrary is less performant. (You can see here howclever
the language is designed, to allow use of And as a container,
untilthe time comes to execute!)This solution can be
improved a little bit; we replace mapping by threadingand
delay execution of the pure functions' body to the
application of Andnow by a different method (since we don't
Map):rectOverlap3[r1 : {p1_, p2_, p3_, p4_}, r2 : {q1_, q2_,
q3_, q4_}] :=! Or[ Block[{v = p2 - p1}, Or[And @@
Thread[Unevaluated[(Unevaluated[v.(# - p1) <= 0] &)[r2]]],
And @@ Thread[Unevaluated[(Unevaluated[v.(# - p2) >=
0]&)[r2]]]]], Block[{v = p4 - p1}, Or[And @@
Thread[Unevaluated[(Unevaluated[v.(# - p1) <= 0] &)[r2]]],
And @@ Thread[Unevaluated[(Unevaluated[v.(# - p4) >=
0]&)[r2]]]]], Block[{v = q2 - q1}, Or[And @@
Thread[Unevaluated[(Unevaluated[v.(# - q1) <= 0] &)[r1]]],
And @@ Thread[Unevaluated[(Unevaluated[v.(# - q2) >=
0]&)[r1]]]]], Block[{v = q4 - q1}, Or[And @@
Thread[Unevaluated[(Unevaluated[v.(# - q1) <= 0] &)[r1]]],
And @@ Thread[Unevaluated[(Unevaluated[v.(# - q4) >=
0]&)[r1]]]]]](testr3 = MapThread[ rectOverlap3, {recs1000,
RotateLeft[recs1000, 1]}]); // Timing{1.442 Second,
Null}testr3 === testr1TrueSo this is fastest by a small
margin, but has lost almost all elegance. Theouter
Unevaluated is necessary to prevent evaluation within Thread,
theinner does the essential trick noted above.Hope this was
of some interest,Hartmut>-----Original Message----->Sent:
Monday, August 26, 2002 10:16 AM> rectangle
intersection>Garry,>>Also note your solution requires
rectangle points to be in clockwise>order (mine doesn't),
but yours works for arbitrary convex polygons
as>written.>>Bobby>>-----Original Message----->RE:
>rectangle intersection>>Garry,>>Here's a solution using
your LeftSide concept; it works perfectly but>takes twice
as much time as my solution. Both solutions look at
every>vertex of both rectangles, but mine uses two sides
from each and yours>requires looking at all four sides of
each rectangle. I'd think yours>should be a triße faster
than this, though. There may be efficiencies>I'm 
missing (in
both solutions).>>ClearAll[cis, rect, pickRect, extent,
cannotIntersect, intersects,>daveRect]>cis[t_] := {Cos@t,
Sin@t}>rect[{pt : {_, _}, angle_, {len1_, len2_}}] :=
Module[{pt2},> {pt, pt2 => pt + len1 cis[angle],> pt2 - len2
cis[angle - Pi/2], pt - len2 cis[angle - Pi/2]}]>daveRect :=
{{Random[], Random[]}, Random[] + Pi/2,
{Random[],>Random[]}}>pickRect := rect@daveRect>extent[r1_,>
r2_] := {Min@#, Max@#} & /@ ((Take[r1, 2] -
r1[[{2,>3}]]).Transpose@r2)>cannotIntersect[{{min1_, max1_},
{min2_,> max2_}}] := max2 < min1 || min2 >
max1>intersects[r1_, r2_] := Catch[> If[cannotIntersect[#],
Throw[False]] & /@>Flatten[Transpose[Outer[extent, >{r1},
{r1, r2}, 1]~Join~Outer[extent, {r2}, {r2, r1}, 1], {1, 3,
2}],>1];>
Throw[True]]>>ClearAll[leftSide,leftIntersects,sides]>sides[a
_List]:=Partition[Join[a,{First@a}],2,1]>leftSide[{a_,b_},{{c
_,d_},{e_,f_}}]:=-b c+a d+b e-d e-a f+c
f>0>leftSide[a:{{_,_}..},b:{{_,_},{_,_}}]:=leftSide[#,b]&/@a
>leftSide[a_List,b:{{{_,_},{_,_}}..}]:=leftSide[a,#]&/@b>
leftIntersects[a_,b_]:=!Or@@(And@@#&/@leftSide[a,sides@b])&&!
>
Or@@(And@@#&/@leftSide[b,sides@a])>>davePairs={daveRect,
daveRect}&/@Range[10000];>rectanglePairs=Map[Reverse@rect[#]&
,davePairs,{2}];>Timing[right=intersects[Sequence@@#]&/@
rectanglePairs;]>Timing[test=leftIntersects[Sequence@@#]&/@
rectanglePairs;]>right[Equal]test>>{3.187999999999999*
Second, Null}>{6.765000000000001*Second, Null}>True>>Bobby
Treat>>-----Original Message----->
rectangle>intersection>>As Daniel Lichtblau pointed out,
the statement below about vertices is >nonsense. Consider two
overlapping rectangles arranged as a cross. You >need to
compute intersections and test them instead of
vertices.>>Begin forwarded message:>>
rectangle>intersection>> Begin forwarded message:>>
Dear colleagues,>> any hints on how to implement a very
fast routine in Mathematica for>> testing if two rectangles
have an intersection area?>> Frank Brand>> Here is one
approach.>> Given three points {x1,y1},{x2,y2},{x3,y3},
switch to homogenous>> coordinates a={1,x1,y1}, b={1,x2,y2},
c={1,x3,y3}. Then>> Sign[Det[{a,b,c}]] is +1 if and only if
the point a lies on your left>as>> you walk along the line
though b and c in the direction from b to c.>> ( If the
result is zero, then a lies on the line.)>> The value of
the determinant is>x2y3-x3y2-x1y3+x3y1+x1y2-x2y1, and the>>
speed of the algorithm depends essentially on how fast this
quantity>can>> be computed. Suppose we write a function
LeftSide[a,{b,c}] that>computes>> the sign of the
determinant.>> Now let {p1,p2, . . ., pn} be a list of
vertices (pi={xi,yi}) of a>> convex polygon traced
counterclockwise. Then a lies within or on the>> boundary of
the polygon if and only if none of the numbers>>
LeftSide[a,{pi,p(i+1)}] are -1. That is, if -1 does not
appear in the>> list LeftSide[a,#]&/@Partition[{p1,p2,. .
.,pn,p1},2,1].>> Now use the fact that if two convex
polynomials overlap, then some>> vertex of one of them must
lie inside or on the boundary of>the other.>> If an
overlap of positive area is required, then the check is
that>only>> +1 appears--not that -1 does not appear.>>
For two rectangles ( or parallelograms) this approach
requires the>> evaluation of 16 determinants, so it may be a
bit expensive. If the>> points have rational coordinates,
then (positive) denominators may be>> cleared in the
homogeneous coordinates and the computations can be>done>>
in integer arithmetic, at the cost of at least three more>>
multiplications per determinant.>Garry Helzer>Department
of Mathematics>University of Maryland>College Park, MD
20742>301-405-5176>gah@math.umd.edu>
====
I'm
looking for a way of finding the approximation for partitial
binomial sum.I'll be pleasant for any
hint..Constantine.Constantine ElsterComputer Science
Dept.Technion I.I.T.Office: Taub 411
====
> Here's my
contestant:> <<DiscreteMath`Combinatorica`>
KTuples[k_Integer, vals_List] :=>
Union[KSubsets[PadRight[vals, k*Length[vals], vals], k]]Dear
mathgroup,Here is another solution; probably less elegant,
but I found it much faster (DiscreteMath's Subsets and
KSubsets are too slow. Besides, it produces each tuple only
once):KTuples2[n_Integer, L_List] := Flatten[Outer[Append,
KTuples2[n-1, L], L, 1], 1] /; n > 1;KTuples2[1, L_List] :=
Transpose[{L}](Both function yields the tuples in
lexicographic order.)Sz. Szikla
====
Here a easy way
:In[4]:=expr=a b x^2 + 5 x^3 + 5 Sqrt[4 -
x^2]In[5]:=FullForm[expr]Out[5]//FullForm=Plus[Times[a,b,
Power[x,2]],Times[5,Power[x,3]],
Times[5,Power[Plus[4,Times[-1,Power[x,2]]],Rational[1,2]]]]
In[9]:=expr[[3,2,1]]Out[9]=4 - x^2Meilleures
salutationsFlorian Jaccardprofesseur de
Math.8ematiquesEICN-HES-----Message
d'origine-----Envoy.8e : mar., 27. ao.9et 2002 08:08è 
:
mathgroup@smc.vnet.netObjet : How do I pick out the
expression under a radical?I'm a poor physicist trying to
figure out how to sort out thephysical from the
non-physical solutions to a problem. To dothat, I need
to be able to look at an expression and pick out
asubexpression, the part under the radical.For example,
say I've got the expressiona b x^2 + 5 x^3 + 5 Sqrt[4 -
x^2]I'd like to pick out 4 - x^2, which would then tell me
that x isbetween +/- 2. I know there has got to be an easy
way to do it, but Ican't find it. Any help would 
be
appreciated.Steve
Beachasb4@psu.eduhttp://www.thebeachfamily.org
====
> 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.>> For
example, say I've got the expression>> a b x^2 + 5 x^3 + 5
Sqrt[4 - x^2]>> I'd like to pick out 4 - x^2, which would
then tell me that x is> between +/- 2. I know there has got
to be an easy way to do it, but I> can't find
it.>Needs[Algebra`InequalitySolve`];expr = a b x^2+5
x^3+5 Sqrt[4-x^2];InequalitySolve[#>=0, x]& /@ Cases[expr,
Sqrt[x_] -> x, Infinity]{-2 <= x <= 2}Bob HanlonChantilly,
VA USA
====
> I would like to know, if is possible to solve>
> Integrate[(exp[-a/x])/(x^2-b^2),{x,0,infinity}] with> a
and b constant ( Real )> using Mathematica.>
Integrate[(Exp[-a/x])/(x^2-b^2),{x,0,Infinity},
GenerateConditions->False]//FullSimplifySqrt[-(1/b^2)]*((1/
2)*Pi*Cosh[a/b] +
CosIntegral[a*Sqrt[-(1/b^2)]]*Sin[a*Sqrt[-(1/b^2)]]) +
(Cosh[a/b]*SinhIntegral[a/b])/bFullSimplify[%,
Element[{a,b},
Reals]](1/(2*b*Abs[b]))*(-2*b*CosIntegral[(I*a)/Abs[b]]*
Sinh[a/Abs[b]] + Cosh[a/b]*(I*b*Pi +
2*Abs[b]*SinhIntegral[a/b]))Bob HanlonChantilly, VA
USA
====
Bob,Table[ Plot[ Sin[n x], {x, 0, Pi}], {n, 1,
5}]andTable[ Show[Graphics[Line[{{0, 0}, {5, n}}]],
PlotRange -> {{0, 6}, {0, 6}}], {n, 1, 5}]They both have the
graphics output cells grouped in the middle and separatefrom
the Input cell.David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/
The reason that this is important for me is that I am doing
ananimation; and after the animation I would like to delete
all thegraphic images; but when I use Show[Graphics[ this
would mean deletingthe input command as well.Could someone
explain this to me and also make a suggestion so how Icould
use Show[Graphics[ ] and have the middle cell marker
*only*encompass the output.
====
Steve,Here is a slightly
more complicated case.expr = a b x^2 + 5 x^3 + 5 Sqrt[4 -
x^2] - 1/Sqrt[5 - 2x^2]The following picks out the
expressions under square roots. Square roots
arerepresented as Power[a,1/2] and if they are in the
denominator they area -1/2 power. We use an Alternative in
the pattern to pick out both.rexprs = Cases[expr, Power[a_,
1/2 | -1/2] -> a, Infinity]{5 - 2*x^2, 4 - x^2}The following
Standard Package is
useful.Needs[Algebra`InequalitySolve`]And @@ (# >= 0 &
/@ rexprs)InequalitySolve[%, x]5 - 2*x^2 >= 0 && 4 - x^2 >=
0-Sqrt[5/2] <= x <= Sqrt[5/2]David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/
Steve
Beachasb4@psu.eduhttp://www.thebeachfamily.org
====

Valdeci,It appears that the integral will not converge if b
is real. Mathematicagives...Integrate[Exp[-a/x]/(x^2 -
b^2), {x, 0, Infinity}]If[Re[a] > 0 && Arg[b^2] != 0,
(1/2)*Sqrt[-(1/b^2)]* (2*CosIntegral[a*Sqrt[-(1/b^2)]]*
Sin[a*Sqrt[-(1/b^2)]] + Cos[a*Sqrt[-(1/b^2)]]* (Pi -
2*SinIntegral[a*Sqrt[-(1/b^2)]])), Integrate[1/(E^(a/x)*(-b^2
+ x^2)), {x, 0, Infinity}]]If b is Real then Arg[b^2]==0 and
Mathematica doesn't solve it. Let's 
definea function that
allows us to test specific values of a and b.f[a_, b_][x_] =
Exp[-a/x]/(x^2 - b^2);Integrate[f[2, 3][x], {x, 0,
Infinity}]Integrate::idiv : Integral of 1/(E^(2/x)*(-9 +
x^2)) does not converge on{0, Infinity}.But if we use an
imaginary value for b...Integrate[f[2, 3I][x], {x, 0,
Infinity}]%//N(1/6)*(2*CosIntegral[2/3]*Sin[2/3] +
Cos[2/3]*(Pi - 2*SinIntegral[2/3]))0.254022David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/
MasterÇs Degree of Applied Physics - Unesp/Rio Claro - 
State
of S.8boPaulo - BrazilLaboratory of Electrical
Measurementsphone : ( 0XX19 ) 526 -
2237*********************************************************
************
====
I am having a strange problem with a
function giving a complex number as a result. I did the
following:- define a function:denom[x_, p_, d_] := Sqrt[1 +
(x*Tan[p]/d)^2]Integrate and simplify it with the assumption
that d is larger than 0:FullSimplify[Integrate[denom[x, p,
d], {x, 0, d}], d > 0]The result of the above line is
d/2*(1+i*Sqrt[2]*d*Cos[p]^2)*Sqrt[Sec[p]^2] where i is
Sqrt[-1]I assing it to a function called peter in the
following way:peter[p_, d_] :=
FullSimplify[Integrate[denom[x, p, d], {x, 0, d}], d>0]and
check the value of the function at [0,1]peter[0,1]and the
result is 1.How is it possible that the result doesn't have
an imaginary part???I would expect the result to be
0.5+Sqrt[2]/2*iPeter
====
Lucas,One way to change the
precedence of CirclePlus is to change the
fileUnicodeCharacters.tr.On my machine the file is 
located
under../4.1/SystemFiles/FrontEnd/TextResourcesOpen up the
file, search for CirclePlus, change the precedence from 450
to420, and then save. Of course, it would be wise to make a
backup copy of thefile before you make any changes. Also,
420 is low enough to get thebehavior you desire, but you
may want to experiment with other precedences.Then, start
mathematica and you will get the behavior you want.Carl
WollPhysics DeptU of Washington>> I'm attempting to
implement an abstract mathematica package in> mathematica
that utilized the [CirclePlus] operator in an unusual>
way. Specifically, the [CirclePlus] has a precidence lower
than +> and introduces barriers in the computation. So, an
expression such as>> a + b [CirclePlus] c + d --> (a+b)
[CirclePlus] (c+d)>> The mathematica ouput of>> a + d + (b
[CirclePlus] c) is incorrect. I've tried playing with the>
PrecedenceForm[] function, but that does not seem able to
produce the> desired effect.>> Also, I would like to
introduce a notation like>> N> [BigCirclePlus] x[[i]] -->
x[[1]] [CirclePlus] x[[2]] [CirclePlus]> ....> i=0>>
analagous to summation, but mathematica does not appear to
offer the> CirclePlus in a large format. to relate this to
the case above, x[1]> = (a + b) and> x[2] = (c + d), so each
indexed element is a subexpression.>> Finally, I would like to
be able to set up the CirclePlus operator> such that the
following algebraic relations hold:> Sum BigCirclePlus
E = BigCirclePlus Sum E> i j ij j i ij>> d d> --
BigCirclePlus E = BigCirclePlus -- E> dx j j j dx j>>
-Lucas Scharenbroich> -MLS Group / JPL>
====
Lucas,I hope
that your question will provoke a number of replies because I
thinkit is an interesting topic.First, it would be nice if
Mathematica had a ShowPrecedence statement toquickly
retrieve a precedence number for any command. It is a bit
timeconsuming to search through the table in Section
A.2.7.Next, it would be nice if the user could set the
precedence for operatorsthat don't have built-in
definitions.Since it appears that you can't do 
that, is it
possible to switch themeanings of Plus and CirclePlus in
your theory? Are you depending on thenumerical behavior of
Plus? Go with the precedence that Mathematica givesyou. Or
you could use something like VerticalBar or RightTee which
havelower precedence than Plus, but perhaps don't have the
look that you want.You could try to use the Notation package
for your CirclePlus sum. But Ialways find the Notation
package difficult to use and like to useStandardForm as much
as possible. One possibility is...!((([Sum]+(i =
1)%5 x[i])) /. Plus -> CirclePlus)These are weak
answers to your question, but maybe they will bring in
morediscussion.David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/
desired effect.Also, I would like to introduce a notation
like N[BigCirclePlus] x[[i]] --> x[[1]] [CirclePlus]
x[[2]] [CirclePlus].... i=0analagous to summation, but
mathematica does not appear to offer theCirclePlus in a
large format. to relate this to the case above, x[1]= (a + b)
andx[2] = (c + d), so each indexed element is a
subexpression.Finally, I would like to be able to set up the
CirclePlus operatorsuch that the following algebraic
relations hold:Sum BigCirclePlus E = BigCirclePlus
Sum E i j ij j i ijd d-- BigCirclePlus E =
BigCirclePlus -- Edx j j j dx j-Lucas Scharenbroich-MLS
Group / JPL
====
I need some help with using FindRoot. I want
to solve a system ofnonlinear equations numerically. Each
equation in the system is anequation with vector variables.
The equations are such that it isdifficult to convert them to
equations involving only the componentsof the vector
variables.I have tried the following two possibilities
without success. In whatfollows I am using a MADE UP
EXAMPLE. In this example it is easy towrite the equations
in terms of only the components. Furthermore thecorrect
solution is obvious. But I made up the example
forillustration
only.----------------------------Try1-----------------------
---------------------q = {{x[1], y[1]}, {x[2],
y[2]}};FindRoot[{q[[1]] + q[[2]] Sqrt[q[[2]].q[[2]]] == {0,
0}, q[[1]] + q[[2]] == {0, 0}}, {q[[1]], {0, 0}}, {q[[2]],
{0, 0}}]This results in the error message
:FindRoot::fddis: Start specification {q[[1]],{0,0}} does
not containdistinctstarting
values.----------------------------Try2---------------------
------------------------FindRoot[{q[[1]] + q[[2]]
Sqrt[q[[2]].q[[2]]] == {0, 0}, q[[1]] + q[[2]] == {0, 0}},
{x[1], 0}, {x[2], 0}, {y[1], 0},{y[2], 0}]This results in
the error message :FindRoot::frnum: Function
{0.,0.},{0.,0.}} is not a length 4 listof numbers at
{x[1],x[2],y[1],y[2]} = {0., 0., 0., 0.}Questions 1) How do
I fix these two methods ? 2) What do the error messages mean
?Please keep in mind that this example is made up and
trivial to solvewithout Mathematica. It is being used for
illustration purpose only.
====
> I need some help with using
FindRoot. I want to solve a system of> nonlinear equations
numerically. Each equation in the system is an> equation
with vector variables. The equations are such that it is>
difficult to convert them to equations involving only the
components> of the vector variables.> I have tried the
following two possibilities without success. In what>
follows I am using a MADE UP EXAMPLE. In this example it is
easy to> write the equations in terms of only the
components. Furthermore the> correct solution is obvious.
But I made up the example for> illustration only.>
----------------------------Try>
1--------------------------------------------> q = {{x[1],
y[1]}, {x[2], y[2]}};> FindRoot[{q[[1]] + q[[2]]
Sqrt[q[[2]].q[[2]]] == {0, 0}, > q[[1]] + q[[2]] == {0, 0}},
{q[[1]], {0, 0}}, {q[[2]], {0, 0}}]> This results in the
error message :> FindRoot::fddis: Start specification
{q[[1]],{0,0}} does not contain> distinct> starting
values.> ----------------------------Try>
2--------------------------------------------->
FindRoot[{q[[1]] + q[[2]] Sqrt[q[[2]].q[[2]]] == {0, 0}, >
q[[1]] + q[[2]] == {0, 0}}, {x[1], 0}, {x[2], 0}, {y[1], 0},>
{y[2], 0}]> This results in the error message :>
FindRoot::frnum: Function {0.,0.},{0.,0.}} is not a
length 4 list> of numbers at {x[1],x[2],y[1],y[2]} = {0.,
0., 0., 0.}> Questions > 1) How do I fix these two methods
? For the second, you can ...o Make sure the first argument
is evaluated (use Evaluate[])o Get rid of the Equal (==)o
Flatten the
vectors.In[1]:=q={{x[1],y[1]},{x[2],y[2]}};FindRoot[
Evaluate[ Flatten[{q[[1]]+q[[2]]
Sqrt[q[[2]].q[[2]]],q[[1]]+q[[2]]}]],{x[1],
0},{x[2],0},{y[1],0},{y[2],0}]Out[2]={x[1] -> 0., x[2] ->
0., y[1] -> 0., y[2] -> 0.}For the first, you will have to
wait until a future version of Mathematica (this works in a
development version now as shown below) which will support
vector variables. It will still not support variables with
head Part (like q[[1]]), so you can doFindRoot[{q1 + q2
Sqrt[q2.q2], q1 + q2},{q1,{0,0}},{q2,{0,0}}]which
returnsIn[1]:=FindRoot[{q1 + q2 Sqrt[q2.q2], q1 +
q2},{q1,{0,0}},{q2,{0,0}}]Out[4]=Note that the Evaluate[]
will no longer be necessary> 2) What do the error messages
mean ?FindRoot::fddis: means FindRoot is looking for
numbers as starting values.FindRoot has a syntax which
accepts two starting values for using
derivativeFindRoot::frnum: If FindRoot cannot resolve a
list of equalities, it looks fora list of something which
evaluates to numbers when the variables take on numerical
values.> Please keep in mind that this example is made up
and trivial to solve> without Mathematica. It is being used
for illustration purpose only.> 
====
If I want to protect
the Mathematica Program, what can I do? Is there any method
to avoid other people's reading my program butit is still
able to run ßuently? Gory
====
You could try obfuscating it,
i.e., scrambling the symbol namesexcept for those you wish
to export.Check out obfuscation tools
underhttp://www.semdesigns.com/Products/Formatters/
index.html.We don't have an obfuscator at this moment for
Mathematica,but our base technology can build obfuscators
for langaugesfor which we have definitions, ... and we
happen to havea definition of Mathematica.--Ira Baxter,
Ph.D. CTO Semantic Designswww.semdesigns.com 512-250-1018>
If I want to protect the Mathematica Program, what can I do?
Is there anymethod to avoid other people's reading my
program but> it is still able to run ßuently?>
Gory>
====
Mathematica version 4.2 has no problem with this
one. All you have todo is spell things
correctly.Integrate[Exp[-a/x]/(x^2 - b^2), {x, 0,
Infinity}]If[Re[a] > 0 && Arg[b^2] != 0,
(1/2)*Sqrt[-(1/b^2)]* (2*CosIntegral[ a*Sqrt[-(1/b^2)]]*
Sin[a*Sqrt[-(1/b^2)]] + Cos[a*Sqrt[-(1/b^2)]]* (Pi -
2*SinIntegral[ a*Sqrt[-(1/b^2)]])), Integrate[1/(E^(a/x)*
(-b^2 + x^2)), {x, 0, Infinity}]]Bobby Treat-----Original
Message-----*************************************************
********************Valdeci Mariano de SouzaMasterÇs
Degree of Applied Physics - Unesp/Rio Claro - State of
S.8boPaulo - BrazilLaboratory of Electrical
Measurementsphone : ( 0XX19 ) 526 -
2237*********************************************************
************
====
FullForm[expr]Plus[Times[a,b,Power[x,2]],
Times[5,Power[x,3]],
Times[5,Power[Plus[4,Times[-1,Power[x,2]]],Rational[1,2]]]]
What you want is the third argument of Plus, the second
argument ofTimes, and the first argument of Power.expr = a
b x^2 + 5 x^3 + 5 Sqrt[4 - x^2];expr[[3]]expr[[3,
2]]expr[[3, 2, 1]]5*Sqrt[4 - x^2]Sqrt[4 - x^2]4 - x^2Bobby
Treat-----Original Message-----can't find it. 
Any help would
be appreciated.Steve
Beachasb4@psu.eduhttp://www.thebeachfamily.org
====
I try
to do the simple task of transposing a matrix.X =
{{a,b},{c,d},{e,f}}whereas Transpose[{{a,b},{c,d},{e,f}}]
works well.What is wrong with writing Transpose[X] ?Terje
Johnsen
====
>-----Original Message----->Sent: Monday, August
26, 2002 10:16 AM>Lucas,>>I hope that your question will
provoke a number of replies >because I think>it is an
interesting topic.>>First, it would be nice if Mathematica
had a ShowPrecedence >statement to>quickly retrieve a
precedence number for any command. It is a bit
time>consuming to search through the table in Section
A.2.7.>>Next, it would be nice if the user could set the
precedence >for operators>that don't have built-in
definitions.>>Since it appears that you can't do 
that, is it
possible to switch the>meanings of Plus and CirclePlus in
your theory? Are you >depending on the>numerical behavior of
Plus? Go with the precedence that >Mathematica gives>you. Or
you could use something like VerticalBar or RightTee which
have>lower precedence than Plus, but perhaps don't have the
look >that you want.>>You could try to use the Notation
package for your CirclePlus >sum. But I>always find the
Notation package difficult to use and like to
use>StandardForm as much as possible. One possibility
is...>>!((([Sum]+(i = 1)%5 x[i])) /. Plus
-> CirclePlus)>>These are weak answers to your question,
but maybe they will >bring in more>discussion.>>David
Park>djmp@earthlink.net>http://home.earthlink.net/~djmp/>
>I'm attempting to implement an abstract mathematica
package in>mathematica that utilized the [CirclePlus]
operator in an unusual>way. Specifically, the [CirclePlus]
has a precidence lower than +>and introduces barriers in the
computation. So, an expression such as>>a + b [CirclePlus]
c + d --> (a+b) [CirclePlus] (c+d)>>The mathematica ouput
of>>a + d + (b [CirclePlus] c) is incorrect. I've tried
playing with the>PrecedenceForm[] function, but that does
not seem able to produce the>desired effect.>>Also, I would
like to introduce a notation like>> N>[BigCirclePlus]
x[[i]] --> x[[1]] [CirclePlus] x[[2]] [CirclePlus]>....>
i=0>>analagous to summation, but mathematica does not appear
to offer the>CirclePlus in a large format. to relate this to
the case above, x[1]>= (a + b) and>x[2] = (c + d), so each
indexed element is a subexpression.>>Finally, I would like to
be able to set up the CirclePlus operator>such that the
following algebraic relations hold:>Sum BigCirclePlus E
= BigCirclePlus Sum E> i j ij j i ij>>d d>--
BigCirclePlus E = BigCirclePlus -- E>dx j j j dx
j>>-Lucas Scharenbroich>-MLS Group / JPL>Lucas,I'm
certainly not competent to give you any answer, and me too,
I'd be eagerto hear such -- hoped Wolfram's to 
react;
anyway, here is my opinion.PrecedenceForm is only for
parenthesizing at output.Although we have MakeExpressions to
add semantic actions to the Mathematicacompiler (if we may
call such the transformation from input string,
i.e.language, to the internal representation as indicated
by FullForm i.e.code), we cannot inßuence the Mathematica
syntax proper, i.e. parsing.The operator precedence rules
however are part of that (esp. needed for boxformation). So
I can't see a way to reach your goal. Now, I 
don't know what
your ordinary Plus shall designate. If it's onlyabstract,
i.e. you don't use it for numeric calculation, you possibly
mightinterchange the roles of Plus and CirclePlusIn[6]:=
a [CirclePlus] b + c [CirclePlus] dOut[6]=
a[CirclePlus]b + c[CirclePlus]dAs for BigCirclePlus (if
you still need it after the reassigned meanings)I'd try to
define a palette, but perhaps this is not possible to the
extendyou desire. I never tried.Another approach would be
to use Union and Intersection for low precedenceplus and
low precedence times (or [Subset]; [Superset] or
[And]; [Or]or [Therefore];[Because] or ...).The
ultima ratio is to write a new front end.But, if I were you,
I'd ignore these kind of problems at first and 
set upthe
package fully functioning (with 1D input), apply it to your
problems andinvestigations, and finally, when time comes to
publishing, at outputformatting you have quite a lot of
choices.I'm sorry, just wanted to keep the discussion alive,
since anyway, this isof much interest.--Hartmut
====
>
Lucas,> I'm certainly not competent to give you any answer,
and me too, I'd be eager> to hear such -- hoped 
Wolfram's to
react; anyway, here is my opinion.> PrecedenceForm is only
for parenthesizing at output.> Although we have
MakeExpressions to add semantic actions to the Mathematica>
compiler (if we may call such the transformation from input
string, i.e.> language, to the internal representation as
indicated by FullForm i.e.> code), we cannot inßuence the
Mathematica syntax proper, i.e. parsing.> The operator
precedence rules however are part of that (esp. needed for
box> formation). So I can't see a way to reach your goal. >
> Now, I don't know what your ordinary Plus shall designate.
If it's only> abstract, i.e. you don't use it 
for numeric
calculation, you possibly might> interchange the roles of
Plus and CirclePlus> In[6]:= a [CirclePlus] b + c
[CirclePlus] d> Out[6]= a[CirclePlus]b +
c[CirclePlus]d> As for BigCirclePlus (if you still need
it after the reassigned meanings)> I'd try to 
define a
palette, but perhaps this is not possible to the extend>
you desire. I never tried.> Another approach would be to
use Union and Intersection for low precedence> plus and
low precedence times (or [Subset]; [Superset] or
[And]; [Or]> or [Therefore];[Because] or ...).>
The ultima ratio is to write a new front end.> But, if I
were you, I'd ignore these kind of problems at 
first and set
up> the package fully functioning (with 1D input), apply it
to your problems and> investigations, and finally, when time
comes to publishing, at output> formatting you have quite a
lot of choices.> I'm sorry, just wanted to keep the
discussion alive, since anyway, this is> of much
interest.is actually quite simple -- just a list of
expressions. If I had anappropriate OutputForm written then
the effect would be:{expr1, expr2, ..., exprN} --> expr1
[CirclePlus] expr2 [CirclePlus]...So certainly I cn get
a functional package, but the final productshould be
notationally similar to the pure mathematical reference. I'm
sure this is a goal shared by many.If anyone is curious about
what exactly I'm up to, I'm attempting 
toimplement a Clocked
Objective Function package based (primarily) onthese
publications:A Lagrangian Formulation of Neural Networks I:
Theory and AnalogDynamicsA Lagrangian Formulation of
Neural Networks II: Clocked ObjectiveFunctions and
ApplicationsBoth are found in Neural, Parallel and
Scientific Computations 6(1998) 297-372, authors are Eric
Mjolsness and Willard L. Miranker-Lucas Scharenbroich-JPL /
MLS Group
====
 > I am writing a Java application that displays
Mathematica output using > MathCanvas. I am having difficulty
in that the internal frames > (JInternalFrame)in my
application are covered by the MathCanvas when > the
internal frames are moved into the MathCanvas area. I tried
> setting the frame's layer to 0, but that 
didn't work. Can
anyone help > me out?I suppose the problem is, that
MathCanvas is an AWT Component, whichcauses problem with
the other Swing JComponents;2 possible solutions: - get
J/Link 2.0, which comes with a Swing version of
MathCanvas(MathGraphicsJPanel if memory serves); - if you
can't do that, write a Swing replacement for
MathCanvasusing Swing; this is pretty simple: you can use
the evaluateToImagemethods, which return a byte array
containg a GIF Image; if you wantto make it very simple,
you just make your control a JPanel thatcontains a JLabel,
and when you want to set a M_-GraphicsExpression (from a
Plot,...) you just set the GIF Image as the JLabels
Icon;something like this (jlPlot being the
JLabel):StdLink.requestTransaction();^Mbyte [] bGifData =
kl.evaluateToImage(sExpression, width,height);if (bGifData
== null || bGifData.length <= 1){
System.out.println(bGifData empty);}Image iPlot =
Toolkit.getDefaultToolkit().createImage(bGifData);
iiPlot.setImage(iPlot);jlPlot.setIcon(iiPlot);
jlPlot.repaint();murphee
====
 > I am writing a Java
application that displays Mathematica output using >
MathCanvas. I am having difficulty in that the internal
frames > (JInternalFrame)in my application are covered by
the MathCanvas when > the internal frames are moved into the
MathCanvas area. I tried > setting the frame's layer to 0,
but that didn't work. Can anyone help > me out?I suppose
the problem is, that MathCanvas is an AWT Component,
whichcauses problem with the other Swing JComponents;2
possible solutions: - get J/Link 2.0, which comes with a
Swing version of MathCanvas(MathGraphicsJPanel if memory
serves); - if you can't do that, write a Swing replacement
for MathCanvasusing Swing; this is pretty simple: you can
use the evaluateToImagemethods, which return a byte array
containg a GIF Image; if you wantto make it very simple,
you just make your control a JPanel thatcontains a JLabel,
and when you want to set a M_-GraphicsExpression (from a
Plot,...) you just set the GIF Image as the JLabels
Icon;something like this (jlPlot being the
JLabel):StdLink.requestTransaction();^Mbyte [] bGifData =
kl.evaluateToImage(sExpression, width,height);if (bGifData
== null || bGifData.length <= 1){
System.out.println(bGifData empty);}Image iPlot =
Toolkit.getDefaultToolkit().createImage(bGifData);
iiPlot.setImage(iPlot);jlPlot.setIcon(iiPlot);
jlPlot.repaint();murphee
====
I am writing a Java
application that displays Mathematica output
usingMathCanvas. I am having difficulty in that the internal
frames(JInternalFrame)in my application are covered by the
MathCanvas whenthe internal frames are moved into the
MathCanvas area. I triedsetting the frame's layer to 0, but
that didn't work. Can anyone helpme out?
====
I wonder if
someone knows a fast way to import data from a tab-separated
file when numbers use coma (instead of point) as a decimal
separator.I know how to do this using:
Import[MyFile.txt,Table,ConversionOptions
->{NumberPoint->,}]However I got to noticed that Import is
far much time and memory consiming than ReadList[].Anyone as
an idea?TIA
====
> I am having a strange problem with a
function giving a complex number as > a result. I did the
following:> - define a function:> denom[x_, p_, d_] := Sqrt[1
+ (x*Tan[p]/d)^2]> Integrate and simplify it with the
assumption that d is larger than 0:>
FullSimplify[Integrate[denom[x, p, d], {x, 0, d}], d > 0]>
The result of the above line is >
d/2*(1+i*Sqrt[2]*d*Cos[p]^2)*Sqrt[Sec[p]^2] where i is
Sqrt[-1]> I assing it to a function called peter in the
following way:> peter[p_, d_] :=
FullSimplify[Integrate[denom[x, p, d], {x, 0, d}], d>0]>
and check the value of the function at [0,1]> peter[0,1]>
and the result is 1.> How is it possible that the result
doesn't have an imaginary part???> I would expect the result
to be 0.5+Sqrt[2]/2*i> You are evaluating the integral each
time that you call peter. Whenyou call it with p=0 denom is
evaluated to 1 before the integration.Define peter with
Evaluate or else don't use a delayed
set.denom[x_,p_,d_]:=Sqrt[1+(x*Tan[p]/d)^2];peter[p_,d_]:=
Evaluate[FullSimplify[Integrate[denom[x,p,d],{x,0,d}],d>0]];
peter2[p_,d_]:=FullSimplify[Integrate[denom[x,p,d],{x,0,d}],d
>0];peter[0,1](1/2)*(1 +
I*Sqrt[2])peter2[0,1]1peter2[10^-6,1]//N0.5 +
0.707107*Ipeter2[-10^-6,1]//N0.5 + 0.707107*IBob
HanlonChantilly, VA USA
====
(1) To discourage alteration of
the program file, save it as a binary file.Look up 
DumpSave in
Help.(2) To discourage alteration of symbols within a
Mathematica session, usethe Locked attributed.(3) To finally
answer your question, use the ReadProtected attribute
onsymbols you would like viewers not to see.Hope this
helps,Tom
====
> I am having a strange problem with a
function giving a complex number as> a result...You should
perhaps examine the integral of denom. Do you really expect
acomplex result? The integral involves multibranched
functions (a fact maskeda bit by the form Tan[p] in the
integrand; try replacing Tan[p] by p to seewhat is going
on). Is your result on the branch you want? I suspect not;
youare probably after a real result. The following
modification produces a realresult: Integrate[denom[x, p,
d], {x, 0, d}, PrincipalValue -> True,GenerateConditions ->
False]But you'll need to take the limit p->0.Hope this
helps,Tom Burton
====
The integral doesn't converge when b
is real, because of the secondorder singularity at Abs[b]
in that case. Arg[b^2]!=0 is used toexpress this because
the issue is whether that singularity is on thepositive
x-axis.Behavior at 0 and Infinity are fine if 
Re[a]>0, but
otherwise those areproblems too, so Mathematica has the
right answer in those terms. Ican't vouch for the formula
it comes up with when the conditions aremet, but it's not
closed-form in the usual sense anyway; but it'sclosed-form
in terms of functions MATHEMATICA is comfortable with!Bobby
Treat-----Original Message----- (Pi -
2*SinIntegral[a*Sqrt[-(1/b^2)]])), Integrate[1/(E^(a/x)*(-b^2
+ x^2)), {x, 0, Infinity}]]If b is Real then Arg[b^2]==0 and
Mathematica doesn't solve it. 
Let'sdefinea function that
allows us to test specific values of a and b.f[a_, b_][x_] =
Exp[-a/x]/(x^2 - b^2);Integrate[f[2, 3][x], {x, 0,
Infinity}]Integrate::idiv : Integral of 1/(E^(2/x)*(-9 +
x^2)) does notconverge on{0, Infinity}.But if we use an
imaginary value for b...Integrate[f[2, 3I][x], {x, 0,
Infinity}]%//N(1/6)*(2*CosIntegral[2/3]*Sin[2/3] +
Cos[2/3]*(Pi - 2*SinIntegral[2/3]))0.254022David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/
MasterÇs Degree of Applied Physics - Unesp/Rio Claro - 
State
of S.8boPaulo - BrazilLaboratory of Electrical
Measurementsphone : ( 0XX19 ) 526 -
2237*********************************************************
************
====
> I solved the epidemic SIR ODE System>
(<http://library.wolfram.com/webMathematica/MSP/Explore/
Biology/Epidemic>> ) numerically using NDSolve:>
approxsolutions=NDSolve[{> sÇ[t]==-a s[t] i[t],> 
iÇ[t]==a
s[t] i[t] - b i[t],> rÇ[t]==b i[t],> s[0]==700, i[0]==1,
r[0]==0}, {i[t], s[t], r[t]}, {t,0,20}];> and this for fixed
values of a and b.> Now I want to fit the solution for
variable a and b to a list of given> points (fitting using
the least squares method and the mathematica fonction>
findminimum). My questions are> 1- Is it possible to solve
the ODE (with mathematica) for not fixed values> of a and b
(so to get a parametric solution that depends on a and b)?>
2- Is it possible in mathematica to use the fonction
findminimum with the> (not explicit) solutions of the ODE, if
not are there some other> alternatives?> In advance, thank
you very much> MajidThere was a discussion on this topic
three years ago. The URL below willtake you to a reply in
that thread that includes remarks and code fromCarl Woll
and myself. I'll also show some code for your
specificexample, in order to indicate potential
problems.http://library.wolfram.com/mathgroup/archive/1999/
Oct/msg00104.htmlBelow is some mildly clumsy code tailored
to your example. It will dothe DE solving given parameter
values, form an objective function(sum-of-squares type)
given actual data and solutions for specificparameters, and
to do the minimization. I did everything at
machineprecision but this is not necessarily a good idea.
Also I do no errorchecking e.g. for cases where the numeric
DE solver fails to solve overthe full range. A related issue
is that, depending on roughly where thecorrect parameter
values live, and how close you are in your initialguess,
this might not be at all a well-behaved
problem.machsoln[a_?NumericQ,b_?NumericQ] := First[NDSolve[{
D[ss[t],t]==-a*ss[t]*ii[t], D[ii[t],t]==a*ss[t]*ii[t]-b*ii[t],
D[rr[t],t]==b*ii[t], ss[0]==700, ii[0]==1, rr[0]==0}, {ii[t],
ss[t], rr[t]}, {t,0,20}]];machsol = machsoln[1/100,1];We'll
generate some data for parameter values a=1/100, b=1.machdata
= Table[{t,ii[t],ss[t],rr[t]} /. machsol, {t,0,20,5}]For the
objective function, given numeric values for {a,b}, we'll
solvethe system, evaluate at the same 't' values 
as are
used for our data,subtract resulting vectors from
corresponding data vectors, take normsquares, and
sum.objfunc[data_,{a_?NumericQ,b_?NumericQ}] := Module[
{iii, sss, rrr, newdata, vals}, {iii,sss,rrr} =
{ii[t],ss[t],rr[t]} /. machsoln[a,b]; newdata =
Map[{iii,sss,rrr} /. t-># &, Map[First,data]]; vals =
Map[#.#&, Map[Rest,data]-newdata]; Apply[Plus, vals] ]For
example, we will get 0 if we use {a,b} = {.01,1}, and
somethingnonzero if we increase both modestly, say by
10%.In[85]:= objfunc[machdata, {.01,1.}]Out[85]=
0.In[86]:= objfunc[machdata, {.011,1.1}]Out[86]=
72.5289With the above objective function we can call
FindMinimum. For this wewill use two starting values for
each parameter so that FIndMinimum canuse a secant method.
An alternative to this, wherein one passes along agradient
black box evaluator, is presented by Carl Woll in his note
atthe URL above.findParams[data_, inits:{a1_,b1_}] :=
Module[ {vals,a,b}, FindMinimum[objfunc[data,{a,b}],
{a,a1,a1+.1*a1}, {b,b1,b1+.1*b1}] ]For example, if I give
{a,b} = {.011,1.1} as initial values, I get aresult
reasonably fast and quite accurate, albeit with many
warningsindicated. They basically indicate In[87]:=
findParams[data, {.011,1.1}]InterpolatingFunction::dmval:
Input value {20} lies outside the range of data in the
interpolating function. Extrapolation will be
used.InterpolatingFunction::dmval: Input value {20} lies
outside the range of data in the interpolating function.
Extrapolation will be used.InterpolatingFunction::dmval:
Input value {20} lies outside the range of data in the
interpolating function. Extrapolation will be
used.General::stop: Further output of
InterpolatingFunction::dmval will be suppressed during this
calculation. -11Out[87]= {8.46495 10 , {a$4965 -> 0.01,
b$4965 -> 1.}}It still does reasonably well when I perturb
by 50% from the correctparameter values. This time the
warning message indicates that we mightbe closer to
trouble.In[89]:= findParams[data,
{.015,1.5}]FindMinimum::fmcv: Failed to converge to the
requested accuracy or precision within 30 iterations.
-10Out[89]= {6.48159 10 , {a$8747 -> 0.01, b$8747 ->
1.}}You might want to play with various options in
FindMinimum such asMaxIterations. You might also need to
use NDSolve options e.g.WorkingPrecision in order to improve
the quality of the result.Daniel LichtblauWolfram
Research
====
I solved the epidemic SIR ODE
System(<http://library.wolfram.com/webMathematica/MSP/
Explore/Biology/Epidemic>) numerically using
NDSolve:approxsolutions=NDSolve[{sÇ[t]==-a s[t]
i[t],iÇ[t]==a s[t] i[t] - b i[t],rÇ[t]==b 
i[t],s[0]==700,
i[0]==1, r[0]==0}, {i[t], s[t], r[t]}, {t,0,20}];and this
for fixed values of a and b.Now I want to fit the 
solution for
variable a and b to a list of givenpoints (fitting using the
least squares method and the mathematica
fonctionfindminimum). My questions are1- Is it possible to
solve the ODE (with mathematica) for not fixed valuesof a
and b (so to get a parametric solution that depends on a and
b)?2- Is it possible in mathematica to use the fonction
findminimum with the(not explicit) solutions of the ODE, if
not are there some otheralternatives?In advance, thank you
very muchMajidReply-To: kjm@KevinMcCann.com
====
Majid,I
separately sent you a nb that addresses the numerical curve
fit of an NDSolve result. I have had this and have used the
technique for a long time; the example, however, did not
originate with me. Apologies to the original author - lost
in the mists of mind and time.Kevin> I solved the
epidemic SIR ODE System>
(<http://library.wolfram.com/webMathematica/MSP/Explore/
Biology/Epidemic>> ) numerically using NDSolve:>
approxsolutions=NDSolve[{> sÇ[t]==-a s[t] i[t],> 
iÇ[t]==a
s[t] i[t] - b i[t],> rÇ[t]==b i[t],> s[0]==700, i[0]==1,
r[0]==0}, {i[t], s[t], r[t]}, {t,0,20}];> and this for fixed
values of a and b.> Now I want to fit the solution for
variable a and b to a list of given> points (fitting using
the least squares method and the mathematica> fonction
findminimum). My questions are> 1- Is it possible to solve
the ODE (with mathematica) for not fixed values> of a and b
(so to get a parametric solution that depends on a and b)?>
2- Is it possible in mathematica to use the fonction
findminimum with the> (not explicit) solutions of the ODE, if
not are there some other> alternatives?> In advance, thank
you very much> Majid
====
Let say you have your data in an
array like so: data = {{5, {685, 12, 4}}, {10, {550, 105,
50}}, {15, {207, 250, 244}}, {20, {69, 169, 463}}}where 5,
10, 15, 20 are the observation times. (These numbers are
perturbed slightly from an actual solution.) The following
defines the least squares error as a function of a and
b:LSE[a_, b_] := Module[ {diff, sir = {s[t],i[t],r[t]} /.
NDSolve[{s'[t]== -a*s[t]*i[t], i'[t]== 
a*s[t]*i[t] - b*i[t],
r'[t]== b*i[t], s[0]== 700, i[0]== 1, r[0]== 0},
{s[t],i[t],r[t]}, {t, 0, 20}]//First },
diff:=(sir/.t->data[[i,1]]) - data[[i,2]]; Sum[diff.diff, {i,
Length[data]}] ]Now...In: FindMinimum[ LSE[a, b], {a, 0,
.1}, {b, 0, 1}]//TimingOut: {5.7 sec, {14.1967, {a ->
0.000999599, b -> 0.200351}}}---Selwyn Hollis> I solved
the epidemic SIR ODE System>
(<http://library.wolfram.com/webMathematica/MSP/Explore/
Biology/Epidemic>> ) numerically using NDSolve:>
approxsolutions=NDSolve[{> sÇ[t]==-a s[t] i[t],> 
iÇ[t]==a
s[t] i[t] - b i[t],> rÇ[t]==b i[t],> s[0]==700, i[0]==1,
r[0]==0}, {i[t], s[t], r[t]}, {t,0,20}];> and this for fixed
values of a and b.> Now I want to fit the solution for
variable a and b to a list of given> points (fitting using
the least squares method and the mathematica fonction>
findminimum). My questions are> 1- Is it possible to solve
the ODE (with mathematica) for not fixed values> of a and b
(so to get a parametric solution that depends on a and b)?>
2- Is it possible in mathematica to use the fonction
findminimum with the> (not explicit) solutions of the ODE, if
not are there some other> alternatives?> In advance, thank
you very much> Majid> 
====
In order to do some
transformations on a tree I need to be able to replacean
expression with head hdA if and if only its parent has head
hdB and itsgrandparent had head hdC. Furthermore, the item
itself and its parent mayhave any number sibling
elements.What I do now is the following. Give the
expression:ttexpr = grandparent[ parent1[grandchild2[],
grandchild1[], grandchild4[], grandchild1[]], grandchild1[],
parent2[grandchild2[], grandchild1[], grandchild4[],
grandchild1[]]];I apply a rule:ttexpr /. {
grandparent[left1___, parent1[left2___, grandchild1[],
right2___],right1___] -> grandparent[left1, parent1[left2,
MATCHED[], right2], right1]}which gives the desired
expression:grandparent[parent1[grandchild2[], MATCHED[],
grandchild4[], grandchild1[]], grandchild1[],
parent2[grandchild2[], grandchild1[], grandchild4[],
grandchild1[]]]But I have the feeling that it should be
possible to do this more elegantly.Does anybody have an
idea in this respect? Sidney Cadot
====
Perhaps you could use
ReadList with the option Word, which will read yourdata in
as strings. Then you may substitute commas with periods and
convertto numbers with ToExpression?Tomas GarzaMexico
City----- Original Message -----> 2,234 2,567 2,89> 3,234
3,567 3,89> I know how to do this using Import:>>
Import[MyFile.txt, Table,> ConversionOptions ->
{NumberPoint -> ,}] ;>> However, Import is not the function
I'd like to use, because it is realy> slow over large 
files.>>
I wonder if someone knows a way to use something faster than
Import (I> namely think of ReadList) with
coma-numbers?>
====
Yes, Import is slow, I suppose because
of it's generality. Your particularcase seems to be close
to one that ReadList can handle efficiently.Therefore I would
recommend the following sequence of steps:(1) Use ReadList to
read into Words instead of Numbers.(2) Use StringReplace to
replace , with . within words.(3) Convert the words to
expressions with ToExpression.This should be easy provided
that you do not encounter scientific notation(1,3e6).Tom
Burton
====
> If I want to protect the Mathematica Program,
what can I do? > Is there any method to avoid other people's
reading my program but> it is still able to run ßuently?
Encode does what you're after
....Dave.
====
====================================== Dr.
David Annetts EM Modelling Analyst Australia
David.Annetts@csiro.au
====
==================================

====
=====I try to solve a three equation linear ODE system
with constant When I try the same thing on Mathematica 4.0
(Solaris) I get a very strange result containing things like
E^x#1 OR #1^2.Am I doing something wrong, or is there a
bug/problem with Mathematica.Neil-- Solving a Three
Equation ODE system (Linear with Constant Coefficients) with
Mathematica 3.0 ----INPUT:DSolve[{y1'[x] + c11*y1[x] +
c12*y2[x] +c13*y3[x] +c1==0,
y2'[x]+c21*y1[x]+c22*y2[x]+c23*y3[x]+c2==0,
y3'[x]+c31*y1[x]+c32*y2[x]+c33*y3[x]+c3==0, y1[0]==y1i,
y2[0]==y2i, y3[0]==y3i}, {y1[x], y2[x], y3[x]},
x]OUTPUT:Eigensystem::eivec: Unable to find eigenvector
for eigenvalue !(Root[((((c13 c22
c31) - (c12 c23 c31) - (c13 c21
c32) + (c11 c23 c32) + ([LeftSkeleton]
20 [RightSkeleton])) &),
1)]).Eigensystem::eivec: Unable to find eigenvector
for eigenvalue !(Root[((((c13 c22
c31) - (c12 c23 c31) - (c13 c21
c32) + (c11 c23 c32) + ([LeftSkeleton]
20 [RightSkeleton])) &),
2)]).Eigensystem::eivec: Unable to find eigenvector
for eigenvalue !(Root[((((c13 c22
c31) - (c12 c23 c31) - (c13 c21
c32) + (c11 c23 c32) + ([LeftSkeleton]
20 [RightSkeleton])) &),
3)]).General::stop: Further output of
!(Eigensystem :: eivec) will be suppressed
during this calculation.--
______________________________________________________Neil
E. Klepeis, UC Berkeley, School of Public Health,and
Lawrence Berkeley National Laboratory,
====
Neil,They are not
a general analitical solution for a system of ODE with 3or
more variables where all coffs. are parameters. I have a
packagefor solving SODE. Perhaps you can find it
useful.Guillermo Sanchez> I try to solve a three equation
linear ODE system with constant > When I try the same thing
on Mathematica 4.0 (Solaris) I get a very > strange result
containing things like E^x#1 OR #1^2.> Am I doing
something wrong, or is there a bug/problem with Mathematica.>
> Neil> -- Solving a Three Equation ODE system (Linear
with Constant > Coefficients) with Mathematica 3.0 ---->
INPUT:> DSolve[{y1'[x] + c11*y1[x] + c12*y2[x] +c13*y3[x]
+c1==0,> y2'[x]+c21*y1[x]+c22*y2[x]+c23*y3[x]+c2==0,>
y3'[x]+c31*y1[x]+c32*y2[x]+c33*y3[x]+c3==0,> y1[0]==y1i,
y2[0]==y2i, y3[0]==y3i},> {y1[x], y2[x], y3[x]}, x]>
OUTPUT:> Eigensystem::eivec:> Unable to find
eigenvector for eigenvalue !(Root[((((c13 >
c22 > c31) - (c12 c23 c31) - (c13
c21 c32) + (c11 c23 c32) + >
([LeftSkeleton] 20 [RightSkeleton])) &),
1)]).> Eigensystem::eivec:> Unable to find
eigenvector for eigenvalue !(Root[((((c13 >
c22 > c31) - (c12 c23 c31) - (c13
c21 c32) + (c11 c23 c32) + >
([LeftSkeleton] 20 [RightSkeleton])) &),
2)]).> Eigensystem::eivec:> Unable to find
eigenvector for eigenvalue !(Root[((((c13 >
c22 > c31) - (c12 c23 c31) - (c13
c21 c32) + (c11 c23 c32) + >
([LeftSkeleton] 20 [RightSkeleton])) &),
3)]).> General::stop:> Further output of
!(Eigensystem :: eivec) will be suppressed >
during this calculation.Reply-To:
nklepeis@uclink4.berkeley.edu
====
Guillermo,indoor air
exposure to particulate matter. Does your package use
Runge-Kutta or some other numerical algorithm? I'm thinking
of something like that, but with a fixed (non-adaptive) step
size.Neil> Neil,> They are not a general analitical
solution for a system of ODE with 3> or more variables where
all coffs. are parameters. I have a package> for solving
SODE. Perhaps you can find it useful.> Guillermo Sanchez>
>>I try to solve a three equation linear ODE system with
constant >>When I try the same thing on Mathematica 4.0
(Solaris) I get a very >>strange result containing things
like E^x#1 OR #1^2.>>Am I doing something wrong, or is
there a bug/problem with Mathematica.>>Neil>>--
Solving a Three Equation ODE system (Linear with Constant
>>Coefficients) with Mathematica 3.0
---->>INPUT:>>DSolve[{y1'[x] + c11*y1[x] + c12*y2[x]
+c13*y3[x] +c1==0,>>
y2'[x]+c21*y1[x]+c22*y2[x]+c23*y3[x]+c2==0,>>
y3'[x]+c31*y1[x]+c32*y2[x]+c33*y3[x]+c3==0,>> y1[0]==y1i,
y2[0]==y2i, y3[0]==y3i},>> {y1[x], y2[x], y3[x]},
x]>>OUTPUT:>>Eigensystem::eivec:>> Unable to find
eigenvector for eigenvalue !(Root[((((c13
>>c22 >>c31) - (c12 c23 c31) - (c13
c21 c32) + (c11 c23 c32) +
>>([LeftSkeleton] 20 [RightSkeleton])) &),
1)]).>>Eigensystem::eivec:>> Unable to find
eigenvector for eigenvalue !(Root[((((c13
>>c22 >>c31) - (c12 c23 c31) - (c13
c21 c32) + (c11 c23 c32) +
>>([LeftSkeleton] 20 [RightSkeleton])) &),
2)]).>>Eigensystem::eivec:>> Unable to find
eigenvector for eigenvalue !(Root[((((c13
>>c22 >>c31) - (c12 c23 c31) - (c13
c21 c32) + (c11 c23 c32) +
>>([LeftSkeleton] 20 [RightSkeleton])) &),
3)]).>>General::stop:>> Further output of
!(Eigensystem :: eivec) will be suppressed
>>during this calculation.> --
______________________________________________________Neil
E. Klepeis, UC Berkeley, School of Public Health,and
Lawrence Berkeley National Laboratory,Berkeley, CA USA.
Voice: 831-768-9510
====
To the
Group,In[104]:=LaplaceTransform[D[f[t],t],t,s]Out[104]=-f[
0] + s
LaplaceTransform[f[t],t,s]In[105]:=LaplaceTransform[
Integrate[f[t],t],t,s]Out[105]=LaplaceTransform[Integrate[f
[t] ,t], t, s]Item 104 is correct, but item 105 is not. The
correct 105 should be:-f[0]/s + LaplaceTransform[f[t],t,s]/s
.If Mathematica can get 104 right, then it should also
produce a correct105. This looks like a bug. Any thoughts on
this?(Note: Number 105 does not look like that on the notebook
page. It showsthe indefinite integral with an integral symbol
followed by f[t] dtinstead of the function Integrate[ ].
I've spelled it out because theVista Research, Inc.
====
I
would like to match a named pattern in an expression and then
square theresult. But my attempt fails.Clear[f, g, x, y,
a]expr = 3*f[x]*g[y] + 2 + x^2;expr /. a:(f[_]*g[_]) :>
a^22 + x^2 + 3*f[x]*g[y]If I drop the name on the pattern,
it matches - but it doesn't do what Iwant.expr /.
f[_]*g[_] :> a^22 + 3*a^2 + x^2How can I name such a pattern
and use it on the rhs of a rule?David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/===
=> I would like to match a named pattern in an expression and
then square the> result. But my attempt fails.> [...] >>
David Park> djmp@earthlink.net> I simplified your example
to see whats the problem. I only take the multiplication of
three symbols a, b, c.In[1]:= Clear[a, b, c, p, x, y]If you
use the unnamed pattern 'a c' in Replace you 
getIn[2]:= a
b c /. a c :> xOut[2]= b xthe replacement will be done
because all possibilites of the multiplication are tested
and the right one will be found.But if you take a named
pattern 'p : a c' your are searching for a 
multiplication of
exactly two variables called 'a' and 
'c' and you are not
successful with the replacement:In[3]:= a b c /. p : a c :>
xOut[3]= a b cIn this case you have to take into account
that the multiplication can consist of more than two terms,
e. g.In[4]:= a b c /. p : a c z___ :> x zOut[4]= b xÂ
Hope that will help you,-- Rainer Gruber
====
I want to get
the coefficient of the Legendre-Gauss Quadrature, I 
find this
method is fast, but when the integration result is big, then
the result isnot accurate enough, I think it is because I
only use 5 coefficient to compute it.Do you think it will be
accurater when I use more coefficients to compute?Is possible
to get 16 coefficients to compute?The 6 
coefficients I use are
from:http://mathworld.wolfram.com/
Legendre-GaussQuadrature.htmlI need more coefficients to get
accurater result. How about the method of the Integration
function built-in Mathematica?Chen
====
The package
NumericalMath`GaussianQuadrature` has what you
want.---Selwyn Hollis> I want to get the coefficient of the
Legendre-Gauss Quadrature, > I find this method is fast, but
when the integration result is big, then the result is> not
accurate enough, I think it is because I only use 5 
coefficient
to compute it.> Do you think it will be accurater when I use
more coefficients to compute?> Is possible to get 16
coefficients to compute?> The 6 coefficients I use 
are from:>
http://mathworld.wolfram.com/Legendre-GaussQuadrature.html>
> I need more coefficients to get accurater result. > How
about the method of the Integration function built-in
Mathematica?> Chen> 
====
The presence of 3* in your
formula prevents the match pattern you want form taking
place
since:In[11]:=FullForm[3*f[x]*g[y]]Out[11]//FullForm=
Times[3,f[x],g[y]]use
instead:In[12]:=expr=3*f[x]*g[y]+2+x^2;In[13]:=expr /.
a:((p_.)*f[_]*g[_]) :> p*a^2Out[13]=2 + x^2 +
27*f[x]^2*g[y]^2Andrzej KozlowskiToyama International
UniversityJAPAN> I would like to match a named pattern in
an expression and then square > the> result. But my attempt
fails.>> Clear[f, g, x, y, a]> expr = 3*f[x]*g[y] + 2 +
x^2;>> expr /. a:(f[_]*g[_]) :> a^2> 2 + x^2 + 3*f[x]*g[y]>>
If I drop the name on the pattern, it matches - but it
doesn't do what > I> want.>> expr /. f[_]*g[_] :> a^2> 2 +
3*a^2 + x^2>> How can I name such a pattern and use it on the
rhs of a rule?>> David Park> djmp@earthlink.net>
http://home.earthlink.net/~djmp/>>
====
I need FrameLabels
of the kind of H^s_z / H^p_Z and tried
somethinglikeFrameLabel->{FontForm{HSuperscript[s]
Subscript[z] ...}}but s is too far away from z. How
can I manage it that s isdirectly above z ? Are there
some backspace-characters to be
used?Harry
====
TryPower[Subscript[H, z], s]>> I need
FrameLabels of the kind of H^s_z / H^p_Z and tried
something> like>>
FrameLabel->{FontForm{HSuperscript[s]Subscript[z]
...}}>> but s is too far away from z. How can I manage
it that s is> directly above z ? Are there some
backspace-characters to be used?>>
Harry>>
====
TryPower[Subscript[H, z], s]>> I need
FrameLabels of the kind of H^s_z / H^p_Z and tried
something> like>>
FrameLabel->{FontForm{HSuperscript[s]Subscript[z]
...}}>> but s is too far away from z. How can I manage
it that s is> directly above z ? Are there some
backspace-characters to be used?>>
Harry>>
====
>-----Original Message----->Sent: Thursday,
August 29, 2002 7:38 AM>I would like to match a named
pattern in an expression and >then square the>result. But my
attempt fails.>>Clear[f, g, x, y, a]>expr = 3*f[x]*g[y] + 2 +
x^2;>>expr /. a:(f[_]*g[_]) :> a^2>2 + x^2 + 3*f[x]*g[y]>>If
I drop the name on the pattern, it matches - but it doesn't
>do what I>want.>>expr /. f[_]*g[_] :> a^2>2 + 3*a^2 +
x^2>>How can I name such a pattern and use it on the rhs of a
rule?>>David
Park>djmp@earthlink.net>http://home.earthlink.net/~djmp/>>
Dear David,just don't insist on a single name!
In[17]:=expr /. (a : f[_])*(b : g[_]) :> (a*b)^2Out[17]=2
+ x^2 + 3*f[x]^2*g[y]^2In[29]:=3*ff[z]*f[x]*g[y] + 2 + x^2
/. (a : f[_])*(b : g[_]) :> (a*b)^2Out[29]=2 + x^2 +
3*f[x]^2*ff[z]*g[y]^2We might tend to understand this
behaviour of the pattern matcher. As Timeshas the Flat,
Orderless attributes, the components of the pattern have to
betaken apart to match separated subexpressions at the
lhs, what should thepattern variable then designate in the
course of this procedure?Look at the FullFormIn[31]:=(a :
f[_])*(b : g[_]) //
FullFormOut[31]//FullForm=Times[Pattern[a, f[Blank[]]],
Pattern[b, g[Blank[]]]]compared to In[12]:=a : f[_]*g[_]
// FullFormOut[12]//FullForm=Pattern[a, Times[f[Blank[]],
g[Blank[]]]]Depending on your real problem...In[43]:=expr
/. a_?NumericQ *b_ :> a*Times[b]^2Out[43]=2 + x^2 +
3*f[x]^2*g[y]^2...might be a more elegant (but risky)
solution (?), or perhaps else (more robust if you know the
names f,g)...In[79]:=expr /. a:(f | g)[___] :>
a^2Out[79]=2 + x^2 + 3*f[x]^2*g[y]^2In[139]:=expr /. a :
h_[___] /; MemberQ[{f, g}, h] :> a^2Out[139]=2 + x^2 +
3*f[x]^2*g[y]^2Perhaps a fine way would beIn[74]:=2 + x^2 +
3 f[x]*g[y] /. HoldPattern[Times[a:(_[___]..)]] :>
Times[a]^2Out[74]=2 + x^2 + 3*f[x]^2*g[y]^2but of course
this only works if you have at least two factors f[] and
g[](and no mixed powers of x and y!) In that case come back
to something likeIn[103]:=2 + x^2*y^2 + 3*x*y^3*f[x]*g[y] +
f[x] /. a : (_[___]) :> a^2 /; FreeQ[a, Power |
Times]Out[103]=2 + x^2*y^2 + f[x]^2 +
3*x*y^3*f[x]^2*g[y]^2--Hartmut
====
version4.1.5.0>> I try
to do the simple task of transposing a matrix.>> X =
{{a,b},{c,d},{e,f}}> whereas Transpose[{{a,b},{c,d},{e,f}}]
works well.>> What is wrong with writing Transpose[X] ?>>
Terje Johnsen>>
====
Andrzej,That doesn't work because I
don't want to square the 3. Of course, I coulddivide by p
but not in my actual example. In my actual example I am
notsquaring but doing a MetricSimplify tensor operation and
the whole purposeis to operate on only two of four factors.
I can do it by giving a specificpattern but not by using a
general pattern.Here is a better example.expr =
f[a]g[b]h[c];This works...expr /. g[b]h[c] :>
op[g[a]h[c]]f[a] op[g[a] h[c]]This even works with a more
general pattern but is a silly way to do it...expr /.
g[_]h[_] :> op[g[a]h[c]]f[a] op[g[a] h[c]]This is what I
want to do, but again it doesn't work:expr /. p_. (sub :
g[_]h[_]) :> p op[sub]f[a] g[b] h[c]I have to be able to
obtain a name for the match to the g[_]h[_] pattern.David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/In
[13]:=expr /. a:((p_.)*f[_]*g[_]) :> p*a^2Out[13]=2 + x^2 +
27*f[x]^2*g[y]^2Andrzej KozlowskiToyama International
UniversityJAPAN> I would like to match a named pattern in
an expression and then square> the> result. But my attempt
fails.>> Clear[f, g, x, y, a]> expr = 3*f[x]*g[y] + 2 +
x^2;>> expr /. a:(f[_]*g[_]) :> a^2> 2 + x^2 + 3*f[x]*g[y]>>
If I drop the name on the pattern, it matches - but it
doesn't do what> I> want.>> expr /. f[_]*g[_] :> a^2> 2 +
3*a^2 + x^2>> How can I name such a pattern and use it on the
rhs of a rule?>> David Park> djmp@earthlink.net>
http://home.earthlink.net/~djmp/>>
====
David,> I would like
to match a named pattern in an expression and then square
the> result. But my attempt fails.> Clear[f, g, x, y, a]>
expr = 3*f[x]*g[y] + 2 + x^2;> expr /. a:(f[_]*g[_]) :> a^2>
2 + x^2 + 3*f[x]*g[y] ...Well, I see that expr /.
a:Times[f[_],g[_],h___]:> a^2works, and I think I understand
why your form above does not work (examineFullForm of
expr--your form does not account for 3), but I
don'tunderstand why eliminating a: from your form allows
the pattern to match.Tom Burton
====
I would expect
thatPlot[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 y value isless than 1 at several
points. Most notibly between x=-1.6 and x=-1.8Is this just
an isolated case? Or does it happen to others? If so -
why?Ken.
====
This results from:a) the fact that Mathematica
implements division as multiplication andreciprocationb)
the use of machine precision for plotting the graphc) the
default scaling of the y axis revealing the resultant
inaccuraciesAlthough Sin[x]/Sin[x] gives 1 when evaluated
symbolically (x undefined),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 small
discrepanciesare 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 y valueis> 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.>
====
I get very small dips (vees?) at the
same spots using 4.1.What is your experience with the
presentation feature of version 4.2 ?That is what is
attracting me to upgrade and I want to hear experiences.> 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 y
valueis> 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.>
====
I
eventually figured this one out:(1) change the default font to
smaller (10 pts vs 12). This solves theproblem, but makes
the help text illegibly small. This is cured by step 2:(2) go
into preferences and change the default zoom to larger (1.25
from 1).> The Windows, the Mathematica Help browser has a
categories section, where> the headings and subheadings of
each help file appear. In version 3.x,those> fonts (by
default) appear nice and small (smaller than the text on
the> menu). In 4.1, the font is so large I cannot read the
entire heading. How> can I revert to the proper display?>
Nicholas>
====
 I am fairly new to Mathematica and once
before asked a question andreceived somevery helpful
responses. This time the problem seems very odd and I may
bemaking a veryelementary error. Here is the context: 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 would beone 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 rhs givesonly 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 put
atthe start of listA. Butthe assignment of this value --
the extended list -- to be the new value oflistA 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 the
functionwith concrete values as aboveis: true, newListA
= {200, 121} Trace doesn't help because all I get back is 
the
function name withits concrete arguments. Any thoughts?
Tom
====
> In order to do some transformations on a tree I
need to be able to replace> an expression with head hdA if
and if only its parent has head hdB and its> grandparent
had head hdC. Furthermore, the item itself and its parent
may> have any number sibling elements.> What I do now is
the following. Give the expression:> ttexpr = grandparent[>
parent1[grandchild2[], grandchild1[], grandchild4[],
grandchild1[]],> grandchild1[],> parent2[grandchild2[],
grandchild1[], grandchild4[], grandchild1[]]];> I apply a
rule:> ttexpr /. {> grandparent[left1___, parent1[left2___,
grandchild1[], right2___],> right1___] ->> grandparent[left1,
parent1[left2, MATCHED[], right2], right1]}> which gives
the desired expression:> grandparent[parent1[grandchild2[],
MATCHED[], grandchild4[], grandchild1[]],> grandchild1[],>
parent2[grandchild2[], grandchild1[], grandchild4[],
grandchild1[]]]> But I have the feeling that it should be
possible to do this more elegantly.> Does anybody have an
idea in this respect?> Sidney CadotI think your
technique is fine for small to medium size trees. For
largeones it might be very slow due to all the work of
patten matching. Ifyou anticipate large inputs you might
thus want to code a simpletree-walk using old-fashioned
procedural code. Any time you match agrandparent head, put
that subtree onto a stack that enters the nextstate,
looking for parent nodes, etc.Daniel LichtblauWolfram
Research
====
with the output your own solution is giving
you. I would have thought not, since only one grandchild1[]
with parent parent1 and grandparent grandparent. is
being matched. You can get a complete match by using
ReplaceRepeated instead of ReplaceAll:In[33]:=ttexpr//
.{grandparent[left1___,parent1[left2___,grandchild1[],right2_
__], right1___]->
grandparent[left1,parent1[left2,MATCHED[],right2],right1]}
Out[33]=grandparent[parent1[grandchild2[],MATCHED[],
grandchild4[],MATCHED[]], grandchild1[],
parent2[grandchild2[],grandchild1[],grandchild4[],grandchild1
[]]]exactly the same result can be achieved in a rather
different way, which may perhaps be seen as more
elegant.In[34]:=ttexpr /. expr_grandparent :> (expr /.
expr1_parent1 :> (expr1 /. expr2_grandchild1 :>
Matched[]))Out[34]=grandparent[parent1[grandchild2[],
Matched[], grandchild4[], Matched[]], grandchild1[],
parent2[grandchild2[], grandchild1[], grandchild4[],
grandchild1[]]]Andrzej KozlowskiToyama International
UniversityJAPAN>> In order to do some transformations on a
tree I need to be able to > replace> an expression with head
hdA if and if only its parent has head hdB and > its>
grandparent had head hdC. Furthermore, the item itself and
its parent > may> have any number sibling elements.>> What I
do now is the following. Give the expression:>> ttexpr =
grandparent[> parent1[grandchild2[], grandchild1[],
grandchild4[], > grandchild1[]],> grandchild1[],>
parent2[grandchild2[], grandchild1[], grandchild4[], >
grandchild1[]]];>> I apply a rule:>> ttexpr /. {>
grandparent[left1___, parent1[left2___, grandchild1[],
right2___],> right1___] ->> grandparent[left1, parent1[left2,
MATCHED[], right2], right1]}>> which gives the desired
expression:>> grandparent[parent1[grandchild2[], MATCHED[],
grandchild4[], > grandchild1[]],> grandchild1[],>
parent2[grandchild2[], grandchild1[], grandchild4[],
grandchild1[]]]> But I have the feeling that it should be
possible to do this more > elegantly.> Does anybody have an
idea in this respect?> Sidney Cadot>>
====
>I'm finding
that the ImageSize option in Export has no effect
when>exporting Cell or Notebook objects. For instance, the
following two>commands produce precisely the same graphic:>>
Export[image1.jpg, Cell[Some cell contents, Text,
FontSize -> 100]]>> Export[image2.jpg, Cell[Some cell
contents, Text, FontSize ->>100], ImageSize -> {576,
288}]>>Has anyone encountered this problem before?>>(This is
with Mathematica 4.1.5 and Mac OS X.)>>---->Selwyn
Hollis>slhollis@mac.comExport takes two options that
control the size of a graphic, ImageSize and
ImageResolution. The difference is sometimes subtle.
ImageSize changes the coordinates while ImageResolution
changes the size of objects. This may sound like the same
thing, but they're actually different.If I have a plot with
a line. Changing ImageSize changes the location of the
points in the line, which may make the line shorter or
longer, but the thickness is unchanged (if it is an
AbsoluteThichness). ImageResolution draws the image with a
different number of pixels, so lengths and thicknesses are
changed. The same is true with fonts.Why does it matter?
Because there is no equivalent to ImageSize for cells or
notebooks. However, Magnification is a good equivalent of
ImageResolution. So ImageSize does nothing, but
ImageResolution does.In[1]:= cont=Cell[Some cell
contents, Text, FontSize -> 100]In[2]:=
Show@ImportString[ExportString[cont, JPEG],
JPEG]In[3]:= Show@ImportString[ExportString[cont,
JPEG, ImageResolution ->72*3/2], JPEG]-Dale
====
I was
wondering if there is a method for resizing Raster
graphics(resizing the actual matrix of pixels, not just the
display size). Iam processing a large number of JPEG
images, and we sometimes need toreduce the image size to
allow data processing algorithms to functionwithout running
out of memory. In the past we simply used a programsuch as
Photoshop to resize them before importing them
intoMathematica. Due to the number of images we are
processing now thisis very inconvenient and it would be
very useful if there was a methodfor accomplishing it in
Mathematica, but I can't find one. I alsothought 
about doing
something simple like sampling every few pixels oraveraging,
but I thought there might be a method with more efficacythan
this. I also tried exporting the graphics with the Export
commandas new JPEGs and manipulating the ImageResolution
and ImageSizeoptions but this seemed to have no effect. Any
help would be muchappreciated.Aaron UrbasReply-To:
kuska@informatik.uni-leipzig.de
====
take the matrix of gray
values or the matrix of the color channels,apply
ListInterpolation[] and generate a resampled table withthe
interpolated function. Jens> 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 Urbas
====
> 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 Urbas>Aaron, with the Digital Image
Processing package see Downsample, Decimate
orResize.Else, the simplest method is to take every k'th
sample, which can be donewith the Take command.This
imports the image and returns the raw data.img =
Import[somefile.jpg][[1,1]];For an image with dimensions
nr x nc, and if you want to take every othersample
use:small = Take[img, {1, nr, 2}, {1, nc, 2}];You can also
perform smoothing using ListConvolve prior to downsampling
toeliminate some of the visual artifacts of
downsampling.Done.--
====
===========================Mariusz
JankowskiUniversity of Southern
Mainemjkcc@usm.maine.edu207-780-5580
====
Yesterday I
posted a question on using named patterns in a rule. I
receiveda number of useful replies and thank all those who
responded. Today I have aquestion that actually generated
yesterday's question. I am using a newsubject heading to
reßect the actual nature of the question.What is the best
way in Mathematica to operate on some, but not all,
levelparts of an expression or subexpression?Suppose I
have the following expression...expr =
f1[a]f2[b]f3[c]f4[d];and I want to do an operation, op,
separately on f1[a]f3[c] and f2[b]f4[d].The operation must
be done on the given pairs and not on all four factors
atonce. One method is to use explicit exact substitution
rules.expr /. f1[a]f3[c] :> op[f1[a]f3[c]] /. f2[b]f4[d] :>
op[f2[b]f4[d]]op[f1[a] f3[c]] op[f2[b] f4[d]]If a,b,c,d
were long expressions, we might not want to type or copy them
in.This raised my question of using named patterns. Hartmut
Wolf pointed outthat each factor must be named to create a
match with ßat expressions. Sowe could use...expr /. (a :
f1[_])(b : f3[_])(c : f2[_])(d : f4[_]) :> op[a b]op[c
d]op[f1[a] f3[c]] op[f2[b] f4[d]]Andrzej Kozlowski
suggested a method using Take, but using Part works
betterhere, so we could use...expr /. a_ :> op[a[[{1,
3}]]]*op[a[[{2, 4}]]]op[f1[a] f3[c]] op[f2[b] f4[d]]All of
the above methods use rules. Is it possible to do it
withReplacePart? I don't think so, but maybe somebody knows
how to do it. Howabout using MapAt? I don't think that works
either in regular Mathematica.Ted Ersek and I did a package,
Algebra`ExpressionManipulation` at my website, that
implements extended positions. An extended position gives
theposition of an expression and a list of the desired
subparts and is packagedin a header eP. So the extended
position of a + c inf[a + b + c + d] is eP[{1},{1,3}]. The
package modifies MapAt to acceptextended positions. Then we
can use...MapAt[op, expr, {eP[{}, {1, 3}], eP[{}, {2,
4}]}]op[f1[a] f3[c]] op[f2[b] f4[d]]However, I don't always
like to drag in the package just to do that. I thinkthat
operating on selected level parts of a subexpression is not
all thatuncommon.Here is another example.1 - Cos[x]^2% //
TrigFactor1 - Cos[x]^2Sin[x]^2But...1 - a - Cos[x]^2% //
TrigFactor1 - a - Cos[x]^2(1/2)*(1 - 2*a - Cos[2*x])I was
hoping for -a + Sin[x]^2. What are the best methods for
handling thiskind of problem in regular Mathematica?David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/
David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/===
=and from kde2 to kde3 my Mathematica 3.0 doesn't work as
usual any more:All the characters have become little
rectangles,I can't read even the error message at the
beginning.I followed the wolfram support pageHow do I
resolve certain font-related error message whenvery
carefully, but the error remains.Markus Koecher
====
I look
for an operator, which has a higher precedence than @ and can
beoberloaded with a new definition.According to the table in
A.2, only PatternTest (?) could be a candidate.But according
to my tests, PartitionTest cannot be overloaded with an
newdefinition.What is your opinion?Hermann Schmitt
====
I
have been trying to get emmathfnt to work for the past few
hours butwhatever I do the output file is slightly smaller
than the input so Iguess it isn't working and emmathfnt
cannot find the font files. Ialso set FONTDIR.I am 
using the
dos commandemmathfnt -d
C:Progra~1Wolfra~1Mathem~14.2System~1Fonts
Type1-o c:output.eps c:input.epsand the following is
outputC:WINDOWS>dir c:*put.eps Volume in drive C has
no label Volume Serial Number is 0508-1ED9 Directory of
C:OUTPUT EPS 26,324 08-30-02 9:57a output.epsINPUT EPS
26,324 08-30-02 10:21a input.eps 2 file(s) 52,648 bytes 0
dir(s) 4,896.69 MB freeand if i run dir
C:Progra~1Wolfra~1Mathem~14.2System~1FontsType1
I see the font files OK.input.eps was produced using the
following Mathematica gr = Plot[Sin[[Alpha]], {[Alpha],
-[Pi], [Pi]}, AxesLabel -> {[Alpha],
Sin([Alpha])}]Export[c:input.eps, gr, EPS]both
input and output set to StandardFormI wonder what I am doing
wrong.C:WINDOWS>SETTMP=C:WINDOWSTEMPTEMP=C:
WINDOWSTEMPPROMPT=$p$gwinbootdir=C:WINDOWSCOMSPEC=C
:WINDOWSCOMMAND.COMPATH=C:DOWNLOADFONTZEMMATH~1;C
:UTILSGHOSTGHOSTGUMGSVIEW;C:TEXMFMIKTEXBIN;C
:WINDOWS;C:WINDOWSCOMMAND;C:MATLAB_SV12BINWIN32
FONTDIR=C:Program
FilesWolframResearchMathematica4.2SystemFiles
FontsType1windir=C:WINDOWSBLASTER=A220 I5 D3
T4CMDLINE=emmathfnt
-dC:Progra~1Wolfra~1Mathem~14.2System~1Fonts
Type1 -oc:output.eps c:input.epsthankyou
====
> 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 y value
is> 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?>4.1 for Mac OS X (November 5,
2001)p = Plot[Sin[x]/Sin[x],{x,-2.5,-1.5}];Note that the
Ticks on the y-axis are all 1, that is, in trying to
findsomethingof interest the adaptive range has zoomed in
and is looking at the machineprecision representations of
1.(First /@(Ticks /.
AbsoluteOptions[p])[[2]])//InputFormZoom out using the
PlotRangePlot[Sin[x]/Sin[x],{x,-2.5,-1.5},
PlotRange->{-0.1,2.1}];Or as stated in the on-line help for
Plot: You should use Evaluate to evaluatethe function to
be plotted if this can safely be done before specific
numericalvalues are
supplied.Plot[Evaluate[Sin[x]/Sin[x]],{x,-2.5,-1.5}];Bob
HanlonChantilly, VA USA
====
I am trying to implement the
inner product in the space ofcomplex-valued, square
integrable functions over [-1/2,1/2], which canbe expressed
in Mathematica code
asinner[f_Function,g_function]:=Integrate[Conjugate[f[x]]*g
[x],{x,-1/2,1/2}]This is simple enough. Problem is,
Mathematica seamingly cannotevaluate the Integral for even
the simplest of
functions:In[10]:=inner[#&,#&]Out[10]:=!([Integral]
_(-(1/2))%(1/2)(x Conjugate[ x])
[DifferentialD]x)As you see, the Integrate returns
unevaluated. It works fine if Iremove the Conjugate.
Unfortunately the Conjugate is needed forpositive
definiteness.Various variants with Composition, Re and Im
etc. don't work either.This should be a So how do I get
Integrate to work with Conjugate? Andreas -- True Pleasure
in this society is more dangerous than bank
robbery.
====
Andreas Dietrich schrieb:> I am trying to
implement the inner product in the space of> complex-valued,
square integrable functions over [-1/2,1/2], which can> be
expressed in Mathematica code as>
inner[f_Function,g_function]:=Integrate[Conjugate[f[x]]*g[x]
,{x,-1/2,1/2}]> This is simple enough. Problem is,
Mathematica seamingly cannot> evaluate the Integral for even
the simplest of functions:> In[10]:=inner[#&,#&]>
Out[10]:=!([Integral]_(-(1/2))%(1/2
)(x Conjugate[> x]) [DifferentialD]x)>
Andreas,Conjugate does not evaluate the expression, if the
variables aren'tknown to be real. Your example
(inner[#&,#&]) works if you define yourfunction using
ComplexExpand:inner[f_,g_]:=Integrate[ComplexExpand[
Conjugate[f[x]]]g[x], ...]You may also use a home-made
Conjugate, I'll call it Konjugiert,
eg:ruKonjugiert={Complex[re_,im_]:>Complex[re,-im]};
Konjugiert[ausdr__]:=ausdr /. ruKonjugiert;Replacing
Conjugate with Konjugiert in your Definition works for
yoursimple Example, but more komplex functions will need a
ComplexExpand toget the wanted real result.Gru¤ Peter--
=--=--=--=--=--=--=--=--=--=--=--=--=--=
http://home.t-online.de/home/phbrf
====
Steve,here...). I
am sending it pretty much in the same way as last time,with
cc to mathgroup@smc.vnet.net (only this time with the intent
thatit not appear
there).Daniel------------------------------------------->
> NDSolve seems to have difficulties with solving integral
equation.> n = 5; NDSolve[{D[[Sigma]norm[z, t], t] ==
3*z*Integrate[[Sigma]norm[z,> t]^n*z, {z, 0, 1}] -
[Sigma]norm[z, t]^n,> [Sigma]norm[z, 0] == 1.5*z,
[Sigma]norm[0, t] == 0}*[Sigma]norm[z,> t], {z, 0.01,
1}, {t, 0.01, 2}]> Mathematica returns a message>
NDSolve::deql: The first argument must have both an
equation and an > initial condition.> which I cannot
understand.> Can anybody tell what's wrong with my attempt?>
> -ToshiI'll show a couple of approaches to this sort of
problem. The first isto iteratively solve for sn[k][z,t] in
terms of sn[k-1][z,t], withsn[0][z,t] initialized to
something appropriate (I used 3/2*z). The sn'sare computed
as interpolating functions based on results from
solvingODEs in t for many fixed values of z. The code below
does this for n=2.The plots at first appear to ßip between
two states but eventuallystabilize. Note that this takes
many minutes to run to completion.n = 2;sn[0][z_,t_] :=
3/2*z;Do [ Do [ sn[k][z,t_] = sn[k][z,t] /.
First[NDSolve[{D[sn[k][z,t], t] ==
3*z*NIntegrate[sn[k-1][w,t]^n*w, {w,0,1}] -sn[k-1][z,t]^n,
sn[k][z,0] == 3/2*z}, sn[k][z,t], {t,0,2}]], {z,0,1,1/100} ];
snew = Interpolation[Flatten[Table[{z,t,sn[k][z,t]},
{z,0,1,1/100}, {t,0,2,1/100}], 1],InterpolationOrder->7];
sn[k] = snew; Print[iteration , k]; Plot3D[sn[k][z,t],
{z,0,1}, {t,0,2}], {k, 1, 15} ]One can test for convergence
as below; it is apparently quite
good.NIntegrate[Abs[sn[15][z,t]-sn[14][z,t]], {z,0,1},
{t,0,2}]A drawback to this method is that it appears to
break down beyond n = 2.Possibly one simply needs a much
better starting function for sn[0], I'mnot sure.Below is
another method that Michael Trott showed me. We expand in a
setof basis functions in z, set up a system of ODEs in t,
and solve them. Itried this using monomials in z for basis
functions and ran into sometrouble with the ODE solving, so
I will show Michael's attempt usingtrig functions for the
basis. Note that we now handle the desired casen=5; another
advantage is that this is alot faster than the methodabove,
though still not exactly blinding in speed. Michael used
theinterval {0,2} for z so as to achieve pointwise (not
just L^2)convergence; otherwise there would be a sharp
drop-off just before z=1as all the trigs vanish there. In
other words, the basis functions areof the form Sin[k/2*Pi]
rather than Sin[k*Pi] for 1<=k<=deg. Cosines areexcluded due
to the vanishing condition at z=0.integrate[p_Plus, {z_, 0,
1}] := Integrate[#, {z, 0, 1}]& /@ p;integrate[p_, {z_, 0,
1}] := Integrate[p, {z, 0, 1}];n = 5;deg = 6;vars[t_] =
Map[#[t]&,Array[a,deg]];zFuns = Sin[Range[deg] Pi/2
z];sn[z_,t_] = vars[t].zFuns;eqs1 =
3*z*integrate[Expand[sn[w,t]^n*w],{w,0,1}] - sn[z,t]^n -
D[sn[z,t],t];eqs2 = integrate[Expand[eqs1 #], {z, 0, 1}]& /@
zFuns;iCs = integrate[Expand[(sn[z,0] - 3/2*z) #], {z, 0,
1}]& /@ zFuns; fulleqns = # == 0& /@ Join[eqs2, iCs];nsd =
NDSolve[fulleqns, vars[t], {t,0,2},
SolveDelayed->True];Plot3D[Evaluate[sn[z,t] /.
nsd[[1]]],{z,0,1},{t,0,2}];an initializer for the
iteration/interpolation method above in anattempt to refine
the solution.Daniel LichtblauWolfram Research
====
>
NDSolve seems to have difficulties with solving integral
equation.> n = 5; NDSolve[{D[[Sigma]norm[z, t], t] ==
3*z*Integrate[[Sigma]norm[z,> t]^n*z, {z, 0, 1}] -
[Sigma]norm[z, t]^n,> [Sigma]norm[z, 0] == 1.5*z,
[Sigma]norm[0, t] == 0}*[Sigma]norm[z,> t], {z, 0.01,
1}, {t, 0.01, 2}]> Mathematica returns a message>
NDSolve::deql: The first argument must have both an
equation and an > initial condition.> which I cannot
understand.> Can anybody tell what's wrong with my attempt?>
> -ToshiI'll show a couple of approaches to this sort of
problem. The first isto iteratively solve for sn[k][z,t] in
terms of sn[k-1][z,t], withsn[0][z,t] initialized to
something appropriate (I used 3/2*z). The sn'sare computed
as interpolating functions based on results from
solvingODEs in t for many fixed values of z. The code below
does this for n=2.The plots at first appear to ßip between
two states but eventuallystabilize. Note that this takes
many minutes to run to completion.n = 2;sn[0][z_,t_] :=
3/2*z;Do [ Do [ sn[k][z,t_] = sn[k][z,t] /.
First[NDSolve[{D[sn[k][z,t], t] ==
3*z*NIntegrate[sn[k-1][w,t]^n*w, {w,0,1}] - sn[k-1][z,t]^n,
sn[k][z,0] == 3/2*z}, sn[k][z,t], {t,0,2}]], {z,0,1,1/100} ];
snew = Interpolation[Flatten[Table[{z,t,sn[k][z,t]},
{z,0,1,1/100}, {t,0,2,1/100}], 1], InterpolationOrder->7];
sn[k] = snew; Print[iteration , k]; Plot3D[sn[k][z,t],
{z,0,1}, {t,0,2}], {k, 1, 15} ]One can test for convergence
as below; it is apparently quite
good.NIntegrate[Abs[sn[15][z,t]-sn[14][z,t]], {z,0,1},
{t,0,2}]A drawback to this method is that it appears to
break down beyond n = 2.Possibly one simply needs a much
better starting function for sn[0], I'mnot sure.Below is
another method that Michael Trott showed me. We expand in a
setof basis functions in z, set up a system of ODEs in t,
and solve them. Itried this using monomials in z for basis
functions and ran into sometrouble with the ODE solving, so
I will show Michael's attempt usingtrig functions for the
basis. Note that we now handle the desired casen=5; another
advantage is that this is alot faster than the methodabove,
though still not exactly blinding in speed. Michael used
theinterval {0,2} for z so as to achieve pointwise (not
just L^2)convergence; otherwise there would be a sharp
drop-off just before z=1as all the trigs vanish there. In
other words, the basis functions areof the form Sin[k/2*Pi]
rather than Sin[k*Pi] for 1<=k<=deg. Cosines areexcluded due
to the vanishing condition at z=0.integrate[p_Plus, {z_, 0,
1}] := Integrate[#, {z, 0, 1}]& /@ p;integrate[p_, {z_, 0,
1}] := Integrate[p, {z, 0, 1}];n = 5;deg = 6;vars[t_] =
Map[#[t]&,Array[a,deg]];zFuns = Sin[Range[deg] Pi/2
z];sn[z_,t_] = vars[t].zFuns;eqs1 =
3*z*integrate[Expand[sn[w,t]^n*w],{w,0,1}] - sn[z,t]^n -
D[sn[z,t],t];eqs2 = integrate[Expand[eqs1 #], {z, 0, 1}]& /@
zFuns;iCs = integrate[Expand[(sn[z,0] - 3/2*z) #], {z, 0,
1}]& /@ zFuns; fulleqns = # == 0& /@ Join[eqs2, iCs];nsd =
NDSolve[fulleqns, vars[t], {t,0,2},
SolveDelayed->True];Plot3D[Evaluate[sn[z,t] /.
nsd[[1]]],{z,0,1},{t,0,2}];an initializer for the
iteration/interpolation method above in anattempt to refine
the solution.Daniel LichtblauWolfram Research
====
On
8/23/02 at 12:25 AM, meshii@mech.fukui-u.ac.jp (Toshiyuki
(Toshi)>NDSolve seems to have difficulties with solving
integral equation.>n = 5; NDSolve[{D[[Sigma]norm[z, t], t]
==>3*z*Integrate[[Sigma]norm[z, t]^n*z, {z, 0, 1}] -
[Sigma]norm[z,>t]^n, [Sigma]norm[z, 0] == 1.5*z,
[Sigma]norm[0, t] ==>0}*[Sigma]norm[z, t], {z, 0.01, 1},
{t, 0.01, 2}]>Mathematica returns a
message>>NDSolve::deql: The first argument must have
both an equation and an> initial condition.A general
solution to a differential equation includes a constant of
integration. It isn't possible to arrive a numerical
solution without providing sufficient information to
determine this constant. Specifying the initial condition
provides this information.Look at the examples using the
help browser to see what NDSolve needs.
====
-----Original
Message----->a number of useful replies and thank all those
who responded. >Today I have a>question that actually
generated yesterday's question. I am using a new>subject
heading to reßect the actual nature of the question.>>What is
the best way in Mathematica to operate on some, but >not all,
level>parts of an expression or subexpression?>>Suppose I
have the following expression...>>expr =
f1[a]f2[b]f3[c]f4[d];>>and I want to do an operation, op,
separately on f1[a]f3[c] >and f2[b]f4[d].>The operation must
be done on the given pairs and not on all >four factors
at>once. One method is to use explicit exact substitution
rules.>>expr /. f1[a]f3[c] :> op[f1[a]f3[c]] /. f2[b]f4[d] :>
op[f2[b]f4[d]]>op[f1[a] f3[c]] op[f2[b] f4[d]]>>If a,b,c,d
were long expressions, we might not want to type or >copy
them in.>This raised my question of using named patterns.
Hartmut Wolf >pointed out>that each factor must be named to
create a match with ßat >expressions. So>we could
use...>>expr /. (a : f1[_])(b : f3[_])(c : f2[_])(d : f4[_])
:> op[a b]op[c d]>op[f1[a] f3[c]] op[f2[b] f4[d]]>>Andrzej
Kozlowski suggested a method using Take, but using >Part
works better>here, so we could use...>>expr /. a_ :>
op[a[[{1, 3}]]]*op[a[[{2, 4}]]]>op[f1[a] f3[c]] op[f2[b]
f4[d]]>>All of the above methods use rules. Is it possible to
do it with>ReplacePart? I don't think so, but maybe somebody
knows how to >do it. How>about using MapAt? I don't think
that works either in regular >Mathematica.>Ted Ersek and I
did a package, Algebra`ExpressionManipulation` >at my
web>site, that implements extended positions. An extended
position >gives the>position of an expression and a list of
the desired subparts >and is packaged>in a header eP. So the
extended position of a + c in>f[a + b + c + d] is
eP[{1},{1,3}]. The package modifies MapAt to accept>extended
positions. Then we can use...>>MapAt[op, expr, {eP[{}, {1,
3}], eP[{}, {2, 4}]}]>op[f1[a] f3[c]] op[f2[b]
f4[d]]>>However, I don't always like to drag in the package
just to do >that. I think>that operating on selected level
parts of a subexpression is >not all that>uncommon.>>Here is
another example.>>1 - Cos[x]^2>% // TrigFactor>1 -
Cos[x]^2>Sin[x]^2>>But...>>1 - a - Cos[x]^2>% //
TrigFactor>1 - a - Cos[x]^2>(1/2)*(1 - 2*a - Cos[2*x])>>I
was hoping for -a + Sin[x]^2. What are the best methods for
>handling this>kind of problem in regular
Mathematica?>>David
Park>djmp@earthlink.net>http://home.earthlink.net/~djmp/>
>David
Park>djmp@earthlink.net>http://home.earthlink.net/~djmp/>
>>Dear David,it's possible that I miss something..., but
look atIn[6]:=1 - a - Cos[x]^2 /. {1 - Cos[x_]^2 :>
Sin[x]^2}Out[6]=-a + Sin[x]^2I don't know what TrigFactor
is doing exactly (or intended to do), itsanswer might well
be consistent with that.if... In[7]:= expr =
f1[a]f2[b]f3[c]f4[d]...whats wrong with..In[8]:=op[#[[{1,
3}]]]op[#[[{2, 4}]]]Take[expr, {5, -1}]
&[expr]Out[8]=op[f1[a] f3[c]] op[f2[b] f4[d]]...?The
problem with this of course is, that you must know the
Sequence of theelements of the expression in advance (at
programming time). The followingneeds not, uses extract and
rebuilds the expression (it is assumed the f<i>are at level
{1}, this must be checked, not well done here, just to pass
theidea):In[7]:=betteropat[ee : head_[__], {e1_, e3_},
{e2_, e4_}, op_] /; (len = Length[ee]) >= 4 := Module[{pos13
= Position[ee, e1[___] | e3[___], {1}], pos24 = Position[ee,
e2[___] | e4[___], {1}], posr}, posr = List /@
Complement[Range[len], Flatten[{pos13, pos24}]]; head @@
Join[{op[head @@ Extract[ee, pos13]]}, {op[ head @@
Extract[ee, pos24]]}, Extract[ee, posr]]
]In[8]:=betteropat[expr, {f1, f3}, {f2, f4},
ox]Out[8]=ox[f1[a] f3[c]] ox[f2[b]
f4[d]]In[9]:=betteropat[[Alpha] f1[a]f4[b][Beta]
f2[c]f3[d], {f1, f3}, {f2, f4}, ox]Out[9]=[Alpha] [Beta]
ox[f1[a] f3[d]] ox[f2[c] f4[b]]If you don't need all of this
functionality, just
reduce.--Hartmut___________________Addendum:Here two
other solutions, one using Part, the other ReplacePart +
Replace:is not possible, since you _must_ reorder your data,
not applying a functionat parts of it). Here now another,
simpler opat version (using Part, insteadof
Extract):In[19]:=mapopat[ee : head_[__], {e1_, e3_}, {e2_,
e4_}, op_] /; (len = Length[ee]) >= 4 := Block[{pos13 =
Flatten[Position[ee, e1[___] | e3[___], {1}]], pos24 =
Flatten[Position[ee, e2[___] | e4[___], {1}]], posr}, posr =
Complement[Range[len], pos13, pos24]; head[op[ee[[pos13]]],
op[ee[[pos24]]], ee[[posr]]] ]Of course this all is most
senseful only for heads with Flat
attribute.In[20]:=mapopat[expr, {f4, f2}, {f1, f3},
ox]Out[20]=ox[f1[a] f3[c]] ox[f2[b]
f4[d]]In[23]:=mapopat[[Alpha] f1[a]f4[b][Beta]
f2[c]f3[d], {f1, f3}, {f2, f4}, ox]Out[23]=[Alpha]
[Beta] ox[f1[a] f3[d]] ox[f2[c]
f4[b]]In[22]:=mapopat[h[f1[a], f4[b], f2[c], f3[d]], {f1,
f3}, {f2, f4}, ox]Out[22]=h[ox[h[f1[a], f3[d]]],
ox[h[f4[b], f2[c]]], h[]]finally another version using
ReplacePart; not all ReplacePart though, butin that
spirit:In[99]:=expr =
f1[a]f2[b]f3[c]f4[d]In[100]:=rplopat[ee : head_[__], {e1_,
e3_}, {e2_, e4_}, op_] /; (len = Length[ee]) >= 4 :=
Block[{head}, Module[{ pos = Join @@ (Position[ee, #, {1}] &)
/@ Through[{e1, e3, e2,e4}[___]], posr, allpos = List /@
Range[len], rr}, posr = Complement[allpos, pos]; rr =
ReplacePart[ee, ee, allpos, Join[pos, posr]]; Replace[rr,
head[a_, b_, c_, d_, r___] :> head[op[head[a, b]], op[head[c,
d]], head[r]]] ]]Blocking head prevents reordering (of rr) in
case head has the
Orderlessattribute.In[101]:=rplopat[expr, {f4, f2}, {f1,
f3}, ox]Out[101]=ox[f1[a] f3[c]] ox[f2[b]
f4[d]]In[103]:=rplopat[[Alpha] f1[a]f4[b][Beta]
f2[c]f3[d], {f1, f3}, {f2, f4}, ox]Out[103]=[Alpha]
[Beta] ox[f1[a] f3[d]] ox[f2[c]
f4[b]]In[104]:=rplopat[h[f1[a], f4[b], f2[c], f3[d]], {f1,
f3}, {f2, f4}, ox]Out[104]=h[ox[h[f1[a], f3[d]]],
ox[h[f2[c], f4[b]]], h[]]--Hartmut
====
>>I need FrameLabels
of the kind of H^s_z / H^p_Z and tried
something>like>>FrameLabel->{FontForm{HSuperscript[s
]Subscript[z] ...}}>>but s is too far away from z.
How can I manage it that s is>directly above z ? Are
there some backspace-characters to be
used?>>HarrySubsuperscript[H, s, z]-Dale
====
>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 y value
is>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.The y axis has a very
narrow range. The differences are very small. To see that,
you can pull out the tick
marks.In[13]:=yticks=FullOptions[gr,Ticks][[2]]In[15]:=
InputForm[First/@yticks]Out[15]//InputForm={
0.9999999999999981, 0.9999999999999991, 1.,
1.000000000000001, 1.000000000000002, 0.9999999999999983,
0.9999999999999986, 0.9999999999999987, 0.9999999999999989,
0.9999999999999993, 0.9999999999999996, 0.9999999999999997,
0.9999999999999999, 1.0000000000000002, 1.0000000000000004,
1.0000000000000007, 1.0000000000000009, 1.0000000000000013,
1.0000000000000016, 1.0000000000000018, 1.000000000000002,
0.9999999999999979, 0.9999999999999977,
1.0000000000000022}The ragged shape is caused by roundoff
error, if you increase the PlotRange of the y-axis you see
what you expect.gr = Plot[Sin[x]/Sin[x], {x, -2.5, -1.5},
PlotRange -> {0, 2}]-Dale
====
Harald,You could try
something like this...Plot[x, {x, 0, 1}, Frame -> True,
FrameLabel -> {z, DisplayForm[ StyleBox[
RowBox[{SubsuperscriptBox[H, z, s], /, SubsuperscriptBox[H,
z, p]}], FontSize -> 16]]}, ImageSize -> 500];where I
enlarged the font size to make the label more
legible.FontForm is an obsolete function and probably
shouldn't be used. David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/

====
Lucas,An addendum to my previous post (by the way, did
you try to changeprecedences as I recommended?).If you are
interested in using something like the normal summation
notation,the summation symbol can't be CirclePlus, as
CirclePlus is not an extensiblecharacter. I don't know all
of the extensible characters, but onepossibility is
[UnionPlus], which looks a bit like CirclePlus, with
anopening on top. Of course, the usual syntax for
[UnionPlus] is as a binaryoperator, and this is not what
we want for our summation notation. So, weneed to
incorporate new syntactical rules. There are three rules
neededhere. A rule to convert 2 dimensional input into a
mathematica internalexpression, a rule to convert the
internal expression into a box structure,and a rule to
convert the internal expression into a regular
CirclePlusexpression. I give these three rules
below:Clear[MakeExpression]MakeExpression[
RowBox[{UnderoverscriptBox[[UnionPlus],RowBox[{i_,=,k_
}],n_],y_}],
StandardForm]:=MakeExpression[RowBox[{BigCirclePlus[,y,,
{,i,,,k,,,n,}]}],StandardForm]Clear[MakeBoxes]
MakeBoxes[BigCirclePlus[y_, {i_, k_, n_}], f_] :=
RowBox[{UnderoverscriptBox[[UnionPlus],
RowBox[{MakeBoxes[i, f], =, MakeBoxes[k, f]}], MakeBoxes[n,
f]], MakeBoxes[y, f]}]BigCirclePlus[y_, {i_, k_Integer,
n_Integer}] := CirclePlus @@ Table[y, {i, k, n}]As you can
see, BigCirclePlus is used in the Mathematica
internalrepresentation. If BigCirclePlus can be converted
into a CirclePlusexpression (when the summation indices are
integers), then the BigCirclePlusrule acts. Here are a
couple of examples:!(([UnionPlus]+(i = 1)%M
g[i]))!(([UnionPlus]+(i = 1)%5
g[i]/(1 + h[i]/5)))Just copy each of the above
expressions into Mathematica and evaluate afterevaluating
the above rules.Of course, if [UnionPlus] is not an
acceptable substitute for an extensibleCirclePlus, then
you will just need to petition Mathematica to include sucha
feature in the future.Carl WollPhysics DeptU of
Washington----- Original Message ----->> Open up the file,
search for CirclePlus, change the precedence from 450 to>
420, and then save. Of course, it would be wise to make a
backup copy ofthe> file before you make any changes. Also,
420 is low enough to get the> behavior you desire, but you
may want to experiment with otherprecedences.> Then, start
mathematica and you will get the behavior you want.>> Carl
Woll> Physics Dept> U of Washington> I'm attempting
to implement an abstract mathematica package in>
mathematica that utilized the [CirclePlus] operator in an
unusual> way. Specifically, the [CirclePlus] has a
precidence lower than +> and introduces barriers in the
computation. So, an expression such as>> a + b
[CirclePlus] c + d --> (a+b) [CirclePlus] (c+d)>> The
mathematica ouput of>> a + d + (b [CirclePlus] c) is
incorrect. I've tried playing with the> PrecedenceForm[]
function, but that does not seem able to produce the>
desired effect.>> Also, I would like to introduce a
notation like>> N> [BigCirclePlus] x[[i]] -->
x[[1]] [CirclePlus] x[[2]] [CirclePlus]> ....> i=0>
>> analagous to summation, but mathematica does not appear
to offer the> CirclePlus in a large format. to relate this
to the case above, x[1]> = (a + b) and> x[2] = (c + d),
so each indexed element is a subexpression.>> Finally, I
would like to be able to set up the CirclePlus operator>
such that the following algebraic relations hold:>
Sum BigCirclePlus E = BigCirclePlus Sum E> i j ij
j i ij>> d d> -- BigCirclePlus E = BigCirclePlus
-- E> dx j j j dx j>> -Lucas Scharenbroich>
-MLS Group / JPL>>
====
I plot several curve depending on
a parameter p and I want the PlotLabel toshow the value of
p.What is wrong with the following
?pl[p_]:=Plot[Sin[x^p],{x,0,Pi},PlotLabel[Rule] p =  p,
DisplayFunction[Rule]Identity]Table[Show[pl[k],
DisplayFunction[Rule]$DisplayFunction],{k,3}]
====
What is
your operating system? How much memory do you have in your
machine?Also, how big is your pagefile? Sounds to me like
the rebuild ran outof room to do its thing in. I just
installed Mathematica 4.2, and had no problem rebuilding the
help index.Subsequently, I ahd no problem using the Help
browser either.Hope that helps!....Terry>I'd start with
the following FAQ.>>
>http://support.wolfram.com/mathematica/interface/
helpbrowser/howrebuildindex.html>>-Dale>> Sorry that i
failed say it immediately in a first place, but of course> i
did try it FAQ at first, and both tried to delete cache and
rebuild> index, but results where the same - whenever i try
to invoke help> browser (or rebuild index, for that matter),
mathematica stops> responding (i did read your answer to the
same question asked> a week ago before - actually that is
why i turned to the FAQ).
====
I too have been unable to get
the help browser functioning. I haveexactly the same
problem. When trying to open the help browser it getsto the
point where it says scanning index file and freezes. I
triedall the suggestions in the faq and the links provided
here to noavail. I'm running Winows 2000 with service pack
3. I have 512mb ofmemory with a 768mb pagefile on a separate
drive, so I don't thinkthat's an issue. 
Mathematica 4.1 ran
just fine on this setup, sojudging by the number of other
people affected by this, I'd say it'ssome kind 
of bug with
4.2. Hopefully Wolfram will address this witheither a link
that will provide a fix that actually works, or a patch.>What
is your operating system? How much memory do you have in your
machine?>Also, how big is your pagefile? Sounds to me like the
rebuild ran out>of room to do its thing in.>> I just
installed Mathematica 4.2, and had no problem rebuilding the
help index.>Subsequently, I ahd no problem using the Help
browser either.>Hope that helps!>....Terry>>I'd start
with the following FAQ.>>
>http://support.wolfram.com/mathematica/interface/
helpbrowser/howrebuildindex.html>>-Dale>> Sorry
that i failed say it immediately in a first place, but of
course>> i did try it FAQ at first, and both tried to delete
cache and rebuild>> index, but results where the same -
whenever i try to invoke help>> browser (or rebuild index,
for that matter), mathematica stops>> responding (i did read
your answer to the same question asked>> a week ago before -
actually that is why i turned to the FAQ).>
====
>What is your
operating system? How much memory do you have in your
machine?>Also, how big is your pagefile? Sounds to me like
the rebuild ran out>of room to do its thing in.I use W2k
Pro + SP2, Athlon 1.33, 260MB RAM (DDR) + 630MB swap file
-doesn't seems to me like a memory problem... 
====
does
anyone know whether there is an effective way of
simulatingpaths of Brownian motion (Wiener process) in
dimensions 1, 2, 3?I could you a random walk approach, but
this seems to be computationallynot very efficient and I am
not sure whether it is even a goodapproximation to the 3-D
BM. Any help on Mathematica codingand in the forms of
algorithms will be greatly appreciated.Janusz Kawczak.
====

What is the problem with this integration. Its keep on
running andnot coming out.[CapitalPsi][r_, [Theta]_,
[Phi]_][4] := r
Exp[-r/2]Sin[[Theta]]Cos[[Phi]];[CapitalPsi][r_,
[Theta]_, [Phi]_][11] := r^2
Exp[-r/2]Sin[[Theta]]Cos[[Theta]]Cos[[Phi]];[
ScriptCapitalH][ r_, [Theta]_, [Phi]_] := (-[HBar]/ 2 m
(1/(r^2 Sin[[Theta]]))(Sin[[Theta]]D[r^2 D[#, r], r]+
D[Sin[[Theta]] D[#, [Theta]], [Theta]] +
1/(Sin[[Theta]])(D[D[#, [Phi]], [Phi]])) + e V
&)Ç_bho = Sum[ Sum[ Integrate[[CapitalPsi][r, [Theta],
[Phi]][ j]([ScriptCapitalH][r, [Theta],
[Phi]])[[CapitalPsi][ r, [Theta], [Phi]][i]], {r, 0,
h}, {[Theta], 0, Pi}, {[Phi], 0, 2 Pi}], {i, 11, 11}],
{j, 4, 4}] Raj Kumar Gupta
====
I know its not a good idea to
run programs as root on a Unix machine and don'tnormally do
so. However, can anyone tell me what is happening here? A
mathematical program 'Mathematica' has been 
installed in
/usr/local on thisSun Ultra 60. The correct password was
entered and it all runs fine with mynormal login (davek),
which has a shell of /bin/tcsh. If I switch user to root,
which has the normal /sbin/sh shell, Mathematica runsfine.
However, if I switch user to 'roottcsh', which 
as a uid of 0,
but a shellof /bin/tcsh, Mathematica fails to run, saying
the password file is invalid. I can't understand 
while
executing a program that works fine as root, fine 
as anormal
user, yet does not when run when execution is attempted as
another userwith root privileges. *** Runs fine as a formal
user 'davek'******wren /export/home/davek % 
mathMathematica
4.0 for SolarisCopyright 1988-1999 Wolfram Research, Inc. --
Motif graphics initialized -- In[1]:= Quit***Mathematica is
run from /usr/local/bin/math, which is a symbolic
link****wren /export/home/davek % which
math/usr/local/bin/mathwren /export/home/davek % ls -l
/usr/local/bin/mathlrwxrwxrwx 1 root other 53 Jun 16 08:18
/usr/local/bin/math
->/usr/local/mathematica-4.0.2/Executables/Solaris/math***
Mathematica runs fine as root****wren /export/home/davek %
suPassword: # /usr/local/bin/mathMathematica 4.0 for
SolarisCopyright 1988-1999 Wolfram Research, Inc. -- Motif
graphics initialized -- In[1]:= Quit***Switch to a user
with uid=0, shell =/bin/tcsh ***wren /export/home/davek % su
- roottcshPassword: Sun Microsystems Inc. SunOS 5.9 Generic
May 2002***Now Mathematica does not run***wren / #
/usr/local/bin/mathMathematica 4.0 for SolarisCopyright
1988-1999 Wolfram Research,
Inc./usr/local/mathematica-4.0.2/Configuration/Licensing/
mathpass:1: Incomplete password entry.No valid
single-machine password entry for Mathematica found.Machine
name: wrenMathID: [deleted by moderator]You will need a
valid license ID and password in orderto proceed. Go to
http://register.wolfram.com orconsult your Getting Started
documentation.Enter the name of your organization: ***Here
are the contents of the password files*******with the contents
of /etc/shadow changed for security resons***wren / # cat
/etc/passwd | grep
rootroot:x:0:1:Super-User:/:/sbin/shroottcsh:x:0:1:Dr.
David Kirkby:/:/bin/tcshwren / # cat /etc/shadow | grep
rootroot:TA9addfsfsdsdfdddMTxHU:11266::::::roottcsh:
dgddsdfsdffsdnSw:11198::::::-- Dr. David Kirkby PhD,web
page: http://www.david-kirkby.co.ukAmateur radio callsign:
G8WRB
====
I managed to post this on alt.math.recreational by
mistake, and confuseeveryone!I am having a bit of trouble
with showing the shapes of pentominoes. Thelist of
pentominoes is 12 x 5 x 2, for 12 pentominoes x 5 squares x
{X,Y}coordinate of the square on a zero-origin basis.Just
for completeness ...pentominoes ={ {{0, 0}, {1, 0}, {1, 1},
{1, 2}, {2, 1}}, (*F*){{0, 0}, {1, 0}, {2, 0}, {3, 0}, {4,
0}}, (*I*){{0, 0}, {1, 0}, {2, 0}, {3, 0}, {0, 1}}, (*L*){{0,
0}, {1, 0}, {2, 0}, {2, 1}, {3, 1}}, (*N*){{0, 0}, {1, 0}, {2,
0}, {0, 1}, {1, 1}}, (*P*){{0, 0}, {1, 0}, {2, 0}, {1, 1}, {1,
2}}, (*T*){{0, 0}, {1, 0}, {2, 0}, {0, 1}, {2, 1}}, (*U*){{0,
0}, {1, 0}, {2, 0}, {0, 1}, {0, 2}}, (*V*){{0, 0}, {1, 0},
{1, 1}, {2, 1}, {2, 2}}, (*W*){{1, 0}, {0, 1}, {1, 1}, {2,
1}, {1, 2}}, (*X*){{0, 0}, {1, 0}, {2, 0}, {3, 0}, {1, 1}},
(*Y*){{0, 0}, {1, 0}, {1, 1}, {1, 2}, {2, 2}} (*Z*)}When I
display them with the code below, the aspect ratio is correct
*within each pentomino* but they are not scaled equivalently.
I would appreciate a pointer to what I have done wrong. The
only solution I have been able to develop so far effectively
places each of the 12 pentominoes at a different location in
the plane, and then draws the whole region in one go.Show[
GraphicsArray[ Partition[ Table[ Graphics[ Map[Rectangle[#,
(# + {1, 1})] &, pentominoes[[i]]], AspectRatio -> Automatic
], {i, 12}], 6] ]]-- Mark R. Diamond
====
> I plot several
curve depending on a parameter p and I want the PlotLabel
to> show the value of p.> What is wrong with the following
?>> pl[p_]:=Plot[Sin[x^p],{x,0,Pi},PlotLabel[Rule] p = 
p,> DisplayFunction[Rule]Identity]>>
Table[Show[pl[k],DisplayFunction[Rule]$DisplayFunction],{k,
3}]>Use ToString and
StringJoinpl[p_]:=Plot[Sin[x^p],{x,0,Pi}, PlotLabel->p =
 <> ToString[p], DisplayFunction->Identity];Alternatively,
you could use StringFormpl[p_]:=Plot[Sin[x^p],{x,0,Pi},
PlotLabel->StringForm[p = `` , p],
DisplayFunction->Identity];Bob HanlonChantilly, VA
USA
====
In a message dated 8/31/02 1:58:36 AM,
berlusconi_pagliusi@fis.unical.it> I'd like to 
use
Mathematica 4.0 to write a function having different>
expressions in different domain's intervals.> 
Let's say:>>
F[x_]= x^2 if 0<x<6> x+1 if x>=6>> I know It's a stupid
syntax problem, but I really do not know how/where to>
search the solution on the Mathematica Book>Just for grins
here are several methods:f1[x_/;x<=0] := 0 ;f1[x_/;0<x<6]
:= x^2 ;f1[x_/;x>=6] := x+1;f2[x_] :=
x^2*UnitStep[x]+(x+1-x^2)*UnitStep[x-6];f3[x_] := Which[
x<=0, 0, 0<x<6, x^2, x>=6, x+1];f4[x_] := If[0<x<6,x^2,
If[x>=6, x+1,0]];f5[x_] := Switch[x, _?(#<=0&), 0,
_?(0<#<6&), x^2, _?(#>=6&), x+1 ]f6[x_?NumericQ] :=
{0,x^2,x+1}[[Position[ {x<=0,0<x<6,x>=6},
True][[1,1]]]];Needs[Calculus`Integration`];(* needed for
definition of Boole *)f7a[x_?NumericQ] := Evaluate[ (Boole /@
{0<x<6,x>=6}). {x^2,x+1}];Off[Part::pspec];f7b[x_?NumericQ]
:= Evaluate[ {0,x^2,x+1}[[1+Tr[Boole /@ {x>0,
x>=6}]]]];f8[x_?NumericQ] := Cases[ {{x<=0, 0}, {0<x<6,
x^2}, {x>=6, x+1}}, {True, z_} :>z][[1]];f9[x_?NumericQ] :=
DeleteCases[ {{x<=0, 0}, {0<x<6, x^2}, {x>=6, x+1}}, {False,
z_}][[1,2]];f10[x_?NumericQ] := Select[ {{x<=0, 0}, {0<x<6,
x^2}, {x>=6, x+1}}, First[#]&][[1,2]];f11[x_?NumericQ] :=
Last[Sort[ {{x<=0, 0}, {0<x<6, x^2}, {x>=6,
x+1}}]][[2]];f12[x_?NumericQ] := Module[{n=1},
While[{x<6,x<0, False}[[n]], n++];
{x+1,x^2,0}[[n]]];Generating some test points:ts =
{Random[Real,{-5,0}],0,
Random[Real,{0,6}],6,Random[Real,{6,15}]};Checking the
different representationsEqual[(# /@ pts)& /@
{f1,f2,f3,f4,f5,f6,f7a,f7b,f8,f9,f10,f11,f12}]TrueTo pick
a favorite, look at how the different definitions behave.Of
the definitions that evaluate with symbolic input, only f2 and
f4simplify with assumptions. For example,FullSimplify[#[x]&
/@ {f2,f4}, 1<x<3]f2 through f5 respond immediately to
differentiation#'[x]& /@ {f2,f3,f4,f5} // Simplify
//ColumnFormOnly f2 responds immediately to
integrationIntegrate[f2[x],x]//SimplifyConsequently, f2
(UnitStep) appears to be the most versatile.Bob
HanlonChantilly, VA USA
====
In a message dated 8/31/02
3:14:33 AM, andreas.dietrich@physik.uni-marburg.d=e> I am
trying to implement the inner product in the space of>
complex-valued, square integrable functions over [-1/2,1/2],
which can> be expressed in Mathematica code as>>
inner[f_Function,g_function]:=Integrate[Conjugate[f[x]]*g[x]
,{x,-1/2,1/2}]>> This is simple enough. Problem is,
Mathematica seamingly cannot> evaluate the Integral for even
the simplest of functions:> In[10]:=inner[#&,#&]>>
Out[10]:=!([Integral]_(-(1/2))%(1/2
)(x Conjugate[> x]) [DifferentialD]x)>> As you
see, the Integrate returns unevaluated. It works fine if I>
remove the Conjugate. Unfortunately the Conjugate is needed
for> positive definiteness.>> Various variants with
Composition, Re and Im etc. don't work either.>> This should
be a So how do I get Integrate to work with
Conjugate?>>Would including Simplify or FullSimplify provide
the results that youn want?inner[f_Function,g_Function]:=
Integrate[Simplify[Conjugate[f[x]]*g[x], Element[x, Reals]],
{x, -1/2, 1/2}];inner[#&,#&]1/12Bob HanlonChantilly, VA
USA
====
I was wondering if there are any scripts/functions
out there to allow me todo decision tree classification with
Mathematica, similar to S+ typeanalysis?--j-- Jonathan
GreenbergGraduate Group in Ecology, U.C.
Davishttp://www.cstars.ucdavis.edu/~jongreenhttp://
www.cstars.ucdavis.eduAIM: jgrn307 or jgrn3007MSN:
jgrn307@msn.com or jgrn3007@msn.com
====
PasKo,You could
look up piecewise functions in the Master Index in
Help.Unfortunately, the book does not have a unified
discussion of methods fordefining piecewise functions.The
first method is to use a set of conditional 
definitions.
Forcompleteness I made f[x] = 0 if one of the conditional
definitions is notmet.f[x_] /; 0 < x < 6 := x^2f[x_] /; x
>= 6 := x + 1f[x_] := 0Plot[f[x], {x, 0, 10}];The second
method is to use a Which statement.Clear[f];f[x_] := Which[
0 < x < 6, x^2, x >= 6, x + 1, True, 0]Plot[f[x], {x, 0,
10}];The above methods are fine in many cases, but if you
want to performfunctions on f, such as differentiation or
integration, you should use theUnitStep function. So our
third method is...Clear[f]f[x_] := x^2(UnitStep[x] -
UnitStep[x - 6]) + (x + 1)UnitStep[x - 6]Plot[f'[x], {x, 0,
10}];Plot[Integrate[f[y], {y, 0, x}] // Evaluate, {x, 0,
10}];David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/
PasKo.
====
> I am trying to implement the inner product in
the space of> complex-valued, square integrable functions
over [-1/2,1/2], which can> be expressed in Mathematica code
as>
inner[f_Function,g_function]:=Integrate[Conjugate[f[x]]*g[x]
,{x,-1/2,1/2}]> This is simple enough. Problem is,
Mathematica seamingly cannot> evaluate the Integral for even
the simplest of functions:> In[10]:=inner[#&,#&]>
Out[10]:=!([Integral]_(-(1/2))%(1/2
)(x Conjugate[> x]) [DifferentialD]x)> As you
see, the Integrate returns unevaluated. It works fine if I>
remove the Conjugate. Unfortunately the Conjugate is needed
for> positive definiteness.> Various variants with
Composition, Re and Im etc. don't work either.> This should
be a So how do I get Integrate to work with Conjugate?>
Andreas> --> True Pleasure in this society is more
dangerous than bank robbery.The deficiency has been addressed
in our development version, at leastto the extent that it
works on the simple example given above. For awork-around
in version 4 one might do as below.inner2[f_Function,
g_Function] := Integrate[(Re[f[x]] - I*Im[f[x]])*g[x], {x,
-1/2, 1/2}] In[2]:= inner2[# &, # &] 1Out[2]= -- 12Daniel
LichtblauWolfram Research
====
Andreas,inner[f_Function,
g_Function] := Integrate[ComplexExpand[Conjugate[f[x]]]*g[x],
{x, -1/2, 1/2}]inner[# &, # &]1/12I'm not too knowledgable
about using complex functions in Mathematica butsometimes I
think that ComplexExpand should be renamed
ComplexSimplify.One very often needs it.David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/
x]) [DifferentialD]x)As you see, the Integrate returns
unevaluated. It works fine if Iremove the Conjugate.
Unfortunately the Conjugate is needed forpositive
definiteness.Various variants with Composition, Re and Im
etc. don't work either.This should be a So how do I get
Integrate to work with Conjugate?Andreas--True Pleasure in
this society is more dangerous than bank robbery.
====
I work
for Gould Academy and am trying to make it really easy for
thescripts which I thought did that really well. But, the
scienceteachers said it was too hard. So, I have been
making it so that Inputboxes prompt the students for the
filename type, regression type, etc.I now have been thinking
that it would be great if there could be anInput box which
asks the student what they want to name there
graph,function, data, and have them simply call that
variable when they wantto use the graph, function, data.
this is Instead of typingVariableName = FilePlot[] which
the science teachers think is forcingthe students to become
mathematica programmers.Charles
====
Use Condition
(/;):In[1]:=f[x_] := x^2 /; 0 < x < 6;f[x_] := x + 1 /; 6
<= x;Plot[f[x], {x, 0, 10}];Avoid using capital initial
letters for your definitions: they are usuallyreserved for
in-built functions in Mathematica.Tomas GarzaMexico
City----- Original Message -----> I know It's a stupid
syntax problem, but I really do not know how/where to>
search the solution on the Mathematica Book>
PasKo.>>
====
try something like this:foo[x_] := Block[{y},
If[x > 0 && x < 6, y = x^2, If[x >= 6, y = x + 1]] ]for
details look at:?And?If?Block----- Original Message ----->
I know It's a stupid syntax problem, but I really do not 
know
how/where to> search the solution on the Mathematica
Book> PasKo.>
====
You can use If, Which,
Conditionf[x_] := If[0 <= x && x <= 6, x^2, x + 1];g[x_]
:= 2*x^2 /; 0 <= x <= 6;g[x_] := 2*x + 1 /; x > 6;h[x_] :=
Which[0 <= x <= 6, 3*x^2, 6 < x, 3*x + 1]Plot[{f[x], g[x],
h[x]}, {x, 0, 10}]> I'd like to use Mathematica 4.0 to write
a function having different> expressions in different
domain's intervals.> Let's say:>> F[x_]= x^2 
if 0<x<6> x+1 if
x>=6>> I know It's a stupid syntax problem, but I really do
not know how/where to> search the solution on the
Mathematica
Book*******************milkcartmilkcart@m17.
alpha-net.ne.jp***********************
====
Dear
MathGroup,I'd like to use Mathematica 4.0 to write a
function having differentexpressions in different domain's
intervals.Let's say:F[x_]= x^2 if 0<x<6 x+1 if x>=6I know
It's a stupid syntax problem, but I really do not know
how/where tosearch the solution on the Mathematica
BookPasKo.
====
> Dear MathGroup,> I'd like to use
Mathematica 4.0 to write a function having different>
expressions in different domain's intervals.> 
Let's say:>>
F[x_]= x^2 if 0<x<6> x+1 if x>=6I'd like to thank everyone.
:-)In the meantime I've found an answer to my problem using
Which statement,but as far as you suggest I'm going to use
the more versatile UnitStepfunction.PasKoP.S.Delete
berlusconi_ to reply.
====
Andreas,One simple approach is
to write:F[x_]:= x^2 /; 0<x<6F[x_]:= x+1 /; x>=6Hope this
helps. You want want to look up /; in the Help Browser.Mark
Westwood> Dear MathGroup,> I'd like to use Mathematica 4.0
to write a function having different> expressions in
different domain's intervals.> Let's say:> 
F[x_]= x^2 if
0<x<6> x+1 if x>=6> I know It's a stupid syntax problem,
but I really do not know how/where to> search the solution
on the Mathematica Book> PasKo.
====
Try the following
(I'm quoting from Tom Wickham-Jones' book on
MathematicaGraphics, p. 41):In[1]:=pl[p_] :=
Plot[Sin[x^p], {x, 0, Pi}, PlotLabel -> StringForm[p =
`1`, p], DisplayFunction ->
Identity]In[12]:=Table[Show[pl[k], DisplayFunction ->
$DisplayFunction], {k, 3}]Tomas GarzaMexico City-----
Original Message ----->>
Table[Show[pl[k],DisplayFunction[Rule]$DisplayFunction],{k,
3}]>>
====
Andre,You should use SequenceForm in your
definition (instead of multiplicationwhich reorders the
items) .pl[p_] := Plot[Sin[x^p], {x, 0, Pi}, PlotLabel ->
SequenceForm[p =  , p], DisplayFunction ->
Identity]David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/===
=> I am trying to implement the inner product in the space
of> complex-valued, square integrable functions over
[-1/2,1/2], which can> be expressed in Mathematica code as>
>
inner[f_Function,g_function]:=Integrate[Conjugate[f[x]]*g[x]
,{x,-1/2,1/2}]> ...Mathematica cannot Integrate through
Conjugate in general. I suspect,however, that most or all
specific cases where you cannot expand Conjugateare not
integrable in closed form anyway. So I suggest that you
attempt toexpand the Conjugate before
integrating:inner[f_Function, g_Function] :=
Integrate[ComplexExpand[f[x] Conjugate[g[x]]], {x, -1/2,
1/2}]Although the above function works for the few simple
cases I tried just now,my experience tells me that the
following, more cumbersome form might bemore robust:
inner[f_Function, g_Function] :=
Integrate[Simplify[ComplexExpand[ f[x] Conjugate[g[x]],
TargetFunctions -> {Re, Im}]], {x, -1/2, 1/2}]Hope this
helps,Tom Burton
====
> I'd like to use Mathematica 4.0 to
write a function having different> expressions in different
domain's intervals.> Let's say:> F[x_]= x^2 if 
0<x<6> x+1
if x>=6> I know It's a stupid syntax problem, but I really
do not know how/where to> search the solution on the
Mathematica Book> Several choices exist. An advantage of
the following choice,F[x_] := If[ x<6, x^2, x+1 ]is that
Mathematica can integrate and different this form
analytically.Tom BurtonReply-To: Peltio
<peltioNOSP@Miname.com.invalid>
====
Sergio Milo ha scritto
nel messaggio ...> Dear Math Group.> Anybody can explain me
what this it means.> Out[22]=!(Null^49)It seems that
Mathematica has produced 49 times in a row a Null result
andinterpreted it as a variable in a multiplication.As far
as I know Mathematica shouldn't compute the powers of Null,
though.And most importantly, it should not output a Null, at
least in standardconditions.Could it be that you
inadvertently redefined Null, after unprotecting it?(yes I
know, it woldn't be much of an inadvertent behaviour...)Does
this happen in a fresh new Mathematica
session?saluti,Peltiopeltio at despammed dot
comantispam address in reply-to
====
Dear group.I am
trying to figure out how I can do multi-dimensional kernel
densityestimation with Mathematica. I know how I can do it
with one variable, butI was wondering if Mathematica
supports kernel density estimation with
3variables.Sincerely,Marcelo-----------------------------
---------------------------------Marcelo MoreiraDepartment
of Economics - Harvard UniversityLittauer M6 - North
Yard1875 Cambridge StreetCambridge, MA
02138http://post.economics.harvard.edu/faculty/moreira/
moreira.html------------------------------------------------
--------------
====
I liked this very much.For n=4 there are
3^n=81 possible combinations, but only 40 positive ones -the
rest would be 0 and each of the positive ones multiplied by
-1. As Bobstated, this is obtained by:(3^n-1)/2=40For
these 40 positive integers to match 1 to 40 correspondingly,
the biggestshould 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 corresponding sum.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 40 positivecombinations. 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), in thesense 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),
so Iwould 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 a
nature>> number,and now I can only make 4 counterweights
to measure out each>> mass of those things.Question:What
mass should the counterweights>> be???>> The answer is
that 1,3,9,27 and I wnat to use mathematica to solve>>
this problem.>> In fact,I think that this physical problem
has various>> answer,ex.2,4,10,28>> this way also
work,because if I have a thing which weight 3 , and I>>
can measure out by comparing 2<3<4 . But,If I want to solve
this math>> problem:>>
{x|x=k1*a+k2*b+k3*c+k4*d}={1,2,3,4,,,,,,40} where a,b,c,d is
nature> numbers.>> and {k1,k2,k3,k4}={1,0,-1}>> How to
solve it ??>> mathematica solving method. appreciate any
idea sharing>> sincerely>> bryan> Just use brute
force.>> Needs[DiscreteMath`Combinatorica`];>> var = {a,
b, c, d}; n = Length[var];>> s = Outer[Times, var, {-1, 0, 1}
];>> f = Flatten[Outer[Plus, Sequence@@s]];>> Since the length
of f is just 3^n then the range of numbers> to be covered
should be {-(3^n-1)/2, (3^n-1)/2}.> Consequently, the largest
of the weights can not exceed> (3^n-1)/2 - (1+2+...+(n-1))
or>> ((3^n-1) - n(n-1))/2>> 34>> Thread[var->#]& /@>> (First
/@ Select[{var,f} /. Thread[var->#]& /@>>
KSubsets[Range[((3^n-1) - n(n-1))/2], n],>> Sort[#[[2]]] ==
Range[-(3^n-1)/2,(3^n-1)/2]&])>> {{a -> 1, b -> 3, c -> 9, d
-> 27}}> Bob Hanlon> Chantilly, VA USA>
====
>I'm sorry
for that my question is not clear,I have correct
below.>>ajvp7h$ibk$1@smc.vnet.net>...> I have a very
interesting math problem:If I have a scales,and I> have 40
things that their mass range from 1~40 which each is a
nature> number,and now I can only make 4 counterweights to
measure out each> mass of those things.Question:What mass
should the counterweights> be???> The answer is that
1,3,9,27 and I wnat to use mathematica to solve> this
problem.> In fact,I think that this physical problem has
various> answer,ex.2,4,10,28> this way also work,because
if I have a thing which weight 3 , and I> can measure out
by comparing 2<3<4 . But,If I want to solve this math>
problem:> {x|x=k1*a+k2*b+k3*c+k4*d}={1,2,3,4,,,,,,40} where
a,b,c,d is nature>numbers.> and {k1,k2,k3,k4}={1,0,-1}>
How to solve it ?? > mathematica solving method. appreciate
any idea sharing> sincerely > bryan>>Just use brute
force.>>Needs[DiscreteMath`Combinatorica`];>var = {a, b,
c, d}; n = Length[var];>s = Outer[Times, var, {-1, 0, 1}
];>f = Flatten[Outer[Plus, Sequence@@s]];>Since the
length of f is just 3^n then the range of numbers>to be
covered should be {-(3^n-1)/2, (3^n-1)/2}.>Consequently, the
largest of the weights can not exceed>(3^n-1)/2 -
(1+2+...+(n-1)) or >>((3^n-1) -
n(n-1))/2>34>Thread[var->#]& /@> (First /@
Select[{var,f} /. Thread[var->#]& /@>
KSubsets[Range[((3^n-1) - n(n-1))/2], n],> Sort[#[[2]]] ==
Range[-(3^n-1)/2,(3^n-1)/2]&])>{{a -> 1, b -> 3, c -> 9, d
-> 27}}>A modification to my earlier response. Since you are
try to cover all of the values up to (3^n-1)/2 then you can
speed up the brute force method by requiring that a+b+c+d ==
(3^n-1)/2Needs[DiscreteMath`Combinatorica`];var =
{a,b,c,d}; n = Length[var]; m = (3^n-1)/2;s = Outer[Times,
var, {-1,0,1} ];f = Flatten[Outer[Plus, Sequence@@s]];k=
Select[KSubsets[Range[m - n(n-1)/2], n], (Plus @@ #) ==
m&];Thread[var->#]& /@ (First /@ Select[{var,f} /.
Thread[var->#]& /@ k, Sort[#[[2]]] == Range[-m,m]&]){{a -> 1,
b -> 3, c -> 9, d -> 27}}Bob HanlonChantilly, VA
USA
====
On 8/31/02 at 1:25 AM,
berlusconi_pagliusi@fis.unical.it (Paskoski)>Dear MathGroup,
I'd like to use Mathematica 4.0 to write a function>having
different expressions in different domain's intervals.
Let's>say:>>F[x_]= x^2 if 0<x<6 x+1 if x>=6Tryf[x_]:= x^2
/; 0<x<6f[x_]:= x+1/; x>=6The /; specfies a conditional
rule. You can define a fuction with as many pieces as you
like in this manner
====
 help,urgent how do you put the
numbers in the circles
====
There are many cases in graphics,
and otherwise, where it is useful toobtain two orthogonal
unit vectors to a given vector. I know a number ofways to
do it, but they all seem to be slightly inelegant. I thought
I wouldpose the problem to MathGroup. Who has the most
elegant Mathematicaroutine...OrthogonalUnitVectors::usage
= OrthogonalUnitVectors[v:{_,_,_}] will returntwo unit
vectors orthogonal to each other and to v.You can assume
that v is nonzero.David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/===
=My 2 cents' worth:OrthogonalUnitVectors[v:{_, _, _}] :=
With[{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 ] }, #/Sqrt[#.#]& /@ {u, Cross[u,v]}]---Selwyn Hollis>
There are many cases in graphics, and otherwise, where it is
useful to> obtain two orthogonal unit vectors to a given
vector. I know a number of> ways to do it, but they all seem
to be slightly inelegant. I thought I would> pose the
problem to MathGroup. Who has the most elegant Mathematica>
routine...> OrthogonalUnitVectors::usage =
OrthogonalUnitVectors[v:{_,_,_}] will return> 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/> 
====
The following
behavior seems to have gone undetected through
severalrevisions of Mathematica:When a SeriesData object
is multiplied by 0, Mathematica (versions 3.0,4.02, 4.1 for
Mac, and 4.0 for Windows) does not give 0. For
example,0*Series[f[x], {x,0,3}]gives O[x]^4, even though
0*x_SeriesData gives 0!Because of this, the following
incorrect result is obtained in
version4:DiagonalMatrix[{a,b}].{{x^2 + O[x]^3, O[x]^2},
{O[x]^2, x^2 + O[x]^3}}gives {{O[x]^2, O[x]^2}, {O[x]^2,
O[x]^2}} !!In this example version 3 gives the correct
result, even though,a*(x^2 + O[x]^3) + 0*O[x]^2gives O[x]^2
- same as in version 4. (Apparently, Dot in version 3
doesnot include in the sum terms multiplied by 0).The
easiest way to correct this bug, without altering Times, is
to useUpSet:Unprotect[SeriesData]; Times[0,x_SeriesData]^=
0; Protect[SeriesData];(ignore the UpSet::write message)I
wonder how many people have obtained incorrect results
because of thisbug...Sotirios
Bonanoshttp://www.inp.demokritos.gr/~sbonano/SB.html
====

In[4]:= f[x_ /; 0 < x < 6] := x^2In[5]:=f[x_/; x >=
6]:=x+1Meilleures salutationsFlorian
JaccardEICN-HES-----Message d'origine-----Envoy.8e :
sam., 31. ao.9et 2002 07:26è : mathgroup@smc.vnet.netObjet
: How to ...?Dear MathGroup,I'd like to use Mathematica 4.0
to write a function having differentexpressions in
different domain's intervals.Let's say:F[x_]= 
x^2 if 0<x<6
x+1 if x>=6I know It's a stupid syntax problem, but I really
do not know how/where tosearch the solution on the
Mathematica BookPasKo.
====
This works :pl[p_] :=
Plot[Sin[x^p], {x, 0, Pi}, DisplayFunction ->
Identity]lab[p_] := Graphics[{Text[p = , {1.5, 1.2}],
Text[p, {1.7, 1.2}]}]Table[Show[{pl[k], lab[k]},
DisplayFunction -> $DisplayFunction], {k, 3}]Meilleures
salutationsFlorian JaccardEICN-HES-----Message
d'origine-----Envoy.8e : sam., 31. ao.9et 2002 07:26è 
:
mathgroup@smc.vnet.netObjet : PlotLabelI plot several
curve depending on a parameter p and I want the PlotLabel
toshow the value of p.What is wrong with the following
?pl[p_]:=Plot[Sin[x^p],{x,0,Pi},PlotLabel[Rule] p =  p,
DisplayFunction[Rule]Identity]Table[Show[pl[k],
DisplayFunction[Rule]$DisplayFunction],{k,3}]
====
David,
Here is my solution, using
NullSpace:OrthogonalUnitVectors[v:{_,_,_}]:=(Needs[
LinearAlgebra`Orthogonalization`];
Map[LinearAlgebra`Orthogonalization`Normalize,
NullSpace[{v,{0,0,0},{0,0,0}}]])One problem with any
solution is that it should never be possible to obtainthe
two output vectors as continuous functions of the input
vector, sincethat would be equivalent to the combing of a
hedgehog in a vortexfree way.Ingolf
Dahlf9aid@fy.chalmers.seChalmers
UniversitySweden-----Original Message-----David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/===
=Adding my two cents to:> There are many cases in graphics,
and otherwise, where it is useful to> obtain two orthogonal
unit vectors to a given vector. I know a number of> ways to
do it, but they all seem to be slightly inelegant. I thought
I would> pose the problem to MathGroup. Who has the most
elegant Mathematica> routine...To this I would like to add
a criterion of smoothness. Armed with a secondvector b not
parallel to the given vector a, it's a trivial matter
toorthogonalize b WRT a by Gram-Schmidt and then form the
third vector c = a xb. (Normalize as needed.)I don't need
more elegance that this, but I would like a scheme to
selectthe vector b that results in a triad {a,b,c} that
various smoothly as avaries over all possible directions.
Each of my attempts to date involve abranched algorithm and
jumps in the resulting triad for certain smallchanges in
a.To David's call for elegance I add a call for
smoothness.Tom Burton
====
Mark,Just add a PlotRange option
and that each pentomino will be plotted to thesame
scale.Show[ GraphicsArray[ Partition[ Table[ Graphics[
Map[Rectangle[#, (# + {1, 1})] &, pentominoes[[i]]],
AspectRatio -> Automatic, PlotRange -> {{0, 5}, {0, 5}} ],
{i, 12}], 6] ]]David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/{{0
, 0}, {1, 0}, {2, 0}, {3, 0}, {4, 0}}, (*I*){{0, 0}, {1, 0},
{2, 0}, {3, 0}, {0, 1}}, (*L*){{0, 0}, {1, 0}, {2, 0}, {2,
1}, {3, 1}}, (*N*){{0, 0}, {1, 0}, {2, 0}, {0, 1}, {1, 1}},
(*P*){{0, 0}, {1, 0}, {2, 0}, {1, 1}, {1, 2}}, (*T*){{0, 0},
{1, 0}, {2, 0}, {0, 1}, {2, 1}}, (*U*){{0, 0}, {1, 0}, {2,
0}, {0, 1}, {0, 2}}, (*V*){{0, 0}, {1, 0}, {1, 1}, {2, 1},
{2, 2}}, (*W*){{1, 0}, {0, 1}, {1, 1}, {2, 1}, {1, 2}},
(*X*){{0, 0}, {1, 0}, {2, 0}, {3, 0}, {1, 1}}, (*Y*){{0, 0},
{1, 0}, {1, 1}, {1, 2}, {2, 2}} (*Z*)}When I display them
with the code below, the aspect ratio is correct
*withineach pentomino* but they are not scaled
equivalently. I would appreciate apointer to what I have
done wrong. The only solution I have been able todevelop so
far effectively places each of the 12 pentominoes at a
differentlocation in the plane, and then draws the whole
region in one go.Show[ GraphicsArray[ Partition[ Table[
Graphics[ Map[Rectangle[#, (# + {1, 1})] &,
pentominoes[[i]]], AspectRatio -> Automatic ], {i, 12}], 6]
]]--Mark R. Diamond
====
Does anybody have Benchmarks for
Mathematica 4.2 to share?If possible the same or similar
machine and different OSes.to the Windows version.I tried
http://www2.staff.fh-vorarlberg.ac.at/~ku/karl/timings40.
htmlvalue of 6.4.---more details--Times =
[InvisibleSpace]{1.18, 1.14, 1.08, 0.55, 1.18, 2.48,
0.74, 0.79, 0.41, 0.11, 0.38, 0.96, 0.97, 1.26, 1.11}Time =
[InvisibleSpace]17.Benchmark =
[InvisibleSpace]6.40784-----------------------------------
-------------I would like to see how Mathematica 4.2
improved in speedcompared to Mathematica 4 or 3.
Armin
====
Can I use Mathematica to find out the volumn of
this 3 dimensional object fromthe equations :z=x^2 +4,
y=4-x^2, y=3xShz Shz
====
Use Text statements. Click of a
coordinate point on the plot, copy it andpaste it into the
Text statement. Here is an example.Show[Graphics[
{Circle[{1, 1}, 2], Text[A, {1.75659, 2.03292}],
Circle[{0, 0}, 2], Text[B, {-1.38818, -0.0197372}],
Circle[{1, -1}, 2], Text[C, {1.57238, -1.99344}]}],
AspectRatio -> Automatic, ImageSize -> 400];To click off a
coordinate:1) Select the plot by putting the cursor in it and
clicking.2) Press and hold Ctrl. When you move the cursor you
will obtain crosshairs.3) Move the cursor to where you want
to put the text and left click.4) Right click and use Copy.
(Or use Ctrl-C or the menu).5) Paste the copied coordinate as
the second argument in the Text statement.David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/===
=On 8/31/02 at 1:25 AM, berlusconi_pagliusi@fis.unical.it
(Paskoski)>Dear MathGroup, I'd like to use Mathematica 4.0 
to
write a function>having different expressions in different
domain's intervals. Let's>say:>>F[x_]= x^2 if 
0<x<6 x+1 if
x>=6TryIn[4]:=F[x_]:=Which[0<x<6,x^2,x>=6,x+1]you can
plot the function withIn[5]:=Plot[F[x],{x,0,8}]Sincerely
yours,Dr. Juan E. Fuentes BetancourtFacultad de Fisica
Universidad de la Habana
====
I have spent an enormous
amount of time (far too much) on this question.Indeed, I
have just completed a program that handles all sorts
ofpiecewise defined functions. I am in the process of
writing it up tooffer it to all of you. In my opinion the
best, cleansest method is toinvoke the UnitStep function as
illustrated in David Park's solutions.To make everything
cleaner, define a characteristic functionChi[x_,a_,b_] :=
UnitStep[x-a]-UnitStep[x-b]This function is continuous on
the right, vanishes outside of [a,b) and isone in the
half-open interval [a,b). Now your answer is x^2 Chi[x,0,6]
+ (x+1) Chi[x,6,Infinity]The integrator handles UnitSteps
easily and smoothly.My program handles these cases easily
and many more, including suchoddities as
Integrate[Abs[x],x] and (for amusement sake only)UnitStep[
Abs[x]-Sign[x]+3]Another virtue of this approach (also
handled in my program) is thepeculiar error messages and
poor answers given by NIntegrate for simplepiecewise
continuous functions at jump discontinuites. The limit
functionalso fails dismally on some examples where it
really shouldn'tI am rather naive about how to transmit
programs to interested users, soif you would like a copy of
my program and about 100 worked examples, Iwill be glad to
send them to you IF YOU TELL ME HOW TO DO IT!Jack Goldberg>
In a message dated 8/31/02 1:58:36 AM,
berlusconi_pagliusi@fis.unical.it> I'd like to 
use
Mathematica 4.0 to write a function having different>
expressions in different domain's intervals.> 
Let's say:>
>> F[x_]= x^2 if 0<x<6> x+1 if x>=6>> I know It's a
stupid syntax problem, but I really do not know how/where
to> search the solution on the Mathematica Book>
Just for grins here are several methods:>> f1[x_/;x<=0] := 0
;>> f1[x_/;0<x<6] := x^2 ;>> f1[x_/;x>=6] := x+1;> f2[x_]
:=>> x^2*UnitStep[x]+(x+1-x^2)*UnitStep[x-6];> f3[x_] :=
Which[>> x<=0, 0,>> 0<x<6, x^2,>> x>=6, x+1];> f4[x_] :=
If[0<x<6,x^2,>> If[x>=6, x+1,0]];> f5[x_] := Switch[x,>>
_?(#<=0&), 0,>> _?(0<#<6&), x^2,>> _?(#>=6&), x+1 ]>
f6[x_?NumericQ] :=>> {0,x^2,x+1}[[Position[>>
{x<=0,0<x<6,x>=6}, True][[1,1]]]];>
Needs[Calculus`Integration`];>> (* needed for definition of
Boole *)> f7a[x_?NumericQ] := Evaluate[>> (Boole /@
{0<x<6,x>=6}).>> {x^2,x+1}];> Off[Part::pspec];>>
f7b[x_?NumericQ] := Evaluate[>> {0,x^2,x+1}[[1+Tr[Boole /@
{x>0, x>=6}]]]];> f8[x_?NumericQ] := Cases[>> {{x<=0, 0},
{0<x<6, x^2}, {x>=6, x+1}},>> {True, z_} :>z][[1]];>
f9[x_?NumericQ] := DeleteCases[>> {{x<=0, 0}, {0<x<6, x^2},
{x>=6, x+1}},>> {False, z_}][[1,2]];> f10[x_?NumericQ] :=
Select[>> {{x<=0, 0}, {0<x<6, x^2}, {x>=6, x+1}},>>
First[#]&][[1,2]];> f11[x_?NumericQ] := Last[Sort[>>
{{x<=0, 0}, {0<x<6, x^2}, {x>=6, x+1}}]][[2]];>
f12[x_?NumericQ] := Module[{n=1},>> While[{x<6,x<0,
False}[[n]], n++];>> {x+1,x^2,0}[[n]]];> Generating some
test points:>> ts = {Random[Real,{-5,0}],0,
Random[Real,{0,6}],6,Random[Real,{6,15}]};> Checking the
different representations>> Equal[(# /@ pts)& /@
{f1,f2,f3,f4,f5,f6,f7a,f7b,f8,f9,f10,f11,f12}]> True>> To
pick a favorite, look at how the different definitions
behave.> Of the definitions that evaluate with symbolic
input, only f2 and f4> simplify with assumptions> . For
example,>> FullSimplify[#[x]& /@ {f2,f4}, 1<x<3]> f2
through f5 respond immediately to differentiation> #'[x]&
/@ {f2,f3,f4,f5} // Simplify //ColumnForm> Only f2
responds immediately to integration>
Integrate[f2[x],x]//Simplify>> Consequently, f2 (UnitStep)
appears to be the most versatile.> Bob Hanlon> Chantilly,
VA USA>>Reply-To: jbrowne@swin.edu.au
====
> There are many
cases in graphics, and otherwise, where it is useful to>
obtain two orthogonal unit vectors to a given vector. I know
a number of> ways to do it, but they all seem to be slightly
inelegant. I thought I would> pose the problem to MathGroup.
Who has the most elegant Mathematica> routine...>>
OrthogonalUnitVectors::usage =
OrthogonalUnitVectors[v:{_,_,_}] will return> two unit
vectors orthogonal to each other and to v.>> You can assume
that v is nonzero.David, here is a solution generating two
random vectors: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]]}]Test:v = {Random[], Random[],
Random[]}{0.864587, 0.727747, 0.669729}{A,B} =
OrthogonalUnitVectors[v]{{0.279985, -0.808701, 0.517311},
{-0.698881, 0.19773, 0.687363}}Chop[{A.v, B.v, A.B, A.A,
B.B}]{0, 0, 0, 1., 1.}John--
_________________________________John BrowneSchool of
Engineering and ScienceSwinburne University of
TechnologyJohn Street, Hawthorn, Victoria,
AustraliaQuantica phone: +613 9431 4007Quantica fax: +613
9431 0940
====
In my previous post, I
proposedOrthogonalUnitVectors[v:{_, _, _}] := With[{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 ] }, #/Sqrt[#.#]& /@ {u, Cross[u,v]}]The trouble with
this is that w ends up being a global variable. The only way
I see around that is to use Module instead of With. (May as
well put in a Return[$Failed]
too.)OrthogonalUnitVectors[v:{_, _, _}] := Module[{u, w}, u
= Which[(w = {0,v[[3]],-v[[2]]}).w != 0, w, (w =
{v[[3]],0,-v[[1]]}).w != 0, w, (w = {v[[2]],-v[[1]],0}).w !=
0, w, True, Return[$Failed]]; #/Sqrt[#.#]& /@ {u, Cross[u,
v]} ]----Selwyn Hollis