A13 ==== I am new with Mathematica, I have one question, I know that the sollution might be very easy, but I wasn't able to find it by now. I would like to draw an ellipse, the formula let's say is as follows: 0.09 x^2 +0.04 x y + 0.06 y^2 = 4 Maciej ==== Although it is not outstanding, I prefer to write your equation in the form: 9*x^2+4*x*y+6*y^2==400 In order to plot the ellipse you can use the package Graphics`ImplicitPlot` in this way: In[1]:= << Graphics`ImplicitPlot` In[2]:= ImplicitPlot[ 9*x^2 + 4*x*y + 6*y^2 == 400, {x, -7, 7}, PlotStyle -> RGBColor[1, 0, 0]]; Germ.87n Buitrago A. ----- Original Message ----- > 0.09 x^2 +0.04 x y + 0.06 y^2 = 4 > Maciej > ==== first this in not an ellipse. the ellipse generic formula is : (x^2)/(a^2)+(y^2)/(b^2) = 0. I guest that you want report a graph of the solution y of the equation: 0.09 x^2 +0.04 x y + 0.06 y^2 = 4 for x varying into a given range. As you can veryfing using: Solve[0.09 x^2 + 0.04 x y + 0.06 y^2 == 4, y] the solutins are two: {{y -> 8.333333333333334*(-0.04*x - 0.1414213562373095*Sqrt[48.00000000000001 + 0.*x - 1.*x^2])}, {y -> 8.333333333333334*(-0.04*x + 0.1414213562373095*Sqrt[48.00000000000001 + 0.*x - 1.*x^2])}} than you can construct a function: function1[x_] = 8.333333333333334*(-0.04*x - 0.1414213562373095*Sqrt[48.00000000000001 - *x^2])}, that is a real value only if (48.00000000000001 - *x^2)>0, i.e. if approximatively -6 True] the same for the other soluztion. good luck -u pimak ha scritto nel messaggio > I am new with Mathematica, I have one question, > I know that the sollution might be very easy, but I wasn't able to > find it by now. > I would like to draw an ellipse, the formula let's say is as follows: 0.09 x^2 +0.04 x y + 0.06 y^2 = 4 > Maciej > Reply-To: murray@math.umass.edu ==== One way is: ContourPlot[0.09 x^2 + 0.04 x y + 0.06 y^2, {x, -10, 10}, {y, -10, 10}, Contours -> {4}, PlotPoints -> 50, ContourShading -> False]; > I am new with Mathematica, I have one question, > I know that the sollution might be very easy, but I wasn't able to > find it by now. > I would like to draw an ellipse, the formula let's say is as follows: 0.09 x^2 +0.04 x y + 0.06 y^2 = 4 > Maciej > -- Murray Eisenberg murray@math.umass.edu Mathematics & Statistics Dept. Lederle Graduate Research Tower phone 413 549-1020 (H) University of Massachusetts 413 545-2859 (W) 710 North Pleasant Street Amherst, MA 01375 Reply-To: kuska@informatik.uni-leipzig.de ==== Needs[Graphics`ImplicitPlot`] ImplicitPlot[0.09 x^2 + 0.04 x y + 0.06 y^2 == 4, {x, -10, 10}, {y, -10, 10}] may help. Jens I am new with Mathematica, I have one question, > I know that the sollution might be very easy, but I wasn't able to > find it by now. > I would like to draw an ellipse, the formula let's say is as follows: 0.09 x^2 +0.04 x y + 0.06 y^2 = 4 > Maciej ==== I pretend to know how many times the function f has to be evaluated when using NDSolve to solve a differential equation. I've tried : cont=0; f[x_]:=(cont++;x^2); NDSolve[ {y''[x]+y[x]==f[x],y[0]==1,y'[0]==0},y,{x,0,6}] I obtain the solution of the ODE but cont returns 1 instead the number of funtion evaluations. How can I obtain it? ==== > I pretend to know how many times the function f has to be evaluated when > using NDSolve to solve a differential equation. > I've tried : cont=0; > f[x_]:=(cont++;x^2); > NDSolve[ {y''[x]+y[x]==f[x],y[0]==1,y'[0]==0},y,{x,0,6}] I obtain the solution of the ODE but cont returns 1 instead the number > of funtion evaluations. > How can I obtain it? > you need to make sure that the function f[x] is evaluated only when it gets a numeric argument. Otherwise, Mathematica expands f[x] symbolically to the expression x^2 before NDSolve is called. Writing your code as follows will solve your problem: cont=0; f[x_?NumericQ]:=(cont++;x^2); NDSolve[ {y''[x]+y[x]==f[x],y[0]==1,y'[0]==0},y,{x,0,6}] Eckhard Hennig ==== > I pretend to know how many times the function f has to be evaluated when > using NDSolve to solve a differential equation. > I've tried : cont=0; > f[x_]:=(cont++;x^2); > NDSolve[ {y''[x]+y[x]==f[x],y[0]==1,y'[0]==0},y,{x,0,6}] I obtain the solution of the ODE but cont returns 1 instead the number > of funtion evaluations. > How can I obtain it? cont=0; f[x_?NumericQ]:=(cont++;x^2) NDSolve[ {y''[x]+y[x]==f[x],y[0]==1,y'[0]==0},y,{x,0,6}] -> cont = 94 -Jim ==== Try the following: Clear[f] cont=0; f[x_?NumericQ]:=(cont++;x^2); NDSolve[ {y''[x]+y[x]==f[x],y[0]==1,y'[0]==0},y,{x,0,6}] The problem you are encountering occurs because NDSolve evaluates it's inputs first, and so NDSolve is trying to solve the differential equation y''[x]+y[x]==x^2 instead of y''[x]+y[x]==f[x]. By restricting the definition of f to evaluate only when it's input is numeric, NDSolve will keep the f[x] in the differential equation. I've included a Clear[f] statement just in case the old definition of f was still there. Carl Woll Physics Dept U of Washington I pretend to know how many times the function f has to be evaluated when > using NDSolve to solve a differential equation. > I've tried : cont=0; > f[x_]:=(cont++;x^2); > NDSolve[ {y''[x]+y[x]==f[x],y[0]==1,y'[0]==0},y,{x,0,6}] I obtain the solution of the ODE but cont returns 1 instead the number > of funtion evaluations. > How can I obtain it? > Reply-To: kuska@informatik.uni-leipzig.de ==== and after cont = 0; f[x_?NumericQ] := (cont++; x^2); NDSolve[{y''[x] + y[x] == f[x], y[0] == 1, y'[0] == 0}, y, {x, 0, 6}] cont is 93. Thats because *you* gave only a blank pattern for f[x_] and NDSolve[] evaluate it to compile the right hand sides or translate it to an internal function. So *your* function is only once when Mathematica can include it into it's internal function for the right hand sides. Jens I pretend to know how many times the function f has to be evaluated when > using NDSolve to solve a differential equation. > I've tried : cont=0; > f[x_]:=(cont++;x^2); > NDSolve[ {y''[x]+y[x]==f[x],y[0]==1,y'[0]==0},y,{x,0,6}] I obtain the solution of the ODE but cont returns 1 instead the number > of funtion evaluations. > How can I obtain it? > ==== Fabio, f[x_ /; IntegerQ[Rationalize[x/(2Pi)]], y] := g[x] f[x_, y_] := h[x, y] f[# Pi, y] & /@ Range[10] {h[Pi, y], g[2*Pi], h[3*Pi, y], g[4*Pi], h[5*Pi, y], g[6*Pi], h[7*Pi, y], g[8*Pi], h[9*Pi, y], g[10*Pi]} f[2.0Pi, y] g[6.28319] f[3.5, y] h[3.5, y] f[4.01Pi, 5] h[12.5978, 5] You can use a second argument in Rationalize to control how small a rationalization error is allowed and there is always the question about what domain around 2nPi you want to include in the first definition. David Park djmp@earthlink.net http://home.earthlink.net/~djmp/ ==== yes, this is a comon problem. You can either do: y=If[x>1000, ToExpression[BigVariable][[x]],x] ] or, maybe more readable, declare at the begging of ThisPackage: bigvar := bigvar = ToExpression[OtherPackage`BigVariable]; and then y=If[x>1000, bigvar, x] (* ************************** *) Alternatively you can create .mx files which load much quicker than .m files. It would be really nice if WRI could investigate why loading larger .m files is so slow ( I suspect it has to do something with the speed of side effects like DownValues declarations etc., but I don't know ). Rolf Mertig Mertig Consulting http://www.mertig.com ================================== DeclareePackage[OtherPackage`, {BigVariable} ]. The idea was to prevent the loading of a large file in routine cases. However, when ThisPackage` defines its functions, inside the Private` area, it includes a conditional call to BigVariable. It turns out that OtherPackage is loaded when that function is defined. I was wondering if there is a way to avoid this. Roughly speaking, here is the setting: BeginPackage[ OtherPackage` ] BigVariable::usage=example Begin[Private`] BigVariable=Table[x y,{x,1000},{y,1000}] End[ ] EndPackage[ ] BeginPackage[ ThisPackage` ] function::usage=example Begin[Private`] function[x_]:=Module[{y}, y=If[x>1000,BigVariable[[x]],x] ] End[ ] EndPackage[ ] I don't want OtherPakage to be loaded unless function[x] is called with x>1000 but it loads when function is defined. Any ideas? Nicholas ==== I was looking for a function that can convert numbers between bases. For example, going from Base 3 to Base 9, or from Base 16 to Base 8. It seems like I have to a few steps and then use BaseForm to convert to the desired end base. Is there a single command or simple function to do such conversions? For example, ConvertBase[from_, to_, num_] would convert from base to base 9 using the number num as the number to convert (it would best be allowed to let users input in that specific base since they'd be going from base 3 to base 9, for example so maybe num is needed at all). ==== Dear NG Mabye this is too simple, but I cant just figure it out I want to get an inverse function for y[t] where y[t_]:=NIntegrate[R[x]^4,{x,0,t}] /. ndsolution[[1]] and R[t] is an interpolatingfunction(R>0 from NDSolve) on the interval 0= Dear NG > Mabye this is too simple, but I cant just figure it out I want to get an inverse function for y[t] where y[t_]:=NIntegrate[R[x]^4,{x,0,t}] /. ndsolution[[1]] > and > R[t] is an interpolatingfunction(R>0 from NDSolve) on the interval 0= NIntegrate[R[InvFunction[y[t]]*Cos[y[t]], {y[t], 0, y[T]}] it works with a constant instead of R[t]. Hope you can help, and that the above is understandable. > Martin Skogstad ==== Martin, In a post a while ago, I introduced a function called NInverse which ought to be able to help you out. The message can be found at http://library.wolfram.com/mathgroup/archive/2001/Jun/msg00125.html The output of NInverse (in your case) is a function which given a value of y returns the corresponding value of t. In your case, you would use NInverse something like the following: NInverse[y, {t0,y0}, {yy, ymin, ymax}] You will probably get some error messages, but the answer should be okay. The function NInverse can be improved (witness the inclusion of yy above), and I will post an improved version sometime. Carl Woll Physics Dept U of Washington > Dear NG > Mabye this is too simple, but I cant just figure it out I want to get an inverse function for y[t] where y[t_]:=NIntegrate[R[x]^4,{x,0,t}] /. ndsolution[[1]] > and > R[t] is an interpolatingfunction(R>0 from NDSolve) on the interval 0= NIntegrate[R[InvFunction[y[t]]*Cos[y[t]], {y[t], 0, y[T]}] it works with a constant instead of R[t]. Hope you can help, and that the above is understandable. > Martin Skogstad > ==== There is a very nice java applet at http://statman.stat.sc.edu/~west/javahtml/classes/ , in which you can include your own data by replacing what is in the Applet with your own, which gives you a real time histogram plot and lets you alter the bin width and see how this effects the histogram....it wasnt until I saw this that i understood the significance of choosing the bin width......jerry blimbaum -----Original Message----- http://www.mathstatica.com for details. Obviously, it is less expensive to write your own functions. Just recently in message Mark Fisher posted code that addresses the empirical CDF. However, in this code you may want to replace 1/n with 1/(n+1) or (j-0.5)/n depending on your application. Note, these will have no significant effect for large data sets. The key issue with an empirical PDF is deciding the bin width. A simple approach would be to use the functions in Statistics`DataManipulation` and BinListCounts. More sophisticated approaches involve kernel methods. These methods will generate smoother estimates for the PDF. Again, the key is bandwidth. There is no apriori choice for bin width or bandwith. Bad choices will obscure significant features in the data set. ==== A while back someone asked if Mathematica could plot data one point at a time....most answers were no, with one exception, where someone showed how to use the same Graphic Cell repeatedly....here is some Mathematica, JLink program that will do just that... << JLink` InstallJava[CommandLine -> c:j2sdk1.4.1binjava.exe] (* or set command line to where your java.exe is *) UseFrontEndForRendering = False; createWindow[] := Module[{frame}, frame = JavaNew[com.wolfram.jlink.MathFrame, Drawing Sine Wave Animation One Point at a Time]; drawArea = JavaNew[com.wolfram.jlink.MathCanvas]; drawArea@setUsesFE[UseFrontEndForRendering]; drawArea@setSize[800, 600]; JavaBlock[frame@setLayout[JavaNew[java.awt.BorderLayout]]; frame@add[drawArea, ReturnAsJavaObject[BorderLayout`CENTER]]; frame@pack[]; points = Range[data // Length] frame@setSize[800, 600]; frame@setLocation[200, 200]; JavaShow[frame]]; frame ] data = Table[{x, Sin[x]}, {x, -0.0000000000001, 2 Pi, .1}]; frame[n_] := ListPlot[Take[data, {1, n}], PlotRange -> {{0, 2 Pi}, {-1.5, 1.5}}, PlotStyle -> {PointSize[.005], Hue[0]}, DisplayFunction -> Identity] drawPoints[ptNow_] := Show[frame[ptNow], AspectRatio -> Automatic]; points = Range[data // Length]; AnimationPlot[pt_List] := JavaBlock[ Block[{frm}, frm = createWindow[]; Map[(obj = drawPoints[#]; drawArea@setMathCommand[obj]; drawArea@repaintNow[]; Pause[0.0001];) &, pt]; ] ] AnimationPlot[points] jerry blimbaum NSWC panama city, fl ==== I never liked the function, but now I have need of it. I hope someone can help me see the way to implement it. For a class I am teaching I want to write a simple function that does a random search for a function's maxima. I feed the function N randomly generated inputs and look at the outputs. Provided N is large enough, the largest of the outputs I obtain should be near the function's maximum value. What I want to be able to do is to identify the input that gave me that largest output. Any ideas? My only idea is this: create a table of pairs {output, input{ and find the element of the table that has the largest first entry. This will give me the input I seek. Is there a slick way of doing this in mathematica? -- Jason Miller, Ph.D. Division of Mathematics and Computer Science Truman State University 100 East Normal St. Kirksville, MO 63501 http://vh216801.truman.edu 660.785.7430 ==== Here's one way: argMax::usage = argMax[f, domain] returns the element of domain for which f of that element is maximal -- breaks ties in favor of first occurrence.; SetAttributes[argMax, HoldFirst]; argMax[f_, dom_List] := Fold[If[f[#1]>=f[#2], #1, #2]&, First[dom], Rest[dom]] And here's another way (defining it in terms of tupleMax)... tupleMax::usage = tupleMax[list] returns the tuple that is lexicographically maximal.; tupleMax[l_List] := Fold[If[OrderedQ[{#1, #2}], #2, #1]&, First[l], Rest[l]] argMax[f_, dom_List] := tupleMax[{f[#],#}& /@ dom][[2]] --- / FROM Jason Miller AT 02.09.18 06:18 (Today) / --- > I never liked the function, but now I have need of it. I hope > someone can help me see the way to implement it. For a class I am teaching I want to write a simple function that does > a random search for a function's maxima. I feed the function N > randomly generated inputs and look at the outputs. Provided N is > large enough, the largest of the outputs I obtain should be near the > function's maximum value. What I want to be able to do is to identify the input that gave me > that largest output. Any ideas? My only idea is this: create a table of pairs {output, input{ and > find the element of the table that has the largest first entry. This > will give me the input I seek. Is there a slick way of doing this in > mathematica? -- -- -- -- -- -- -- -- -- -- -- -- Daniel Reeves http://ai.eecs.umich.edu/people/dreeves/ In science it often happens that scientists say, 'You know that's a really good argument; my position is mistaken,' and then they would actually change their minds and you never hear that old view from them again. They really do it. It doesn't happen as often as it should, because scientists are human and change is sometimes painful. But it happens every day. I cannot recall the last time something like that happened in politics or religion. -- Carl Sagan, 1987 CSICOP Keynote Address Reply-To: kuska@informatik.uni-leipzig.de ==== fun[x_, y_] := x^2 + y^2 RandomMin[n_Integer] := First[Sort[ {#, fun @@ #} & /@ Table[{Random[], Random[]}, {n}], Last[#1] < Last[#2] &]] should do it. Jens I never liked the function, but now I have need of it. I hope > someone can help me see the way to implement it. For a class I am teaching I want to write a simple function that does > a random search for a function's maxima. I feed the function N > randomly generated inputs and look at the outputs. Provided N is > large enough, the largest of the outputs I obtain should be near the > function's maximum value. What I want to be able to do is to identify the input that gave me > that largest output. Any ideas? My only idea is this: create a table of pairs {output, input{ and > find the element of the table that has the largest first entry. This > will give me the input I seek. Is there a slick way of doing this in > mathematica? -- > Jason Miller, Ph.D. > Division of Mathematics and Computer Science > Truman State University > 100 East Normal St. > Kirksville, MO 63501 > http://vh216801.truman.edu > 660.785.7430 Reply-To: murray@math.umass.edu ==== The same principles that allow particular cases for defining a function of a single variable would apply here, because Mathematica applies particular rules before it applies general ones. For example: f[0, y_] := 0 f[2 Pi, y_] := y - 2 f[4 Pi, y_] := y - 4 f[x_, y_] := x^2 + y^3 This will do exactly what it looks like it does! If you have a general family of particular cases, say at all even integral multiples of Pi, then you could use something like the following in place of the first three lines above: f[k_ Pi, y_] := y - k /; IntegerQ[k] && EvenQ[k] There are variants as to where to place the condition IntegerQ[k] && EvenQ[k], for example: f[k_ Pi , y_] /; IntegerQ[k] && EvenQ[k] := y - k f[k_ Pi /; IntegerQ[k] && EvenQ[k], y_] := y - k > I have quite an easy and annoying problem with mathematica: > I need to define a function f(x,y) which takes some values for > x=0,2pi,4pi (indepently of y) and has a different expression for all > the other values of y. This is easily done for one-dimensional > functions but I am in serious troubles for my two-dimensional problem: > any suggestion? > Fabio > -- Murray Eisenberg murray@math.umass.edu Mathematics & Statistics Dept. Lederle Graduate Research Tower phone 413 549-1020 (H) University of Massachusetts 413 545-2859 (W) 710 North Pleasant Street Amherst, MA 01375 ==== >>and has a different expression for all the other values of y. Do you mean all the other values of x? You can't expect to make yourself understood to Mathematica, if you don't say what you mean. Mathematica can't guess, like we can. If that's what you meant, it's really easy: f[0 | N[0], y_] := y f[2*Pi | N[2*Pi], y_] := y^2 f[4*Pi | 4.*Pi, y_] := y^3 f[x_, y_] := y^4 If the first three of those expressions are supposed to be the same, you can substitute f[0 | N[0] | 2*Pi | N[2*Pi] | 4*Pi | 4.*Pi, y_] := y^3 Bobby Treat -----Original Message----- Fabio Reply-To: kuska@informatik.uni-leipzig.de ==== a) you must change the fonts, because the Mathematica symbols are designed to fit to Times/Helvetica/Courier and not to Computer Modern ! Mixing Computer Modern and the Mathematica fonts look not very nice because Comuter Modern has smaler strokes. You need atleast wrisym.sty, i.e. usepackage[monospacemath]{wrisym} but this will switch the main document font to Times and the mono-space font to courier. With Mathematica 4.2 you can use Adobe-Garamond and Janson as document fonts too, with usepackage[monospacemath,garamond]{wrisym} but this are commercial fonts and you have to buy it from Adobe. b) If you like the narrow monotype fonts similar to Computer Modern you can use the CMTT fonts, now included in Mathematica 4.2 Just use the package option usepackage[cmtt]{wrisym} c) I have a TeX frontend that use only the wrisym package and a preprocessor. It run fine with the most style files But you have to use TeX notebooks that are essential TeX files with mathinput environments. The TeX frontend send the contents of the mathinput environments to the kernel and paste the output into the final TeX file. It also does some fancy formating with the Mathematica input and replace -> with Rule, :> with RuleDelayed .. You can have the program and the style files if you like. Jens Dear all, I want to include some Mathematica code in some lecture notes I am writing. > For that purpose, I use the LaTeX style provided with Mathematica. It seems > to be working fine when I produce a document that is just Mathematica code. The problem - as expected - is that when I include the Mathematica code as > a part of a larger document, there are > (1) clashes with existing packages > (2) some LaTeX elements, such as the equations, start to use the > Mathematica fonts too. My objective would be to typeset a document in > standard LaTeX, where only the Mathematica code has the Mathematica feeling. Does anyone know of any style available that is completely seperated from > all other LaTeX definitions? Best Kyriakos _____+**+____+**+___+**+__+**+_ Kyriakos Chourdakis > Lecturer in Financial Economics URL: http://www.qmw.ac.uk/~te9001 > tel: (++44) (+20) 7882 5086 > Dept of Economics > University of London, QM > London E1 4NS > U.K. Reply-To: kuska@informatik.uni-leipzig.de ==== does help you ? Jens > I often run into this difficulty: when designing a program, say as a > module, and testing it for various inputs, I get wrong answers. What to > do? I use a method that works for me but may not be the best available. > I want to show my method and then ask a question about how it can be > imporved. (Oh yes, I abandoned Trace a long time ago!) myFunction[f_] := Module[ {L1,L2,L3}, L1 = ... ; > L2 = ... ; > l3 = ... ; > final step ] To see what went wrong, I use (* *) selectively as follows: Stage 1 myFunction[f_] := Module[ {L1,L2,L3}, L1 = ... (*; > L2 = ... ; > L3 = ... ; > final step *) ] Thus I see if L1 worked as expected. The next step is to put (* after L2 > and see if this works. I continue this til the bitter end and I usually > find my errors. My question; The process of moving (* *) step by step through the > program is quite tedious when the code has lots more lines. What I would > like is a meta-program which (like FoldList) does this job for me. The > output of this meta-program is the list of outputs of each line in the > module, probably best printed as a column. > This sounds like Trace but my problem with Trace is it is terribly > difficult to read. For the not-so-subtle programming I do, the only thing > I need is what expression is returned line by line. Any advice? All remarks are appreciated! Jack ==== I've enhanced the code for PartialEvaluation (see my post, which I incorrectly attaced to someone else's reply). It is somewhat more flexible now and does more error checking. It finds the rule in DownValues that matches f[args] (allowing for more than one rule in DownValues); it allows the user to specify the position of the main CompoundExpression; and it allows the user to specify an expression that gets appended to the truncated CompoundExpression (and hence evaluated and returned). The package is just a bit too large to post here. It can be downloaded from my web site at http://www.markfisher.net/~mefisher/mma/mathematica.html Nevertheless, I can give an outline of the code here (absent the error checking stuff). PartialEvaluation[f[args], n, expr]: (* get the DownValues and turn them off *) dv = DownValues[f]; DownValues[f] = {}; (* find the rule that matches *) matches = Position[MatchQ[f[args], #]& /@ dv[[All, 1]], True]; match = dv[[ matches[[1, 1]] ]]; (* find the main CompoundExpression *) ppos = Position[match, HoldPattern[CompoundExpression[__]]]; pos = First[Sort[ppos]]; (* extract, truncate, append to, and reinsert it *) held = Extract[match, pos, Hold]; held = ReplacePart[held, Sequence, {1, 0}]; held = Take[held, n]; held = Join[held, Hold[expr]]; held = ReplacePart[Hold[Evaluate[held]], CompoundExpression, {1, 0}]; match = ReplacePart[match, held, pos, 1]; (* apply the modified rule and restore DownValues *) result = f[args] /. match; DownValues[f] = dv; result --Mark I often run into this difficulty: when designing a program, say as a > module, and testing it for various inputs, I get wrong answers. What to > do? I use a method that works for me but may not be the best available. > I want to show my method and then ask a question about how it can be > imporved. (Oh yes, I abandoned Trace a long time ago!) myFunction[f_] := Module[ {L1,L2,L3}, L1 = ... ; > L2 = ... ; > l3 = ... ; > final step ] To see what went wrong, I use (* *) selectively as follows: Stage 1 myFunction[f_] := Module[ {L1,L2,L3}, L1 = ... (*; > L2 = ... ; > L3 = ... ; > final step *) ] Thus I see if L1 worked as expected. The next step is to put (* after L2 > and see if this works. I continue this til the bitter end and I usually > find my errors. My question; The process of moving (* *) step by step through the > program is quite tedious when the code has lots more lines. What I would > like is a meta-program which (like FoldList) does this job for me. The > output of this meta-program is the list of outputs of each line in the > module, probably best printed as a column. > This sounds like Trace but my problem with Trace is it is terribly > difficult to read. For the not-so-subtle programming I do, the only thing > I need is what expression is returned line by line. Any advice? All remarks are appreciated! Jack Reply-To: kuska@informatik.uni-leipzig.de ==== BeginPackage[ OtherPackage` ] BigVariable::usage=example Begin[`Private`] BigVariable=Table[x y,{x,1000},{y,1000}] End[ ] EndPackage[ ] Print[Loading ...]; and BeginPackage[ ThisPackage` ] function::usage=example Begin[`Private`] function[x_]:=Module[{y}, y=If[x>1000, Needs[OtherPackage`]; OtherPackage`BigVariable[[x]], x] ] End[ ] EndPackage[ ] should do that. Jens DeclarePackage[OtherPackage`, {BigVariable} ]. The idea was to prevent > the loading of a large file in routine cases. However, when ThisPackage` > defines its functions, inside the Private` area, it includes a conditional > call to BigVariable. It turns out that OtherPackage is loaded when that > function is defined. I was wondering if there is a way to avoid this. > Roughly speaking, here is the setting: BeginPackage[ OtherPackage` ] > BigVariable::usage=example > Begin[Private`] > BigVariable=Table[x y,{x,1000},{y,1000}] > End[ ] > EndPackage[ ] BeginPackage[ ThisPackage` ] > function::usage=example > Begin[Private`] > function[x_]:=Module[{y}, > y=If[x>1000,BigVariable[[x]],x] ] > End[ ] > EndPackage[ ] I don't want OtherPakage to be loaded unless function[x] is called with > x>1000 but it loads when function is defined. Any ideas? > Nicholas ==== Dynamic programming might help: replace the line defining BigVariable with BigVariable:=BigVariable=Table[x y,{x,1000},{y,1000}] Then the table will not be calculated unless it is needed. (See the section on Functions that Remember Values They Have Found, 2.4.9 in the Mathematica Book). John Jowett > DeclarePackage[OtherPackage`, {BigVariable} ]. The idea was to prevent > the loading of a large file in routine cases. However, when ThisPackage` > defines its functions, inside the Private` area, it includes a conditional > call to BigVariable. It turns out that OtherPackage is loaded when that > function is defined. I was wondering if there is a way to avoid this. > Roughly speaking, here is the setting: BeginPackage[ OtherPackage` ] > BigVariable::usage=example > Begin[Private`] > BigVariable=Table[x y,{x,1000},{y,1000}] > End[ ] > EndPackage[ ] BeginPackage[ ThisPackage` ] > function::usage=example > Begin[Private`] > function[x_]:=Module[{y}, > y=If[x>1000,BigVariable[[x]],x] ] > End[ ] > EndPackage[ ] > I don't want OtherPakage to be loaded unless function[x] is called with > x>1000 but it loads when function is defined. Any ideas? Nicholas > Reply-To: kuska@informatik.uni-leipzig.de ==== Clear[f] f[0, _] := q f[Pi, _] := p f[2Pi, _] := r f[x_, y_] := x^2*y^2 ??? Jens I have quite an easy and annoying problem with mathematica: > I need to define a function f(x,y) which takes some values for > x=0,2pi,4pi (indepently of y) and has a different expression for all > the other values of y. This is easily done for one-dimensional > functions but I am in serious troubles for my two-dimensional problem: > any suggestion? > Fabio ==== > I have quite an easy and annoying problem with mathematica: > I need to define a function f(x,y) which takes some values for > x=0,2pi,4pi (indepently of y) and has a different expression for all > the other values of y. This is easily done for one-dimensional > functions but I am in serious troubles for my two-dimensional problem: > any suggestion? > Fabio There were some typos in my last message. It should have read: Let the known values of the function at x={0,2Pi,4Pi} be {f0,f2,f4}. Your conditions can be met by f(x,y)=x(x-2Pi)(x-4Pi)g(y)+h(x), where {h(0),h(2Pi),h(4Pi)}={f0,f2,f4}. An h(x) can be found by quadratic interpolation: h[x]= a + bx + cx^2; Solve[{a==f0,a+2Pi b +(2Pi)^2 c==f2, a+4Pi b+(4Pi)^2 c==f4},{a,b,c}] {{b -> -(3*f0 - 4*f2 + f4)/(4*Pi), c -> -(-f0 + 2*f2 - f4)/(8*Pi^2), a -> f0}} ==== > I have quite an easy and annoying problem with mathematica: > I need to define a function f(x,y) which takes some values for > x=0,2pi,4pi (indepently of y) and has a different expression for all > the other values of y. This is easily done for one-dimensional > functions but I am in serious troubles for my two-dimensional problem: > any suggestion? > Fabio Let the known values of the function at x={0,2Pi,4Pi} be {f0,f2,f4}. Your conditions can be met by f(x,y)=x(x-2Pi)(x-4Pi)g(y)+h(x), where {h(0),h(2Pi),h(4Pi)}={f0,f2,f4}. An h(x) can be found by quadratic interpolation: h[x]= a + bx + cx^2; Solve[{a==f0,a+2Pi b +(2Pi)^2+(2Pi)^2 c==f2, a+4Pi b+(4Pi)^2 c==f4] {{b -> -(3*f0 - 4*f2 + f4)/(4*Pi), c -> -(-f0 + 2*f2 - f4)/(8*Pi^2), a -> f0}} Maurice ==== > I have quite an easy and annoying problem with mathematica: > I need to define a function f(x,y) which takes some values for > x=0,2pi,4pi (indepently of y) and has a different expression for all > the other values of y. This is easily done for one-dimensional > functions but I am in serious troubles for my two-dimensional problem: > any suggestion? > Fabio Something like this? In[1]:= f[x_,y_] := If[Mod[x,2*Pi] === 0, Exp[x], Sin[y]+x] In[2]:= f[4*Pi, y] 4 Pi Out[2]= E In[3]:= f[1, y] Out[3]= 1 + Sin[y] -- Bhuvanesh, Wolfram Research. ==== > I have quite an easy and annoying problem with mathematica: > I need to define a function f(x,y) which takes some values for > x=0,2pi,4pi (indepently of y) and has a different expression for all > the other values of y. This is easily done for one-dimensional > functions but I am in serious troubles for my two-dimensional problem: > any suggestion? > Fabio > f[0, y_] := 5 f[2 Pi, y_] := 7 f[4 Pi, y_] := -3 f[x_ /; x < 0, y_] := x y f[x_ /; 0 < x < 2 Pi, y_] := x + y f[x_ /; 2Pi < x < 4 Pi, y_] := x - y f[x_ /; 4Pi < x, y_] := 2x y ==== >-----Original Message----- >Sent: Monday, September 16, 2002 6:34 AM >I have quite an easy and annoying problem with mathematica: >I need to define a function f(x,y) which takes some values for >x=0,2pi,4pi (indepently of y) and has a different expression for all >the other values of y. This is easily done for one-dimensional >functions but I am in serious troubles for my two-dimensional problem: >any suggestion? >Fabio It's difficult to guess what you wanted to attain and what your problems were. Perhaps this example might help you: In[11]:= f[x_, _] /; Mod[x, 2Pi] == 0 := x/(2Pi) In[12]:= f[x_, y_] := x + y In[13]:= epsilon = $MachineEpsilon Out[13]= 2.220446049250313*^-16 In[14]:= f[6Pi(1 + epsilon Sin[#]), #] & /@ Range[0, 4Pi, Pi 10^-1] Out[14]= {3, 3., 19.4779, 19.792, 20.1062, 20.4204, 20.7345, 21.0487, 21.3628, 3., 3, 22.3053, 22.6195, 22.9336, 23.2478, 23.5619, 23.8761, 24.1903, 24.5044, 24.8186, 3, 3., 25.7611, 26.0752, 26.3894, 26.7035, 27.0177, 27.3319, 27.646, 3., 3, 28.5885, 28.9027, 29.2168, 29.531, 29.8451, 30.1593, 30.4734, 30.7876, 31.1018, 3} Look close at the values returned (compare with epsilon Sin[Range[0, 4Pi, Pi 10^-1]]), also to recognize the dangers of such a definition. -- Hartmut ==== How can I invert a funcion? For example In:=f[h_]:=Exp[-h](a*h+b*h^2); I want to invert it as h as a function of f How can I do that? Please suggest. Raj ==== Well, you can just solve for h. However, in this case there isn't a symbolic solution. How can I invert a funcion? For example In:=f[h_]:=Exp[-h](a*h+b*h^2); I want to invert it as h as a function of f > How can I do that? Please suggest. Raj > Reply-To: kuska@informatik.uni-leipzig.de ==== Solve[ f=Exp[-h](a*h+b*h^2),h] will not help you but you can try to invert the series expansion as long as the function is monoton ser = Series[Exp[-h](a*h + b*h^2), {h, 0, 4}]; InverseSeries[ser, f] Jens > How can I invert a funcion? For example In:=f[h_]:=Exp[-h](a*h+b*h^2); I want to invert it as h as a function of f How can I do that? Please suggest. Raj ==== I don't have any idea about how to solve next problem. Back space key when I am in the front end don't work correctly. It advances instead of go back and there is pointer mark of the mouse looks like a hand instead of an arrow, and this don't work fine too. Configuration one? How can I solve this? Note: This problem seems to appear just in my workspace i mean this not happen for other users. Cesar. __________________________________________________ Do you Yahoo!? Yahoo! News - Today's headlines http://news.yahoo.com ==== > I don't have any idea about how to solve next problem. Back space key when I am in the front end don't work > correctly. It advances instead of go back and there is > pointer mark of the mouse looks like a hand instead > of an arrow, and this don't work fine too. > Configuration one? How can I solve this? > Note: This problem seems to appear just in my > workspace > i mean this not happen for other users. Cesar. > __________________________________________________ > Do you Yahoo!? > Yahoo! News - Today's headlines > http://news.yahoo.com > Pressing the 'Num'-Key maybe solves your problem: If you switch off the 'Num Lock' your 'hand pointer mark' should turn into an 'arrow pointer mark' and you can place the cursor in a notebook. You get more details on -- Rainer Gruber ==== this thing. Here is what I'm using now: In[1]:= a=ReadList[197-tst.txt,Word,RecordLists->True, WordSeparators->None]; In[2]:= flds={{25,31}, {32,39},{40,46},{47,54},{55,62},{63,70},{71,78},{79,86},{87, 94},{95,102},{103,110}}; Define this function: In[3]:= f[x_]:=ToExpression[StringTake[x[[1]],#]]&/@flds Now, map this function onto a: In[8]:= f/@a Out[8]= {{5935.8,5946.66,27.06,-1281.9,-229.,321.,317.,367.,-115.,126.,146.},{5935.8 .... ************ The data is now in a matrix of numbers and each variable is in a column. Now I just transpose this to get each variable in a row for easy access (apparently Mathematica has no way to directly access a column in a matrix, you gotta transpose I think) Rob > I've got to extract some numbers from a file that are in lines of text. Since the line contents are not numbers, I presume I must pull the line out as a string. Here I start by pulling out just one line: inFile = OpenRead[197-tst.txt] > y = ReadList[inFile, String, 1, RecordLists -> True] > Close[inFile]; This appears to pull in a line. Now I want to take characters 25 to 110 to get just the stuff I want: > y1=StringTake[y ,{25,110}]; Here's the output. StringTake doesn't seem to work. > It doesn't take loading another package as far > as I can tell from the help. I'm thinking that it doesn't work because it's trying to work on a list > rather than a string. I've tried Flatten, and other stuff to try to get to just a string and not a list but > nothing has worked so far. I'm a long way from getting to those numbers in there but heck, I > can't even get to the string. Can anyone point me in the right direction? > ==== > The data is now in a matrix of numbers and each variable is in a column. > Now I just transpose this to get each variable in a row for easy access > (apparently Mathematica has no way to directly access a column in a matrix, > you gotta transpose I think) You can use All: In[1]:= mat = {{a,b,c},{d,e,f}}; In[2]:= mat[[All,2]] Out[2]= {b, e} -- Bhuvanesh, Wolfram Research. ==== Help! After using SetOptions[Graphics3D, ...] I am trying to set the default options of Graphics3D back to what they were when I started up Mathematica. I cannot find a command to do this. Is this possible? ==== o = Options[Graphics3D]; (* save the original options *) SetOptions[Graphics3D, ...]; (* mess with them *) SetOptions[Graphics3D, Sequence@@o]; (* restore them *) --- / FROM David Sagan AT 02.09.18 06:29 (Today) / --- > Help! After using SetOptions[Graphics3D, ...] I am trying to set the default > options of Graphics3D back to what they were when I started up > Mathematica. I cannot find a command to do this. Is this possible? > -- -- -- -- -- -- -- -- -- -- -- -- Daniel Reeves http://ai.eecs.umich.edu/people/dreeves/ The idea of programming in a low level language like C will seem as specialized and esoteric as programming in microcode or assembler seems today. -- Stephen Wolfram, creator of Mathematica Reply-To: kuska@informatik.uni-leipzig.de ==== Hmm, oldOptions=Options[Graphics3D]; SetOptions[Graphics3D,Boxed->False] (* do something with it *) SetOptions[Graphics3D, Sequence @@ oldOptions] works fine. Jens Help! After using SetOptions[Graphics3D, ...] I am trying to set the default > options of Graphics3D back to what they were when I started up > Mathematica. I cannot find a command to do this. Is this possible? > ==== This code is almost identical to yours in principle, yet 30% faster: (compared with the code in your notebook, not the code in the post > below) ClearAll[emptyTree, treeInsert]; > emptyTree = {}; treeInsert[emptyTree, elem_] := {emptyTree, elem, emptyTree} > treeInsert[tree_, elem_] /; SameQ[tree[[2]], elem] := tree > treeInsert[tree_, elem_] /; OrderedQ[{tree[[2]], elem}] := > {First[tree], tree[[2]], treeInsert[Last[tree], elem]} > treeInsert[tree_, elem_] := {treeInsert[First[tree], elem], tree[[2]], > Last[tree]} Here's an even simpler version, same speed: ClearAll[emptyTree, treeInsert]; > emptyTree = {}; treeInsert[emptyTree, elem_] := {emptyTree, elem, emptyTree} > treeInsert[tree_, elem_] := Which[ > SameQ[tree[[2]], elem], tree, OrderedQ[{tree[[2]], elem}], > {First@tree, tree[[2]], treeInsert[Last@tree, elem]}, > True, {treeInsert[First@tree, elem], tree[[2]], Last@tree}] I had tried variations on this but they all seemed ever so slightly slower. Might be version an OS dependent. > I see one obvious advantage of your algorithm over Husain's and my > earlier code; though I'm not sure it explains its better behavior. That > is, while my algorithm made pattern matching more burdensome, yours > virtually eliminated it. Also, the biggest improvement I've seen comes simply from replacing Null > with anything else in Husain's code. Again, that seems to suggest that > time spent on pattern matching is the problem. All the algorithms do > approximately the same real work on the tree structure. I really like how Flatten changes your tree into a sorted list. Very > nice! Goes by the name of tree sort. Using Flatten is basically a shortcut for walking the tree left-to-right. > Bobby Treat It seems I did not look hard enough at the original post or the follow-ups. In point of fact, regarding speed, there IS no obvious advantage of my code over that first version. The advantage is there, of course, as timing checks indicate. But it is quite far from obvious. The reason is related to the (rare) need for Update. It is pointed out in the manual that infinite evaluation is not in fact possible (no surprise thus far). A consequence is that Mathematica requires internal optimization features to keep reevaluation to a minimum. Update may be required in cases where such optimizations are overzealous. What we have in the case of this example is, roughly, the opposite. The code that determines need for reevaluation is simply not working sufficiently well. It does make the correct determination, but only after looking over the entire structure that is being evaluated. As we have a structure growing linearly in the number of iterations, and it gets mostly traversed each iteration, the algorithmic complexity becomes at least O(n^2). Not a disaster unless the appropriate complexity is smaller, which was the case for this example. This is not a bug insofar as it is a (regretably) necessary consequence of Mathematica evaluation semantics. The problems it can cause are all the same quite undesirable. We've made some inroads on this problem by enlarging substantially the class of expressions for which reevaluation may be quickly short-circuited, with the idea of fixing all but the most pathological of examples. The status of that work is unfortunately not known to me at present, though I'll try to find out about it. Daniel Lichtblau Wolfram Research ==== Daniel, I'd say your algorithm (as an ALGORITHM) is precisely the same as Husain's. The implementation is different, but the logical steps are the same (except perhaps for <= vs. <), and so is the logical structure of the tree produced. Even the storage method is pretty much the same. Using {} instead of Null and List instead of 'node', and putting values in the middle position, doesn't add up to a change that could affect algorithmic complexity. Even the fact that Flatten gives a sorted List with your implementation is something you get for free, and the advantage comes only when it's time to output a sorted list. It has nothing to do with the work of building the tree. Algorithmic complexity (absent hidden factors under the kernel's control) is precisely the same for your solution and Husain's (mine too, I think). The speed differences we're seeing have to do with nuances of the kernel, pattern matching costs, etc. Our three nearly equivalent algorithms simply incur different hidden costs. That's the case with many of the problems that turn into speed wars among us. We compare implementation speed (including hidden costs), not pure algorithmic complexity. Compiled code or packed arrays often make all the difference, and it usually happens in the background where we can't see it. As you've hinted below, the vagaries of Update can make a linear algorithm quadratic, but may NOT so plague a slightly different implementation of the very same algorithm. So... I'm suggesting we might want to be careful when we leap to conclusions about algorithmic complexity based on a few timed experiments. Bobby Treat -----Original Message----- > ClearAll[emptyTree, treeInsert]; > emptyTree = {}; treeInsert[emptyTree, elem_] := {emptyTree, elem, emptyTree} > treeInsert[tree_, elem_] /; SameQ[tree[[2]], elem] := tree > treeInsert[tree_, elem_] /; OrderedQ[{tree[[2]], elem}] := > {First[tree], tree[[2]], treeInsert[Last[tree], elem]} > treeInsert[tree_, elem_] := {treeInsert[First[tree], elem], tree[[2]], > Last[tree]} Here's an even simpler version, same speed: ClearAll[emptyTree, treeInsert]; > emptyTree = {}; treeInsert[emptyTree, elem_] := {emptyTree, elem, emptyTree} > treeInsert[tree_, elem_] := Which[ > SameQ[tree[[2]], elem], tree, OrderedQ[{tree[[2]], elem}], > {First@tree, tree[[2]], treeInsert[Last@tree, elem]}, > True, {treeInsert[First@tree, elem], tree[[2]], Last@tree}] I had tried variations on this but they all seemed ever so slightly slower. Might be version an OS dependent. > I see one obvious advantage of your algorithm over Husain's and my > earlier code; though I'm not sure it explains its better behavior. That > is, while my algorithm made pattern matching more burdensome, yours > virtually eliminated it. Also, the biggest improvement I've seen comes simply from replacing Null > with anything else in Husain's code. Again, that seems to suggest that > time spent on pattern matching is the problem. All the algorithms do > approximately the same real work on the tree structure. I really like how Flatten changes your tree into a sorted list. Very > nice! Goes by the name of tree sort. Using Flatten is basically a shortcut for walking the tree left-to-right. > Bobby Treat It seems I did not look hard enough at the original post or the follow-ups. In point of fact, regarding speed, there IS no obvious advantage of my code over that first version. The advantage is there, of course, as timing checks indicate. But it is quite far from obvious. The reason is related to the (rare) need for Update. It is pointed out in the manual that infinite evaluation is not in fact possible (no surprise thus far). A consequence is that Mathematica requires internal optimization features to keep reevaluation to a minimum. Update may be required in cases where such optimizations are overzealous. What we have in the case of this example is, roughly, the opposite. The code that determines need for reevaluation is simply not working sufficiently well. It does make the correct determination, but only after looking over the entire structure that is being evaluated. As we have a structure growing linearly in the number of iterations, and it gets mostly traversed each iteration, the algorithmic complexity becomes at least O(n^2). Not a disaster unless the appropriate complexity is smaller, which was the case for this example. This is not a bug insofar as it is a (regretably) necessary consequence of Mathematica evaluation semantics. The problems it can cause are all the same quite undesirable. We've made some inroads on this problem by enlarging substantially the class of expressions for which reevaluation may be quickly short-circuited, with the idea of fixing all but the most pathological of examples. The status of that work is unfortunately not known to me at present, though I'll try to find out about it. Daniel Lichtblau Wolfram Research ==== I'm curious, why the built-in boolean functions And and Or aren't commutative? Attributes/@{Or,And}//ColumnForm {Flat, HoldAll, OneIdentity, Protected} {Flat, HoldAll, OneIdentity, Protected} However, the arithmetical functions are Orderless: Attributes/@{Plus,Times}//ColumnForm {Flat, Listable, NumericFunction, OneIdentity, Orderless, Protected} {Flat, Listable, NumericFunction, OneIdentity, Orderless, Protected} ------------------- Evgeni Trifonov Vladivostok, Russia ==== Evgeni, Note that both And and Or can return a value without evaluating all of their arguments. For example, if the first argument to And is False, there is no reason to look at the other arguments. Suppose a user knows that one of the arguments is usually false, and so would like to look at that argument before any of the others, to avoid unneeded computations of the other arguments. He would put that argument first. If Mathematica turned around and sorted the arguments (which is what happens when a function is orderless), then that argument might end up being evaluated last. If the arguments take a significant amount of time to compute, then sorting first may cause the function to take much longer to evaluate. At any rate, if the above is not a concern for you, you may always change the attributes of And and Or to anything you want. Carl Woll Physics Dept U of Washington ----- Original Message ----- > However, the arithmetical functions are Orderless: Attributes/@{Plus,Times}//ColumnForm {Flat, Listable, NumericFunction, OneIdentity, Orderless, Protected} > {Flat, Listable, NumericFunction, OneIdentity, Orderless, Protected} ------------------- > Evgeni Trifonov > Vladivostok, Russia > ==== I have defined two functions (Math 4.1) In[1]:= integrate[c_ y_,t_]:=c*integrate[y,t]/;FreeQ[c,t]; In[2]:= integrate[Exp[(a_.)*tau], t_]:=(E^(a*t)-1)/a/; FreeQ[a,t]&&FreeQ[a,tau]; when I applied these functions the solution is different if the coefficient is the letter d or lower (solution fine) or f or higher (solution wrong) . Here is an example: Using letter f In[3]:=integrate[f Exp[-0.8316*tau], t] Out[3]:=integrate[f, t]/E^(0.8316*tau) (*wrong*) Using letter a (the solution is rigth In[4]:=integrate[a Exp[-0.8316*tau], t] Out[4]:=-1.20250*a*(-1 + E^(-0.8316*t)) What it wrong? Guillermo Sanchez --------------------------------------------- This message was sent using Endymion MailMan. ==== Neither of your solutions are wrong according to your rules. Lets step through them In[3]:=integrate[f Exp[-0.8316*tau], t] lexographical (sp) order as In[3]:=integrate[Exp[-0.8316*tau] f, t] It then applies rule 1: integrate[c_ y_,t_]:=c*integrate[y,t]/;FreeQ[c,t] Well, The term Exp[-0.8316*tau] contains no t, so according to your rule it is equal to Exp[-0.8316*tau] integrate[f, t] At this point none of your rules apply, so the kernel stops. When you use a instead of f the ordering of the constant and the exponential are reversed, and you get the behaviour that you actually desired. A simple fix is to replace the condition in rule 1 by FreeQ[c,t]&&FreeQ[c,tau] You may however want to think carefully about exactly what you want your function integrate to do. For example, with the rules given, the variable t and the variable tau seem to be playing the same role. Erich > I have defined two functions (Math 4.1) > In[1]:= integrate[c_ y_,t_]:=c*integrate[y,t]/;FreeQ[c,t]; > In[2]:= integrate[Exp[(a_.)*tau], t_]:=(E^(a*t)-1)/a/; > FreeQ[a,t]&&FreeQ[a,tau]; when I applied these functions the solution is different if the coefficient is > the letter d or lower (solution fine) or f or higher (solution wrong) . Here is > an example: Using letter f > In[3]:=integrate[f Exp[-0.8316*tau], t] Out[3]:=integrate[f, t]/E^(0.8316*tau) (*wrong*) Using letter a (the solution is rigth In[4]:=integrate[a Exp[-0.8316*tau], t] Out[4]:=-1.20250*a*(-1 + E^(-0.8316*t)) What it wrong? Guillermo > Sanchez --------------------------------------------- > This message was sent using Endymion MailMan. ==== << Graphics`ImplicitPlot` does the job. Matthias Bode Sal. Oppenheim jr. & Cie. KGaA Koenigsberger Strasse 29 D-60487 Frankfurt am Main GERMANY Mobile: +49(0)172 6 74 95 77 Internet: http://www.oppenheim.de -----UrsprÌ.b9ngliche Nachricht----- Gesendet: Mittwoch, 18. September 2002 08:09 An: mathgroup@smc.vnet.net Betreff: Drawing an ellipse I am new with Mathematica, I have one question, I know that the sollution might be very easy, but I wasn't able to find it by now. I would like to draw an ellipse, the formula let's say is as follows: 0.09 x^2 +0.04 x y + 0.06 y^2 = 4 Maciej ==== Solve[eqn, var] and InverseFunction are designed to do what you want, in principle. Try Solve[(a*h+b*h^2) == f, h] For your eqn, however, it is impossible to solve for h since h appears as a factor in a sum *and* as an exponent. Try Solve[Exp[-h]*(a*h+b*h^2) == f, h] and study Mathematica's message. Matthias Bode Sal. Oppenheim jr. & Cie. KGaA Koenigsberger Strasse 29 D-60487 Frankfurt am Main GERMANY Mobile: +49(0)172 6 74 95 77 Internet: http://www.oppenheim.de -----UrsprÌ.b9ngliche Nachricht----- Gesendet: Mittwoch, 18. September 2002 08:10 An: mathgroup@smc.vnet.net Betreff: Invert a function How can I invert a funcion? For example In:=f[h_]:=Exp[-h](a*h+b*h^2); I want to invert it as h as a function of f How can I do that? Please suggest. Raj ==== > Map[(obj = drawMembrane[#]; > drawArea@setMathCommand[obj]; > drawArea@repaintNow[]; > Pause[.0001];) &, t ] This is the exact code you use? Maybe I am missing something here, but you seem to pass only the String obj to the setMathCommand method, not the content of the obj variable; > public class My_DisplayGraphicsViaJava { > public void displayGraphics(String cmds[]) { > for (int i = 0; i < cmds.length-1; i++) > drawArea.setMathCommand(cmds[i]); > drawArea.repaintNow(); > Thread.sleep(200);} } er... this won't work; you are using the drawArea variable, without ever declaring or assigning something to it! pff... hm... you could try this: import com.wolfram.jlink.MathCanvas; public class My_DisplayGraphicsViaJava { public static void displayGraphics(String cmds[], MathCanvas drawArea){ for (int i = 0; i < cmds.length-1; i++){ drawArea.setMathCommand(cmds[i]); drawArea.repaintNow(); Thread.sleep(200); } } } save this to a file called My_DisplayGraphicsViaJava.java and compile it. Put the class file in a directory that is in your CLASSPATH (or look up how the CLASSPATH works in J/Link 2.0 in the J/Link documentation, its properly explained there); If all that has worked, load the class: LoadJavaClass[My_DisplayGraphicsViaJava]; then use the static method displayGraphics, by passing it the String array and the drawArea variablen (which is the MathCanvas object, you have instantiated in your Mathematica Notebook): My_DisplayGraphicsViaJava`displayGraphics[obj, drawArea]; obj being the String array (or string List in M_) that holds your Graphics Expressions (or what you want to show); > But needless to say this dont work....LoadJavaClass responds by saying Class > not Found.....so what should I do to correct this? Also , on the java > routine, was i supposed to add: Extends MathFrame, etc....how will my > java routine know what drawArea.setMathCommand is, etc..... > am I also supposed to compile my java routine? Yes! ... use the Java code I proposed above, that should at least compile properly; Place the compiled class file in a directory on your CLASSPATH (consult J/Link docu if you don't know about CLASSPATH) and then try the Mathematica code again. I hope this was not too confusing; I don't have a working Mathematica here, so I can't verify, if my proposed code would really work; there migth be typos/errors in my code. murphee ==== I am trying to write an animation using JLink.....basically as follows...(using Mathematica 4.1, JLink 2.01 and java 1.4.1)... frm = createWindow[]; (* basically to create a MathFrame and a drawArea = MathCanvas *) Here is my drawing routine.... Map[(obj = drawMembrane[#]; drawArea@setMathCommand[obj]; drawArea@repaintNow[]; Pause[.0001];) &, t ] t = time list ={0,.25,etc}.... and drawMembrane is a function that uses the time value to make a graphics plot......unfortunately, this routine draws the next time plot very slowly, plus I also notice that without the Pause[] statement, Mathematica only draws the graph for the final time value... Following an example in the JLink documentation, I have written the following java routine....which I saved in the same dir as MathFrame, etc.... public class My_DisplayGraphicsViaJava { public void displayGraphics(String cmds[]) { // declare String Array for Graphics Objects for (int i = 0; i < cmds.length-1; i++) drawArea.setMathCommand(cmds[i]); drawArea.repaintNow(); Thread.sleep(200);} } and then in Mathematica I have LoadJavaClass[My_DisplayGraphicsViaJava]; My_DisplayGraphicsViaJava`displayGraphics[obj]; (* where obj is the array of Graphics plots created *) But needless to say this dont work....LoadJavaClass responds by saying Class not Found.....so what should I do to correct this? Also , on the java routine, was i supposed to add: Extends MathFrame, etc....how will my java routine know what drawArea.setMathCommand is, etc.....am I also supposed to compile my java routine? thanks....jerry blimbaum NSWC panama city, fl ==== Dear colleagues, I'm trying to plot the next exponential function Plot[Exp[1/(Abs[x-1]-Abs[x-2])],Range] If the Range is positive there aren«t problems. However if the Range is negative Mathematica plots a strange beast... Have anyone found anything of this sort? I would apreciate to know how to resolve this problem. Juan Egea Garcia Bullas - Murcia - Espa.96a ==== The function you are plotting is constant for negative x. You might want to specify a plotrange, such as PlotRange->{0,2}, or Mathematica may try to zoom in on the line and show you a picture of numerical noise. Erich > Dear colleagues, I'm trying to plot the next exponential function Plot[Exp[1/(Abs[x-1]-Abs[x-2])],Range] If the Range is positive there aren«t problems. However if the Range is > negative Mathematica plots a strange beast... Have anyone found anything of this sort? I would apreciate to know how to > resolve this problem. Juan Egea Garcia > Bullas - Murcia - Espa.96a > ==== I need to solve algebraic-differential-equations (some of index 1, maybe some of higher index as well) from within Mathematica 4.2. Is there an implementation of numerical (like DASSL) or symbolic algorithms capable of doing that around? I know that some numerical software systems which may have DASSL-routines that do what I want. Has anybody suggestions on which of these systems is easy to interface with Mathematica? Is there a C++ library that contains such routines and can be interfaced with Mathematica? Yours, Reinhard Oldenburg ==== > I need to solve algebraic-differential-equations > (some of index 1, maybe some of higher index as well) > from within Mathematica 4.2. > Is there an implementation of numerical (like DASSL) > or symbolic algorithms capable of doing that around? I know that some numerical software systems which > may have DASSL-routines that do what > I want. Has anybody suggestions on which of these > systems is easy to interface with Mathematica? > Is there a C++ library that contains such routines and can be > interfaced with Mathematica? Reinhard, have you tried the NDAESolve package that comes with the circuit analysis toolbox Analog Insydes? You can get a free evaluation version of Analog Insydes from www.analog-insydes.de. Eckhard Hennig ==== Dear Colleagues, I intend to make an animation in which ball A rolls down on an inclined plane from the left whilst ball B - starting from the same height - rolls down Cosh[t]'s path from the right. x-axis is time t, y-axis is height h. Ball A is fine; ball B - which should arrive at h=0 before A - is beyond my means. Matthias Bode Sal. Oppenheim jr. & Cie. KGaA Koenigsberger Strasse 29 D-60487 Frankfurt am Main GERMANY Mobile: +49(0)172 6 74 95 77 Internet: http://www.oppenheim.de ==== Matthias, The simplest way to get the equation of motion is to set up the Lagrangian. Let's assume a 1 kg mass. Then the kinetic energy is KE = Simplify[(1/2)*(x'[t]^2 + D[Cosh[x[t]],t]^2)] and the potential energy is PE = 9.8*Cosh[x[t]] The Lagrangian is L = KE - PE and the equation of motion is diffeq = Simplify[ D[D[L, x'[t]], t] ] == Simplify[ D[L, x[t]] ] Now solve and animate ... xx[t_] = x[t]/. First[ NDSolve[{diffeq, x[0] == -1, x'[0] == 0}, x[t], {t, 0, 5}]] curve = Plot[Cosh[x], {x, -1, 1}] Do[ Show[curve, Graphics[Disk[{xx[t], Cosh[xx[t]]}, 0.025]], PlotRange -> {{-1.2, 1.2}, {0.9, 1.65}}, AspectRatio -> Automatic, Axes->None], {t, 0, 5, 0.1}] ---- Selwyn Hollis > Dear Colleagues, I intend to make an animation in which ball A rolls down on an inclined plane from the left whilst ball B - starting from the same height - rolls down Cosh[t]'s path from the > right. x-axis is time t, y-axis is height h. Ball A is fine; ball B - which should arrive at h=0 before A - is beyond my > means. > Matthias Bode > Sal. Oppenheim jr. & Cie. KGaA > Koenigsberger Strasse 29 > D-60487 Frankfurt am Main > GERMANY > Mobile: +49(0)172 6 74 95 77 > Internet: http://www.oppenheim.de ==== Having noticed your statement ... BEYOND MY MEANS I thing you aren't yet familiar with Lagrangian formalism. It's quite easy to derive a general equation of motion for a point mass, subjected to gravity and to moving on a curve f = y(x) (i.e. f = Cosh[#]&). 1) I'll leave re-deriving equation to you, here is what I've got (just copy paste it).: !(getEq[ f_] := [IndentingNewLine](x'')[ t] + (x')[t]^2 (2 (f')[x[t]] (f'')[x[t]])/(1 + (f' )[x[t]]^2) + (g (f')[x[t]])/(1 + (f')[x[t]]^2) == 0 /. g -> 1) 2) Next, you integrate it .: getSol[f_, x0_] := Module[{tStop}, NDSolve[{getEq[f], x[0] == x0, x'[0] == 0} , x , {t, 0, 10} , MaxStepSize -> 1/100 , StoppingTest :> If[x0 < 0, x > 0, x < 0] ] // First ] 3) Here follows animation code, specially for linear versus cosh case, apply my initial conditions (below) makeDuo[f_, g_, x0_, fpt_] := Module[{ solf = getSol[f, x0], solg = getSol[g, x0], tf, tg, maxT, minT }, tf = solf[[1, 2, 1, 1, 2]]; tg = solg[[1, 2, 1, 1, 2]]; maxT = Max[{tf, tg}]; minT = Min[{tf, tg}]; Do[ Plot[{f@x, g@x} , {x, 0, x0} , AspectRatio -> Automatic , Frame -> True , Axes -> False , Epilog -> { AbsolutePointSize[10], Hue[0], Point[{x[t], f@x[t]} /. solf], Hue[.6], Point[{x[t], g@x[t]} /. solg] } ] , {t, 0, minT, minT/fpt} ] ] 4) My initial conditions. In my opinion, you weren't true on this. Saying STARTING FROM THE SAME HEIGHT is not enough - you should specify x as well, thus specifing starting POINT and not just height y. makeDuo[(Cosh[1.] - 1) # &, Cosh[#] - 1 &, 1, 23] Feel free to modify the code, it should be easy. bye, Borut | Dear Colleagues, | | I intend to make an animation in which | | ball A rolls down on an inclined plane from the left whilst | | ball B - starting from the same height - rolls down Cosh[t]'s path from the | right. | | x-axis is time t, y-axis is height h. | | Ball A is fine; ball B - which should arrive at h=0 before A - is beyond my | means. | | | Matthias Bode | Sal. Oppenheim jr. & Cie. KGaA | Koenigsberger Strasse 29 | D-60487 Frankfurt am Main | GERMANY | Mobile: +49(0)172 6 74 95 77 | Internet: http://www.oppenheim.de | | ==== As I derived a generalization for a 3D parameterized curve yesterday, I'd noticed a mistake in my equation posted below, a factor '2' in expression involving x'[t]^2, should be '1'. Since this forum is of an alt. type, I've published the whole notebook at http://www2.arnes.si/~gljpoljane22/math/FallingCurve3D.nb Bye, Borut p.s. A 'fill-the-gap' riddle for those interested in physics lore. Richard Feynman once said: Science is like _ _ _, sometimes something useful comes out, but that is not the reason why we are doing it. | ... | 1) I'll leave re-deriving equation to you, here is what I've got (just copy | paste it).: | | !(getEq[ | f_] := [IndentingNewLine](x'')[ | t] + (x')[t]^2 (2 (f')[x[t]] (f'')[x[t]])/(1 + | (f' | )[x[t]]^2) + (g (f')[x[t]])/(1 + (f')[x[t]]^2) == 0 /. g -> | 1) | ... ==== David, There is no command such as RestoreDefaults. But you can do the following: saveGraphics3DOptions = Options[Graphics3D]; SetOptions[Graphics3D, Boxed -> False]; (say) Make your plots. SetOptions[Graphics3D, Sequence @@ saveGraphics3DOptions]; The example in the Help for SetOptions shows a similar example. My own personal preference would be to not change the Graphics3D options, but to add the option change to the plot itself. David Park djmp@earthlink.net http://home.earthlink.net/~djmp/ Sender: steve@smc.vnet.net Approved: Steven M. Christensen , Moderator ==== I have the following code and am trying to include a legend. The BarChart appears, but with no legend. I have followed the 'help files', but I think I'm missing something. In my code, SortTime and SessTime are value pairs. Photog is a string array containing labels. /* *Begin code here */ ShowLegend[BarChart[SortTime, SessTime, BarSpacing -> -.3,, BarGroupSpacing -> .5, BarLabels -> Photog, BarStyle -> {RGBColor[1, 0, 0], RGBColor[0.5, 0.5, 1]}, PlotLabel -> Photographer Time , DefaultFont -> {Helvetica, 9}], {{{RGBColor[1, 0, 0], Sort Time }, {RGBColor[0.5, 0.5, 1], Session Time }}, LegendLabel -> Mean Time}] pete ==== How's this: maxima[f_, x_List] := Last@Sort@({f[#], #} & /@ x) maxima[Sin,Range[0,2Pi,0.1]] {0.999574,1.6} That gives you both f[x] and x at the maximum. Bobby Treat -----Original Message----- that largest output. Any ideas? My only idea is this: create a table of pairs {output, input{ and find the element of the table that has the largest first entry. This will give me the input I seek. Is there a slick way of doing this in mathematica? -- Jason Miller, Ph.D. Division of Mathematics and Computer Science Truman State University 100 East Normal St. Kirksville, MO 63501 http://vh216801.truman.edu 660.785.7430 ==== want base 12 numbers, for instance, stored in -- a List of digits? A character string? A list of digits, highest power to lowest, is easy to handle: ClearAll[convertBase] convertBase[(from_Integer)? (#1 > 1 & ), (to_Integer)?(#1 > 1 & ), n:{_Integer..}] /; Max[n] < from := from], to] convertBase[5, 3, {1, 2}] {2, 1} Bobby Treat -----Original Message----- example, ConvertBase[from_, to_, num_] would convert from base to base 9 using the number num as the number to convert (it would best be allowed to let users input in that specific base since they'd be going from base 3 to base 9, for example so maybe num is needed at all). ==== This came up some time ago. Can't remember exactly what my reply was but I discovered later that it is because the number lock is on. Turn it off and the hand will disappear. May even fix the back space trouble. Yas > I don't have any idea about how to solve next problem. Back space key when I am in the front end don't work > correctly. It advances instead of go back and there is > pointer mark of the mouse looks like a hand instead > of an arrow, and this don't work fine too. > Configuration one? How can I solve this? > Note: This problem seems to appear just in my > workspace > i mean this not happen for other users. Cesar. > __________________________________________________ > Do you Yahoo!? > Yahoo! News - Today's headlines > http://news.yahoo.com ==== I believe that's not the adjacency matrix Thomas asked for. It doesn't have zeroes on the diagonal (it isn't square) and it doesn't have ones to indicate that two actors are associated with the same event. Instead, it shows connections between actors and events, which is actually more useful, as I'll demonstrate. (In addition -- though it doesn't really matter -- Jens-Peer switched the 'actors' and 'events' nomenclature within the function.) If you multiply it by its transpose, you get something else that's useful: lst = {{1, A}, {1, B}, {2, B}, {3, C}, {3, D}, {1, D}, {1, C}}; AdjacenceMatrix[lst : {{_, _} ..}] := Module[{actors, events, adj}, {actors, events} = Union /@ Transpose[lst]; adj = Table[0, {Length[actors]}, {Length[events]}]; Scan[(Part[adj, Sequence @@ #] = 1) &, lst /. MapIndexed[Rule[#1, First[#2]] &, events]]; adj] MatrixForm[a = AdjacenceMatrix[lst]] MatrixForm[b = a.Transpose[a]] Matrix 'b' records how many events two actors have in common. On the diagonal, it shows the total number of events each actor is connected to. It's easy to put zeroes on the diagonal: MatrixForm[c = b (1 - IdentityMatrix[Length[b]])] To get the originally intended incidence matrix, this works: d = c /. {_?Positive -> 1} However, I think matrices 'a' and 'b' are actually more useful, and 'a' easily leads to all the others. Bobby -----Original Message----- AdjacenceMatrix[lst : {{_, _} ..}] := Module[ {actors,events adj}, {events, actors} = Union /@ Transpose[lst]; adj = Table[0, {Length[events]}, {Length[actors]}]; Scan[(Part[adj, Sequence @@ #] = 1) &, lst /. MapIndexed[Rule[#1, First[#2]] &, actors]]; adj ] you get In[]:=AdjacenceMatrix[lst] Out[]={{1, 1, 0, 1}, {0, 1, 0, 0}, {0, 0, 1, 1}} Jens I need to create an adjacency matrix from my data, which is currently in > the form of a .txt file and is basically a two column incidence list. > For example: 1 A > 1 B > 2 B > 3 C > . . > . . > . . > m n Where 1 to m represent actors and A to n represent events. My goal is to > have an (m x m) matrix where cell i,j equals 1 if two actors are > incident to the same event (in the sample above, 1 and 2 are both > incident to B) and 0 otherwise (w/ zeros on the diagonal). I'm new to Mathmatica, and so I'm on the steep part of the learning > curve ... All I've been able to figure out so far is how to get my > incidence list into the program using Import[filename.txt]. But then > what? How do I convert to the adjacency matrix? I've found the > ToAdjacencyMatrix[] command in DiscreteMath`Combinatorica`, but I can't > seem to get it to work ... > Tom ********************************************** > Thomas P. Moliterno > Graduate School of Management > University of California, Irvine > tmoliter@uci.edu > ********************************************** ==== Maybe this will help. I had to do some guessing, and I fabricated my own R function (r). I never start user-defined symbols with capital letters. ClearAll[r] t = Pi/2.; r = Interpolation[{#, Abs@Cos@#} & /@ Range[0, N@t, Pi/100.]]; y[z_] := NIntegrate[r[x]^4, {x, 0, z}] Timing[yInverse = Interpolation[{y[#], #} & /@ Range[0, N@t, Pi/100.]];] {0.030999999999998806*Second, Null} a = NIntegrate[r[yInverse[z]*Cos[z]], {z, 0, y[t]}]/y[t] 0.9249048118568042 Bobby Treat -----Original Message----- a = (1/y[T])* NIntegrate[R[InvFunction[y[t]]*Cos[y[t]], {y[t], 0, y[T]}] it works with a constant instead of R[t]. Hope you can help, and that the above is understandable. Martin Skogstad ==== << Graphics`ImplicitPlot` eqn = 0.09*x^2 + 0.04*x*y + 0.06*y^2 == 4; maxX = 1.1*Sqrt[4/0.09]; ImplicitPlot[eqn, {x, -maxX, maxX}]; You could also use ParametricPlot: polarEqn = eqn /. {x -> r Sin[t], y -> r Cos[t]} radius[t_] = r /. Last@Solve[polarEqn, r] ParametricPlot[radius[x]{Cos[x], Sin[x]}, {x, 0, 2Pi}, AspectRatio -> 1]; I tried and failed with InequalityGraphics: << Graphics`InequalityGraphics` InequalityPlot[{LessEqual @@ eqn}, {x, -xMax, xMax}, {y, -yMax, yMax}] InequalityPlot::region:The region defined by [LeftSkeleton]1[RightSkeleton] could not be broken down into cylinders. InequalityPlot[ 0.09*x^2 + 0.04*x*y + 0.06*y^2 <= 4, {x, -xMax, xMax}, {y, -yMax, yMax}] Plot can be used very neatly, if you first figure out the true extent of the x-values: positiveX[y_] = x /. Last@Solve[eqn, x] Solve[positiveX'[y] == 0] xMax = First@(positiveX[y] /. %) Plot[Evaluate[y /. Solve[eqn, y]], {x, -xMax, xMax}, AspectRatio -> 1]; Bobby Treat -----Original Message----- ==== That function isn't easily inverted and, in fact, has no inverse unless you limit the range of h. In this plot, for instance, you see that the same f[h] values are taken on for two different values of h: Plot[f@h /. {a -> 3, b -> 0.2}, {h, 0, 10}] The peak of the graph depends on the parameters a and b, so the domain of an inverse (if it exists) also depends on a and b. That's exactly the kind of situation Solve can't handle. For specific a, b and y, FindRoot can find h for you: FindRoot[Evaluate[f[h] == 1 /. {a -> 3, b -> 0.2}], {h, 2}] {h -> 1.799292219454869} FindRoot[Evaluate[f[h] == 1 /. {a -> 3, b -> 0.2}], {h, 0.5}] {h -> 0.5654110956582447} Bobby Treat -----Original Message----- ==== I'd say the applet demonstrates that histograms are useless unless we choose a pretty small bin-width. Choosing a bin-width of significant size that doesn't yield a misleading histogram is an unlikely accident. Bobby -----Original Message----- understood the significance of choosing the bin width......jerry blimbaum -----Original Message----- very nice package that does all of the above and more is mathStatica. See http://www.mathstatica.com for details. Obviously, it is less expensive to write your own functions. Just recently in message Mark Fisher posted code that addresses the empirical CDF. However, in this code you may want to replace 1/n with 1/(n+1) or (j-0.5)/n depending on your application. Note, these will have no significant effect for large data sets. The key issue with an empirical PDF is deciding the bin width. A simple approach would be to use the functions in Statistics`DataManipulation` and BinListCounts. More sophisticated approaches involve kernel methods. These methods will generate smoother estimates for the PDF. Again, the key is bandwidth. There is no apriori choice for bin width or bandwith. Bad choices will obscure significant features in the data set. ==== > I'm curious, why the built-in boolean functions And and Or > aren't commutative? The on-line help explains why. (Look up And or Or and then link A.4.2.) You are assured that the argments of And or Or are evaluated left to right, stopping on the first False (for And) or True (for Or). Tom Burton ==== True, but in this case, excitement means being misled far too often. I've never favored histograms much, except when there are lots of bars and little apparent noise. Otherwise, unless you're very lucky, a histogram doesn't even locate the mode well. Bobby Treat -----Original Message----- Bobby -----Original Message----- blimbaum -----Original Message----- very nice package that does all of the above and more is mathStatica. See http://www.mathstatica.com for details. Obviously, it is less expensive to write your own functions. Just recently in message Mark Fisher posted code that addresses the empirical CDF. However, in this code you may want to replace 1/n with 1/(n+1) or (j-0.5)/n depending on your application. Note, these will have no significant effect for large data sets. The key issue with an empirical PDF is deciding the bin width. A simple approach would be to use the functions in Statistics`DataManipulation` and BinListCounts. More sophisticated approaches involve kernel methods. These methods will generate smoother estimates for the PDF. Again, the key is bandwidth. There is no apriori choice for bin width or bandwith. Bad choices will obscure significant features in the data set. ==== Could someone calculate the number Pi to 67,108,864 (2^26) decimal places I made the calculation in another program and would like to verify its ==== > Could someone calculate the number Pi to 67,108,864 (2^26) decimal places > I made the calculation in another program and would like to verify its Does it really matter what program is used to verify it? If not, here's the digits (computed with the fastest pi crunching program for a PC): 33863220896223409803 ==== >I am trying to write an animation using JLink.....basically as >follows...(using Mathematica 4.1, JLink 2.01 and java 1.4.1)... > >frm = createWindow[]; (* basically to create a MathFrame and a drawArea >= MathCanvas *) > >Here is my drawing routine.... Map[(obj = drawMembrane[#]; >drawArea@setMathCommand[obj]; > drawArea@repaintNow[]; > Pause[.0001];) &, t ] > >t = time list ={0,.25,etc}.... and drawMembrane is a function that uses the >time value to make a graphics plot......unfortunately, this routine draws >the next time plot very slowly, plus I also notice that without the Pause[] >statement, Mathematica only draws the graph for the final time value... Jerry, You did a nice job with this one-frame-at-a-time plot. (I'm replying here to this message and your previous one where you presented the full code.) The main reason that it is slow drawing the frames is that the Pause[] function only works with integer second intervals. If you say Pause[.0001] it pauses for a whole second, which inserts a 1-second delay between each frame. If I take out the Pause, I get about 5 frames per second on my laptop (for your Sin plot). This still seems slow to me, but it turns out that Java is just slow at taking GIF data and making it into pixels to be displayed. That step consumes virtually all the time for the animation. If you make the image smaller, the animation runs much faster. Try making your window size 400 x 300 instead of 800 x 600. At that size, the animation runs at a good speed on my machine. If you want to insert a delay between frames, you need to use Java instead of calling Pause[]: LoadJavaClass[java.lang.Thread]; Map[(obj = drawMembrane[#]; drawArea@setMathCommand[obj]; drawArea@repaintNow[]; (* Add a 100 ms delay *) Thread`sleep[100];) &, t ] You said that without the Pause[] you only get the last frame drawn, but that shouldn't have any effect. Perhaps you took out the call to repaintNow() along with the Pause? >Following an example in the JLink documentation, I have written the >following java routine....which I saved in the same dir as MathFrame, >etc.... >public class My_DisplayGraphicsViaJava { > public void displayGraphics(String cmds[]) { // declare String Array >for Graphics Objects > for (int i = 0; i < cmds.length-1; i++) > drawArea.setMathCommand(cmds[i]); > drawArea.repaintNow(); > Thread.sleep(200);} } >and then in Mathematica I have LoadJavaClass[My_DisplayGraphicsViaJava]; >My_DisplayGraphicsViaJava`displayGraphics[obj]; (* where obj is the >array of Graphics plots created *) >But needless to say this dont work....LoadJavaClass responds by saying Class >not Found.....so what should I do to correct this? Also , on the java >routine, was i supposed to add: Extends MathFrame, etc....how will my >java routine know what drawArea.setMathCommand is, etc.....am I also >supposed to compile my java routine? You will definitely need to compile any Java classes you write! But writing this in Java won't change anything because the delay is in Java itself, not in J/Link or the MathLink communication. This Java class would need a lot of work before it would compile. You would need to either create drawArea in the Java class rather than in Mathematica code, or pass it as an argument to displayGraphics() (which needs to be static, by the way). Anyway, there is no point in thinking about any of that, since this displayGraphics() method just duplicates functionality that you have already (much more conveniently) scripted in Mathematica code. One final general note for J/Link programmers: It's OK to create Java names with underscores in them, but you have to remember how they are handled in Mathematica. When you refer to a Java name in Mathematica code as a string, you keep the underscore: LoadJavaClass[My_DisplayGraphicsViaJava]; But when a name appears as a symbol in Mathematica, J/Link maps it to a U because _ is not legal in a symbol name: MyUDisplayGraphicsViaJava`displayGraphics[....] It's often easiest to just avoid using _ in the names of Java classes and methods that you write for use with Mathematica. Todd Gayley Wolfram Research ==== >-----Original Message----- >Sent: Wednesday, September 18, 2002 8:09 AM >I am new with Mathematica, I have one question, >I know that the sollution might be very easy, but I wasn't able to >find it by now. >I would like to draw an ellipse, the formula let's say is as follows: 0.09 x^2 +0.04 x y + 0.06 y^2 = 4 >Maciej > Two methods: (1) solve for explit functions y = f[x] and plot those: In[2]:= eqn = 0.09 x^2 + 0.04 x y + 0.06 y^2 == 4 ... just for convenience; In[5]:= sols = Solve[eqn, y] In[6]:= y /. sols ...these are the two functions for y expressed by x. To find the minimum and maximum values for x, we spot the discriminant, and solve for x: In[12]:= {xmin, xmax} = x /. Solve[ Cases[y /. sols[[1]], Sqrt[disc_] :> disc, Infinity] == 0, x] ...now we may use Plot: In[18]:= Plot[Evaluate[y /. sols], {x, xmin, xmax}, AspectRatio -> Automatic] (2) guess approximate values for xmin, xmax and let Mathematica do all that work: In[1]:= << Graphics`ImplicitPlot` ...load the package In[19]:= ImplicitPlot[eqn, {x, -7, 7}] ...done. -- hw ==== b = 2; r = 4; n = 10101111 p = Reverse[Characters[n]] {1, 1, 1, 1, 0, 1, 0, 1} z = Reverse[Partition[p, r]] {{0, 1, 0, 1}, {1, 1, 1, 1}} (* this is okay *) b = 2; r = 3; z = Reverse[Partition[p, r]] {{1, 0, 1}, {1, 1, 1}} (* this is wrong, want output to be {{0,1,0},{1,0,1},{1,1,1}}*) What I want is a function that takes a string as input, parses each character, partitions that to whatever length I want and Prepends zeros to the input as needed to create an appropriate length array. What is the correct syntax for Partition (or is there a better way?). ==== > b = 2; r = 4; > n = 10101111 > p = Reverse[Characters[n]] > {1, 1, 1, 1, 0, 1, 0, 1} > z = Reverse[Partition[p, r]] > {{0, 1, 0, 1}, {1, 1, 1, 1}} (* this is okay *) > b = 2; r = 3; > z = Reverse[Partition[p, r]] > {{1, 0, 1}, {1, 1, 1}} (* this is wrong, want output to be > {{0,1,0},{1,0,1},{1,1,1}}*) What I want is a function that takes a string as input, parses each > character, partitions that to whatever length I want and Prepends zeros to > the input as needed to create an appropriate length array. What is the correct syntax for Partition (or is there a better way?). z = Reverse[Partition[p, r, r, {1,1}, 0] pads 0 on the right before partitioning. This doesn't completely do what you want because when you reverse, those digits are still on the right in the first element (i.e., they are appended instead of prepended. To have them prepended, just add z[[1]] = RotateRight[z[[1]], r Length[z] - Length[p]]; and z will have what you want. Here is an example with Range (easier to see what is happening) In[1]:= p = Range[8]; b = 2; r = 3; In[2]:= z = Reverse[Partition[p, r, r, {1, 1}, 0]] Out[2]= {{7, 8, 0}, {4, 5, 6}, {1, 2, 3}} In[3]:= z[[1]] = RotateRight[z[[1]], r Length[z] - Length[p]]; In[4]:= z Out[4]= {{0, 7, 8}, {4, 5, 6}, {1, 2, 3}} Rob Knapp > ==== Has anyone had any luck getting Mathematica 4.2 to work on a Sun running Solaris 6? I get the following error message when starting Mathematica: ./Mathematica ld.so.1: ./Mathematica: fatal: librt.so.1: open failed: No such file or directory Killed But math works. Any thoughts on this would be appreciated - even if it's just Solaris 6 sucks - Reinstall with Solaris 8/9. I'd rather not go to the trouble of reinstalling the box unless absolutely necessary (It is a 270mhz Ultra 5) -- Alex Dawson 08 9380 1587 School of Electrical, Electronic and Computer Engineering Systems Administrator http://www.ee.uwa.edu.au/~alex ==== > Has anyone had any luck getting Mathematica 4.2 to work > on a Sun running Solaris 6? I get the following error message when starting Mathematica: ./Mathematica > ld.so.1: ./Mathematica: fatal: librt.so.1: open failed: No such file or > directory > Killed But math works. Any thoughts on this would be appreciated - even if it's just > Solaris 6 sucks - Reinstall with Solaris 8/9. I'd rather not go to the > trouble of reinstalling the box unless absolutely necessary (It is a > 270mhz Ultra 5) I'd be tempted to run 'ldd' and see what libraries it needs. On my Solaris 9 box, with Mathematica 4.2 I see: wren /usr/local/stow/Mathematica-4.2/SystemFiles/FrontEnd/Binaries/Solaris % ldd ../Mathematica libXt.so.4 => /usr/lib/libXt.so.4 libXext.so.0 => /usr/lib/libXext.so.0 libXmu.so.4 => /usr/lib/libXmu.so.4 libX11.so.4 => /usr/lib/libX11.so.4 libnsl.so.1 => /usr/lib/libnsl.so.1 libsocket.so.1 => /usr/lib/libsocket.so.1 libc.so.1 => /usr/lib/libc.so.1 libucb.so.1 => /usr/ucblib/libucb.so.1 librt.so.1 => /usr/lib/librt.so.1 libpthread.so.1 => /usr/lib/libpthread.so.1 libSM.so.6 => /usr/lib/libSM.so.6 libICE.so.6 => /usr/lib/libICE.so.6 libm.so.1 => /usr/lib/libm.so.1 libdl.so.1 => /usr/lib/libdl.so.1 libmp.so.2 => /usr/lib/libmp.so.2 libelf.so.1 => /usr/lib/libelf.so.1 libaio.so.1 => /usr/lib/libaio.so.1 libmd5.so.1 => /usr/lib/libmd5.so.1 /usr/platform/SUNW,Ultra-60/lib/libc_psr.so.1 libthread.so.1 => /usr/lib/libthread.so.1 /usr/platform/SUNW,Ultra-60/lib/libmd5_psr.so.1 You should be able to find the libraries it's using, and hopefully it might just be looking in the wrong place, in which case you may be able to create a symbolic link. However, if Solaris 2.6 is not supported (I don't know if it is) it is quite possible it wants a library you don't have, in which case you are stuck unless you re-install a later OS. -- Dr. David Kirkby, Senior Research Fellow, Department of Medical Physics, University College London, 11-20 Capper St, London, WC1E 6JA. Internal telephone: ext 46408 ==== One of these days the company I work for will buy one Mathematica 4.2 versions of the product. window manager. Our Windows systems are running on top of VMWare in Windows XP Professional. In testing using an evaluation version I found that the VMWare/XP version is performing pretty well so that wouldn't be an issue. My main concern (and probably the grounds on which we will base our decision) is the quality of the front-end. The Windows frontend works flawlessly; however, in test-driving earlier versions of Mathematica (3.0 satisfactorily. I didn't spend much time to hunt down the source of the problem, but it is probable that some X key-mapping was wrong making, for example, keyboard shortcuts for 'Copy' and 'Paste' malfunctioning. On the Windows frontend. to full satisfaction? I'd be willing to spend a couple of hours to fix some settings, but only if I'm sure it could be done. The bottom line is that I I not if that would mean putting up with a quircky front-end. Sidney Cadot ==== > One of these days the company I work for will buy one Mathematica 4.2 > versions of the product. window manager. Our Windows systems are running on top of VMWare in Windows > XP Professional. > In testing using an evaluation version I found that the VMWare/XP version is > performing pretty well so that wouldn't be an issue. My main concern (and probably the grounds on which we will base our > decision) is the quality of the front-end. The Windows frontend works > flawlessly; however, in test-driving earlier versions of Mathematica (3.0 > satisfactorily. I didn't spend much time to hunt down the source of the > problem, but it is probable that some X key-mapping was wrong making, for > example, keyboard shortcuts for 'Copy' and 'Paste' malfunctioning. On the > Windows frontend. to full satisfaction? I'd be willing to spend a couple of hours to fix some > settings, but only if I'm sure it could be done. The bottom line is that I I > not if that would mean putting up with a quircky front-end. For what it is worth, I have no problems with the front end under Solaris - another varient of Unix. -- Dr. David Kirkby PhD, web page: http://www.david-kirkby.co.uk Amateur radio callsign: G8WRB ==== > One of these days the company I work for will buy one Mathematica 4.2 > Windows versions of the product. window manager. Our Windows systems are running on top of VMWare in > Windows XP Professional. > In testing using an evaluation version I found that the VMWare/XP version > is performing pretty well so that wouldn't be an issue. My main concern (and probably the grounds on which we will base our > decision) is the quality of the front-end. The Windows frontend works > flawlessly; however, in test-driving earlier versions of Mathematica (3.0 > satisfactorily. I didn't spend much time to hunt down the source of the > problem, but it is probable that some X key-mapping was wrong making, for > example, keyboard shortcuts for 'Copy' and 'Paste' malfunctioning. On the > Windows frontend. working to full satisfaction? I'd be willing to spend a couple of hours > to fix some settings, but only if I'm sure it could be done. The bottom > VMWare/Windows, but not if that would mean putting up with a quircky > front-end. For what it is worth, I have no problems with the front end under Solaris > - another varient of Unix. I sincerely wish WRI were more relaxed about allowing people to use the same license on the same hardware, but with a different OS. My understanding is that I would have to buy two licenses if I wanted to switch back and forth That's how open source is paid for. Not everybody has the same amount of spare blood in their veins, or willingness to spill it. For those who release of Mathematica, it would be nice to have a dual boot option. Perhaps it's simply a support issue. It would potentially cause the same customer to require more support if he were switching back and forth between OSs. http://public.globalsymmetry.com/proprietary/com/wri/ch05.html http://66.92.149.152/proprietary/com/wri/ch05.html I still have some tweaks to add to this, such as M-k -> C-k, but it at least demonstrates that (all) things aren't cast in stone. STH ==== Sidney, (or other Unix) for some years so perhaps my experience could help with your decision. Most of the time I run on Windows (2000, not yet XP) because: 1. The Front End is more stable. 2. The Front End is more comfortable in the sense that the keys for deleting, copying, pasting etc. are the ones my fingers expect them to be. 3. Command completion via Ctrl-k, which I use all the time, does not work 4. Generally speaking it takes more mouse clicks and keystrokes to do things I think these are general differences between the user interfaces on these operating systems rather than something specific to the Mathematica implementations. I have tried to overcome them at various time without success. Since I switched from Windows 95 to Windows NT several years ago, 2. I need to have access to certain files created on that system by other people. 3. I want to run long calculations on additional CPUs, leaving my desktop computer free for other things. You can do this between Windows machines kernel via Mathlink. This is very simple and more reliable and robust than Front End and it is easy, say, to transfer input or output betwen systems within a single Front End. I should add that in our lab we can access the the additional convenience of being able to save a notebook from the Windows Front End in the same directory as other files the kernel may be working with. You may not have this option and, of course, you need Mathematica on both systems. These are my personal opinions based on practical experience, I really do not want to enter into any controversy about operating systems. John Jowett One of these days the company I work for will buy one Mathematica 4.2 Windows > versions of the product. window manager. Our Windows systems are running on top of VMWare in Windows > XP Professional. > In testing using an evaluation version I found that the VMWare/XP version is > performing pretty well so that wouldn't be an issue. My main concern (and probably the grounds on which we will base our > decision) is the quality of the front-end. The Windows frontend works > flawlessly; however, in test-driving earlier versions of Mathematica (3.0 > satisfactorily. I didn't spend much time to hunt down the source of the > problem, but it is probable that some X key-mapping was wrong making, for > example, keyboard shortcuts for 'Copy' and 'Paste' malfunctioning. On the > Windows frontend. > working > to full satisfaction? I'd be willing to spend a couple of hours to fix some > settings, but only if I'm sure it could be done. The bottom line is that I I but > not if that would mean putting up with a quircky front-end. > Sidney Cadot > ==== One of these days the company I work for will buy one Mathematica 4.2 > Windows versions of the product. The Redmond chaos code is IMHO no system at all (SCNR). I have only one problem with my Mathematica 4.0 for students, which is that the Export function does not work for gif and other graphics formats. I have posted a question about this into this group but nobody has answered, so I guess there is no fix for that, but on the other hand 4.2 is much newer and should work. > My main concern (and probably the grounds on which we will base our > decision) is the quality of the front-end. The Windows frontend works > flawlessly; however, in test-driving earlier versions of Mathematica (3.0 > satisfactorily. I didn't spend much time to hunt down the source of the > problem, but it is probable that some X key-mapping was wrong making, for > example, keyboard shortcuts for 'Copy' and 'Paste' malfunctioning. On the > Windows frontend. I have no problem concerning the frontend. One has only to adjust the font settings (if you need it, I can send you a pdf file explaining how). -- Hendrik van Hees Fakult.8at f.9fr Physik http://theory.gsi.de/~vanhees/ D-33615 Bielefeld ==== > What I want is a function that takes a string as input, parses each > character, partitions that to whatever length I want and Prepends zeros to > the input as needed to create an appropriate length array. What is the correct syntax for Partition (or is there a better way?). Although you can't see it in your particular example, I think your two Reverse functions would lead to reversed partitions, so I got rid of them. Then it's a matter of following directions for Partition in Help. The only issue is computing kL. The following way is a bit complicated. Perhaps someone will post a simpler way. n = 10101111; r = 3; p = Characters[n]; kL = r + 1 - Mod[Length[p], r, 1]; kR = r; z = Partition[p, r, r, {kL, kR}, 0] Tom Burton ==== Although And and Or don't have to evaluate all their arguments, I find that this is useless in many situations I run into. For instance, I might have a long list 'boolList' of Boolean expressions and I want to know if ANY of them are False, so I calculate And@@boolList In this situation, ALL the expressions in 'boolList' are calculated, even though the first may be False. Here's an example in which I overcome that: boolList = Hold[Print@#; PrimeQ@#] & /@ Range[3, 1000, 2]; ReleaseHold[And @@ boolList] 3 5 7 9 False Bobby Treat -----Original Message----- argument before any of the others, to avoid unneeded computations of the other arguments. He would put that argument first. If Mathematica turned around and sorted the arguments (which is what happens when a function is orderless), then that argument might end up being evaluated last. If the arguments take a significant amount of time to compute, then sorting first may cause the function to take much longer to evaluate. At any rate, if the above is not a concern for you, you may always change the attributes of And and Or to anything you want. Carl Woll Physics Dept U of Washington ----- Original Message ----- > However, the arithmetical functions are Orderless: Attributes/@{Plus,Times}//ColumnForm {Flat, Listable, NumericFunction, OneIdentity, Orderless, Protected} > {Flat, Listable, NumericFunction, OneIdentity, Orderless, Protected} ------------------- > Evgeni Trifonov > Vladivostok, Russia > ==== Maybe this will work for you: groupDigits[n_String, k_Integer] := Partition[ PadLeft[Characters[n], k Ceiling[StringLength[n]/k], 0], k] Bobby Treat -----Original Message----- {{1, 0, 1}, {1, 1, 1}} (* this is wrong, want output to be {{0,1,0},{1,0,1},{1,1,1}}*) What I want is a function that takes a string as input, parses each character, partitions that to whatever length I want and Prepends zeros to the input as needed to create an appropriate length array. What is the correct syntax for Partition (or is there a better way?). ==== After I load a package using < b = 2; r = 4; > n = 10101111 > p = Reverse[Characters[n]] > {1, 1, 1, 1, 0, 1, 0, 1} > z = Reverse[Partition[p, r]] > {{0, 1, 0, 1}, {1, 1, 1, 1}} (* this is okay *) > b = 2; r = 3; > z = Reverse[Partition[p, r]] > {{1, 0, 1}, {1, 1, 1}} (* this is wrong, want output to be > {{0,1,0},{1,0,1},{1,1,1}}*) What I want is a function that takes a string as input, parses each > character, partitions that to whatever length I want and Prepends zeros to > the input as needed to create an appropriate length array. What is the correct syntax for Partition (or is there a better way?). ==== Dear Group, It is a pleasure write to you again. I have three questions to ask. Alexandre Costa Question One: The below code is working fine but I think there are more elegant ways of doing so. Any suggestions are VERY WELCOME. << DiscreteMath`Permutations` individual1 = RandomPermutation[12] individual2 = RandomPermutation[12] CutPosition = 3 ; CutLength = 4 ; CutSequence = Take[Drop[individual1, CutPosition], CutLength] SonOne = individual2; Do[ If[MemberQ[SonOne, CutSequence[[i]]], SonOne = Delete[SonOne, Position[SonOne, CutSequence[[i]]]]], {i, 1, CutLength}]; SonOne SonOne = RotateLeft[SonOne, 1] Do[SonOne = Insert[SonOne, Reverse[CutSequence][[i]], CutPosition + 1], {i, 1, CutLength}] SonOne Question Two: How can I change points properties (such as Size and Color) for the plot below? The PlotStyle Option does not work for this LabeledListPlot << Graphics`Graphics` listcities = {{1, 5}, {4, 6}, {7, 5}, {5, 4}, {9, 4}, {2, 3}, {4, 2}, {6, 2}, {1, 1}, {5, 1}, {3, 0}, {9, 0}}; LabeledListPlot[listcities, Axes -> None, Frame -> True, DisplayFunction -> $DisplayFunction] Question Three: Why the Goto statement below is not working? q = 2; Label[start]; q = 3; Label[begin]; Print[q]; q += 1; If[q < 6, Goto[begin], Goto[start]] Reply-To: murray@math.umass.edu ==== I presume packagename is a context name, e.g., Graphics`Colors` . Then evaluating the expression Names[packagename*] containing the wildcard symbol * will probably do what you want. It may tell you more than you want to know unless the package was decently written. By decently I mean that names not needing to be known to the outside world have been enclosed within a Private context within the package's context. > After I load a package using > < the source file? -- Murray Eisenberg murray@math.umass.edu Mathematics & Statistics Dept. Lederle Graduate Research Tower phone 413 549-1020 (H) University of Massachusetts 413 545-2859 (W) 710 North Pleasant Street Amherst, MA 01375 ==== Jose, I generally use the Needs statement to load a package. It avoids double loading and all the error messages that result. Say Needs[Graphics`Graphics`] Then to obtain information on the routines in the package just use... ?Graphics`Graphics`* names in the package. If you click on any of the names you obtain the usage message for the name. (On earlier Mathematica versions you only obtain a list of the names and you have to use ?name to obtain the usage message.) More link. If you click on that it will bring you to the relevant Help page. (Private packages may not have that feature.) David Park djmp@earthlink.net http://home.earthlink.net/~djmp/ Sender: steve@smc.vnet.net Approved: Steven M. Christensen , Moderator Reply-To: ==== Maybe there's a faster way -- I hope so! -- but here's my answer: n = 2^26 Timing[RealDigits[N[Pi - 3, n], 10, 20, 19 - n]] 67108864 {15027.922*Second, {{3, 3, 8, 6, 3, 2, 2, 0, 8, 9, 6, 2, 2, 3, 4, 0, 9, 8, 0, 3}, -67108844}} hrMinSec[15027.922] {4*hours, 10*minutes, 27.92200000000048*seconds} Bobby Treat -----Original Message----- ==== Could you tell me the CPU you used and its speed etc...i am curious, other programs out there. > Maybe there's a faster way -- I hope so! -- but here's my answer: n = 2^26 > Timing[RealDigits[N[Pi - 3, n], 10, 20, 19 - n]] 67108864 > {15027.922*Second, > {{3, 3, 8, 6, 3, 2, 2, 0, 8, 9, 6, 2, 2, 3, 4, 0, 9, 8, 0, 3}, > -67108844}} hrMinSec[15027.922] > {4*hours, 10*minutes, 27.92200000000048*seconds} Bobby Treat -----Original Message----- > 4 for me? Could someone calculate the number Pi to 67,108,864 (2^26) decimal > places > I made the calculation in another program and would like to verify its > ==== > Could you tell me the CPU you used and its speed etc...i am curious, > other programs out there. I used one processor of a dual 1GH Mac and got the same answer with the following speed: 4.2 for Mac OS X (June 4, 2002) oldmax = $MaxPrecision 6 1. 10 $MaxPrecision = Infinity Infinity With[{n = 2^26}, Timing[ pd = RealDigits[N[Pi, n + 1], 10, 20, 19 - n]; ]] {28794.1 Second, Null} MaxMemoryUsed[] 512055204 pd {{3, 3, 8, 6, 3, 2, 2, 0, 8, 9, 6, 2, 2, 3, 4, 0, 9, 8, 0, 3}, -67108844} Tom Burton -- ==== So would it take about the same amont of time for the complete printout of digits? Of course it would take a few additional seconds to format the output... Or does Mathematica take alot less time when it truncates the output? > Could you tell me the CPU you used and its speed etc...i am curious, > other programs out there. I used one processor of a dual 1GH Mac and got the same answer with the > following speed: 4.2 for Mac OS X (June 4, 2002) > oldmax = $MaxPrecision > 6 > 1. 10 > $MaxPrecision = Infinity > Infinity > With[{n = 2^26}, Timing[ > pd = RealDigits[N[Pi, n + 1], 10, 20, > 19 - n]; ]] > {28794.1 Second, Null} > MaxMemoryUsed[] > 512055204 > pd > {{3, 3, 8, 6, 3, 2, 2, 0, 8, 9, 6, 2, 2, 3, > > 4, 0, 9, 8, 0, 3}, -67108844} Tom Burton ==== Here's an answer to Question 1: individual1 = RandomPermutation[12] individual2 = RandomPermutation[12] cutPosition = 3; cutLength = 4; cutSequence = Take[Drop[individual1, cutPosition], cutLength] sonOne = Complement[individual2, cutSequence] sonOne = RotateLeft[sonOne, 1] sonOne = Join[Take[sonOne, cutLength - 1], cutSequence, Drop[sonOne, cutLength - 1]] Bobby Treat -----Original Message----- << DiscreteMath`Permutations` individual1 = RandomPermutation[12] individual2 = RandomPermutation[12] CutPosition = 3 ; CutLength = 4 ; CutSequence = Take[Drop[individual1, CutPosition], CutLength] SonOne = individual2; Do[ If[MemberQ[SonOne, CutSequence[[i]]], SonOne = Delete[SonOne, Position[SonOne, CutSequence[[i]]]]], {i, 1, CutLength}]; SonOne SonOne = RotateLeft[SonOne, 1] Do[SonOne = Insert[SonOne, Reverse[CutSequence][[i]], CutPosition + 1], {i, 1, CutLength}] SonOne Question Two: How can I change points properties (such as Size and Color) for the plot below? The PlotStyle Option does not work for this LabeledListPlot << Graphics`Graphics` listcities = {{1, 5}, {4, 6}, {7, 5}, {5, 4}, {9, 4}, {2, 3}, {4, 2}, {6, 2}, {1, 1}, {5, 1}, {3, 0}, {9, 0}}; LabeledListPlot[listcities, Axes -> None, Frame -> True, DisplayFunction -> $DisplayFunction] Question Three: Why the Goto statement below is not working? q = 2; Label[start]; q = 3; Label[begin]; Print[q]; q += 1; If[q < 6, Goto[begin], Goto[start]] ==== I would like to build my Mathematica notebooks in manner which allows me to carry out two overlapping purposes: 1) prototype an algorithm completely within Mathematica, 2) use Mathematica as a partial evaluator to splice derived expressions into a C language version of the algorithm ( with the aid of the Format package from MathSource). The key complication in doing this is that for the Mathematica prototype I want everything to be evaluated, while for splicing I want the results from some of the functions to be retained as data in temporary variables. A first step towards these purposes is simple: write my component functions so they operate on lists (the natural form of the prototype data), then provide lists of arrays for arguments which I would like to supply as data in the resulting C language output forms. An example: -------------------------------------------------- In[1]:= t[x_] := {x[[4]]-x[[3]], x[[2]]-x[[1]]} In[2]:= f[x_,y_] := {y, 1}.t[x] In[3]:= f[{1,2,3,4},2] Out[3]= 3 In[4]:= f[Array[x,4],2] Out[4]= -x[1] + x[2] + 2 (-x[3] + x[4]) In[5]:= CAssign[fcnval, f[Array[x,4],2], AssignToArray->{x}] Out[5]//OutputForm= fcnval=-x[1]+x[2]+2.*(-x[3]+x[4]); -------------------------------------------------- For various reasons, however, I would like certain expressions of my symbolic derivation to be treated as data for the C language version. In the above example, for instance, I would like the array that function t produces to be a data array. The modification of the above example: -------------------------------------------------- In[6]:= tc[x_] := Array[tt,2] In[7]:= fc[x_,y_] := {y, 1}.tc[x] In[8]:= CAssign[tt, t[Array[x,4]], AssignToArray->{x}] Out[8]//OutputForm= tt[0]=-x[3]+x[4]; tt[1]=-x[1]+x[2]; In[9]:= CAssign[fcnval, fc[Array[x,4],2], AssignToArray->{tt}] Out[9]//OutputForm= fcnval=2.*tt[1]+tt[2]; -------------------------------------------------- But I don't want to carry around two versions of everything, nor do I really want to thread all of the supporting functions through my definitions in order to choose the correct function for the purpose I would rather define the C language alternatives only for functions like t and tc in the example, with f using the appropriate one depending on some evaluation flag that I set at the highest level. For example, it would be nice to be able to sprinkle in the C language alternatives with a construct like In[10]:= t[x_] := {x[[4]]-x[[3]], x[[2]]-x[[1]]} In[11]:= DefineTemporaryForm[t[x_]] := Array[tt,Length[x]/2] then be able to write In[8]:= CAssign[tt, t[Array[x,4]], AssignToArray->{x}] Out[8]//OutputForm= tt[0]=-x[3]+x[4]; tt[1]=-x[1]+x[2]; In[9]:= CAssign[fcnval, UseTemporaryForm[f[Array[x,4],2]], AssignToArray- >{tt}] Out[9]//OutputForm= fcnval=2.*tt[1]+tt[2]; I would appreciate any pointers on a good, clean and hopefully simple way to do this within Mathematica! Alex ==== I've been using Mathematica 4.1 on Win98 as a word processor for math-related documents, but often people that need to see the documents don't have Mathematica, and for whatever reason on my computer the HTML saves don't work at all. I'd like to export to PDF format. I can export images to PDF format no problem using, for example Export[c:docsplot3.pdf, Plot[Sin[x],{x,-2Pi,2Pi}]], and I can export cells correctly to GIF, JPEG, and WMF formats (probably more, those are the only ones I tested) using, for example Export[c:docscell4.gif, Cell[ <<...(copied cell data from Edit->Copy As->Cell Expression)...> ]] When I change the filename to a .PDF and evaluate the cell, the program displays 'Running...' for a second and gives the 'Out[n] = c:docscell4.pdf ' message as if a file was created, but no file is created anywhere with any name that I could find with Start->Find->[All files and folders created in the previous day]. Is there a limitation to PDF exporting I don't know about? Do I need to upgrade to 4.2? Am I doing something wrong with the Export[] command? Do I need a faster computer? A patch? Something else? ==== I recently communicated with technical support about precisely this issue. Here's their reply: ``PDF and AI export use psrender, which is a MathLink program that interfaces with the kernel. Since the kernel has no knowledge of how cells are formatted, export cannot generate PDF and AI for cells, just Graphics. Note that this is contrary to the documentation, which says: ``All graphics formats in Export can handle any type of 2D or 3D Mathematica graphics. ... They can also handle Notebook and Cell objects. --- Selwyn Hollis > I've been using Mathematica 4.1 on Win98 as a word processor for > math-related documents, but often people that need to see the documents > don't have Mathematica, and for whatever reason on my computer the HTML > saves don't work at all. I'd like to export to PDF format. I can export > images to PDF format no problem using, for example > Export[c:docsplot3.pdf, Plot[Sin[x],{x,-2Pi,2Pi}]], > and I can export cells correctly to GIF, JPEG, and WMF formats (probably > more, those are the only ones I tested) using, for example > Export[c:docscell4.gif, Cell[ <<...(copied cell data from Edit->Copy > As->Cell Expression)...> ]] > When I change the filename to a .PDF and evaluate the cell, the program > displays 'Running...' for a second and gives the 'Out[n] = c:docscell4.pdf > ' message as if a file was created, but no file is created anywhere with any > name that I could find with Start->Find->[All files and folders created in > the previous day]. > Is there a limitation to PDF exporting I don't know about? Do I need to > upgrade to 4.2? Am I doing something wrong with the Export[] command? Do I > need a faster computer? A patch? Something else? > ==== Here's a smoother animation, taking into account the period and cutting the step size in half (without using more frames): Do[Show[curve, Graphics[Disk[{ xx[t], Cosh[xx[t]]}, 0.035]], PlotRange -> {{-1.2, 1.2}, {0.9, 1.65}}, AspectRatio -> Automatic, Axes -> None], {t, 0, 2.3, 0.05}] SelectionMove[EvaluationNotebook[], All, GeneratedCell] FrontEndTokenExecute[OpenCloseGroup] FrontEndTokenExecute[SelectionAnimate] Bobby Treat -----Original Message----- and the equation of motion is diffeq = Simplify[ D[D[L, x'[t]], t] ] == Simplify[ D[L, x[t]] ] Now solve and animate ... xx[t_] = x[t]/. First[ NDSolve[{diffeq, x[0] == -1, x'[0] == 0}, x[t], {t, 0, 5}]] curve = Plot[Cosh[x], {x, -1, 1}] Do[ Show[curve, Graphics[Disk[{xx[t], Cosh[xx[t]]}, 0.025]], PlotRange -> {{-1.2, 1.2}, {0.9, 1.65}}, AspectRatio -> Automatic, Axes->None], {t, 0, 5, 0.1}] ---- Selwyn Hollis > Dear Colleagues, I intend to make an animation in which ball A rolls down on an inclined plane from the left whilst ball B - starting from the same height - rolls down Cosh[t]'s path from the > right. x-axis is time t, y-axis is height h. Ball A is fine; ball B - which should arrive at h=0 before A - is beyond my > means. > Matthias Bode > Sal. Oppenheim jr. & Cie. KGaA > Koenigsberger Strasse 29 > D-60487 Frankfurt am Main > GERMANY > Mobile: +49(0)172 6 74 95 77 > Internet: http://www.oppenheim.de ==== I would like to build my Mathematica notebooks in manner which allows me to carry out two overlapping purposes: 1) prototype an algorithm completely within Mathematica, 2) use Mathematica as a partial evaluator to splice derived expressions into a C language version of the algorithm (with the aid of the Format package from MathSource). The key complication in doing this is that for the Mathematica prototype I want everything to be evaluated, while for splicing I want the results from some of the functions to be retained as data in temporary variables. A first step towards these purposes is simple: write my component functions so they operate on lists (the natural form of the prototype data), then provide lists of arrays for arguments which I would like to supply as data in the resulting C language output forms. An example: -------------------------------------------------- In[1]:= t[x_] := {x[[4]]-x[[3]], x[[2]]-x[[1]]} In[2]:= f[x_,y_] := {y, 1}.t[x] In[3]:= f[{1,2,3,4},2] Out[3]= 3 In[4]:= f[Array[x,4],2] Out[4]= -x[1] + x[2] + 2 (-x[3] + x[4]) In[5]:= CAssign[fcnval, f[Array[x,4],2], AssignToArray->{x}] Out[5]//OutputForm= fcnval=-x[1]+x[2]+2.*(-x[3]+x[4]); -------------------------------------------------- For various reasons, however, I would like certain expressions of my symbolic derivation to be treated as data for the C language version. In the above example, for instance, I would like the array that function t produces to be a data array. The modification of the above example: -------------------------------------------------- In[6]:= tc[x_] := Array[tt,2] In[7]:= fc[x_,y_] := {y, 1}.tc[x] In[8]:= CAssign[tt, t[Array[x,4]], AssignToArray->{x}] Out[8]//OutputForm= tt[0]=-x[3]+x[4]; tt[1]=-x[1]+x[2]; In[9]:= CAssign[fcnval, fc[Array[x,4],2], AssignToArray->{tt}] Out[9]//OutputForm= fcnval=2.*tt[1]+tt[2]; -------------------------------------------------- But I don't want to carry around two versions of everything, nor do I really want to thread all of the supporting functions through my definitions in order to choose the correct function for the purpose I would rather define the C language alternatives only for functions like t and tc in the example, with f using the appropriate one depending on some evaluation flag that I set at the highest level. For example, it would be nice to be able to sprinkle in the C language alternatives with a construct like In[10]:= t[x_] := {x[[4]]-x[[3]], x[[2]]-x[[1]]} In[11]:= DefineTemporaryForm[t[x_]] := Array[tt,Length[x]/2] then be able to write In[8]:= CAssign[tt, t[Array[x,4]], AssignToArray->{x}] Out[8]//OutputForm= tt[0]=-x[3]+x[4]; tt[1]=-x[1]+x[2]; In[9]:= CAssign[fcnval, UseTemporaryForm[f[Array[x,4],2]], AssignToArray->{tt}] Out[9]//OutputForm= fcnval=2.*tt[1]+tt[2]; I would appreciate any pointers on a good, clean and hopefully simple way to do this within Mathematica! Alex ==== To find out what loaded contexts are related to the package, use: Contexts[Integrate`*] {Integrate`,Integrate`Elliptic`} To see the symbols exported by the context, execute: Names[Integrate`*] {(* too many to list here *)} Names[Integrate`*`*] {Integrate`Elliptic`Elliptic} To find out what context a symbol comes from: Context[Integrate] System` Bobby Treat -----Original Message----- Sender: steve@smc.vnet.net Approved: Steven M. Christensen , Moderator Reply-To: ==== Oops! That should be individual1 = RandomPermutation[12] individual2 = RandomPermutation[12] cutPosition = 3; cutLength = 4; cutSequence = Take[Drop[individual1, cutPosition], cutLength] sonOne = DeleteCases[individual2, _?(MemberQ[cutSequence, #] &)] sonOne = RotateLeft[sonOne, 1] sonOne = Join[Take[ sonOne, cutLength - 1], cutSequence, Drop[sonOne, cutLength - 1]] Complement returns a sorted result, and you didn't want that. Bobby Treat -----Original Message----- sonOne = Join[Take[sonOne, cutLength - 1], cutSequence, Drop[sonOne, cutLength - 1]] Bobby Treat -----Original Message----- << DiscreteMath`Permutations` individual1 = RandomPermutation[12] individual2 = RandomPermutation[12] CutPosition = 3 ; CutLength = 4 ; CutSequence = Take[Drop[individual1, CutPosition], CutLength] SonOne = individual2; Do[ If[MemberQ[SonOne, CutSequence[[i]]], SonOne = Delete[SonOne, Position[SonOne, CutSequence[[i]]]]], {i, 1, CutLength}]; SonOne SonOne = RotateLeft[SonOne, 1] Do[SonOne = Insert[SonOne, Reverse[CutSequence][[i]], CutPosition + 1], {i, 1, CutLength}] SonOne Question Two: How can I change points properties (such as Size and Color) for the plot below? The PlotStyle Option does not work for this LabeledListPlot << Graphics`Graphics` listcities = {{1, 5}, {4, 6}, {7, 5}, {5, 4}, {9, 4}, {2, 3}, {4, 2}, {6, 2}, {1, 1}, {5, 1}, {3, 0}, {9, 0}}; LabeledListPlot[listcities, Axes -> None, Frame -> True, DisplayFunction -> $DisplayFunction] Question Three: Why the Goto statement below is not working? q = 2; Label[start]; q = 3; Label[begin]; Print[q]; q += 1; If[q < 6, Goto[begin], Goto[start]] ==== I would appreciate help with these problems: 1. I'm plotting several thousand points, which I can do either with something like this (this is a test): w = Table[Point[{Random[], Random[]}], {i, 1, 4096}]; x = Table[Point[{Random[], Random[]}], {i, 1, 4096}]; y = Table[Point[{Random[], Random[]}], {i, 1, 4096}]; z = Table[Point[{Random[], Random[]}], {i, 1, 4096}]; dw = Graphics[{PointSize[0.01], RGBColor[ 1, 0, 0], w}]; dx = Graphics[{PointSize[0.01], RGBColor[.8, .8, .8], x}]; dy = Graphics[{PointSize[0.01], RGBColor[ 0, .5, .9], y}]; dz = Graphics[{PointSize[0.01], RGBColor[.8, .8, 0], z}]; Show[dw, dx, dy, dz, AspectRatio -> Automatic, PlotRange -> {{0, 1}, {0, 1}}, Axes -> Automatic, Frame -> True, Background -> GrayLevel[.026], This gives me dots in 4 colors for distinguishing different kinds of points in my real application. This works fine but needs the Point structure. Or, I can do (this is for one kind of point), t = Table[{Random[], Random[]}, {i, 1, 1024}]; ListPlot[t, AspectRatio -> Automatic, Axes -> Automatic, Frame -> True, Background -> GrayLevel[.026] ]; which seems simpler and may fit into the rest of the program more easily. 1. How do I get the RGBColor Rule or the equivalent into the latter? The RGBColor[] call is not a rule. 2. In the latter case I also want to plot 4 types of points. How do I get ListPlot to put down 4 plots superimposed? Or can't I? 3. In either case, I need to make the whole plot area about twice as big. That is, it now occupies about a 4 square. To see details better in my real plot, and because with 16k points the small plot just looks almost like a solid blur, I want to make it more like 8 square, or as big as will fit the screen (without changing the plot range or anything else). There must be a scale factor somewhere. ==== Steve, You could try something like this: Needs[Graphics`Colors`] Clear[t]; Do[t[i] = Table[{Random[], Random[]}, {i, 1, 1024}], {i, 4}]; Show[Graphics[ {PointSize[0.005], RoyalBlue, Point /@ t[1], OrangeRed, Point /@ t[2], SpringGreen, Point /@ t[3], CadmiumLemon, Point /@ t[4]}], AspectRatio -> Automatic, Frame -> True, PlotRegion -> {{0.02, 0.96}, {0, 1}}, Background -> IvoryBlack, ImageSize -> 700]; The overall size of the plot can be controlled with the ImageSize option. ListPlot is more a hindrance than a help - throw it in the ash can. Just Map Point onto the lists of point coordinates. Give the color before each set of points. I adjusted the PlotRegion to obtain some black margin on all sides of the frame. David Park djmp@earthlink.net http://home.earthlink.net/~djmp/ dy = Graphics[{PointSize[0.01], RGBColor[ 0, .5, .9], y}]; dz = Graphics[{PointSize[0.01], RGBColor[.8, .8, 0], z}]; Show[dw, dx, dy, dz, AspectRatio -> Automatic, PlotRange -> {{0, 1}, {0, 1}}, Axes -> Automatic, Frame -> True, Background -> GrayLevel[.026], This gives me dots in 4 colors for distinguishing different kinds of points in my real application. This works fine but needs the Point structure. Or, I can do (this is for one kind of point), t = Table[{Random[], Random[]}, {i, 1, 1024}]; ListPlot[t, AspectRatio -> Automatic, Axes -> Automatic, Frame -> True, Background -> GrayLevel[.026] ]; which seems simpler and may fit into the rest of the program more easily. 1. How do I get the RGBColor Rule or the equivalent into the latter? The RGBColor[] call is not a rule. 2. In the latter case I also want to plot 4 types of points. How do I get ListPlot to put down 4 plots superimposed? Or can't I? 3. In either case, I need to make the whole plot area about twice as big. That is, it now occupies about a 4 square. To see details better in my real plot, and because with 16k points the small plot just looks almost like a solid blur, I want to make it more like 8 square, or as big as will fit the screen (without changing the plot range or anything else). There must be a scale factor somewhere. ==== Bobby, Your ODE is different from both mine and Borut's. You do however get the same as mine with your approach, if you differentiate the total energy rather than the Lagrangian. KE = Simplify[(x'[t]^2 + D[f[x[t]], t]^2)/2]; PE = g*f[x[t]]; totalE = KE + PE; treat = D[totalE, t] == 0 x''[t] - (x''[t] /. First[Solve[treat, x''[t]]]//Apart) == 0 Borut's ODE differs only in his factor of 2 in the first-order term. I suspect that might have been a typo... Borut? By the way, some nice animations of mechanical systems (some based on the same sort of approach) can be seen here: http://www.math.armstrong.edu/faculty/hollis/DEmovies/ --Selwyn > The two of you derived slightly different ODE's. I think Selwyn's is > correct, but I only had one physics course, 30 years ago. Here's a notebook expression showing how Selwyn's approach can be used > to derive the ODE for y[t] = f[x[t]], with arbitrary f. It shows that > solution in a form that is easily compared with Borut's ODE, and then it > shows my own solution (same as Borut's, by a slightly simpler method). Bobby Treat > KE = Simplify[(x'[t]^2 + D[f[x[t]], t]^2)/2]; > PE = g*f[x[t]]; > L = KE - PE; > treat = D[L, t] == 0 > ==== I've modified Selwyn's solution to make it more general. In particular, the height can be specified (up to about 35 meters). The differential equation is solved for t=0 to 5 first. The quarter-period is computed by finding a zero; then the differential equation is solved again for t=0 to the quarter-period, and the solution is extended using reflection and periodicity. This yields a higher-precision solution. Then I graph the solution, with a stepsize equal to period/40, from t=0 to 'period', labeling each frame with the values of t, x[t], and y[t]. Using a step size that divides period/4 guarantees the lowest point is reached ON a frame when appropriate. Here's the solution as a notebook expression: Notebook[{ Cell[CellGroupData[{ Cell[Borut L's solution:, Subsubtitle], Cell[TextData[StyleBox[Having noticed your statement ... BEYOND MY MEANS I thing you aren't yetnfamiliar with Lagrangian formalism. It's quite easy to derive a generalnequation of motion for a point mass, subjected to gravity and to moving on ancurve f = y(x) (i.e. f = Cosh[#]&).nn1) I'll leave re-deriving equation to you, here is what I've got (just copynpaste it).:, FontFamily->Courier New, FontSize->10, CharacterEncoding->WindowsANSI]], Text], Cell[BoxData[ RowBox[{selwyn, =, RowBox[{First, [, RowBox[{ RowBox[{ RowBox[{ RowBox[{x, ''}], [, t, ]}], /., RowBox[{Solve, [, RowBox[{diffeq, ,, RowBox[{ RowBox[{x, ''}], [, t, ]}]}], ]}]}], //, Simplify}], ]}]}]], Input], Cell[BoxData[ RowBox[{borut, =, RowBox[{First, [, RowBox[{ RowBox[{ RowBox[{x, ''}], [, t, ]}], /., RowBox[{Solve, [, RowBox[{ RowBox[{getEq, [, Cosh, ]}], ,, RowBox[{ RowBox[{x, ''}], [, t, ]}]}], ]}]}], ]}]}]], Input], Cell[BoxData[ RowBox[{ RowBox[{getEq, [, f_, ]}], :=, [IndentingNewLine], RowBox[{Simplify, [, RowBox[{ RowBox[{ RowBox[{ RowBox[{x, ''}], [, t, ]}], +, RowBox[{ SuperscriptBox[ RowBox[{ RowBox[{x, '}], [, t, ]}], 2], , FractionBox[ RowBox[{2, , RowBox[{ RowBox[{f, '}], [, RowBox[{x, [, t, ]}], ]}], , RowBox[{ RowBox[{f, ''}], [, RowBox[{x, [, t, ]}], ]}]}], RowBox[{1, +, SuperscriptBox[ RowBox[{ RowBox[{f, '}], [, RowBox[{x, [, t, ]}], ]}], 2]}]]}], +, FractionBox[ RowBox[{g, , RowBox[{ RowBox[{f, '}], [, RowBox[{x, [, t, ]}], ]}]}], RowBox[{1, +, SuperscriptBox[ RowBox[{ RowBox[{f, '}], [, RowBox[{x, [, t, ]}], ]}], 2]}]]}], ==, 0}], ]}]}]], Input], Cell[TextData[{ StyleBox[2) Next, you integrate it .:, FontFamily->Courier New, FontSize->10, CharacterEncoding->WindowsANSI], }], Text], Cell[BoxData[ RowBox[{ RowBox[{getSol, [, RowBox[{f_, ,, RowBox[{h0_, ?, Positive}], ,, RowBox[{x0_, ?, Positive}]}], ]}], :=, RowBox[{Module, [, RowBox[{ RowBox[{{, tStop, }}], ,, [IndentingNewLine], RowBox[{First, @, RowBox[{NDSolve, [, [IndentingNewLine], RowBox[{ RowBox[{{, RowBox[{ RowBox[{getEq, [, f, ]}], ,, RowBox[{ RowBox[{x, [, 0, ]}], [Equal], h0}], ,, RowBox[{ RowBox[{ RowBox[{x, '}], [, 0, ]}], [Equal], 0}]}], }}], ,, [IndentingNewLine], x, ,, [IndentingNewLine], RowBox[{{, RowBox[{t, ,, 0, ,, 10}], }}], ,, [IndentingNewLine], RowBox[{MaxStepSize, [Rule], RowBox[{1, /, 100}]}], ,, RowBox[{StoppingTest, [RuleDelayed], RowBox[{If, [, RowBox[{ RowBox[{h0, <, 0}], ,, RowBox[{x, >, 0}], ,, RowBox[{x, <, 0}]}], ]}]}]}], [IndentingNewLine], ]}]}]}], [IndentingNewLine], ]}]}]], Input], Cell[< 3) Here follows animation code, specially for linear versus cosh case, apply my initial conditions (below) >, Text], Cell[BoxData[ RowBox[{ RowBox[{makeDuo, [, RowBox[{f_, ,, g_, ,, h0_}], ]}], :=, RowBox[{Module, [, RowBox[{ RowBox[{{, RowBox[{ RowBox[{solf, =, RowBox[{getSol, [, RowBox[{f, ,, h0}], ]}]}], ,, RowBox[{solg, =, RowBox[{getSol, [, RowBox[{g, ,, h0}], ]}]}], ,, tf, ,, tg, ,, maxT, ,, minT}], }}], ,, [IndentingNewLine], RowBox[{ RowBox[{tf, =, RowBox[{solf, [, RowBox[{[, RowBox[{ 1, ,, 2, ,, 1, ,, 1, ,, 2}], ]}], ]}]}], ;, [IndentingNewLine], RowBox[{tg, =, RowBox[{solg, [, RowBox[{[, RowBox[{ 1, ,, 2, ,, 1, ,, 1, ,, 2}], ]}], ]}]}], ;, [IndentingNewLine], RowBox[{maxT, =, RowBox[{Max, [, RowBox[{{, RowBox[{tf, ,, tg}], }}], ]}]}], ;, [IndentingNewLine], RowBox[{minT, =, RowBox[{Min, [, RowBox[{{, RowBox[{tf, ,, tg}], }}], ]}]}], ;, [IndentingNewLine], RowBox[{Do, [, RowBox[{ RowBox[{Plot, [, RowBox[{ RowBox[{{, RowBox[{ RowBox[{f, @, x}], ,, RowBox[{g, @, x}]}], }}], ,, RowBox[{{, RowBox[{x, ,, 0, ,, RowBox[{ArcCosh, @, h0}]}], }}], ,, RowBox[{AspectRatio, [Rule], Automatic}], ,, RowBox[{Frame, [Rule], True}], ,, RowBox[{Axes, [Rule], False}], ,, RowBox[{Epilog, [Rule], RowBox[{{, RowBox[{ RowBox[{ AbsolutePointSize, [, 10, ]}], ,, RowBox[{Hue, [, 0, ]}], ,, RowBox[{Point, [, RowBox[{ RowBox[{{, RowBox[{ RowBox[{x, [, t, ]}], ,, RowBox[{f, @, RowBox[{x, [, t, ]}]}]}], }}], /., solf}], ]}], ,, RowBox[{Hue, [, .6, ]}], ,, RowBox[{Point, [, RowBox[{ RowBox[{{, RowBox[{ RowBox[{x, [, t, ]}], ,, RowBox[{g, @, RowBox[{x, [, t, ]}]}]}], }}], /., solg}], ]}]}], }}]}]}], ]}], ,, RowBox[{{, RowBox[{t, ,, 0, ,, minT, ,, minT}], }}]}], ]}]}]}], ]}]}]], Input], Cell[TextData[{ StyleBox[4) My initial conditions. In my opinion, you weren't true on this. Saying nSTARTING FROM THE SAME HEIGHT is not enough - you should specify x asnwell, thus specifing starting POINT and not just height y., FontFamily->Courier New, FontSize->10, CharacterEncoding->WindowsANSI], }], Text], Cell[BoxData[{ RowBox[{makeDuo, [, RowBox[{ RowBox[{ RowBox[{ RowBox[{Cosh, [, 1., ]}], , #}], &}], ,, RowBox[{ RowBox[{Cosh, [, #, ]}], &}], ,, 23}], ]}], [IndentingNewLine], RowBox[{SelectionMove, [, RowBox[{ RowBox[{EvaluationNotebook, [, ]}], ,, All, ,, GeneratedCell}], ]}], n, RowBox[{ FrontEndTokenExecute, [, , ]}], n, RowBox[{ FrontEndTokenExecute, [, , ]}]}], Input] }, Open ]] }, ScreenRectangle->{{0, 1024}, {0, 711}}, WindowSize->{815, 569}, WindowMargins->{{0, Automatic}, {Automatic, -1}}, ShowSelection->True ] Bobby Treat -----Original Message----- and the equation of motion is diffeq = Simplify[ D[D[L, x'[t]], t] ] == Simplify[ D[L, x[t]] ] Now solve and animate ... xx[t_] = x[t]/. First[ NDSolve[{diffeq, x[0] == -1, x'[0] == 0}, x[t], {t, 0, 5}]] curve = Plot[Cosh[x], {x, -1, 1}] Do[ Show[curve, Graphics[Disk[{xx[t], Cosh[xx[t]]}, 0.025]], PlotRange -> {{-1.2, 1.2}, {0.9, 1.65}}, AspectRatio -> Automatic, Axes->None], {t, 0, 5, 0.1}] ---- Selwyn Hollis > Dear Colleagues, I intend to make an animation in which ball A rolls down on an inclined plane from the left whilst ball B - starting from the same height - rolls down Cosh[t]'s path from the > right. x-axis is time t, y-axis is height h. Ball A is fine; ball B - which should arrive at h=0 before A - is beyond my > means. > Matthias Bode > Sal. Oppenheim jr. & Cie. KGaA > Koenigsberger Strasse 29 > D-60487 Frankfurt am Main > GERMANY > Mobile: +49(0)172 6 74 95 77 > Internet: http://www.oppenheim.de ==== Try this: << Graphics`Colors` n = 4096; data = {w, x, y, z} = Array[Random[] &, {4, n, 2}]; g@{a_, b_} := ListPlot[a, PlotStyle -> {PointSize[0.005], b}, DisplayFunction -> Identity, Show[g /@ Transpose[{data, {Red, Blue, Yellow, White}}], PlotRange -> {{0, 1}, {0, 1}}, Axes -> Automatic, Frame -> True, DisplayFunction -> $DisplayFunction, Background -> GrayLevel[.026], ImageSize -> 500]; Bobby Treat -----Original Message----- dx = Graphics[{PointSize[0.01], RGBColor[.8, .8, .8], x}]; dy = Graphics[{PointSize[0.01], RGBColor[ 0, .5, .9], y}]; dz = Graphics[{PointSize[0.01], RGBColor[.8, .8, 0], z}]; Show[dw, dx, dy, dz, AspectRatio -> Automatic, PlotRange -> {{0, 1}, {0, 1}}, Axes -> Automatic, Frame -> True, Background -> GrayLevel[.026], This gives me dots in 4 colors for distinguishing different kinds of points in my real application. This works fine but needs the Point structure. Or, I can do (this is for one kind of point), t = Table[{Random[], Random[]}, {i, 1, 1024}]; ListPlot[t, AspectRatio -> Automatic, Axes -> Automatic, Frame -> True, Background -> GrayLevel[.026] ]; which seems simpler and may fit into the rest of the program more easily. 1. How do I get the RGBColor Rule or the equivalent into the latter? The RGBColor[] call is not a rule. 2. In the latter case I also want to plot 4 types of points. How do I get ListPlot to put down 4 plots superimposed? Or can't I? 3. In either case, I need to make the whole plot area about twice as big. That is, it now occupies about a 4 square. To see details better in my real plot, and because with 16k points the small plot just looks almost like a solid blur, I want to make it more like 8 square, or as big as will fit the screen (without changing the plot range or anything else). There must be a scale factor somewhere. ==== The same command worked for me, insofar as creating the plot is concerned. I did get an error message opening the file, saying that Adobe Acrobat was Unable to find or create the font 'Mathematica1Mono-Bold'. Some characters may not display or print correctly. The plot looks fine, though. Bobby Treat -----Original Message----- Export[c:docscell4.gif, Cell[ <<...(copied cell data from Edit->Copy As->Cell Expression)...> ]] When I change the filename to a .PDF and evaluate the cell, the program displays 'Running...' for a second and gives the 'Out[n] = c:docscell4.pdf ' message as if a file was created, but no file is created anywhere with any name that I could find with Start->Find->[All files and folders created in the previous day]. Is there a limitation to PDF exporting I don't know about? Do I need to upgrade to 4.2? Am I doing something wrong with the Export[] command? Do I need a faster computer? A patch? Something else? ==== I'm trying to export a table from Mathematica 4.0 to Notepad, for examples, or something similar, but it's impossible for me! It always exports the cell with all its rubbish, or an image (if I use Word). How can I export just plain _numbers_? Thanx Fip ==== You can use Export[file.dat, expr] where expr is a two-dimensional array. Or for more flexibility study the following. Assuming data has n rows and 2 columns. file = OpenWrite[file.dat]; Map[ ( (* the N may be redundant haven't tested the code w/o it *) str = ToString[PaddedForm[N[#[[1]]],{10,6}]]<> <> ToString[PaddedForm[N[#[[2]]],{10,6}]]; (* make numbers e notation: the *^ notation has given me fits, so I just put this in as a precaution *) str = StringReplace[str, *^ -> e]; WriteString[file, str, n] )&, data]; Close[file] Hope this helps, Lawrence > I'm trying to export a table from Mathematica 4.0 to Notepad, for examples, > or something similar, but it's impossible for me! It always exports the cell > with all its rubbish, or an image (if I use Word). > How can I export just plain _numbers_? Thanx > Fip -- Lawrence A. Walker Jr. http://www.kingshonor.com ==== I'm trying to export a table from Mathematica 4.0 to Notepad, for examples, > or something similar, but it's impossible for me! It always exports the cell > with all its rubbish, or an image (if I use Word). > How can I export just plain _numbers_? Try with this: WriteSimpleTableForm[file_String, data_List, opts___] := Module[{str}, str = OpenWrite[file]; WriteString[str, ToString[TableForm[data, opts]]]; Close[str] ] This is my small solution for this problem. marek ==== I put the following in .../.../KeyEventTranslations.tr: Item[KeyEvent[<,Modifiers->{Control}], FrontEndExecute[{ FrontEnd`NotebookWrite[FrontEnd`SelectedNotebook[], [LeftDoubleBracket][RightDoubleBracket],After]}]], to make it easy to insert the double-brackets. This works fine, but I want the cursor to be placed between the brackets not after them. How can this be done? Gru137 Peter -- =--=--=--=--=--=--=--=--=--=--=--=--=--= http://home.t-online.de/home/phbrf ~ ~ ~ ==== >I've been using Mathematica 4.1 on Win98 as a word processor for >math-related documents, but often people that need to see the documents >don't have Mathematica, and for whatever reason on my computer the HTML >saves don't work at all. I'd like to export to PDF format. I can export >images to PDF format no problem using, for example >Export[c:docsplot3.pdf, Plot[Sin[x],{x,-2Pi,2Pi}]], >and I can export cells correctly to GIF, JPEG, and WMF formats (probably >more, those are the only ones I tested) using, for example >Export[c:docscell4.gif, Cell[ <<...(copied cell data from Edit->Copy >As->Cell Expression)...> ]] >When I change the filename to a .PDF and evaluate the cell, the program >displays 'Running...' for a second and gives the 'Out[n] = c:docscell4.pdf >' message as if a file was created, but no file is created anywhere with any >name that I could find with Start->Find->[All files and folders created in >the previous day]. >Is there a limitation to PDF exporting I don't know about? Do I need to >upgrade to 4.2? Am I doing something wrong with the Export[] command? Do I >need a faster computer? A patch? Something else? This is a limitation of PDF export. The mechanism for exporting GIF, JPEG, and other raster formats is completely different than the system used for PDF export. Because of this PDF export is limited to graphics expressions. Cells and Notebooks cannot be converted via Export. However, there is a way to generate PDFs using the frontend. See http://support.wolfram.com/mathematica/graphics/export/convertpdfghostscript .html http://support.wolfram.com/mathematica/graphics/export/convertpdfdistiller.h tml -Dale ==== Dear Mathematica friends, how can Mathematica 4.1 be used to combine sound and graphics? In particular, I would like to prepare a demo video about differential equations. I can Plot the solution and I can Play the solution. How to combine the 2 results into a single file that can be played back using xine or DivX, like ordinary video can? Best wishes from Prague -- Pavel Pokorny Math Dept, Prague Institute of Chemical Technology http://staffold.vscht.cz/mat/Pavel.Pokorny ==== Hoi, it depends a bit what you want to do. E.g.: If you want to write applications for clients then go with that systems your clients use (probably Windows). If you just use it for yourself, for development: use what you like more. anymore, like there were in 3.0 times. The copy and paste problems are gone if you switch off the KDE Klipper. On the other hand, there are a few OS- (or better Window-manager-specific) 1.: you cannot rotate text (i.e., FrameLabel settings will look weird (vertically arranged horizontal letters), you have to use RotateLabel -> False generally, or play with the Fonts settings such that horizontal tick marks still fit) 2.: If you work with bigger graphics in notebooks I suggest Windows (or MacOS X) since least on my XFree 4.2 installation with a not too modern graphics card). Also I find resizing of larger notebooks somewhat slow. 3.: If you like to work with keyboard shortcuts: Windows is better, clearly. 4.: There are a couple of Font issues which are better on Windows since not all fonts engels, nederlands, duits of spaans) Rolf Mertig Mertig Consulting Berlin ==== I'd like to add a JLink animation of the rolling ball based on Selwyn's solution: UseFrontEndForRendering = False; createWindow[] := Module[{frame}, frame = JavaNew[com.wolfram.jlink.MathFrame, Doppler Animation]; drawArea = JavaNew[com.wolfram.jlink.MathCanvas]; drawArea@setUsesFE[UseFrontEndForRendering]; drawArea@setSize[800, 600]; JavaBlock[frame@setLayout[JavaNew[java.awt.BorderLayout]]; frame@add[drawArea, ReturnAsJavaObject[BorderLayout`CENTER]]; frame@pack[]; frame@setSize[800, 600]; frame@setLocation[200, 200]; JavaShow[frame]]; frame ] drawRoll[t_] := Show[curve, Graphics[{RGBColor[1, 0, 0], Disk[{xx[t], Cosh[xx[t]]}, 0.05]}], PlotRange -> {{-1.2, 1.2}, {0.9, 1.65}}, AspectRatio -> Automatic, Axes -> None, DisplayFunction -> Identity]; times = Range[0, 5, .05]; LoadJavaClass[java.lang.Thread]; AnimationPlot[t_List] := JavaBlock[Block[{frm}, frm = createWindow[]; Map[(obj = drawRoll[#]; drawArea@setMathCommand[obj]; drawArea@repaintNow[]; Thread@sleep[5];) &, t ] ]] AnimationPlot[times] jerry blimbaum panama city, fl -----Original Message----- and the equation of motion is diffeq = Simplify[ D[D[L, x'[t]], t] ] == Simplify[ D[L, x[t]] ] Now solve and animate ... xx[t_] = x[t]/. First[ NDSolve[{diffeq, x[0] == -1, x'[0] == 0}, x[t], {t, 0, 5}]] curve = Plot[Cosh[x], {x, -1, 1}] Do[ Show[curve, Graphics[Disk[{xx[t], Cosh[xx[t]]}, 0.025]], PlotRange -> {{-1.2, 1.2}, {0.9, 1.65}}, AspectRatio -> Automatic, Axes->None], {t, 0, 5, 0.1}] ---- Selwyn Hollis > Dear Colleagues, I intend to make an animation in which ball A rolls down on an inclined plane from the left whilst ball B - starting from the same height - rolls down Cosh[t]'s path from the > right. x-axis is time t, y-axis is height h. Ball A is fine; ball B - which should arrive at h=0 before A - is beyond my > means. > Matthias Bode > Sal. Oppenheim jr. & Cie. KGaA > Koenigsberger Strasse 29 > D-60487 Frankfurt am Main > GERMANY > Mobile: +49(0)172 6 74 95 77 > Internet: http://www.oppenheim.de ==== I noticed that n Mathematica 3.0 , IntegerDigits function is giving wrong results. This problem is not found in Mathematica 4.1. Whether any body else has also noted any such problem. For example IntegerDigits[10^18+7] will give the digits 0 and 7 , omitting 1. Shyam Sunder Gupta ==== You're right; I wasn't exporting a Cell. I misunderstood his post and successfully executed this: Export[c:docsplot3.pdf, Plot[Sin[x],{x,-2Pi,2Pi}]] but that was working for him, already. Sorry for the confusion. Bobby -----Original Message----- Sender: steve@smc.vnet.net Approved: Steven M. Christensen , Moderator ==== > w = Table[Point[{Random[], Random[]}], {i, 1, 4096}]; > x = Table[Point[{Random[], Random[]}], {i, 1, 4096}]; > y = Table[Point[{Random[], Random[]}], {i, 1, 4096}]; > z = Table[Point[{Random[], Random[]}], {i, 1, 4096}]; > 2. In the latter case I also want to plot 4 types of points. > How do I get ListPlot to put down 4 plots superimposed? > Or can't I? You can, but you have to do it via Graphics`MulitpleListPlot`. Load the requisite packages Needs[Graphics`Colors`] Needs[Graphics`MultipleListPlot`] Define cols = {Black, Red, Green, Blue}; pnts = {PlotSymbol[Triangle], PlotSymbol[Box], PlotSymbol[Diamond], PlotSymbol[Star]}; Then grf = MultipleListPlot[w, x, y, z, PlotStyle->cols, SymbolShape->pnts, SymbolsStyle->cols ]; Documentation shows how to incorporate legends and to define other symbols. Dave. ========================================== Dr. David Annetts EM Modelling Analyst Australia David.Annetts@csiro.au =========================================== ==== >Dear Group, It is a pleasure write to you again. I have three questions to ask. > Alexandre Costa > >Question Two: > >How can I change points properties (such as Size and Color) for the plot >below? The PlotStyle Option does not work for this LabeledListPlot << Graphics`Graphics` > >listcities = {{1, 5}, {4, 6}, {7, 5}, {5, 4}, {9, 4}, {2, 3}, {4, 2}, >{6, 2}, {1, 1}, {5, 1}, {3, 0}, {9, 0}}; > >LabeledListPlot[listcities, Axes -> None, Frame -> True, > DisplayFunction -> $DisplayFunction] It's not a built-in feature of LabeledListPlot, so you'll have to do it manually. gr=LabeledListPlot[listcities, Axes -> None, Frame -> True, DisplayFunction -> $DisplayFunction] You can add graphics directives to the points and text with a replacement rule. Show[gr/.x_Point|x_Text->{RGBColor[1,0,0],x}] >Question Three: Why the Goto statement below is not working? > >q = 2; >Label[start]; >q = 3; >Label[begin]; >Print[q]; >q += 1; If[q < 6, Goto[begin], Goto[start]] First, let me say that noone should ever use Goto. You should always use a loop or some other process instead. With that said . . . When you type semicolon separated input into the frontend, each command is treated as a separate input (as if they were in separate input cells). So the Labels and the Gotos are evaluated separately and there's no way to jump from one to the other. Instead, the commands need to be within the same expression. This can be done by wrapping the command in a CompoundExpression ( q = 2; Label[start]; q = 3; Label[begin]; Print[q]; q += 1; If[q < 6, Goto[begin], Goto[start]] ) Or some other expression, like a Module. -------------------------------------------------------------- Omega Consulting The final answer to your Mathematica needs Spend less time searching and more time finding. http://www.wz.com/internet/Mathematica.html ==== >I would appreciate help with these problems: 1. I'm plotting several thousand points, which I can do >either with something like this (this is a test): > >w = Table[Point[{Random[], Random[]}], {i, 1, 4096}]; >x = Table[Point[{Random[], Random[]}], {i, 1, 4096}]; >y = Table[Point[{Random[], Random[]}], {i, 1, 4096}]; >z = Table[Point[{Random[], Random[]}], {i, 1, 4096}]; >dw = Graphics[{PointSize[0.01], RGBColor[ 1, 0, 0], w}]; >dx = Graphics[{PointSize[0.01], RGBColor[.8, .8, .8], x}]; >dy = Graphics[{PointSize[0.01], RGBColor[ 0, .5, .9], y}]; >dz = Graphics[{PointSize[0.01], RGBColor[.8, .8, 0], z}]; >Show[dw, dx, dy, dz, AspectRatio -> Automatic, > PlotRange -> {{0, 1}, {0, 1}}, > Axes -> Automatic, > Frame -> True, > Background -> GrayLevel[.026], > >This gives me dots in 4 colors for distinguishing different >kinds of points in my real application. This works fine but >needs the Point structure. Or, I can do (this is for one >kind of point), > >t = Table[{Random[], Random[]}, {i, 1, 1024}]; >ListPlot[t, AspectRatio -> Automatic, > Axes -> Automatic, > Frame -> True, > Background -> GrayLevel[.026] > ]; > >which seems simpler and may fit into the rest of the program >more easily. 1. How do I get the RGBColor Rule or the equivalent into >the latter? The RGBColor[] call is not a rule. 2. In the latter case I also want to plot 4 types of points. >How do I get ListPlot to put down 4 plots superimposed? >Or can't I? 3. In either case, I need to make the whole plot area about >twice as big. That is, it now occupies about a 4 square. To >see details better in my real plot, and because with 16k points >the small plot just looks almost like a solid blur, I want to >make it more like 8 square, or as big as will fit the screen >(without changing the plot range or anything else). There >must be a scale factor somewhere. > You can plot more than one list with MultipleListPlot. <Automatic,Axes->Automatic,Frame->True, SymbolShape->{ColorPoint[RGBColor[1,0,0]], ColorPoint[RGBColor[.8, .8, .8]],ColorPoint[RGBColor[ 0, .5, .9]], ColorPoint[RGBColor[.8, .8, 0]]}, SymbolStyle->PointSize[0.01],Background->GrayLevel[.026] ] -------------------------------------------------------------- Omega 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 to use a graphics of mathematica with powerpoint for a little >Y-axes and line of function) are bold or more visible: how can I do >that? Use the PlotStyle and AxesStyle options as in Plot[x,{x,0,5},PlotStyle->Thickness[0.015],AxesStyle->Thickness[0.015]] ==== Dear Mathematica friends, how can Mathematica 4.1 be used to combine sound and graphics? In particular, I would like to prepare a demo video about differential equations. I can Plot the solution and I can Play the solution. How to combine the 2 results into a single file that can be played back using xine or DivX, like ordinary video can? Best wishes from Prague -- Pavel Pokorny Math Dept, Prague Institute of Chemical Technology http://staffold.vscht.cz/mat/Pavel.Pokorny ==== I'd like to add a JLink animation of the rolling ball based on Selwyn's solution: UseFrontEndForRendering = False; createWindow[] := Module[{frame}, frame = JavaNew[com.wolfram.jlink.MathFrame, Doppler Animation]; drawArea = JavaNew[com.wolfram.jlink.MathCanvas]; drawArea@setUsesFE[UseFrontEndForRendering]; drawArea@setSize[800, 600]; JavaBlock[frame@setLayout[JavaNew[java.awt.BorderLayout]]; frame@add[drawArea, ReturnAsJavaObject[BorderLayout`CENTER]]; frame@pack[]; frame@setSize[800, 600]; frame@setLocation[200, 200]; JavaShow[frame]]; frame ] drawRoll[t_] := Show[curve, Graphics[{RGBColor[1, 0, 0], Disk[{xx[t], Cosh[xx[t]]}, 0.05]}], PlotRange -> {{-1.2, 1.2}, {0.9, 1.65}}, AspectRatio -> Automatic, Axes -> None, DisplayFunction -> Identity]; times = Range[0, 5, .05]; LoadJavaClass[java.lang.Thread]; AnimationPlot[t_List] := JavaBlock[Block[{frm}, frm = createWindow[]; Map[(obj = drawRoll[#]; drawArea@setMathCommand[obj]; drawArea@repaintNow[]; Thread@sleep[5];) &, t ] ]] AnimationPlot[times] jerry blimbaum panama city, fl -----Original Message----- and the equation of motion is diffeq = Simplify[ D[D[L, x'[t]], t] ] == Simplify[ D[L, x[t]] ] Now solve and animate ... xx[t_] = x[t]/. First[ NDSolve[{diffeq, x[0] == -1, x'[0] == 0}, x[t], {t, 0, 5}]] curve = Plot[Cosh[x], {x, -1, 1}] Do[ Show[curve, Graphics[Disk[{xx[t], Cosh[xx[t]]}, 0.025]], PlotRange -> {{-1.2, 1.2}, {0.9, 1.65}}, AspectRatio -> Automatic, Axes->None], {t, 0, 5, 0.1}] ---- Selwyn Hollis > Dear Colleagues, I intend to make an animation in which ball A rolls down on an inclined plane from the left whilst ball B - starting from the same height - rolls down Cosh[t]'s path from the > right. x-axis is time t, y-axis is height h. Ball A is fine; ball B - which should arrive at h=0 before A - is beyond my > means. > Matthias Bode > Sal. Oppenheim jr. & Cie. KGaA > Koenigsberger Strasse 29 > D-60487 Frankfurt am Main > GERMANY > Mobile: +49(0)172 6 74 95 77 > Internet: http://www.oppenheim.de ==== >I've been using Mathematica 4.1 on Win98 as a word processor for >math-related documents, but often people that need to see the documents >don't have Mathematica, and for whatever reason on my computer the HTML >saves don't work at all. I'd like to export to PDF format. I can export >images to PDF format no problem using, for example >Export[c:docsplot3.pdf, Plot[Sin[x],{x,-2Pi,2Pi}]], >and I can export cells correctly to GIF, JPEG, and WMF formats (probably >more, those are the only ones I tested) using, for example >Export[c:docscell4.gif, Cell[ <<...(copied cell data from Edit->Copy >As->Cell Expression)...> ]] >When I change the filename to a .PDF and evaluate the cell, the program >displays 'Running...' for a second and gives the 'Out[n] = c:docscell4.pdf >' message as if a file was created, but no file is created anywhere with any >name that I could find with Start->Find->[All files and folders created in >the previous day]. >Is there a limitation to PDF exporting I don't know about? Do I need to >upgrade to 4.2? Am I doing something wrong with the Export[] command? Do I >need a faster computer? A patch? Something else? This is a limitation of PDF export. The mechanism for exporting GIF, JPEG, and other raster formats is completely different than the system used for PDF export. Because of this PDF export is limited to graphics expressions. Cells and Notebooks cannot be converted via Export. However, there is a way to generate PDFs using the frontend. See http://support.wolfram.com/mathematica/graphics/export/convertpdfghostscript .html http://support.wolfram.com/mathematica/graphics/export/convertpdfdistiller.h tml -Dale ==== I noticed that n Mathematica 3.0 , IntegerDigits function is giving wrong results. This problem is not found in Mathematica 4.1. Whether any body else has also noted any such problem. For example IntegerDigits[10^18+7] will give the digits 0 and 7 , omitting 1. Shyam Sunder Gupta ==== You're right; I wasn't exporting a Cell. I misunderstood his post and successfully executed this: Export[c:docsplot3.pdf, Plot[Sin[x],{x,-2Pi,2Pi}]] but that was working for him, already. Sorry for the confusion. Bobby -----Original Message----- Sender: steve@smc.vnet.net Approved: Steven M. Christensen , Moderator ==== > w = Table[Point[{Random[], Random[]}], {i, 1, 4096}]; > x = Table[Point[{Random[], Random[]}], {i, 1, 4096}]; > y = Table[Point[{Random[], Random[]}], {i, 1, 4096}]; > z = Table[Point[{Random[], Random[]}], {i, 1, 4096}]; > 2. In the latter case I also want to plot 4 types of points. > How do I get ListPlot to put down 4 plots superimposed? > Or can't I? You can, but you have to do it via Graphics`MulitpleListPlot`. Load the requisite packages Needs[Graphics`Colors`] Needs[Graphics`MultipleListPlot`] Define cols = {Black, Red, Green, Blue}; pnts = {PlotSymbol[Triangle], PlotSymbol[Box], PlotSymbol[Diamond], PlotSymbol[Star]}; Then grf = MultipleListPlot[w, x, y, z, PlotStyle->cols, SymbolShape->pnts, SymbolsStyle->cols ]; Documentation shows how to incorporate legends and to define other symbols. Dave. ========================================== Dr. David Annetts EM Modelling Analyst Australia David.Annetts@csiro.au =========================================== ==== >Dear Group, It is a pleasure write to you again. I have three questions to ask. > Alexandre Costa > >Question Two: > >How can I change points properties (such as Size and Color) for the plot >below? The PlotStyle Option does not work for this LabeledListPlot << Graphics`Graphics` > >listcities = {{1, 5}, {4, 6}, {7, 5}, {5, 4}, {9, 4}, {2, 3}, {4, 2}, >{6, 2}, {1, 1}, {5, 1}, {3, 0}, {9, 0}}; > >LabeledListPlot[listcities, Axes -> None, Frame -> True, > DisplayFunction -> $DisplayFunction] It's not a built-in feature of LabeledListPlot, so you'll have to do it manually. gr=LabeledListPlot[listcities, Axes -> None, Frame -> True, DisplayFunction -> $DisplayFunction] You can add graphics directives to the points and text with a replacement rule. Show[gr/.x_Point|x_Text->{RGBColor[1,0,0],x}] >Question Three: Why the Goto statement below is not working? > >q = 2; >Label[start]; >q = 3; >Label[begin]; >Print[q]; >q += 1; If[q < 6, Goto[begin], Goto[start]] First, let me say that noone should ever use Goto. You should always use a loop or some other process instead. With that said . . . When you type semicolon separated input into the frontend, each command is treated as a separate input (as if they were in separate input cells). So the Labels and the Gotos are evaluated separately and there's no way to jump from one to the other. Instead, the commands need to be within the same expression. This can be done by wrapping the command in a CompoundExpression ( q = 2; Label[start]; q = 3; Label[begin]; Print[q]; q += 1; If[q < 6, Goto[begin], Goto[start]] ) Or some other expression, like a Module. -------------------------------------------------------------- Omega Consulting The final answer to your Mathematica needs Spend less time searching and more time finding. http://www.wz.com/internet/Mathematica.html ==== Mario: Use, for example, PlotStyle->Thickness[0.01] for plot and AxesStyle->Thickness[0.01]. Other way is using AbsoluteThickness instead of Thickness. See The Mathematica Book: Section 2.9.3. Germ.87n Buitrago A. ----- Original Message ----- Sender: steve@smc.vnet.net Approved: Steven M. Christensen , Moderator ==== Mario: Use, for example, PlotStyle->Thickness[0.01] for plot and AxesStyle->Thickness[0.01]. Other way is using AbsoluteThickness instead of Thickness. See The Mathematica Book: Section 2.9.3. Germ.87n Buitrago A. ----- Original Message ----- Sender: steve@smc.vnet.net Approved: Steven M. Christensen , Moderator ==== I have to use a graphics of mathematica with powerpoint for a little Y-axes and line of function) are bold or more visible: how can I do that? ==== >I have to use a graphics of mathematica with powerpoint for a little >Y-axes and line of function) are bold or more visible: how can I do >that? Use the PlotStyle and AxesStyle options as in Plot[x,{x,0,5},PlotStyle->Thickness[0.015],AxesStyle->Thickness[0.015]] ==== >I would appreciate help with these problems: 1. I'm plotting several thousand points, which I can do >either with something like this (this is a test): > >w = Table[Point[{Random[], Random[]}], {i, 1, 4096}]; >x = Table[Point[{Random[], Random[]}], {i, 1, 4096}]; >y = Table[Point[{Random[], Random[]}], {i, 1, 4096}]; >z = Table[Point[{Random[], Random[]}], {i, 1, 4096}]; >dw = Graphics[{PointSize[0.01], RGBColor[ 1, 0, 0], w}]; >dx = Graphics[{PointSize[0.01], RGBColor[.8, .8, .8], x}]; >dy = Graphics[{PointSize[0.01], RGBColor[ 0, .5, .9], y}]; >dz = Graphics[{PointSize[0.01], RGBColor[.8, .8, 0], z}]; >Show[dw, dx, dy, dz, AspectRatio -> Automatic, > PlotRange -> {{0, 1}, {0, 1}}, > Axes -> Automatic, > Frame -> True, > Background -> GrayLevel[.026], > >This gives me dots in 4 colors for distinguishing different >kinds of points in my real application. This works fine but >needs the Point structure. Or, I can do (this is for one >kind of point), > >t = Table[{Random[], Random[]}, {i, 1, 1024}]; >ListPlot[t, AspectRatio -> Automatic, > Axes -> Automatic, > Frame -> True, > Background -> GrayLevel[.026] > ]; > >which seems simpler and may fit into the rest of the program >more easily. 1. How do I get the RGBColor Rule or the equivalent into >the latter? The RGBColor[] call is not a rule. 2. In the latter case I also want to plot 4 types of points. >How do I get ListPlot to put down 4 plots superimposed? >Or can't I? 3. In either case, I need to make the whole plot area about >twice as big. That is, it now occupies about a 4 square. To >see details better in my real plot, and because with 16k points >the small plot just looks almost like a solid blur, I want to >make it more like 8 square, or as big as will fit the screen >(without changing the plot range or anything else). There >must be a scale factor somewhere. > You can plot more than one list with MultipleListPlot. <Automatic,Axes->Automatic,Frame->True, SymbolShape->{ColorPoint[RGBColor[1,0,0]], ColorPoint[RGBColor[.8, .8, .8]],ColorPoint[RGBColor[ 0, .5, .9]], ColorPoint[RGBColor[.8, .8, 0]]}, SymbolStyle->PointSize[0.01],Background->GrayLevel[.026] ] -------------------------------------------------------------- Omega Consulting The final answer to your Mathematica needs Spend less time searching and more time finding. http://www.wz.com/internet/Mathematica.html ==== For the life of me I am not sure why the following is not working in my v. 4.2: ru[a]=a->x; f[x_]:=(a+b) /. ru[a]; Why do I get f[c] = b+x and not f[c] = b+c? What gives? Lawrence -- Lawrence A. Walker Jr. http://www.kingshonor.com Reply-To: kuska@informatik.uni-leipzig.de ==== because te right hand side of SetDelayed[] is not evaluate. Try ru[a] = a -> x; f[x_] := (a + b) /. ru[a]; f1[x_] := Evaluate[(a + b) /. ru[a]]; and f1[] does what you expect In[]:={f[c], f1[c]} Out[]={b + x, b + c} Jens > For the life of me I am not sure why the following is not working in my > v. 4.2: ru[a]=a->x; > f[x_]:=(a+b) /. ru[a]; Why do I get > f[c] = b+x and not > f[c] = b+c? What gives? Lawrence -- > Lawrence A. Walker Jr. > http://www.kingshonor.com ==== Sir, we are having FreeBSD Server, In this we connected a heavy duty Dot matrix printer locally. While on taking outputs, username and file name are printing as BANNER TYPE. Instead of this, we would like to take the print out as username and file name should be printed as header through out the file. if u are having any scripts like that please send us Raj Mohan System Administrator Reply-To: jmt@dxdydz.net ==== I would like to add : Some characters, at least with a french keyboard, are not directly available typed through their Mathematica entities : [EHat] , [OHat], etc. While this is not an issue when programming, it can become quite painful when writing documentation. > Hoi, > it depends a bit what you want to do. > E.g.: If you want to write applications for clients then go with that > systems your clients use (probably Windows). > If you just use it for yourself, for development: use what you like more. > anymore, like > there were in 3.0 times. > The copy and paste problems are gone if you switch off the KDE Klipper. On the other hand, there are a few OS- (or better Window-manager-specific) > 1.: you cannot rotate text (i.e., FrameLabel settings will look weird > (vertically arranged > horizontal letters), you have to use > RotateLabel -> False generally, or play with the Fonts settings > such that horizontal > tick marks still fit) 2.: If you work with bigger graphics in notebooks I suggest Windows (or > MacOS X) since > least on my XFree 4.2 > installation with a not too modern graphics card). > Also I find resizing of larger notebooks somewhat slow. 3.: If you like to work with keyboard shortcuts: Windows is better, clearly. 4.: There are a couple of Font issues which are better on Windows since > not all fonts engels, nederlands, duits of spaans) Rolf Mertig Mertig Consulting > Berlin > ==== Hoi, it depends a bit what you want to do. E.g.: If you want to write applications for clients then go with that systems your clients use (probably Windows). If you just use it for yourself, for development: use what you like more. anymore, like there were in 3.0 times. The copy and paste problems are gone if you switch off the KDE Klipper. On the other hand, there are a few OS- (or better Window-manager-specific) 1.: you cannot rotate text (i.e., FrameLabel settings will look weird (vertically arranged horizontal letters), you have to use RotateLabel -> False generally, or play with the Fonts settings such that horizontal tick marks still fit) 2.: If you work with bigger graphics in notebooks I suggest Windows (or MacOS X) since least on my XFree 4.2 installation with a not too modern graphics card). Also I find resizing of larger notebooks somewhat slow. 3.: If you like to work with keyboard shortcuts: Windows is better, clearly. 4.: There are a couple of Font issues which are better on Windows since not all fonts engels, nederlands, duits of spaans) Rolf Mertig Mertig Consulting Berlin ==== I am considering the following integral W[m_,n_]:=Integrate[BesselJ[m, x]*BesselJ[n, x], {x, 0, Infinity}] where m,n are reals >=0. With Mathematica 4.1 I obtain: If[Re[m+n]>-1, -Cos[(m-n)Pi/2]/(2 Pi)* (2 EulerGamma + Log[4] + PolyGamma[0, 1/2(1 + m - n)] + PolyGamma[0, 1/2(1 - m + n)] + 2PolyGamma[0, 1/2(1 + m + n)]) and so using this answer as a definition I obtain W[0,0]=-(2 EulerGamma + Log[4] + 4 PolyGamma[0, 1/2])/(2 Pi)=0.84564 I suspect that these integrals are divergent (*). So I try the numerical integration: NW[m_,n_]:=NIntegrate[BesselJ[m, x]*BesselJ[n, x], {x, 0, Infinity}] so that NW[0,0]=11.167 Othe couples are W[1,0]=Indeterminate NW[1,0]=0.597973 W[0,1.5]=0.537095 NW[0,1.5]=-5.79306 W[1,1]=0.20902 NW[1,1]=17.5425 W[2,0]=0.427599 NW[2,0]=-6.83464 W[2,1]=Indeterminate NW[2,1]=4.69013 (*) The integral is a particular case of the Weber-Schafheitlin integrals (Abramowitz, 11.4.33). Any explanation about the analytical expression will be gratefully accepteed. Roberto. Roberto Brambilla CESI Via Rubattino 54 20134 Milano tel +39.02.2125.5875 fax +39.02.2125.5492 rlbrambilla@cesi.it ==== WRI Tech Support sent me an answer to the font problem. Copying the Type 1 (not True Type) fonts made the error go away: A workaround for the font-related issue that you encountered, is to place copies of the Mathematica Type 1 fonts from C:Program FilesWolfram ResearchMathematica4.2SystemFilesFontsType1 into C:Program FilesAdobeAcrobat 5.0ResourceFont Sincerely, George Kambouroglou Technical Support Wolfram Research, Inc. support@wolfram.com -----Original Message----- I've been using Mathematica 4.1 on Win98 as a word processor for math-related documents, but often people that need to see the documents don't have Mathematica, and for whatever reason on my computer the HTML saves don't work at all. I'd like to export to PDF format. I can export images to PDF format no problem using, for example Export[c:docsplot3.pdf, Plot[Sin[x],{x,-2Pi,2Pi}]], and I can export cells correctly to GIF, JPEG, and WMF formats (probably more, those are the only ones I tested) using, for example Export[c:docscell4.gif, Cell[ <<...(copied cell data from Edit->Copy As->Cell Expression)...> ]] When I change the filename to a .PDF and evaluate the cell, the program displays 'Running...' for a second and gives the 'Out[n] = c:docscell4.pdf ' message as if a file was created, but no file is created anywhere with any name that I could find with Start->Find->[All files and folders created in the previous day]. Is there a limitation to PDF exporting I don't know about? Do I need to upgrade to 4.2? Am I doing something wrong with the Export[] command? Do I need a faster computer? A patch? Something else? ==== SSG> I noticed that n Mathematica 3.0 , IntegerDigits function is SSG> giving wrong results. This problem is not found in Mathematica SSG> 4.1. Whether anybody else has also noted any such problem. SSG> For example IntegerDigits[10^18+7] will give the digits 0 SSG> and 7, omitting 1. Having made totally 4,000,000 attemtps, I was not able to reproduce your example. What version of Mathematica do you use? ...................................................... 4.2 for Microsoft Windows (February 28, 2002) 4.1 for Microsoft Windows (November 2, 2000) 4.0 for Microsoft Windows (April 21, 1999) Microsoft Windows 3.0 (April 25, 1997) Windows 387 2.2 (April 9, 1993) IntegerDigits[10^18 + 7] IntegerDigits[10^18 + 7] IntegerDigits[10^18 + 7] IntegerDigits[10^18 + 7] IntegerDigits[10^18 + 7] {1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,7} {1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,7} {1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,7} {1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,7} {1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,7} ...................................................... Best wishes, Vladimir Bondarenko Mathematical Director Symbolic Testing Group Web : http://www.CAS-testing.org/ http://maple.bug-list.org/VER2/ (under tuning) http://maple.bug-list.org/VER3/ (under tuning) http://maple.bug-list.org/VER1/ (under tuning) Voice : (380)-652-447325 Mon-Fri 9 a.m.-6 p.m. Mail : 76 Zalesskaya Str, Simferopol, Crimea, Ukraine ==== Just tested it with no errors, version 3.0 John A. Velling -----Original Message----- ==== I would appreciate any info on these issues. 1. I have a symbol slback. I was getting tag times lback protected (or something) errors. (A typically uninformative error message.) I suspected there was a separation between the s and l, and they looked too far apart, but when I spaced past them with the arrow keys, there did not appear to be anything like a space in between. I retyped the symbol and all was ok. This has happened before. What's going on ??? 2. I'm using Raster to plot data points in a 256x256 array. Two problems: First, regardless of the ImageSize setting (I need the display as big as possible), the individual data cells when examined closely at 300% vary in size by almost 2:1. This makes detailed inspection of the data values difficult. Surely there is some setting of something which would make each data cell an exact number of screen pixels? Second, I'm using Show[ Graphics[Raster[ rescol, ColorFunction -> Hue]], AspectRatio -> Automatic, ImageSize -> 700]; but the Hues don't allow enough easily distinguishable shades to visually recognize even 6 data values easily. (I'm slightly colorblind.) It would be nice to have black and white available for two of the data values, but Hue does not allow this. I don't understand what the documentation says about RasterArray. 3. I wanted to get this raster image into a format where I could dissect it with Photoshop or equivalent. After much fooling around, I found that I can export the selection as an html file, read it into Navigator, do File> Edit Page, which brings up Netscape Composer, right click the image which, allows saving it as a GIF, which I can finally work on with a photo editor. Maybe there is an easier way, or maybe this description will help someone with the same need. Reply-To: kuska@informatik.uni-leipzig.de ==== > I would appreciate any info on these issues. 1. I have a symbol slback. I was getting > tag times lback protected (or something) errors. > (A typically uninformative error message.) > I suspected there was a separation between the s > and l, and they looked too far apart, but when I > spaced past them with the arrow keys, there did > not appear to be anything like a space in between. > I retyped the symbol and all was ok. This has > happened before. What's going on ??? As long as you don't send us the *exact* input we can't help you. You should also send the mathematica version you are using. But typical this error comes from a equation a*b==c where the user has mixed up Equal[] and Set[] 2. I'm using Raster to plot data points > in a 256x256 array. Two problems: First, > regardless of the ImageSize setting (I need the > display as big as possible), the individual data cells > when examined closely at 300% vary in size by > almost 2:1. This makes detailed inspection of the > data values difficult. Surely there is some setting > of something which would make each data cell an > exact number of screen pixels? Second, I'm > using > Show[ Graphics[Raster[ rescol, > ColorFunction -> Hue]], > AspectRatio -> Automatic, > ImageSize -> 700]; > but the Hues don't allow enough easily distinguishable > shades to visually recognize even 6 data values easily. > (I'm slightly colorblind.) The most humans can distinguish 160 gray levels >It would be nice to have black > and white available for two of the data values, but Hue > does not allow this. And something like: mycolor[i_] := Switch[Round[i], 0, RGBColor[0, 0, 0], 1, RGBColor[1, 0, 0], 2, RGBColor[1, 1, 0], 3, RGBColor[0, 1, 0], 4, RGBColor[0, 1, 1], 5, RGBColor[0, 0, 1], _, RGBColor[1, 1, 1]] Show[Graphics[ Raster[Table[Random[Integer, {0, 6}], {16}, {16}], ColorFunction -> mycolor, ColorFunctionScaling -> False]]] does not help ? > I don't understand what the > documentation says about RasterArray. If you can't be more specific *what* you not understand we can not help you. 3. I wanted to get this raster image into a format > where I could dissect it with Photoshop or equivalent. And ? waht does Export[] do ? it write the expression in a desired format, TIFF, PNG, PPM are all lossless compressed bitmap formats, that Mathematica can export and that can be imported into PhotoShop > After much fooling around, I found that I can export the > selection as an html file, read it into Navigator, do File> > Edit Page, which brings up Netscape Composer, right > click the image which, allows saving it as a GIF, which > I can finally work on with a photo editor. Maybe > there is an easier way, or maybe this description will > help someone with the same need. May be that thhis description help someone who can't read the fancy documation on Import[] and Export[]. Jens ==== [snip] > 2. I'm using Raster to plot data points > in a 256x256 array. Two problems: First, > regardless of the ImageSize setting (I need the > display as big as possible), the individual data cells > when examined closely at 300% vary in size by > almost 2:1. This makes detailed inspection of the > data values difficult. Surely there is some setting > of something which would make each data cell an > exact number of screen pixels? GRAY: (No answer received.) I can get around this by making the images bigger but this is not a complete solution. > Second, I'm using > Show[ Graphics[Raster[ rescol, > ColorFunction -> Hue]], > AspectRatio -> Automatic, > ImageSize -> 700]; > but the Hues don't allow enough easily distinguishable > shades to visually recognize even 6 data values easily. > (I'm slightly colorblind.) The most humans can distinguish 160 gray levels >It would be nice to have black > and white available for two of the data values, but Hue > does not allow this. And something like: mycolor[i_] := > Switch[Round[i], > 0, RGBColor[0, 0, 0], > 1, RGBColor[1, 0, 0], > 2, RGBColor[1, 1, 0], > 3, RGBColor[0, 1, 0], > 4, RGBColor[0, 1, 1], > 5, RGBColor[0, 0, 1], > _, RGBColor[1, 1, 1]] Show[Graphics[ > Raster[Table[Random[Integer, {0, 6}], {16}, {16}], > ColorFunction -> mycolor, ColorFunctionScaling -> False]]] GRAY: Sounds good. I'll try it. Interesting that the help does not contain this in a form I could easily find. > 3. I wanted to get this raster image into a format > where I could dissect it with Photoshop or equivalent. And ? what does Export[] do ? it write the expression > in a desired format, TIFF, PNG, PPM are all lossless > compressed bitmap formats, that Mathematica can export > and that can be imported into PhotoShop GRAY: I foolishly thought something like Export would be under the File menu. > After much fooling around, I found that I can export the > selection as an html file, read it into Navigator, do File> > Edit Page, which brings up Netscape Composer, right > click the image which, allows saving it as a GIF, which > I can finally work on with a photo editor. Maybe > there is an easier way, or maybe this description will > help someone with the same need. GRAY: I just found that simply Copying the image and Pasting it into Paint Shop Pro (or no doubt lots of other bitmap editors) works. For outputting a Raster noninteractively, there is Export, but I haven't tried it yet. Jens, thank you for your reply. ==== explain to me why I get this behavior and what I can do to fix it. When I run Mathematica in X with the graphical user interface by typing mathematica, I can run the following commands and get the expected output, namely a plot with the plot label in a big font. However, when I run the math command and get the text interface and run the same code I do not get the label in a big font. cc = Plot[Sin[x], {x, 0, Pi}, {PlotLabel -> StyleForm[Label, FontSize -> 60]}] Export[test.eps, cc, eps] I believe that this is due to the different method in which the math and mathematica commands setup fonts. Is there anyway to get the behavior that I want? As a side note when I use the old syntax of cc=Plot[Sin[x],{x,0,Pi},{PlotLabel->FontForm[Label,{Courier,60}]}] I get it to work. However, I would like not to use this syntax as it has several limitations. Any suggestions would be greatly appreciated. ==== I have a dual processor Dell computer...unfortunately, I can only access one of the processors...I purchased Wolfram's parallel processing toolkit which , because of the very poor documentation, never did me any good...i'm told that another option for accessing the dual processors is to write C code etc...I can't C program, so i'm wondering this...duzz anyone have C code, both source and binary, that they could give me for accomplishing this....in addition, i would like to access this C code with JavaNativeCode, etc...can anyone help me with this? thanks... jerry blimbaum NSWC panama city, fl Reply-To: kuska@informatik.uni-leipzig.de ==== you can have a parallel implicit Runge-Kutta method programmed in C *and* with the Parallel Computing Toolkit both on the base of the MathLink protocol. Since I have a SGI I can't help you with the binary for a Dell what ever computer but if you like the source ... You can also have a native MPI source of the code but you will need a running MPI for your computer. Jens I have a dual processor Dell computer...unfortunately, I can only access one > of the processors...I purchased Wolfram's parallel processing toolkit which > , because of the very poor documentation, never did me any good...i'm told > that another option for accessing the dual processors is to write C code > etc...I can't C program, so i'm wondering this...duzz anyone have C code, > both source and binary, that they could give me for accomplishing this....in > addition, i would like to access this C code with JavaNativeCode, etc...can > anyone help me with this? thanks... jerry blimbaum NSWC panama city, fl ==== I have a set of inequalities that I solve with InequalitySolve. But then it gives a complete set of solutions, but not in the way I would like it to be! :-) For example, the simple following calculation will give: In[1]:= ineq = {y4 >= -1, y5 >= -1, y6 + y4 >= y5 - 1, y5 >= y6, y6 >= -1}; InequalitySolve[ineq,{y4,y6,y5}] Out[1]:= y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6 the result is good, but I would like it to be in the simpler but equivalent form y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6 How can I tell InequalitySolve to do that? It is simple for this example, but for a large set of simple inequalities InequalitySolve gives lines and lines of results instead of a simple result. Vincent Bouchard ==== f[c] == (c+b)/.ru[c] == (c+b)/.c->x == x+b == b+x Bobby Treat -----Original Message----- f[c] = b+c? What gives? Lawrence -- Lawrence A. Walker Jr. http://www.kingshonor.com ==== Sorry; disregard my earlier answer. ru[c] isn't defined, so that obviously wasn't the correct sequence of events. Here's the right one: f[c] == HoldPattern[(a+b)/.ru[a]]/.x->c == (a+b)/.ru[a] == (a+b)/.a->x == x + b == b + x The substitution of c for x occurs before the rule ru[a] is evaluated, so there's no x in the expression to replace. Instead, there's an 'a' to replace with x. If the other sequence had been correct, you have no rule for evaluating ru[c], so it would have remained just that -- ru[c]. When the kernel tried to apply it as a rule to (a+b), there would have been an error. That didn't happen, so that wasn't the sequence of events. Bobby Treat -----Original Message----- f[c] = b+c? What gives? Lawrence -- Lawrence A. Walker Jr. http://www.kingshonor.com ==== You really should read about the difference between := (SetDelayed) and = (Set). When you enter your definition f[x_]:=(a+b) /. ru[a] the right hand side is not evaluated. So next when you evaluate f[c] you get (a+b)/.ru[a] and only now ru[a] is evaluated, thus giving you (a+b)/.a->x which is b+x There are several ways to get what you want. One is to use = instead of :=, another to force evaluation of the right hand side with f[x_]:=Evaluate[(a+b) /. ru[a]] yet another to insert the actual rule in the definition of f: f[x_]:=(a+b) /. a->x On Wednesday, September 25, 2002, at 02:50 PM, Lawrence A. Walker Jr. For the life of me I am not sure why the following is not working in my > v. 4.2: ru[a]=a->x; > f[x_]:=(a+b) /. ru[a]; Why do I get > f[c] = b+x and not > f[c] = b+c? What gives? Lawrence -- > Lawrence A. Walker Jr. > http://www.kingshonor.com > ==== Well, I have written the following notebook with Mathematica In[78]:= a=Sqrt[4*Pi/Sqrt[3]] In[79]:= fcom[k_,mu_]:=(( 1+Exp[-I*mu*2*Pi/a]+Exp[-I*mu*4*Pi/a])*(1-Exp[-I*k*(a-mu)])/(I*k)+ Exp[-I*mu*2* Pi/a]*((1-Exp[-I*(k+2*Pi/a)*(a-mu)])/(I*(k+2*Pi/a))+( 1-Exp[-I*(k-2*Pi/a)*(a-mu)])/(I*(k-2*Pi/a)))+ Exp[-I*mu*4* Pi/a]*((1-Exp[-I*(k+4*Pi/a)*(a-mu)])/(I*(k+4*Pi/a))+( 1-Exp[-I*(k+2*Pi/a)*(a-mu)])/(I*(k+2*Pi/a)))+( 1-Exp[-I*(k-2*Pi/a)*(a-mu)])/(I*(k-2*Pi/a))+( 1-Exp[-I*(k-4*Pi/a)*(a-mu)])/(I*(k-4*Pi/a)))/(3*a) In[80]:= f0[k_,mu_]:=((1+Exp[-I*mu*2*Pi/a]+Exp[-I*mu*4*Pi/a])*(a-mu)+ Exp[-I*mu*2* Pi/a]*((1-Exp[-I*2*Pi/a*(a-mu)])/(I*(2*Pi/a))+( 1-Exp[-I*(-2*Pi/a)*(a-mu)])/(I*(-2*Pi/a)))+ Exp[-I*mu*4* Pi/a]*((1-Exp[-I*(4*Pi/a)*(a-mu)])/(I*(4*Pi/a))+( 1-Exp[-I*(2*Pi/a)*(a-mu)])/(I*(2*Pi/a)))+( 1-Exp[-I*(-2*Pi/a)*(a-mu)])/(I*(-2*Pi/a))+( 1-Exp[-I*(-4*Pi/a)*(a-mu)])/(I*(-4*Pi/a)))/(3*a) In[81]:= fp1[k_,mu_]:=(( 1+Exp[-I*mu*2*Pi/a]+Exp[-I*mu*4*Pi/a])*(1-Exp[-I*2*Pi/a*(a-mu)])/( I*2*Pi/a)+ Exp[-I*mu*2*Pi/a]*((1-Exp[-I*(4*Pi/a)*(a-mu)])/(I*(4*Pi/a))+(a-mu))+ Exp[-I*mu*4* Pi/a]*((1-Exp[-I*(6*Pi/a)*(a-mu)])/(I*(6*Pi/a))+( 1-Exp[-I*(4*Pi/a)*(a-mu)])/(I*(4*Pi/a)))+( a-mu)+(1-Exp[-I*(-2*Pi/a)*(a-mu)])/(I*(-2*Pi/a)))/(3*a) In[82]:= fm1[k_,mu_]:=(( 1+Exp[-I*mu*2*Pi/a]+Exp[-I*mu*4*Pi/a])*( 1-Exp[I*2*Pi/a*(a-mu)])/(-I*2*Pi/a)+ Exp[-I*mu*2*Pi/a]*((a-mu)+(1-Exp[-I*(-4*Pi/a)*(a-mu)])/(I*(-4*Pi/a)))+ Exp[-I*mu*4*Pi/a]*((1-Exp[-I*(2*Pi/a)*(a-mu)])/(I*(2*Pi/a))+(a-mu))+( 1-Exp[-I*(-4*Pi/a)*(a-mu)])/(I*(-4*Pi/a))+( 1-Exp[-I*(-6*Pi/a)*(a-mu)])/(I*(-6*Pi/a)))/(3*a) In[83]:= fp2[k_,mu_]:=(( 1+Exp[-I*mu*2*Pi/a]+Exp[-I*mu*4*Pi/a])*(1-Exp[-I*4*Pi/a*(a-mu)])/( I*4*Pi/a)+ Exp[-I*mu*2* Pi/a]*((1-Exp[-I*(6*Pi/a)*(a-mu)])/(I*(6*Pi/a))+(1-Exp[-I*( 2*Pi/a)*(a-mu)])/(I*(2*Pi/a)))+ Exp[-I*mu*4* Pi/a]*((1-Exp[-I*(8*Pi/a)*(a-mu)])/(I*(8*Pi/a))+( 1-Exp[-I*(6*Pi/a)*(a-mu)])/(I*(6*Pi/a)))+( 1-Exp[-I*(2*Pi/a)*(a-mu)])/(I*(2*Pi/a))+(a-mu))/(3*a) In[84]:= fm2[k_,mu_]:=(( 1+Exp[-I*mu*2*Pi/a]+Exp[-I*mu*4*Pi/a])*( 1-Exp[I*4*Pi/a*(a-mu)])/(-I*4*Pi/a)+ Exp[-I*mu*2* Pi/a]*((1-Exp[-I*(-2*Pi/a)*(a-mu)])/(I*(-2*Pi/a))+( 1-Exp[-I*(-6*Pi/a)*(a-mu)])/(I*(-6*Pi/a)))+ Exp[-I*mu*4* Pi/a]*((a-mu)+(1-Exp[-I*(-2*Pi/a)*(a-mu)])/(I*(-2*Pi/a)))+( 1-Exp[-I*(-6*Pi/a)*(a-mu)])/(I*(-6*Pi/a))+( 1-Exp[-I*(-8*Pi/a)*(a-mu)])/(I*(-8*Pi/a)))/(3*a) In[85]:= Lp[0|N[0],mu_]:=f0[0,mu] In[86]:= Lp[2*Pi/a|N[2*Pi/a],mu_]:=fp1[2*Pi/a,mu] In[87]:= Lp[-2*Pi/a|-N[2*Pi/a],mu_]:=fm1[2*Pi/a,mu] In[88]:= Lp[4*Pi/a|N[4*Pi/a],mu_]:=fp2[4*Pi/a,mu] In[89]:= Lp[-4*Pi/a|-N[4*Pi/a],mu_]:=fm2[-4*Pi/a,mu] In[90]:= Lp[k_,mu_]:=fcom[k,mu] In[91]:= Ll[k_,mu_]:=0/;mu>=a In[92]:= Ll[k_,mu_]:=0/;mu<=-a In[93]:= Ll[k_,mu_]:=Lp[k,mu]/;0<=muMonteCarlo,MaxPoints->100000000,Compiled->False] but this is not enough to ensure convergence of the integration. Notice that I have inserted some points in the integration path in order to avoid problems with numerical divergences which Mathematica detects in fcom[k,mu] (but these divergences do not really exist, analitically) Does somebody has a smart suggestion to perform this computation? Fabio ==== In a presentation I wish to use Plot to generate a sequence of frames and then animate them. The problem is that the audience sees the animation twice. Once when the frames are being generated and then again after I have closed the group and double clicked on the top graphic. However, the first showing during generation is enough (but uncontrolled). Is it possible to tidy up the generation of the graphic so that it becomes the animation? I have tried the following Do[Plot[Sin[t]*Sin[x], {x, 0, Pi}, PlotRange -> {{0, Pi}, {-1, 1}}, ImageSize -> 400]; SelectionMove[EvaluationNotebook[], All, GeneratedCell]; FrontEndExecute[{FrontEnd`SelectionAnimate[0.1]}]; FrontEndExecute[{FrontEndToken[Clear]}], {t, 0, 15, 0.1}] This works but the cell dividing line flashes on and off spoiling the animation and if there is anything in the cell below this jumps up and down. Is there a proper way of doing this? Hugh Goyder ==== I need to program the following recursion scheme for time series forecasting (The Innovation Algorithm). I will write it in pseudo-Mathematica notation. K is the given m x m autocovariance (numerical) matrix of the process v[0] = K[[1,1]]; H[n,n-k] = (v[k])^(-1) (K[[n+1,k+1]] - Sum[(H[k,k-j] H[n,n-j] v[j]), {j,0,k-}]) for k=0,1,...,n-1 v[n] = K[[n+1,n+1]] -Sum[(H[n,n-j])^2 v[j],{j,0,n-1}] The scheme should be solved in the order v[0], H[1,1], v[1], H[2,2], H[2,1], v[2], H[3,3], H[3,2], H[3,1], ... I have already tried to program it in a straght-forward way, but as have no experience with recursive functions with two variables, it doesn't seem to work properly and is also very slow. Any help would be Robert Reply-To: murray@math.umass.edu ==== I don't understand the problem. Say I enter the 2D form for the reciprocal of the square root of a^2 + b^2, but I enter it in Standard Form (i.e., using the Control key with / to form the fraction, and the Control key with 2 to get the square root symbol). If then I click anywhere within the subexpression inside the square root, then with 2 to three clicks the entire expression under the square root will be highlighted. I can then do a copy-and-paste to put that into any cell, in the usual way -- with keyboard key combinations, right-click context menu, or Mathematica Edit menu. Isn't that GUI enough? Would you REALLY want to be able to DRAG the highlighted expression to a new place? Think of how much of a mess this could cause through inadvertent movement of the mouse after highlighting some subexpression. In fact, I hate Microsoft Word's doing just that! >I'm a poor physicist trying to figure out how to sort out the >physical from the non-physical solutions to a problem. To do >>that, I need to be able to look at an expression and pick out a >>subexpression, the part under the radical. > GRAY: > This points up the need, which I've been aware of for years, to be > able to select any part of an expression and drag it to a new line for > further processing. Mathematica and the other CAS I'm familiar with are still > pretty much stuck with a command line interface. They need a true > GUI with extensive interaction. > When I can see on the screen exactly what I want to do, why > should I have to type a bunch of stuff in to access what I want? > Maybe we'll see this in version 5? 6? Never? > -- Murray Eisenberg murray@math.umass.edu Mathematics & Statistics Dept. Lederle Graduate Research Tower phone 413 549-1020 (H) University of Massachusetts 413 545-2859 (W) 710 North Pleasant Street Amherst, MA 01375 ==== > I'm a poor physicist trying to figure out how to sort out the > physical from the non-physical solutions to a problem. To do > that, I need to be able to look at an expression and pick out a > subexpression, the part under the radical. GRAY: This points up the need, which I've been aware of for years, to be able to select any part of an expression and drag it to a new line for further processing. Mathematica and the other CAS I'm familiar with are still pretty much stuck with a command line interface. They need a true GUI with extensive interaction. When I can see on the screen exactly what I want to do, why should I have to type a bunch of stuff in to access what I want? Maybe we'll see this in version 5? 6? Never? Reply-To: murray@math.umass.edu ==== I get the same result in 4.2 (under Windows). Not sure why. But you could try this modification: PlaySeq[L_, dur_] := Do[PlayTone[L[[i]], dur]; Pause[dur], {i, Length[L]}] Then PlaySeq[{224, 256, 384}, 3] seems to work as expected, prolonging the notes before the last one to their full expected duration. > PlayTone[F_, dur_] := Play[Sin[2*Pi*F*t], {t, 0, dur}] > allows me to play a sine wave of frequency F and duration dur PlaySeq[L_, dur_] := Do[PlayTone[L[[i]], dur], {i, Length[L]}] should allow me to play a sequence of tones > with a given list of frequencies L > and all the same duration dur. It does not work at least in 4.1 all tones but the last one are much shorter than dur. any help? > -- Murray Eisenberg murray@math.umass.edu Mathematics & Statistics Dept. Lederle Graduate Research Tower phone 413 549-1020 (H) University of Massachusetts 413 545-2859 (W) 710 North Pleasant Street Amherst, MA 01375 ==== this is almost the solution. but it creates some artificial rests. when the sounds are created, there is some silence between the different notes. when after creation one clicks the play sound icon of the group with all the sounds, these rests disappear. is there a way of Play creating ghte waves, but NOT playing them, and then, doubleclicking the sound icon, hearing the sound without rests immediately? > I get the same result in 4.2 (under Windows). Not sure why. But you > could try this modification: PlaySeq[L_, dur_] := Do[PlayTone[L[[i]], dur]; > Pause[dur], {i, Length[L]}] Then PlaySeq[{224, 256, 384}, 3] seems to work as expected, prolonging the notes before the last one to > their full expected duration. >>PlayTone[F_, dur_] := Play[Sin[2*Pi*F*t], {t, 0, dur}] >>allows me to play a sine wave of frequency F and duration dur >>PlaySeq[L_, dur_] := Do[PlayTone[L[[i]], dur], {i, Length[L]}] >>should allow me to play a sequence of tones >>with a given list of frequencies L >>and all the same duration dur. >>It does not work at least in 4.1 >>all tones but the last one are much shorter than dur. >>any help? -- -- Erich Neuwirth, Computer Supported Didactics Working Group Visit our SunSITE at http://sunsite.univie.ac.at ==== PlayTone[F_, dur_] := Play[Sin[2*Pi*F*t], {t, 0, dur}] allows me to play a sine wave of frequency F and duration dur PlaySeq[L_, dur_] := Do[PlayTone[L[[i]], dur], {i, Length[L]}] should allow me to play a sequence of tones with a given list of frequencies L and all the same duration dur. It does not work at least in 4.1 all tones but the last one are much shorter than dur. any help? -- -- Erich Neuwirth, Computer Supported Didactics Working Group Visit our SunSITE at http://sunsite.univie.ac.at ==== using Set (=) instead of SetDelayed (:=) in f[x_]:=(a+b) /. ru[a]; will give you what you desire. Matthias Bode. -----UrsprÌ.b9ngliche Nachricht----- Gesendet: Mittwoch, 25. September 2002 07:51 An: mathgroup@smc.vnet.net Betreff: Strange ReplaceAll behavior For the life of me I am not sure why the following is not working in my v. 4.2: ru[a]=a->x; f[x_]:=(a+b) /. ru[a]; Why do I get f[c] = b+x and not f[c] = b+c? What gives? Lawrence -- Lawrence A. Walker Jr. http://www.kingshonor.com ==== If you want to do it with the mouse, it's not that difficult. I click within the part I want and double-click until the part is selected. (Each double-click expands the selection by one level of nesting.) Then push Ctrl-C, click where you want to put the result, and push Ctrl-V. Does that help any? Bobby -----Original Message----- able to select any part of an expression and drag it to a new line for further processing. Mathematica and the other CAS I'm familiar with are still pretty much stuck with a command line interface. They need a true GUI with extensive interaction. When I can see on the screen exactly what I want to do, why should I have to type a bunch of stuff in to access what I want? Maybe we'll see this in version 5? 6? Never? ==== I met a critical problem for running Mathematica 4.1 on MacOS X 10.2. (OSX native version) I hope such a problem is suitable to be posted to this ML. What happens is the following. When Mathematica 4.1 installed in HDD is started, a window quickly appears saying that A serious error has occured while Mathematica was starting up. Mathematica will probably not function properly until this problem is resolved. ... The Mathematica fonts are not properly installed in your system. Without these fonts, typeset mathematical expressions cannot be displayed properly. This message is correct. If I chose Continue Anyway, mathematical expressins are totally broken. What is strange is that when I start Mathematica 4.1 by double clicking Mathematica in CDROM, it works without any problem. Further more, Once Mathematica 4.1 is correctly started in this way, I can restart Mathematica directly from double clicking Mathematica in HDD, which is the installed one. Once CDROM is ejected, the error appears again. It is clearly font related problem but I do not know how to solve. If someone know how to manage, please let me know. -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- I met a critical problem for running Mathematica 4.1 on MacOS X 10.2. (OSX native version) I hope such a problem is suitable to be posted to this ML. What happens is the following. When Mathematica 4.1installed in HDD is started, a window quickly appears saying that A serious error has occured while Mathematica was starting up. Mathematica will probably not function properly until this problem is resolved. ... The Mathematica fonts are not properly installed in your system. Without these fonts, typeset mathematical expressions cannot be displayed properly. This message is correct. If I chose Continue Anyway, mathematical expressins are totally broken. What is strange is that when I start Mathematica 4.1 by double clicking Mathematica in CDROM, it works without any problem. Further more, Once Mathematica 4.1 is correctly started in this way, I can restart Mathematica directly from double clicking Mathematica in HDD, which is the installed one. Once CDROM is ejected, the error appears again. It is clearly font related problem but I do not know how to solve. If someone know how to manage, please let me know. -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- --Apple-Mail-2-977368608-- ==== Toshinao, Try reinstalling Mathematica from the CD. --- Selwyn Hollis I met a critical problem for running Mathematica 4.1 on MacOS X 10.2. > (OSX native version) I hope such a problem is suitable to be posted to > this ML. What happens is the following. When Mathematica 4.1 installed in HDD > is started, a window quickly appears saying that A serious error has occured while Mathematica was starting up. > Mathematica will probably not function properly until this problem is > resolved. > ... > The Mathematica fonts are not properly installed in your system. > Without these fonts, typeset mathematical expressions cannot be > displayed properly. This message is correct. If I chose Continue Anyway, mathematical > expressins are totally broken. What is strange is that when I start Mathematica 4.1 by double clicking > Mathematica in CDROM, it works without any problem. Further more, > Once Mathematica 4.1 is correctly started in this way, I can restart > Mathematica > directly from double clicking Mathematica in HDD, which is the > installed one. > Once CDROM is ejected, the error appears again. It is clearly font related problem but I do not know how to solve. If > someone > know how to manage, please let me know. -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- > I met a critical problem for running Mathematica 4.1 on MacOS X 10.2. (OSX native version) I hope such a problem is suitable to be posted to this ML. > What happens is the following. When Mathematica 4.1installed in HDD is started, a window quickly appears saying that > A serious error has occured while Mathematica was starting up. Mathematica will probably not function properly until this problem is resolved. ... The Mathematica fonts are not properly installed in your system. Without these fonts, typeset mathematical expressions cannot be displayed properly. > This message is correct. If I chose Continue Anyway, mathematical expressins are totally broken. > What is strange is that when I start Mathematica 4.1 by double clicking Mathematica in CDROM, it works without any problem. Further more, Once Mathematica 4.1 is correctly started in this way, I can restart > Mathematica directly from double clicking Mathematica in HDD, which is the > installed one. Once CDROM is ejected, the error appears again. > It is clearly font related problem but I do not know how to solve. If > someone know how to manage, please let me know. > -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- --Apple-Mail-2-977368608-- > ==== The reason why InequalitySolve returns it's answer in what sometimes turns out to be unnecessarily complicated form is that the underlying algorithm, Cylindrical Agebraic Decomposition (CAD) returns its answers in this form. Unfortunately it seems to me unlikely that a simplification of the kind you need can be can be accomplished in any general way. To see why observe the following. First of all: In[1]:= FullSimplify[x > 0 || x == 0] Out[1]= x >= 0 This is fine. However: In[2]:= FullSimplify[x > 0 && x < 2 || x == 0 && x < 2] Out[2]= x == 0 || 0 < x < 2 Of course what you would like is simply 0 <= x < 2. One reason why you can't get it is that while Mathematica can perform a LogicalExpand, as in: In[3]:= LogicalExpand[(x > 0 || x == 0) && x < 2] Out[3]= x == 0 && x < 2 || x > 0 && x < 2 There i no LogicalFactor or anything similar that would reverse what LogicalExpand does. if there was then you could perform the sort of simplifications you need for: In[4]:= FullSimplify[(x > 0 || x == 0) && x < 2] Out[4]= 0 <= x < 2 However, it does not seem to me very likely that such logical factoring can be performed by a general enough algorithm (though I am no expert in this field). In any case, certainly Mathematica can't do this. I also noticed that Mathematica seems unable to show that the answer it returns to your problem is actually equivalent to your simpler one. In fact this looks like a possible bug in Mathematica. Let's first try the function ImpliesQ from the Experimental context: << Experimental` Now Mathematica correctly gives: In[6]:= ImpliesQ[y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6, y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6] Out[6]= True However: In[7]:= ImpliesQ[y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6, y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6] Out[7]= False That simply means that ImpliesQ cannot show the implication, not that it does not hold. ImpliesQ relies on CAD, as does FullSimplify. Switching to FullSimplify we see that: In[8]:= FullSimplify[y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6, y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6] Out[8]= True while In[9]:= FullSimplify[y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6, y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6] Out[9]= y4 >= -1 && y6 <= y5 <= 1 + y4 + y6 On the other hand, taking just the individual summands of Or as hypotheses; In[10]:= FullSimplify[y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6, y4 > -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6] Out[10]= True In[11]:= FullSimplify[y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6, y4 == -1 && y6 >= -1 && y5 == y6 ] Out[11]= True In fact FullSimplify is unable to use Or in assumptions, which can be demonstrated on an abstract example: In[12]:= FullSimplify[C,(A||B)&&(C)] Out[12]= True In[13]:= FullSimplify[C,LogicalExpand[(A||B)&&(C)]] Out[13]= C This could be fixed by modifying FullSimplify: In[14]:= Unprotect[FullSimplify]; In[14]:= FullSimplify[expr_,Or[x_,y__]]:=Or[FullSimplify[expr,x],FullSimplify[exp r,y]]; In[15]:= Protect[FullSimplify]; Now at least we get as before: In[16]:= FullSimplify[y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6, y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6] Out[16]= True but also: In[17]:= FullSimplify[y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6, y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6] Out[17]= True This seems to me a possible worthwhile improvement in FullSimplify, though of course not really helpful for your problem. > I have a set of inequalities that I solve with InequalitySolve. But > then > it gives a complete set of solutions, but not in the way I would like > it > to be! :-) For example, the simple following calculation will give: In[1]:= ineq = {y4 >= -1, y5 >= -1, y6 + y4 >= y5 - 1, y5 >= y6, y6 >= > -1}; > InequalitySolve[ineq,{y4,y6,y5}] Out[1]:= y4 == -1 && y6 >= -1 && y5 == y6 || > y4 > -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6 the result is good, but I would like it to be in the simpler but > equivalent form y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6 How can I tell InequalitySolve to do that? It is simple for this > example, > but for a large set of simple inequalities InequalitySolve gives lines > and > lines of results instead of a simple result. > Vincent Bouchard ==== I have been trying to integrate the following : Integrate[Cosh[2 Abs[x-y]] 2 y, {y,0,1/2}, Assumptions->{Im[x]==0,x>0}] However, Mathematica chokes and simply returns the integral as it is. However, if I split up the integral into two portions, it quickly gives me an answer for the parts. Is there something implicit that I am missing in the Assumptions ? MS. -- 12:02am up 5:06, 1 user, load average: 0.54, 0.22, 0.08 ==== The modification to FullSimplify that I sent earlier works correctly only for assumptions of the form Or[a,b] (and even then not is not always what one would like). For what it's worth here is a better (but slow) version: In[1]:= Unprotect[FullSimplify]; In[2]:= FullSimplify[expr_, x_ || y__] := FullSimplify[ FullSimplify[expr, x] || FullSimplify[expr, Or[y]]]; In[3]:= Protect[FullSimplify]; For example: In[4]:= FullSimplify[Sqrt[(x - 1)^2] + Sqrt[(x - 2)^2] + Sqrt[(x - 3)^2], x > 1 || x > 2 || x > 3] Out[4]= -1 + x + Abs[-3 + x] + Abs[-2 + x] || -3 + 2*x + Abs[-3 + x] || 3*(-2 + x) Andrzej Kozlowski Toyama International University JAPAN > The reason why InequalitySolve returns it's answer in what sometimes > turns out to be unnecessarily complicated form is that the underlying > algorithm, Cylindrical Agebraic Decomposition (CAD) returns its > answers in this form. Unfortunately it seems to me unlikely that a > simplification of the kind you need can be can be accomplished in any > general way. To see why observe the following. First of all: In[1]:= > FullSimplify[x > 0 || x == 0] Out[1]= > x >= 0 This is fine. However: In[2]:= > FullSimplify[x > 0 && x < 2 || x == 0 && x < 2] Out[2]= > x == 0 || 0 < x < 2 Of course what you would like is simply 0 <= x < 2. One reason why you > can't get it is that while Mathematica can perform a LogicalExpand, > as in: > In[3]:= > LogicalExpand[(x > 0 || x == 0) && x < 2] Out[3]= > x == 0 && x < 2 || x > 0 && x < 2 There i no LogicalFactor or anything similar that would reverse what > LogicalExpand does. if there was then you could perform the sort of > simplifications you need for: In[4]:= > FullSimplify[(x > 0 || x == 0) && x < 2] Out[4]= > 0 <= x < 2 However, it does not seem to me very likely that such logical > factoring can be performed by a general enough algorithm (though I am > no expert in this field). In any case, certainly Mathematica can't do > this. I also noticed that Mathematica seems unable to show that the answer > it returns to your problem is actually equivalent to your simpler one. > In fact this looks like a possible bug in Mathematica. Let's first try > the function ImpliesQ from the Experimental context: << Experimental` Now Mathematica correctly gives: In[6]:= > ImpliesQ[y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6, > y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1 && y6 <= y5 > <= 1 + y4 + y6] Out[6]= > True However: In[7]:= > ImpliesQ[y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1 && > y6 <= y5 <= 1 + y4 + y6, y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + > y4 + y6] Out[7]= > False That simply means that ImpliesQ cannot show the implication, not that > it does not hold. ImpliesQ relies on CAD, as does FullSimplify. > Switching to FullSimplify we see that: In[8]:= > FullSimplify[y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1 && > y6 <= y5 <= 1 + y4 + y6, y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + > y4 + y6] Out[8]= > True while In[9]:= > FullSimplify[y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6, > y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1 && y6 <= y5 > <= 1 + y4 + y6] Out[9]= > y4 >= -1 && y6 <= y5 <= 1 + y4 + y6 On the other hand, taking just the individual summands of Or as > hypotheses; > In[10]:= > FullSimplify[y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6, > y4 > -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6] Out[10]= > True In[11]:= > FullSimplify[y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6, > y4 == -1 && y6 >= -1 && y5 == y6 ] Out[11]= > True In fact FullSimplify is unable to use Or in assumptions, which can be > demonstrated on an abstract example: > In[12]:= > FullSimplify[C,(A||B)&&(C)] Out[12]= > True In[13]:= > FullSimplify[C,LogicalExpand[(A||B)&&(C)]] Out[13]= > C This could be fixed by modifying FullSimplify: In[14]:= > Unprotect[FullSimplify]; In[14]:= > FullSimplify[expr_,Or[x_,y__]]:=Or[FullSimplify[expr,x],FullSimplify[ex > pr,y]]; In[15]:= > Protect[FullSimplify]; Now at least we get as before: In[16]:= > FullSimplify[y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1 && > y6 <= y5 <= 1 + y4 + y6, y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + > y4 + y6] Out[16]= > True but also: In[17]:= > FullSimplify[y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6, > y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1 && y6 <= y5 > <= 1 + y4 + y6] Out[17]= > True This seems to me a possible worthwhile improvement in FullSimplify, > though of course not really helpful for your problem. > Andrzej Kozlowski > Toyama International University > JAPAN > I have a set of inequalities that I solve with InequalitySolve. But > then > it gives a complete set of solutions, but not in the way I would like > it > to be! :-) For example, the simple following calculation will give: > In[1]:= ineq = {y4 >= -1, y5 >= -1, y6 + y4 >= y5 - 1, y5 >= y6, y6 >= -1}; > InequalitySolve[ineq,{y4,y6,y5}] > Out[1]:= y4 == -1 && y6 >= -1 && y5 == y6 || > y4 > -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6 > the result is good, but I would like it to be in the simpler but > equivalent form > y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6 > How can I tell InequalitySolve to do that? It is simple for this > example, > but for a large set of simple inequalities InequalitySolve gives > lines and > lines of results instead of a simple result. > Vincent Bouchard > > ==== >-----Original Message----- >Sent: Wednesday, September 25, 2002 7:51 AM > >For the life of me I am not sure why the following is not >working in my >v. 4.2: > >ru[a]=a->x; >f[x_]:=(a+b) /. ru[a]; > >Why do I get >f[c] = b+x > >and not >f[c] = b+c? > >What gives? > >Lawrence -- >Lawrence A. Walker Jr. >http://www.kingshonor.com > Lawrence, in your definition of f, x doesn't show up explicitely. So, in the evaluation sequence, when the definition for f[c] is applied, no x appears at rhs i.e. (a + b) /. ru[a] and such c cannot be inserted. The result is the same as directly executing In[11]:= (a + b) /. ru[a] Out[11]= b + x If you don't like this, you have to make explicit the Value of ru[a] in the definiton of f. One way to do so is to use Set instead of SetDelayed: In[9]:= f[x_] = (a + b) /. ru[a] Out[9]= b + x In[10]:= f[c] Out[10]= b + c The drawback of this that not only the value of ru[a] is inserted but also the whole expression including ReplaceAll is evaluated. If this is not wanted, you have to insert the value of ru[a] into the unevaluated rhs at the definition. The general means for this are function application, With or Replace: In[7]:= (g[x_] := (a + b) /. #) &[ru[a]] In[8]:= g[c] Out[8]= b + c In[16]:= Clear[g] In[20]:= Unevaluated[g[x_] := (a + b) /. rule] /. rule -> ru[a] In[21]:= g[c] Out[21]= b + c Here we have to prevent evaluation of the defintion before our rule is inserted, this is achieved by Unevaluated. With is a bit more complicated, since the scoping rules for SetDelayed would not allow the substition of an expression at rhs containing a pattern variable (the pattern variable is renamed in this case). A simple answer to this is to also substitute the argument variable (the pattern): In[31]:= Clear[g] In[32]:= With[{rule = ru[a], arg = x_}, g[arg] := (a + b) /. rule] In[33]:= g[c] Out[33]= b + c -- Hartmut Wolf ==== Hartmut, I add an explicit illustration to your ingenious solution using With. Hartmut's solution: Clear[g]; ru[a] = a -> x; With[{rule = ru[a], arg = x_}, g[arg] := a + b /. rule]; g[ c] b + c Why is arg = x_ needed? Without it we get Clear[g]; With[{rule=ru[a]}, g[x_]:=a+b/.rule]; g [c] b+x The reason for this shows in ?g Global`g g[x$_] := a + b /. a -> x The x in x_ has been changed to x$ and there is no x$ on the right side. This is a general feature of scoping. Taking it further we get Clear[g]; With[{rule=ru[a]},g[x_]:=a+x/.rule]; g [c] c+x ?g Global`g g[x$_] := a + x$ /. a -> x -- Allan --------------------- Allan Hayes Mathematica Training and Consulting Leicester UK www.haystack.demon.co.uk hay@haystack.demon.co.uk Voice: +44 (0)116 271 4198 >-----Original Message----- >Sent: Wednesday, September 25, 2002 7:51 AM > >For the life of me I am not sure why the following is not >working in my >v. 4.2: > >ru[a]=a->x; >f[x_]:=(a+b) /. ru[a]; > >Why do I get >f[c] = b+x > >and not >f[c] = b+c? > >What gives? > >Lawrence > >-- >Lawrence A. Walker Jr. >http://www.kingshonor.com Lawrence, in your definition of f, x doesn't show up explicitely. So, in the > evaluation sequence, when the definition for f[c] is applied, no x appears > at rhs i.e. > (a + b) /. ru[a] and such c cannot be inserted. The result is the same as > directly executing In[11]:= (a + b) /. ru[a] > Out[11]= b + x If you don't like this, you have to make explicit the Value of ru[a] in the > definiton of f. One way to do so is to use Set instead of SetDelayed: In[9]:= f[x_] = (a + b) /. ru[a] > Out[9]= b + x In[10]:= f[c] > Out[10]= b + c The drawback of this that not only the value of ru[a] is inserted but also > the whole expression including ReplaceAll is evaluated. If this is not > wanted, you have to insert the value of ru[a] into the unevaluated rhs at > the definition. The general means for this are function application, With or > Replace: In[7]:= (g[x_] := (a + b) /. #) &[ru[a]] > In[8]:= g[c] > Out[8]= b + c In[16]:= Clear[g] > In[20]:= > Unevaluated[g[x_] := (a + b) /. rule] /. rule -> ru[a] > In[21]:= g[c] > Out[21]= b + c Here we have to prevent evaluation of the defintion before our rule is > inserted, this is achieved by Unevaluated. With is a bit more complicated, since the scoping rules for SetDelayed would > not allow the substition of an expression at rhs containing a pattern > variable (the pattern variable is renamed in this case). A simple answer to > this is to also substitute the argument variable (the pattern): > In[31]:= Clear[g] > In[32]:= > With[{rule = ru[a], arg = x_}, g[arg] := (a + b) /. rule] > In[33]:= g[c] > Out[33]= b + c -- > Hartmut Wolf ==== Try In[1]:= M={{1,2,3},{4,5,6},{7,8,9}}; In[2]:= Export[test.csv,M,CSV]; Also, avoid using names with capital initial, since these are usually reserved for Mathematica functions. Tomas Garza Mexico City ----- Original Message ----- > M={{1,2,3},{4,5,6},{7,8,9}}; > TableForm[M] 1 2 3 > 4 5 6 > 7 8 9 The commands like Put or Write create files with a standard Mathematica > matrix notation with braces, I get the the same result with Save As/Copy Robert ==== I would appreciate if someone could tell me how to put my output for example a matrix in TableForm as a space-delimited ASCII file in Mathematica 3.0. for example: M={{1,2,3},{4,5,6},{7,8,9}}; TableForm[M] 1 2 3 4 5 6 7 8 9 The commands like Put or Write create files with a standard Mathematica matrix notation with braces, I get the the same result with Save As/Copy Robert ==== Much longer. At least 10 times as long. Not sure because I quit. As I read Bobby Treat's response, one of my processors had been laboring for more than three days to generate all the digits. (I had not considered to display or print all of them. Why on Earth would I want to do that?) So I started over, asking RealDigits to skip most of the digits. Tom Burton > So would it take about the same amont of time for the complete printout > of digits? Of course it would take a few additional seconds to format > the output... Or does Mathematica take alot less time when it truncates the output? > Could you tell me the CPU you used and its speed etc...i am curious, > other programs out there. I used one processor of a dual 1GH Mac and got the same answer with the > following speed: 4.2 for Mac OS X (June 4, 2002) > oldmax = $MaxPrecision > 6 > 1. 10 > $MaxPrecision = Infinity > Infinity > With[{n = 2^26}, Timing[ > pd = RealDigits[N[Pi, n + 1], 10, 20, > 19 - n]; ]] > {28794.1 Second, Null} > MaxMemoryUsed[] > 512055204 > pd > {{3, 3, 8, 6, 3, 2, 2, 0, 8, 9, 6, 2, 2, 3, > > 4, 0, 9, 8, 0, 3}, -67108844} Tom Burton > ==== Often posters to MathGroup copy and paste in the complete cell expression, including the In and Out numbers, when posting to MathGroup. I wonder if this is the best method because one can't then just copy out all the statements and paste them into a Mathematica notebook. All the statement numbers have to be edited out and if there are many statement definitions this is an extended task for any responder. This, of course, decreases the chances for a response. A better method is for the poster to just copy and paste the CONTENTS of each cell. This is more work for the poster, but it may pay off in better responses. David Park djmp@earthlink.net http://home.earthlink.net/~djmp/ ==== David, I agree with you that it makes responding and using easier if the In and Out numbers are not included. An easier way to do this, rather than copying the contents of the cells, is Usually a little manual reformating is needed to distinguish text, input and output --- I try to use one tab indent for input and two tabs indent for output, plus some blank line adjustment. I wonder if anyone has a way of automatically achieving this reformating. -- 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 > Often posters to MathGroup copy and paste in the complete cell expression, > including the In and Out numbers, when posting to MathGroup. I wonder if this is the best method because one can't then just copy out all > the statements and paste them into a Mathematica notebook. All the statement > numbers have to be edited out and if there are many statement definitions > this is an extended task for any responder. This, of course, decreases the > chances for a response. A better method is for the poster to just copy and > paste the CONTENTS of each cell. This is more work for the poster, but it > may pay off in better responses. David Park > djmp@earthlink.net > http://home.earthlink.net/~djmp/ ==== copying and pasting is straightforward. I always run my notebooks this way, because showing the In[x], Out[x] does not seem to provide me with much beyond clutter. Kevin > Often posters to MathGroup copy and paste in the complete cell expression, > including the In and Out numbers, when posting to MathGroup. I wonder if this is the best method because one can't then just copy out all > the statements and paste them into a Mathematica notebook. All the statement > numbers have to be edited out and if there are many statement definitions > this is an extended task for any responder. This, of course, decreases the > chances for a response. A better method is for the poster to just copy and > paste the CONTENTS of each cell. This is more work for the poster, but it > may pay off in better responses. David Park > djmp@earthlink.net > http://home.earthlink.net/~djmp/ ==== >In a presentation I wish to use Plot to generate a sequence of frames and >then animate them. The problem is that the audience sees the animation >twice. Once when the frames are being generated and then again after I have >closed the group and double clicked on the top graphic. However, the first >showing during generation is enough (but uncontrolled). > >Hugh Goyder This creates a graphics cell from a graphics expression. GraphicCell[graphics_] := Cell[GraphicsData[PostScript, DisplayString[graphics]],Graphics] cellgroup. Block[{$DisplayFunction=Identity, graphs}, graphs = Table[GraphicCell[ Plot[Sin[t]*Sin[x], {x, 0, Pi}, PlotRange -> {{0, Pi}, {-1, 1}}, ImageSize -> 400]], {t,0,15,.1}]; NotebookWrite[EvaluationNotebook[],Cell[CellGroupData[graphs,Closed]]]; SelectionMove[EvaluationNotebook[], All, GeneratedCell]; FrontEndExecute[{FrontEndToken[EvaluationNotebook[], SelectionAnimate]}] ] -------------------------------------------------------------- Omega Consulting The final answer to your Mathematica needs Spend less time searching and more time finding. http://www.wz.com/internet/Mathematica.html ==== Actually, the reason why ImpliesQ (and FullSimplify) fail to prove the implication is not that the hypothesis is a disjunction. To use the cylindrical algebraic decomposition algorithm they need to know that the assumptions imply that all variables are real. The assumptions mechanism infers variable domains in a purely syntactical way, i.e. v is assumed to be real if there is an Element[v, Reals] statement or v appears in an inequality. It does not attempt to analyze assumptions further, to figure out that, say y6 >= -1 implies that y6 is real, and then if we have y5 == y6 then y5 must be real too. Doing such an analysis in general would require solving the assumptions over complex numbers, and then finding out which variables need to be real. This would be in general too time consuming to do, but analyzing linear dependencies like the ones in your example is a possible future improvement. ImpliesQ cannot prove the implication here, because it knows only that y6 is real. In[1]:= <= -1 && y5 == y6, y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6] Out[2]= False If we add an explicit assumption that y4 and y5 are real, ImpliesQ (and FullSimplify) can prove this implication, and the full version of your example. In[3]:= ImpliesQ[Element[y4|y5, Reals] && y4 == -1 && y6 >= -1 && y5 == y6, y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6] Out[3]= True In[4]:= ImpliesQ[Element[y4|y5, Reals] && (y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6), y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6] Out[4]= True In[5]:= FullSimplify[y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6, Element[y4|y5, Reals] && (y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6)] Out[5]= True Adam Strzebonski Wolfram Research On second thoughts I realized that there seems to be an inherent > ambiguity about what one coudl mean by using alternatives (statements > joned by Or) assumptions. In fact it now seems to me that the > reasonable intertpretation for ImpliesQ and FullSimplify ought to > perhaps be different. It seems to me that ImpliesQ[Or[a,b],c] ought to > return True if aand only if ImpliesQ[a,c] and ImpliesQ[b,c] both return > True. If so this could be acomplished by adding the rule > ImpliesQ[Or[a,b],c] = And[ImpliesQ[a,c],ImpliesQ[b,c]]. That could then > be used in proving that the two answers to the system of inequalities > that of Vincent's original posting are equivalent. On the other hand > probably FullSimplify[a, Or[p,q]] ought to return > Or[FullSimplify[a,p],FullSimplify[a,q]] (or do nothing as it doe snow). > The first approach would seem to be consistent with the way > FullSimplify works with domain specifications but would however have > the strange effect of returning True if just one of the alternatives > were true and the other false. So perhaps after all it is best to > leave FullSimplify as it is. However, it seems to me that ImpliesQ > shoud be able to handle such cases (?) Andrzej Kozlowski > Toyama International University > JAPAN > The modification to FullSimplify that I sent earlier works correctly > only for assumptions of the form Or[a,b] (and even then not is not > always what one would like). For what it's worth here is a better (but > slow) version: > In[1]:= > Unprotect[FullSimplify]; > In[2]:= > FullSimplify[expr_, x_ || y__] := FullSimplify[ > FullSimplify[expr, x] || FullSimplify[expr, Or[y]]]; > In[3]:= > Protect[FullSimplify]; > For example: > In[4]:= > FullSimplify[Sqrt[(x - 1)^2] + Sqrt[(x - 2)^2] + > Sqrt[(x - 3)^2], x > 1 || x > 2 || x > 3] > Out[4]= > -1 + x + Abs[-3 + x] + Abs[-2 + x] || > -3 + 2*x + Abs[-3 + x] || 3*(-2 + x) > Andrzej Kozlowski > Toyama International University > JAPAN > > The reason why InequalitySolve returns it's answer in what sometimes > turns out to be unnecessarily complicated form is that the underlying > algorithm, Cylindrical Agebraic Decomposition (CAD) returns its > answers in this form. Unfortunately it seems to me unlikely that a > simplification of the kind you need can be can be accomplished in any > general way. To see why observe the following. First of all: > In[1]:= > FullSimplify[x > 0 || x == 0] > Out[1]= > x >= 0 > This is fine. However: > In[2]:= > FullSimplify[x > 0 && x < 2 || x == 0 && x < 2] > Out[2]= > x == 0 || 0 < x < 2 > Of course what you would like is simply 0 <= x < 2. One reason why > you can't get it is that while Mathematica can perform a > LogicalExpand, as in: > In[3]:= > LogicalExpand[(x > 0 || x == 0) && x < 2] > Out[3]= > x == 0 && x < 2 || x > 0 && x < 2 > There i no LogicalFactor or anything similar that would reverse > what LogicalExpand does. if there was then you could perform the sort > of simplifications you need for: > In[4]:= > FullSimplify[(x > 0 || x == 0) && x < 2] > Out[4]= > 0 <= x < 2 > However, it does not seem to me very likely that such logical > factoring can be performed by a general enough algorithm (though I > am no expert in this field). In any case, certainly Mathematica can't > do this. > I also noticed that Mathematica seems unable to show that the answer > it returns to your problem is actually equivalent to your simpler > one. In fact this looks like a possible bug in Mathematica. Let's > first try the function ImpliesQ from the Experimental context: > << Experimental` > Now Mathematica correctly gives: > In[6]:= > ImpliesQ[y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6, > y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1 && y6 <= y5 > <= 1 + y4 + y6] > Out[6]= > True > However: > In[7]:= > ImpliesQ[y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1 && > y6 <= y5 <= 1 + y4 + y6, y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + > y4 + y6] > Out[7]= > False > That simply means that ImpliesQ cannot show the implication, not that > it does not hold. ImpliesQ relies on CAD, as does FullSimplify. > Switching to FullSimplify we see that: > In[8]:= > FullSimplify[y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1 > && > y6 <= y5 <= 1 + y4 + y6, y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + > y4 + y6] > Out[8]= > True > while > In[9]:= > FullSimplify[y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6, > y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1 && y6 <= y5 > <= 1 + y4 + y6] > Out[9]= > y4 >= -1 && y6 <= y5 <= 1 + y4 + y6 > On the other hand, taking just the individual summands of Or as > hypotheses; > In[10]:= > FullSimplify[y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6, > y4 > -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6] > Out[10]= > True > In[11]:= > FullSimplify[y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6, > y4 == -1 && y6 >= -1 && y5 == y6 ] > Out[11]= > True > In fact FullSimplify is unable to use Or in assumptions, which can be > demonstrated on an abstract example: > In[12]:= > FullSimplify[C,(A||B)&&(C)] > Out[12]= > True > In[13]:= > FullSimplify[C,LogicalExpand[(A||B)&&(C)]] > Out[13]= > C > This could be fixed by modifying FullSimplify: > In[14]:= > Unprotect[FullSimplify]; > In[14]:= > FullSimplify[expr_,Or[x_,y__]]:=Or[FullSimplify[expr,x],FullSimplify[e > xpr,y]]; > In[15]:= > Protect[FullSimplify]; > Now at least we get as before: > In[16]:= > FullSimplify[y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1 > && > y6 <= y5 <= 1 + y4 + y6, y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + > y4 + y6] > Out[16]= > True > but also: > In[17]:= > FullSimplify[y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6, > y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1 && y6 <= y5 > <= 1 + y4 + y6] > Out[17]= > True > This seems to me a possible worthwhile improvement in FullSimplify, > though of course not really helpful for your problem. > Andrzej Kozlowski > Toyama International University > JAPAN > I have a set of inequalities that I solve with InequalitySolve. But > then > it gives a complete set of solutions, but not in the way I would > like it > to be! :-) For example, the simple following calculation will give: > In[1]:= ineq = {y4 >= -1, y5 >= -1, y6 + y4 >= y5 - 1, y5 >= y6, y6 >= -1}; > InequalitySolve[ineq,{y4,y6,y5}] > Out[1]:= y4 == -1 && y6 >= -1 && y5 == y6 || > y4 > -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6 > the result is good, but I would like it to be in the simpler but > equivalent form > y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6 > How can I tell InequalitySolve to do that? It is simple for this > example, > but for a large set of simple inequalities InequalitySolve gives > lines and > lines of results instead of a simple result. > Vincent Bouchard > ==== On second thoughts I realized that there seems to be an inherent ambiguity about what one could mean by using alternatives (statements joned by Or) assumptions. In fact it now seems to me that the reasonable intertpretation for ImpliesQ and FullSimplify ought to perhaps be different. It seems to me that ImpliesQ[Or[a,b],c] ought to return True if aand only if ImpliesQ[a,c] and ImpliesQ[b,c] both return True. If so this could be acomplished by adding the rule ImpliesQ[Or[a,b],c] = And[ImpliesQ[a,c],ImpliesQ[b,c]]. That could then be used in proving that the two answers to the system of inequalities that of Vincent's original posting are equivalent. On the other hand probably FullSimplify[a, Or[p,q]] ought to return Or[FullSimplify[a,p],FullSimplify[a,q]] (or do nothing as it doe snow). The first approach would seem to be consistent with the way FullSimplify works with domain specifications but would however have the strange effect of returning True if just one of the alternatives were true and the other false. So perhaps after all it is best to leave FullSimplify as it is. However, it seems to me that ImpliesQ shoud be able to handle such cases (?) > The modification to FullSimplify that I sent earlier works correctly > only for assumptions of the form Or[a,b] (and even then not is not > always what one would like). For what it's worth here is a better (but > slow) version: In[1]:= > Unprotect[FullSimplify]; In[2]:= > FullSimplify[expr_, x_ || y__] := FullSimplify[ > FullSimplify[expr, x] || FullSimplify[expr, Or[y]]]; In[3]:= > Protect[FullSimplify]; For example: In[4]:= > FullSimplify[Sqrt[(x - 1)^2] + Sqrt[(x - 2)^2] + > Sqrt[(x - 3)^2], x > 1 || x > 2 || x > 3] Out[4]= > -1 + x + Abs[-3 + x] + Abs[-2 + x] || > -3 + 2*x + Abs[-3 + x] || 3*(-2 + x) Andrzej Kozlowski > Toyama International University > JAPAN > The reason why InequalitySolve returns it's answer in what sometimes > turns out to be unnecessarily complicated form is that the underlying > algorithm, Cylindrical Agebraic Decomposition (CAD) returns its > answers in this form. Unfortunately it seems to me unlikely that a > simplification of the kind you need can be can be accomplished in any > general way. To see why observe the following. First of all: > In[1]:= > FullSimplify[x > 0 || x == 0] > Out[1]= > x >= 0 > This is fine. However: > In[2]:= > FullSimplify[x > 0 && x < 2 || x == 0 && x < 2] > Out[2]= > x == 0 || 0 < x < 2 > Of course what you would like is simply 0 <= x < 2. One reason why > you can't get it is that while Mathematica can perform a > LogicalExpand, as in: > In[3]:= > LogicalExpand[(x > 0 || x == 0) && x < 2] > Out[3]= > x == 0 && x < 2 || x > 0 && x < 2 > There i no LogicalFactor or anything similar that would reverse > what LogicalExpand does. if there was then you could perform the sort > of simplifications you need for: > In[4]:= > FullSimplify[(x > 0 || x == 0) && x < 2] > Out[4]= > 0 <= x < 2 > However, it does not seem to me very likely that such logical > factoring can be performed by a general enough algorithm (though I > am no expert in this field). In any case, certainly Mathematica can't > do this. > I also noticed that Mathematica seems unable to show that the answer > it returns to your problem is actually equivalent to your simpler > one. In fact this looks like a possible bug in Mathematica. Let's > first try the function ImpliesQ from the Experimental context: > << Experimental` > Now Mathematica correctly gives: > In[6]:= > ImpliesQ[y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6, > y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1 && y6 <= y5 > <= 1 + y4 + y6] > Out[6]= > True > However: > In[7]:= > ImpliesQ[y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1 && > y6 <= y5 <= 1 + y4 + y6, y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + > y4 + y6] > Out[7]= > False > That simply means that ImpliesQ cannot show the implication, not that > it does not hold. ImpliesQ relies on CAD, as does FullSimplify. > Switching to FullSimplify we see that: > In[8]:= > FullSimplify[y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1 > && > y6 <= y5 <= 1 + y4 + y6, y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + > y4 + y6] > Out[8]= > True > while > In[9]:= > FullSimplify[y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6, > y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1 && y6 <= y5 > <= 1 + y4 + y6] > Out[9]= > y4 >= -1 && y6 <= y5 <= 1 + y4 + y6 > On the other hand, taking just the individual summands of Or as > hypotheses; > In[10]:= > FullSimplify[y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6, > y4 > -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6] > Out[10]= > True > In[11]:= > FullSimplify[y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6, > y4 == -1 && y6 >= -1 && y5 == y6 ] > Out[11]= > True > In fact FullSimplify is unable to use Or in assumptions, which can be > demonstrated on an abstract example: > In[12]:= > FullSimplify[C,(A||B)&&(C)] > Out[12]= > True > In[13]:= > FullSimplify[C,LogicalExpand[(A||B)&&(C)]] > Out[13]= > C > This could be fixed by modifying FullSimplify: > In[14]:= > Unprotect[FullSimplify]; > In[14]:= > FullSimplify[expr_,Or[x_,y__]]:=Or[FullSimplify[expr,x],FullSimplify[e > xpr,y]]; > In[15]:= > Protect[FullSimplify]; > Now at least we get as before: > In[16]:= > FullSimplify[y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1 > && > y6 <= y5 <= 1 + y4 + y6, y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + > y4 + y6] > Out[16]= > True > but also: > In[17]:= > FullSimplify[y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6, > y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1 && y6 <= y5 > <= 1 + y4 + y6] > Out[17]= > True > This seems to me a possible worthwhile improvement in FullSimplify, > though of course not really helpful for your problem. > Andrzej Kozlowski > Toyama International University > JAPAN > I have a set of inequalities that I solve with InequalitySolve. But > then > it gives a complete set of solutions, but not in the way I would > like it > to be! :-) For example, the simple following calculation will give: > In[1]:= ineq = {y4 >= -1, y5 >= -1, y6 + y4 >= y5 - 1, y5 >= y6, y6 >= -1}; > InequalitySolve[ineq,{y4,y6,y5}] > Out[1]:= y4 == -1 && y6 >= -1 && y5 == y6 || > y4 > -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6 > the result is good, but I would like it to be in the simpler but > equivalent form > y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6 > How can I tell InequalitySolve to do that? It is simple for this > example, > but for a large set of simple inequalities InequalitySolve gives > lines and > lines of results instead of a simple result. > Vincent Bouchard > > ==== Hallo!, My name is Nagesh and pursuing research studies in Refrigeration. At present I am writing a Dynamic Refrigeration System Simulation Package. I am using Mathematica as a programming language for the same since last one year. I don't have any programming experience before this. I have following querries:- 1. Is any body here have expertise or information about the capability of Mathematica as a system simulation tool? 2. Is is possible to program a user friendly interface for my system simulation package with Mathematica or I have to use some other software? 3. My refrigeration system simulation package is likely to have approximately 60 First order Differential equations. Is is possible to solve these in Mathematica ? If yes then can anybody here guide me about this further. I am explaining below in short about the objectives I want to fulfill from coding out of my main input file 1. Example from Main Input File ( this will contain about 200-250 variables which will be entered by the user of this package) Below is examples of two variables entered into this file, which will be used in other analysis files for further evaluation. 2. Example from other analysis file ( there will be about 20-25 other such component analysis files ) where the above mentioned variables from main input file will be used for further evaluations:- Below is one example from this file explaining how the variables from main input file will be used in other files. I hope that this short information will be useful for guiding me to solve the following problems that I am facing. I am facing follwing problems or objectives:- 1. My 1st Objective:- The user of this package must be able to change only the value of the variable in the main input file but he must not be able to change the name of the variable itself. For example he must be able to change the value of the variable but he must not be able to change the name of this variable itself. Here our problem is how to achieve or program it so that our objective will be fullfilled. 2. My 2nd Objective:- How I can program the main input file so that it will be user friendly in terms of its visuals and satisfying the constraint mentioned above in objective1. 3. My 3rd Objective:- How can I program the optional values for each variable in the main input file ? so that there will be always a value assigned to each variable listed in main input file whenever the user opens up this file. If user want to change the values of some variables then he can change them and run the simulation otherwise the simulation run will be done with optional values assigned to each variable in the input file. 4. My 4th Objective:- How can I program the check for correctness of the input values supplied by the package user ? I will be very greatful to you for helping and suggesting some technique or method to solve some of my above mentioned programming problems. Can anybody here guide me about the above mentioned programming problems ? Nagesh Rajepandhare ==== I want to add a further suggestion: Probably each differential equation logically belongs to one of the defined classes in the sense, that it helps to define the behaviour of the component. In this case, each differential equation should be stored in one of theses classes as textstring in a form appropriate for the execution in Mathematica. Hermann Schmitt ----- Original Message ----- Is any body here have expertise or information about the capability of > Mathematica as a system simulation tool? > Mr. Kuska answers: > Since the most system simulation tools are simply solving a system of > ordinary differntial equations it is simple to do this with NDSolve[]. My comment: That is: He sees the simulation system merely as a set of differential > equations. The question of Mr. Nagesh: My 4th Objective:- How can I program the check for correctness of the > input values supplied by the package user ? > The answer of Mr. Kuska is: > And @@ (NumericQ /@ {aListOfAllYourNumericParameters}) My comment: This is a nice command and shows the knowledge of Mr.Kuska. But does Mr. > Nagesh understand it and is it sufficient to check, if all inputs are > numerical? Additionally I think, it is not userfriendly to see the input merely as a > set of 200-250 numbers. My suggestion is, that JLink is used, a suggestion Mr. Kusk takes into > consideration, too. But further I suggest, that classes are defined in Java, which represent the > parts of the system. Constructors of the classes should build objects with default values. > Graphical user interfaces should give the opportunity to change the data fields in the objects and > check the input for correctness. The system should give the opportunity, to store the objects on harddisk > (serialization). accessed directly. > My name is Nagesh and pursuing research studies in Refrigeration. At > present I am writing a Dynamic Refrigeration System Simulation Package. > I > am using Mathematica as a programming language for the same since last > one > year. I don't have any programming experience before this. I have > following > querries:- > 1. Is any body here have expertise or information about the capability > of > Mathematica as a system simulation tool? > Since the most system simulation tools are simply solving a system of > ordinary differntial equations it is simple to do this with NDSolve[]. > 2. Is is possible to program a user friendly interface for my system > simulation package with Mathematica or I have to use some other > software? > Write a MathLink or J/Link frontend that launch the kernel. But you > should keep > in mind that the user interface is typical 80-90 % of your code. > If you just whant to solve some ode's it is probably easyer to > include one of the excelent ode-solvers from netlib in your C-code > than to call Mathematica to do that. As long as you dont wish to change > the ode's very often (than Mathematica is more flexible) you should > not use Mathematica. > 3. My refrigeration system simulation package is likely to have > approximately 60 First order Differential equations. Is is possible to > solve these in Mathematica ? > Sure. > If yes then can anybody here guide me about > this further. > Write down the equations and call NDSolve[]. > > I am explaining below in short about the objectives I want to fulfill > from > coding out of my main input file > 1. Example from Main Input File ( this will contain about 200-250 > variables > which will be entered by the user of this package) > This sounds like a *very* userfiendly interface ;-) > > Below is examples of two variables entered into this file, which will be > used in other analysis files for further evaluation. > 2. Example from other analysis file ( there will be about 20-25 other > such > component analysis files ) where the above mentioned variables from main > input file will be used for further evaluations:- > Below is one example from this file explaining how the variables from > main > input file will be used in other files. > I hope that this short information will be useful for guiding me to > solve > the following problems that I am facing. I am facing follwing problems > or > objectives:- > 1. My 1st Objective:- The user of this package must be able to change > only > the value of the variable in the main input file but he must not be able > to > change the name of the variable itself. For example he must be able to > change the value of the variable but he must not be able to change > the > name of this variable itself. > Here our problem is how to achieve or program it so that our objective > will > be fullfilled. > Options with defaulf values ? or something like > {aParam,bParam}={ODEParameter1,ODEParameter2} /. > userRules /. > {ODEParameter1->1,ODEParameter2->2} > > 2. My 2nd Objective:- How I can program the main input file so that it > will > be user friendly in terms of its visuals and satisfying the constraint > mentioned above in objective1. > What is *userfiendly* in a file with 250 variables ??? > > 3. My 3rd Objective:- How can I program the optional values for each > variable in the main input file ? so that there will be always a value > assigned to each variable listed in main input file whenever the user > opens > up this file. If user want to change the values of some variables then > he > can change them and run the simulation otherwise the simulation run will > be > done with optional values assigned to each variable in the input file. > See above. > > 4. My 4th Objective:- How can I program the check for correctness of the > input values supplied by the package user ? > And @@ (NumericQ /@ {aListOfAllYourNumericParameters}) > Jens ==== I have developed one package where it is included a functions for solving System of ODE. You can downloaded from: http://web.usal.es/~guillerm/biokmod. The functions are: SystemDSolve and SystemNDSolve Guillermo Sanchez > I want to add a further suggestion: > Probably each differential equation logically belongs to one of the defined > classes in the sense, that it helps to define the behaviour of the > component. > In this case, each differential equation should be stored in one of theses > classes as textstring in a form appropriate for the execution in > Mathematica. > Hermann Schmitt > ----- Original Message ----- > I agree with Mr. Kuska, that the system Mr Nagesh describes is not > userfriendly. But I think, the suggestions of Mr. Kuska do not make it > more > userfriendly, rather the opposite is true. > Mr. Nagesh asks > Is any body here have expertise or information about the capability of > Mathematica as a system simulation tool? > Mr. Kuska answers: > Since the most system simulation tools are simply solving a system of > ordinary differntial equations it is simple to do this with NDSolve[]. > My comment: > That is: He sees the simulation system merely as a set of differential > equations. > The question of Mr. Nagesh: > My 4th Objective:- How can I program the check for correctness of the > input values supplied by the package user ? > The answer of Mr. Kuska is: > And @@ (NumericQ /@ {aListOfAllYourNumericParameters}) > My comment: > This is a nice command and shows the knowledge of Mr.Kuska. But does Mr. > Nagesh understand it and is it sufficient to check, if all inputs are > numerical? > Additionally I think, it is not userfriendly to see the input merely as a > set of 200-250 numbers. > > My suggestion is, that JLink is used, a suggestion Mr. Kusk takes into > consideration, too. > But further I suggest, that classes are defined in Java, which represent > the > parts of the system. > Constructors of the classes should build objects with default values. > Graphical user interfaces > should give the opportunity to change the data fields in the objects and > check the input for correctness. > The system should give the opportunity, to store the objects on harddisk > (serialization). > accessed directly. > > > My name is Nagesh and pursuing research studies in Refrigeration. At > present I am writing a Dynamic Refrigeration System Simulation > Package. > I > am using Mathematica as a programming language for the same since last > one > year. I don't have any programming experience before this. I have > following > querries:- > 1. Is any body here have expertise or information about the capability > of > Mathematica as a system simulation tool? > Since the most system simulation tools are simply solving a system of > ordinary differntial equations it is simple to do this with NDSolve[]. > 2. Is is possible to program a user friendly interface for my system > simulation package with Mathematica or I have to use some other > software? > Write a MathLink or J/Link frontend that launch the kernel. But you > should keep > in mind that the user interface is typical 80-90 % of your code. > If you just whant to solve some ode's it is probably easyer to > include one of the excelent ode-solvers from netlib in your C-code > than to call Mathematica to do that. As long as you dont wish to change > the ode's very often (than Mathematica is more flexible) you should > not use Mathematica. > 3. My refrigeration system simulation package is likely to have > approximately 60 First order Differential equations. Is is possible to > solve these in Mathematica ? > Sure. > If yes then can anybody here guide me about > this further. > Write down the equations and call NDSolve[]. > > I am explaining below in short about the objectives I want to fulfill > from > coding out of my main input file > 1. Example from Main Input File ( this will contain about 200-250 > variables > which will be entered by the user of this package) > This sounds like a *very* userfiendly interface ;-) > > Below is examples of two variables entered into this file, which will > be > used in other analysis files for further evaluation. > 2. Example from other analysis file ( there will be about 20-25 other > such > component analysis files ) where the above mentioned variables from > main > input file will be used for further evaluations:- > Below is one example from this file explaining how the variables from > main > input file will be used in other files. > I hope that this short information will be useful for guiding me to > solve > the following problems that I am facing. I am facing follwing problems > or > objectives:- > 1. My 1st Objective:- The user of this package must be able to change > only > the value of the variable in the main input file but he must not be > able > to > change the name of the variable itself. For example he must be able to > change the value of the variable but he must not be able to > change > the > name of this variable itself. > Here our problem is how to achieve or program it so that our objective > will > be fullfilled. > Options with defaulf values ? or something like > {aParam,bParam}={ODEParameter1,ODEParameter2} /. > userRules /. > {ODEParameter1->1,ODEParameter2->2} > > 2. My 2nd Objective:- How I can program the main input file so that it > will > be user friendly in terms of its visuals and satisfying the constraint > mentioned above in objective1. > What is *userfiendly* in a file with 250 variables ??? > > 3. My 3rd Objective:- How can I program the optional values for each > variable in the main input file ? so that there will be always a value > assigned to each variable listed in main input file whenever the user > opens > up this file. If user want to change the values of some variables then > he > can change them and run the simulation otherwise the simulation run > will > be > done with optional values assigned to each variable in the input file. > See above. > > 4. My 4th Objective:- How can I program the check for correctness of > the > input values supplied by the package user ? > And @@ (NumericQ /@ {aListOfAllYourNumericParameters}) > > Jens > ==== I agree with Mr. Kuska, that the system Mr Nagesh describes is not userfriendly. But I think, the suggestions of Mr. Kuska do not make it more userfriendly, rather the opposite is true. Mr. Nagesh asks Is any body here have expertise or information about the capability of Mathematica as a system simulation tool? Mr. Kuska answers: Since the most system simulation tools are simply solving a system of ordinary differntial equations it is simple to do this with NDSolve[]. My comment: That is: He sees the simulation system merely as a set of differential equations. The question of Mr. Nagesh: My 4th Objective:- How can I program the check for correctness of the input values supplied by the package user ? The answer of Mr. Kuska is: And @@ (NumericQ /@ {aListOfAllYourNumericParameters}) My comment: This is a nice command and shows the knowledge of Mr.Kuska. But does Mr. Nagesh understand it and is it sufficient to check, if all inputs are numerical? Additionally I think, it is not userfriendly to see the input merely as a set of 200-250 numbers. My suggestion is, that JLink is used, a suggestion Mr. Kusk takes into consideration, too. But further I suggest, that classes are defined in Java, which represent the parts of the system. Constructors of the classes should build objects with default values. Graphical user interfaces should give the opportunity to change the data fields in the objects and check the input for correctness. The system should give the opportunity, to store the objects on harddisk (serialization). accessed directly. > My name is Nagesh and pursuing research studies in Refrigeration. At > present I am writing a Dynamic Refrigeration System Simulation Package. I > am using Mathematica as a programming language for the same since last one > year. I don't have any programming experience before this. I have following > querries:- > 1. Is any body here have expertise or information about the capability of > Mathematica as a system simulation tool? Since the most system simulation tools are simply solving a system of > ordinary differntial equations it is simple to do this with NDSolve[]. > 2. Is is possible to program a user friendly interface for my system > simulation package with Mathematica or I have to use some other software? Write a MathLink or J/Link frontend that launch the kernel. But you > should keep > in mind that the user interface is typical 80-90 % of your code. > If you just whant to solve some ode's it is probably easyer to > include one of the excelent ode-solvers from netlib in your C-code > than to call Mathematica to do that. As long as you dont wish to change > the ode's very often (than Mathematica is more flexible) you should > not use Mathematica. > 3. My refrigeration system simulation package is likely to have > approximately 60 First order Differential equations. Is is possible to > solve these in Mathematica ? Sure. > If yes then can anybody here guide me about > this further. Write down the equations and call NDSolve[]. > I am explaining below in short about the objectives I want to fulfill from > coding out of my main input file > 1. Example from Main Input File ( this will contain about 200-250 variables > which will be entered by the user of this package) This sounds like a *very* userfiendly interface ;-) > Below is examples of two variables entered into this file, which will be > used in other analysis files for further evaluation. > 2. Example from other analysis file ( there will be about 20-25 other such > component analysis files ) where the above mentioned variables from main > input file will be used for further evaluations:- > Below is one example from this file explaining how the variables from main > input file will be used in other files. > I hope that this short information will be useful for guiding me to solve > the following problems that I am facing. I am facing follwing problems or > objectives:- > 1. My 1st Objective:- The user of this package must be able to change only > the value of the variable in the main input file but he must not be able to > change the name of the variable itself. For example he must be able to > change the value of the variable but he must not be able to change the > name of this variable itself. > Here our problem is how to achieve or program it so that our objective will > be fullfilled. Options with defaulf values ? or something like {aParam,bParam}={ODEParameter1,ODEParameter2} /. > userRules /. > {ODEParameter1->1,ODEParameter2->2} > 2. My 2nd Objective:- How I can program the main input file so that it will > be user friendly in terms of its visuals and satisfying the constraint > mentioned above in objective1. What is *userfiendly* in a file with 250 variables ??? > 3. My 3rd Objective:- How can I program the optional values for each > variable in the main input file ? so that there will be always a value > assigned to each variable listed in main input file whenever the user opens > up this file. If user want to change the values of some variables then he > can change them and run the simulation otherwise the simulation run will be > done with optional values assigned to each variable in the input file. See above. > 4. My 4th Objective:- How can I program the check for correctness of the > input values supplied by the package user ? And @@ (NumericQ /@ {aListOfAllYourNumericParameters}) Jens > ==== See my comments in the following text! Hermann Schmitt ----- Original Message ----- > Is any body here have expertise or information about the capability of > Mathematica as a system simulation tool? > Mr. Kuska answers: > Since the most system simulation tools are simply solving a system of > ordinary differntial equations it is simple to do this with NDSolve[]. > My comment: > That is: He sees the simulation system merely as a set of differential > equations. hmm, since the original poster write My refrigeration system simulation package is likely to have > approximately 60 First order Differential equations. it seems not completly wrong to assume that the system consists of > of ode's .. > Yout should not ignore the word merely. It is not enough to have a set 60 differential equations and a set of 200-250 numbers. That is not simulation system, which can be used by users with the exception, perhaps, of the programmer of the system himself. How does e.g. the user know what meaning a number in the set has, ought he count the numbers from the beginning? Your nice command shows only, if there is an input, which is not a number. But I think the user would like to know, which of the 200 elements are not numbers. The only good of your command is, that it looks nice and shows your knowledge! > The question of Mr. Nagesh: > My 4th Objective:- How can I program the check for correctness of the > input values supplied by the package user ? > The answer of Mr. Kuska is: > And @@ (NumericQ /@ {aListOfAllYourNumericParameters}) > My comment: > This is a nice command and shows the knowledge of Mr.Kuska. But does Mr. > Nagesh understand it and is it sufficient to check, if all inputs are > numerical? It seems you have a deeper knowlege about the things that Mr. Nagesh > understand. You know him personally ? It is not very polite to > make speculations *what* a other person understand. And no it is not sufficent to check that all parameters are numbers. > Typical paramters described by intervals, where the assumptions of > a model are valid. But for this checks one needs more knowlege > about the meaning of the parameters. And one needs the knowlege about > the differntial equations, to find out the eigenvalues of the jacobian > ... > Additionally I think, it is not userfriendly to see the input merely as a > set of 200-250 numbers. > My suggestion is, that JLink is used, a suggestion Mr. Kusk takes into > consideration, too. That will be fast as lightning ! > But further I suggest, that classes are defined in Java, which represent the > parts of the system. That is notable nonsense! When the differntial equations should be > solved with Mathematica, the parts can't be Java classes. Mathematica's > NDSolve[] need a explicit expression to integrate the equations. That's not nonsense, the Mathematica program does not fetch the classes but the numbers in the classes (or better in the objects). the Java classes in textform. They can then be fetched from the Mathematica program and transformed into expressions by the Command ToExpression. The aim is, to have a clear separation of the system into components, which are manageable and understandable. OK you can call a Java class member from Mathematica but this will > be incredible slow when the right hand side is evaluated 200-300 > times and every evaluation make several callbacks to the Java source. > Event handler of the Java main program (without some modification in > the event loop) while it is evaluating an other command. My idea is to fetch the values once from the Java objects at the beginning. 200-250 numbers is not so much.. > Constructors of the classes should build objects with default values. That's a great idea. If a simulation run should be documented, one > always > need the full listing of the Java source to find the actual parameter > settings. That is not my opinion. I think not every user of the simulation system should have to know Java and Mathematica. The user must look for the values in the objects. And the values are in the objects, if they come from the constructor or from the graphical user interface. I think, there should be listings of the objects including names of the variables. In the objects the values are in an meaningful environment. > Graphical user interfaces > should give the opportunity to change the data fields in the objects and > check the input for correctness. *and* what has a GUI for 200-250 variables to do with Mathematica ? > BTW one has typical much less variables because many parameters > are fixed and it make no sense to change, for example, material > constants of materials that can't exchanged > The system should give the opportunity, to store the objects on harddisk > (serialization). > accessed directly. Can you be so kind, to explain *how* your posting help a person that > ask How can I build a simulation system with Mathematica ? You *can* say Don't use Mathematica, use Java! but this has nothing > to do with the question or with my reply. It is you, who proposes to solve the problem with C/C++ and not to use Mathematica (see below!) My point of view is: Use Mathematica, for what Mathemtica is good, and Java, for what Java is good. Mathematica is not so good as Java for data entry and Java is better than Mathematica to represent the simulation system (by objects). But I still would suggest to use C/C++ and a numerical > ode-solver, make a fancy GUI/Script > interface and don't use Mathematica for such a simple task. > The ode-solver is the smallest part of such a simulation system. Jens > > My name is Nagesh and pursuing research studies in Refrigeration. At > present I am writing a Dynamic Refrigeration System Simulation Package. > I > am using Mathematica as a programming language for the same since last > one > year. I don't have any programming experience before this. I have > following > querries:- > 1. Is any body here have expertise or information about the capability > of > Mathematica as a system simulation tool? > Since the most system simulation tools are simply solving a system of > ordinary differntial equations it is simple to do this with NDSolve[]. > 2. Is is possible to program a user friendly interface for my system > simulation package with Mathematica or I have to use some other > software? > Write a MathLink or J/Link frontend that launch the kernel. But you > should keep > in mind that the user interface is typical 80-90 % of your code. > If you just whant to solve some ode's it is probably easyer to > include one of the excelent ode-solvers from netlib in your C-code > than to call Mathematica to do that. As long as you dont wish to change > the ode's very often (than Mathematica is more flexible) you should > not use Mathematica. > 3. My refrigeration system simulation package is likely to have > approximately 60 First order Differential equations. Is is possible to > solve these in Mathematica ? > Sure. > If yes then can anybody here guide me about > this further. > Write down the equations and call NDSolve[]. > > I am explaining below in short about the objectives I want to fulfill > from > coding out of my main input file > 1. Example from Main Input File ( this will contain about 200-250 > variables > which will be entered by the user of this package) > This sounds like a *very* userfiendly interface ;-) > > Below is examples of two variables entered into this file, which will be > used in other analysis files for further evaluation. > 2. Example from other analysis file ( there will be about 20-25 other > such > component analysis files ) where the above mentioned variables from main > input file will be used for further evaluations:- > Below is one example from this file explaining how the variables from > main > input file will be used in other files. > I hope that this short information will be useful for guiding me to > solve > the following problems that I am facing. I am facing follwing problems > or > objectives:- > 1. My 1st Objective:- The user of this package must be able to change > only > the value of the variable in the main input file but he must not be able > to > change the name of the variable itself. For example he must be able to > change the value of the variable but he must not be able to change > the > name of this variable itself. > Here our problem is how to achieve or program it so that our objective > will > be fullfilled. > Options with defaulf values ? or something like > {aParam,bParam}={ODEParameter1,ODEParameter2} /. > userRules /. > {ODEParameter1->1,ODEParameter2->2} > > 2. My 2nd Objective:- How I can program the main input file so that it > will > be user friendly in terms of its visuals and satisfying the constraint > mentioned above in objective1. > What is *userfiendly* in a file with 250 variables ??? > > 3. My 3rd Objective:- How can I program the optional values for each > variable in the main input file ? so that there will be always a value > assigned to each variable listed in main input file whenever the user > opens > up this file. If user want to change the values of some variables then > he > can change them and run the simulation otherwise the simulation run will > be > done with optional values assigned to each variable in the input file. > See above. > > 4. My 4th Objective:- How can I program the check for correctness of the > input values supplied by the package user ? > And @@ (NumericQ /@ {aListOfAllYourNumericParameters}) > > Jens > Reply-To: kuska@informatik.uni-leipzig.de ==== > I agree with Mr. Kuska, that the system Mr Nagesh describes is not > userfriendly. But I think, the suggestions of Mr. Kuska do not make it more > userfriendly, rather the opposite is true. Mr. Nagesh asks Is any body here have expertise or information about the capability of > Mathematica as a system simulation tool? > Mr. Kuska answers: > Since the most system simulation tools are simply solving a system of > ordinary differntial equations it is simple to do this with NDSolve[]. My comment: That is: He sees the simulation system merely as a set of differential > equations. hmm, since the original poster write My refrigeration system simulation package is likely to have approximately 60 First order Differential equations. it seems not completly wrong to assume that the system consists of of ode's .. The question of Mr. Nagesh: My 4th Objective:- How can I program the check for correctness of the > input values supplied by the package user ? > The answer of Mr. Kuska is: > And @@ (NumericQ /@ {aListOfAllYourNumericParameters}) My comment: This is a nice command and shows the knowledge of Mr.Kuska. But does Mr. > Nagesh understand it and is it sufficient to check, if all inputs are > numerical? It seems you have a deeper knowlege about the things that Mr. Nagesh understand. You know him personally ? It is not very polite to make speculations *what* a other person understand. And no it is not sufficent to check that all parameters are numbers. Typical paramters described by intervals, where the assumptions of a model are valid. But for this checks one needs more knowlege about the meaning of the parameters. And one needs the knowlege about the differntial equations, to find out the eigenvalues of the jacobian ... Additionally I think, it is not userfriendly to see the input merely as a > set of 200-250 numbers. My suggestion is, that JLink is used, a suggestion Mr. Kusk takes into > consideration, too. That will be fast as lightning ! But further I suggest, that classes are defined in Java, which represent the > parts of the system. That is notable nonsense! When the differntial equations should be solved with Mathematica, the parts can't be Java classes. Mathematica's NDSolve[] need a explicit expression to integrate the equations. OK you can call a Java class member from Mathematica but this will be incredible slow when the right hand side is evaluated 200-300 times and every evaluation make several callbacks to the Java source. Event handler of the Java main program (without some modification in the event loop) while it is evaluating an other command. Constructors of the classes should build objects with default values. That's a great idea. If a simulation run should be documented, one always need the full listing of the Java source to find the actual parameter settings. > Graphical user interfaces > should give the opportunity to change the data fields in the objects and > check the input for correctness. *and* what has a GUI for 200-250 variables to do with Mathematica ? BTW one has typical much less variables because many parameters are fixed and it make no sense to change, for example, material constants of materials that can't exchanged The system should give the opportunity, to store the objects on harddisk > (serialization). accessed directly. > Can you be so kind, to explain *how* your posting help a person that ask How can I build a simulation system with Mathematica ? You *can* say Don't use Mathematica, use Java! but this has nothing to do with the question or with my reply. But I still would suggest to use C/C++ and a numerical ode-solver, make a fancy GUI/Script interface and don't use Mathematica for such a simple task. The ode-solver is the smallest part of such a simulation system. Jens > > My name is Nagesh and pursuing research studies in Refrigeration. At > present I am writing a Dynamic Refrigeration System Simulation Package. > I > am using Mathematica as a programming language for the same since last > one > year. I don't have any programming experience before this. I have > following > querries:- > 1. Is any body here have expertise or information about the capability > of > Mathematica as a system simulation tool? > Since the most system simulation tools are simply solving a system of > ordinary differntial equations it is simple to do this with NDSolve[]. > 2. Is is possible to program a user friendly interface for my system > simulation package with Mathematica or I have to use some other > software? > Write a MathLink or J/Link frontend that launch the kernel. But you > should keep > in mind that the user interface is typical 80-90 % of your code. > If you just whant to solve some ode's it is probably easyer to > include one of the excelent ode-solvers from netlib in your C-code > than to call Mathematica to do that. As long as you dont wish to change > the ode's very often (than Mathematica is more flexible) you should > not use Mathematica. > 3. My refrigeration system simulation package is likely to have > approximately 60 First order Differential equations. Is is possible to > solve these in Mathematica ? > Sure. > If yes then can anybody here guide me about > this further. > Write down the equations and call NDSolve[]. > > I am explaining below in short about the objectives I want to fulfill > from > coding out of my main input file > 1. Example from Main Input File ( this will contain about 200-250 > variables > which will be entered by the user of this package) > This sounds like a *very* userfiendly interface ;-) > > Below is examples of two variables entered into this file, which will be > used in other analysis files for further evaluation. > 2. Example from other analysis file ( there will be about 20-25 other > such > component analysis files ) where the above mentioned variables from main > input file will be used for further evaluations:- > > Below is one example from this file explaining how the variables from > main > input file will be used in other files. > I hope that this short information will be useful for guiding me to > solve > the following problems that I am facing. I am facing follwing problems > or > objectives:- > 1. My 1st Objective:- The user of this package must be able to change > only > the value of the variable in the main input file but he must not be able > to > change the name of the variable itself. For example he must be able to > change the value of the variable but he must not be able to change > the > name of this variable itself. > Here our problem is how to achieve or program it so that our objective > will > be fullfilled. > Options with defaulf values ? or something like > {aParam,bParam}={ODEParameter1,ODEParameter2} /. > userRules /. > {ODEParameter1->1,ODEParameter2->2} > > 2. My 2nd Objective:- How I can program the main input file so that it > will > be user friendly in terms of its visuals and satisfying the constraint > mentioned above in objective1. > What is *userfiendly* in a file with 250 variables ??? > > 3. My 3rd Objective:- How can I program the optional values for each > variable in the main input file ? so that there will be always a value > assigned to each variable listed in main input file whenever the user > opens > up this file. If user want to change the values of some variables then > he > can change them and run the simulation otherwise the simulation run will > be > done with optional values assigned to each variable in the input file. > See above. > > 4. My 4th Objective:- How can I program the check for correctness of the > input values supplied by the package user ? > And @@ (NumericQ /@ {aListOfAllYourNumericParameters}) > Jens > Reply-To: kuska@informatik.uni-leipzig.de ==== > My name is Nagesh and pursuing research studies in Refrigeration. At > present I am writing a Dynamic Refrigeration System Simulation Package. I > am using Mathematica as a programming language for the same since last one > year. I don't have any programming experience before this. I have following > querries:- > 1. Is any body here have expertise or information about the capability of > Mathematica as a system simulation tool? Since the most system simulation tools are simply solving a system of ordinary differntial equations it is simple to do this with NDSolve[]. > 2. Is is possible to program a user friendly interface for my system > simulation package with Mathematica or I have to use some other software? Write a MathLink or J/Link frontend that launch the kernel. But you should keep in mind that the user interface is typical 80-90 % of your code. If you just whant to solve some ode's it is probably easyer to include one of the excelent ode-solvers from netlib in your C-code than to call Mathematica to do that. As long as you dont wish to change the ode's very often (than Mathematica is more flexible) you should not use Mathematica. > 3. My refrigeration system simulation package is likely to have > approximately 60 First order Differential equations. Is is possible to > solve these in Mathematica ? Sure. > If yes then can anybody here guide me about > this further. Write down the equations and call NDSolve[]. I am explaining below in short about the objectives I want to fulfill from > coding out of my main input file 1. Example from Main Input File ( this will contain about 200-250 variables > which will be entered by the user of this package) This sounds like a *very* userfiendly interface ;-) Below is examples of two variables entered into this file, which will be > used in other analysis files for further evaluation. 2. Example from other analysis file ( there will be about 20-25 other such > component analysis files ) where the above mentioned variables from main > input file will be used for further evaluations:- Below is one example from this file explaining how the variables from main > input file will be used in other files. I hope that this short information will be useful for guiding me to solve > the following problems that I am facing. I am facing follwing problems or > objectives:- 1. My 1st Objective:- The user of this package must be able to change only > the value of the variable in the main input file but he must not be able to > change the name of the variable itself. For example he must be able to > change the value of the variable but he must not be able to change the > name of this variable itself. > Here our problem is how to achieve or program it so that our objective will > be fullfilled. Options with defaulf values ? or something like {aParam,bParam}={ODEParameter1,ODEParameter2} /. userRules /. {ODEParameter1->1,ODEParameter2->2} 2. My 2nd Objective:- How I can program the main input file so that it will > be user friendly in terms of its visuals and satisfying the constraint > mentioned above in objective1. What is *userfiendly* in a file with 250 variables ??? 3. My 3rd Objective:- How can I program the optional values for each > variable in the main input file ? so that there will be always a value > assigned to each variable listed in main input file whenever the user opens > up this file. If user want to change the values of some variables then he > can change them and run the simulation otherwise the simulation run will be > done with optional values assigned to each variable in the input file. See above. 4. My 4th Objective:- How can I program the check for correctness of the > input values supplied by the package user ? And @@ (NumericQ /@ {aListOfAllYourNumericParameters}) > Jens ==== You would need to make assumptions about y, and you can't. The function and the limits have to take care of that, and when you try to do that, you end up with an expression that has different antiderivatives on different regions, for different values of x. So, you have to break it up. The assumption x > 0 implies that x is real, so Im[x]==0 is unnecessary. one = Integrate[Cosh[2 (y - x)] 2 y, {y, Min[x, 1/2], 1/2}, Assumptions -> {x > 0}]; two = Integrate[Cosh[2 (x - y)] 2 y, {y, 0, Min[x, 1/2]}, Assumptions -> {x > 0}]; one + two // FullSimplify Plot[{one, two, one + two}, {x, 0, 1/2}]; (1/4)*E^(-1 - 2*x)*(E + (-2 + E)*E^(4*x)) Bobby Treat -----Original Message----- there something implicit that I am missing in the Assumptions ? MS. -- 12:02am up 5:06, 1 user, load average: 0.54, 0.22, 0.08 Reply-To: spammers-go-here@yahoo.com ==== On Friday 27 September 2002 04:18 am, DrBob, as drbob@bigfoot.com, held forth the following in comp.soft-sys.math.mathematica () : > You would need to make assumptions about y, and you can't. The function > and the limits have to take care of that, and when you try to do that, > you end up with an expression that has different antiderivatives on > different regions, for different values of x. So, you have to break it > up. The assumption x > 0 implies that x is real, so Im[x]==0 is > unnecessary. one = Integrate[Cosh[2 (y - x)] > 2 y, {y, Min[x, 1/2], 1/2}, Assumptions -> {x > 0}]; > two = Integrate[Cosh[2 (x - y)] 2 > y, {y, 0, Min[x, 1/2]}, Assumptions -> {x > 0}]; > one + two // FullSimplify > Plot[{one, two, one + two}, {x, 0, 1/2}]; (1/4)*E^(-1 - 2*x)*(E + (-2 + E)*E^(4*x)) Bobby Treat -----Original Message----- I have been trying to integrate the following : Integrate[Cosh[2 Abs[x-y]] 2 y, {y,0,1/2}, > Assumptions->{Im[x]==0,x>0}] However, Mathematica chokes and simply returns the integral as > it is. However, if I > split up the integral into two portions, it quickly gives me an answer > for the parts. Is > there something implicit that I am missing in the Assumptions ? > MS. > ? Anyways, I will try what you suggest. -- 12:53pm up 27 min, 1 user, load average: 0.19, 0.15, 0.09 ==== Put this after the Plot statement, in the same cell: SelectionMove[EvaluationNotebook[], All, GeneratedCell] FrontEndTokenExecute[OpenCloseGroup] FrontEndTokenExecute[SelectionAnimate] Bobby Treat -----Original Message----- Do[Plot[Sin[t]*Sin[x], {x, 0, Pi}, PlotRange -> {{0, Pi}, {-1, 1}}, ImageSize -> 400]; SelectionMove[EvaluationNotebook[], All, GeneratedCell]; FrontEndExecute[{FrontEnd`SelectionAnimate[0.1]}]; FrontEndExecute[{FrontEndToken[Clear]}], {t, 0, 15, 0.1}] This works but the cell dividing line flashes on and off spoiling the animation and if there is anything in the cell below this jumps up and down. Is there a proper way of doing this? Hugh Goyder ==== >>So would it take about the same amont of time for the complete printout of digits? Of course it would take a few additional seconds to format the output... I think it would take FAR more time for a complete printout, and might crash the Front End. I was thinking about the fact that I calculated all those digits and then threw them away. I could save them with Save or DumpSave, and read them in with Get the next time I wanted any of them, although the file would be close to 70 MB (if not more). I may do that, in fact -- I have plenty of disk space. The next step would be to somehow reuse the stored digits if I wanted MORE digits. But how? The Bailey-Borwein-Plouffe Pi algorithm is an avenue of attack, since it can calculate digits far from the decimal point, without calculating those in between. Unfortunately, it calculates hexadecimal digits in that way, not decimal digits. (That's true for the version I've seen, anyway.) Still, I could take the stored digits, convert to hexadecimal, add more hexadecimal digits with the B-B-P algorithm, and then convert back to decimal. In both conversions, I'd have to be very cognizant of how much precision I end up with, but that shouldn't be too difficult. It might go faster if I store hexadecimal digits, as well as decimal digits, to eliminate one of those conversions at each increase in the number of digits. The next step would be to set up an application that allowed anyone to ping for digits across the Internet, and would return them if they're stored. Hasn't someone already done that? It seems as if someone would have. Bobby Treat -----Original Message----- > Could you tell me the CPU you used and its speed etc...i am curious, to > other programs out there. I used one processor of a dual 1GH Mac and got the same answer with the > following speed: 4.2 for Mac OS X (June 4, 2002) > oldmax = $MaxPrecision > 6 > 1. 10 > $MaxPrecision = Infinity > Infinity > With[{n = 2^26}, Timing[ > pd = RealDigits[N[Pi, n + 1], 10, 20, > 19 - n]; ]] > {28794.1 Second, Null} > MaxMemoryUsed[] > 512055204 > pd > {{3, 3, 8, 6, 3, 2, 2, 0, 8, 9, 6, 2, 2, 3, > > 4, 0, 9, 8, 0, 3}, -67108844} Tom Burton ==== I've been trying to get my Mathematica 4.1 properly configured. I set: ################################################################### /usr/local/mathematica/SystemFiles/FrontEnd/TextResources/X/Specific.tr: @@resource maxForXListFonts 10000 # xlsfonts | wc -l 5572 /etc/X11/XF86Config: FontPath /usr/X11R6/lib/X11/fonts/100dpi:unscaled FontPath /usr/X11R6/lib/X11/fonts/75dpi:unscaled FontPath /usr/X11R6/lib/X11/fonts/CID FontPath /usr/X11R6/lib/X11/fonts/Speedo FontPath /usr/X11R6/lib/X11/fonts/Type1 FontPath /usr/X11R6/lib/X11/fonts/URW FontPath /usr/X11R6/lib/X11/fonts/kwintv:unscaled FontPath /usr/X11R6/lib/X11/fonts/latin2/Type1 FontPath /usr/X11R6/lib/X11/fonts/local/mma/Type1 FontPath /usr/X11R6/lib/X11/fonts/local/mma/X:unscaled FontPath /usr/X11R6/lib/X11/fonts/misc:unscaled FontPath /usr/X11R6/lib/X11/fonts/misc/sgi:unscaled FontPath /usr/X11R6/lib/X11/fonts/truetype FontPath /usr/X11R6/lib/X11/fonts/uni:unscaled # ls -R /usr/X11R6/lib/X11/fonts/ | grep / /usr/X11R6/lib/X11/fonts/: /usr/X11R6/lib/X11/fonts/100dpi: /usr/X11R6/lib/X11/fonts/75dpi: /usr/X11R6/lib/X11/fonts/CID: /usr/X11R6/lib/X11/fonts/Speedo: /usr/X11R6/lib/X11/fonts/Type1: /usr/X11R6/lib/X11/fonts/URW: /usr/X11R6/lib/X11/fonts/encodings: /usr/X11R6/lib/X11/fonts/encodings/large: /usr/X11R6/lib/X11/fonts/kwintv: /usr/X11R6/lib/X11/fonts/latin2: /usr/X11R6/lib/X11/fonts/latin2/Type1: /usr/X11R6/lib/X11/fonts/local: /usr/X11R6/lib/X11/fonts/local/mma: /usr/X11R6/lib/X11/fonts/local/mma/Type1: /usr/X11R6/lib/X11/fonts/local/mma/X: /usr/X11R6/lib/X11/fonts/misc: /usr/X11R6/lib/X11/fonts/misc/sgi: /usr/X11R6/lib/X11/fonts/truetype: /usr/X11R6/lib/X11/fonts/uni: /usr/X11R6/lib/X11/fonts/util: ######################################################## When I open the Mathematica Book Reference Guide in the Help Browser, I get a beep and the message says: Unable to find font with family Helvetica, weight Bold, slant Plain, and size 26. Substituting Courier. Compared to the things which *were* broken, this is a minor problem. I can live with the beep. What I would now like to know is how to tell Mathematica what fonts to use by default. This seemingly simple question seems to have no simple answer. Could someone please help me. TIA, ^L Reply-To: kuska@informatik.uni-leipzig.de ==== I expect that the Helvetica request comes from the Frontend menues and this is lited I your X resources $TopDirectory/SystemFiles/FontEnd/SystemResources/X/XMathematica request the helvetica font: !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! ! ! GENERAL GUI STYLE / COLOR SETTINGS ! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! ! Choose your GUI widget background color. ! CDE: Comment this out. *background: #c0c0c0 ! Choose a font for buttons/text fields, etc. ! You can use xfontsel to get a listing of fonts. !*Menu*fontList: -*-helvetica-medium-r-*-*-*-100-*-*-*-*-*-* ! Or just set all the fonts at once. *fontList: -*-helvetica-medium-r-*-*-*-100-*-*-*-*-*-* Jens I've been trying to get my Mathematica 4.1 properly configured. I set: > ################################################################### > /usr/local/mathematica/SystemFiles/FrontEnd/TextResources/X/Specific.tr: > @@resource maxForXListFonts > 10000 # xlsfonts | wc -l > 5572 /etc/X11/XF86Config: > FontPath /usr/X11R6/lib/X11/fonts/100dpi:unscaled > FontPath /usr/X11R6/lib/X11/fonts/75dpi:unscaled > FontPath /usr/X11R6/lib/X11/fonts/CID > FontPath /usr/X11R6/lib/X11/fonts/Speedo > FontPath /usr/X11R6/lib/X11/fonts/Type1 > FontPath /usr/X11R6/lib/X11/fonts/URW > FontPath /usr/X11R6/lib/X11/fonts/kwintv:unscaled > FontPath /usr/X11R6/lib/X11/fonts/latin2/Type1 > FontPath /usr/X11R6/lib/X11/fonts/local/mma/Type1 > FontPath /usr/X11R6/lib/X11/fonts/local/mma/X:unscaled > FontPath /usr/X11R6/lib/X11/fonts/misc:unscaled > FontPath /usr/X11R6/lib/X11/fonts/misc/sgi:unscaled > FontPath /usr/X11R6/lib/X11/fonts/truetype > FontPath /usr/X11R6/lib/X11/fonts/uni:unscaled # ls -R /usr/X11R6/lib/X11/fonts/ | grep / > /usr/X11R6/lib/X11/fonts/: > /usr/X11R6/lib/X11/fonts/100dpi: > /usr/X11R6/lib/X11/fonts/75dpi: > /usr/X11R6/lib/X11/fonts/CID: > /usr/X11R6/lib/X11/fonts/Speedo: > /usr/X11R6/lib/X11/fonts/Type1: > /usr/X11R6/lib/X11/fonts/URW: > /usr/X11R6/lib/X11/fonts/encodings: > /usr/X11R6/lib/X11/fonts/encodings/large: > /usr/X11R6/lib/X11/fonts/kwintv: > /usr/X11R6/lib/X11/fonts/latin2: > /usr/X11R6/lib/X11/fonts/latin2/Type1: > /usr/X11R6/lib/X11/fonts/local: > /usr/X11R6/lib/X11/fonts/local/mma: > /usr/X11R6/lib/X11/fonts/local/mma/Type1: > /usr/X11R6/lib/X11/fonts/local/mma/X: > /usr/X11R6/lib/X11/fonts/misc: > /usr/X11R6/lib/X11/fonts/misc/sgi: > /usr/X11R6/lib/X11/fonts/truetype: > /usr/X11R6/lib/X11/fonts/uni: > /usr/X11R6/lib/X11/fonts/util: ######################################################## When I open the Mathematica Book Reference Guide in the Help Browser, I get > a beep and the message says: Unable to find font with family Helvetica, weight Bold, slant Plain, and > size 26. Substituting Courier. Compared to the things which *were* broken, this is a minor problem. I can > live with the beep. What I would now like to know is how to tell Mathematica what > fonts to use by default. This seemingly simple question seems to have no > simple answer. Could someone please help me. TIA, ^L ==== I expect that the Helvetica request comes from the > Frontend menues and this is lited I your X resources $TopDirectory/SystemFiles/FontEnd/SystemResources/X/XMathematica $PreferencesDirectory? This is where I expected to find this kind of thing in the first place. The idea of chaning user preferences in a system file bothers me. request the helvetica font: !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! > ! > ! GENERAL GUI STYLE / COLOR SETTINGS > ! > !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! ! Choose your GUI widget background color. > ! CDE: Comment this out. > *background: #c0c0c0 ! Choose a font for buttons/text fields, etc. > ! You can use xfontsel to get a listing of fonts. > !*Menu*fontList: -*-helvetica-medium-r-*-*-*-100-*-*-*-*-*-* ! Or just set all the fonts at once. > *fontList: -*-helvetica-medium-r-*-*-*-100-*-*-*-*-*-* I still don't know why Mathematica would ship with the font defaults arranged in such a way as to cause it to give me errors when viewing the help files. It makes me think something is still amiss. Were they expecting my system to have these already installed? Should I have other fonts installed? Where would I get them, if I wanted them? Jens > STH ==== I have spent far more time attempting to get Mathematica configured than I have using it. I've found the the product to be exceptionally difficult to As an example, I spent several hours trying to figure out how to tell Mathematica to understand the delete key in the way most contemporary systems understand it. I wanted to avoid modifying system configuration files such as: /usr/local/mathematica/SystemFiles/FrontEnd/TextResources/KeyEventTranslatio ns.tr I expected to be able to change something in my own ~/.Mathematic directory, but I could not figure out an obvious way to affect this modification. I want to adjust the font size used in the widgets, but again, I see no ovbious means of modifying these attributes. I suspect it can be accomplished by modifying the ~/.Mathematica/4.1/FrontEnd/init.m. Perhaps to an experienced Mathematica user, the syntax and semantics of this file are obvious. They aren't to me. I also find the overall look & feel of the interface to be archaic. I understand that Mathematica 4.1 was written years ago, and much of the desktops for Unix which exist today did not exist when it was written. I should be receiving Mathematica 4.2 in a few days, so perhaps these concerns will prove undounded. Something tells me that not a lot has changed in this respect. What have others observed? Reply-To: kuska@informatik.uni-leipzig.de ==== I can't understand your criticism. Open a terminal window and type >math the delete key work perfect, due to your terminal settings and a command line is the most modern interface I can imagine. It still comes with Mathematica 4.2 and is perfect as before. Jens I have spent far more time attempting to get Mathematica configured than I > have using it. I've found the the product to be exceptionally difficult to As an example, I spent several hours trying to figure out how to tell Mathematica to > understand the delete key in the way most contemporary systems understand > it. I wanted to avoid modifying system configuration files such as: > /usr/local/mathematica/SystemFiles/FrontEnd/TextResources/KeyEventTranslation s.tr I expected to be able to change something in my own ~/.Mathematic directory, > but I could not figure out an obvious way to affect this modification. I > want to adjust the font size used in the widgets, but again, I see no > ovbious means of modifying these attributes. I suspect it can be > accomplished by modifying the ~/.Mathematica/4.1/FrontEnd/init.m. Perhaps > to an experienced Mathematica user, the syntax and semantics of this file are > obvious. They aren't to me. I also find the overall look & feel of the interface to be archaic. I > understand that Mathematica 4.1 was written years ago, and much of the desktops for > Unix which exist today did not exist when it was written. I should be > receiving Mathematica 4.2 in a few days, so perhaps these concerns will prove > undounded. Something tells me that not a lot has changed in this respect. What have others observed? ==== I can't understand your criticism. Open a terminal window > and type >math the delete key work perfect, due to your terminal settings and > a command line is the most modern interface I can imagine. > It still comes with Mathematica 4.2 and is perfect as before. Jens > If I could use bash to edit the command line, recall the command history, and other sorts of things that bash is good at, that would make the command line attractive. There *are* certain advantages to the gui side of things. For example the M-k code completion. That can really speed things up with languages such as Java and C++. I don't believe there is any fundamental reason such a thing could not exist with the command line. IMHO, a well conceived XEmacs lisp package could provide a wonderful frontend. I played around with mma.el for a few minutes, but it wasn't acting like a normal XEmacs package, so I switched back to cursing the Mod1<->Ctrl inversion of the Mathematica frontend. gui front end, get in the way of what really matters. The last thing I Mathematica. What I want to do is improve the overal experience for all of us. STH ==== Dear friends I have the following problem with Legend and LogPlot and LogPlotPlot: Needs[Graphics`Graphics`] Needs[Graphics`Legend`] {q1[t_],q2[t_],q3[t_]}={0.1 Exp[-0.02 t], 0.2 Exp[-0.025 t], 0.4 Exp[-0.028 t]}; (*With Plot legend works fine*) Plot[{q1[t],q2[t],q3[t]}, {t, 0, 100},PlotStyle[Rule]{ {AbsoluteThickness[0.5], AbsoluteDashing[{4,4}]}, AbsoluteThickness[1.5], {AbsoluteThickness[2], AbsoluteDashing[{1,8}]}}, AxesLabel[Rule]{Y, X}, PlotLabel[Rule]Title, PlotLegend[Rule]{1,3,5}, LegendPosition[Rule] {0.5,0}] (*However with LogPlot or LogLogPlot the legend desappear*) LogLogPlot[{q1[t],q2[t],q3[t]}, {t, 0, 100},PlotStyle[Rule]{ {AbsoluteThickness[0.5], AbsoluteDashing[{4,4}]}, AbsoluteThickness[1.5], {AbsoluteThickness[2], AbsoluteDashing[{1,8}]}}, AxesLabel[Rule]{Y, X}, PlotLabel[Rule]Title, PlotLegend[Rule]{1,3,5}, LegendPosition[Rule] {0.5,0}] I have shown a particular case, but I has this problem always with Legend and LogPlot and LogPlotPlot. I will appreciate any help. Guillermo Sanchez --------------------------------------------- This message was sent using Endymion MailMan. ==== IIRC, there is a way to get a list of all the symbols defined in the currently running session. I can't seem to find the reference to that command. Could somone point me in the direction of documentation which will tell me how to get information about the current session? TIA, Reply-To: kuska@informatik.uni-leipzig.de ==== Mathematica's PostScript does not need a 3d graphics card and will not use it. Only the RealTime3D` package make use of the 3d graphics and it's memory. Since it make an off screen rendering it will be slow and since one must be mad to use it for 10^5 polygons it does not matter how much graphics memory you have. The tiny polygon counts of Plot3D[] & friends can be handled wit 8 MByte :-) MathGL3d will run fine (not fast) with 8 MByte. For the most MathGL3d applications 64 MByte sufficient. With more than 10^6 polygons or many huge (1024^2) textures you may run into problems. For the MathGL3d development I have GeForce 3/4 cards with 128 MByte RAM. BTW the RAM of the 3d graphics cards is for textures and Mathematica does not know what a texture is or how usefull it is for scientific visualization. Jens We are about to order new PCs for a university student lab in which > Mathematica will be installed. Of course they will be using 2D and 3D > graphics -- plots of surfaces, e.g. Sooner or later students will want > to rotate such plots, too. An unresolved issue is how much graphics RAM to get. On existing > machines we typically have 64MB. But for the PCs we are looking at, > manufacturer's limits on graphics RAM, rather than cost, seems to limit > us to 32 MB. Is 32 MB adequate not just now, but likely to be adequate as well for > the near future (say, a 3- to 5-year equipment lifetime)? -- > Murray Eisenberg murray@math.umass.edu > Mathematics & Statistics Dept. > Lederle Graduate Research Tower phone 413 549-1020 (H) > University of Massachusetts 413 545-2859 (W) > 710 North Pleasant Street > Amherst, MA 01375 Reply-To: kuska@informatik.uni-leipzig.de ==== Names[Global`*] ?? Jens IIRC, there is a way to get a list of all the symbols defined in the > currently running session. I can't seem to find the reference to that > command. Could somone point me in the direction of documentation which > will tell me how to get information about the current session? TIA, ==== To get a complete list of all variables that have been defined for the current Mathematica session, please try: ?`* which should return a list of all variables that have been defined for the current Mathematica session. Steven Shippee ==== > IIRC, there is a way to get a list of all the symbols defined in the > currently running session. I can't seem to find the reference to that > command. Could somone point me in the direction of documentation which > will tell me how to get information about the current session? > TIA, > Let's begin with a fresh session Quit Make some entries, notice that b, x and y have no definitions - they are simply created. a=3; b; p= 3x +1; f[y_]:=y^2; We can find all the symbols we have created so far. Names[`*] {a,b,f,p,x,y} Actually, they are the strings of the symbols (otherwise, for example, a would immediately evaluate to 3). InputForm[%] {a, b, f, p, x, y} How can we take out the strings of the undefined symbols? I make a function that test if the symbol has been defined (or has an attribute assigned): SetAttributes[definedQ, HoldFirst]; definedQ[x_String]:= Or[DownValues@@#=!={}, UpValues@@#=!={},OwnValues@@#=!={}, SubValues@@#=!={},DefaultValues@@#=!={},NValues@@#=!={}, Attributes@@#=!={}]&[ ToExpression[x, InputForm, Hold]] definedQ[x_]:= definedQ[Evaluate[ToString[Unevaluated[x]]]] Using this we get Select[Names[`*], definedQ] {a,definedQ,f,p} To get information about the symbols Information/@Select[Names[`*], definedQ]; a a = 3 definedQ Attributes[definedQ] = {HoldFirst} definedQ[x_String] := (DownValues @@ #1 =!= {} || UpValues @@ #1 =!= {} || OwnValues @@ #1 =!= {} || SubValues @@ #1 =!= {} || DefaultValues @@ #1 =!= {} || NValues @@ #1 =!= {} & )[ToExpression[x, InputForm, Hold]] definedQ[x_] := definedQ[Evaluate[ ToString[Unevaluated[x]]]] f f[y_] := y^2 p p = 1 + 3*x I have assumed that you are interested only in the symbols in the symbols in the default context Global`. Other context can be provided for. -- 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 ==== There are several ways (note that newly defined things are by default in Global` context) , one would be ?@ bye, Borut | IIRC, there is a way to get a list of all the symbols defined in the | currently running session. I can't seem to find the reference to that | command. Could somone point me in the direction of documentation which | will tell me how to get information about the current session? | | | TIA, | ==== Dear all, As far as I understand, FindMinimum can be requested to use the Berndt-Hall-Hall-Hausman method, if the option Method->QuasiNewton is specified. I am pretty convinced that there is no way to tell mathematica to produce as an output the Jacobian vector and the Hessian matrix at every step. In addition, I have not found a way to specify the Davidon-Fletcher-Powell algorithm to be used. Before I start to program it myself, I was wondering if anybody has written some [non-commercial] code that produces such an output for BFGS and even better for DFP optimization. Kyriakos _____+**+____+**+___+**+__+**+_ Kyriakos Chourdakis Lecturer in Financial Economics URL: http://www.qmw.ac.uk/~te9001 tel: (++44) (+20) 7882 5086 Dept of Economics University of London, QM London E1 4NS U.K. ==== Both Simplify and ImpliesQ need the assumption that all variables are real in order to use the CAD algorithm. However, Simplify uses many other methods trying to simplify each subexpression of the input, while ImpliesQ only tries to prove that the whole input is implied by the assumptions. Here is the series of simplifications that Simplify does in this example. - An inequality simplification heuristic using the inequality assumption y6 >= -1 is applied to the whole system. We get simplification: y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6 -> 1 + y4 >= 0 && y5 >= y6 && 1 + y4 + y6 >= y5 - We go into simplification of subexpressions, and get these simplifications using reduction modulo a GroebnerBasis of equation assumptions y4 == -1 and y5 == y6 (Simplify keeps the result of this transformation even if it has the same complexity as the original expression.) 1 + y4 -> 0 y5 -> y6 1 + y4 + y6 -> y6 y5 -> y6 - As the result of subexpression simplifications the three inequalities become then: 1 + y4 >= 0 -> 0 >= 0 y5 >= y6 -> y6 >= y6 1 + y4 + y6 >= y5 -> y6 >= y6 - 0 >= 0 evaluates to True, y6 >= y6 is simplified to True using transformation x >= y -> x - y >= 0. Adam Strzebonski Wolfram Research That makes everything clear, except for just one small mystery: In[1]:= > << Experimental` In[2]:= > FullSimplify[y4 >= -1 && y6 >= -1 && > y6 <= y5 <= 1 + y4 + y6, y4 == -1 && y6 >= -1 && y5 == y6] Out[2]= > True In[3]:= > ImpliesQ[y4 == -1 && y6 >= -1 && y5 == y6, > y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6] Out[3]= > False I now understand why the last one returns False, but why does the > second one return True? Should not the same arument apply in both > cases? Or is it because FullSimplify does not actually need the > assumption that the variables are real while ImliesQ does? Andrzej > Actually, the reason why ImpliesQ (and FullSimplify) fail to > prove the implication is not that the hypothesis is a disjunction. > To use the cylindrical algebraic decomposition algorithm they > need to know that the assumptions imply that all variables are > real. > The assumptions mechanism infers variable domains in a purely > syntactical way, i.e. v is assumed to be real if there is > an Element[v, Reals] statement or v appears in an inequality. > It does not attempt to analyze assumptions further, to figure > out that, say y6 >= -1 implies that y6 is real, and then if > we have y5 == y6 then y5 must be real too. Doing such an analysis > in general would require solving the assumptions over complex > numbers, and then finding out which variables need to be real. > This would be in general too time consuming to do, but analyzing > linear dependencies like the ones in your example is a possible > future improvement. > ImpliesQ cannot prove the implication here, because it knows only > that y6 is real. > In[1]:= < In[2]:= ImpliesQ[y4 == -1 && y6 >= -1 && y5 == y6, > y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6] > Out[2]= False > If we add an explicit assumption that y4 and y5 are real, ImpliesQ > (and FullSimplify) can prove this implication, and the full version > of your example. > In[3]:= ImpliesQ[Element[y4|y5, Reals] && y4 == -1 && y6 >= -1 && y5 == > y6, > y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6] > Out[3]= True > In[4]:= ImpliesQ[Element[y4|y5, Reals] && (y4 == -1 && y6 >= -1 && > y5 == y6 || y4 > -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6), > y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6] > Out[4]= > True > In[5]:= FullSimplify[y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6, > Element[y4|y5, Reals] && (y4 == -1 && y6 >= -1 && y5 == y6 || > y4 > -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6)] > Out[5]= > True > > Adam Strzebonski > Wolfram Research > On second thoughts I realized that there seems to be an inherent > ambiguity about what one coudl mean by using alternatives (statements > joned by Or) assumptions. In fact it now seems to me that the > reasonable intertpretation for ImpliesQ and FullSimplify ought to > perhaps be different. It seems to me that ImpliesQ[Or[a,b],c] ought to > return True if aand only if ImpliesQ[a,c] and ImpliesQ[b,c] both > return > True. If so this could be acomplished by adding the rule > ImpliesQ[Or[a,b],c] = And[ImpliesQ[a,c],ImpliesQ[b,c]]. That could > then > be used in proving that the two answers to the system of inequalities > that of Vincent's original posting are equivalent. On the other hand > probably FullSimplify[a, Or[p,q]] ought to return > Or[FullSimplify[a,p],FullSimplify[a,q]] (or do nothing as it doe > snow). > The first approach would seem to be consistent with the way > FullSimplify works with domain specifications but would however have > the strange effect of returning True if just one of the alternatives > were true and the other false. So perhaps after all it is best to > leave FullSimplify as it is. However, it seems to me that ImpliesQ > shoud be able to handle such cases (?) > Andrzej Kozlowski > Toyama International University > JAPAN > The modification to FullSimplify that I sent earlier works correctly > only for assumptions of the form Or[a,b] (and even then not is not > always what one would like). For what it's worth here is a better > (but > slow) version: > In[1]:= > Unprotect[FullSimplify]; > In[2]:= > FullSimplify[expr_, x_ || y__] := FullSimplify[ > FullSimplify[expr, x] || FullSimplify[expr, Or[y]]]; > In[3]:= > Protect[FullSimplify]; > For example: > In[4]:= > FullSimplify[Sqrt[(x - 1)^2] + Sqrt[(x - 2)^2] + > Sqrt[(x - 3)^2], x > 1 || x > 2 || x > 3] > Out[4]= > -1 + x + Abs[-3 + x] + Abs[-2 + x] || > -3 + 2*x + Abs[-3 + x] || 3*(-2 + x) > Andrzej Kozlowski > Toyama International University > JAPAN > On Thursday, September 26, 2002, at 11:14 AM, Andrzej Kozlowski > The reason why InequalitySolve returns it's answer in what sometimes > turns out to be unnecessarily complicated form is that the > underlying > algorithm, Cylindrical Agebraic Decomposition (CAD) returns its > answers in this form. Unfortunately it seems to me unlikely that a > simplification of the kind you need can be can be accomplished in > any > general way. To see why observe the following. First of all: > In[1]:= > FullSimplify[x > 0 || x == 0] > Out[1]= > x >= 0 > This is fine. However: > In[2]:= > FullSimplify[x > 0 && x < 2 || x == 0 && x < 2] > Out[2]= > x == 0 || 0 < x < 2 > Of course what you would like is simply 0 <= x < 2. One reason why > you can't get it is that while Mathematica can perform a > LogicalExpand, as in: > In[3]:= > LogicalExpand[(x > 0 || x == 0) && x < 2] > Out[3]= > x == 0 && x < 2 || x > 0 && x < 2 > There i no LogicalFactor or anything similar that would reverse > what LogicalExpand does. if there was then you could perform the > sort > of simplifications you need for: > In[4]:= > FullSimplify[(x > 0 || x == 0) && x < 2] > Out[4]= > 0 <= x < 2 > However, it does not seem to me very likely that such logical > factoring can be performed by a general enough algorithm (though I > am no expert in this field). In any case, certainly Mathematica > can't > do this. > I also noticed that Mathematica seems unable to show that the answer > it returns to your problem is actually equivalent to your simpler > one. In fact this looks like a possible bug in Mathematica. Let's > first try the function ImpliesQ from the Experimental context: > << Experimental` > Now Mathematica correctly gives: > In[6]:= > ImpliesQ[y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6, > y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1 && y6 <= > y5 > <= 1 + y4 + y6] > Out[6]= > True > However: > In[7]:= > ImpliesQ[y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1 && > y6 <= y5 <= 1 + y4 + y6, y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + > y4 + y6] > Out[7]= > False > That simply means that ImpliesQ cannot show the implication, not > that > it does not hold. ImpliesQ relies on CAD, as does FullSimplify. > Switching to FullSimplify we see that: > In[8]:= > FullSimplify[y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1 > && > y6 <= y5 <= 1 + y4 + y6, y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + > y4 + y6] > Out[8]= > True > while > In[9]:= > FullSimplify[y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6, > y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1 && y6 <= > y5 > <= 1 + y4 + y6] > Out[9]= > y4 >= -1 && y6 <= y5 <= 1 + y4 + y6 > On the other hand, taking just the individual summands of Or as > hypotheses; > In[10]:= > FullSimplify[y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6, > y4 > -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6] > Out[10]= > True > In[11]:= > FullSimplify[y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6, > y4 == -1 && y6 >= -1 && y5 == y6 ] > Out[11]= > True > In fact FullSimplify is unable to use Or in assumptions, which can > be > demonstrated on an abstract example: > In[12]:= > FullSimplify[C,(A||B)&&(C)] > Out[12]= > True > In[13]:= > FullSimplify[C,LogicalExpand[(A||B)&&(C)]] > Out[13]= > C > This could be fixed by modifying FullSimplify: > In[14]:= > Unprotect[FullSimplify]; > In[14]:= > FullSimplify[expr_,Or[x_,y__]]:=Or[FullSimplify[expr,x],FullSimplify > [e > xpr,y]]; > In[15]:= > Protect[FullSimplify]; > Now at least we get as before: > In[16]:= > FullSimplify[y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1 > && > y6 <= y5 <= 1 + y4 + y6, y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + > y4 + y6] > Out[16]= > True > but also: > In[17]:= > FullSimplify[y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6, > y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1 && y6 <= > y5 > <= 1 + y4 + y6] > Out[17]= > True > This seems to me a possible worthwhile improvement in FullSimplify, > though of course not really helpful for your problem. > Andrzej Kozlowski > Toyama International University > JAPAN > On Wednesday, September 25, 2002, at 02:51 PM, Vincent Bouchard > I have a set of inequalities that I solve with InequalitySolve. But > then > it gives a complete set of solutions, but not in the way I would > like it > to be! :-) For example, the simple following calculation will give: > In[1]:= ineq = {y4 >= -1, y5 >= -1, y6 + y4 >= y5 - 1, y5 >= y6, y6 > = -1}; > InequalitySolve[ineq,{y4,y6,y5}] > Out[1]:= y4 == -1 && y6 >= -1 && y5 == y6 || > y4 > -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6 > the result is good, but I would like it to be in the simpler but > equivalent form > y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6 > How can I tell InequalitySolve to do that? It is simple for this > example, > but for a large set of simple inequalities InequalitySolve gives > lines and > lines of results instead of a simple result. > Vincent Bouchard >> Andrzej Kozlowski > Toyama International University > JAPAN ==== That makes everything clear, except for just one small mystery: In[1]:= << Experimental` In[2]:= FullSimplify[y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6, y4 == -1 && y6 >= -1 && y5 == y6] Out[2]= True In[3]:= ImpliesQ[y4 == -1 && y6 >= -1 && y5 == y6, y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6] Out[3]= False I now understand why the last one returns False, but why does the second one return True? Should not the same arument apply in both cases? Or is it because FullSimplify does not actually need the assumption that the variables are real while ImliesQ does? Andrzej > Actually, the reason why ImpliesQ (and FullSimplify) fail to > prove the implication is not that the hypothesis is a disjunction. > To use the cylindrical algebraic decomposition algorithm they > need to know that the assumptions imply that all variables are > real. The assumptions mechanism infers variable domains in a purely > syntactical way, i.e. v is assumed to be real if there is > an Element[v, Reals] statement or v appears in an inequality. > It does not attempt to analyze assumptions further, to figure > out that, say y6 >= -1 implies that y6 is real, and then if > we have y5 == y6 then y5 must be real too. Doing such an analysis > in general would require solving the assumptions over complex > numbers, and then finding out which variables need to be real. > This would be in general too time consuming to do, but analyzing > linear dependencies like the ones in your example is a possible > future improvement. ImpliesQ cannot prove the implication here, because it knows only > that y6 is real. In[1]:= <= -1 && y5 == y6, > y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6] Out[2]= False If we add an explicit assumption that y4 and y5 are real, ImpliesQ > (and FullSimplify) can prove this implication, and the full version > of your example. In[3]:= ImpliesQ[Element[y4|y5, Reals] && y4 == -1 && y6 >= -1 && y5 == > y6, > y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6] Out[3]= True In[4]:= ImpliesQ[Element[y4|y5, Reals] && (y4 == -1 && y6 >= -1 && > y5 == y6 || y4 > -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6), > y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6] Out[4]= > True In[5]:= FullSimplify[y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6, > Element[y4|y5, Reals] && (y4 == -1 && y6 >= -1 && y5 == y6 || > y4 > -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6)] Out[5]= > True Adam Strzebonski > Wolfram Research > On second thoughts I realized that there seems to be an inherent > ambiguity about what one coudl mean by using alternatives (statements > joned by Or) assumptions. In fact it now seems to me that the > reasonable intertpretation for ImpliesQ and FullSimplify ought to > perhaps be different. It seems to me that ImpliesQ[Or[a,b],c] ought to > return True if aand only if ImpliesQ[a,c] and ImpliesQ[b,c] both > return > True. If so this could be acomplished by adding the rule > ImpliesQ[Or[a,b],c] = And[ImpliesQ[a,c],ImpliesQ[b,c]]. That could > then > be used in proving that the two answers to the system of inequalities > that of Vincent's original posting are equivalent. On the other hand > probably FullSimplify[a, Or[p,q]] ought to return > Or[FullSimplify[a,p],FullSimplify[a,q]] (or do nothing as it doe > snow). > The first approach would seem to be consistent with the way > FullSimplify works with domain specifications but would however have > the strange effect of returning True if just one of the alternatives > were true and the other false. So perhaps after all it is best to > leave FullSimplify as it is. However, it seems to me that ImpliesQ > shoud be able to handle such cases (?) > Andrzej Kozlowski > Toyama International University > JAPAN > The modification to FullSimplify that I sent earlier works correctly > only for assumptions of the form Or[a,b] (and even then not is not > always what one would like). For what it's worth here is a better > (but > slow) version: > In[1]:= > Unprotect[FullSimplify]; > In[2]:= > FullSimplify[expr_, x_ || y__] := FullSimplify[ > FullSimplify[expr, x] || FullSimplify[expr, Or[y]]]; > In[3]:= > Protect[FullSimplify]; > For example: > In[4]:= > FullSimplify[Sqrt[(x - 1)^2] + Sqrt[(x - 2)^2] + > Sqrt[(x - 3)^2], x > 1 || x > 2 || x > 3] > Out[4]= > -1 + x + Abs[-3 + x] + Abs[-2 + x] || > -3 + 2*x + Abs[-3 + x] || 3*(-2 + x) > Andrzej Kozlowski > Toyama International University > JAPAN > On Thursday, September 26, 2002, at 11:14 AM, Andrzej Kozlowski > The reason why InequalitySolve returns it's answer in what sometimes > turns out to be unnecessarily complicated form is that the > underlying > algorithm, Cylindrical Agebraic Decomposition (CAD) returns its > answers in this form. Unfortunately it seems to me unlikely that a > simplification of the kind you need can be can be accomplished in > any > general way. To see why observe the following. First of all: > In[1]:= > FullSimplify[x > 0 || x == 0] > Out[1]= > x >= 0 > This is fine. However: > In[2]:= > FullSimplify[x > 0 && x < 2 || x == 0 && x < 2] > Out[2]= > x == 0 || 0 < x < 2 > Of course what you would like is simply 0 <= x < 2. One reason why > you can't get it is that while Mathematica can perform a > LogicalExpand, as in: > In[3]:= > LogicalExpand[(x > 0 || x == 0) && x < 2] > Out[3]= > x == 0 && x < 2 || x > 0 && x < 2 > There i no LogicalFactor or anything similar that would reverse > what LogicalExpand does. if there was then you could perform the > sort > of simplifications you need for: > In[4]:= > FullSimplify[(x > 0 || x == 0) && x < 2] > Out[4]= > 0 <= x < 2 > However, it does not seem to me very likely that such logical > factoring can be performed by a general enough algorithm (though I > am no expert in this field). In any case, certainly Mathematica > can't > do this. > I also noticed that Mathematica seems unable to show that the answer > it returns to your problem is actually equivalent to your simpler > one. In fact this looks like a possible bug in Mathematica. Let's > first try the function ImpliesQ from the Experimental context: > << Experimental` > Now Mathematica correctly gives: > In[6]:= > ImpliesQ[y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6, > y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1 && y6 <= > y5 > <= 1 + y4 + y6] > Out[6]= > True > However: > In[7]:= > ImpliesQ[y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1 && > y6 <= y5 <= 1 + y4 + y6, y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + > y4 + y6] > Out[7]= > False > That simply means that ImpliesQ cannot show the implication, not > that > it does not hold. ImpliesQ relies on CAD, as does FullSimplify. > Switching to FullSimplify we see that: > In[8]:= > FullSimplify[y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1 > && > y6 <= y5 <= 1 + y4 + y6, y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + > y4 + y6] > Out[8]= > True > while > In[9]:= > FullSimplify[y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6, > y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1 && y6 <= > y5 > <= 1 + y4 + y6] > Out[9]= > y4 >= -1 && y6 <= y5 <= 1 + y4 + y6 > On the other hand, taking just the individual summands of Or as > hypotheses; > In[10]:= > FullSimplify[y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6, > y4 > -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6] > Out[10]= > True > In[11]:= > FullSimplify[y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6, > y4 == -1 && y6 >= -1 && y5 == y6 ] > Out[11]= > True > In fact FullSimplify is unable to use Or in assumptions, which can > be > demonstrated on an abstract example: > In[12]:= > FullSimplify[C,(A||B)&&(C)] > Out[12]= > True > In[13]:= > FullSimplify[C,LogicalExpand[(A||B)&&(C)]] > Out[13]= > C > This could be fixed by modifying FullSimplify: > In[14]:= > Unprotect[FullSimplify]; > In[14]:= > FullSimplify[expr_,Or[x_,y__]]:=Or[FullSimplify[expr,x],FullSimplify > [e > xpr,y]]; > In[15]:= > Protect[FullSimplify]; > Now at least we get as before: > In[16]:= > FullSimplify[y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1 > && > y6 <= y5 <= 1 + y4 + y6, y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + > y4 + y6] > Out[16]= > True > > but also: > In[17]:= > FullSimplify[y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6, > y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1 && y6 <= > y5 > <= 1 + y4 + y6] > Out[17]= > True > This seems to me a possible worthwhile improvement in FullSimplify, > though of course not really helpful for your problem. > Andrzej Kozlowski > Toyama International University > JAPAN > On Wednesday, September 25, 2002, at 02:51 PM, Vincent Bouchard > I have a set of inequalities that I solve with InequalitySolve. But > then > it gives a complete set of solutions, but not in the way I would > like it > to be! :-) For example, the simple following calculation will give: > In[1]:= ineq = {y4 >= -1, y5 >= -1, y6 + y4 >= y5 - 1, y5 >= y6, y6 > = -1}; > InequalitySolve[ineq,{y4,y6,y5}] > Out[1]:= y4 == -1 && y6 >= -1 && y5 == y6 || > y4 > -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6 > the result is good, but I would like it to be in the simpler but > equivalent form > y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6 > How can I tell InequalitySolve to do that? It is simple for this > example, > but for a large set of simple inequalities InequalitySolve gives > lines and > lines of results instead of a simple result. > Vincent Bouchard >> ==== inside a program I need to solve this linear equation in terms of p1. However something odds happens. Sometimes the solution is computed and sometimes the result is empty [I mean no output...]. Is this a bug of the solve command or am I doing something wrong? The problem is robust to: changing name to the variables and other makeups.. David ps: Sorry for the stupid way in which I copied the command... Solve[(x^2*((-0.9*x^7*(p^2*(-1 - 5.8*x^6 - 14.010000000000002*x^12 - 18.04*x^18 - 13.06*x^24 - 5.040000000000001*x^30 - 0.81*x^36) + x*(7.777777777777779 - 9.074074074074076*x + 30.333333333333336*x^6 - 21.51851851851852*x^7 - 16.333333333333336*x^8 + 44.33333333333334*x^12 + 3.188888888888883*x^13 - 65.68333333333332*x^14 + 28.777777777777786*x^18 + 47.937037037037044*x^19 - 100.10000000000002*x^20 + 7.*x^24 + 45.6037037037037*x^25 - 69.53333333333333*x^26 + 13.299999999999999*x^31 - 19.833333333333332*x^32 - 1.0499999999999996*x^38) + p*(-6 + 8.296296296296296*x - 28.799999999999997*x^6 + 32.785185185185185*x^7 + 9.333333333333336*x^8 - 55.260000000000005*x^12 + 49.04777777777776*x^13 + 38.38333333333334*x^14 - 52.980000000000004*x^18 + 34.20518518518518*x^19 + 60.20000000000001*x^20 - 25.380000000000003*x^24 + 11.736296296296294*x^25 + 43.63333333333334*x^26 - 4.86*x^30 + 2.8999999999999986*x^31 + 13.533333333333333*x^32 + 0.81*x^37 + 1.0499999999999996*x^38)))/(x + 1.9*x^7 + 0.9*x^13)^2 - ((-1 + p - 7*x^6 + p*x^6 + 6*x^7)*(1.2962962962962965 - 3.111111111111112*x^6 + 9.333333333333336*x^7 - 10.111111111111114*x^12 + 22.05*x^13 - 5.703703703703705*x^18 + 17.15*x^19 + 5.483333333333331*x^25 + 1.0499999999999996*x^31 + p1*x^5*(7.000000000000002 - 7.000000000000002*x + 14.000000000000004*x^6 - 14.000000000000004*x^7 + 7.000000000000002*x^12 - 6.999999999999998*x^13) - 1.166666666666667*p*x^4*x1 - 3.500000000000001*p*x^10*x1 - 1.0500000000000003*p*x^11*x1 - 3.500000000000001*p*x^16*x1 - 3.150000000000001*p*x^17*x1 - 1.166666666666667*p*x^22*x1 - 3.150000000000001*p*x^23*x1 - 1.0500000000000003*p*x^29*x1))/((1 + 0.9*x^6)^2*(1 + x^6)^2)))/(p^2*(1 + x^6)^3) == 0, p1] ==== You're right; I misunderstood your problem. First of all, never use Do -- forget it exists -- and don't put the SelectionMove, etc. commands inside a loop. That's your biggest problem. This should help: Table[Plot[ Sin[t]*Sin[x], {x, 0, Pi}, PlotRange -> {{0, Pi}, {-1, 1}}], {t, 0, 2Pi - Pi/32, Pi/32}]; SelectionMove[EvaluationNotebook[], All, GeneratedCell] FrontEndTokenExecute[OpenCloseGroup] FrontEndTokenExecute[SelectionAnimate] The first time through the animation is a bit slow because the frames are being generated, but then the cell group collapses and things are better. I used an increment that's an exact divisor of the period in order to catch the max and min values of Sin[t], as well as the zero value, on each swing. Use the period MINUS the step-size as the upper limit in order to avoid having a last frame identical to the first. I'm a bit annoyed at the tendency for the plot to jump near zero and Pi, but that's because the derivative of Sin is higher there, and we're not compensating by picking more points there. If we do pick more points there, however, we won't perceive t as time. If that's not a consideration, you could do it this way: f = Which[ -1 ? # ? 1, #, 1 < # ? 3, 2 - #, True, f@Mod[#, 4, -1] ] &; Table[Plot[ f[t]*Sin[x], {x, 0, Pi}, PlotRange -> {{0, Pi}, {-1, 1}}], { t, -1, 2.9, 0.1}]; SelectionMove[EvaluationNotebook[], All, GeneratedCell] FrontEndTokenExecute[OpenCloseGroup] FrontEndTokenExecute[SelectionAnimate] Bobby -----Original Message----- Put this after the Plot statement, in the same cell: SelectionMove[EvaluationNotebook[], All, GeneratedCell] FrontEndTokenExecute[OpenCloseGroup] FrontEndTokenExecute[SelectionAnimate] Bobby Treat -----Original Message----- Is it possible to tidy up the generation of the graphic so that it becomes the animation? I have tried the following Do[Plot[Sin[t]*Sin[x], {x, 0, Pi}, PlotRange -> {{0, Pi}, {-1, 1}}, ImageSize -> 400]; SelectionMove[EvaluationNotebook[], All, GeneratedCell]; FrontEndExecute[{FrontEnd`SelectionAnimate[0.1]}]; FrontEndExecute[{FrontEndToken[Clear]}], {t, 0, 15, 0.1}] This works but the cell dividing line flashes on and off spoiling the animation and if there is anything in the cell below this jumps up and down. Is there a proper way of doing this? Hugh Goyder ==== That's nice because it avoids watching the frames being slowly created, and the group doesn't always collapse as it should, the way I've been doing it. However, a couple of small changes give a smoother animation with 64 frames rather than 151, while GENERATING only 33 frames. Block[{$DisplayFunction = Identity, half, graphs, step = Pi/32}, half = Table[GraphicCell[Plot[Sin[t]*Sin[x], {x, 0, Pi}, PlotRange -> {{0, Pi}, {-1, 1}}, ImageSize -> 400]], {t, -Pi/2, Pi/2 - step, step}]; graphs = Rest@Join[half, Rest@Reverse@half]; NotebookWrite[EvaluationNotebook[], Cell[CellGroupData[graphs, Closed]]]; SelectionMove[EvaluationNotebook[], All, GeneratedCell]; FrontEndExecute[{FrontEndToken[EvaluationNotebook[], SelectionAnimate]}]] Bobby -----Original Message----- >showing during generation is enough (but uncontrolled). > >Hugh Goyder This creates a graphics cell from a graphics expression. GraphicCell[graphics_] := Cell[GraphicsData[PostScript, DisplayString[graphics]],Graphics] cellgroup. Block[{$DisplayFunction=Identity, graphs}, graphs = Table[GraphicCell[ Plot[Sin[t]*Sin[x], {x, 0, Pi}, PlotRange -> {{0, Pi}, {-1, 1}}, ImageSize -> 400]], {t,0,15,.1}]; NotebookWrite[EvaluationNotebook[],Cell[CellGroupData[graphs,Closed]]]; SelectionMove[EvaluationNotebook[], All, GeneratedCell]; FrontEndExecute[{FrontEndToken[EvaluationNotebook[], SelectionAnimate]}] ] -------------------------------------------------------------- Omega Consulting The final answer to your Mathematica needs Spend less time searching and more time finding. http://www.wz.com/internet/Mathematica.html ==== I'm woking on a kind of a Mathematica cheat-sheet. So I don't have to repeat the same learning process if I get pulled away for another 6 months. I've attempted to get my domain name to resolve to my IP address, but it seems Verisign and I have different ideas about what 24 hours is. The site is supposed to be www.globalsymmetry.com, but that will not currently resolve. Here's the IP and path: http://66.92.149.152/proprietary/com/wri/index.html This is not a literary masterpiece. It's probably proof that giving just anybody the power to publish is, perhaps, not a guaranty that more quality publication will take place. If anybody has answers to the questions I've come up with, or comments about the answeres, etc. I'd be happy to know. ==== >>I believe the complexity is O(n log n), so this should be good enough. Umm ...good enough? I understand the words individually, but the phrase makes no sense to me. Bobby Treat -----Original Message----- > crash the Front End. I was thinking about the fact that I calculated > all those digits and then threw them away. I could save them with Save > or DumpSave, and read them in with Get the next time I wanted any of > them, although the file would be close to 70 MB (if not more). I may do > that, in fact -- I have plenty of disk space. The next step would be to somehow reuse the stored digits if I wanted > MORE digits. But how? The Bailey-Borwein-Plouffe Pi algorithm is an avenue of attack, since it > can calculate digits far from the decimal point, without calculating > those in between. Unfortunately, it calculates hexadecimal digits in > that way, not decimal digits. (That's true for the version I've seen, > anyway.) Still, I could take the stored digits, convert to hexadecimal, > add more hexadecimal digits with the B-B-P algorithm, and then convert > back to decimal. In both conversions, I'd have to be very cognizant of > how much precision I end up with, but that shouldn't be too difficult. > It might go faster if I store hexadecimal digits, as well as decimal > digits, to eliminate one of those conversions at each increase in the > number of digits. The next step would be to set up an application that allowed anyone to > ping for digits across the Internet, and would return them if they're > stored. Hasn't someone already done that? It seems as if someone would have. Bobby Treat If you're interested in decimal digits, I don't think the BBP algorithm is the way to go. In order to get the nth decimal digit of Pi you need to compute the previous n-1 digits, since base conversion is global, not local. The algorithm Mathematica uses for computing Pi is quite fast - I believe the complexity is O(n log n), so this should be good enough. David > -----Original Message----- calculation in So would it take about the same amont of time for the complete printout > of digits? Of course it would take a few additional seconds to format > the output... Or does Mathematica take alot less time when it truncates the output? > > Could you tell me the CPU you used and its speed etc...i am curious, performance > to > other programs out there. > I used one processor of a dual 1GH Mac and got the same answer with > the > following speed: > 4.2 for Mac OS X (June 4, 2002) > oldmax = $MaxPrecision > 6 > 1. 10 > $MaxPrecision = Infinity > Infinity > With[{n = 2^26}, Timing[ > pd = RealDigits[N[Pi, n + 1], 10, 20, > 19 - n]; ]] > {28794.1 Second, Null} > MaxMemoryUsed[] > 512055204 > pd > {{3, 3, 8, 6, 3, 2, 2, 0, 8, 9, 6, 2, 2, 3, > 4, 0, 9, 8, 0, 3}, -67108844} > Tom Burton ==== Edit ->Preferences -> Font Options In Preferences you will find everything you need to configure your Mathematica environment. Also you may want to look up Style Sheets in the book or the on line help. Yas > I've been trying to get my Mathematica 4.1 properly configured. I set: > ################################################################### > /usr/local/mathematica/SystemFiles/FrontEnd/TextResources/X/Specific.tr: > @@resource maxForXListFonts > 10000 # xlsfonts | wc -l > 5572 /etc/X11/XF86Config: > FontPath /usr/X11R6/lib/X11/fonts/100dpi:unscaled > FontPath /usr/X11R6/lib/X11/fonts/75dpi:unscaled > FontPath /usr/X11R6/lib/X11/fonts/CID > FontPath /usr/X11R6/lib/X11/fonts/Speedo > FontPath /usr/X11R6/lib/X11/fonts/Type1 > FontPath /usr/X11R6/lib/X11/fonts/URW > FontPath /usr/X11R6/lib/X11/fonts/kwintv:unscaled > FontPath /usr/X11R6/lib/X11/fonts/latin2/Type1 > FontPath /usr/X11R6/lib/X11/fonts/local/mma/Type1 > FontPath /usr/X11R6/lib/X11/fonts/local/mma/X:unscaled > FontPath /usr/X11R6/lib/X11/fonts/misc:unscaled > FontPath /usr/X11R6/lib/X11/fonts/misc/sgi:unscaled > FontPath /usr/X11R6/lib/X11/fonts/truetype > FontPath /usr/X11R6/lib/X11/fonts/uni:unscaled # ls -R /usr/X11R6/lib/X11/fonts/ | grep / > /usr/X11R6/lib/X11/fonts/: > /usr/X11R6/lib/X11/fonts/100dpi: > /usr/X11R6/lib/X11/fonts/75dpi: > /usr/X11R6/lib/X11/fonts/CID: > /usr/X11R6/lib/X11/fonts/Speedo: > /usr/X11R6/lib/X11/fonts/Type1: > /usr/X11R6/lib/X11/fonts/URW: > /usr/X11R6/lib/X11/fonts/encodings: > /usr/X11R6/lib/X11/fonts/encodings/large: > /usr/X11R6/lib/X11/fonts/kwintv: > /usr/X11R6/lib/X11/fonts/latin2: > /usr/X11R6/lib/X11/fonts/latin2/Type1: > /usr/X11R6/lib/X11/fonts/local: > /usr/X11R6/lib/X11/fonts/local/mma: > /usr/X11R6/lib/X11/fonts/local/mma/Type1: > /usr/X11R6/lib/X11/fonts/local/mma/X: > /usr/X11R6/lib/X11/fonts/misc: > /usr/X11R6/lib/X11/fonts/misc/sgi: > /usr/X11R6/lib/X11/fonts/truetype: > /usr/X11R6/lib/X11/fonts/uni: > /usr/X11R6/lib/X11/fonts/util: ######################################################## When I open the Mathematica Book Reference Guide in the Help Browser, I get > a beep and the message says: Unable to find font with family Helvetica, weight Bold, slant Plain, and > size 26. Substituting Courier. Compared to the things which *were* broken, this is a minor problem. I can > live with the beep. What I would now like to know is how to tell Mathematica what > fonts to use by default. This seemingly simple question seems to have no > simple answer. Could someone please help me. TIA, ^L ==== > Edit ->Preferences -> Font Options In Preferences you will find everything you need to configure your > Mathematica environment. Also you may want to look up Style Sheets in the > book or the on line help. > Yas > I went into the preferences browser, and it was not clear to me what I was modifying. At one point I clicked on a field filled with text. I had inteded to edit it, and all the text vanished. It didn't bother me as much as such things use to, because I believe I know a backout strategy. It's been a while since I looked at this stuff, and I have to admit it seems far more tractible than it did a year ago. I'll look at the discussion again, and see if it makes more sence to me now. STH ==== > As an example, I spent several hours trying to figure out how to tell > Mathematica to understand the delete key in the way most contemporary > systems understand it. I wanted to avoid modifying system > configuration files such as: > /usr/local/mathematica/SystemFiles/FrontEnd/TextResources/KeyEventTranslation s.tr I expected to be able to change something in my own ~/.Mathematic > directory, but I could not figure out an obvious way to affect this > modification. If you could post a precise description of what you expect the Delete key to do when depressed, we could probably provide you with a clear cut answer of what needs to be done. > I want to adjust the font size used in the widgets, but again, I see > no ovbious means of modifying these attributes. I suspect it can be > accomplished by modifying the ~/.Mathematica/4.1/FrontEnd/init.m. > Perhaps to an experienced Mathematica user, the syntax and semantics > of this file are obvious. They aren't to me. The size of fonts in user interface elements is not specified through the Mathemtica init.m file. It is set through an X resource. If you are not familiar with resources, you may want to track down an introductory text on the X Window System. Information on application-specific resource settings can be found in the Mathematica Getting Started Guide: http://documents.wolfram.com/v4/GettingStarted/TroubleshootingUnixX.html The setting that you would need to adjust is XMathematica*fontList. The value of the resource is an X Logical Font Description field. > I also find the overall look & feel of the interface to be archaic. That's because Mathematica relies on the Motif library for user interface elements. http://www.opengroup.org/desktop/motif.html The appearance of these elements, such as the menu and scroll bars, would be the same for any other Motif application, such as the DDD debugger or releases of Netscape prior to verison 4. -- User Interface Programmer paulh@wolfram.com Wolfram Research, Inc. ==== hi, > I sholdn't have to. If I start messing with X resource settings for my > user environment, I am sure to break something else which is configured > based on the current settings. There should either be a GUI interface, or a > clearly documented, and easily accessible configuration file to modify such > properties as the size of the fonts in the GUI widgets. This is > functionality which is rightfully expected of a modern desktop UI. [snip] > And I'm sure there is some configuration file in which I could place that, > and hope that what you think will be read by my system *will* in fact be > read, and not subsequently overridden during xsession startup. Things > aren't the way they used to be back in the 1980s. The modern Unix desktop > has moved beyond the paradigm of openlook and motif. See for example > http://www.trolltech.com, http://www.gnome.org, and http://www.kde.org > moving the frontend over to QT would have some neat side effects: consistent look & feel with the modern linux gui, themeability, source code truetype fonts as QT supports Xrender and Xft (looks great - see KDE3). i think all of those points are of value, but the most important might be source compatibility. ONE frontend for MOST (or ALL) platforms - sounds like a dream :-)) gerald -- ************************************* Gerald Roth M@th Desktop Development ************************************* ==== > If you could post a precise description of what you expect the Delete key > to do when depressed, we could probably provide you with a clear cut > answer of what needs to be done. Item[KeyEvent[Delete], DeleteNext] 'Most' means 'more than half.' > I want to adjust the font size used in the widgets, but again, I see > no ovbious means of modifying these attributes. I suspect it can be > accomplished by modifying the ~/.Mathematica/4.1/FrontEnd/init.m. > Perhaps to an experienced Mathematica user, the syntax and semantics > of this file are obvious. They aren't to me. The size of fonts in user interface elements is not specified through the > Mathemtica init.m file. It is set through an X resource. If you are not > familiar with resources, you may want to track down an introductory text > on the X Window System. I sholdn't have to. If I start messing with X resource settings for my user environment, I am sure to break something else which is configured based on the current settings. There should either be a GUI interface, or a clearly documented, and easily accessible configuration file to modify such properties as the size of the fonts in the GUI widgets. This is functionality which is rightfully expected of a modern desktop UI. > Information on application-specific resource > settings can be found in the Mathematica Getting Started Guide: http://documents.wolfram.com/v4/GettingStarted/TroubleshootingUnixX.html It should be in a clear and easy to access configuraton interface, or at least be redily available through the help system in such a way that reasonable queries will locate it. Changing fonts does not belong in a section on trouble shooting, unless this is an acknowledgement that the UI is broken. > The setting that you would need to adjust is XMathematica*fontList. The > value of the resource is an X Logical Font Description field. And I'm sure there is some configuration file in which I could place that, and hope that what you think will be read by my system *will* in fact be read, and not subsequently overridden during xsession startup. Things aren't the way they used to be back in the 1980s. The modern Unix desktop has moved beyond the paradigm of openlook and motif. See for example http://www.trolltech.com, http://www.gnome.org, and http://www.kde.org > I also find the overall look & feel of the interface to be archaic. That's because Mathematica relies on the Motif library for user interface > elements. http://www.opengroup.org/desktop/motif.html The appearance of these elements, such as the menu and scroll bars, would > be the same for any other Motif application, such as the DDD debugger or > releases of Netscape prior to verison 4. My point exactly. ==== Awk! Legends! Basically, the answer to your question is that the PlotLegend option works ONLY for the Plot command and does not work for other types of plots. For other types of plots you have to use ShowLegend. And ShowLegend is not all that easy to use, especially since WRI does not give an example for multiple curves in the Help. Needs[Graphics`Graphics`] Needs[Graphics`Legend`] {q1[t_], q2[t_], q3[t_]} = {0.1 Exp[-0.02 t], 0.2 Exp[-0.025 t], 0.4 Exp[-0.028 t]}; Let's look at your first plot. Plot[{q1[t], q2[t], q3[t]}, {t, 0, 100}, PlotStyle -> {{AbsoluteThickness[0.5], AbsoluteDashing[{4, 4}]}, AbsoluteThickness[1.5], {AbsoluteThickness[2], AbsoluteDashing[{1, 8}]}}, AxesLabel -> {Y, X}, PlotLabel -> Title, PlotLegend -> {1, 3, 5}, LegendPosition -> {0.5, 0}, ImageSize -> 500]; The legend is almost as big as the plot. It distracts from the real information you are trying to convey. Furthermore, the order of the curves in the legend is the reverse of their order in the plot. The following shows how to put the legend in a LogLogPlot, or other types of plots. I defined the plot styles independently because they are used in several places. I made the legend much smaller and put it in an empty area of the plot. I also reversed the order of the keys so they would match the order of the curves in the plot. styles={{AbsoluteThickness[0.5], AbsoluteDashing[{4,4}]},{AbsoluteThickness[1.5]},{AbsoluteThickness[ 2],AbsoluteDashing[{1,8}]}}; ShowLegend[ LogLogPlot[{q1[t], q2[t], q3[t]}, {t, 0, 100}, PlotStyle -> styles, AxesLabel -> {Y, X}, PlotLabel -> Title, ImageSize -> 500, DisplayFunction -> Identity], {MapThread[{Graphics[{Sequence @@ #1, Line[{{0, 0}, {1, 0}}]}], #2} &, {styles, {1, 3, 5}}] // Reverse, LegendPosition -> {-0.7, -0.4}, LegendSize -> {0.2, 0.3}, LegendShadow -> {0.02, -0.02}, LegendSpacing -> 0.5} ]; But why use a legend at all? After all, a legend is nothing but another plot in which you have put labels on the curves. Why not put the labels directly on the curves in the real plot in the first place? LogLogPlot[{q1[t], q2[t], q3[t]}, {t, 0, 100}, PlotStyle -> styles, AxesLabel -> {Y, X}, PlotLabel -> Title, ImageSize -> 500, Epilog -> MapThread[ Text[SequenceForm[Case , #1], {Log[10, 0.01], Log[10, #2[0.01]]}, {0, -1}] &, {{1, 2, 3}, {q1, q2, q3}}]]; In the legend you have keyed the curves to numbers 1, 3 and 5. (Perhaps you just used these as examples and meant to use something different in the real plots?) But these don't seem to have any obvious relation to your functions. I suppose the reader will have to look at another table or look into the text of your paper or notebook to find out what 1, 3 and 5 mean. So the reader has to go from the graph to the legend then to the text and then mentally transfer the meaning of the curve back to the main plot. It is so much nicer to put the meaning right on the curve if you can. For the most part, legends are just computer junk and not even easy to nicely construct. When the legend urge comes over you - try to resist. David Park djmp@earthlink.net http://home.earthlink.net/~djmp/ AbsoluteThickness[1.5], {AbsoluteThickness[2], AbsoluteDashing[{1,8}]}}, AxesLabel[Rule]{Y, X}, PlotLabel[Rule]Title, PlotLegend[Rule]{1,3,5}, LegendPosition[Rule] {0.5,0}] (*However with LogPlot or LogLogPlot the legend desappear*) LogLogPlot[{q1[t],q2[t],q3[t]}, {t, 0, 100},PlotStyle[Rule]{ {AbsoluteThickness[0.5], AbsoluteDashing[{4,4}]}, AbsoluteThickness[1.5], {AbsoluteThickness[2], AbsoluteDashing[{1,8}]}}, AxesLabel[Rule]{Y, X}, PlotLabel[Rule]Title, PlotLegend[Rule]{1,3,5}, LegendPosition[Rule] {0.5,0}] I have shown a particular case, but I has this problem always with Legend and LogPlot and LogPlotPlot. I will appreciate any help. Guillermo Sanchez --------------------------------------------- This message was sent using Endymion MailMan. ==== Using the Front End as a interface with the kernel I was running some calulations when suddenly pressing Shift+Enter causes the contents of the cell being evaluated to transform to the next text underlined with a red line: NotebookObject[FrontEndObject[LinkObject[dd8,1,1]],8] foollowed by the next messages: An unknown box name (NotebookObject) was sent as the BoxForm for the expression. Check the format rules for the expression. An unknown box name (FrontEndObject) was sent as the BoxForm for the expression. Check the format rules for the expression. An unknown box name (LinkObject) was sent as the BoxForm for the expression. Check the format rules for the expression. An invalid typeset structure was generated: Missing BoxFormData. Any suggestions will be very aprreciated. Cesar ==== I have an odd problem. I need to use and simplify functions that have been provided by a piece of software that insists on outputing the functional results of a data mining proceedure, using e when outputing numbers in scientific notation. I'm having difficultly using Replace, Hold, etc. to correctly evaluate these types of function formats. For example, y = 5e+5x1+2e-1x2, should be transcribed into 5 10^5 x1 + 0.2 x2. Chuck Reply-To: kuska@informatik.uni-leipzig.de ==== str = 5e+5x1+2e-1x2; StringJoin[Characters[str] /. e -> *10^] // ToExpression ?? Work fine for me. But this type of output is typical generated by a C/FORTRAN Program and you should rewrite the formating rules in the code that produce this output. Jens > I have an odd problem. I need to use and simplify functions that have > been provided by a piece of software that insists on outputing the > functional results of a data mining proceedure, using e when > outputing numbers in scientific notation. I'm having difficultly using Replace, Hold, etc. to correctly evaluate > these types of function formats. For example, y = 5e+5x1+2e-1x2, > should be transcribed into 5 10^5 x1 + 0.2 x2. > Chuck ==== RB> I am considering the following integral RB> W[m_,n_]:=Integrate[BesselJ[m, x]*BesselJ[n, x], {x, 0, Infinity}] RB> where m,n are reals >=0. With Mathematica 4.1 I obtain: RB> If[Re[m+n]>-1, -Cos[(m-n)Pi/2]/(2 Pi)* RB> (2 EulerGamma + Log[4] + RB> PolyGamma[0, 1/2(1 + m - n)] + RB> PolyGamma[0, 1/2(1 - m + n)] + RB> 2PolyGamma[0, 1/2(1 + m + n)]) RB> Any explanation about the analytical expression will be RB> gratefully accepteed. The expression for W[m_,n_] returned by Mathematica is wrong. To prove, just substitute m = n = 0 which is exactly what you had done and observe that the output you had had W[0,0]=-(2 EulerGamma + Log[4] + 4 PolyGamma[0, 1/2])/(2 Pi) = 0.84564 was incorrect. The correct answer is 1/2. Mathematica can handle the numeric integration successfully In[1] := NIntegrate[BesselJ[1, x]*BesselJ[0, x], {x, 0, Infinity}, Method -> Oscillatory] (* The warnings are skipped *) Out[1] = 0.5 Using NIntegrate[BesselJ[0, x]*BesselJ[0, x], {x, 0, Infinity}] without Method -> Oscillatory is not the optimal choice as the integrand oscillates fairly rapidly over the integration region. RB> I suspect that these integrals are divergent (*). In fact, not exactly. Integrate[BesselJ[1, x]*BesselJ[0, x], {x, 0, Infinity}] is equal to 1/2, and Mathematica 4.1 for Microsoft Windows (November 2, 2000) does it correctly, while Mathematica 4.2 for Microsoft Windows (February 28, 2002) concocts a strange mixture of a wrong divergence message and the warning that it cannot check the convergence [should I trust to the second warning? or the first?] As a matter of fact, Integrate[BesselJ[1, x]*BesselJ[0, x], {x, 0, Infinity}] converges because the integrand is regular at x=0, bounded over the whole right semi-axis, and decays as 2*Cos[Pi/4 - x]*Cos[(3*Pi)/4 - x]/(Pi*x) + o(1/x) at x -> Infinity . Say, calculate Normal[Series[BesselJ[1, x], {x, Infinity, 1}]] Normal[ Series[BesselJ[0, x], {x, Infinity, 1}]] // InputForm -> (2*(Cos[Pi/4 - x] - Sin[Pi/4 - x]/(8*x))*(Cos[(3*Pi)/4 - x] + (3*Sin[(3*Pi)/4 - x])/(8*x)))/(Pi*x) then Plot[%,{x,1,10}] and Plot[BesselJ[1,x]*BesselJ[0,x],{x,1,10}] and you could hardly see the difference. Generally, to get to the convergence domain for W in terms of m and n is easy via the asymtotics of the Bessel functions (use something like Expand[Normal[Series[BesselJ[m, x], {x, Infinity, 1}]]Normal[ Series[BesselJ[n, x], {x, Infinity, 1}]]] then analyze the main term). Best wishes, Vladimir Bondarenko Mathematical Director Symbolic Testing Group Web : http://www.CAS-testing.org/ http://maple.bug-list.org/VER2/ (under tuning) http://maple.bug-list.org/VER3/ (under tuning) http://maple.bug-list.org/VER1/ (under tuning) http://www.beautyriot.com/ (teamwork) http://www.ohaha.com/ (teamwork) Voice: (380)-652-447325 Mon-Fri 9 a.m. - 6 p.m. Mail : 76 Zalesskaya Str, Simferopol, Crimea, Ukraine ==== > inside a program I need to solve this linear equation in terms of p1. > However something odds happens. Sometimes the solution is computed and > sometimes the result is empty [I mean no output...]. Is this a bug of the > solve command or am I doing something wrong? The problem is robust to: > changing name to the variables and other makeups.. > David That's the weirdest bug I've seen in weeks. As it happens, it's mine. At least the inconsistent behavior, that is. I'll fix it, and maybe also try to address the issue of how to handle approximate numbers in testing subexpressions for zero. I've excised your code and put in place a substantially smaller example that I believe is responsible. The table will tend to give erratic results. zz = (-1.*x^7*(-1. + p - 7.*x^6 + p*x^6 + 6.*x^7)* (7.000000000000002 - 7.000000000000002*x + 14.000000000000004*x^6 - 14.000000000000004*x^7 + 7.000000000000002*x^12 - 6.999999999999998*x^13))/ (p^2*(1. + 0.9*x^6)^2*(1. + x^6)^5); One workaround would be to use exact input, say by preprocessing with Rationalize. Daniel Lichtblau Wolfram Research ==== >inside a program I need to solve this linear equation in terms of p1. >However something odds happens. Sometimes the solution is computed and >sometimes the result is empty [I mean no output...]. Is this a bug of the >solve command or am I doing something wrong? The problem is robust to: >changing name to the variables and other makeups.. >David > >ps: Sorry for the stupid way in which I copied the command... > >Solve[(x^2*((-0.9*x^7*(p^2*(-1 - 5.8*x^6 - 14.010000000000002*x^12 - > 18.04*x^18 - 13.06*x^24 - > 5.040000000000001*x^30 - 0.81*x^36) + > x*(7.777777777777779 - 9.074074074074076*x + > 30.333333333333336*x^6 - > 21.51851851851852*x^7 - > 16.333333333333336*x^8 + > 44.33333333333334*x^12 + > 3.188888888888883*x^13 - > 65.68333333333332*x^14 + > 28.777777777777786*x^18 + > 47.937037037037044*x^19 - > 100.10000000000002*x^20 + 7.*x^24 + > 45.6037037037037*x^25 - > 69.53333333333333*x^26 + > 13.299999999999999*x^31 - > 19.833333333333332*x^32 - > 1.0499999999999996*x^38) + > p*(-6 + 8.296296296296296*x - > 28.799999999999997*x^6 + > 32.785185185185185*x^7 + > 9.333333333333336*x^8 - > 55.260000000000005*x^12 + > 49.04777777777776*x^13 + > 38.38333333333334*x^14 - > 52.980000000000004*x^18 + > 34.20518518518518*x^19 + > 60.20000000000001*x^20 - > 25.380000000000003*x^24 + > 11.736296296296294*x^25 + > 43.63333333333334*x^26 - 4.86*x^30 + > 2.8999999999999986*x^31 + > 13.533333333333333*x^32 + 0.81*x^37 + > 1.0499999999999996*x^38)))/(x + 1.9*x^7 + > 0.9*x^13)^2 - ((-1 + p - 7*x^6 + p*x^6 + > 6*x^7)*(1.2962962962962965 - 3.111111111111112*x^6 >+ > 9.333333333333336*x^7 - 10.111111111111114*x^12 >+ > 22.05*x^13 - 5.703703703703705*x^18 + 17.15*x^19 >+ > 5.483333333333331*x^25 + 1.0499999999999996*x^31 >+ > p1*x^5*(7.000000000000002 - 7.000000000000002*x >+ > 14.000000000000004*x^6 - > 14.000000000000004*x^7 + > 7.000000000000002*x^12 - > 6.999999999999998*x^13) - > 1.166666666666667*p*x^4*x1 - > 3.500000000000001*p*x^10*x1 - > 1.0500000000000003*p*x^11*x1 - > 3.500000000000001*p*x^16*x1 - > 3.150000000000001*p*x^17*x1 - > 1.166666666666667*p*x^22*x1 - > 3.150000000000001*p*x^23*x1 - > 1.0500000000000003*p*x^29*x1))/((1 + 0.9*x^6)^2*(1 >+ > x^6)^2)))/(p^2*(1 + x^6)^3) == 0, p1] > You might find it more robust (and the results cleaner) if you Simplify the equation prior to using Solve. Such as Solve[eqn // Rationalize // Simplify, p1] However, if you are assigning values to p or x prior to using Solve, there may not be a solution. That is, for whenever the numerator of the expression for p1 would be zero, e.g., p = (-6 x^7 + 7 x^6 +1)/(x^6 + 1). Bob Hanlon ==== I prefer to delete all output and then Copy As>Notebook expression. It Bobby -----Original Message----- and output --- I try to use one tab indent for input and two tabs indent for output, plus some blank line adjustment. I wonder if anyone has a way of automatically achieving this reformating. -- 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 > Often posters to MathGroup copy and paste in the complete cell expression, > including the In and Out numbers, when posting to MathGroup. I wonder if this is the best method because one can't then just copy out all > the statements and paste them into a Mathematica notebook. All the statement > numbers have to be edited out and if there are many statement definitions > this is an extended task for any responder. This, of course, decreases the > chances for a response. A better method is for the poster to just copy and > paste the CONTENTS of each cell. This is more work for the poster, but it > may pay off in better responses. David Park > djmp@earthlink.net > http://home.earthlink.net/~djmp/ ==== Could someone explain what is going on here, please? In[1]:= a = 77617.; b = 33096.; In[2]:= SetAccuracy[a, Infinity]; SetAccuracy[b, Infinity]; SetPrecision[a, Infinity]; SetPrecision[b, Infinity]; In[4]:= f := 333.75*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) + 5.5*b^8 + a/(2*b) In[5]:= SetAccuracy[f, Infinity]; SetPrecision[f, Infinity]; In[6]:= f Out[6]= -1.1805916207174113*^21 In[7]:= a = 77617; b = 33096; In[8]:= g := (33375/100)*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) + (55/10)*b^8 + a/(2*b) In[9]:= g Out[9]= -(54767/66192) In[10]:= N[%] Out[10]= -0.8273960599468214 PK ==== Peter, I hope that the following example will help - it is a matter or when things evaluate. The a in SetAccuracy[a, Infinity], below, evaluates before SetAccuracy acts, so we get SetAccuracy[2.3, Infinity]. The value of a is not changed. a = 2.3; aa = SetAccuracy[a, Infinity] 2589569785738035/1125899906842624 But, a 2.3 Whereas aa 2589569785738035/1125899906842624 -- 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 > Could someone explain what is going on here, please? In[1]:= > a = 77617.; b = 33096.; In[2]:= > SetAccuracy[a, Infinity]; SetAccuracy[b, Infinity]; > SetPrecision[a, Infinity]; SetPrecision[b, Infinity]; In[4]:= > f := 333.75*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) + 5.5*b^8 + a/(2*b) In[5]:= > SetAccuracy[f, Infinity]; SetPrecision[f, Infinity]; In[6]:= > f Out[6]= > -1.1805916207174113*^21 In[7]:= > a = 77617; b = 33096; In[8]:= > g := (33375/100)*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) + (55/10)*b^8 + a/(2*b) In[9]:= > g Out[9]= > -(54767/66192) In[10]:= > N[%] Out[10]= > -0.8273960599468214 > PK > ==== In receiving notebooks from many different people I have noticed that beginners often do not know how to use Text cells and write all of their comments as Input cells. I have even run across some extremely advanced users who did not know the easy method for entering Text cells. A good notebook is usually a blend of Text cells, Input/Output cells and graphics cells. Text cells are very useful for documenting what you are doing and passing information to other people. Since many people do not know how to use Text cells, I thought I would write a little explanation for beginners who are followers of MathGroup. The very easiest method for entering a Text cell is to put the insertion point where you want the new cell to be (at the end of the notebook or between two existing cells) and then type Alt-7. Then just start typing and you will have a Text cell. Alternatively you can use MenuFormatStyleText to start a new Text cell. Often, it is useful to put the ToolBar at the top of the notebook. Use MenuFormatShow ToolBar. The drop-down menu on the ToolBar has the various kinds of cells available for the current style of the notebook. You can select Text (or any other style) from there. Some users may hesitate to use Text cells because they want to include a mathematical expression in the comments. However, that is also very easy. Just use an Inline cell within the text cell. At the point within the text cell where you want to include a mathematical expression, start an Inline cell by typing Ctrl-(. A selection placeholder will appear on a pink background. You can type a Mathematica expression there just as in an Input cell. Use Ctrl-) to complete the Inline cell, or Shift-Space. You can even select an Inline cell and evaluate it with Shift-Ctrl-Enter. Putting comments in Text cells is far better than using Input cells (or cell group header cells). Mathematica won't try to evaluate Text cells, the text will wrap properly and adjust better to the notebook width if you change it. You can also check the spelling of words by putting the cursor after a word and using Ctrl-K. (In an Input cell Mathematica doesn't use the dictionary, but uses the table of symbols instead and hence it won't check spelling.) David Park djmp@earthlink.net http://home.earthlink.net/~djmp/ ==== This is because inline cells are not in StandardForm by default, but TraditionalForm. Use the menu item Cell -> Default Inline Format Type -> StandardForm. >David Park's posting reminded me of a frequent annoyance when I am >trying to include some Mathematica expressions within text cells -- a >Mathematica input expression in Standard Form that involves use of a >Control-key combination to form a superscript, square-root, or built-up >fraction: > >For example, suppose I want to include within a text cell a Standard >Form expression for the square of x, with the exponent 2 raised. If I >type the x first, even if I immediately highlight it and change it to >Courier (to match the default font for Input cells in Standard Form), as >soon as I press the Control-^ key combination, an Inline cell is created >beginning with the x, and then when I type the exponent 2 everything in >that Inline cell is now in Times, and the x is Italic. To change both >characters to Courier is not so easy: it seems to require separately >the entire Inline cell and selecting Courier does not change the exponent!) > >So to avoid this annoyance I normally must first type the desired >expression in a separate Input cell, then copy the contents of that cell >to the desired point in the Text cell. > >Any suggestions on a more efficient method for handling this? > In receiving notebooks from many different people I have noticed that > beginners often do not know how to use Text cells .... > Some users may hesitate to use Text cells because they want to include a > mathematical expression in the comments.... > Just use an Inline cell within the text cell.... -- >Murray Eisenberg murray@math.umass.edu >Mathematics & Statistics Dept. >Lederle Graduate Research Tower phone 413 549-1020 (H) >University of Massachusetts 413 545-2859 (W) >710 North Pleasant Street >Amherst, MA 01375 -------------------------------------------------------------- Omega Consulting The final answer to your Mathematica needs Spend less time searching and more time finding. http://www.wz.com/internet/Mathematica.html Reply-To: murray@math.umass.edu ==== David Park's posting reminded me of a frequent annoyance when I am trying to include some Mathematica expressions within text cells -- a Mathematica input expression in Standard Form that involves use of a Control-key combination to form a superscript, square-root, or built-up fraction: For example, suppose I want to include within a text cell a Standard Form expression for the square of x, with the exponent 2 raised. If I type the x first, even if I immediately highlight it and change it to Courier (to match the default font for Input cells in Standard Form), as soon as I press the Control-^ key combination, an Inline cell is created beginning with the x, and then when I type the exponent 2 everything in that Inline cell is now in Times, and the x is Italic. To change both characters to Courier is not so easy: it seems to require separately the entire Inline cell and selecting Courier does not change the exponent!) So to avoid this annoyance I normally must first type the desired expression in a separate Input cell, then copy the contents of that cell to the desired point in the Text cell. Any suggestions on a more efficient method for handling this? > In receiving notebooks from many different people I have noticed that > beginners often do not know how to use Text cells .... Some users may hesitate to use Text cells because they want to include a > mathematical expression in the comments.... > Just use an Inline cell within the text cell.... -- Murray Eisenberg murray@math.umass.edu Mathematics & Statistics Dept. Lederle Graduate Research Tower phone 413 549-1020 (H) University of Massachusetts 413 545-2859 (W) 710 North Pleasant Street Amherst, MA 01375 ==== correctly, all you need to do is to make sure the default inline cell format is StandardForm. Go to the menu item Cell, select Default Inline FormatType, and change it to StandardForm. Carl Woll Physics Dept U of Washington ----- Original Message ----- > type the x first, even if I immediately highlight it and change it to > Courier (to match the default font for Input cells in Standard Form), as > soon as I press the Control-^ key combination, an Inline cell is created > beginning with the x, and then when I type the exponent 2 everything in > that Inline cell is now in Times, and the x is Italic. To change both > characters to Courier is not so easy: it seems to require separately > the entire Inline cell and selecting Courier does not change the exponent!) So to avoid this annoyance I normally must first type the desired > expression in a separate Input cell, then copy the contents of that cell > to the desired point in the Text cell. Any suggestions on a more efficient method for handling this? > In receiving notebooks from many different people I have noticed that > beginners often do not know how to use Text cells .... > Some users may hesitate to use Text cells because they want to include a > mathematical expression in the comments.... > Just use an Inline cell within the text cell.... -- > Murray Eisenberg murray@math.umass.edu > Mathematics & Statistics Dept. > Lederle Graduate Research Tower phone 413 549-1020 (H) > University of Massachusetts 413 545-2859 (W) > 710 North Pleasant Street > Amherst, MA 01375 Reply-To: kuska@informatik.uni-leipzig.de ==== just one comment: the meaning of the Alt-7 key depend on the style sheet that is in use. The TMJ style use Alt-8 for text and one has to learn new key short-cuts for every style sheet ! Jens In receiving notebooks from many different people I have noticed that > beginners often do not know how to use Text cells and write all of their > comments as Input cells. I have even run across some extremely advanced > users who did not know the easy method for entering Text cells. A good > notebook is usually a blend of Text cells, Input/Output cells and graphics > cells. Text cells are very useful for documenting what you are doing and > passing information to other people. Since many people do not know how to > use Text cells, I thought I would write a little explanation for beginners > who are followers of MathGroup. The very easiest method for entering a Text cell is to put the insertion > point where you want the new cell to be (at the end of the notebook or > between two existing cells) and then type Alt-7. Then just start typing and > you will have a Text cell. Alternatively you can use MenuFormatStyleText to start a new Text cell. Often, it is useful to put the ToolBar at the top of the notebook. Use > MenuFormatShow ToolBar. The drop-down menu on the ToolBar has the various > kinds of cells available for the current style of the notebook. You can > select Text (or any other style) from there. Some users may hesitate to use Text cells because they want to include a > mathematical expression in the comments. However, that is also very easy. > Just use an Inline cell within the text cell. At the point within the text > cell where you want to include a mathematical expression, start an Inline > cell by typing Ctrl-(. A selection placeholder will appear on a pink > background. You can type a Mathematica expression there just as in an Input > cell. Use Ctrl-) to complete the Inline cell, or Shift-Space. You can even > select an Inline cell and evaluate it with Shift-Ctrl-Enter. Putting comments in Text cells is far better than using Input cells (or cell > group header cells). Mathematica won't try to evaluate Text cells, the text > will wrap properly and adjust better to the notebook width if you change it. > You can also check the spelling of words by putting the cursor after a word > and using Ctrl-K. (In an Input cell Mathematica doesn't use the dictionary, > but uses the table of symbols instead and hence it won't check spelling.) David Park > djmp@earthlink.net > http://home.earthlink.net/~djmp/ ==== Solve[youre equation, p1, VerifySolutions->True] will return a solution. So will Solve[Rationalize[your equation],p1]. inside a program I need to solve this linear equation in terms of p1. > However something odds happens. Sometimes the solution is computed and > sometimes the result is empty [I mean no output...]. Is this a bug of > the > solve command or am I doing something wrong? The problem is robust to: > changing name to the variables and other makeups.. > David ps: Sorry for the stupid way in which I copied the command... Solve[(x^2*((-0.9*x^7*(p^2*(-1 - 5.8*x^6 - 14.010000000000002*x^12 - > 18.04*x^18 - 13.06*x^24 - > 5.040000000000001*x^30 - 0.81*x^36) + > x*(7.777777777777779 - 9.074074074074076*x + > 30.333333333333336*x^6 - > 21.51851851851852*x^7 - > 16.333333333333336*x^8 + > 44.33333333333334*x^12 + > 3.188888888888883*x^13 - > 65.68333333333332*x^14 + > 28.777777777777786*x^18 + > 47.937037037037044*x^19 - > 100.10000000000002*x^20 + 7.*x^24 + > 45.6037037037037*x^25 - > 69.53333333333333*x^26 + > 13.299999999999999*x^31 - > 19.833333333333332*x^32 - > 1.0499999999999996*x^38) + > p*(-6 + 8.296296296296296*x - > 28.799999999999997*x^6 + > 32.785185185185185*x^7 + > 9.333333333333336*x^8 - > 55.260000000000005*x^12 + > 49.04777777777776*x^13 + > 38.38333333333334*x^14 - > 52.980000000000004*x^18 + > 34.20518518518518*x^19 + > 60.20000000000001*x^20 - > 25.380000000000003*x^24 + > 11.736296296296294*x^25 + > 43.63333333333334*x^26 - 4.86*x^30 + > 2.8999999999999986*x^31 + > 13.533333333333333*x^32 + 0.81*x^37 + > 1.0499999999999996*x^38)))/(x + 1.9*x^7 + > 0.9*x^13)^2 - ((-1 + p - 7*x^6 + p*x^6 + > 6*x^7)*(1.2962962962962965 - > 3.111111111111112*x^6 + > 9.333333333333336*x^7 - > 10.111111111111114*x^12 + > 22.05*x^13 - 5.703703703703705*x^18 + > 17.15*x^19 + > 5.483333333333331*x^25 + > 1.0499999999999996*x^31 + > p1*x^5*(7.000000000000002 - > 7.000000000000002*x + > 14.000000000000004*x^6 - > 14.000000000000004*x^7 + > 7.000000000000002*x^12 - > 6.999999999999998*x^13) - > 1.166666666666667*p*x^4*x1 - > 3.500000000000001*p*x^10*x1 - > 1.0500000000000003*p*x^11*x1 - > 3.500000000000001*p*x^16*x1 - > 3.150000000000001*p*x^17*x1 - > 1.166666666666667*p*x^22*x1 - > 3.150000000000001*p*x^23*x1 - > 1.0500000000000003*p*x^29*x1))/((1 + > 0.9*x^6)^2*(1 + > x^6)^2)))/(p^2*(1 + x^6)^3) == 0, p1] > Reply-To: murray@math.umass.edu ==== For all names, perhaps: Names[*`*] For names you defined at a normal session (without changing to some other context than the default Global`): Names[Global`*] > IIRC, there is a way to get a list of all the symbols defined in the > currently running session. I can't seem to find the reference to that > command. Could somone point me in the direction of documentation which > will tell me how to get information about the current session? > TIA, > -- Murray Eisenberg murray@math.umass.edu Mathematics & Statistics Dept. Lederle Graduate Research Tower phone 413 549-1020 (H) University of Massachusetts 413 545-2859 (W) 710 North Pleasant Street Amherst, MA 01375 ==== > For all names, perhaps: Names[*`*] For names you defined at a normal session (without changing to some > other context than the default Global`): Names[Global`*] > IIRC, there is a way to get a list of all the symbols defined in the > currently running session. I can't seem to find the reference to that > command. Could somone point me in the direction of documentation which > will tell me how to get information about the current session? > TIA, I think I asked the wrong question. I'll have to look at things some more. What you gave me resulted in far more than I was looking for. http://public.globalsymmetry.com/proprietary/com/wri/system-symbols.html http://66.92.149.152/proprietary/com/wri/system-symbols.html I think I really hosed the code for generating that table. I used 5 lines. I probably didn't need more than two, but I'm too tired right now to think about it. Mathematica is totaly awesome when it comes to what it was intended for. They really need to rent a Troll for a few months and fix this interface. Qt will run on just about anything. Heck, my Win-XP partition runs XFree86, with the KDE, or it did when I booted into XP a month ago. STH ==== ?* does the trick. You can limit it to Global variables with ?Global`* Bobby -----Original Message----- Sender: steve@smc.vnet.net Approved: Steven M. Christensen , Moderator ==== Well, first of of all, your using SetAccuracy and SetPrecision does nothing at all here, since they do not change the value of a or b. You should use a = SetAccuracy[a, Infinity] etc. But even then you won't get the same answer as when you use exact numbers because of the way you evaluate f. Here is the order of evaluation that will give you the same answer, and should explain what is going on: f = SetAccuracy[333.75*b^6 + a^2*(11*a^2*b^2 - b^6 - 121* b^4 - 2) + 5.5*b^8 + a/(2*b), Infinity]; a = 77617.; b = 33096.; a = SetAccuracy[a, Infinity]; b = SetAccuracy[b, Infinity]; f 54767 -(-----) 66192 > Could someone explain what is going on here, please? In[1]:= > a = 77617.; b = 33096.; In[2]:= > SetAccuracy[a, Infinity]; SetAccuracy[b, Infinity]; > SetPrecision[a, Infinity]; SetPrecision[b, Infinity]; In[4]:= > f := 333.75*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) + 5.5*b^8 + > a/(2*b) In[5]:= > SetAccuracy[f, Infinity]; SetPrecision[f, Infinity]; In[6]:= > f Out[6]= > -1.1805916207174113*^21 In[7]:= > a = 77617; b = 33096; In[8]:= > g := (33375/100)*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) + > (55/10)*b^8 + a/(2*b) In[9]:= > g Out[9]= > -(54767/66192) In[10]:= > N[%] Out[10]= > -0.8273960599468214 > PK > ==== Andrzej, Bobby, Peter It looks as if using SetAccuracy succeeds here because the inexact numbers that occur have finite binary representations. If we change them slightly to avoid this then we have to use Rationalize: 1) Using SetAccuracy Clear[a,b,f] f=SetAccuracy[333.74*b^6+a^2*(11*a^2*b^2-b^6-121*b^4-2)+5.4*b^8+a/(2*b), Infinity]; a=77617.1; b=33096.1; a=SetAccuracy[a,Infinity];b=SetAccuracy[b,Infinity]; f -1564032114908486835197494923989618867972540153873951942813115514949 3891236234 5250077191686937045911977601879880463043614978691991293196257430102923631246 75 /108671061439707605510003578275547938881981431359756495796079898677435728240 16 0653953612982932181371232436367739737604096 2) Rewriting as fractions a=776171/10; b=330961/10; f=33374/100*b^6+a^2*(11*a^2*b^2-b^6-121*b^4-2)+54/10*b^8+a/(2*b) -(5954133808997234115690303589909929091649391296257/ 41370125000000) 3) Using Rationalize Clear[a,b,f] f=Rationalize[333.74*b^6+a^2*(11*a^2*b^2-b^6-121*b^4-2)+5.4*b^8+a/(2*b),0]; a=77617.1; b=33096.1; a=Rationalize[a,0];b=Rationalize[b,0]; f -(5954133808997234115690303589909929091649391296257/ 41370125000000) I use Rationalize[. , 0] besause of results like Rationalize[3.1415959] 3.1416 Rationalize[3.1415959,0] 31415959/10000000 -- 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 > Well, first of of all, your using SetAccuracy and SetPrecision does > nothing at all here, since they do not change the value of a or b. You > should use a = SetAccuracy[a, Infinity] etc. But even then you won't > get the same answer as when you use exact numbers because of the way > you evaluate f. Here is the order of evaluation that will give you the > same answer, and should explain what is going on: f = SetAccuracy[333.75*b^6 + a^2*(11*a^2*b^2 - b^6 - 121* > b^4 - 2) + 5.5*b^8 + a/(2*b), Infinity]; a = 77617.; > b = 33096.; a = SetAccuracy[a, Infinity]; b = SetAccuracy[b, Infinity]; f 54767 > -(-----) > 66192 Andrzej Kozlowski > Toyama International University > JAPAN > Could someone explain what is going on here, please? > In[1]:= > a = 77617.; b = 33096.; > In[2]:= > SetAccuracy[a, Infinity]; SetAccuracy[b, Infinity]; > SetPrecision[a, Infinity]; SetPrecision[b, Infinity]; > In[4]:= > f := 333.75*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) + 5.5*b^8 + > a/(2*b) > In[5]:= > SetAccuracy[f, Infinity]; SetPrecision[f, Infinity]; > In[6]:= > f > Out[6]= > -1.1805916207174113*^21 > In[7]:= > a = 77617; b = 33096; > In[8]:= > g := (33375/100)*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) + > (55/10)*b^8 + a/(2*b) > In[9]:= > g > Out[9]= > -(54767/66192) > In[10]:= > N[%] > Out[10]= > -0.8273960599468214 > PK ==== > Integrate[BesselJ[1, x]*BesselJ[0, x], {x, 0, Infinity}] is equal to 1/2, and Mathematica 4.1 for Microsoft Windows > (November 2, 2000) does it correctly, while Mathematica 4.2 > for Microsoft Windows (February 28, 2002) concocts a strange > mixture of a wrong divergence message and the warning that > it cannot check the convergence [should I trust to the second > warning? or the first?] In[6]:= Integrate[BesselJ[1, x]*BesselJ[0, x], {x, 0, Infinity}] Out[6]= 1/2 In[7]:= Out[7]= 4.2 for Mac OS X (June 4, 2002) ==== That should read ...denominator of the expression for p1... Bob >You might find it more robust (and the results cleaner) if you Simplify >the >equation prior to using Solve. Such as > >Solve[eqn // Rationalize // Simplify, p1] > >However, if you are assigning values to p or x prior to using Solve, there > >may not be a solution. That is, for whenever the numerator of the expression > >for p1 would be zero, e.g., > >p = (-6 x^7 + 7 x^6 +1)/(x^6 + 1). > Bob Hanlon ==== > The expression for W[m_,n_] returned by Mathematica is wrong. To prove, just substitute m = n = 0 which is exactly what you had done and observe that the output you had had W[0,0]=-(2 EulerGamma + Log[4] + 4 PolyGamma[0, 1/2])/(2 Pi) = 0.84564 was incorrect. The correct answer is 1/2. Mathematica can handle the numeric integration successfully In[1] := NIntegrate[BesselJ[1, x]*BesselJ[0, x], {x, 0, Infinity}, > Method -> Oscillatory] (* The warnings are skipped *) Out[1] = 0.5 You'll find that W[m=1,n=0]=1/2, so Mathematica gets that right. W[0,0] diverges. Mathematica gets that wrong. I note that Mathematica yields a result for Integrate[BesselJ[m, a*x]*BesselJ[n, b*x], {x, 0, Infinity}] that appears to agree with formula 6.512(1) of Gradshteyn and Ryshik (4th ed., 1965), including the condition b True] If the zi's need to be scaled, then you can do something like this: colorfn = Hue[.67#]&; With[{ m = {Min[#],Max[#]}&@vals }, Show[Graphics[ {colorfn[#[[3]]], PointSize[.01], Point[{#[[1]],#[[2]]}]}]& /@ (data /. {x_,y_,z_} -> {x, y, (z-m[[1]])/(m[[2]]-m[[1]])}), Axes -> True]] --- Selwyn Hollis > I have a list of points l1={xi,yi,zi} how can I make a 2D list plot of