1345 ==== Hallo, I have the following problem that I am trying to solve with Mathematica. I am actually creating simulations. I simulate summers: four months of summer-May,June,July,August. I use the following formula: Temperature today=a*(The average temperature for this date during the last 30 yrs)+b*(The temperature yesterday) + c*(Random historical difference between the temperature today and the temperature yesterday:I have thirty years of data , therefore, 30 differences for every day); a+b=1 and c is approaching 1. Based on this formula and changing the differences I create 1000 temperatures for every day. I want to make the historical average for a certain date to be equal to the simulated average for this date and the same with the standard deviation and this by changing the values for a and c. I tried to do it with the month of May but it gives me an error message: It requires non-algebraic computation(which is true). So I am stuck there. I have Mathematica 4.1 version. I would like to assign number values to to a and c and then create my summers and then change the values I assigned to the variables so that I make the simulated averages and standard deviations equal to the historical ones. I would highly appreciate the help and thank you in advance. Gratefully, Hristina Georgieva. ****************************DISCLAIMER*********************************** passed on to, or made available for use by any person other than the addressee(s). Fortis rules out any and every liability resulting from any electronic transmission. ****************************************************************************\ *** ==== David Park wanted to fix it so that (-1/2)(a+b+c) wouldn't return (1/2)(-a-b-c) I don't know why it works that way, but the following definitions will fix the problem. The second definition puts the minus sign in the numerator of a rational number. --------------------------- MakeBoxes(a_*b:Plus[_?Negative*_, (_?Negative*_)..], form:(StandardForm|TraditionalForm)]:= RowBox[{MakeBoxes@@{-a,form}, RowBox[{\(\, MakeBoxes@@{-b, form}, \ \)\}] }] MakeBoxes[p_Rational?Negative, form:(StandardForm|TraditionalForm) ]:= FractionBox[ MakeBoxes@@{Numerator[p],form}, MakeBoxes@@{Denominator[p],form} ] ----------------------------- This fixes situations you didn't address. Now (-1/2)(a+2.3 b+c) returns itself. ---------------------------- Ted Ersek Get Mathematica tips, tricks from http://www.verbeia.com/mathematica/tips/Tricks.html ==== I can't get Mathematica to display Helvetica Narrow or Arial Narrow as part of a Graphics object, although these fonts show and print properly if they are used in cell text. How can I troubleshoot this problem? bill ==== > > This matches all but the first term of your series --- I think! (You > didn't have things in Mathematica notation, so I found it a little > confusing.) > > ClearAll[term, sum] > term[3] = 2/5; > term[n_Integer] /; n > 3 := term[n] = term[n - 1](Prime[n - 1] - 2) > /Prime[n] > term[n_] := term@Round@n (* for plots *) > sum[3] = term[3]; > sum[n_Integer] /; n > 3 := sum[n] = sum[n - 1] + N@term[n] > sum[n_] := sum@Round@n (* for plots *) > term /@ Range[3, 8] > Short[sum /@ Range[3, 5002], 5] > > {2/5, 6/35, 6/77, 54/1001, 54/1547, 810/29393} > > {0.4, 0.571429, 0.649351, 0.703297, \ \\[LeftSkeleton]4992\\[RightSkeleton], > 0.978568, 0.978569, 0.97857, 0.97857} > > (Add 1/5 to get the result with your first term.) > > The ratio test is inconclusive: > > Plot[(Prime[Round[n]] - 2)/ > Prime[Round[n] + 1], {n, 4, 1000}]; > > a constant times > 1/(n*Log[n]^2.7), > > Plot[n*Log[n]^2.7*term[n], {n, 3, 1000}]; > > but I don't have a ready proof of it, despite the plot. > > Likewise, it seems that THAT series converges. > > Sum[1/(n*Log[n]^2.7), {n, 2, 100000}] > > 1.98995 > > It certainly does if the following is correct: > > NSum[1/(n*Log[n]^2), {n, 2, Infinity}] > > 2.109742801134059 + 0.*I > > but I wouldn't bet a gold mine on it, without a theoretical proof. > > Bobby > > > > I have a tricky infinite sum over the primes and want to see whether it > > converges or diverges. I am having difficulty programming it (in > > Mathematica 4), and would appreciate some help. The sum is as follows: > > > > 3/5 + (3/5)(2/7) + (3/5)(5/7)(2/11) + (3/5)(5/7)(9/11)(2/13) + . . . > > > > + [(Prime[3]-2)(Prime[4]-2). . .(Prime[n-1]-2)/(Prime[3])(Prime[4]) . . > > .(Prime[n-1])] x [2/(Prime[n])] . . . > > > > and I want to see what happens as n --> Infinity. > > > > > > Allan Haley > > > > > > > > -- > majort@cox-internet.com > Bobby R. Treat Here is a nonrigorous approach that, I believe, could be made rigorous with appropriate majorization. Among other things it gives an asymptotic estimate for terms that is close to what you found numerically. I'll use 'k' to denote generic constants that may change from one step to another. The idea is to approximate Prime[j] by j*Log[j]. A rigorous argument would need to find appropriate upper and lower bounds to this approximation, among other things. The nth term, for large n, may be approximated as k * Product[1-2/(j*Log[j]),{j,m,n}] * 2/(n*Log[n]) where m is some suitably large fixed integer (and k is dependent on m). The logarithm of this term is thus described as k + Sum[Log[1-2/(j*Log[j]),{j,m,n}] - (Log[n] + Log[Log[n]]) This is approximated as k - Sum[2/(j*Log[j]),{j,m,n}] - (Log[n] + Log[Log[n]]) which in turn is approximated as k - 2*Log[Log[n]] - (Log[n] + Log[Log[n]]) Hence the nth term is approximated by the exponential of this, or k * 1/n * 1/(Log[n])^3 This converges by the integral test. Daniel Lichtblau Wolfram Research ==== There was an error in the first MakeBoxes statement: \(\ instead of \[\. \ So I think it should have been: MakeBoxes[a_*b : Plus[_?Negative*_, (_?Negative*_) ..], form : (StandardForm | TraditionalForm)] := RowBox[{MakeBoxes @@ {-a, form}, RowBox[{\(\, MakeBoxes @@ {-b, form}, \)\}]}] MakeBoxes[p_Rational?Negative, form : (StandardForm | TraditionalForm) ] := FractionBox[ MakeBoxes @@ {Numerator[p], form}, MakeBoxes @@ {Denominator[p], form} ] But what about -(a + b + c) or (-1/2)(a - b - c) Actually, I think this is not a simple question because the user will often have various desires about which term in the sum he wants to be positive \ and which factors he wants taken out. There is a similar problem in lists and arrays. Sin[x]*E^y*{{1, 1}, {-1, 1}} {{E^y*Sin[x], E^y*Sin[x]}, {(-E^y)*Sin[x], E^y*Sin[x]}} leading to a lot of duplicated expressions. It would certainly be nice if Mathematica had a controlled factoring of expressions from sums and arrays. For sums it's the minus sign that is the problem and for arrays it is everything. Of course, without introducing HoldForm (if possible). David Park djmp@earthlink.net http://home.earthlink.net/~djmp/ RowBox[{MakeBoxes@@{-a,form}, RowBox[{\(\, MakeBoxes@@{-b, form}, \ \)\}] }] MakeBoxes[p_Rational?Negative, form:(StandardForm|TraditionalForm) ]:= FractionBox[ MakeBoxes@@{Numerator[p],form}, MakeBoxes@@{Denominator[p],form} ] ----------------------------- This fixes situations you didn't address. Now (-1/2)(a+2.3 b+c) returns itself. ---------------------------- Ted Ersek Get Mathematica tips, tricks from http://www.verbeia.com/mathematica/tips/Tricks.html ==== My way of \rationalising\ this way of using ParametricPlot3D is to \ imagine that the fourth component represents an internal coordinate, so you have a 2-dimensional surface embedded in a 4-dimensional space rather than a 3-dimensional space. One way to represent this internal coordinate is to \ use various graphics tricks (e.g. colour, etc). Turning this back to front, you might as well place any parameter-dependent (or parameter-independent) graphics in the fourth component. However, I am perplexed that PlotStyle is not available (for parameter-independent graphics) in addition to the above 4-dimensional trick. -- Steve Luttrell West Malvern, UK > Ahh.. my own favorite, long-standing complaint: No PlotStyle option for > ParametricPlot3D. Why this DESIGN FLAW still persists after all these > years/versions is beyond me. > > Anyway, here's the arcane way to work around it: > > ParametricPlot3D[{Sin[t],Cost[t],t/10, {Hue[0],Thickness[.01]}}, > {t,0,4Pi}] > > ----- > Selwyn Hollis > http://www.math.armstrong.edu/faculty/hollis > > > > > GentleBeings > > I have a ParametricPlot3D which is a line in space > > I want to change the color and thickness of the line > > in Plot this is PlotStyle > > but Prolog and Shaw do not seem to work > > kenf > > > > > ==== something like this : plt = Graphics3D[ Plot3D[Sin[x^2 + y^2], {x, -Pi, Pi}, {y, -Pi, Pi}, PlotPoints -> 30, DisplayFunction -> Identity]] Show[plt /. Polygon[pnts_] :> {{EdgeForm[], Polygon[pnts]}, Line /@ Part[Transpose[{pnts, RotateLeft[pnts]}], {1, 3}]}, DisplayFunction -> $DisplayFunction] Jens > > Is it possible to create 3D graphics, such as ListPlot3D, which will > show the mesh lines in one direction only? > > Something akin to > > MeshStyle->{ {mesh style for x lines}, {mesh style for y lines} } ? > > [I'm trying to do convey an effect similar to what StackGraphics does, > but I have other problems with it.] ==== try to put CheckAbort[] into your computation and return the last result in the second argument of CheckAbort[] Jens > > I was wondering if anyone knew a way to interrupt a calculation that \ would > produce an answer. > > More specifically, I am currently running a FindMinimum calculation with > MaxIterations -> 3000 and WorkingPrecision -> 32. With the complexity of \ my > problem this takes quite a long time. When my MaxIterations were lower it > would fail out with a 'failed to converge after 30 iterations' error and > produce the last iteration. Is there a way to produce the last fully \ solved > iteration the FindMinimum has found? I'm pretty sure my solution will not > fully converge but I would like to see where it is at so I can do some \ more > calculations. > > > Jason ==== > I was wondering if anyone knew a way to interrupt a calculation that \ would > produce an answer. > > More specifically, I am currently running a FindMinimum calculation with > MaxIterations -> 3000 and WorkingPrecision -> 32. With the complexity of \ my > problem this takes quite a long time. When my MaxIterations were lower it > would fail out with a 'failed to converge after 30 iterations' error and > produce the last iteration. Is there a way to produce the last fully \ solved > iteration the FindMinimum has found? I'm pretty sure my solution will not > fully converge but I would like to see where it is at so I can do some \ more > calculations. > In Mathematica 5, you can use the StepMonitor option to keep track of the position of the iterations. To get a result, you will probably want to use \ it inside of CheckAbort, as in slowf[x_?NumberQ, y_] := CompoundExpression[Pause[1], (x - 1)^2 + (x^2 - \ y)^2] CheckAbort[ FindMinimum[f[x, y], {{x, -1.2}, {y, 1}}, StepMonitor :> (laststep = {x, y})], laststep] If the problem runs to completion, you'll get the result from FindMinimum, or if you abort the computation, you will get the position at the last step \ taken in the iteration. Rob Knapp Wolfram Research ==== I have a huge problem with the computation of the correlation coefficient and I thought somebody can help me with it. I compute that in many \ different ways. I even used the Time-series package to do it. But I still think there is a better way to do that computation. What is the best way to compute the correlation coefficient and the related integral timescale and Taylor Micro timescale from an experimental data. ==== I have to confess to being totally clueless re Mathematica so I am hoping for some nice guiding advice to help me solve my problem ... I'm writing an optimization routine and need to take derivatives (first and second) of the objective function. This isn't really my area so although I hope to do it analytically I would like to use the symbolic subroutines to verify my argument. Anyway, here is the problem: Objective function F(Y)=tr[(X'BX)^-1 X'AX] I need dF(Y) and ddF(Y). To get dF(Y) I used the implicit definition where tr[V'dF(Y)]=(d/dt)F(Y(t)) evaluated at t=0 for Y(t)=Y+tV. After some manipulations I get that dF(Y)=-2(BX(X'BX)^-1)X'AX(X'BX)^-1+2AX(X'BX)^-1 now I guess I could just take the straight forward derivatives with respect to X and it will be messy but presumably correct. Then I can use these in the numerical part. It's just that I'm not so sure of my matrix calculus and would like Mathematica to give it some added credibility. ==== My impression is that MathReader is basically the front end without the kernel. I agree wholeheartedly that the hypothetical MathReader you described would be vastly superior. But it's hard to imagine WRI giving it (or anything other than cripple-ware) away for free, and I doubt many people would want to pay for it. ----- Selwyn Hollis http://www.math.armstrong.edu/faculty/hollis > Dear MathGroup and WRI, > > I am interesting in knowing more precisely what can be done with > MathReader. > I suspect that it doesn't do what I really want but the documentation > is a > bit vague. > > This is what I would like MathReader to do: > > 1) The MathReader user should be able to open a Mathematica notebook > with > all output and graphics deleted and then evaluate the notebook and > generate > the graphics and animation. > > 2) The MathReader user should be able to install third party packages > and > applications, which could be utilized by the notebooks he reads. > > 3) But the MathReader user could not edit or change any notebook. He > could > not type in his own statements. He would be strictly limited to > evaluating > notebooks created by full Mathematica owners. > > If MathReader cannot evaluate a notebook, then all the graphics, > animation > and traditional form output must be embedded and this means that for > all > practical purposes the notebooks are too large for routine > transmission. In > that case MathReader can only be used for trivial purposes. > > David Park > djmp@earthlink.net > http://home.earthlink.net/~djmp/ > > ==== << Graphics`ParametricPlot3D` SphericalPlot3D[{1 + 0.25*Re[SphericalHarmonicY[8, 2, th, phi]], EdgeForm[]}, {th, 0, Pi}, {phi, 0, 2Pi}, PlotPoints -> {32, 64}] Jens > > Is it possible to remove the mesh lines in SphericalPlot3D? There is no > \mesh\ option. > > Also I was playing around with RealTime3D recently. Is their any way to \ get > colours displayed when using this? > > > Mike ==== Needs[\Graphics`ParametricPlot3D`\]; SphericalPlot3D[{2, EdgeForm[]}, {theta, 0, Pi},{phi, 0, 2 Pi}, PlotPoints -> 36]; Bob Hanlon << Is it possible to remove the mesh lines in SphericalPlot3D? There is no \mesh\ option. ==== why it is named \MathCode C++\ http://www.wolfram.com/products/applications/mathcode/ because it translate to ALGOL 60 ?? Jens > > It is learnt that any expression or program written in Mathematica can > be converted to C or Fortran programming languages, as > well as MathML, TeX, HTML, and Java. > How is actually conversion into C or C++ done ? > ==== \David Park\ schrieb im Newsbeitrag > If MathReader cannot evaluate a notebook, then all the graphics, \ animation > and traditional form output must be embedded and this means that for all > practical purposes the notebooks are too large for routine transmission. In > that case MathReader can only be used for trivial purposes. it's mainly the bitmap image and cell cache data that makes Mathematica notebooks too fat for transmission. However, this data can be safely discarded before transmitting a notebook file because it will be restored automatically by the Mathematica frontend upon loading the notebook. You \ can use Tim Wichmann's rmnbcache tool to strip cache data from notebooks (see the following URL). http://www.itwm.fhg.de/as/asemployees/wichmann/nbcache.html Eckhard -- Dr.-Ing. Eckhard Hennig www.kaninkolo.de/ai aidev \\at kaninkolo \\dot de ==== it is a woderfull idea to turn Mathematica into free software, because everyone can use a ASCII editor and create a notebook with Notebook[{Cell[TextData[\< > Dear MathGroup and WRI, > > I am interesting in knowing more precisely what can be done with \ MathReader. > I suspect that it doesn't do what I really want but the documentation is \ a > bit vague. > > This is what I would like MathReader to do: > > 1) The MathReader user should be able to open a Mathematica notebook with > all output and graphics deleted and then evaluate the notebook and \ generate > the graphics and animation. > > 2) The MathReader user should be able to install third party packages and > applications, which could be utilized by the notebooks he reads. > > 3) But the MathReader user could not edit or change any notebook. He \ could > not type in his own statements. He would be strictly limited to \ evaluating > notebooks created by full Mathematica owners. > > If MathReader cannot evaluate a notebook, then all the graphics, \ animation > and traditional form output must be embedded and this means that for all > practical purposes the notebooks are too large for routine transmission. \ In > that case MathReader can only be used for trivial purposes. > > David Park > djmp@earthlink.net > http://home.earthlink.net/~djmp/ ==== Perhaps you are really asking for a third product, between MathReader and Mathematica in functionality -- with pricing somewhere between those two. > Dear MathGroup and WRI, > > I am interesting in knowing more precisely what can be done with \ MathReader. > I suspect that it doesn't do what I really want but the documentation is \ a > bit vague. > > This is what I would like MathReader to do: > > 1) The MathReader user should be able to open a Mathematica notebook with > all output and graphics deleted and then evaluate the notebook and \ generate > the graphics and animation. > > 2) The MathReader user should be able to install third party packages and > applications, which could be utilized by the notebooks he reads. > > 3) But the MathReader user could not edit or change any notebook. He \ could > not type in his own statements. He would be strictly limited to \ evaluating > notebooks created by full Mathematica owners. > > If MathReader cannot evaluate a notebook, then all the graphics, \ animation > and traditional form output must be embedded and this means that for all > practical purposes the notebooks are too large for routine transmission. \ In > that case MathReader can only be used for trivial purposes. > > David Park > djmp@earthlink.net > http://home.earthlink.net/~djmp/ > > -- 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 fax 413 545-1801 Amherst, MA 01003-9305 ==== As MathReader doesn't include the Mathematica kernel, I don't see how it \ can possibly do any of the things you want. (Except perhaps locking the notebook.) As for .nb size issues, I have found that using WinRAR (http://www.rarlab.com/rar_archiver.htm available for just about all platforms) at the highest compression settings can compress image and animation heavy notebooks more than 20:1. --Urijah Kaplan > Dear MathGroup and WRI, > > I am interesting in knowing more precisely what can be done with MathReader. > I suspect that it doesn't do what I really want but the documentation is \ a > bit vague. > > This is what I would like MathReader to do: > > 1) The MathReader user should be able to open a Mathematica notebook with > all output and graphics deleted and then evaluate the notebook and generate > the graphics and animation. > > 2) The MathReader user should be able to install third party packages and > applications, which could be utilized by the notebooks he reads. > > 3) But the MathReader user could not edit or change any notebook. He \ could > not type in his own statements. He would be strictly limited to \ evaluating > notebooks created by full Mathematica owners. > > If MathReader cannot evaluate a notebook, then all the graphics, \ animation > and traditional form output must be embedded and this means that for all > practical purposes the notebooks are too large for routine transmission. In > that case MathReader can only be used for trivial purposes. > > David Park > djmp@earthlink.net > http://home.earthlink.net/~djmp/ > ==== (-1/2)(a + b + c) /. a_*b_Plus /; a < 0 :> (-a)*Expand[-b] Jens > > Dear MathGroup, > > How would one simplify > > (-1/2)(a + b + c) > 1/2(-a - b - c) > > so the minus sign was in front? Why does it get pulled within in the \ first > place? > > David Park > djmp@earthlink.net > http://home.earthlink.net/~djmp/ ==== David was right about a typo in my code. I don't know how that got in there. I added the first definition below to cover things like (-a-b-c) Notice I could have chaged MakeBoxes[a_*b: ] to MakeBoxes[(a_.)*b: ] but then we would get -1(a+b+c) when we really want -(a+b+c) So I made another definition to handle that case. ---------------------------- MakeBoxes[b:Plus[_?Negative*_, (_?Negative*_)..], form:(StandardForm | TraditionalForm)] := RowBox[{\-\, RowBox[{\(\, MakeBoxes @@ {-b, form}, \)\}]}] MakeBoxes[a_*b:Plus[_?Negative*_, (_?Negative*_)..], form:(StandardForm | TraditionalForm)] := RowBox[{MakeBoxes @@ {-a, form}, RowBox[{\(\, MakeBoxes @@ {-b, form}, \)\}]}] MakeBoxes[p_Rational?Negative, form:(StandardForm | TraditionalForm) ] := FractionBox[ MakeBoxes @@ {Numerator[p], form}, MakeBoxes @@ {Denominator[p], form} ] -------------------- For lots of other cases I recommend my HoldTemporary package posted at http://library.wolfram.com/infocenter/MathSource/705/ The package is really neat. In the package I define a function called HoldTemporary that works like HoldForm, but the hold is automatically released. You can use HoldTemporary to prevent mapping of Times across an array. ---------------------------- Ted Ersek Get Mathematica tips, tricks from http://www.verbeia.com/mathematica/tips/Tricks.html ==== With the previous version of Mathematica I was able to import data like this easily. \Raw Data\ \Sweep Frequency: Control Voltage\ Commentary Description FileLocation \Frequency\ 0,3,0,1,10000,0.0075 63 \ Freq(Hz) Ampl Bias Time(Sec) Z'(a) Z''(b) GD Err Range\ 1.000000E+04, 2.5000E-01, -1.7000E-02, 3.450000E+00, 3.0328E+04, -2.6945E+05, 0.0000E+00, 0, 6 7.943282E+03, 2.5000E-01, -1.7000E-02, 4.160000E+00, 4.0364E+04, -3.3784E+05, 0.0000E+00, 0, 6 This data file consists of 11 lines of header followed by 63 rows of data consisting of 9 columns. (Note: disregard the word wrapping in the above data) I would use a function like this: ReadData[datafile_] := Module[{fn, data}, fn = Import[datafile, \Table\]; data = Drop[fn, 11]; data] If is listed the data it woul look as expected: {1.000000E+04, 2.5000E-01, -1.7000E-02, 3.450000E+00, 3.0328E+04, -2.6945E+05, 0.0000E+00, 0,6}, ...etc. commas between each column. {{1.000000E+04,, 2.5000E-01,, -1.7000E-02,, 3.450000E+00,, 3.0328E+04,, -2.6945E+05,, 0.0000E+00,, 0,, 6}, ...etc. Any suggestions on a workaround? Sincerely, Charles Koehler ==== in general you can't, because Mathematica evaluates inner expressions first. In D[expression,z] the expression is evaluated first and when expression get a value different from R[z] it will be used later and your definition for Derivative[_]{R][_] is never found or searched. You must hinder Mathematica to evaluate R[z] for a symbolic argument and Derivative[m_][R][z_] := D[E^(-z^4), {z, m - 1}] R[z_?NumericQ] := (2 UnitStep[z] - 1) (Gamma[1 + 1/n] - Gamma[1/n, Abs[z]^n]/n); may work, if it fit to your computation. Jens > > I wish to define a function and its derivatives. The function expression \ is > complicated, but the derivatives are fairly simple (full cut-and-paste > notebook expression below). The problem is that the definition of the > function itself seems to shadow the definition of the derivatives. How can \ I > make Mathematica prefer my definition of the derivative over the one \ derived > from the definition of the funtion? > > In[1]:= n=4; > In[2]:= Derivative[m_][R][z_] := D[E^(-z^4), {z, m - 1}] > In[3]:= {R'[z],R''[z],R'''[z]} > Out[3]= {E^(-z^4), -4 E^(-z^4) z^3, -12 E^(-z^4) z^2 + 16 E^(-z^4) z^6} > In[4]:= R[z_] = (2 UnitStep[z] - 1) (Gamma[1 + 1/n] - Gamma[1/n, > Abs[z]^n]/n); > In[5]:= {R'[z],R''[z],R'''[z]} > Out[3]= > > Sincerely, > > Martin Manscher ==== > >>Thus, to summarize, if a,b, and c are coplanar (and not 0), then there \ exists some relation of the >>form: >>k1 a + k2 b + k3 c = 0 > > > Which brings up a good point about logic: > IF [points are coplanar] THEN [formula holds] > does not necessarily mean: > IF [formula holds] THEN [points are coplanar] > > Does anyone know how to test this situation? > > You're right, but in this case: [points are coplanar] IF AND ONLY IF [formula holds (with at least one k_i =!= 0)] You might have read \iff\ sometimes. That means \if and only if\, which \ is equivalence (often written as \<=>\). Peter ==== If the formula holds with k1,k2,k3 not all zero, then the three vectors (from the origin to each point) are linearly dependent. Since any two non-collinear vectors define a plane, and the third is a linear combination of these two, the three vectors are coplanar. The statement is true in either direction. Kevin > > > > Thus, to summarize, if a,b, and c are coplanar (and not 0), then there exists some relation of the > > form: > > k1 a + k2 b + k3 c = 0 > > Which brings up a good point about logic: > IF [points are coplanar] THEN [formula holds] > does not necessarily mean: > IF [formula holds] THEN [points are coplanar] > > Does anyone know how to test this situation? > ==== I have just installed Mathematica for students 4.2 on Mandrake 9.1. The installation worked without errors, and all parts of the distribution are in the folder they're supposed to be. However, the binaries \math\ and \mathematica\ exit with a \ \segmentation fault\ after(!) echoing the banner: me@mycomputer:~ > math Copyright 1988-2002 Wolfram Research, Inc. Segmentation fault me@mycomputer:~ > Does anyone know anything about this problem? I did already delete the entire distribution from /usr/local/Wolfram, and the binaries in /usr/local/bin and the program /usr/bin/mcc (all locations are the default!), and re-installed it (three times in total). The only problem I encountered was that the program \uncompress\ didn't exist. I \ solved this problem by linking \/bin/uncompress\ to \/bin/gunzip\, which \ worked without error-message. One more info: the exact same message (quoted above) was the reason why I re-installed Mathematica; so, it doesn't seem to be a problem of the re- installation, but rather of the binary under my current kernel \ (2.4.21-0.13mdk). Please answer directly to \loskamp@math.wisc.edu\ (I'm not a regular \ reader of this newsgroup, sorry). Marco ==== > It is learnt that any expression or program written in Mathematica can > be converted to C or Fortran programming languages, as > well as MathML, TeX, HTML, and Java. > How is actually conversion into C or C++ done ? In[1]:= ?CForm \CForm[expr] prints as a C language version of expr.\* Button[More\.89¥Ï, ButtonData :> \CForm\, Active \ -> True, ButtonStyle -> \RefGuideLink\] In[2]:= x^3+3 x +3 +1//CForm Out[2]//CForm= 4 + 3*x + Power(x,3)