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Re: Re: 3d plots in mathematica 5.0 #4

Hi David,

<snippage>

Very nice!  

Begs the question as to why anyone would bother with ContourPlot3D ....

Regards,

Dave

0
8/1/2005 5:12:52 AM
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Re: Re: 3d plots in mathematica 5.0 #5
Hi David, I suppose because it is not always possible to simply parametrize the contour surfaces. David Park djmp@earthlink.net http://home.earthlink.net/~djmp/ From: David Annetts [mailto:davidannetts@aapt.net.au] Hi David, <snippage> Very nice! Begs the question as to why anyone would bother with ContourPlot3D .... Regards, Dave ...

Re: Re: 3d plots in mathematica 5.0
Hi Fiz, > > Here is the function > > w=exp(-2(x^2+y^2/r0^2))*exp(-2(z^2/z0^2)) > > r0=1.4 and z0=1.4 The function is Exp, not exp, and it needs square brackets not parentheses. r0 = 1.4; z0 = 1.4; w = Exp[(-2(x^2 + y^2/r0^2))]*Exp[(-2(z^2/z0^2))] // FullSimplify The function you're after is ContourPlot3D, and you use it like Needs["Graphics`ContourPlot3D`"] ContourPlot3D[w, {x, -1, 1}, {y, -1, 1}, {z, -1, 1}, Contours -> {.75}, \ PlotRange -> All, Axes -> True, BoxRatios -> {1, 1, 1}]; This produces "something",...

Re: Re: 3d plots in mathematica 5.0 #3
In this particular case we can parametrize the contour surfaces. Looking for the constant values surface, we obtain a simplification to ellipsoids. Exp[-2*(x^2 + y^2/r0^2)]*Exp[-2*(z^2/z0^2)] == c Simplify[%] % /. x^2 + y^2/r0^2 + z^2/z0^2 -> k2 Solve[%, k2][[2,1]] krule = % /. k2 -> k^2 Write the equation for the constant value ellipsoid in terms of contour value c. x^2 + y^2/r0^2 + z^2/z0^2 == k^2 % /. krule (Distribute[#1/(-Log[c]/2)] & ) /@ % Standard parametrization of an ellipsoid (from Alfred Gray, Modern Differential Geometry).... ellipsoid[a_, b_, c_...

Re: 3d plots in mathematica 5.0
Hi, I would like to plot a 3d gaussian function but I'm having trouble getting started. Any help gratefully appreciated. Fiz. Here is the function w=exp(-2(x^2+y^2/r0^2))*exp(-2(z^2/z0^2)) r0=1.4 and z0=1.4 Needs["Graphics`"]; r0=1.4;z0=1.4; w[x_,y_,z_]:= E^(-2(x^2+y^2/r0^2))*E^(-2(z^2/z0^2)); DisplayTogetherArray[ Partition[ Table[ Plot3D[ w[x,y,z], {x,-1.5r0,1.5r0}, {y,-1.5r0,1.5r0}, PlotRange->{0,1}, AxesLabel->{"x","y","w "}, ...

Re: 3d plots in mathematica 5.0 #2
Needs["Graphics`"]; r0=1.4;z0=1.4; w[x_,y_,z_]:= E^(-2(x^2+y^2/r0^2))*E^(-2(z^2/z0^2)); DisplayTogetherArray[ Partition[ Table[ Plot3D[ w[x,y,z], {x,-1.5r0,1.5r0}, {y,-1.5r0,1.5r0}, PlotRange->{0,1}, AxesLabel->{"x","y","w "}, PlotLabel->"z = " <> ToString[z]], {z,0,z0,z0/7}], 2], ImageSize->468]; Bob Hanlon > > From: jfitzp@gmail.com > Date: 2005/07/29 Fri AM 12:42:10 EDT > Subject...

Re: 3d plots in mathematica 5.0 #3
You have a function of three variables x, y, z. Are you sure of this? Tomas Garza Mexico City >From: jfitzp@gmail.com >Subject: Re: 3d plots in mathematica 5.0 >Date: Fri, 29 Jul 2005 00:42:10 -0400 (EDT) > >Hi, > >I would like to plot a 3d gaussian function but I'm having trouble >getting started. Any help gratefully appreciated. > >Fiz. > >Here is the function > >w=exp(-2(x^2+y^2/r0^2))*exp(-2(z^2/z0^2)) > >r0=1.4 and z0=1.4 > ...

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Manfred, There are a number of ways to combine two plots with different domains. Let's define your two functions. f1[x_] := 1/2x^2 - x f2[x_] := -1/2x^2 + x We could combine them into one function using UnitSteps and make a regular Plot. f[x_] := f1[x](UnitStep[x + 3] - UnitStep[x]) + f2[x](UnitStep[x] - UnitStep[x - 3]) Plot[f[x], {x, -4, 4}, Frame -> True, PlotRange -> All]; But if you want to deal with them as separate functions you could combine two plots with the following statement. Block[{$DisplayFunction = Identity}, g1 = Plot[f1[x],...

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I have been considering making a switch from Windows XP to a Mac for a while. The arrival of Vista makes the decsion a little harder, and I have just been told that Mac Office 2008 will not include VBA which means some of my spreadsheets won't work under Leopard. I visited the Apple Mac store this past weekend to see a demo - and was a little shocked that the demo machine crashed four times in one hour. It seemed like iCal, Time Machine and iPhoto are not stable and likely to be a huge pain. I also note there is a major bug which results in lost files during copy and move operatio...

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Hi, The keyboard problems on Linux appear when the NumLock is on. yehuda > > From: Daniel Arteaga <darteaga@NOSPAMPLEASE.ub.edu.INVALID> > Date: 2004/12/16 � AM 10:40:26 GMT+02:00 > To: mathgroup@smc.vnet.net > Subject: Re: Mathematica 5.0 on Mandrake 10.1? > > Hi, > > > Hi, > > > > Is it possible to use Mathematica 5.0 on Mandrake 10.0/10.1? > > > > I can install it without problems, but when I try to run it I get a lot > > of errors about fonts, colors etc. > > > > Anyone with hints? >...

Re: Re: Re: Re: Newly Released Mathematica 5.1 Delivers Unmatched Performance for Handling Data
John, I was thinking of angular measurement. The following works... Needs["Miscellaneous`Units`"] Convert[(Pi/4)*Radian, �] 45 � but the following, using an approximate value, does not work because Degree immediately multiplies out undoing the conversion. Convert[1.35 Radian, �] 1.35 The ExtendUnits package at my web site fixes this but at the expense of putting a HoldForm on the degree symbol. Needs["Miscellaneous`V4ExtendUnits`"] 1.35 Radian // ToUnit[�] % // FullForm 77.3493 � Times[77.34930234266115`, HoldForm[Degree]] Since I generally...

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Re: Re: Re: Re: Newly Released Mathematica 5.1 Delivers Unmatched Performance for Handling Data #2
Speaking of colors, are any of you aware of a standard color palette or utility that electronically automates conversion of a given color scheme legend to a palette viewable by students who have trichromatic or dichromatic colorblindness? For example, to such people the area below the light blue in this graph http://www.wolfram.com/products/mathematica/usersanduses/experience/images/s urna3dvolume.th.jpg might be viewed as brown instead of a continuous green-yellow-red gradient. While colors are used as teaching tools in math and geography, colorblindness (like nearsightedness) is ...

AW: Re: Politically incorrect response to Re: a bug in RealDigits? Mathematica 5.0
Hi, both is right. If you calculate 1/3 * 6.4 -> 2.13333 % // Precision -> MachinePrecision But SetPrecision[2./3* 6.4, 100] -> 4.26666666666666660745477201999165117740631103515625`100. You can see, that the `100 in the output 4.2...`100. shows, that the precision of that number is 100. The $MachinePrecision is 53*Subscript[log, 10]*2 an depend on the machine. If you define a higher precision, it depends on the definition you use (and the amount of memory, you have in your computer; and the amount of time you have to wait for the result ;-) ). Regards...

AW: Re: Politically incorrect response to Re: a bug in RealDigits? Mathematica 5.0 #2
Hi, of course are Mathematica programs hardware dependent, if they use the processor arithmetic units. And there are differences, if you use a 16, 32 or 64 bit processor. or a 586, ppc or a xyz processor Mathematica runs on (...for 8 bit Mathematica was not implemented, as far as I know...). If you define a arbitrary precision, Mathematica is not (so much...) hardware depended. Regards Peter -----Urspr�ngliche Nachricht----- Von: Harold Noffke [mailto:Harold.Noffke@wpafb.af.mil] Gesendet: Sonntag, 28. Dezember 2003 11:11 An: mathgroup@smc.vnet.net Betreff: Re: Politically ...

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