A42 === Do you refer here to the Option Inspector? If so, what option name are you referring to?-- Murray Eisenberg murray@math.umass.eduMathematics & Statistics Dept.Lederle Graduate Research Tower phone 413 549-1020 (H)University of Massachusetts 413 545-2859 (W)710 North Pleasant StreetAmherst, MA 01375 ==== 4.2.1 embeds the required Mathematica fonts into the EPS graphic by *default*. So, you don't have to fiddle with any options.But if you want to turn font embedding off, the option name is IncludeSpecialFonts under the ToPostScriptOptions category (there's another IncludeSpecialFonts in the PrintingOptions category which does not affect EPS, but is related to printing under X)..Sincerely,John Fultzjfultz@wolfram.comUser Interface GroupWolfram Research, Inc. ==== Be careful ! Sqrt[a]*Sqrt[b] is not equal to Sqrt[a*b] if a and b are notreal positive !If you consider a and b real positive, you can tell Mathematica that it isso :You can also use PowerExpandMeilleures salutationsFlorian Jaccard-----Message d'origine-----Envoy.8e : mar., 3. d.8ecembre 2002 10:35è : mathgroup@smc.vnet.netObjet : Hi!Does somebody know an algorithm to simplify the product Sqrt[a]*Sqrt[b] ?With many thanksGernot ==== Apply Simplify or FullSimplify with Assumptions.Generally speaking, over the complex plane, Sqrt[a]*Sqrt[b] cannot besimplified further, and Mathematica knows it. In[1] := FullSimplify[Sqrt[x]Sqrt[y], x [Element] Complexes] Out[1] = Sqrt[x]*Sqrt[y]The same holds for the real plane. In[2] := FullSimplify[Sqrt[x] Sqrt[y], x [Element] Reals] Out[2] = Sqrt[x]*Sqrt[y]Sometimes, if your work over the real plane, things might look simpler. Out[3] = Sqrt[x*y] In[4] := Simplify[Sqrt[x] Sqrt[y], x < 0 && y < 0] Out[4] = -Sqrt[x*y]and so on.Best wishes,Vladimir BondarenkoMathematical and Production DirectorSymbolic Testing Group http://www.CAS-testing.org/ GEMM Project (95% ready)Mail : 76 Zalesskaya Str, Simferopol, Crimea, Ukraine ==== Dear Hartmut,It may be that introducing a test prevents internal compiling.One more simplification - and this points up the value of using Alternativesto reduce the problem to simple pattern matching without conditions ortests: Random[Integer,{1,2000}]},{6000}]; Random[Integer,{1,2000}]},{12000}];In my second posting I ended with s1[[Flatten[Position[ s1[[All,1]], Alternatives@@Union[s2[[All,1]]]]]]];//Timing {3.24 Second,Null}Using Cases gives a simpler and slightly faster version: Cases[ s1, {Alternatives@@Union[s2[[All,1]]],_}];//Timing {3.13 Second,Null}How useful Union or, as below, Intersection are will be, depends of courseon the particular inputs, but for the current ones I get Cases[ s1, {Alternatives@@s2[[All,1]],_}];//Timing {5.32 Second,Null} Cases[ s1, {Alternatives@@Intersection[s1[[All,1]],s2[[All,1]]],_}];// Timing {3.41 Second,Null}In the last one, Cases may be repeating some of the work that Intersectionhas done.Perhaps you can build spme of this in your developments.Allan---------------------Allan HayesMathematica Training and ConsultingLeicester UKwww.haystack.demon.co.ukhay@haystack.demon.co.ukVoice: +44 (0)116 271 4198----- Original Message -----believedkillstill{4}]],Iand ==== I don't know the answer, but my simple mind is telling me the integer 1 is thesame as 1.0 is that right or no?what we can also do to look at another perspective is this:Limit 1^nanyone?RAyRAy =) hehehe ==== Yes, in my opinion. Indeed, I would, arguing similarly, say that1^x should be 1 for _all_ x, including ComplexInfinity. It is, inessence, the same idea that I proposed in the recent threadindeterminate expression. (Below my signature, I've copied the mostrelevant post from that thread.)At least I can say that Mathematica's returning Indeterminate for both0^0 and 1^Infinity indicates that it is following a _consistent_philosophy. But I disagree with that philosophy. David-- -------------------- http://NewsReader.Com/ --------------------Usenet Newsgroup Service New Rate! $9.95/Month 50GB ==== Isn't this a bit like trying to divide by zero?Infinity + Infinity = Infinity; however, one can not subtract infinity frominfinity or or divide by it.Steven Shippeeshippee@jcs.mil ==== It is astonishing that debates like this keep coming up. To anyone who disagrees with the notion that 1^Infinity is indeterminate, I suggest that you write something called a new kind of calculus. But you might want to learn the old kind first.---Selwyn Hollis ==== Mathematica generally uses indeterminate for expressions which are by nature ambiguous, which means they can be represented as limits leading to different answers. This is the case with 1^Infinity. For example, consider(2^(1/n))^n and let n tend to Infinity. This can be thought of as 1^Infinity yet it remains all the time equal to 2.I assume the same is true with other expressions where Indeterminate is returned. Of course this is not mathematically the only valid approach, one adapt various conventions, for after all these are but conventions. But personally I find it rather convenient and have got programming examples which only work because Indeterminate rather than a number are returned in similar cases. I personally think this answer is on the whole more ßexible than any alternative, and I would vote for keeping it.Andrzej Kozlowski ==== I've been trying to run some modestly complicated animations consisting of several ParametricPlot3D curves. I can generate the sequence of images, but when I try to use ShowAnimation, Mathematica starts using all the CPU, and if there is sufficient number of frames, It starts swapping. What that happens I either have to reboot the system, or wait several minutes for Mathematica to let me have a few cycles so I can kill it. I've seen what I can do with Java 3D, so I know such animations can be run on this box.What is actually happening when the animations are run? Even if I get them to start, they never run smoothly. The always freze momentarily about every 7 frames or so, and sometimes they will stall for a fairly long period.I really think Mathematica is a wonderful program for handling mathematical problems. It's ability to graph mathematical functions is fantastic. I am interested in studying the time evolution of physical systems, and believe that 3D animation is the best way to understand the mathematical expressions used to describe these systems. When I get beyond the simplest systems, Mathematic seems unable to support this aspect of my modeling.I don't have time to write my own mapping between Java 3D and Mathematica, unless WRI wants to fund the effort. Is there a good way to handle this kind of situation using Mathematica?-- STHHatton's Law: There is only One inviolable Law.Reply-To: kuska@informatik.uni-leipzig.de ==== Hi,Mathematica must keep all the frames of you animation inmemory and this is probably to much.You can make simple animations with MathGL3d http://phong.informatik.uni-leipzig.de/~kuska/mathgl3dv3/as explained in the MVManual.nb. But it depend a bit on thecomplexity of your scene and the actual speeed of MathLink. Jens ==== If you have a finalized set of images, you can use automatedExport[...] commands to create a numbered sequence of images in asubdirectory.Something like pic001.jpg,pic002.jpg,...,pic100.jpg.Then you can use the free Bink tool from www.radgames.com to make ananimation. It compresses the animation, plus you can turn it into astandalone .exe file for people who don't have Mathematica.This solution is great for finalized large animations.Orestis ==== can you include or send me the notebook with these parametic plots. I would beinterested in looking at them and see how they run on my machine. ==== I guess I never learn anything the easy way. I've been told this before, but until I actually started poking around in the data from Mathematica 3D animations, I didn't understand what all this meant. I had been creating my When I looked at how they were represented, I realised how much useless data was being generated. Anyhow, just about the time I was ready to put this on the web, I had the first hard drive failure in my life (on my own system). What luck! It was the drive holind my web-site. Fortunately, I have backups of everything. Nonetheless, It took a while to get everything back up. It was already well past the end of the day. I'm too tired to add much explanation to the I made extensive use of Dr. Míóder.81[YAcute] CSM`Classes` package. It's not quite Java level OO, but It certainly is usable. http://public.globalsymmetry.com/projects/lorenz3d/ lorenz3D-scene1.nbThe notebook is just a front-end. I'm not ready to publish the code yet.-- STHHatton's Law: There is only One inviolable Law. ==== I have measured the absolute value of the electric field in severalpoints of a room.I would like to make a 3D plot of the room, showing each one of thepoints measured. Each point should be accompanied by its electricfield value or, preferably, by a color associated with the value.After that, I wanted to make a 3D interpolation of those points toobtain a colorful plot that showed how the electric field varies alongthat room (a 3D density plot).Any ideas are welcome, even if you don't know how to solve the wholeproblem!Reply-To: kuska@informatik.uni-leipzig.de ==== Hi,a) you can't draw a field {fx,fy,fz} with DensityPlot[]b) make a Delauny triangulation (in 3d it is a tetrahedisation) of you points and serach for every point the tetrahedron and interpolate the field components from the measured corner values Jens ==== Leonardo,Here I make a sample function to show how you might do it.f[x_, y_, z_] := Abs[Sin[x]Cos[y]Sin[z]];The following creates a sample set of data.data = Flatten[Table[{f[x, y, z], {x, y, z}}, {x, 0, Pi/2, Pi/20}, {y, 0, Pi/2, Pi/20}, {z, 0, Pi/2, Pi/20}], 2];Now we create a set of plotting points by transforming each data point.In our data, f goes from 0 to 1. I made the Hue go to less than 1 because 1looks almost like 0 and doesn't distinguish. For your actual function youwould have to design a more sophisticated color function. Here is a sampleplot point...Part[plotpoints, 654]{Hue[0.2241661706546524], Point[{Pi/4, Pi/5, Pi/5}]}Now, the plot is easy...plot1 = Show[Graphics3D[ {plotpoints}],You can get a better picture by spinning it aroundNeeds[Graphics`Animation`]SpinShow[plot1] SelectionMove[EvaluationNotebook[], All, GeneratedCell]FrontEndTokenExecute[OpenCloseGroup]; Pause[0.01];0.25,The whole thing looks perfectly awful!! It is very difficult to make a good3D density plot. The problem is that the points get in each other's way.Another effect makes things even worse. The ordered points create variouspatterns, depending upon the particular viewpoint. These patterns havenothing to do with the function or data that you are trying to show, butonly with the arrangement of data points. Even if the points are randomlyplaced, you will still obtain various spatial patterns. It is what EdwardTufte calls the 1+1=3 effect in graphics. The various elements combine toproduce unintended visual elements that grab the viewers attention at theexpense of the real information.So, is there a better approach? A lot depends upon the nature of yourfunction or data, and also on what aspect of the function you are trying toemphasize. You could, for example, do 2D contour plots for various slices ofthe room. You could also try cut-a-way plots of various level surfaces in3D. A multiple combination of images may best show whatever it is you aretrying to show.David Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/problem !Reply-To: francoisNO@SPAMit-c.dk ==== Hi,Hope I won't be too imprecise, I just started with mathematicalast Friday.1)I want to define a function, say f, with among others the following propertyf[f[a,b,c],d,e] should be evaluated as f[f[a,b,1/2],b,1/2] in caseb = d and c = d = 12) from a list, say {a,b,c}, I would like to generateI guess I have understood the iterative way of doing that,but what about a more functionqal form?Francois-- Francois LauzeThe IT University of Copenhagen, Glentevej 672400 Kbh NV, Denmark ==== Francois,1) I wonder if b=d should be b = e. If so f[f[a_,b_,1],1,b_] :=f[f[a,b,1/2],b,1/2];--Allan---------------------Allan HayesMathematica Training and ConsultingLeicester UKwww.haystack.demon.co.ukhay@haystack.demon.co.ukVoice: +44 (0)116 271 4198property ==== Francois1) The naive answer is simply to include the linef[f[a_,1,1],1,e] := f[f[a,1,1/2],1,1/2]in your definition of the function f. (Note that by your definitionb=c=d=1). However, I'm not sure I've fully understood your question sothis naive answer may not go far enough.2) Again, I'm not entirely sure what you want to do. But here's oneway of transforming your sample input into your sample output. First,load the Combinatorica package, we need the KSubset function:<