Subject: Re: Exporting Cells to PDF >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 === Subject: How to combime sound + graphics? 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 === Subject: Re: Choosing between Mathematica for Windows and Linux 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. As for stability: the 4.2 Linux FE ist very stable, there are no crashes 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) limitations of Linux: 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 scrolling through larger notebooks is noticably slower on Linux (at 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 (like some Helvetica) are available for free on Linux. If you have more specific questions, feel free to email me (in het engels, nederlands, duits of spaans) Rolf Mertig Mertig Consulting Berlin === Subject: FW: Path 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----- === Subject: Path 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 > 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 > Tel.: +49(0)69 71 34 53 80 > Mobile: +49(0)172 6 74 95 77 > Fax: +49(0)69 71 34 95 380 > E-mail: matthias.bode@oppenheim.de > Internet: http://www.oppenheim.de === Subject: Mathematica 3.0 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 guptass@rediffmail.com === Subject: RE: pdf export 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----- === Subject: pdf export Bobby, Are you sure you're exporting a Cell object? If so, this must be something new in version 4.2 ?? --Selwyn P.S. Sorry if this is a duplicate message; I've been getting delivery failed messages all morning. === Subject: RE: Questions on Plot Hi Steve, > 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 CSIRO DEM Tel: +612 9490 5416 North Ryde Fax: +612 9490 5467 Australia David.Annetts@csiro.au === Subject: Re: List operation, LabeledListPlot, Goto Statement > 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, > GridLines -> Automatic, PlotRange -> All, > 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, GridLines -> Automatic, PlotRange -> All, 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 === Subject: Re: Questions on Plot >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], > GridLines -> Automatic]; >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 === Subject: Re: how to plot with BOLD LINE >I have to use a graphics of mathematica with powerpoint for a little >conference. I'd like that the plot's line (I mean the line of X-axes >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]] === Subject: How to combime sound + graphics? 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 === Subject: FW: Path 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----- === Subject: Path 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 > 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 > Tel.: +49(0)69 71 34 53 80 > Mobile: +49(0)172 6 74 95 77 > Fax: +49(0)69 71 34 95 380 > E-mail: matthias.bode@oppenheim.de > Internet: http://www.oppenheim.de === Subject: Re: Exporting Cells to PDF >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 === Subject: Mathematica 3.0 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 guptass@rediffmail.com === Subject: RE: pdf export 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----- === Subject: pdf export Bobby, Are you sure you're exporting a Cell object? If so, this must be something new in version 4.2 ?? --Selwyn P.S. Sorry if this is a duplicate message; I've been getting delivery failed messages all morning. === Subject: RE: Questions on Plot Hi Steve, > 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 CSIRO DEM Tel: +612 9490 5416 North Ryde Fax: +612 9490 5467 Australia David.Annetts@csiro.au === Subject: Re: List operation, LabeledListPlot, Goto Statement > 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, > GridLines -> Automatic, PlotRange -> All, > 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, GridLines -> Automatic, PlotRange -> All, 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 === Subject: Re: how to plot with BOLD LINE with mathematica? 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 ----- === Subject: how to plot with BOLD LINE with mathematica? > I have to use a graphics of mathematica with powerpoint for a little > conference. I'd like that the plot's line (I mean the line of X-axes > Y-axes and line of function) are bold or more visible: how can I do > that? === Subject: Re: how to plot with BOLD LINE with mathematica? 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 ----- === Subject: how to plot with BOLD LINE with mathematica? > I have to use a graphics of mathematica with powerpoint for a little > conference. I'd like that the plot's line (I mean the line of X-axes > Y-axes and line of function) are bold or more visible: how can I do > that? === Subject: Re: how to plot with BOLD LINE >I have to use a graphics of mathematica with powerpoint for a little >conference. I'd like that the plot's line (I mean the line of X-axes >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]] === Subject: Re: Questions on Plot >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], > GridLines -> Automatic]; >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 === Subject: 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 === Subject: Re: Strange ReplaceAll behavior 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 === Subject: Re: Could someone verify a long Pi calculation in Version 4 for me? > 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: $Version 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 -- === Subject: Re: Could someone verify a long Pi calculation in Version 4 for me? 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: > $Version > 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 === Subject: Printing Scripts 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 === Subject: Re: Choosing between Mathematica for Windows and Linux Sidney, I have been running Mathematica regularly on both Windows and Linux (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 properly on Linux. Neither do other extremely useful operations such as 4. Generally speaking it takes more mouse clicks and keystrokes to do things on Linux. Many things that you can do with a keyboard shortcut on Windows require a sequence of mouse-clicks on Linux. 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, I have never seen any sign of less stability of the Kernel on Windows. Nowadays, I run Mathematica on Linux in the following situations: 1. I need to run other Linux programs from my Mathematica session. 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 too but in our lab there happen to be clusters of Linux machines available. What I do then is run the Front End on Windows and connect to a Linux kernel via Mathlink. This is very simple and more reliable and robust than running purely on Linux. You get all the comfort and speed of the Windows 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 (AFS) file system used by our Linux clusters directly in Windows. So I have 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 > Hi group, > One of these days the company I work for will buy one Mathematica 4.2 > license. Now we will have to choose between either the Linux or the Windows > versions of the product. > Our Linux systems are Debian based, running 2.4.19 kernels, and the KDE-2 > 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 > and 4.0) on Linux, I found that the front-end was working a lot less > 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 > whole, I'd say the Linux front-end simply felt less robust than does the > Windows frontend. > Can anybody please tell me whether they got the Linux 4.2 Front-end 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 > would much prefer to run Mathematica on Linux instead of VMWare/Windows, but > not if that would mean putting up with a quircky front-end. > Sidney Cadot === Subject: Re: Re: Choosing between Mathematica for Windows and Linux I would like to add : Some characters, at least with a french keyboard, are not directly available with the Linux FrontEnd, for instance some accented characters have to be 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. > As for stability: the 4.2 Linux FE ist very stable, there are no crashes > 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) > limitations of Linux: > 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 > scrolling through larger notebooks is noticably slower on Linux (at > 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 > (like some Helvetica) are available for free on Linux. > If you have more specific questions, feel free to email me (in het > engels, nederlands, duits of spaans) > Rolf Mertig > Mertig Consulting > Berlin === Subject: Re: Choosing between Mathematica for Windows and Linux 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. As for stability: the 4.2 Linux FE ist very stable, there are no crashes 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) limitations of Linux: 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 scrolling through larger notebooks is noticably slower on Linux (at 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 (like some Helvetica) are available for free on Linux. If you have more specific questions, feel free to email me (in het engels, nederlands, duits of spaans) Rolf Mertig Mertig Consulting Berlin === Subject: A Bessel integral Hi to all Mathematica friend! 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 === Subject: RE: Exporting Cells to PDF WRI Tech Support sent me an answer to the font problem. Copying the Type 1 (not True Type) fonts made the error go away: Hello Dr. Treat, 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 George Kambouroglou Technical Support Wolfram Research, Inc. support@wolfram.com -----Original Message----- === Subject: RE: Exporting Cells to PDF 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----- === Subject: Exporting Cells to PDF 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? === Subject: Re: Mathematica 3.0 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 Email : vvb@mail.strace.net 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 === Subject: RE: Mathematica 3.0 Just tested it with no errors, version 3.0 John A. Velling -----Original Message----- === Subject: Mathematica 3.0 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 guptass@rediffmail.com === Subject: Re: How to export plain text numbers!? Hi Soxos, 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 === Subject: Re: How to export plain text numbers!? > 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 === Subject: Miscellaneous questions 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. === Subject: Re: Miscellaneous questions > 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 === Subject: Re: Miscellaneous questions > 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. === Subject: Exporting Fonts Mathematica 4.2 Linux I am running Mathematica 4.2 for students under Redhat Linux 7.2. Can someone 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. === Subject: C code for dual processor machine 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 jerry blimbaum NSWC panama city, fl === Subject: Re: C code for dual processor machine 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 > jerry blimbaum NSWC panama city, fl === Subject: Simplifying inequalities 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 DPHil student in theoretical physics in University of Oxford === Subject: RE: Strange ReplaceAll behavior f[c] == (c+b)/.ru[c] == (c+b)/.c->x == x+b == b+x Bobby Treat -----Original Message----- === Subject: 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 === Subject: RE: Strange ReplaceAll behavior 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----- === Subject: 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 === Subject: Re: Strange ReplaceAll behavior 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 Andrzej Kozlowski Toyama International University JAPAN 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 === Subject: a numerical integration 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 === Subject: Real Time Animation 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 === Subject: The Innovation Algorithm Hi there! 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 === Subject: Re: Re: How do I pick out the expression under a radical? 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 === Subject: Re: How do I pick out the expression under a radical? > I'm a poor physicist trying to figure out how to sort out the > physical from the 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? === Subject: Re: timing with Play 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 === Subject: timing with Play 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 Phone: +43-1-4277-38624 Fax: +43-1-4277-9386 === Subject: AW: Strange ReplaceAll behavior Hello Lawrence, using Set (=) instead of SetDelayed (:=) in f[x_]:=(a+b) /. ru[a]; will give you what you desire. Matthias Bode. -----UrsprÌ.b9ngliche Nachricht----- Von: Lawrence A. Walker Jr. [mailto:lwalker701@earthlink.net] 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 === Subject: RE: Re: How do I pick out the expression under a radical? 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----- === Subject: Re: How do I pick out the expression under a radical? > I'm a poor physicist trying to figure out how to sort out the > physical from the 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? === Subject: Mathematica 4.1 and MacOSX 10.2 Hello. 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. -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- Toshinao Ishii mailto: ici@ici.mine.nu Hello. 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. -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- Hiragino Kaku Gothic ProToshinao Ishii mailto: ici@ici.mine.nu --Apple-Mail-2-977368608-- === Subject: Re: Mathematica 4.1 and MacOSX 10.2 Toshinao, Try reinstalling Mathematica from the CD. --- Selwyn Hollis > Hello. > 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. > -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- > Toshinao Ishii mailto: ici@ici.mine.nu > Hello. > 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. > -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- > Hiragino Kaku Gothic ProToshinao Ishii > mailto: ici@ici.mine.nu > --Apple-Mail-2-977368608-- === Subject: Re: Simplifying inequalities 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. 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 > DPHil student in theoretical physics in University of Oxford === Subject: Why can't Mathematica do this simple integral ? Hi 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. -- Linux 12:02am up 5:06, 1 user, load average: 0.54, 0.22, 0.08 === Subject: Re: Simplifying inequalities 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]= > 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 >> DPHil student in theoretical physics in University of Oxford === Subject: RE: Strange ReplaceAll behavior >-----Original Message----- === >Subject: 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 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 === Subject: Re: Strange ReplaceAll behavior 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 Fax: +44 (0)870 164 0565 >-----Original Message----- === >Subject: 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 > 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 === Subject: Re: Exporting as space-delimited ASCII file in 3.0? 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 ----- === Subject: Exporting as space-delimited ASCII file in 3.0? > Hi! > 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 === Subject: Exporting as space-delimited ASCII file in 3.0? Hi! 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 === Subject: Re: Could someone verify a long Pi calculation in Version 4 for me? 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: >> $Version >> 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 === Subject: Posting Mathematica Input to MathGroup 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/ === Subject: Re: Posting Mathematica Input to MathGroup 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 to use menu>Kernel>Show In/Out Names. 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 Fax: +44 (0)870 164 0565 > 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/ === Subject: Re: Posting Mathematica Input to MathGroup Actually, it is easily done if you uncheck Kernel->Show In/Out Names. Then 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/ === Subject: Re: Real Time Animation >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 === Subject: Re: Simplifying inequalities 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 > DPHil student in theoretical physics in University of Oxford > > > > >> >> === Subject: Re: Simplifying inequalities 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 (?) 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 > DPHil student in theoretical physics in University of Oxford === Subject: Request for Mathematica Programming help. 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 === Subject: Re: Request for Mathematica Programming help. 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 === Subject: Re: Request for Mathematica Programming help. > 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 === Subject: RE: Why can't Mathematica do this simple integral ? 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----- === Subject: Why can't Mathematica do this simple integral ? Hi 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. -- Linux 12:02am up 5:06, 1 user, load average: 0.54, 0.22, 0.08 === Subject: RE: Real Time Animation Put this after the Plot statement, in the same cell: SelectionMove[EvaluationNotebook[], All, GeneratedCell] FrontEndTokenExecute[OpenCloseGroup] FrontEndTokenExecute[SelectionAnimate] Bobby Treat -----Original Message----- === Subject: Real Time Animation 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 === Subject: RE: Re: Could someone verify a long Pi calculation in Version 4 for me? >>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----- === Subject: Re: Could someone verify a long Pi calculation in Version 4 for me? 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, to > other programs out there. > I used one processor of a dual 1GH Mac and got the same answer with the > following speed: > $Version > 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 === Subject: Mathematica 4.1 Config:Fonts 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 === Subject: Re: Mathematica 4.1 Config:Fonts 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 === Subject: Re: Mathematica 4.1 Config:Fonts > 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 === Subject: Are configuration & UI better in 4.2? 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 configure. My platform is SuSE Linux on Intel. 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? === Subject: Re: Are configuration & UI better in 4.2? 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 > configure. My platform is SuSE Linux on Intel. > 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? === Subject: Problems using Legend whith LogPlot and LogPlotPlot 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. http://www.endymion.com/products/mailman/ === Subject: How to get a listing of currently defined symbols 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, === Subject: Re: How to get a listing of currently defined symbols 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, === Subject: How to get a listing of currently defined symbols 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 mailto:shippee@jcs.mil === Subject: Re: How to get a listing of currently defined symbols > 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 Fax: +44 (0)870 164 0565 === Subject: Re: How to get a listing of currently defined symbols 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, | === Subject: Davidon-Fletcher-Powell Algorithm 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. === Subject: Re: Simplifying inequalities 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 > > > > >> 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 > DPHil student in theoretical physics in University of Oxford > > > > >> >> > > > Andrzej Kozlowski > Toyama International University > JAPAN === Subject: Re: Simplifying inequalities 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 >> 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 > DPHil student in theoretical physics in University of Oxford > > > > >> >> Andrzej Kozlowski Toyama International University JAPAN === Subject: Bug in Solve? 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] === Subject: RE: Real Time Animation 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----- === Subject: RE: Real Time Animation You still get one show from the generation of the frames which you don't want the audience to see. The grouping and closing tidies up the frames but now it is too late. Hugh -----Original Message----- === Subject: RE: Real Time Animation Put this after the Plot statement, in the same cell: SelectionMove[EvaluationNotebook[], All, GeneratedCell] FrontEndTokenExecute[OpenCloseGroup] FrontEndTokenExecute[SelectionAnimate] Bobby Treat -----Original Message----- === Subject: Real Time Animation 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 === Subject: RE: Real Time Animation 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----- === Subject: Real Time Animation >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 === Subject: Some General Questions about Mathematica 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. === Subject: Re: Mathematica 4.1 Config:Fonts 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 === Subject: Re: Mathematica 4.1 Config:Fonts > 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 === Subject: Re: Are configuration & UI better in 4.2? > 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. -- P.J. Hinton User Interface Programmer paulh@wolfram.com Wolfram Research, Inc. Disclaimer: Opinions expressed herein are those of the author alone. === Subject: Re: Are configuration & UI better in 4.2? > 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. === Subject: RE: Problems using Legend whith LogPlot and LogPlotPlot 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/ 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. http://www.endymion.com/products/mailman/ === Subject: Strange Behavior of Mathematica under Windows Hi MathGroup, 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. After that the FrontEnd is still linked to the Kernel but all results are totally strange, for example I evaluated the bad cell and it gave the correct result then I tried to evaluate next cell and no results are given. This is not the first time it happens. After 100 or less inputs are evaluated there is great probability that it occurs. I used Mathematica 4.1 for Windows. Any suggestions will be very aprreciated. Cesar === Subject: Need to evaluate functions of the form y=5e+5 x1+2e-1x2, etc., were e means 10^ 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 === Subject: Re: A Bessel integral 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 Email: vvb@mail.strace.net 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 === Subject: Re: Bug in Solve? > 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); Table [Developer`ClearCache[]; Developer`ZeroQ[zz], {10}] One workaround would be to use exact input, say by preprocessing with Rationalize. Daniel Lichtblau Wolfram Research === Subject: Re: Bug in Solve? >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 === Subject: Re: Could someone verify a long Pi calculation in Version 4 for me? > Could someone calculate the number Pi to 67,108,864 (2^26) decimal places > for me in Version 4 and post here or email me the last 20 digits or so. > I would do it but I only have Version 3 and it would take way to long.. > 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 === Subject: RE: Re: Could someone verify a long Pi calculation in Version 4 for me? >>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----- === Subject: Re: Re: Could someone verify a long Pi calculation in Version 4 for me? >>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 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----- === > Subject: Re: Could someone verify a long Pi calculation in > Version 4 for me? > 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: > $Version > 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 === Subject: RE: Re: Posting Mathematica Input to MathGroup I prefer to delete all output and then Copy As>Notebook expression. It looks ugly in the e-mail, but it's easily pasted into Mathematica. Bobby -----Original Message----- === Subject: Re: Posting Mathematica Input to MathGroup 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 to use menu>Kernel>Show In/Out Names. 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 Fax: +44 (0)870 164 0565 > 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/ === Subject: Accuracy and Precision 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 === Subject: Re: Accuracy and Precision 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 Fax: +44 (0)870 164 0565 > 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]:= > 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]:= > Out[9]= > -(54767/66192) > In[10]:= > N[%] > Out[10]= > -0.8273960599468214 > PK === Subject: Using Text Cells 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/ === Subject: Re: Bug in Solve? Solve[youre equation, p1, VerifySolutions->True] will return a solution. So will Solve[Rationalize[your equation],p1]. Andrzej Kozlowski Toyama International University JAPAN > 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] === Subject: Re: How to get a listing of currently defined symbols 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 === Subject: RE: How to get a listing of currently defined symbols ?* does the trick. You can limit it to Global variables with ?Global`* Bobby -----Original Message----- === Subject: How to get a listing of currently defined symbols 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, === Subject: Re: Mathematica 4.2 on Solaris 6? > Hi All, > 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. Tel: 020 7679 6408 Fax: 020 7679 6269 Internal telephone: ext 46408 e-mail davek@medphys.ucl.ac.uk === Subject: Re: Accuracy and Precision 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]:= > 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]:= > Out[9]= > -(54767/66192) > In[10]:= > N[%] > Out[10]= > -0.8273960599468214 > PK === Subject: Re: A Bessel integral > 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]:= $Version Out[7]= 4.2 for Mac OS X (June 4, 2002) Andrzej Kozlowski Toyama International University JAPAN === Subject: Re: Bug in Solve? 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 === Subject: Re: A Bessel integral > 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 *10^] 5*10^+5x1+2*10^-1x2 ToExpression[%] 500000*x1 + x2/5 Bobby Treat -----Original Message----- === Subject: Need to evaluate functions of the form y=5e+5 x1+2e-1x2, etc., were e means 10^ 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 === Subject: Re: Accuracy and Precision Neither SetAccuracy[expr,n] nor SetPrecisions[expr,n] modify expr. These functions modify the prinout not the internal representation. So, the first computation of f is done with approximate numbers and doesn't result in a correct answer due to approximate arithmetic. By assigning a rational expression to each of the variables, you have made them exact numbers and Mathematica responds with an exact solution. >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 === Subject: Determination of a double integral Hallo, I have the problem, that I want to determine the numerical solution of a double integral, like e.g. the following: << Statistics`ContinuousDistributions` ndist = NormalDistribution[0, 1]; N[Sigma^2/(Abs[Mu_1^2 - Mu_2^2]) Integrate[(Integrate[1/y_1 1/(y - y_1) PDF[ndist, d*Log[y_1] + c*Log[y - y_1] + f]* PDF[ndist, Sigma*(c*Log[y_1] + d*Log[y - y_1] + g)], {y_1,0,y}]), {y,0,Lambda}]] whereby Lambda := 3; Mu_1 := 4; Mu_2 := 5; Sigma := 0.25; a = 1/2 (Mu_1^2 + Mu_2^2/Sigma^2) b = 1/2 (Mu_2^2 + Mu_1^2/Sigma^2) c = -Mu_2/(Mu_1^2 - Mu_2^2) d = -Mu_1/(Mu_2^2 - Mu_1^2) f = d*a + c*b g = c*a + d*b Sorry for the poor code... The problem is, that Mathematica doesn't give me a result (after waiting 2 hours I turned my machine off). Thus, the question is, if there is still a possibility of solving such complicated expressions... TIA, Sven. === Subject: Solve simultaneous eqns I am trying to solve a system of simultaneous equations with 26 variables and 14 equations (the 12 free variables can be any of the 26 from the eqns.. preferably ones that will minimize the computation time for the other 14). These equation are not linearly related.. the highest degree in any one eqn is degree 4 i believe.. and there are some cross terms in the equations but not every equation depends on every variable.. (some are actually rather simple eqns). Any ideas on how to get started with this using mathematica (any ideas for algorithms).. Anything will be helpful.. I can be reached at ngupta2@seas.upenn.edu Nachi === Subject: Swapping Ctlr and Alt on Linux This may seem like a trivial issue, but I find it very frustrating. I use several different packages on my Linux system. I also work in a wide variety of intellectual domains. In every package (JBuilder, KMail, Emacs, XEmacs, Mozilla, xterm, Konsole, etc.) the keyboar mapping is a bit differnt from the other. There are certain idioms which I find to be fairly invaraint between these different packages. I tend to use this common subset more than the package specific idioms. Switching from one package to the next can be a very disorienting experience. It can be even trickier to try and copy and paste from one to the next. I also use 'special' characters in certain domains, ç,?,.81,?,[CapitalYAcute],[CapitalThorn], ß, ¯, etc. Add to all of this, that I run beta code for just about everything. The function of my keyboard changed like the weather. I have spent hours trying to figure out why I can no longer type '.9a'. This doesn't even address the problems of switching between character encodings, or keyboard compose modes. The last thing I want to start doing is messing with the key mappings in my user environment. I want to control the way my keyboard works with Mathematica from within the Mathematica runtime environment. That is, the configuration should be loaded when Mathematica loads, and should not impact the rest of my X environment. If I have come across as a bit jaded regarding this issue, there are reasons. There is a history. I don't find keyboard configuration issues interesting. I want my fine keyboard to just work, the way I want it to work. Most of the software I use on my SuSE Linux box now uses the Shift+point movement = select text. Ctrl+Insert = copy Shift+Insert = paste Ctrl+c = copy Ctrl+x = cut Ctrl+v = paste Shift+End = select to end of line. etc. Yes, I said I use XEmacs, and Emacs. Yes (X)Emacs is different, but adding yet another alteration with Mathematica is just too much. Is there a way around this? STH === Subject: Kernel API Hi all, Is it possible to get a document of the description of how to interface with the kernel? Kind of what should an interface say to the kernel and how to connect to it. Paulo === Subject: Re: Question KeyEventTranslation >FrontEndExecute[{ > FrontEnd`NotebookWrite[FrontEnd`SelectedNotebook[], > [LeftDoubleBracket][RightDoubleBracket],After]}] FrontEndExecute[{ FrontEnd`NotebookWrite[FrontEnd`SelectedNotebook[], [LeftDoubleBracket][RightDoubleBracket],After], FrontEnd`SelectionMove[FrontEnd`SelectedNotebook[], Previous, Character]}] or FrontEndExecute[{ FrontEnd`NotebookWrite[FrontEnd`SelectedNotebook[], [LeftDoubleBracket][SelectionPlaceholder][RightDoubleBracket], Placeholder]}] -------------------------------------------------------------- Omega Consulting The final answer to your Mathematica needs Spend less time searching and more time finding. http://www.wz.com/internet/Mathematica.html === Subject: Linux Users? I'm trying to address the special issues related to using Mathematica on the Linux platform. I use SuSE Linux. I hit a lot of snags which I suspect I would not if I used a Windows system. These are typically not all that big if a problem _once I figure out what's going on_. What I hope to do is collect all such matters and document them effectively in something like a Mathematica Linux survival guide. If anybody is familiar with gotchas which I haven't discussed here http://66.92.149.152/proprietary/com/wri/index.html I'm interested in hearing what you have to say. Of particular interest are the issues faced by a person who is not familiar with the technical aspects of Linux and Unix. What does the former Windows user experience? What would help make this less painful? STH