> 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 certainly
not 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
freeware
button tools from http://www.mertig.com/mathdepot ) and you get the button.
With the HP and VP you can easily and quickly generate (horizontally
or
verically) 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
Buttondesign 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
X
product from http://www.episoft.com
Rolf Mertig
Mertig Consulting
http://www.mertig.com
====
try
In[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,abc+d,a}
Out[5]=a+b
In[6]:=g[a_,b_] =s01
Out[6]=a+b
In[7]:=g[1,3]
Out[7]=2
*NEVER* use capital letters at the beginning of a variable's name! Never!
Matthias Bode
Sal. Oppenheim jr. & Cie. KGaA
Koenigsberger Strasse 29
D60487 Frankfurt am Main
GERMANY
Mobile: +49(0)172 6 74 95 77
Internet: http://www.oppenheim.de
UrsprÌ.b9ngliche Nachricht
Gesendet: Freitag, 23. August 2002 06:25
An: mathgroup@smc.vnet.net
Betreff: := 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 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,ABC+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?
====
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 Kozlowski
Toyama International University
JAPAN
On 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,ABC+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]];
?a
Global`a
a[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 RHS
a[A_,B_]:=
Evaluate[LinearSolve[X,Y][[1]]];
?a
Global`a
a[A_, B_] := A + B
a[1,3]
2
Bob Hanlon
Chantilly, 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]^n
which is supposed to hold true for every z>0. Rewrite it as
Integrate[[Sigma]norm[s]^n*s, {s, 0, d}]/d^3 =[Sigma]norm[z]^n/3z
for 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 equation
In[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]=
True
That means you can take g to be an arbitrary function of d.
Andrzej Kozlowski
Toyama International University
JAPAN
On Friday, August 23, 2002, at 05:25 AM, Toshiyuki ((Toshi)) Meshii
> How can I solve the following integral equation?
> Mathematica seems not to work.
> Is there any way?
DSolve[((3*z)/d^3)*Integrate[[Sigma]norm[z]^n*z, {z, 0, d}] ==
> [Sigma]norm[z]^n, [Sigma]norm[z], z]
note: z>0 & n>1
I know that the answer is simple and
> $B&R(Bnorm[z_] = (1 + 1/(2*n))*(z/d)^(1/n)
Toshi
>
====
But suddenly the following Errormessage appears:
/usr/bin/local/mathematica:
file or directory not found
Mandrakeupdates ... but so far everything except
of Mathematica seems to work fine.)
Of course I checked for the File (it is indeed there)
and (a hint from a Unixusegroup) 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 would
automatically open a browser page, and give my username and pwd to log me
into a https:.... site? The server's login page has a script resetting
fields:
and the relevant input cells are
Nicholas
====
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
singleargument function that holds two of those arguments constant (i.e.
without creating that singleargument 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 radix10 multiplier7 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
====
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/howrebuildinde
x.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).
====
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
ShiftEnter, which selects a cell for evaluation after all current
evaluations have finished.
See
http://support.wolfram.com/mathematica/kernel/interface/selectionevaluate.ht
ml
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 realtime 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,ABC+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. So
a[A_,B_] := LinearSolve[X,Y][[1]]
a[1,3]
is similar to doing
ReleaseHold[
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+B
In[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 Consulting
The final answer to your Mathematica needs
Spend 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 headscratching, 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]
]//DisplayForm
Some random things to note:
 ButtonEvaluator>Automatic. This says use the kernel to implement the
ButtonFunction.
 Buttons only create sideeffects. 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 sideeffect 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 sideeffect 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.nb
http://library.wolfram.com/conferences/devconf2001/horton2/horton2.nb

Omega Consulting
The final answer to your Mathematica needs
Spend 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 would
like to use DigitBlock>3 option of NumberForm to format large numbers
appearing 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 x2y3x3y2x1y3+x3y1+x1y2x2y1, 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 appearsnot 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
3014055176
gah@math.umd.edu
====
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
====
>
>
> 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
during the Replaceall (/.) Mathematica first evaluate the rule
SomeHead[q_] > Flatten[q] . q is stilla symbol then and will be
replaces 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)
GottliebDaimlerStrasse, Geb. 7^2=49/313 D67663 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}, Mathematica
complains:
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 generated
saying essentially nothing to flatten.
*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, >, since
Flatten[q] would not then be evaluated.
z:=SomeHead[{{1,2},{3,4}}]
z/.{SomeHead[q_]:>Flatten[q]}
{1,2,3,4}

Allan

Allan Hayes
Mathematica Training and Consulting
Leicester UK
www.haystack.demon.co.uk
hay@haystack.demon.co.uk
Voice: +44 (0)116 271 4198
====
Never mind my question, I should have used a delayed rule there (:> instead
of >).
Bit silly of me.
====
Suppose f is a function of n real variables, and returns a
vector of n real variables. What is the correct syntax to find a root
of f using FindRoot?
For instance, the following works with n=7, but I would need to change
the 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
>singleargument function that holds two of those arguments constant (i.e.
>without creating that singleargument 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 radix10 multiplier7 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]
True
Bob Hanlon
Chantilly, 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 RuleDelayed
z/.{SomeHead[q_]:>Flatten[q]}
{1, 2, 3, 4}
Bob Hanlon
Chantilly, VA USA
====
Can anyone help me with this problem.
If I have an nelement 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.
====
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 Hayes
Mathematica Training and Consulting
Leicester UK
www.haystack.demon.co.uk
hay@haystack.demon.co.uk
Voice: +44 (0)116 271 4198
> Can anyone help me with this problem.
If I have an nelement 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 nelement 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)
GottliebDaimlerStrasse, Geb. 7^2=49/313 D67663 Kaiserslautern /
====
some function with three arguments, and I want to syntactically describe
the
singleargument function that holds two of those arguments constant (i.e.
without creating that singleargument 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 radix10 multiplier7 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
====
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 Park
djmp@earthlink.net
http://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 Kozlowski
Toyama International University
JAPAN
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 write
In[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 write
NestList[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 was
undefined before you tried this, so Flatten[q]'s evaluation yielded itself.
Later, in the process of substitution, with q set as you intended, Flatten
evaluated again, this time yielding the desired result.
Try it again after setting q to something. Then you get not only the
complaint but also the wrong answer.
To avoid these issues, use the delayed operator :> instead of the prompt
operator >. A rule of thumb is: Use delayed operators := and :> when the
LHS (SomeHead[q_] in your case) contains a blank (_).
Tom Burton
====
Mark:
In: ?K
Out: K is a default generic name for a summation index in a symbolic sum.
Turns out that there are seven singleletter 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 Mathgroup
I have convert pde to ode like this
du_i/dt = 14 u_i + .02 (u_i1 ö 2 u_i + u i+1)/(delta (x))^3
+ (u_i)^3
v_i
dv_i /dt = 3 u_i + .02 (v_i1 ö 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]
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 help
Khaled
_________________________________________________________________
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 = 14 u_i + .02 (u_i1 ? 2 u_i + u i+1)/(delta (x))^3 + (u_i)^3
> v_i
>
> dv_i /dt = 3 u_i + .02 (v_i1 ? 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 use
NDSolve[{
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 use
NDSolve[{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_i1 ? 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_i1 ? 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 use
NDSolve[eqns, Join[U[t], V[t]], {t, 0, 10}, Method>RungeKutta]
This uses the RungeKuttaFehlberg 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 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?
Please suggest.
Raj
====
Is there an easy (elegant?) way to generate the set of all ktuples
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}}.
====
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 Hayes
Mathematica Training and Consulting
Leicester UK
www.haystack.demon.co.uk
hay@haystack.demon.co.uk
Voice: +44 (0)116 271 4198
> Is there an easy (elegant?) way to generate the set of all ktuples
> 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:
< Is there an easy (elegant?) way to generate the set of all ktuples
> 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 ktuples
> 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 the
integration of above?
In particular I am working with a 4 gaussian process with joint pdf
p(x1,y1,x2,y2) transformed to> p(r1,r2,theta1,theta2), the processes
have nonzero mean and I would like to integrate the joint pdf twice
(over r1 and r2) and obtain an expression for the pdf of the phase
angles between p(theta1,theta2). I could assume that the means are the
same (to make life easier). Has anyone come accross this  it doesn't
seem that unusual to me. I am asking because the integrals are
becoming 'interesting'.
thankyou for any tips.
====
Start Excel and go to Tools > Macro > Security..., then on the
Security
Level 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 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 (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 MartinSerrano
>
>
====
Download the newest patch from wolfram research  this should theoretically
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 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 (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 MartinSerrano
>
====
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 should
theoretically
> 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 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 (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 MartinSerrano
>
>
====
>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:
>> {xx=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^n1)/2, (3^n1)/2}.
Consequently, the largest of the weights can not exceed
(3^n1)/2  (1+2+...+(n1)) or
((3^n1)  n(n1))/2
34
Thread[var>#]& /@
(First /@ Select[{var,f} /. Thread[var>#]& /@
KSubsets[Range[((3^n1)  n(n1))/2], n],
Sort[#[[2]]] == Range[(3^n1)/2,(3^n1)/2]&])
{{a > 1, b > 3, c > 9, d > 27}}
Bob Hanlon
Chantilly, 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]:= Infinity
It should be 0
Jonathan Rockmann
mtheory@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 Hollis
slhollis@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.html
that WRI advertises comes with 4.2?
Jonathan
mtheory@msn.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.
====
G'day,
Looks like you forgot to define base in your Machine function and when
using pure functions, don't forget the ampersand.
Thus
NestList[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
> singleargument function that holds two of those arguments constant (i.e.
> without creating that singleargument 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 radix10 multiplier7 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 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_] := uv
Calling it with g[{x1,y1},{x2,y2}] one would expect the answer {x1x2,
y1y2} 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).
====
I do not know why you are getting a message, using your own code I get the
result 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_] := uv
> Calling it with g[{x1,y1},{x2,y2}] one would expect the answer {x1x2,
> y1y2} 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_] := uv
> Calling it with g[{x1,y1},{x2,y2}] one would expect the answer {x1x2,
> y1y2} but instead one gets error messages
Works fine for me.
Please check that you have no existing definitions for the symbols used.
Clear[g,x1,x2,y1,y2];
g[u_,v_]:=uv
g[{x1,y1},{x2,y2}]
{x1x2,y1y2}

Allan

Allan Hayes
Mathematica Training and Consulting
Leicester UK
www.haystack.demon.co.uk
hay@haystack.demon.co.uk
Voice: +44 (0)116 271 4198
> 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_] := uv
> Calling it with g[{x1,y1},{x2,y2}] one would expect the answer {x1x2,
> y1y2} 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_] := uv
> Calling it with g[{x1,y1},{x2,y2}] one would expect the answer {x1x2,
> y1y2} 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 You
really use the minus character? Maybe some erroneous code was assigned
to g before. Restart Mathematica or use Remove[g] to really start up
from the beginning.
Sincerly
Alexander

/ Alexander Dreyer, Dipl.Math.  Abteilung Adaptive Systeme
/ Fraunhofer Institut fuer Techno und Wirtschaftsmathematik (ITWM)
GottliebDaimlerStrasse, Geb. 7^2=49/313 D67663 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 trifle 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 ed ea 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 x2y3x3y2x1y3+x3y1+x1y2x2y1, 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 appearsnot 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
3014055176
gah@math.umd.edu
====
I would like to have a list of all the directives that can be used for
command in the Windows Start Menu). I am especially interested in a
directive with which you can appoint a file which has to be evaluated by
the
kernel after it has been launched. I imagine that such a directive looks
like 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
straightforward
to write a function that uses
> 3) Plot3D the data
This will be tricky  ListSurfacePlot3D plots f[x, y]. ListContourPlot3D
which 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 function
to 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 couple
of comments.
The following is an example of yet another type of button that you can use
as a model. Copy and paste the following code into a notebook. When you
evaluate it, it will create a button within the same notebook which when
pressed evaluates the entire notebook. Sort of like selecting from the
menu
is something in the notebook already to evaluate. So input 1+2 (and don't
hit enter) somewhere in the notebook so that you can see what
the 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
this
happen 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,
useful
programs that can serve as models when programming. If you look at
some of the MathGroup archived threads you will find similar complaints
that the Mathematica documentation is confusing.
Jonathan Rockmann
mtheory@msn.com
 Original Message 
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 bee
n
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.
====
This was REALLY interesting. Here's a solution that looks only at the
7,560 relevant combinations. It first chooses three numerators. Then
it chooses two denominators for the first fraction. Then two
denominators for the second fraction. The last fraction is determined
at 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, #]] & /@ e
j[{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 is
1/(3*6)+5/(8*9)+7/(2*4)
Bobby Treat
Original Message
University Professor of Philanthropy and the Law
Director, National Center on Philanthropy and the Law
New York University School of Law
Room 206A
110 West 3rd Street
New York, N.Y. 100121074
Original Message
byran
________________________________________________________________________
service. For more information on a proactive antivirus service working
around the clock, around the globe, visit http://www.messagelabs.com
________________________________________________________________________
====
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 trifle 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 ed ea 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 x2y3x3y2x1y3+x3y1+x1y2x2y1, 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 appearsnot 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
3014055176
gah@math.umd.edu
====
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
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 ed ea 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 x2y3x3y2x1y3+x3y1+x1y2x2y1, 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 appearsnot 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
3014055176
gah@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 using
Windows 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]
Infinity
Bobby 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]
Infinity
That limit is zero. For the Sieve to be polynomial, we only need the
sequence 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 test
vertices only. I haven't coded it yet, but the LeftSide idea seems
like a good one.
It is sufficient to test whether all vertices of one convex polygon are
on the left (out) side of some side of the second polygon (both polygons
in clockwise order). If that happens for any side of either polygon,
the polygons don't intersect. In the cross example, some vertices are
to 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 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 x2y3x3y2x1y3+x3y1+x1y2x2y1, 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 appearsnot 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
3014055176
gah@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 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},tendt]},{t,tstart,tend,tstep}]],AspectRatio>Automatic,Axes>Tru
e,PlotLabel>Mach
Sqrt(vx^2+vy^2)];
doppcone[vx_,vy_,tend_,tstep_]:=Show[Graphics[Table[{Hue[t/tend],Circle
[{vx*t,vy*t},tendt]},{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]]]*(tendtstep)*sqrt[vx^2+vy^2]

tstep*Sin[If[vx==0,0,ArcTan[vy/vx]]],Sin[If[vx==0,0,ArcTan[vy/vx]]]*Sign[vx]*
(tendtstep)*sqrt[vx^2+vy^2]+tstep*Cos[If[vx==0,0,ArcTan[vy/vx]]]}}]],Graphic
s[Line[{{tend*Sin[If[vx==0,0,ArcTan[vy/vx]]],tend*Cos[If[vx==0,0,ArcTan[vy/v
x]]]},{Sign[vx]*Cos[If[vx==0,0,ArcTan[vy/vx]]]*(tendtstep)*sqrt[vx^2+vy^2]+t
step*Sin[If[vx==0,0,ArcTan[vy/vx]]],Sign[vx]*Sin[If[vx==0,0,ArcTan[vy/vx]]]*(
tendtstep)*s
rt[vx^2+vy^2]tstep*Cos[If[vx==0,0,ArcTan[vy/vx]]]}}]],AspectRatio>Automati
c,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
image
size; 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},tendt]},
{t,0,tend,tstep}],
Text[Mach <>ToString@Sqrt[vx^2+vy^2]
,{0,.5+tend},{1,0}]
}],AspectRatio>Automatic,Axes>True,
PlotRange> {{tendmargin,margin+Max[tend,(vx tend)+.5]},
{tendmargin,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},tendt]},{t,tstart,tend,tstep}]],AspectRatio>Automatic,Axes>Tr
ue,PlotLabel>Mach
> Sqrt(vx^2+vy^2)];
doppcone[vx_,vy_,tend_,tstep_]:=Show[Graphics[Table[{Hue[t/tend],Circle
>
[{vx*t,vy*t},tendt]},{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]]]*(tendtstep)*sqrt[vx^2+vy^2]
> 
tstep*Sin[If[vx==0,0,ArcTan[vy/vx]]],Sin[If[vx==0,0,ArcTan[vy/vx]]]*Sign[vx]
*(tendtstep)*sqrt[vx^2+vy^2]+tstep*Cos[If[vx==0,0,ArcTan[vy/vx]]]}}]],Graph
ics[Line[{{tend*Sin[If[vx==0,0,ArcTan[vy/vx]]],tend*Cos[If[vx==0,0,ArcTan[v
y/vx]]]},{Sign[vx]*Cos[If[vx==0,0,ArcTan[vy/vx]]]*(tendtstep)*sqrt[vx^2+vy^
2]+tstep*Sin[If[vx==0,0,ArcTan[vy/vx]]],Sign[vx]*Sin[If[vx==0,0,ArcTan[vy/vx
]]]*(tendtstep)*s
>
rt[vx^2+vy^2]tstep*Cos[If[vx==0,0,ArcTan[vy/vx]]]}}]],AspectRatio>Automati
c,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 how
functions 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
> ShiftEnter, which selects a cell for evaluation
> after all current
> evaluations have finished.
>
> See
>
http://support.wolfram.com/mathematica/kernel/interface/selectionevaluate.ht
ml
>
> Dale
>
__________________________________________________
Do You Yahoo!?
Yahoo! Finance  Get realtime stock quotes
http://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^27*x+10, x]
{{x > (1/2)*(7  Sqrt[9 + 4*y])}, {x > (1/2)*(7 + Sqrt[9 + 4*y])}}
x^27*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 == (x3)/(x+2), x]
{{x > (3  2*y)/(1 + y)}}
(x3)/(x+2) /. % // Simplify
{y}
Bob Hanlon
Chantilly, 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 as
g /@ (#[r,theta,phi]& /@ {f1,f2,f3})
{g[f1[r, theta, phi]], g[f2[r, theta, phi]], g[f3[r, theta, phi]]}
Bob Hanlon
Chantilly, VA USA
====
> Is there an easy (elegant?) way to generate the set of all ktuples
> 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}]],n1];
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}]]
1024
Bob Hanlon
Chantilly, VA USA
====
> If I have an nelement 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 Sequence
kTuples[n_Integer?Positive,s_List]:=
Flatten[Outer[List,Sequence@@Table[s,{n}]],n1];
s = {a,b,c,d,e}; n =3;
Length[kTuples[n,s]] == Length[s]^n
True
Bob Hanlon
Chantilly, 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 one
big list
and Outer is expecting a sequence of N separate lists after
SomeFunction.
====
f[x_] = x^2  7*x + 10
g[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 the
time. 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 {x1x2,
y1y2} 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[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 Garza
Mexico 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.net
Approved: Steven M. Christensen , Moderator
====
The trick is to use the Sequence Function. Try the following code and you
will 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 of
arguments to outer. You need the third argument to operate on the parts of
the list if this is your goal.
Read the description of Sequence. It also works in contexts other than
Outer.
Richard Palmer
> Can anyone help me with this problem.
>
> If I have an nelement 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 Group
I would like to know, if is possible to solve
Integrate[(exp[a/x])/(x^2b^2),{x,0,infinity}] with
a and b constant ( Real )
using Mathematica.
Valdeci Mariano
*********************************************************************
Valdeci Mariano de Souza
Master«s Degree of Applied Physics  Unesp/Rio Claro 
State of S.8bo
Paulo  Brazil
Laboratory of Electrical Measurements
phone : ( 0XX19 ) 526  2237
*********************************************************************
====
> Is there an easy (elegant?) way to generate the set of all ktuples
> 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 Pratt
Department of Operations Research
http://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 Graphics
Expressions 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 like
this:
(* This code is necessary for setting up J/Link *)
< {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 comanumbers?
====
I'm a poor physicist trying to figure out how to sort out the
physical from the nonphysical 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
====
I had looked at using Cases, but had gotten tripped up by
forgettting 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 nonphysical 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[4x^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 all
levels including level 0, which is the whole expression. The search
will
not 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 Hayes
Mathematica Training and Consulting
Leicester UK
www.haystack.demon.co.uk
hay@haystack.demon.co.uk
Voice: +44 (0)116 271 4198
> I'm a poor physicist trying to figure out how to sort out the
> physical from the nonphysical 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
fairly
common. The first command is Apply (@@ in prefix form) and the second
command 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 @@ arglist
f[{a, b}, c, {d, e}, 3, Report > True]
If you want to insert them into another function, h, that also has other
arguments, 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 Park
djmp@earthlink.net
http://home.earthlink.net/~djmp/
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.
====
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 flatten it to
get
it down to a twolevel array.
David Park
djmp@earthlink.net
http://home.earthlink.net/~djmp/
Sender: steve@smc.vnet.net
Approved: Steven M. Christensen , Moderator
====
Raj,
You could include an index for the function in the argument list, or make a
clear separation between parameters and variables as follows (I just
use
two 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, by
Sum[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 to
the 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 functions
with 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]^2
If possible, try to steer away from For and Do loops and use functional
programming as much as you can. At first it may seem strange, but it is
much
more powerful and easier once you get used to it. Ask further questions to
MathGroup with SPECIFIC examples and you will get a lot of help on how to
use functional programming.
David Park
djmp@earthlink.net
http://home.earthlink.net/~djmp/
Sender: steve@smc.vnet.net
Approved: Steven M. Christensen , Moderator
====
David,
Except in a few cases, Mathematica will not automatically generate the
inverse function for you. You will have to Solve the equations and
construct
the inverse function yourself. In two of your three cases there are
actually
multiple inverse functions.
Taking the easy one first.
Clear[x];
Solve[f[x] == f, x][[1, 1]] // Simplify
x[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 this
right)
Clear[x];
x[1, n_][f_] = (ArcSin[f] + 2 Pi n)/3
x[2, n_][f_] = (Pi + ArcSin[f] + 2Pi n)/3
Here 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}] //
Sort
f /@ %
{((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 some
work.
David Park
djmp@earthlink.net
http://home.earthlink.net/~djmp/
Sender: steve@smc.vnet.net
Approved: Steven M. Christensen , Moderator
====
g[u_, v_] := u  v
g[{x1, y1}, {x2, y2}]
{x1  x2, y1  y2}
David Park
djmp@earthlink.net
http://home.earthlink.net/~djmp/
y1y2} 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).
====
>If I have an nelement 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,
with
the 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 also
includes the input command?
The reason that this is important for me is that I am doing an
animation; and after the animation I would like to delete all the
graphic images; but when I use Show[Graphics[ this would mean deleting
the input command as well.
Could someone explain this to me and also make a suggestion so how I
could 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 was
only for rectangles oriented parallel to the X and Y axis, as this is
perhaps the predominant application.
The idea was, to decide whether all vertices of a rectangle lie to one side
of the other. This idea expands to the general case. And effectively it is
the same idea as of your suggestion, as well as of the improved version
after Garry Helzer, i.e. the constructive solutions, as opposed to
those
involving more abstract mathematical reasoning as of David Park and Andrzej
Kozlowski. And these, esp. Andrzej's, are very interesting, as they show
how
to derive a solution just from this kind of reasoning.
First some preliminaries: random rectangles
I had chosen to describe the rectangles as a list of counterclockwise
vertices. 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 it
elegant:
offSide[r2_][{p1_, p2_}] := And @@ ((p2  p1).(#  p1) <= 0 &) /@
r2
rectOverlap[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 oriented
rectangles to the general case. I first tried oblique coordinates, sighting
along the sides of the rectangles, then saw that giving them in the dual
base is computationally simpler, to recognize i.e. just the distance of a
point from the side's (straight line).
Correctness:
rectOverlap[rect[1], rect[2]]
True
rectOverlap[rect[1], rx]
False
rectOverlap[rect[1], rxx]
True
Performance:
(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 from
different reasoning.)
rightSide[a_, {b_, c_}] := Det[Prepend[#, 1] & /@ {a, b, c}] < 0
vertexExcluded[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 === testr4
True
of quite similar performance.
Bobby Treat gave a different implementation of this (as seen below) which
is
marginally faster.
Now here Bobby's solution, rewritten in my style (which makes it slightly
faster):
extent[r1_, r2_] := {Min[#], Max[#]} & /@
((r1[[{1, 2}]]  r1[[{2, 3}]]).Transpose[r2])
cannotIntersect[{{min1_, max1_}, {min2_, max2_}}] :=
max2 < min1  max1 < min2
intersects[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 === testr1
True
the tests for all sides. Such we come to use of the nonstrict evaluation
of
And[ ] 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 === testr1
True
Equal in performance to Bobby's solution. The use of And with ((p2  p1).(#
 p1) <= 0 &) /@ And @@ r2, etc. is a bit tricky: it effectively prevents
the evaluation of (p2  p1).(q1  p1) <= 0 etc., such that this is
evaluated
within And (nonstandard evaluation) after mapping. And @@ ((p2  p1).(# 
p1) <= 0 &) /@ r2 to the contrary is less performant. (You can see here how
clever the language is designed, to allow use of And as a container, until
the time comes to execute!)
This solution can be improved a little bit; we replace mapping by threading
and delay execution of the pure functions' body to the application of And
now 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 === testr1
True
So this is fastest by a small margin, but has lost almost all elegance.
The
outer Unevaluated is necessary to prevent evaluation within Thread, the
inner 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 trifle 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 ed ea 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
>x2y3x3y2x1y3+x3y1+x1y2x2y1, 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 appearsnot 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
>3014055176
>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 Elster
Computer Science Dept.
Technion I.I.T.
Office: Taub 411
====
> Here's my contestant:
>
> <
> 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[n1, 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^2
Meilleures salutations
Florian Jaccard
professeur de Math.8ematiques
EICNHES
Message d'origine
Envoy.8e : mar., 27. ao.9et 2002 08:08
Ë : mathgroup@smc.vnet.net
Objet : How do I pick out the expression under a radical?
I'm a poor physicist trying to figure out how to sort out the
physical from the nonphysical 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
====
> I'm a poor physicist trying to figure out how to sort out the
> physical from the nonphysical 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[4x^2];
InequalitySolve[#>=0, x]& /@
Cases[expr, Sqrt[x_] > x, Infinity]
{2 <= x <= 2}
Bob Hanlon
Chantilly, VA USA
====
> I would like to know, if is possible to solve
>
> Integrate[(exp[a/x])/(x^2b^2),{x,0,infinity}] with
>
> a and b constant ( Real )
>
> using Mathematica.
>
Integrate[(Exp[a/x])/(x^2b^2),{x,0,Infinity},
GenerateConditions>False]//FullSimplify
Sqrt[(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])/b
FullSimplify[%, 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 Hanlon
Chantilly, VA USA
====
Bob,
Table[ Plot[ Sin[n x], {x, 0, Pi}], {n, 1, 5}]
and
Table[ 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 separate
from the Input cell.
David Park
djmp@earthlink.net
http://home.earthlink.net/~djmp/
The reason that this is important for me is that I am doing an
animation; and after the animation I would like to delete all the
graphic images; but when I use Show[Graphics[ this would mean deleting
the input command as well.
Could someone explain this to me and also make a suggestion so how I
could 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
are
represented as Power[a,1/2] and if they are in the denominator they are
a 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 Park
djmp@earthlink.net
http://home.earthlink.net/~djmp/
Steve Beach
asb4@psu.edu
http://www.thebeachfamily.org
====
Valdeci,
It appears that the integral will not converge if b is real. Mathematica
gives...
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
define
a 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]))
0.254022
David Park
djmp@earthlink.net
http://home.earthlink.net/~djmp/
Master«s Degree of Applied Physics  Unesp/Rio Claro 
State of S.8bo
Paulo  Brazil
Laboratory of Electrical Measurements
phone : ( 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*i
Peter
====
Lucas,
One way to change the precedence of CirclePlus is to change the file
UnicodeCharacters.tr.
On my machine the file is located under
../4.1/SystemFiles/FrontEnd/TextResources
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 of
the
file before you make any changes. Also, 420 is low enough to get the
behavior you desire, but you may want to experiment with other precedences.
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
>
====
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 builtin 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/
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 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 ?
2) What do the error messages mean ?
Please keep in mind that this example is made up and trivial to solve
without 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 do
FindRoot[{q1 + q2 Sqrt[q2.q2], q1 + q2},{q1,{0,0}},{q2,{0,0}}]
which returns
In[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
derivative
FindRoot::frnum: If FindRoot cannot resolve a list of equalities, it
looks for
a 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 but
it is still able to run fluently?
Gory
====
You could try obfuscating it, i.e., scrambling the symbol names
except for those you wish to export.
Check out obfuscation tools under
http://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 langauges
for which we have definitions, ... and we happen to have
a definition of Mathematica.

Ira Baxter, Ph.D. CTO Semantic Designs
www.semdesigns.com 5122501018
> 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 fluently?
> Gory
>
====
Mathematica version 4.2 has no problem with this one. All you have to
do 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 Souza
Master«s Degree of Applied Physics  Unesp/Rio Claro 
State of S.8bo
Paulo  Brazil
Laboratory of Electrical Measurements
phone : ( 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 of
Times, 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^2
Bobby Treat
Original Message
can't find it. Any help would be appreciated.
Steve Beach
asb4@psu.edu
http://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 builtin 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
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 influence 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.

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 influence 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 an
appropriate 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 product
should 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 to
implement a Clocked Objective Function package based (primarily) on
these publications:
A Lagrangian Formulation of Neural Networks I: Theory and Analog
Dynamics
A Lagrangian Formulation of Neural Networks II: Clocked Objective
Functions and Applications
Both are found in Neural, Parallel and Scientific Computations 6
(1998) 297372, 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, which
causes 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 MathCanvas
using Swing; this is pretty simple: you can use the evaluateToImage
methods, which return a byte array containg a GIF Image; if you want
to make it very simple, you just make your control a JPanel that
contains 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();^M
byte [] 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, which
causes 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 MathCanvas
using Swing; this is pretty simple: you can use the evaluateToImage
methods, which return a byte array containg a GIF Image; if you want
to make it very simple, you just make your control a JPanel that
contains 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();^M
byte [] 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 wonder if someone knows a fast way to import data from a tabseparated
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. When
you 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]
1
peter2[10^6,1]//N
0.5 + 0.707107*I
peter2[10^6,1]//N
0.5 + 0.707107*I
Bob Hanlon
Chantilly, 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, use
the Locked attributed.
(3) To finally answer your question, use the ReadProtected attribute on
symbols 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 a
complex result? The integral involves multibranched functions (a fact
masked
a bit by the form Tan[p] in the integrand; try replacing Tan[p] by p to see
what is going on). Is your result on the branch you want? I suspect not;
you
are probably after a real result. The following modification produces a
real
result:
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 second
order singularity at Abs[b] in that case. Arg[b^2]!=0 is used to
express this because the issue is whether that singularity is on the
positive xaxis.
Behavior at 0 and Infinity are fine if Re[a]>0, but otherwise those are
problems too, so Mathematica has the right answer in those terms. I
can't vouch for the formula it comes up with when the conditions are
met, but it's not closedform in the usual sense anyway; but it's
closedform 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's
define
a 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]))
0.254022
David Park
djmp@earthlink.net
http://home.earthlink.net/~djmp/
Master«s Degree of Applied Physics  Unesp/Rio Claro 
State of S.8bo
Paulo  Brazil
Laboratory of Electrical Measurements
phone : ( 0XX19 ) 526  2237
*********************************************************************
====
>
>
> I solved the epidemic SIR ODE System
> (
> approxsolutions=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}, {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
There was a discussion on this topic three years ago. The URL below will
take you to a reply in that thread that includes remarks and code from
Carl Woll and myself. I'll also show some code for your specific
example, in order to indicate potential problems.
http://library.wolfram.com/mathgroup/archive/1999/Oct/msg00104.html
Below is some mildly clumsy code tailored to your example. It will do
the DE solving given parameter values, form an objective function
(sumofsquares type) given actual data and solutions for specific
parameters, and to do the minimization. I did everything at machine
precision but this is not necessarily a good idea. Also I do no error
checking e.g. for cases where the numeric DE solver fails to solve over
the full range. A related issue is that, depending on roughly where the
correct parameter values live, and how close you are in your initial
guess, this might not be at all a wellbehaved 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 solve
the system, evaluate at the same 't' values as are used for our data,
subtract resulting vectors from corresponding data vectors, take norm
squares, 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 something
nonzero 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]= 72.5289
With the above objective function we can call FindMinimum. For this we
will use two starting values for each parameter so that FIndMinimum can
use a secant method. An alternative to this, wherein one passes along a
gradient black box evaluator, is presented by Carl Woll in his note at
the 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 a
result reasonably fast and quite accurate, albeit with many warnings
indicated. 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.
11
Out[87]= {8.46495 10 , {a$4965 > 0.01, b$4965 > 1.}}
It still does reasonably well when I perturb by 50% from the correct
parameter values. This time the warning message indicates that we might
be closer to trouble.
In[89]:= findParams[data, {.015,1.5}]
FindMinimum::fmcv:
Failed to converge to the requested accuracy or precision within 30
iterations.
10
Out[89]= {6.48159 10 , {a$8747 > 0.01, b$8747 > 1.}}
You might want to play with various options in FindMinimum such as
MaxIterations. You might also need to use NDSolve options e.g.
WorkingPrecision in order to improve the quality of the result.
Daniel Lichtblau
Wolfram Research
====
I solved the epidemic SIR ODE System
(
) numerically using NDSolve:
approxsolutions=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}, {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
ReplyTo: 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
> (
> approxsolutions=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}, {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}]//Timing
Out: {5.7 sec, {14.1967, {a > 0.000999599, b > 0.200351}}}

Selwyn Hollis
>
> I solved the epidemic SIR ODE System
> (
> approxsolutions=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}, {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 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
====
Perhaps you could use ReadList with the option Word, which will read your
data in as strings. Then you may substitute commas with periods and convert
to numbers with ToExpression?
Tomas Garza
Mexico 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 comanumbers?
>
====
Yes, Import is slow, I suppose because of it's generality. Your particular
case 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 fluently?
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 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.
ReplyTo: nklepeis@uclink4.berkeley.edu
====
Guillermo,
indoor air exposure to particulate matter. Does your package use
RungeKutta or some other numerical algorithm? I'm thinking of
something like that, but with a fixed (nonadaptive) 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: 8317689510
====
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 correct
105. This looks like a bug. Any thoughts on this?
(Note: Number 105 does not look like that on the notebook page. It shows
the indefinite integral with an integral symbol followed by f[t] dt
instead of the function Integrate[ ]. I've spelled it out because the
Vista Research, Inc.
====
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 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 get
In[2]:= a b c /. a c :> x
Out[2]= b x
the 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 :> x
Out[3]= a b c
In 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 z
Out[4]= b x¬
Hope that will help you,

Rainer Gruber
====
I want to get the coefficient of the LegendreGauss 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/LegendreGaussQuadrature.html
I need more coefficients to get accurater result.
How about the method of the Integration function builtin Mathematica?
Chen
====
The package NumericalMath`GaussianQuadrature` has what you want.

Selwyn Hollis
> I want to get the coefficient of the LegendreGauss 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/LegendreGaussQuadrature.html
>
> I need more coefficients to get accurater result.
> How about the method of the Integration function builtin 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^2
Out[13]=
2 + x^2 + 27*f[x]^2*g[y]^2
Andrzej Kozlowski
Toyama International University
JAPAN
> 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 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 backspacecharacters to be used?
Harry
====
Try
Power[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 backspacecharacters to be
used?
Harry
====
Try
Power[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 backspacecharacters 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)^2
Out[17]=
2 + x^2 + 3*f[x]^2*g[y]^2
In[29]:=
3*ff[z]*f[x]*g[y] + 2 + x^2 /. (a : f[_])*(b : g[_]) :> (a*b)^2
Out[29]=
2 + x^2 + 3*f[x]^2*ff[z]*g[y]^2
We might tend to understand this behaviour of the pattern matcher. As Times
has the Flat, Orderless attributes, the components of the pattern have to
be
taken apart to match separated subexpressions at the lhs, what should the
pattern variable then designate in the course of this procedure?
Look at the FullForm
In[31]:=
(a : f[_])*(b : g[_]) // FullForm
Out[31]//FullForm=
Times[Pattern[a, f[Blank[]]], Pattern[b, g[Blank[]]]]
compared to
In[12]:=
a : f[_]*g[_] // FullForm
Out[12]//FullForm=
Pattern[a, Times[f[Blank[]], g[Blank[]]]]
Depending on your real problem...
In[43]:=
expr /. a_?NumericQ *b_ :> a*Times[b]^2
Out[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^2
Out[79]=
2 + x^2 + 3*f[x]^2*g[y]^2
In[139]:=
expr /. a : h_[___] /; MemberQ[{f, g}, h] :> a^2
Out[139]=
2 + x^2 + 3*f[x]^2*g[y]^2
Perhaps a fine way would be
In[74]:=
2 + x^2 + 3 f[x]*g[y] /.
HoldPattern[Times[a:(_[___]..)]] :> Times[a]^2
Out[74]=
2 + x^2 + 3*f[x]^2*g[y]^2
but 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 like
In[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
====
version
4.1.5.0
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 could
divide by p but not in my actual example. In my actual example I am not
squaring but doing a MetricSimplify tensor operation and the whole purpose
is to operate on only two of four factors. I can do it by giving a specific
pattern 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 Park
djmp@earthlink.net
http://home.earthlink.net/~djmp/
In[13]:=
expr /. a:((p_.)*f[_]*g[_]) :> p*a^2
Out[13]=
2 + x^2 + 27*f[x]^2*g[y]^2
Andrzej Kozlowski
Toyama International University
JAPAN
> 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^2
works, and I think I understand why your form above does not work (examine
FullForm of expryour form does not account for 3), but I don't
understand why eliminating a: from your form allows the pattern to
match.
Tom Burton
====
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.
====
This results from:
a) the fact that Mathematica implements division as multiplication and
reciprocation
b) the use of machine precision for plotting the graph
c) the default scaling of the y axis revealing the resultant inaccuracies
Although Sin[x]/Sin[x] gives 1 when evaluated symbolically (x undefined),
substituting a floatingpoint value for x results in composite division
being performed numerically, with consequent inaccuracy. This can be
verified 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 discrepancies
are visible. The PlotRange option may be given to explicitly specify the
range 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 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.
>
====
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 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.
>
====
I eventually figured this one out:
(1) change the default font to smaller (10 pts vs 12). This solves the
problem, 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
and
received some
very helpful responses. This time the problem seems very odd and I may be
making a very
elementary 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 exhausive
tests, each test with its own
action that modifies the concrete list subsituted for listA_.
For instance, if there were only two tests the rhs above would be
one that puts the newEntry
at 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, ...]. The
output should be the appropriately modified list. But my actual output
reproduces the input form,
i.e., it is simply f[...listA,newEntry,...].
Yet, the program does do something, as the use of evaluate in
placeshows:
Suppose that test1 is satisfied by the particular values of the
arguments. Then when
I apply evaluate in place to the lhs of the action that is supposed to
occur when test1 is true,
in fact the value of Insert[listA,newEntry,1] is correct, but the rhs gives
only the initial value of
listA and not the modified value that is on the rhs.
Concretely, suppose that in the function argument listA is {121} and
newEntry is 200.
Then after evaluation of the function (input to the kernel), that line
reads, according to
evaluate in place:
true, {121} = {200, 121}
So it seems that only part of the correct action was taken: 200 was put at
the start of listA. But
the assignment of this value  the extended list  to be the new value of
listA did not take place.
I then tried using a new name for the modified list, substituting this
program line for test1:
test1, newListA = Insert[listA,
newEntry,
1]
What I get here, applying evaluate in place after evaluating the
function
with concrete values as above
is:
true, newListA = {200, 121}
Trace doesn't help because all I get back is the function name with
its 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 Cadot
I think your technique is fine for small to medium size trees. For large
ones it might be very slow due to all the work of patten matching. If
you anticipate large inputs you might thus want to code a simple
treewalk using oldfashioned procedural code. Any time you match a
grandparent head, put that subtree onto a stack that enters the next
state, looking for parent nodes, etc.
Daniel Lichtblau
Wolfram 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 Kozlowski
Toyama International University
JAPAN
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.com
Export 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). 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
ReplyTo: kuska@informatik.unileipzig.de
====
take the matrix of gray values or the matrix of the color channels,
apply ListInterpolation[] and generate a resampled table with
the 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
or
Resize.
Else, the simplest method is to take every k'th sample, which can be done
with 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 other
sample use:
small = Take[img, {1, nr, 2}, {1, nc, 2}];
You can also perform smoothing using ListConvolve prior to downsampling to
eliminate some of the visual artifacts of downsampling.
Done.

===============================
Mariusz Jankowski
University of Southern Maine
mjkcc@usm.maine.edu
2077805580
====
Yesterday I posted a question on using named patterns in a rule. I received
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 reflect 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 flat 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/
====
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 page
How do I resolve certain fontrelated error message when
very carefully, but the error remains.
Markus Koecher
====
I look for an operator, which has a higher precedence than @ and can be
oberloaded 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 new
definition.
What is your opinion?
Hermann Schmitt
====
I have been trying to get emmathfnt to work for the past few hours but
whatever I do the output file is slightly smaller than the input so I
guess it isn't working and emmathfnt cannot find the font files. I
also set FONTDIR.
I am using the dos command
emmathfnt d
C:Progra~1Wolfra~1Mathem~14.2System~1FontsType1
o c:output.eps c:input.eps
and the following is output
C:WINDOWS>dir c:*put.eps
Volume in drive C has no label
Volume Serial Number is 05081ED9
Directory of C:
OUTPUT EPS 26,324 083002 9:57a output.eps
INPUT EPS 26,324 083002 10:21a input.eps
2 file(s) 52,648 bytes
0 dir(s) 4,896.69 MB free
and 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 StandardForm
I wonder what I am doing wrong.
C:WINDOWS>SET
TMP=C:WINDOWSTEMP
TEMP=C:WINDOWSTEMP
PROMPT=$p$g
winbootdir=C:WINDOWS
COMSPEC=C:WINDOWSCOMMAND.COM
PATH=C:DOWNLOADFONTZEMMATH~1;C:UTILSGHOSTGHOSTGUMGSVIEW;C:TE
XMFMIKTEXB
IN;C:WINDOWS;C:WINDOWSCOMMAND;C:MATLAB_SV12BINWIN32
FONTDIR=C:Program FilesWolfram
ResearchMathematica4.2SystemFilesFontsType
1
windir=C:WINDOWS
BLASTER=A220 I5 D3 T4
CMDLINE=emmathfnt d
C:Progra~1Wolfra~1Mathem~14.2System~1FontsType1 o
c:output.eps c:input.eps
thankyou
====
> 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 yaxis are all 1, that is, in trying to find
something
of interest the adaptive range has zoomed in and is looking at the machine
precision representations of 1.
(First /@(Ticks /. AbsoluteOptions[p])[[2]])//InputForm
Zoom out using the PlotRange
Plot[Sin[x]/Sin[x],{x,2.5,1.5}, PlotRange>{0.1,2.1}];
Or as stated in the online help for Plot: You should use Evaluate to
evaluate
the function to be plotted if this can safely be done before specific
numerical
values are supplied.
Plot[Evaluate[Sin[x]/Sin[x]],{x,2.5,1.5}];
Bob Hanlon
Chantilly, VA USA
====
I am trying to implement the inner product in the space of
complexvalued, 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.
====
Andreas Dietrich schrieb:
>
> I am trying to implement the inner product in the space of
> complexvalued, 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't
known to be real. Your example (inner[#&,#&]) works if you define your
function using ComplexExpand:
inner[f_,g_]:=Integrate[ComplexExpand[Conjugate[f[x]]]g[x], ...]
You may also use a homemade 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 your
simple Example, but more komplex functions will need a ComplexExpand to
get the wanted real result.
Gru§ Peter

============== http://home.tonline.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 that
it 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?
>
> Toshi
I'll show a couple of approaches to this sort of problem. The first is
to iteratively solve for sn[k][z,t] in terms of sn[k1][z,t], with
sn[0][z,t] initialized to something appropriate (I used 3/2*z). The sn's
are computed as interpolating functions based on results from solving
ODEs in t for many fixed values of z. The code below does this for n=2.
The plots at first appear to flip between two states but eventually
stabilize. 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[k1][w,t]^n*w, {w,0,1}] 
sn[k1][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'm
not sure.
Below is another method that Michael Trott showed me. We expand in a set
of basis functions in z, set up a system of ODEs in t, and solve them. I
tried this using monomials in z for basis functions and ran into some
trouble with the ODE solving, so I will show Michael's attempt using
trig functions for the basis. Note that we now handle the desired case
n=5; another advantage is that this is alot faster than the method
above, though still not exactly blinding in speed. Michael used the
interval {0,2} for z so as to achieve pointwise (not just L^2)
convergence; otherwise there would be a sharp dropoff just before z=1
as all the trigs vanish there. In other words, the basis functions are
of the form Sin[k/2*Pi] rather than Sin[k*Pi] for 1<=k<=deg. Cosines are
excluded 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 an
attempt to refine the solution.
Daniel Lichtblau
Wolfram 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?
>
> Toshi
I'll show a couple of approaches to this sort of problem. The first is
to iteratively solve for sn[k][z,t] in terms of sn[k1][z,t], with
sn[0][z,t] initialized to something appropriate (I used 3/2*z). The sn's
are computed as interpolating functions based on results from solving
ODEs in t for many fixed values of z. The code below does this for n=2.
The plots at first appear to flip between two states but eventually
stabilize. 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[k1][w,t]^n*w, {w,0,1}] 
sn[k1][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'm
not sure.
Below is another method that Michael Trott showed me. We expand in a set
of basis functions in z, set up a system of ODEs in t, and solve them. I
tried this using monomials in z for basis functions and ran into some
trouble with the ODE solving, so I will show Michael's attempt using
trig functions for the basis. Note that we now handle the desired case
n=5; another advantage is that this is alot faster than the method
above, though still not exactly blinding in speed. Michael used the
interval {0,2} for z so as to achieve pointwise (not just L^2)
convergence; otherwise there would be a sharp dropoff just before z=1
as all the trigs vanish there. In other words, the basis functions are
of the form Sin[k/2*Pi] rather than Sin[k*Pi] for 1<=k<=deg. Cosines are
excluded 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 an
attempt to refine the solution.
Daniel Lichtblau
Wolfram Research
====
On 8/23/02 at 12:25 AM, meshii@mech.fukuiu.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 reflect 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 flat
>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 at
In[6]:=
1  a  Cos[x]^2 /. {1  Cos[x_]^2 :> Sin[x]^2}
Out[6]=
a + Sin[x]^2
I don't know what TrigFactor is doing exactly (or intended to do), its
answer 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 the
elements of the expression in advance (at programming time). The following
needs not, uses extract and rebuilds the expression (it is assumed the f
are at level {1}, this must be checked, not well done here, just to pass
the
idea):
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
function
at parts of it). Here now another, simpler opat version (using Part,
instead
of 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, but
in 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 Orderless
attribute.
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 backspacecharacters to be used?
Harry
Subsuperscript[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 yaxis 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 Park
djmp@earthlink.net
http://home.earthlink.net/~djmp/
====
Lucas,
An addendum to my previous post (by the way, did you try to change
precedences 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
extensible
character. I don't know all of the extensible characters, but one
possibility is [UnionPlus], which looks a bit like CirclePlus, with an
opening on top. Of course, the usual syntax for [UnionPlus] is as a
binary
operator, and this is not what we want for our summation notation. So, we
need to incorporate new syntactical rules. There are three rules needed
here. A rule to convert 2 dimensional input into a mathematica internal
expression, a rule to convert the internal expression into a box structure,
and a rule to convert the internal expression into a regular CirclePlus
expression. 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,}]}
],Standard
Form]
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 internal
representation. If BigCirclePlus can be converted into a CirclePlus
expression (when the summation indices are integers), then the
BigCirclePlus
rule 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 after
evaluating the above rules.
Of course, if [UnionPlus] is not an acceptable substitute for an
extensible
CirclePlus, then you will just need to petition Mathematica to include such
a feature in the future.
Carl Woll
Physics Dept
U 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 of
the
> file before you make any changes. Also, 420 is low enough to get the
> behavior you desire, but you may want to experiment with other
precedences.
> 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 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}]
====
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).
====
I too have been unable to get the help browser functioning. I have
exactly the same problem. When trying to open the help browser it gets
to the point where it says scanning index file and freezes. I tried
all the suggestions in the faq and the links provided here to no
avail. I'm running Winows 2000 with service pack 3. I have 512mb of
memory with a 768mb pagefile on a separate drive, so I don't think
that's an issue. Mathematica 4.1 ran just fine on this setup, so
judging by the number of other people affected by this, I'd say it's
some kind of bug with 4.2. Hopefully Wolfram will address this with
either 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 simulating
paths of Brownian motion (Wiener process) in dimensions 1, 2, 3?
I could you a random walk approach, but this seems to be computationally
not very efficient and I am not sure whether it is even a good
approximation to the 3D BM. Any help on Mathematica coding
and in the forms of algorithms will be greatly appreciated.
Janusz Kawczak.
====
What is the problem with this integration. Its keep on running and
not 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
&)«_b
ho = 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't
normally do so. However, can anyone tell me what is happening here?
A mathematical program 'Mathematica' has been installed in /usr/local on
this
Sun Ultra 60. The correct password was entered and it all runs fine with my
normal login (davek), which has a shell of /bin/tcsh.
If I switch user to root, which has the normal /sbin/sh shell, Mathematica
runs
fine. However, if I switch user to 'roottcsh', which as a uid of 0, but a
shell
of /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 a
normal user, yet does not when run when execution is attempted as another
user
with root privileges.
*** Runs fine as a formal user 'davek'******
wren /export/home/davek % math
Mathematica 4.0 for Solaris
Copyright 19881999 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/math
wren /export/home/davek % ls l /usr/local/bin/math
lrwxrwxrwx 1 root other 53 Jun 16 08:18 /usr/local/bin/math
>
/usr/local/mathematica4.0.2/Executables/Solaris/math
***Mathematica runs fine as root****
wren /export/home/davek % su
Password:
# /usr/local/bin/math
Mathematica 4.0 for Solaris
Copyright 19881999 Wolfram Research, Inc.
 Motif graphics initialized 
In[1]:= Quit
***Switch to a user with uid=0, shell =/bin/tcsh ***
wren /export/home/davek % su  roottcsh
Password:
Sun Microsystems Inc. SunOS 5.9 Generic May 2002
***Now Mathematica does not run***
wren / # /usr/local/bin/math
Mathematica 4.0 for Solaris
Copyright 19881999 Wolfram Research, Inc.
/usr/local/mathematica4.0.2/Configuration/Licensing/mathpass:1:
Incomplete password entry.
No valid singlemachine password entry for Mathematica found.
Machine name: wren
MathID: [deleted by moderator]
You will need a valid license ID and password in order
to proceed. Go to http://register.wolfram.com or
consult 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 root
root:x:0:1:SuperUser:/:/sbin/sh
roottcsh:x:0:1:Dr. David Kirkby:/:/bin/tcsh
wren / # cat /etc/shadow  grep root
root:TA9addfsfsdsdfdddMTxHU:11266::::::
roottcsh:dgddsdfsdffsdnSw:11198::::::

Dr. David Kirkby PhD,
web page: http://www.davidkirkby.co.uk
Amateur radio callsign: G8WRB
====
I managed to post this on alt.math.recreational by mistake, and confuse
everyone!
I am having a bit of trouble with showing the shapes of pentominoes. The
list of pentominoes is 12 x 5 x 2, for 12 pentominoes x 5 squares x {X,Y}
coordinate of the square on a zeroorigin 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 StringJoin
pl[p_]:=Plot[Sin[x^p],{x,0,Pi},
PlotLabel>p = <> ToString[p],
DisplayFunction>Identity];
Alternatively, you could use StringForm
pl[p_]:=Plot[Sin[x^p],{x,0,Pi},
PlotLabel>StringForm[p = `` , p],
DisplayFunction>Identity];
Bob Hanlon
Chantilly, 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+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=6] := x+1;
f2[x_] :=
x^2*UnitStep[x]+(x+1x^2)*UnitStep[x6];
f3[x_] := Which[
x<=0, 0,
0=6, x+1];
f4[x_] := If[0=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=6}, True][[1,1]]]];
Needs[Calculus`Integration`];
(* needed for definition of Boole *)
f7a[x_?NumericQ] := Evaluate[
(Boole /@ {0=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=6, x+1}},
{True, z_} :>z][[1]];
f9[x_?NumericQ] := DeleteCases[
{{x<=0, 0}, {0=6, x+1}},
{False, z_}][[1,2]];
f10[x_?NumericQ] := Select[
{{x<=0, 0}, {0=6, x+1}},
First[#]&][[1,2]];
f11[x_?NumericQ] := Last[Sort[
{{x<=0, 0}, {0=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 I am trying to implement the inner product in the space of
> complexvalued, 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[#&,#&]
1/12
Bob Hanlon
Chantilly, VA USA
====
I was wondering if there are any scripts/functions out there to allow me to
do decision tree classification with Mathematica, similar to S+ type
analysis?
j

Jonathan Greenberg
Graduate Group in Ecology, U.C. Davis
http://www.cstars.ucdavis.edu/~jongreen
http://www.cstars.ucdavis.edu
AIM: jgrn307 or jgrn3007
MSN: 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 for
defining piecewise functions.
The first method is to use a set of conditional definitions. For
completeness I made f[x] = 0 if one of the conditional definitions is not
met.
f[x_] /; 0 < x < 6 := x^2
f[x_] /; x >= 6 := x + 1
f[x_] := 0
Plot[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 perform
functions on f, such as differentiation or integration, you should use the
UnitStep 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 Park
djmp@earthlink.net
http://home.earthlink.net/~djmp/
PasKo.
====
>
>
> I am trying to implement the inner product in the space of
> complexvalued, 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 least
to the extent that it works on the simple example given above. For a
workaround 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[# &, # &]
1
Out[2]= 
12
Daniel Lichtblau
Wolfram Research
====
Andreas,
inner[f_Function, g_Function] :=
Integrate[ComplexExpand[Conjugate[f[x]]]*g[x], {x, 1/2, 1/2}]
inner[# &, # &]
1/12
I'm not too knowledgable about using complex functions in Mathematica but
sometimes I think that ComplexExpand should be renamed
ComplexSimplify.
One very often needs it.
David Park
djmp@earthlink.net
http://home.earthlink.net/~djmp/
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.
====
I work for Gould Academy and am trying to make it really easy for the
scripts which I thought did that really well. But, the science
teachers said it was too hard. So, I have been making it so that Input
boxes prompt the students for the filename type, regression type, etc.
I now have been thinking that it would be great if there could be an
Input box which asks the student what they want to name there graph,
function, data, and have them simply call that variable when they want
to use the graph, function, data. this is Instead of typing
VariableName = FilePlot[] which the science teachers think is forcing
the 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 usually
reserved for inbuilt functions in Mathematica.
Tomas Garza
Mexico 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, Condition
f[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+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
*******************
milkcart
milkcart@m17.alphanet.ne.jp
***********************
====
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=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.
====
> 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+1 if x>=6
I'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 UnitStep
function.
PasKo
P.S.
Delete berlusconi_ to reply.
====
Andreas,
One simple approach is to write:
F[x_]:= x^2 /; 0=6
Hope 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+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 WickhamJones' book on Mathematica
Graphics, 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 > $DisplayFunction],
{k, 3}]
Tomas Garza
Mexico City
 Original Message 
Table[Show[pl[k],DisplayFunction[Rule]$DisplayFunction],{k,3}]
====
Andre,
You should use SequenceForm in your definition (instead of multiplication
which reorders the items) .
pl[p_] := Plot[Sin[x^p], {x, 0, Pi}, PlotLabel > SequenceForm[p = ,
p],
DisplayFunction > Identity]
David Park
djmp@earthlink.net
http://home.earthlink.net/~djmp/
====
> I am trying to implement the inner product in the space of
> complexvalued, 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 Conjugate
are not integrable in closed form anyway. So I suggest that you attempt to
expand 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 be
more 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+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 Burton
ReplyTo: Peltio
====
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 and
interpreted 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 standard
conditions.
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,
Peltio
peltio at despammed dot com
antispam address in replyto
====
Dear group.
I am trying to figure out how I can do multidimensional kernel density
estimation with Mathematica. I know how I can do it with one variable, but
I was wondering if Mathematica supports kernel density estimation with 3
variables.
Sincerely,
Marcelo

Marcelo Moreira
Department of Economics  Harvard University
Littauer M6  North Yard
1875 Cambridge Street
Cambridge, MA 02138
http://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 Bob
stated, this is obtained by:
(3^n1)/2=40
For these 40 positive integers to match 1 to 40 correspondingly, the
biggest
should 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> 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:
>> {xx=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^n1)/2, (3^n1)/2}.
> Consequently, the largest of the weights can not exceed
> (3^n1)/2  (1+2+...+(n1)) or
((3^n1)  n(n1))/2
34
Thread[var>#]& /@
(First /@ Select[{var,f} /. Thread[var>#]& /@
KSubsets[Range[((3^n1)  n(n1))/2], n],
Sort[#[[2]]] == Range[(3^n1)/2,(3^n1)/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:
>>> {xx=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^n1)/2, (3^n1)/2}.
>Consequently, the largest of the weights can not exceed
>(3^n1)/2  (1+2+...+(n1)) or
((3^n1)  n(n1))/2
>34
>Thread[var>#]& /@
>
> (First /@ Select[{var,f} /. Thread[var>#]& /@
>
> KSubsets[Range[((3^n1)  n(n1))/2], n],
>
> Sort[#[[2]]] == Range[(3^n1)/2,(3^n1)/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^n1)/2 then you can speed up the brute force method by
requiring that a+b+c+d == (3^n1)/2
Needs[DiscreteMath`Combinatorica`];
var = {a,b,c,d}; n = Length[var]; m = (3^n1)/2;
s = Outer[Times, var, {1,0,1} ];
f = Flatten[Outer[Plus, Sequence@@s]];
k= Select[KSubsets[Range[m  n(n1)/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 Hanlon
Chantilly, 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=6
Try
f[x_]:= x^2 /; 0=6
The /; 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 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/
====
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 several
revisions 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 version
4:
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]^2
gives O[x]^2  same as in version 4. (Apparently, Dot in version 3 does
not include in the sum terms multiplied by 0).
The easiest way to correct this bug, without altering Times, is to use
UpSet:
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 this
bug...
Sotirios Bonanos
http://www.inp.demokritos.gr/~sbonano/SB.html
====
In[4]:= f[x_ /; 0 < x < 6] := x^2
In[5]:=f[x_/; x >= 6]:=x+1
Meilleures salutations
Florian Jaccard
EICNHES
Message d'origine
Envoy.8e : sam., 31. ao.9et 2002 07:26
Ë : mathgroup@smc.vnet.net
Objet : How to ...?
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=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.
====
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 > $DisplayFunction], {k, 3}]
Meilleures salutations
Florian Jaccard
EICNHES
Message d'origine
Envoy.8e : sam., 31. ao.9et 2002 07:26
Ë : mathgroup@smc.vnet.net
Objet : PlotLabel
I plot several curve depending on a parameter p and I want the PlotLabel to
show the value of p.
What is wrong with the following ?
pl[p_]:=Plot[Sin[x^p],{x,0,Pi},PlotLabel[Rule] p = p,
DisplayFunction[Rule]Identity]
Table[Show[pl[k],DisplayFunction[Rule]$DisplayFunction],{k,3}]
====
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 obtain
the two output vectors as continuous functions of the input vector, since
that would be equivalent to the combing of a hedgehog in a vortexfree way.
Ingolf Dahl
f9aid@fy.chalmers.se
Chalmers University
Sweden
Original Message
David Park
djmp@earthlink.net
http://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 second
vector b not parallel to the given vector a, it's a trivial matter to
orthogonalize b WRT a by GramSchmidt and then form the third vector c = a
x
b. (Normalize as needed.)
I don't need more elegance that this, but I would like a scheme to select
the vector b that results in a triad {a,b,c} that various smoothly as a
varies over all possible directions. Each of my attempts to date involve a
branched algorithm and jumps in the resulting triad for certain small
changes 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 the
same scale.
Show[
GraphicsArray[
Partition[
Table[
Graphics[
Map[Rectangle[#, (# + {1, 1})] &, pentominoes[[i]]],
AspectRatio > Automatic,
PlotRange > {{0, 5}, {0, 5}}
], {i, 12}],
6]
]
]
David Park
djmp@earthlink.net
http://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
*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
====
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.fhvorarlberg.ac.at/~ku/karl/timings40.html
value 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 speed
compared to Mathematica 4 or 3.
Armin
====
Can I use Mathematica to find out the volumn of this 3 dimensional object
from
the equations :
z=x^2 +4, y=4x^2, y=3x
Shz Shz
====
Use Text statements. Click of a coordinate point on the plot, copy it and
paste 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 cross
hairs.
3) Move the cursor to where you want to put the text and left click.
4) Right click and use Copy. (Or use CtrlC or the menu).
5) Paste the copied coordinate as the second argument in the Text
statement.
David Park
djmp@earthlink.net
http://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=6
Try
In[4]:=
F[x_]:=Which[0=6,x+1]
you can plot the function with
In[5]:=
Plot[F[x],{x,0,8}]
Sincerely yours,
Dr. Juan E. Fuentes Betancourt
Facultad 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 of
piecewise defined functions. I am in the process of writing it up to
offer it to all of you. In my opinion the best, cleansest method is to
invoke the UnitStep function as illustrated in David Park's solutions.
To make everything cleaner, define a characteristic function
Chi[x_,a_,b_] := UnitStep[xa]UnitStep[xb]
This function is continuous on the right, vanishes outside of [a,b) and is
one in the halfopen 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 such
oddities 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 the
peculiar error messages and poor answers given by NIntegrate for simple
piecewise continuous functions at jump discontinuites. The limit function
also fails dismally on some examples where it really shouldn't
I am rather naive about how to transmit programs to interested users, so
if you would like a copy of my program and about 100 worked examples, I
will 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+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=6] := x+1;
> f2[x_] :=
x^2*UnitStep[x]+(x+1x^2)*UnitStep[x6];
> f3[x_] := Which[
x<=0, 0,
0=6, x+1];
> f4[x_] := If[0=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=6}, True][[1,1]]]];
> Needs[Calculus`Integration`];
(* needed for definition of Boole *)
> f7a[x_?NumericQ] := Evaluate[
(Boole /@ {0=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=6, x+1}},
{True, z_} :>z][[1]];
> f9[x_?NumericQ] := DeleteCases[
{{x<=0, 0}, {0=6, x+1}},
{False, z_}][[1,2]];
> f10[x_?NumericQ] := Select[
{{x<=0, 0}, {0=6, x+1}},
First[#]&][[1,2]];
> f11[x_?NumericQ] := Last[Sort[
{{x<=0, 0}, {0=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 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
ReplyTo: 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 Browne
School of Engineering and Science
Swinburne University of Technology
John Street, Hawthorn, Victoria, Australia
Quantica phone: +613 9431 4007
Quantica fax: +613 9431 0940
====
In my previous post, I proposed
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]}]
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