d-11

I was wondering whether there's a way to provide mathematica with x
and y coordinates of a point from a c++ program (with mathlink or
otherwise) and have it graph it out point by point, perhaps even
erasing the previous image as it goes along. For example, if the
correct points are supplied as output, can mathematica show me the
movement of a planet in its orbit or SHM, etc. (whatever the code is
meant to generate). Currently I output the points to a file in the
ListPlot form, but then I just get the orbit drawn out instead of the
groups, but I read somewhere that realtime plotting of points can't be
done on mathematica. is this true? and if it can than how?
Any help would be greatly appreciated

====
ok, it is not perfect but what is with
DynamicFrame[t_Integer, range_] := Module[{arg, gr},
    arg = 2Pi*t/18;
    gr = Graphics[{{Hue[Mod[t/18, 1]], AbsolutePointSize[8.0],
            Point[{Cos[arg], Sin[arg]}]},
          Line[{{0, 0}, {Cos[arg], Sin[arg]}}]}, PlotRange -> range, 
        AspectRatio -> Automatic];
    DisplayString[gr]
    ]
and
nb = NotebookCreate[];
Do[
  NotebookFind[nb, MyAnimationFrame, Previous, CellTags];
NotebookDelete[nb];
  NotebookWrite[nb, 
    Cell[GraphicsData[PostScript, 
        DynamicFrame[i, {{-1.5, 1.5}, {-1.5, 1.5}}]
        ]
      , Graphics, CellTags -> MyAnimationFrame]
    ], {i, 18}]
  
> 
> I was wondering whether there's a way to provide mathematica with x
> and y coordinates of a point from a c++ program (with mathlink or
> otherwise) and have it graph it out point by point, perhaps even
> erasing the previous image as it goes along. For example, if the
> correct points are supplied as output, can mathematica show me the
> movement of a planet in its orbit or SHM, etc. (whatever the code is
> meant to generate). Currently I output the points to a file in the
> ListPlot form, but then I just get the orbit drawn out instead of the
> groups, but I read somewhere that realtime plotting of points can't be
> done on mathematica. is this true? and if it can than how?
> Any help would be greatly appreciated
> aatish
> aatish@mac.com
====
1) With the up/down keys I can cycle repeatedly in either direction with no
problem, without having to add any extra cells.
2) However, the control strip keys behave erratically for single stepping.
Or else, I don't know the proper way to use them for single stepping. 
Adding
a Text cell after the animation frames did not improve it.
David Park
djmp@earthlink.net
http://home.earthlink.net/~djmp/
take you up to the k=4 plot, but will NOT loop back down to the k=1 plot
(although the free-running animation does seem to loop around OK).
Adding another cell (even a header or text cell) at the end of the
notebook, however, so that there is another cell after the k=4 plot,
will enable cyclic single-step looping in either direction around the
loop.
[Mac PB G3, OS 9.1, Mathematica 4.1]
====
Is something like this what you're after:
In[1]:= soln = y /.First[NDSolve[{y'[t]==y[t], y[0]==1},
        y, {t,0,10}, StoppingTest->(y[t]>2)]]
Out[1]= InterpolatingFunction[{{0., 0.696495}}, <>]
In[2]:= soln[[1, 1, 2]]
Out[2]= 0.696495
In[3]:= soln[soln[[1, 1, 2]]]
Out[3]= 2.00672
?
---

htp://www.math.armstrong.edu/faculty/hollis
> I am using StoppingTest in NDSolve, as a consequence the
> terminal value of the independent variable is unknown prior
> to NDSolve finishing. How does one extract this number from
> the Interpolation object?

====
findTimeRange[expr_] := 
  First[Union[
      Cases[expr, InterpolatingFunction[___][_], Infinity] /. 
        InterpolatingFunction[a_, b__][_] :> First[a]]]
ff = NDSolve[{y''[x] == -y[x], y[0] == 1, y'[0] == 0}, y[x], {x, 0,
2Pi}]
and the endpoint is:
Last[findTimeRange[ff]]
  
> 
> I am using StoppingTest in NDSolve, as a consequence the
> terminal value of the independent variable is unknown prior
> to NDSolve finishing. How does one extract this number from
> the Interpolation object?
====
This works:
polyCoeff = CoefficientList[(-m + x)^5*(a + b*x + 
   c*x^2 + d*x^3 + e*x^4 + f*x^5),{x}];
Sum[polyCoeff[[k]]Integrate[Exp[-const (x-m)^2]
x^(k-1),{x,-Infinity,Infinity},Assumptions->const>0],{k,1,11}]
Maurice
I'm trying to compute the 5th moment of a tweak to the normal pdf. 
Here's what I have: F[a_, b_, c_, d_, e_, f_, X_] := (a + b X + c X^2
+ d X^3 + e X^4 + f X^5)* > Exp[-(((X - m)/sd)^2)/2]/(Sqrt[2Pi]sd).
The strange thing, and I'd appreciate someone shedding some light on
this,  is that this:Integrate[F[p_0, p_1, p_2, p_3, p_4, p_5, >
X]*((((X - m))/sd))^5, {X, -Infinity, Infinity},  Assumptions ->
{sd > 0}]) (I cut and pasted that it looks like it works) Anyway, THAT
gives me an answer really quickly... (within 5 minutes on > my box)
JUST changing it from p0..5 to a,b,c,d,e,f like so: Integrate[F[a, b,
c, d, e, f, X]*((X - m)/sd)^5, {X, -Infinity, > Infinity},Assumptions
-> {sd > 0}] Causes mathematica to go into an infinite loop (seemingly
after 6 hours). This ... sucks. How am I to know if a certain equation
is solvable, if  only I choose the right variables??
Binesh Bannerjee 
does sem to be a bug here, but it is not quite what you think. It's
the first answer that you get that is probably wrong. On the other
hand, in the second case it is very unlikely that Mathematica enters
an infinite loop, rather it is still trying to arrive at the answer
and there is no guarantee that it will reach one after, say a week or
a month. As for first case, the reason why the answer is probably
this. Evaluate the formula: formula=Integrate[F[Subscript[p, 0],
Subscript[p, 1], Subscript[p, 2], Subscript[p, 3], Subscript[p, 4],
Subscript[p, 5], X]*((X - m)/sd)^5, {X, -Infinity, Infinity}] Now set
Evaluate[Table[Subscript[p, i], {i, 1, 5}]] = Table[Random[], {5}] and
also sd = Random[]; m = Random[]; Now evaluate again
Integrate[F[Subscript[p, 0], Subscript[p, 1], Subscript[p, 2],
Subscript[p, 3], Subscript[p, 4], Subscript[p, 5], X]*((X - m)/sd)^5,
{X, -Infinity, Infinity}] and formula You will almost certainly get
different answers, while they clearly ought to be the same. It seems
that it is nto the fact that the names you are suing are different
that leads to different results in both of your integrals but the fact
that the names of the parameters in the first case are not symbols.
Officially there is no reason why they should be, but in practice
using non-symbols in formulas makes them more complicated and is more
likely to result in errors. The really bad news as far as your problem
is concerned is that it is the 6 hour fruitless computation that
appears to be the correct one ... Andrzej Kozlowski Toyama
International University JAPAN http://platon.c.u-tokyo.ac.jp/andrzej/
====

> According your theory, I think, klasse1 should be defined in Global` 
and
> klasse only in Rechteck But this is not the case. Therefore - excuse
me -
> I doubt, that your theory is correct.
I have added some comments to your code about how I think that this 
happens:
oo`hfield = klasse1;
oo`newContext[ctxp_] := ($ContextPath = Join[ctxp, $ContextPath];
    BeginPackage[ctxp[[1]]];
    If[Length[ctxp] > 1, $ContextPath = Join[ctxp, Rest[$ContextPath]];])
oo`resumeContext[] := (EndPackage[];
    $ContextPath = Drop[$ContextPath, 1];)
oo`new11[x_] := (Print[Anfang new11];
    klassnam = x;
    tsx = ToString[klassnam];
    oo`dat = D:Math_OOclasses <> tsx <> .cl;
    ctx = klassnam <> `;
    oo`newContext[{ctx}];)
oo`new12[] := (Print[Name klasse - Anfang];
      Print[$Context: , $Context];
      ToExpression[oo`hfield];
      (* the symbol klasse1 is created when oo`new12[] is evaluated*)
      Print[Context von klasse: , Context[klasse]];
      (*klasse is created as Global`klasse when this rule is read*)
      oo`resumeContext[];
      Print[Name klasse - Ende];);
Print[Anfang Programm:];
Print[$Context: , $Context];
oo`new11[Rechteck];
Print[nach new11:];
Print[$Context: , $Context];
oo`hfield = klasse;
oo`new12[];(*klasse and klass1 are created as in context Rechteck` when 

this is evaluated, because context,
  contextpath are Rechteck`, {Rechteck`, System`}*)
Print[nach new12:];
Print[AH3];
Print[$Context: , $Context];
--

> I have a new program:
 oo`hfield = klasse1;
> oo`newContext[ctxp_] := (
   $ContextPath = Join[ctxp, $ContextPath];
>       BeginPackage[ctxp[[1]]];
>    If[Length[ctxp]>1,
>        $ContextPath = Join[ctxp, Rest[$ContextPath]];
>     ]
>     )
> oo`resumeContext[] := (
>         EndPackage[];
>        $ContextPath = Drop[$ContextPath, 1];
> )
> oo`new11[x_] := (
> Print[Anfang new11];
> klassnam = x;
> tsx = ToString[klassnam];
> oo`dat = D:Math_OOclasses <> tsx <> .cl;
> ctx = klassnam <> `;
> oo`newContext[{ctx}];
> )
> oo`new12[] := (
> Print[Name klasse - Anfang];
>  Print[$Context: , $Context];
>  ToExpression[oo`hfield];
>  Print[Context von klasse: , Context[klasse]];
>  oo`resumeContext[];
>  Print[Name klasse - Ende];
>  );
>   Print[Anfang Programm:];
>   Print[$Context: , $Context];
>  oo`new11[Rechteck];
>   Print[nach new11:];
>   Print[$Context: , $Context];
>  oo`hfield = klasse;
>  oo`new12[];
>    Print[nach new12:];
>    Print[$Context: , $Context];
 and the results:
 Anfang Programm:
> $Context: Global`
> Anfang new11
> nach new11:
> $Context: Rechteck`
> Name klasse - Anfang
> $Context: Rechteck`
> klasse
> klasse::shdw: Symbol klasse appears in multiple contexts {Rechteck`,
> Global`}
>     ; definitions in context Rechteck`
>      may shadow or be shadowed by other definitions.Context von klasse:
> Global`
> Name klasse - Ende
> nach new12:
> $Context: Global`
> In[19]:=
 According your theory, I think, klasse1 should be defined in Global` 
and
> klasse only in Rechteck But this is not the case. Therefore - excuse
me -
> I doubt, that your theory is correct.

 ----- Original Message -----

>  > My question is:
> Why is the variable klasse also defined in the context Global?
>  > klasse is created in the context Global` when the assignment
>  > oo`new12[] := (Print[Name klasse - Begin];
>     Print[$Context: , $Context];
>     ToExpression[klasse];
>     Print[Context of klasse: , Context[klasse]];
>     oo`resumeContext[];
>     Print[Name klasse - End];)
>  > is converted into internal form (with contexted symbols) prior to
> evaluation.
>  > This is before
>     oo`new11[Rechteck];
> is evaluated
>  > --
> 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 have the following program:
>  > oo`newContext[ctxp_] := (
>     BeginPackage[ctxp[[1]]];
>     )
> oo`resumeContext[] := (
>         EndPackage[];
>        $ContextPath = Drop[$ContextPath, 1];
>      )
> oo`new11[x_] := (
> Print[Begin new11];
> klassnam = x;
> ctx = klassnam <> `;
> oo`newContext[{ctx}];
> );
> oo`new12[] := (
> Print[Name klasse - Begin];
>  Print[$Context: , $Context];
>  ToExpression[klasse];
> Print[Context of klasse: , Context[klasse]];
>  oo`resumeContext[];
>  Print[Name klasse - End];
>  );
>   Print[Begin Programm:];
>   Print[$Context: , $Context];
>  oo`new11[Rechteck];
>  oo`new12[];
>    Print[after new12:];
>    Print[$Context: , $Context];
>  > I get the following results:
>  > Begin Programm:
> $Context: Global`
> Begin new11
> Name klasse - Begin
> $Context: Rechteck`
> klasse
> klasse::shdw: Symbol klasse appears in multiple contexts {Rechteck`,
> Global`}
>     ; definitions in context Rechteck`
>      may shadow or be shadowed by other definitions.Context of 
klasse:
> Global`
> Name klasse - End
> after new12:
> $Context: Global`
> In[14]:=
>  > My question is:
> Why is the variable klasse also defined in the context Global?

====
according to Allan's an my explanation
Mathematica *must* parse the definition of oo`new12[] *before*
it is evaluated. Because you don't like local variables you
write
oo`new12[] := (
Print[Name klasse - Anfang];
 Print[$Context: , $Context];
 ToExpression[oo`hfield];
 Print[Context von klasse: , Context[klasse]];
 oo`resumeContext[];
 Print[Name klasse - Ende];
 );
and parsing this create a global symbol klasse because
the parser see the line
 Print[Context von klasse: , Context[klasse]];
If you wish to avoid this you can use
oo`new12[] := Module[{UsingObjectsIsDaft`class},
      Print[Name klasse - Anfang];
      Print[$Context: , $Context];
      UsingObjectsIsDaft`class = ToExpression[oo`hfield];
      Print[Context von klasse: , 
        Context[Evaluate[UsingObjectsIsDaft`class]]];
      oo`resumeContext[];
      Print[Name klasse - Ende];
      ];
  
> 
> I have a new program:
> 
> oo`hfield = klasse1;
> oo`newContext[ctxp_] := (
  $ContextPath = Join[ctxp, $ContextPath];
>       BeginPackage[ctxp[[1]]];
>    If[Length[ctxp]>1,
>        $ContextPath = Join[ctxp, Rest[$ContextPath]];
>     ]
>     )
> oo`resumeContext[] := (
>         EndPackage[];
>        $ContextPath = Drop[$ContextPath, 1];
> )
> oo`new11[x_] := (
> Print[Anfang new11];
> klassnam = x;
> tsx = ToString[klassnam];
> oo`dat = D:Math_OOclasses <> tsx <> .cl;
> ctx = klassnam <> `;
> oo`newContext[{ctx}];
> )
> oo`new12[] := (
> Print[Name klasse - Anfang];
>  Print[$Context: , $Context];
>  ToExpression[oo`hfield];
>  Print[Context von klasse: , Context[klasse]];
>  oo`resumeContext[];
>  Print[Name klasse - Ende];
>  );
>   Print[Anfang Programm:];
>   Print[$Context: , $Context];
>  oo`new11[Rechteck];
>   Print[nach new11:];
>   Print[$Context: , $Context];
>  oo`hfield = klasse;
>  oo`new12[];
>    Print[nach new12:];
>    Print[$Context: , $Context];
> 
> and the results:
> 
> Anfang Programm:
> $Context: Global`
> Anfang new11
> nach new11:
> $Context: Rechteck`
> Name klasse - Anfang
> $Context: Rechteck`
> klasse
> klasse::shdw: Symbol klasse appears in multiple contexts {Rechteck`,
> Global`}
>     ; definitions in context Rechteck`
>      may shadow or be shadowed by other definitions.Context von klasse:
> Global`
> Name klasse - Ende
> nach new12:
> $Context: Global`
> In[19]:=
> 
> According your theory, I think, klasse1 should be defined in Global` 
and
> klasse only in Rechteck But this is not the case. Therefore - excuse 
me -
> I doubt, that your theory is correct.
> 
> 
----- Original Message -----
> 
> Hermann,
>  > My question is:
> Why is the variable klasse also defined in the context Global?
>  > klasse is created in the context Global` when the assignment
>
> oo`new12[] := (Print[Name klasse - Begin];
>     Print[$Context: , $Context];
>     ToExpression[klasse];
>     Print[Context of klasse: , Context[klasse]];
>     oo`resumeContext[];
>     Print[Name klasse - End];)
>  > is converted into internal form (with contexted symbols) prior to
> evaluation.
>  > This is before
>     oo`new11[Rechteck];
> is evaluated
>  > --
> 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 have the following program:
>  > oo`newContext[ctxp_] := (
>     BeginPackage[ctxp[[1]]];
>     )
> oo`resumeContext[] := (
>         EndPackage[];
>        $ContextPath = Drop[$ContextPath, 1];
>      )
> oo`new11[x_] := (
> Print[Begin new11];
> klassnam = x;
> ctx = klassnam <> `;
> oo`newContext[{ctx}];
> );
> oo`new12[] := (
> Print[Name klasse - Begin];
>  Print[$Context: , $Context];
>  ToExpression[klasse];
> Print[Context of klasse: , Context[klasse]];
>  oo`resumeContext[];
>  Print[Name klasse - End];
>  );
>   Print[Begin Programm:];
>   Print[$Context: , $Context];
>  oo`new11[Rechteck];
>  oo`new12[];
>    Print[after new12:];
>    Print[$Context: , $Context];
>  > I get the following results:
>  > Begin Programm:
> $Context: Global`
> Begin new11
> Name klasse - Begin
> $Context: Rechteck`
> klasse
> klasse::shdw: Symbol klasse appears in multiple contexts {Rechteck`,
> Global`}
>     ; definitions in context Rechteck`
>      may shadow or be shadowed by other definitions.Context of 
klasse:
> Global`
> Name klasse - End
> after new12:
> $Context: Global`
> In[14]:=
>  > My question is:
> Why is the variable klasse also defined in the context Global?
>  > 
>  >  >    >
Reply-To: <drbob@bigfoot.com>
====
I agree.  I had a graduate math course in non-standard analysis years
ago, and I recall that it was just as formally logical as the usual way.
Integrals became finite (transfinite) sums and the usual proofs of
calculus became much simpler, once the groundwork was laid.
It could be tricky to incorporate it in Mathematica, but no more so than
intervals or complex numbers.  The synthesis of integration with
summation could be very powerful, too, as each can build on the other.
Ditto for differentiation and finite differences, differential equations
and difference equations, etc.
Bobby Treat
-----Original Message-----
symbolic algebra.
(For more see Abraham Robinson, Non-standard Analysis, Princeton 
Landmarks in Mathematics, 1996).
Andrzej Kozlowski
Toyama International University
JAPAN
http://platon.c.u-tokyo.ac.jp/andrzej/
 Actually on second thoughts I began to suspect that this question is
> related to another one posted by Heather, concerning simplifying
> expressions in which x is much larger than y. I am not at all sure
if
> a sensible calculus of this kind can be developed but obviously 
> Simplify
>  will not do this.
 It seems to me that this is essentially a (capital-C) Calculus 
> problem,
> and unless a simple /.y->0 is what's wanted, the correct tool is 
> Limit[].
> Berkeley's critique of 18th century Calculus applies here: while it
was
> essentially antiscientific, his reasoning was flawless and should warn
> us
> against trying to solve this sort of problem by mindless algebra.
 Of course, Limit[] is a tricky and somewhat unreliable power tool,
> requiring caution. This reflects the mathematical subtlety of this
kind
> of problem. It is generally essential to formulate the problem in such
a
> way that the direction of the approach to the limit is unambiguous.
 --
> | John Doty           You can't confuse me, that's my job.
> | Home: jpd@w-d.org
> | Work: jpd@space.mit.edu

>
====
Since switching to Mac OSX almost a year ago, I have suffered with ugly
on-screen FrameLabels on 2D graphs. The ordinates' labels are supposed to
rotate, but on screen they don't. Wolfram places the blame on your 
system's
default font manager and singled out Macintosh systems,
http://support.wolfram.com/mathematica/graphics/textfonts/rotatedtext.html
in a FAQ response last updated January 2000. So this is a long-standing
issue. The most recent helpful discussion I found in the mathgroup archives
is far older,
http://library.wolfram.com/mathgroup/archive/1997/Mar/msg00257.html
and the status of the issue apparently remains unchanged since then.
But as on-screen documents become more important and post-script documents
(which do rotate text) less important, this issue is getting increasingly
serious. The failing contaminates exported PDF files (on screen), too. So I
am (without much hope) raising this issue once again: Wolfram's FAQ 
response
cited above refers the failing of the DEFAULT font manager. Can I perhaps
install another font manager that would correct the problem? Or shall I
suffer the fate of poor Mr. Zacher in 1997, who on a wild goose chase
installed various font managers on Mac without success?
Tom Burton
====
I tried to evaluate the expression with j ranging from 0 to 20000 step of 
1.
Something obviously periodic seems to be going on (just plot the sim data).
anyone interested) a flat text file with the output of Mark's program (the
sim data for j ranging from 0 to 20000 in steps of 1).
It is my understanding that the period of the rule 30-cellular automata RNG
(if any) used by MATHEMATICA has not been extensively studied. Could we use
the data geenrated to do so ?
Christos Argyropoulos
> There is some real weirdness with Random[Integer, j] for some large
> values of j but not others (for both 4.1 and 4.2 versions).
 <<Statistics`
> n = 1000;
> sim = Table[{j, (2/2^j) * Mean[Table[Random[Integer, 2^j], {n}]]},
>       {j, 45, 65, 1/8}];
> ListPlot[sim];
 All the points should be near one if the random number generator were
> working properly. What gives? Is this just a (very stark) symptom of
> what Daniel Lichtblau mentioned in his May 2000 post?
 > The other failure only is apparent in very large big
> integers, and I do not know exactly how to categorize it.
 I just find this very troubling.
 --Mark.
>
Reply-To: <drbob@bigfoot.com>
====
That IS bizarre.  The following works much better, though it may have
other statistical problems:
sim = Table[{j, Mean[Table[Ceiling[Random[]2^j], {n}]]
              2^(1 - j)}, {j, 45, 65, 1/8}];
ListPlot[sim];
Bobby Treat
-----Original Message-----
working properly. What gives? Is this just a (very stark) symptom of
what Daniel Lichtblau mentioned in his May 2000 post?
> The other failure only is apparent in very large big
> integers, and I do not know exactly how to categorize it.
I just find this very troubling.
--Mark.
====
I understand you, the line
 Print[Context von klasse: , Context[klasse]];
causes the definition of klasse to be created in the Context Global`
I thank you for your help! I exerted much effort without finding the cause!
I do dislike local variables because they mean more writing effort, I think
they are often not necessary for tests.
This test had the aim to define klasse in the context Rechteck`, 
where
the context will be a variable in the final program..

----- Original Message -----
> oo`new12[] := (
> Print[Name klasse - Anfang];
>  Print[$Context: , $Context];
>  ToExpression[oo`hfield];
>  Print[Context von klasse: , Context[klasse]];
>  oo`resumeContext[];
>  Print[Name klasse - Ende];
>  );
 and parsing this create a global symbol klasse because
> the parser see the line
  Print[Context von klasse: , Context[klasse]];
 If you wish to avoid this you can use
 oo`new12[] := Module[{UsingObjectsIsDaft`class},
>       Print[Name klasse - Anfang];
>       Print[$Context: , $Context];
>       UsingObjectsIsDaft`class = ToExpression[oo`hfield];
>       Print[Context von klasse: ,
>         Context[Evaluate[UsingObjectsIsDaft`class]]];
>       oo`resumeContext[];
>       Print[Name klasse - Ende];
>       ];

>   

>  > I have a new program:
>  > oo`hfield = klasse1;
> oo`newContext[ctxp_] := (
>  >   $ContextPath = Join[ctxp, $ContextPath];
>       BeginPackage[ctxp[[1]]];
>    If[Length[ctxp]>1,
>        $ContextPath = Join[ctxp, Rest[$ContextPath]];
>     ]
>     )
> oo`resumeContext[] := (
>         EndPackage[];
>        $ContextPath = Drop[$ContextPath, 1];
> )
> oo`new11[x_] := (
> Print[Anfang new11];
> klassnam = x;
> tsx = ToString[klassnam];
> oo`dat = D:Math_OOclasses <> tsx <> .cl;
> ctx = klassnam <> `;
> oo`newContext[{ctx}];
> )
> oo`new12[] := (
> Print[Name klasse - Anfang];
>  Print[$Context: , $Context];
>  ToExpression[oo`hfield];
>  Print[Context von klasse: , Context[klasse]];
>  oo`resumeContext[];
>  Print[Name klasse - Ende];
>  );
>   Print[Anfang Programm:];
>   Print[$Context: , $Context];
>  oo`new11[Rechteck];
>   Print[nach new11:];
>   Print[$Context: , $Context];
>  oo`hfield = klasse;
>  oo`new12[];
>    Print[nach new12:];
>    Print[$Context: , $Context];
>  > and the results:
>  > Anfang Programm:
> $Context: Global`
> Anfang new11
> nach new11:
> $Context: Rechteck`
> Name klasse - Anfang
> $Context: Rechteck`
> klasse
> klasse::shdw: Symbol klasse appears in multiple contexts {Rechteck`,
> Global`}
>     ; definitions in context Rechteck`
>      may shadow or be shadowed by other definitions.Context von klasse:
> Global`
> Name klasse - Ende
> nach new12:
> $Context: Global`
> In[19]:=
>  > According your theory, I think, klasse1 should be defined in 
Global`
and
> klasse only in Rechteck But this is not the case. Therefore - 
excuse
me -
> I doubt, that your theory is correct.
>  > 
>  > ----- Original Message -----
>  > Hermann,
>  > My question is:
> Why is the variable klasse also defined in the context Global?
>  > klasse is created in the context Global` when the assignment
>  > oo`new12[] := (Print[Name klasse - Begin];
>     Print[$Context: , $Context];
>     ToExpression[klasse];
>     Print[Context of klasse: , Context[klasse]];
>     oo`resumeContext[];
>     Print[Name klasse - End];)
>  > is converted into internal form (with contexted symbols) prior to
> evaluation.
>  > This is before
>     oo`new11[Rechteck];
> is evaluated
>  > --
> 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 have the following program:
>  > oo`newContext[ctxp_] := (
>     BeginPackage[ctxp[[1]]];
>     )
> oo`resumeContext[] := (
>         EndPackage[];
>        $ContextPath = Drop[$ContextPath, 1];
>      )
> oo`new11[x_] := (
> Print[Begin new11];
> klassnam = x;
> ctx = klassnam <> `;
> oo`newContext[{ctx}];
> );
> oo`new12[] := (
> Print[Name klasse - Begin];
>  Print[$Context: , $Context];
>  ToExpression[klasse];
> Print[Context of klasse: , Context[klasse]];
>  oo`resumeContext[];
>  Print[Name klasse - End];
>  );
>   Print[Begin Programm:];
>   Print[$Context: , $Context];
>  oo`new11[Rechteck];
>  oo`new12[];
>    Print[after new12:];
>    Print[$Context: , $Context];
>
> I get the following results:
>  > Begin Programm:
> $Context: Global`
> Begin new11
> Name klasse - Begin
> $Context: Rechteck`
> klasse
> klasse::shdw: Symbol klasse appears in multiple contexts 
{Rechteck`,
> Global`}
>     ; definitions in context Rechteck`
>      may shadow or be shadowed by other definitions.Context of
klasse:
> Global`
> Name klasse - End
> after new12:
> $Context: Global`
> In[14]:=
>  > My question is:
> Why is the variable klasse also defined in the context Global?
>  > 
>  >  >  >  >  > 
====
Sorry for 'refreshing' this message but does anyone know if I can
define the assumption below.
I don't want to write specific assumptions for each term in Limit
since I have many different variables and forms of a. Is it possible
to generalise the assumption?
it would really help me (and I really don't want to have to buy any
additional maths software).
>    I am also interesting in the solution of how to make an asumption
> in Mathematica. I can't find any method in Mathematica. If anybody has
> 
> I know that this should be 0 but why can't I get mathematica to think
> likewise.

> In[4]:= Limit[Sin[a*x]/(a*x),x->Infinity]
              Sin[a x]
> Out[4]= Limit[--------, x -> Infinity]
>                 a x
> 
> Is the problem a? How can I specify the properties of or assumptions
> that may be made about a?
====
I used to use the add-on package Utilities`CleanSlate` all the time, but
it seems to be missing from version 4.2 of Mathematica (at least, I
can't find it anywhere). Has it been replaced by something else 
with the same functionality? 
-Tom
====
Is there a Mathematica function or package that generates random vectors
according to a n-dimensional normal distribution ?
Etienne
====
In[1]:=
x[t_] := 1 + Cos(2000*Pi*t) + Sin(4000*Pi*t)
In[2]:=
Plot[x, {t, 0, 8000*Pi}]
Plot::plnr: x is not a machine-size real number at t = 
0.0010471975511965976
Plot::plnr: x is not a machine-size real number at t = 
1019.5597016296903
Plot::plnr: x is not a machine-size real number at t = 
2131.4776207838036
General::stop: Further output of Plot :: plnr will be suppressed 
during this calculation
I am trying to plot the function x(t) defined above, same thing
happens when i try to plot other similar functions always get an error
saying that the machine size number is nor real, I have tried to find
an answer in this message group, as well as the mathematica help
online and built in, if someone knows what is going on please help me.
My computer is an AMD 1Gig with 256MG of ram, if that is of
consequence.  Windows XP pro
Any help would be appreciated
====
Rick,
You need square brackets not round ones - standard Mathematica syntax with
functions:
    x[t_] := 1 + Cos[2000*Pi*t] + Sin[4000*Pi*t]
This can be tracked down by evaluating
    x[0.0010471975511965976]
--

> In[1]:=
> x[t_] := 1 + Cos(2000*Pi*t) + Sin(4000*Pi*t)
 In[2]:=
> Plot[x, {t, 0, 8000*Pi}]
 Plot::plnr: x is not a machine-size real number at t =
> 0.0010471975511965976
 Plot::plnr: x is not a machine-size real number at t =
> 1019.5597016296903
 Plot::plnr: x is not a machine-size real number at t =
> 2131.4776207838036
 General::stop: Further output of Plot :: plnr will be suppressed
> during this calculation
 I am trying to plot the function x(t) defined above, same thing
> happens when i try to plot other similar functions always get an error
> saying that the machine size number is nor real, I have tried to find
> an answer in this message group, as well as the mathematica help
> online and built in, if someone knows what is going on please help me.
 My computer is an AMD 1Gig with 256MG of ram, if that is of
> consequence.  Windows XP pro
 Any help would be appreciated

====
> 
> In[1]:=
> x[t_] := 1 + Cos(2000*Pi*t) + Sin(4000*Pi*t)
x[t_] := 1 + Cos[2000*Pi*t] + Sin[4000*Pi*t]
marek
====
> MultCol[f_,x_,y_]:=
>      Select[ Import[f, Table][[All, {x, y}]], NumberQ[First[#]]& ]
> should work.
(when being online again) I didn't recocnize Allan's previous post with the 

really good idea to use n__ instead of x_, y_. So sorry for repeating 
solutions.

-- 
====
Mechnical Engineering fans 
will be able to tell me where to find a ready-made for a vibrating circular 

membrane. I need to be able to change locally (as a function of radius) the 

stiffness and mass of the membrane and visualise the vibrational modes.
dr.Lass
====
Here's is a simple example where the wave speed is 1 and the radius is 2:
<< NumericalMath`BesselZeros`;
r12= .5 BesselJZeros[1,2][[2]];
r31= .5 BesselJZeros[3,1][[1]];
u[t_,r_,theta_]:= Cos[r12*t] BesselJ[1,r12*r] Cos[theta] +
        Sin[r31*t] BesselJ[3,r31*r] Cos[3theta];
Do[
ParametricPlot3D[Evaluate[{r*Cos[theta], r*Sin[theta], u[t,r,theta]}], 
{r, 0, 2}, {theta, 0, 2*Pi},
PlotPoints->{10, 45}, PlotRange->{-1,1}, Ticks->None],
{t, 0, 4, 0.1}];
----

slhollis@mac.com
http://www.math.armstrong.edu/faculty/hollis
> Mechnical Engineering fans 
> will be able to tell me where to find a ready-made for a vibrating 
circular 
> membrane. I need to be able to change locally (as a function of radius) 
the 
> stiffness and mass of the membrane and visualise the vibrational modes.


====
Is there a way to produce text that rotates with a graphic in Mathematica?
I can only imagine of producing a 2-D text graphic, then displaying it in a
rectangle in 3D, and rotating the result. This works only as long as you
stay in front of (on the picture side) of the rectangle.
Any ideas?
Nicholas
====
When writing a series of equations in mathematica (in a text cell) is there
any way to align the equations at the = similar to what can be done in 
the
equation editor made by math type (used in word etc.)?
thanks
Mike
====
> When writing a series of equations in mathematica (in a text cell) is
there
> any way to align the equations at the = similar to what can be done 
in
the
> equation editor made by math type (used in word etc.)?

Two ways
1) Start an inline cell in your text cell (menu>Edit>Expression Input>Start
Inline Cell)
Type in your equations with the alignment marker (Esc am Esc) in after each
equal sign.
Close the inline cell (menu>Edit>Expression Input>End Inline Cell).
Select the text cell (or the inline cell, though this is a delicate
operation, do it by repeatedly double clicking in it}.
Use menu> Format>Text Alignment>On Alignment Marker.
-- you will find that all single letters are now italic - you can select
individual letters and change this - alternatively you can select the 
inline
cell and use the option inspector (menu>Format>Option Inspector) to set
SingleLetterItalics->False)
-- you can put existing text in an inline cell by selcting it and using
menu>Edit>Expression Input>Start Inline Cell, but you  may have to adjust
the line breaks after this.
2) use a grid box with a column for the equal sign and set the column
alignments appropriately - but this maay give problems if you want text to
run over columns.
--

====
Umm...
Use tab or count the number of characters to each = sign.
Yas
> When writing a series of equations in mathematica (in a text cell) is 
there
> any way to align the equations at the = similar to what can be done in 
the
> equation editor made by math type (used in word etc.)?
 thanks


====
x := 1 + Cos[2000 Pi t] + Sin[2000 Pi t]
Plot[x, {t, 0, 8000 Pi}]
works fine for me.
Notes :
- use [ and ] for functions ;
- x and x[t] are different !
> x[t_] := 1 + Cos(2000*Pi*t) + Sin(4000*Pi*t)
> 
> In[2]:=
> Plot[x, {t, 0, 8000*Pi}]
====
You have to use [ ] , not ( ) , for the argument of the function. And you
have to tell him to draw x[t], not just x...
In[1]:=
x[t_] := 1 + Cos[2000*Pi*t] + Sin[4000*Pi*t]
In[2]:=
Plot[x[t], {t, 0, 8000*Pi}]
works...
But 8000*Pi is a bit to much... You should draw until Pi/2000.
Much more interesting !

-----Message d'origine-----

Objet :  machine size real numbers?
In[1]:=
x[t_] := 1 + Cos(2000*Pi*t) + Sin(4000*Pi*t)
In[2]:=
Plot[x, {t, 0, 8000*Pi}]
Plot::plnr: x is not a machine-size real number at t =
0.0010471975511965976
Plot::plnr: x is not a machine-size real number at t =
1019.5597016296903
Plot::plnr: x is not a machine-size real number at t =
2131.4776207838036
General::stop: Further output of Plot :: plnr will be suppressed
during this calculation
I am trying to plot the function x(t) defined above, same thing
happens when i try to plot other similar functions always get an error
saying that the machine size number is nor real, I have tried to find
an answer in this message group, as well as the mathematica help
online and built in, if someone knows what is going on please help me.
My computer is an AMD 1Gig with 256MG of ram, if that is of
consequence.  Windows XP pro
Any help would be appreciated
====
I am looking for a feature in the mathematica front end editor that behaves
like e.g vi in case of an OS crash:
vi offers the option to recover the currently edited file at the time of
the crash. So additionaly to your last saved version you can recover the 
latest changes that you incoperated and have not saved so far. Is there a
similar feature in the mathematica front end, and how would I activate
this? 
Simply using save is irritating since at times I wish to rearrange greater
pieces of code without saving the change until the code does what it is
supposed to.
thanks for any help and hints,

====
This works:
x[t_] := 1 + Cos[2000*Pi*t] + Sin[4000*Pi*t]
(with square brackets for function arguments)
followed by
Plot[x[t], {t, 0, 8000*Pi}]
(with the function argument specified).
The plot looks like white noise, but do not blame Mathematica for that! If
you change the options for Plot, you surely could produce a printout that
tests the blackness of your printer toner :)
Use
Plot[x[t], {t, 0, 0.002}]
instead. Then you will see 2 periods of your function.

-----Original Message-----
Plot::plnr: x is not a machine-size real number at t =
2131.4776207838036
General::stop: Further output of Plot :: plnr will be suppressed
during this calculation
I am trying to plot the function x(t) defined above, same thing
happens when i try to plot other similar functions always get an error
saying that the machine size number is nor real, I have tried to find
an answer in this message group, as well as the mathematica help
online and built in, if someone knows what is going on please help me.
My computer is an AMD 1Gig with 256MG of ram, if that is of
consequence.  Windows XP pro
Any help would be appreciated
====
>In[1]:=
>x[t_] := 1 + Cos(2000*Pi*t) + Sin(4000*Pi*t)
>
>In[2]:=
>Plot[x, {t, 0, 8000*Pi}]
>
>Plot::plnr: x is not a machine-size real number at t = 
>0.0010471975511965976
>
>Plot::plnr: x is not a machine-size real number at t = 
>1019.5597016296903
>
>Plot::plnr: x is not a machine-size real number at t = 
>2131.4776207838036
>
>General::stop: Further output of Plot :: plnr will be suppressed 
>during this calculation
>
>I am trying to plot the function x(t) defined above, same thing
>happens when i try to plot other similar functions always get an error
>saying that the machine size number is nor real, I have tried to find
>an answer in this message group, as well as the mathematica help
>online and built in, if someone knows what is going on please help me.
>
>My computer is an AMD 1Gig with 256MG of ram, if that is of
>consequence.  Windows XP pro
x[t_] := 1 + Cos[2000*Pi*t] + Sin[4000*Pi*t];
Plot[x[t], {t, 0, 8000*Pi}];
Although for this many cycles you need to increase the PlotPoints to get an 

accurate Plot at which point it would be quicker to draw a filled 
rectangle.
Bob Hanlon
Chantilly, VA  USA
====
Square brackets around the Sin and Cos or indeed around any other
function (e.g. Tan[x], ArcSin[2.0*t], f[x])
e.g. x[t_]:= 1 + Cos[2000*Pi*t] + Sin[4000*Pi*t]
and   vvv
Plot[x[t], {t, 0, 8000*Pi}]
      ^^^
Yas
> In[1]:=
> x[t_] := 1 + Cos(2000*Pi*t) + Sin(4000*Pi*t)
 In[2]:=
> Plot[x, {t, 0, 8000*Pi}]
 Plot::plnr: x is not a machine-size real number at t =
> 0.0010471975511965976
 Plot::plnr: x is not a machine-size real number at t =
> 1019.5597016296903
 Plot::plnr: x is not a machine-size real number at t =
> 2131.4776207838036
 General::stop: Further output of Plot :: plnr will be suppressed
> during this calculation
 I am trying to plot the function x(t) defined above, same thing
> happens when i try to plot other similar functions always get an error
> saying that the machine size number is nor real, I have tried to find
> an answer in this message group, as well as the mathematica help
> online and built in, if someone knows what is going on please help me.
 My computer is an AMD 1Gig with 256MG of ram, if that is of
> consequence.  Windows XP pro
 Any help would be appreciated

>
====
How do I set up a function that behaves like the Mathematica function Sum 
vis-a-vis
its treatment of dummy variables?  For example I can define i=6 and then
enter 
Sum[f[i],{i,1,n}]
and mathematica will not make the substitution i->6 even though the output
generally still involves the dummy variable i.  I figure this has something
to do with Modules, Blocks, and Holds but I'm not sure how to put it all
together.  Any help would be greatly appreciated.
-chris
====
I have a question about the structure of the binary files generated by the 
DumpSave command.
Because the speed with which binary files generated with the DumpSave 
command can be read into Mathematica in comparison with the text files 
generated 
by the Save command, I would like to be able to let my own C program 
generate binary files with the same structure as the DumpSave generated 
binary files. For that, I have been looking for a description of the 
binary file structure. Unfortunately, I could not find such a description 
in the Mathematica book or the Wolfram website. 
Therefore, I would like to know if there exists a document outlining the 
structure of the binary file as generated by DumpSave (assuming that this 
is an 'open' file format). The version of Mathematica that I am using is 4.1 

Windows, on a Windows NT system. 

-- 




====

You should consider working through as much of Part I of The Mathematica
Book as you can to become familiar with Mathematica syntax.
In Mathematica the function arguments are always enclosed in square
brackets, not parentheses.
x[t_] := 1 + Cos[2000*Pi*t] + Sin[4000*Pi*t]
In your plot statement you must plot x[t] and not just x. The following 
will
plot, but I don't think it is what you intended.
Plot[x[t], {t, 0, 8000*Pi}];
You probably want something more like...
Plot[x[t], {t, 0, Pi/1000}];
David Park
djmp@earthlink.net
http://home.earthlink.net/~djmp/
2131.4776207838036
General::stop: Further output of Plot :: plnr will be suppressed
during this calculation
I am trying to plot the function x(t) defined above, same thing
happens when i try to plot other similar functions always get an error
saying that the machine size number is nor real, I have tried to find
an answer in this message group, as well as the mathematica help
online and built in, if someone knows what is going on please help me.
My computer is an AMD 1Gig with 256MG of ram, if that is of
consequence.  Windows XP pro
Any help would be appreciated
====

At present this is a deficiency in Mathematica 3D graphics. You can write a
Text statement and attach it to a 3D point. If you rotate the graphics, say
with SpinShow, the text does rotate with the object. But Mathematica always
renders it, so the text still shows even when the point it is attached to 
is
hidden behind a surface. It would be nice if Mathematica turned the text
rendering on and off according as to whether the point it was attached to
was hidden or not.
One solution might be to use colored points to mark features and then
identify them by text outside the rotating plot.

====
In Mathematica 4.2, Help says in part ...Limit therefore makes no
explicit assumptions about symbolic functions. Same for the Package
NLimit.  Therefore, I think we're out of luck here.
As an newbe observation, the only option for Limit (For Mathematica 4.2)
is Direction.  However, if you type ??Limit, there is another Option
listed that appears not to be documented. That option is Analytic.  I
thought maybe this would solve this type of problem, but it does not. I
do not know for sure what this option does.  Anyone else??
Information[Limit, LongForm -> True]
Limit[expr, x->x0] finds the limiting value of 
expr when x approaches x0...
Attributes[Limit] = {Listable, Protected}
Options[Limit] = {Analytic -> False, Direction -> Automatic}
As a possible workaround, would substituting many values for a work?
Here is what I am thinking...
v = Table[Random[Real, {0, 2*Pi}], {100}];
Union[(Limit[Sin[#1*x]/(#1*x), x -> Infinity] & ) /@ v]
I get all zero's like you predicted for many values of a.
{0}
If you try Integers, then I get an Indeterminate because the limit
does not like a being 0.
v = Table[Random[Integer, {-4, 4}], {10}];
Union[(Limit[Sin[#1*x]/(#1*x), x -> Infinity] & ) /@ v]
{0, Indeterminate}
--

Windows XP  & Mathematica 4.1
= = = = = = = = = = = = = = = = =
> Sorry for 'refreshing' this message but does anyone know if I can 
> define the assumption below.
> 
> I don't want to write specific assumptions for each term in Limit 
> since I have many different variables and forms of a. Is it possible 
> to generalise the assumption?
> 
> it would really help me (and I really don't want to have to buy any 
> additional maths software).

>    I am also interesting in the solution of how to make an asumption
> in Mathematica. I can't find any method in Mathematica. If anybody 
> you all ..
> 
> I know that this should be 0 but why can't I get mathematica to 
> think likewise.

> In[4]:= Limit[Sin[a*x]/(a*x),x->Infinity]
              Sin[a x]
> Out[4]= Limit[--------, x -> Infinity]
>                 a x
> 
> Is the problem a? How can I specify the properties of or 
> assumptions that may be made about a?
> 
Reply-To: <drbob@bigfoot.com>
====
There are many values of a for which the limit doesn't exist... a==I and
a==0, for instance.  You can eliminate the first possibility by setting
a=Re[b] where b is unknown, but that still leaves the second
possibility, so Mathematica still doesn't find the limit you want.
Here's an answer, in which we assume a is greater than or equal to some
small delta (surrogate for greater than zero).
delta = 10^(-25); 
a = Abs[b] + delta; 
Limit[Sin[a*x]/(a*x), x -> Infinity]
0
This also works, though it shouldn't:
delta = 10^-25;
a = Re[b] + delta;
Limit[Sin[a*x]/(a*x), x -> Infinity]
0
Perhaps THIS is the best way to assume a>0:
a = Exp[Re[b]]; 
Limit[Sin[a*x]/(a*x), 
  x -> Infinity]
0

-----Original Message-----
>    I am also interesting in the solution of how to make an asumption
> in Mathematica. I can't find any method in Mathematica. If anybody has
all ..
> 
> I know that this should be 0 but why can't I get mathematica to
think
> likewise.

> In[4]:= Limit[Sin[a*x]/(a*x),x->Infinity]
              Sin[a x]
> Out[4]= Limit[--------, x -> Infinity]
>                 a x
> 
> Is the problem a? How can I specify the properties of or assumptions
> that may be made about a?
Reply-To: <drbob@bigfoot.com>
====
x isn't a number; it's a function.  x[t] is a number.
Bobby
-----Original Message-----
Plot::plnr: x is not a machine-size real number at t = 
2131.4776207838036
General::stop: Further output of Plot :: plnr will be suppressed 
during this calculation
I am trying to plot the function x(t) defined above, same thing
happens when i try to plot other similar functions always get an error
saying that the machine size number is nor real, I have tried to find
an answer in this message group, as well as the mathematica help
online and built in, if someone knows what is going on please help me.
My computer is an AMD 1Gig with 256MG of ram, if that is of
consequence.  Windows XP pro
Any help would be appreciated
====
Header says it all.
====
The line below is from the 'Extra Examples' in the help browser for the
BitAnd function.  
On my computer the resulting graphic flickers rapidly.  
Do others have the same results, and does anyone have an explanation for 
the
flickering?  
I suspect this is demonstrating a limitation of my monitor/graphics card.
In[1] := 
   ListDensityPlot[Table[Sign[BitAnd[x,y]],{x,127},{y,127}]];
 
====
>
>I downloaded the SpreadOptions package and modified it so that it will 
work
>like a standard package. The package file is attached. (Other readers can
>modify the package as I describe below.)
>
>To use it, create a new Finance folder in ExtraPackages. Put the
>SpreadOption.m file there. Then you can simply load the package with:
>
>Needs[Finance`SpreadOption`]
>
>All I did was change the name of the package in the BeginPackage statement
>to Finance`SpreadOption. Anyone can do this by editing the package 
file.
>
>This, it seems to me, is the method that should be used for third party
>packages. If this method is used, the packages behave just like standard
>packages. However, probably because it is not clearly presented in the WRI
>documentation, very few people use the method. The result is a lot of
>confusion in connecting to packages. Also, it seems to me that packages
>should be in a standard place and the StandardPackages and ExtraPackges
>folders that WRI provides are those places.
>
>David Park
>djmp@earthlink.net
>http://home.earthlink.net/~djmp/
ExtraPackages is a fine place to put packages if your package consists 
solely of *.m files.
However, in general I would recommend the Applications directory. The front 

end has its own sets of paths (for finding palettes, stylesheets, and 
documentation). With the default settings, the front end looks in the 
Applications and Autoload directories, not the ExtraPackages directory.
--------------------------------------------------------------
Omega Consulting
The final answer to your Mathematica needs
Spend less time searching and more time finding.
http://www.wz.com/internet/Mathematica.html
====
>
>I have a problem with using the function ToExpression to make a string
>into an expression that can then be evaluated. Something goes wrong with
>the evaluation - when I type the same expression in by hand, then it
>evaluates properly. Either I am totally off in my understanding of how
>ToExpression works, or there is a potential bug (I am using version
>(sheubac@calstatela.edu) rather than to the mathgroup, as I am
>travelling and cannot subscribe to the mathgroup right now - it would
>
>Silvia
********************************
>
>i = {1, 0, 0}; j = {0, 1, 0}; k = {0, 0, 1}; zero = {0, 0, 0};
>a = {i, j, k}
>
>str1 = (((..)(..))(..))

>
>The function insertx is to convert this string of parentheses and dots
>into a cross product, where each dot represents a different variable,
>and the parentheses indicate the groupings of the cross product.

>
>insertx[str_] := Module[{foo = str, vars, n, pos},
>     foo =
>       StringReplace[
>         foo, {)( -> ),(, .( -> .,(, ). -> ),., 
.. -.,.}];
>     foo = StringReplace[foo, {) -> ], ( -> Cross[}];
>          n = Length[StringPosition[foo, .]];
>     Do[pos = StringPosition[foo, .][[1]];
>           foo = StringReplacePart[foo, ToString[Slot[i]], pos]
>          (*Print[foo]*) , {i, n}];
>     ToExpression[foo]]

>
>The function properly converts the string into the associated cross
>product, but does not evaluate correctly. Checking whether what is
>returned by the function is still a string gives a false, and it should,
>as the function should return an expression. (the output is indented)
>
>insertx[str1]
    ((#1[Cross]#2)[Cross](#3[Cross]#4))[Cross](#5[Cross]#6)
>
>insertx[str1] &[j, k, i, k, k, i]
    ((#1[Cross]#2)[Cross](#3[Cross]#4))[Cross](#5[Cross]#6)
>
>StringQ[insertx[str1]]
    False

>
>Typing in the corresponding cross product by hand and evaluating it
>returns the correct result.

>
>Cross[Cross[Cross[#1, #2], Cross[#3, #4]], Cross[#5, #6]] &[j, k, i, k,
>k, i]
    {1, 0, 0}
>
>What is wrong??? I have spent many hours on trying to understand this,
>so either I am completely off or there is a bug.
This is an improper use of Slot. Instead, I would specify the vectors in 
the definition. Then you can use the vectors with Part:
In[1]:=
insertx[str_][lst_] := Module[{foo = str, n, pos},
     foo =
       StringReplace[
         foo, {)( -> ),(, .( -> .,(, ). -> ),., 
.. ->
.,.}];
     foo = StringReplace[foo, {) -> ], ( -> Cross[}];
          n = Length[StringPosition[foo, .]];
     Do[pos = StringPosition[foo, .][[1]];
           foo = StringReplacePart[foo, ToString[lst[[i]]], pos]
          (*Print[foo]*) , {i, n}];
     ToExpression[foo]]
In[2]:=
insertx[str1][{j, k, i, k, k, i}]
Out[2]=
{1,0,0}
Or with a list operation like Fold:
In[3]:=
myinsertx[str_][ lst_] := Module[{foo = str},
     foo =
       StringReplace[
         foo, {)( -> ),(, .( -> .,(, ). -> ),., 
.. ->
.,.}];
     foo = StringReplace[foo, {) -> ], ( -> Cross[}];
     foo = Fold[replace,foo,lst];
         ToExpression[foo]]
In[4]:=
replace[str_, item_]:=
   Module[{pos},
     pos=First[StringPosition[str,.]];
     StringReplacePart[str, ToString[item], pos]
     ]
In[5]:=
insertx[str1][ {j, k, i, k, k, i}]
Out[5]=
{1,0,0}
--------------------------------------------------------------

The final answer to your Mathematica needs
Spend less time searching and more time finding.
http://www.wz.com/internet/Mathematica.html
====
I am glad that some discussion of this is being generated.
In general I disagree with your recommendation. If palettes and StyleSheets
go along with a package, why not put them in 
ConfigurationFrontEndPalettes
and ConfigurationFrontEndStyleSheets, which seem to be the places made 
for
them? Then they automatically appear in the menu listings. (And you as a
package writer would also know where to expect to find them.)
I am not certain if I completely understand your approach and would
appreciate further explanation. If you have a set of packages, stylesheets,
palettes and documentation, where precisely would you put them? At the top
level of the Applications directory? Or would you put them in a 
subdirectory
within the Applications folder? If they are in a subdirectory, will
Mathematica automatically find them? What if you want to call them from
another subdirectory in the Applications folder? What about having several
documentation folders for different packages? Can you have two
DocumentationEnglish folders in one directory? I don't think so. Suppose
you provide a package with documentation and I supply a package with
documentation. Is the user supposed to merge the BrowserCatergories files
together? Would he know how to do it?
In providing software to a user, I think it is a great advantage if the 
user
does not have to specify or alter any paths. That may be simple for you, 
but
it is a major hurdle to many users. If one puts the various objects in the
places that WRI provided for them, it is never neccessary to specify paths.
(Perhaps you want some commands in your package to automatically open a
palette. If the palette has been placed in 
ConfigurationFrontEndPalettes,
you as the programmer can automatically write the path to open it. The user
shouldn't have to worry about it. And he could also open the palette from
the menu.)
I don't claim to be the greatest expert on this subject. I would be glad to
learn more. But, so far, all the arguments I have heard for putting packages
somewhere other than an ExtraPackages subfolder have been unconvincing to
me.

http://www.wz.com/internet/Mathematica.html
====
1. There is no way to make assumptions inside Limit. I very much doubt 
that buying any other software will help.
2. There is no need to use Mathematica or any other computer program to 
do solve this problem. It is exactly the kind of thing that is trivial 
to do with a little knowledge of mathematics and a bit of intelligence 
but is not suitable for a computer. Computer software should be used to 
deal with problems that are *computationally* difficult but do not 
require any intelligence. There is no computer software today (and there 
won't be for quite a long time) that can replace the latter .

http://platon.c.u-tokyo.ac.jp/andrzej/
> Sorry for 'refreshing' this message but does anyone know if I can
> define the assumption below.
 I don't want to write specific assumptions for each term in Limit
> since I have many different variables and forms of a. Is it possible
> to generalise the assumption?
 it would really help me (and I really don't want to have to buy any
> additional maths software).

>    I am also interesting in the solution of how to make an asumption
> in Mathematica. I can't find any method in Mathematica. If anybody has
> all ..
 I know that this should be 0 but why can't I get mathematica to think
> likewise.

> In[4]:= Limit[Sin[a*x]/(a*x),x->Infinity]
               Sin[a x]
> Out[4]= Limit[--------, x -> Infinity]
>                 a x
 Is the problem a? How can I specify the properties of or assumptions
> that may be made about a?

>
====
If we want to contract two tensor arrays S and T in Mathematica we can 
write
S.T. We could also use the Inner product. This will contract the last level
of S with the first level of T.
I would like to know how to contract more than one pair of levels. Suppose 
S
and T are 3rd order tensors and I want to contract the last two levels of S
with the first two levels of T.  How can I do that?
Or suppose I wanted to contract all three levels. For example with...
Smat = Array[S, {3, 3, 3}]
Tmat = Array[T, {3, 3, 3}]
I would like to obtain the answer...
Smat Transpose[Tmat, {3, 2, 1}]
Plus @@ Flatten[%]
but by a more general method. That is, I would like to contract any number
of adjacent levels in two tensors.
David Park
djmp@earthlink.net
http://home.earthlink.net/~djmp/

====
You need to copy:
<Drive Letter>:<Path to Wolfram Research Files>Wolfram 
ResearchMathematica4.1SystemFilesFontsType1Math1Mono.pfb
to
<Drive Letter>:<Path to MikTeX Files>fontstype1adobeMath1Mono.pfb
You then need to click on:
Start Menu -> All Programs -> MikTeX 2 -> MikTeX Options
and then click Refresh Now under the General tab to refresh the file 

name database.
Vince
> I am a new user of Mathematica and I have currently encountered a
> problem.  In fact, I plotted a graph which contains arrows ( for this I
> called up on the <Graphics` arrows      package) and then i exported it
> to Tex as an EPS file but when compiling in Miktex, I am getting the
> following error message:  Header file Math1Mono.pfb couldn't be found.
> Is there a way to solve the problem?

====
I am new myself, and trying to learn.  I went to 
http://search.wolfram.com and searched Trig->True under The Document
Center.
There was only 1 result.  It pointed to a Mathematica 3 example.
Denominator[Cos[3*x]/Sin[2*x]]
1
Denominator[Cos[3*x]/Sin[2*x], Trig -> True]
Sin[2*x]
Apparently, this can make a big difference on what one is expecting for
an answer.
I think it is because Mathematica is not using Denominator 
directly,
but Times & Power instead.
 
-- 
Dana 
Windows XP  & Mathematica 4.1
= = = = = = = = = = = = = = = = = 
The usage message for Trig says:
Trig is an option for algebraic manipulation functions which specifies
> whether trigonometric functions should be treated as rational
functions of
> exponentials.
> ---------
The built-in functions that have the Trig option are:  
> Apart, ApartSquareFree, Cancel, Coefficient, CoefficientList, Collect,
> Denominator, Expand, ExpandAll, ExpandDenominator, ExpandNumerator,
> Exponent, Factor, FactorList, FactorSquareFree, FactorSquareFreeList,
> FactorTerms, FactorTermsList, FullSimplify, Numerator, PolynomialGCD,
> PolynomialLCM, PolynomialMod, Resultant, Simplify, Together.

> After searching all available documentation I can't find a single
example of
> where 
> f[expr, Trig->True]   would give a different result than 
> f[expr, Trig->False], and I can't come up with one on my own. Can you
give
> some examples. An example for each function above is not necessary.
---------
>   Ted Ersek
>   Get Mathematica tips, tricks from
>   http://www.verbeia.com/mathematica/tips/Tricks.html
====
When exporting graphs to EPS, I find that the Micro character (upright 
Mu, used as an abbreviation for the micro in micrometre) isn't 
displayed.  I'm loathe to replace [Micro] in my graph labels with 
[Mu], because I want to be consistent and reserve italicized symbols 
for math variables.
Does anyone know which Mathematica font package contains the Micro 
character, or some other way I can export [Micro] to EPS?

====
the [Micro] in Math3 at 0xD0  
but it's optical difference to [Mu]
is only visible for a trainied eye :-)
  
When exporting graphs to EPS, I find that the Micro character (upright
> Mu, used as an abbreviation for the micro in micrometre) isn't
> displayed.  I'm loathe to replace [Micro] in my graph labels with
> [Mu], because I want to be consistent and reserve italicized symbols
> for math variables.
Does anyone know which Mathematica font package contains the Micro
> character, or some other way I can export [Micro] to EPS?

> Vince
Reply-To: <John_C_Erb@prodigy.net>
====
One could use the Rectangle function.  An example of this
is given in Section 5.2 (pages 114-116) of the book Applied Mathematica,
Getting
It Started Getting It Done, by Shaw and Tigg.
An example:
Plot1=Plot[Sin[x],{x,0,2Pi}];
Plot2=Plot[2x^2+5x, {x,0,2Pi}];
Show[Graphics[{Rectangle[{0,0},{1,1},Plot1],
        Rectangle[{0.55,0.45},{0.95,0.9},Plot2]}]];

-----Original Message-----

====
I have the following function and it is called many many times in a program
I have written.  This is a bracket operator for a certain type of Lie
Algebra I am looking at and it must be called over and over to test whether
or not the Jacobi is actually 0 as it should be.  With very low numbers for
n, my program runs in a second or less.  If n is around 20+ or if the
coefficients are large, the program takes nearer a minute or two.  That's
not a long time, but I have a feeling that it's having to calculate the 
same
Bracket many times in the program and I'm hoping to get rid of that 
somehow.
However, I don't understand the differences between := and = enough.  I've
read the help files, but I don't understand the subtle differences such as 
I
can call
Bracket[stuff, something]:= somethingelse
more than once, but it doesn't seem as if I can use = more than once.
Bracket[a_.*e[i_], b_.*e[j_]] :=
    Which[
      i + j > n, 0,
      i == j, 0,
      i == 1, a*b*e[j + 1],
      i == 2 && j == 3, 14a*b*e[7],
      i == 3 && j == 4, 0,
      i < j, -a*b*Bracket[e[i + 1], e[j - 1]] +
        Bracket[e[1], Bracket[e[i], e[j - 1]]],
      i > j, -a*b*Bracket[e[j], e[i]]
      ];
(* bilinear function *)
Bracket[mul_, expr_Plus] := Map[Bracket[mul, #] &, expr];
Bracket[expr_Plus, mul_] := Map[Bracket[#, mul] &, expr];
Bracket[0, x_] := 0
Bracket[x_, 0] := 0
Any help would be much appreciated.
Geoff Tims
====

How about this?
t = Table[Sin[(x + y)], {x, 0, 100}, {y, 0, 100}];
plot1 = First[
      Graphics[ListDensityPlot[t, Mesh -> False,
          DisplayFunction -> Identity]]];
plot2 =
    Show[Graphics[plot1],
      AspectRatio -> Automatic,
      PlotRange -> {{-0.25, 101}, {0.25, 101}},
      ImageSize -> 500];
Try exporting plot2. ListDensityPlot doesn't allow you to adjust the
PlotRange directly. At least, it didn't work for me. By converting from
DensityGraphics to Graphics we can adjust the PlotRange to just meet the
edges of the grid. (The values I used seem a little strange, but that is
what seemed to work best. I made the image size larger to more easily see
what is happening.)
This can be done more directly with the DrawGraphics package.
Needs[DrawGraphics`DrawingMaster`]
Draw2D[
    {ListDensityDraw[t, Mesh -> False]},
    AspectRatio -> Automatic,
    PlotRange -> {{-0.25, 101}, {0.25, 101}},
    ImageSize -> 500];
David Park
djmp@earthlink.net
http://home.earthlink.net/~djmp/

> I want to calculate some patterns but I always get artifacts. The reason
> seems to be the white border around the plots. If I calculate 100
> x 100 points
> and plot or export them (see example below), the picture is
> really 100 x 100
> pixels but with a white border. So my calculations are 'squeezed'
> (some lines
> and rows are missing!).
> Does anyone have an idea how to get rid of this white border or will I
> really have to export my tables to a textfile and use another
> software zu produce
> a bitmap-file??

> Example:
> t = Table[Sin[(x + y)], {x, 0, 100}, {y, 0, 100}];
> Export[test.tif,
>   ListDensityPlot[t, Frame -> False, Mesh -> False, ImageSize -> {100,
> 100}]]

====
I hava some trouble with Mathematica 4 image import. Usually one should 
import an image with:
myImage = Import[image.jpg]
The image is a valid jpg and can be found via $Path. 
But Mathematica can't load it. Wenn I  type
$InputFormats
I get
$InputFormats
as the answer. This should be something like
{AIFF, AU, BMP, Dump, EPS, EPSI, EPSTIFF, 
Expression, 
PNM, PPM, PSImage, RawBitmap, SND, Table, Text, 
TIFF, 
UnicodeText, WAV, Words, XBitmap}
I guess. The help says these formats should be build in. So why they aren't visible?

-- 

====
Mathematica search packages in the $Path and
*not* an arbitary file. 
Give the full path an all will work fine.
  

> I hava some trouble with Mathematica 4 image import. Usually one should
> import an image with:
myImage = Import[image.jpg]
The image is a valid jpg and can be found via $Path.
But Mathematica can't load it. Wenn I  type
$InputFormats
I get
$InputFormats
as the answer. This should be something like
{AIFF, AU, BMP, Dump, EPS, EPSI, EPSTIFF, 
Expression,
> PNM, PPM, PSImage, RawBitmap, SND, Table, 
Text, TIFF,
> UnicodeText, WAV, Words, XBitmap}
I guess. The help says these formats should be build in. So why they 
aren't visible?

====

I want to calculate some patterns but I always get artifacts. 
> The reason
> seems to be the white border around the plots. If I calculate 
> 100 x 100 points
> and plot or export them (see example below), the picture is 
> really 100 x 100
> pixels but with a white border. So my calculations are 
> 'squeezed' (some lines
> and rows are missing!).
> Does anyone have an idea how to get rid of this white border or will I
> really have to export my tables to a textfile and use another 
> software zu produce
> a bitmap-file??

> Example:
> t = Table[Sin[(x + y)], {x, 0, 100}, {y, 0, 100}];
> Export[test.tif, 
>   ListDensityPlot[t, Frame -> False, Mesh -> False, ImageSize -> {100,
> 100}]]

I don't know whether this is of any help. The white margin you see is the
rest of the plot region not filled by your density plot. You may recognize
this, when you specify Background -> Hue[0] (now beeing red).
Deplorably the Option Background -> None is only effective for Text 
objects.
So I played for a while  with the option PlotRange to have filled all of 
the
range but nothing cut off from your plot, e.g.
In[91]:=
graph2 = ListDensityPlot[t, Frame -> False, Mesh -> False, 
        PlotRegion -> {{0 - #1, 1 + #2}, {0 - #3, 1 + #4}}, 
        Background -> Hue[0]] &[0.019, 0.019, 0.3, 0.3]
The effects where not predictable (to me), especially I couldn't get rid 
off
a small margin at the top or at the bottom. If #3 and #4 are great enough
and equal (you can make them arbitrarily great) there remains a small 
margin
at top and bottom as seen from Mathematica rendering. Exporting this as a
.tif and looking at that from Microsoft Photo Editor, I saw the margins at
bottom and right.
Exporting to .eps and displaying with Ghostview gave a similar appearance 
to
rendering within Mathematica . A short inspection of the .eps file makes me
guess that there is some slight imprecision with the calculation of the
bounding box and/or the scaling factor. Perhaps it is possible to fix the
postscript and convert it to another format. But then it may appear to be
easier to extract the gray values form the RasterGraphics and generate your
graphic file by different means (perhaps not too difficult in your case).
Possibly you will be content with the smallest margin possible (when you
color it white, or better adapt it to the background of you final
application).

====
Dear all,
I hope someone can give me advice concerning the following Mathematica
I am trying to produce an animation of a circular walk around a 3D
plot. I get the animation, but it looks like sometimes I get nearer to
the plot and sometimes the plot looks more distant, so that I don't
get the impression of walking on a circle.
My guess is that I have not yet understood how the options
ViewCenter and ViewPoint work.
For [alpha = 0.0, alpha < 2 Pi, alpha = alpha + 0.1,
  Plot3D[ Exp[-(x^2 + y^2)], {x, -2, 2}, {y, -2, 2}, 
          {ViewCenter -> {0, 0, 0}, 
           ViewPoint -> {5 Cos[alpha], 5 Sin[alpha], 0.5}}]]
So all viewpoints should lie on a circle of radius 5 around the origin
and with a constant height of 0.5. But the visual impression is
different, as decribed above. Anyone knows what's wrong?
  Tobi
====