=== Subject: Mandelbrot Sets and Sierpinski-Menger Sponge Code and Graphic Mandelbrot Sets and Sierpinski-Menger Sponge Code and Graphic can be found on Wolfram's site. See, for example, URL: http://library.wolfram.com/infocenter/Demos/4662/ Steven Shippee shippee@jcs.mil > I'm searching for a simple programm, compiled with mathematica 5.0, that reproduce fractals! > Mandelbrot sets, Julia sets, Menger sponge..... > Please, could you help me to find it? === Subject: Re: Re: Length of actual parameters in a function call. >> A few weeks ago, reviewing some tutorials on pattern matching and >> error >> handling, I came across an undocumented function to monitor the >> number of >> parameters submitted to a function in a call; but somehow I lost its >> track >> later. I guess that it must be something well-known by our >> specialists. Any >> help about where to locate it? >> The undocumented feature helps in tracking situations as when we >> define >> MyFunction[a_,b_,c_]:= something and then it is called as >> myFunction[a] >> leaving the function unevaluated. > MyFunction[a_, b_, c_] := something > MyFunction[args___] := Null /; Length[{args}] != 3 && > Message[MyFunction::argx, MyFunction, Length[{args}]] > MyFunction[a] > Paul > -- > Paul Abbott Phone: +61 8 9380 2734 > School of Physics, M013 Fax: +61 8 9380 1014 > The University of Western Australia (CRICOS Provider No 00126G) > 35 Stirling Highway > Crawley WA 6009 mailto:paul@physics.uwa.edu.au > AUSTRALIA http://physics.uwa.edu.au/~paul This does not seem to produce the correct message; I think you need something like: MyFunction[a_, b_, c_] := something MyFunction::argx = MyFunction called with `1` arguments, 3 arguments are expected. MyFunction[args___] := Null /; Length[{args}] != 3 && Message[MyFunction::argx, Length[{args}]] MyFunction[a] Andrzej Kozlowski Chiba, Japan http://www.mimuw.edu.pl/~akoz/ === Subject: Re: exporting fonts in postscript Is this a good case of stating one's version of Mathematica along with the problem statement, e.g., if one is using version 4.2 or 5.x it would be a different issue as URL provided below deals with versions prior to version 4.2? Steven Shippee > http://support.wolfram.com/mathematica/graphics/export/includefonts.html > > Mathematica shows Greek and other fonts in its frontend, but > when I Export into .eps format, these fonts are lost. How do I > get the Mathematica fonts to appear in my postscipt files? === Subject: Re: NDSolve Repeated convergence test failure Paul, can identify the singularity, and solve this problem in a two part piecewise fashion. It would still be nice if NDSolve identified this transition from a second order to first order ODE and correctly integrated the entire solution, but perhaps there are technical reasons that it doesn't. So, for now, I have an effective workaround in place. Seeing the complete plot, I also realized I had a mistake in the control law, which I have corrected. The updated notebook is available here in case you are interested. http://students.cs.tamu.edu/sro7516/double-pendulum-2.nb Steve === Subject: RE: Building a list Mark, xlist = {x1, x2, x3}; Outer[f[a, b, #1, #2] &, xlist, xlist] {{f[a, b, x1, x1], f[a, b, x1, x2], f[a, b, x1, x3]}, {f[a, b, x2, x1], f[a, b, x2, x2], f[a, b, x2, x3]}, {f[a, b, x3, x1], f[a, b, x3, x2], f[a, b, x3, x3]}} David Park djmp@earthlink.net http://home.earthlink.net/~djmp/ I've got a relatively simple function f[a_,b_,c_,d_], where a and b are integers and c and d are lists (of reals), and f returns a real. I need to evaluate f on (potentially long) sets of lists and calculate a square matrix from the possible permutations of the elements in the lists. For instance If xlist={x1,x2,x3} I need to calculate {{f[a,b,x1,x1],f[a,b,x1,x2],f[a,b,x1,x3]},{f[a,b,x2,x1],f[a,b,x2,x2],f[a ,b,x2,x3]},{f[a,b,x3,x1],f[a,b,x3,x2],f[a,b,x3,x3]}} Note that f[a,b,c,d] does not equal f[a,b,d,c]. My first reaction was that Outer[] might do this, but I cannot seem to get it to work. I'd appreciate any suggestions. Mark === Subject: Re: Yet another Mathematica docs whopper David: Re: It would be nice if it had a search engine. I think that would be nice, too. I've been able to take a small step in that direction by copying most of the individual notebooks in the Ver. 5.0.1 MainBook Windows folder into a single large MainBook.nb file which can be searched fairly rapidly for phrases or expressions I want to find, but which are not indexed by the Help Browser. I can also highlight and bookmark various topics with ad-hoc search strings (+++, ///, etc.), and insert notes and text clips from this newsgroup. My MainBook.nb file size is 40MB. Harold === Subject: Re: Yet another Mathematica docs whopper Yasvir: Re: Something such as this is probably best sent to Wolfram. I do not have Premier Service, and my organization is not about to buy it for me, as I'm the only person using Mathematica now. Everyone else uses a competitor's products. So I don't have big bucks invested in Ver 5.0.1. I would think the Wolfram folks read this newsgroup every day, and pull out notes and problem answers for their own future use. On the other hand, my Mathematica 5.0.1 About... doc lists Contacts as follows ... Bug reports support@wolfram.com http://www.wolfram.com/support Comments and feedback comments@wolfram.com http://www.wolfram.com/comments One person's comment is another's bug, I suppose. Am I expecting too much from Wolfram when I post here and don't send them a separate comment or bug report? Harold === Subject: Re: Yet another Mathematica docs whopper I'm REALLY not meaning to be a knucklhead here, but if I click on HELP in Mathematica then click on HELP BROWSER, I get a text box where I can type in a term to do a search on. What do you all (more experienced users) mean, remembering I am a novice user, by it would be nice if help were searchable? Steven Shippee > I've posted about the poor quality of Mathematica's documentation > before, and received several e-mails challenging my claim that this > documentation is awash in errors and woefully incomplete. Well, > for those who doubt, here's the latest example I just bumped into: > In[1]:= ??AspectRatio > AspectRatio is an option for Show and related functions which specifies the > ^^^^^^^^^^^^^^^^^^^^^ > ratio of height to width for a plot. > Attributes[AspectRatio] = {Protected} > In[2]:= SetOptions[Show, AspectRatio->Automatic]; > SetOptions::optnf: AspectRatio is not a known option for Show. > ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ > Faced with this sort of nonsense, what is a user to do? > I am furious, because I spend an unreasonable amount of time > resolving inadequacies like this one in the Mathematica documentation. > Considering the exorbitant sum we paid for Mathematica, I expect > something much better documented than this. > jill > -- > To s&e^n]d me m~a}i]l r%e*m?ov[e bit from my a|d)d:r{e:s]s. === Subject: RE: Re: Yet another Mathematica docs whopper Documentation can be bad for different reasons. In this case it seems like there is a conceptual fuzziness. Functions can either have options or transmit options to objects. This is not clear in the documentation or the error messages. The functions Options[] and SetOptions[] display or change the default values of options that the function has. A function like Show only transmits options. Show is an overloaded function, taking and generating various types of objects: Graphics, Graphics3D, SurfaceGraphics, ContourGraphics, DensityGraphics and GraphicsArray. The transmitted options are applied to all the arguments of Show. As the various types of Graphics objects allow different options, the options that Show transmits depend on the argument. Show[x, Boxed -> True] is correct if x is a SurfaceGraphics object, but not if it is a Graphics object. Another fuzziness in the documentation is in the concept of a Symbol. The first thing usually specified in programming languages is what constitutes a variable, ie what is a valid name. Mathematica is extremely sloppy in this respect. According to appendix A.1.2 The name of a symbol must be a sequence of letters, letter like forms and digits, not starting with a digit. The problem is that letter like forms is not defined. In addition, in the relevant section in the main text, section 3.10.2, both [Infinity] and OverBar[x] are listed as symbols. The latter is not a symbol however. In[1]:= !(x = 1[IndentingNewLine] x&_) Out[1]= 1 Out[2]= !(1&_) None of the objects in the table Creating objects with annotated names. are symbols. The fact that the package Utilities`Notation` can make them symbols does not make the text less erroneous. Jonas > -----Original Message----- === > Subject: Re: Yet another Mathematica docs whopper > I've posted about the poor quality of Mathematica's documentation > before, and received several e-mails challenging my claim that this > documentation is awash in errors and woefully incomplete. > Well, for > those who doubt, here's the latest example I just bumped into: > > In[1]:= ??AspectRatio > AspectRatio is an option for Show and related functions > which specifies the > ^^^^^^^^^^^^^^^^^^^^^ > ratio of height to width for a plot. > > Attributes[AspectRatio] = {Protected} > > In[2]:= SetOptions[Show, AspectRatio->Automatic]; > > SetOptions::optnf: AspectRatio is not a known option for Show. > ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ > > Faced with this sort of nonsense, what is a user to do? > Interesting example -- and a little subtle because, in fact, both > statements are true! If you enter > Options[Show] > you will see that it has no built-in options. However, > Show[Plot[Sin[x], {x, 0, 2 Pi}], AspectRatio -> Automatic] > does work. (Since Show has no options, I assume that the > option must be > getting passed to the object produced by the Plot command, > which in this > case is a -Graphics- object). > To set AspectRatio->Automatic for all Plots, one can do > SetOptions[Plot, AspectRatio -> Automatic] > Plot[Sin[x], {x, 0, 2 Pi}] > Similarly, to set it for Graphics (and to add a frame) > SetOptions[Graphics, AspectRatio -> Automatic, Frame -> True] > Show[Graphics[Point /@ Table[Random[], {100}, {2}]]] > I am furious, because I spend an unreasonable amount of > time resolving > inadequacies like this one in the Mathematica documentation. > Considering the exorbitant sum we paid for Mathematica, I expect > something much better documented than this. > Agreed -- but the documentation of Mathematica is already > overwhelmingly > large ... > Paul > -- > Paul Abbott Phone: +61 8 9380 2734 > School of Physics, M013 Fax: +61 8 9380 1014 > The University of Western Australia (CRICOS Provider No > 00126G) > 35 Stirling Highway > Crawley WA 6009 mailto:paul@physics.uwa.edu.au > AUSTRALIA http://physics.uwa.edu.au/~paul === Subject: RE: Re: help with integration and warning message I don't know if this is directly related, but there is a bug filed with 5.x with local variables in summations and integrations. Mathematica does not consistently treat iterators as local variables. In the first sum i is a local variable, in the second it is not: In[1]:= i=1; Sum[i, {i, 1, N}] Sum[f[i], {i, 1, N}] Out[1]= !(1/2 N ((1 + N))) Out[2]= N f[1] Version 4.1 does not have this problem. Jonas > -----Original Message----- > [mailto:astanoff_otez_ceci@yahoo.fr] === > Subject: Re: help with integration and warning message > I am running a code which can be downloaded from:- > http://www.itee.uq.edu.au/~dsalman/_downloads/Pbtest.nb > I am getting the warning message, related to the intergal > defining the > symbol [Chi]: > Unique::usym : 15 is not a symbol name or a valid symbol name > Can anyone please suggest how to correct the script and remove this > warning. > Don't use 'K'. > K is a generic index name used by Mathematica. > Names that start with a capital letter may conflict with > built-in Mathematica > names... > -- > 0% de pub! Que du bonheur et des vrais adh.8erents ! > Vous aussi inscrivez-vous sans plus tarder!! > Message post.8e .88 partir de http://www.gyptis.org, BBS actif > depuis 1995. === Subject: Options in Plot Why isn't this working ? rt = {PlotStyle -> RGBColor[1, 1, 0]}; ParametricPlot[{Cos[s], Sin[s]}, {s, 0, 1}, rt] Sebastien === Subject: Re: Building a list For instance... f[a_,b_,c_List]:=Outer[f[a,b,#1,#2]&,c,c] f[a,b,{x,y,z}] {{f[a,b,x,x],f[a,b,x,y],f[a,b,x,z]},{f[a,b,y,x],f[a,b,y,y],f[a,b,y,z]},{ f[a,b,z,x],f[a,b,z,y],f[a,b,z,z]}} Bobby > I've got a relatively simple function f[a_,b_,c_,d_], where a and b are > integers and c and d are lists (of reals), and f returns a real. > I need to evaluate f on (potentially long) sets of lists and calculate > a square matrix from the possible permutations of the elements in the > lists. For instance > If > xlist={x1,x2,x3} > I need to calculate > {{f[a,b,x1,x1],f[a,b,x1,x2],f[a,b,x1,x3]},{f[a,b,x2,x1],f[a,b,x2,x2],f[a > ,b,x2,x3]},{f[a,b,x3,x1],f[a,b,x3,x2],f[a,b,x3,x3]}} > Note that f[a,b,c,d] does not equal f[a,b,d,c]. > My first reaction was that Outer[] might do this, but I cannot seem to > get it to work. I'd appreciate any suggestions. > Mark === Subject: Re: Building a list Hi Mark, Perhaps this will help: In[1]:= xlist = {x1, x2, x3}; In[2]:= mat = Outer[List, xlist, xlist] Out[2]= {{{x1,x1},{x1,x2},{x1,x3}},{{x2,x1},{x2,x2},{x2,x3}},{{x3,x1},{x3,x2},{x3, x3}}} In[3]:= ((f[a, b, Sequence @@ #] & ) /@ # & ) /@ mat Out[3]= {{f[a,b,x1,x1],f[a,b,x1,x2],f[a,b,x1,x3]},{f[a,b,x2,x1],f[a,b,x2,x2], f[a,b,x2,x3]},{f[a,b,x3,x1],f[a,b,x3,x2],f[a,b,x3,x3]}} Tomas Garza Mexico City ----- Original Message ----- === Subject: Building a list > I've got a relatively simple function f[a_,b_,c_,d_], where a and b are > integers and c and d are lists (of reals), and f returns a real. > I need to evaluate f on (potentially long) sets of lists and calculate > a square matrix from the possible permutations of the elements in the > lists. For instance > If > xlist={x1,x2,x3} > I need to calculate > {{f[a,b,x1,x1],f[a,b,x1,x2],f[a,b,x1,x3]},{f[a,b,x2,x1],f[a,b,x2,x2],f[a > ,b,x2,x3]},{f[a,b,x3,x1],f[a,b,x3,x2],f[a,b,x3,x3]}} > Note that f[a,b,c,d] does not equal f[a,b,d,c]. > My first reaction was that Outer[] might do this, but I cannot seem to > get it to work. I'd appreciate any suggestions. > Mark === Subject: Re: Building a list > I've got a relatively simple function f[a_,b_,c_,d_], where a and b are > integers and c and d are lists (of reals), and f returns a real. > I need to evaluate f on (potentially long) sets of lists and calculate > a square matrix from the possible permutations of the elements in the > lists. For instance > If > xlist={x1,x2,x3} > I need to calculate > {{f[a,b,x1,x1],f[a,b,x1,x2],f[a,b,x1,x3]},{f[a,b,x2,x1],f[a,b,x2,x2],f[ > ,b,x2,x3]},{f[a,b,x3,x1],f[a,b,x3,x2],f[a,b,x3,x3]}} > Note that f[a,b,c,d] does not equal f[a,b,d,c]. > My first reaction was that Outer[] might do this, but I cannot seem to > get it to work. I'd appreciate any suggestions. Your first reaction was fine: Outer[f[a, b, ##] &, xlist, xlist] works. You could also explicitly use Array, for example Array[f[a, b, xlist[[#]], xlist[[#2]]] &, {Length[xlist], Length[xlist]}] also works. Ssezi === Subject: Re: Building a list > I've got a relatively simple function f[a_,b_,c_,d_], where a and b are > integers and c and d are lists (of reals), and f returns a real. > I need to evaluate f on (potentially long) sets of lists and calculate > a square matrix from the possible permutations of the elements in the > lists. For instance > If > xlist={x1,x2,x3} > I need to calculate > {{f[a,b,x1,x1],f[a,b,x1,x2],f[a,b,x1,x3]},{f[a,b,x2,x1],f[a,b,x2,x2],f[a > ,b,x2,x3]},{f[a,b,x3,x1],f[a,b,x3,x2],f[a,b,x3,x3]}} > Note that f[a,b,c,d] does not equal f[a,b,d,c]. > My first reaction was that Outer[] might do this, but I cannot seem to > get it to work. I'd appreciate any suggestions. Is Outer[f[a, b, #1, #2] &, {x1, x2, x3}, {x1, x2, x3}] what fits your need? -- Roland Franzius === Subject: Re: Building a list Outer is what you want xlist={x1,x2,x3}; Outer[f[a,b,#1,#2]&, xlist, xlist ] {{f[a, b, x1, x1], f[a, b, x1, x2], f[a, b, x1, x3]}, {f[a, b, x2, x1], f[a, b, x2, x2], f[a, b, x2, x3]}, {f[a, b, x3, x1], f[a, b, x3, x2], f[a, b, x3, x3]}} Bob Hanlon << I've got a relatively simple function f[a_,b_,c_,d_], where a and b are integers and c and d are lists (of reals), and f returns a real. I need to evaluate f on (potentially long) sets of lists and calculate a square matrix from the possible permutations of the elements in the lists. For instance If xlist={x1,x2,x3} I need to calculate {{f[a,b,x1,x1],f[a,b,x1,x2],f[a,b,x1,x3]},{f[a,b,x2,x1],f[a,b,x2,x2],f[a ,b,x2,x3]},{f[a,b,x3,x1],f[a,b,x3,x2],f[a,b,x3,x3]}} Note that f[a,b,c,d] does not equal f[a,b,d,c]. My first reaction was that Outer[] might do this, but I cannot seem to get it to work. I'd appreciate any suggestions. === Subject: Re: Building a list > I've got a relatively simple function f[a_,b_,c_,d_], where a and b are > integers and c and d are lists (of reals), and f returns a real. > I need to evaluate f on (potentially long) sets of lists and calculate > a square matrix from the possible permutations of the elements in the > lists. For instance > If > xlist={x1,x2,x3} > I need to calculate > {{f[a,b,x1,x1],f[a,b,x1,x2],f[a,b,x1,x3]},{f[a,b,x2,x1],f[a,b,x2,x2],f[a > ,b,x2,x3]},{f[a,b,x3,x1],f[a,b,x3,x2],f[a,b,x3,x3]}} > Note that f[a,b,c,d] does not equal f[a,b,d,c]. > My first reaction was that Outer[] might do this, but I cannot seem to > get it to work. I'd appreciate any suggestions. Outer will do the job. Try Outer[f[a, b, #1, #2] &, {x1, x2, x3}, {x1, x2, x3}] -- Roland Franzius === Subject: Re: Building a list > I've got a relatively simple function f[a_,b_,c_,d_], where a and b are > integers and c and d are lists (of reals), and f returns a real. > I need to evaluate f on (potentially long) sets of lists and calculate > a square matrix from the possible permutations of the elements in the > lists. For instance > If > xlist={x1,x2,x3} > I need to calculate > {{f[a,b,x1,x1],f[a,b,x1,x2],f[a,b,x1,x3]},{f[a,b,x2,x1],f[a,b,x2,x2],f[ > ,b,x2,x3]},{f[a,b,x3,x1],f[a,b,x3,x2],f[a,b,x3,x3]}} > Note that f[a,b,c,d] does not equal f[a,b,d,c]. > My first reaction was that Outer[] might do this, but I cannot seem to > get it to work. I'd appreciate any suggestions. > Mark You were quite right, Outer is certainly a natural function to use here: xlist={x1,x2,x3} Apply[f[a,b,##1]&,Outer[List,xlist,xlist],{2}] {{f[a,b,x1,x1],f[a,b,x1,x2],f[a,b,x1,x3]},{f[a,b,x2,x1], f[a,b,x2,x2],f[a,b,x2,x3]},{f[a,b,x3,x1],f[a,b,x3,x2],f[a,b,x3,x3]}} Andrzej Kozlowski Chiba, Japan http://www.mimuw.edu.pl/~akoz/ === Subject: Re: Building a list xlist = {x1, x2, x3}; Map[f[a, b, Sequence @@ #] &, Outer[List, xlist, xlist], {2}] > I've got a relatively simple function f[a_,b_,c_,d_], where a and b are > integers and c and d are lists (of reals), and f returns a real. > I need to evaluate f on (potentially long) sets of lists and calculate > a square matrix from the possible permutations of the elements in the > lists. For instance > If > xlist={x1,x2,x3} > I need to calculate > {{f[a,b,x1,x1],f[a,b,x1,x2],f[a,b,x1,x3]},{f[a,b,x2,x1],f[a,b,x2,x2],f[a > ,b,x2,x3]},{f[a,b,x3,x1],f[a,b,x3,x2],f[a,b,x3,x3]}} > Note that f[a,b,c,d] does not equal f[a,b,d,c]. > My first reaction was that Outer[] might do this, but I cannot seem to > get it to work. I'd appreciate any suggestions. > Mark === Subject: Re: usage / More... There's a menu option for this, too: Input>Create Button>(choose a button type). Bobby > I wonder how the More... hyperlinks appended to usage info are > generated as I want to have that feature for our own packages. I found a > posting by Bobby Treat dt 2002 suggesting to copy and modify the full > form of some existing usage text, but looking at the source of standard > Mathematica packages shows that the More... hyperlink is actually not > coded into the usage text itself. I also failed to find a hint in an > init.m file. I guess that the frontend generates this hyperlink > automatically, but based on what information? What do I have to do to > get those links to our own help notebooks? > Felix > Here is how the More... button can be generated for the Plot usage: > Button[More?, RuleDelayed[ButtonData, Plot], Rule[Active, True], > Rule[ButtonStyle, RefGuideLink]] > If you evaluate this, you will get a clickable More... button. > Ken === Subject: solving nonlinear equation with mathematica !! I face root finding problem solving nonlinear equation I have system nonlinear equation x^2-y-1 = 0; 10x exp(-x)-y = 0 ; If you draw the graph you see two intersecting point . I have found a point [Using FindRoot function ] In (-0.1,-0.5) the point is (-0.0905897,-0.991794 )---it is OK But In (2.0,1.0) the point is (1.94463,2.7816) But the graph says the x->2.something ,y->1.somthing I have used same FindRoot function to find the root in (2.0,1.0 ). Please help me ASAP . .. I want(exam)to use other methods(bisection/secant etc). If any one have workout examples of nonlinear equation please let me know . ..... === Subject: Re: Building a list >I've got a relatively simple function f[a_,b_,c_,d_], where a and b >are integers and c and d are lists (of reals), and f returns a >real. >I need to evaluate f on (potentially long) sets of lists and >calculate a square matrix from the possible permutations of the >elements in the lists. For instance >xlist={x1,x2,x3} >I need to calculate >{{f[a,b,x1,x1],f[a,b,x1,x2],f[a,b,x1,x3]},{f[a,b,x2,x1],f[a,b,x2,x2 >],f[a ,b,x2,x3]},{f[a,b,x3,x1],f[a,b,x3,x2],f[a,b,x3,x3]}} You can get what you want with Table e.g. Table[f[a, b, xlist[[i]], xlist[[j]]],{i, Length@xlist}, {j, Length@xlist}] -- To reply via email subtract one hundred and four === Subject: Yet another simple question. As i'm new to this amazing world of Mathematica, i'm not able yet to do simple things that for an experienced user seem ridiculous. So i want to ask a simple question. If: +++++++++++++++++++++++++++++++++++++++++++++++ y,m,day : integer positive variables yr,mr : integer positive variables which depend on y and m fff[day_] = Mod[day + 2*mr + Quotient[3*(mr + 1), 5] + yr + 2 + Quotient[yr, 4] + Quotient[yr, 400] - Quotient[yr, 100], 7]; +++++++++++++++++++++++++++++++++++++++++++++++ Now for a specific day=D i want to find the value of the integer variable Q which belongs to (D,D+7], so fff[Q]=1 would be true. So i write: Do[Q = Which[fff[x] == 1, x], {x, D + 1, D + 7, 1}] and i expect Q to take the appropriate value. But this doesn't work. Can anyone show me the way to the light? === Subject: Re: Infrequent Mathematica User If you actually see a way to apply one of the standard inequalities to this sum, please share. Nothing obvious springs to mind. Bobby >A Math NG posed the problem: > Let x1,x2,...,xn be real numbers. Prove > x1/(1+x1^2) + x2/(1+x1^2+x2^2) +...+ xn/(1+x1^2+...+xn^2) < sqrt(n) > To prove this, I would do a search for inequalites, e.g, > http://mathworld.wolfram.com/ChebyshevSumInequality.html > Also, see > Hardy, G. H.; Littlewood, J. E.; and P.97lya, G. > Inequalities, 2nd ed. Cambridge, England: > Cambridge University Press, pp. 43-44, 1988. > To investigate using Mathematica, I like to use subscripted variables: > s[n_] := Sum[Subscript[x,i]/(1 + Sum[Subscript[x,j]^2, {j, i}]), {i, n}] > If you enter this expression into Mathematica and > do Cell | Convert to StandardForm (or > TraditionalForm) you will get a nicely formatted > expression for the n-th left-hand side of the > inequality. > Note that you can prove the inequality on a > case-by-case basis using CylindricalDecomposition > (in Version 5.0): > CylindricalDecomposition[s[1] > 1, {Subscript[x, 1]}] > CylindricalDecomposition[s[2] > Sqrt[2], {Subscript[x, 1], Subscript[x, 2]}] > and so on. This may not seem convincing, but see > what happens if the you change the inequality: > CylindricalDecomposition[s[2] > 1/2, {Subscript[x, 1], Subscript[x, 2]}] >To get a feel for the problem, and maybe spark >an idea, I hoped to look at some few early >maximum values. However, these proved difficult >to come by. > NMaximize is the way to go: > Table[NMaximize[s[n], Table[{Subscript[x, i], -5, 5}, {i, n}]], {n, 6}] > Paul === Subject: Re: Disappearing Help Browser Categories Harold, you said: My Help Browser listing of Notation topics (under Demos) just disappeared, and I have not been able to get it back. All the files used before are still intact in C: ....Mathematica5.0AddOnsExtraPackagesUtilitiesDocumentationEng lish I have a default installation of Version Number: 5.0.1.0 and my paths look like: C:Program FilesWolfram ResearchMathematica5.0DocumentationEnglishDemosSoundGallery.nb C:Program FilesWolfram ResearchMathematica5.0DocumentationEnglishDemosDataFiles C:Program FilesWolfram ResearchMathematica5.0DocumentationEnglishDemosNotations C:Program FilesWolfram ResearchMathematica5.0DocumentationEnglishDemosNotebookDemos C:Program FilesWolfram ResearchMathematica5.0DocumentationEnglishDemosPalettes C:Program FilesWolfram ResearchMathematica5.0DocumentationEnglishDemos2DGraphics.nb C:Program FilesWolfram ResearchMathematica5.0DocumentationEnglishDemos3DGraphics.nb C:Program FilesWolfram ResearchMathematica5.0DocumentationEnglishDemosAnimations.nb C:Program FilesWolfram ResearchMathematica5.0DocumentationEnglishDemosArtImages.nb C:Program FilesWolfram ResearchMathematica5.0DocumentationEnglishDemosBrowserCategories .m C:Program FilesWolfram ResearchMathematica5.0DocumentationEnglishDemosBrowserIndex.nb C:Program FilesWolfram ResearchMathematica5.0DocumentationEnglishDemosColorCharts.nb C:Program FilesWolfram ResearchMathematica5.0DocumentationEnglishDemosDiagrams.nb C:Program FilesWolfram ResearchMathematica5.0DocumentationEnglishDemosFormulaGallery.nb C:Program FilesWolfram ResearchMathematica5.0DocumentationEnglishDemosPalettes.nb Could you have accidently moved your files or directory without knowing it? Steven Shippee === Subject: When You Hear The Heavy Accent & The Poor Phone Connection... HANG UP!! ____ iMkhUM6p1W X-Comment: Brought to you by the Pet Peeve Usenet Network X-Comment: Resistance Is Futile! X-Comment: Oh_andBTW... F_R_E_E___M_A_R_T_H_A!!! When You Hear The Heavy Accent & The Poor Phone Connection... HANG UP!! - I think you must know what Iâm talking about. Together, a heavy foreign accent, coupled with a lousy phone connection can only mean one thing... An outsourced operation, in a place like India, China, The Philippines, etc.; where some greedy American corporation is saving a few pieces of Silver and displacing American workers in the process. - The best thing you can do is hang up... look for the companiesâ on-shore counterpart and complain! 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