https://community.wolfram.com RSS Feed for Wolfram Community showing any discussions tagged with Mathematics sorted by most viewed. https://community.wolfram.com/groups/-/m/t/366628 POSTED BY: ======== **Marvin Ray Burns, and distinguished colleagues** ---------- My exciting experiences using Wolfram technologies! --------------------------------------------------- ---------- The MRB constant, a fascinating mathematical anomaly, has intrigued researchers and enthusiasts alike for decades. Defined as the limiting value of a unique alternating series, this enigmatic constant showcases the beauty of numerical exploration and convergence. Despite its relatively recent emergence, the MRB constant reveals unexpected connections to various fields within mathematics and computational analysis. In this post, we dive into its origins, properties, and the ongoing quest to uncover its more profound significance. The MRB constant is an anomaly because it emerges from an alternating series with unusual convergence behavior. Unlike many well-known mathematical constants, the MRB constant has no closed-form expression nor a known exact nature—whether it is algebraic, transcendental, or even irrational. Additionally, the sequence of partial sums that define the MRB constant oscillates between two limit points, creating a bounded yet divergent behavior. This oscillatory nature distinguishes it from more conventional mathematical constants, which typically exhibit straightforward convergence. Its mysterious properties continue to intrigue mathematicians as they explore its deeper connections to number theory and computational analysis. ---------- ---------- CMRB ![If you see this instead of an image, reload the page][1] **is the MRB constant.** Without solicitation, GPT echoed one of this discussion's contributors and gave a shoutout to Mathematica's accomplishments by saying: ------------------- > > **Why Computing Digits of the MRB Constant Is Interesting** > > At first glance, the computation of ever more digits of a numerical > constant may appear to be a sterile exercise, offering little insight > beyond the digits themselves. For the MRB constant, this > interpretation is profoundly misleading. The interest lies not in the > digits, but in the act of computing them. > > The MRB series occupies a delicate numerical regime: it is convergent, > yet only marginally so; its terms are elementary in form, yet hostile > to naive numerical evaluation; and its alternating structure conceals > severe cancellation effects. As a result, each additional digit of > accuracy is not merely an extension of a decimal expansion, but a > validation of the mathematical and algorithmic framework used to > obtain it. > > In this sense, the digits serve as a stress test for mathematics > itself. They expose the difference between formal convergence and > effective convergence, between theoretical existence and computational > feasibility. Techniques that are elegant on paper often fail > catastrophically in practice, while successful methods must balance > asymptotics, numerical stability, precision control, and computational > cost. > > Moreover, the computation of the MRB constant forces the > reconciliation of disparate mathematical ideas. Alternating series, > logarithmic asymptotics, convergence acceleration, special functions, > and analytic continuation must all agree numerically. Any hidden > inconsistency—whether conceptual or computational—reveals itself > immediately in the failure of the digits to stabilize. > > Thus, computing digits of the MRB constant transforms mathematics from > a static collection of statements into a living experiment. The > resulting digits are not trophies; they are evidence. They measure the > effectiveness of our methods, the soundness of our assumptions, and > the practical reach of our theory. > > In short, the digits themselves do not matter. What matters is that > they *can* be computed at all—and how difficult it is to do so. > That difficulty is precisely what makes the MRB constant > mathematically interesting. **In general, we will see:** ![enter image description here][7] ---------- ---------- ---------- [1]: https://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2024-11-25023042.png&userId=366611 [2]: https://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2026-01-01235656.png&userId=366611 [3]: https://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2026-01-01235710.png&userId=366611 [7]: https://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2026-02-18190718.png&userId=366611 Marvin Ray Burns A.G.S. (cum laude) 2014-10-09T18:08:49Z https://community.wolfram.com/groups/-/m/t/1106375 Greetings, I am working on a very complicated equation that needs to be solved. To simplify the function I use a limit in which x is much less than D or ( x<<D). I cannot find much less/greater condition in Mathematica. I need to solve that equation symbolically and I know that I should get four solutions. By taking a limit manually I got something like A*x^4-B*x^2+C=0. Then using solve I was able to get four solution, but with coefficients that were very long. (In manual calculations, I was able to get shorter coefficients) How do I solve equation with much less condition? Just for the sake of example let say I want to solve f(x)= sqrt( D^2 - ( x - z )^2 ) + x * cos(a) + x + z after limit f(x)= sqrt( D^2 - ( z )^2 ) + x * cos(a) + x + z Is there a command in Mathematica to selectively simplify/factor/expand some terms instead all terms in equations? Adam Szewczyk 2017-05-25T01:56:46Z https://community.wolfram.com/groups/-/m/t/1323951 This has been one of my favorite Mathematica projects! Here are a couple of Ai generated outlines of my progress of computing the MKB constant digits: ![enter image description here][1] ![enter image description here][2] [1]: https://community.wolfram.com//c/portal/getImageAttachment?filename=4506unnamed.png&userId=366611 [2]: https://community.wolfram.com//c/portal/getImageAttachment?filename=unnamed%281%29.png&userId=366611 Marvin Ray Burns A.G.S. (cum laude) 2018-04-20T12:06:18Z https://community.wolfram.com/groups/-/m/t/1104811 I'm wondering how to define a net function that will multiply matrices inside a tensor. For example, if $s$ and $t$ are tensors with shapes $\{a,b,c\}$ and $\{a,c,d\}$ respectively, for each $i\leq a$ the subtensors in $s$ and $t$ with first index $i$ form matrices of dimensions $b\times c$ and $c\times d$ respectively. Multiplying corresponding matrices in $s$ and $t$ would produce a new tensor with shape $\{a,b,d\}$. Is there currently a way to do this in Mathematica using nn layers? DotLayer and ThreadingLayer have depth limitations that seem to prevent it. Andrew Dabrowski 2017-05-23T19:26:30Z https://community.wolfram.com/groups/-/m/t/1108310 Below is a code snippet to plot a handful of curves with some plot labels. For some reason only the first label is used and not the other four. What am I missing here? I've successfully used PlotLabels for parametric curves before but they seem to fail me here. Got any ideas??: lmt=10 qq[zets_]=1/Sqrt[1+zets^2]^3 zets[rr_,eps_,eta_]=eps (1-rr^2) (1+eta (1+rr^2)) demcurves=Array[qq[zets[rr,2 #/lmt,0]]&,lmt/2] Plot[demcurves,{rr,0,1}] demlabels = Map[ToString,Array[qqzz[N[2/lmt]#]&,lmt/2]] Plot[demcurves,{rr,0,1},PlotLabels -> Placed[demlabels,Left]] Anthony DeGance 2017-05-25T17:19:36Z https://community.wolfram.com/groups/-/m/t/1108667 Hello! I want to plot a simple function, but on the OX axis, the output labels are by default 1,2,3, etc. I want to have them in the format **x/2**. Instead of 1,2,3, I want 2/2, 4/2, 6/2. Is this possible? But only on the OX axis. Thank you in advance! Robert Poenaru 2017-05-26T09:12:30Z https://community.wolfram.com/groups/-/m/t/1108617 Hi i'm trying to plot disks of radius =(1-f[x]), so that it makes a sphere, as we know that the volume of a cylinder having height=2 radius - the volume of the cone having radius r, is always equal to the volume of a sphere having radius r. but i want to show that using disks. Muhammad Afzal 2017-05-26T06:59:19Z https://community.wolfram.com/groups/-/m/t/1108324 ![Möbius transformations of the circle][1] **Back and Forth** One fact I've known for a while but never really dived into is that Möbius transformations of the circle can be realized by inverse stereographic projecting to the sphere (here I'm thinking of the circle as the equator of the sphere, so inverse stereographic projection is just the identity in this case), rotating the sphere in space (say, around the south pole), and then stereographically projecting from the new "north pole" back to the circle. The animation shows what happens when you do this to 15 equally-spaced points on the circle, where the sphere is being rotated by an angle of $\pi/3$ around the axis $(\cos \psi, \sin \psi, 0)$ anchored at $(0,0,-1)$, and we let $\psi$ vary from 0 to $2\pi$. &[Wolfram Notebook][2] [1]: http://community.wolfram.com//c/portal/getImageAttachment?filename=mobius12.gif&userId=610054 [2]: https://www.wolframcloud.com/obj/545113ac-8325-4e63-82ae-d770422f55f5 Clayton Shonkwiler 2017-05-25T19:34:31Z https://community.wolfram.com/groups/-/m/t/2286246 $\int_0^\infty e^{i \pi x} \left(1-(x+1)^{\frac{1}{x+1}}\right) dx$ ![enter image description here][1] ![enter image description here][2] ![enter image description here][3] f[x_] = E^(I*Pi*x)*(1 - (x + 1)^(1/(x + 1))); g[x_] = x^(1/x); u := t/(1 - t); sub = Im[NIntegrate[(f[(-I t)] - f[( I t)])/(Exp[2 Pi t] - 1), {t, 0, Infinity}, WorkingPrecision -> 100]] 0.1170836031505383167089899122239912286901483986967757585888318959258587743002\ 7817712246477316693025869 m = NSum[f[( t)] , {t, 0, Infinity}, WorkingPrecision -> 100, Method -> "AlternatingSigns"] 0.1878596424620671202485179340542732300559030949001387861720046840894772315646\ 6021370329665443217278 m - sub 0.0707760393115288035395280218302820013657546962033630275831727881636184572643\ 8203658083188126524252 Is the same as {Re[NIntegrate[f[t], {t, 0, Infinity}]], and, Re[NIntegrate[f[t], {t, 0, Infinity I}, WorkingPrecision -> 100]]} NIntegrate::deodiv: DoubleExponentialOscillatory returns a finite integral estimate, but the integral might be divergent. NIntegrate::deodiv: DoubleExponentialOscillatory returns a finite integral estimate, but the integral might be divergent. {0.070776, and, \ 0.0707760393115288035395280218302820013657546962033630275831727881636184572643\ 8203658083188126617723821} To be continued. ---------------- [1]: https://community.wolfram.com//c/portal/getImageAttachment?filename=01.jpg&userId=366611 [2]: https://community.wolfram.com//c/portal/getImageAttachment?filename=02.jpg&userId=366611 [3]: https://community.wolfram.com//c/portal/getImageAttachment?filename=03.jpg&userId=366611 Marvin Ray Burns A.G.S. (cum laude) 2021-06-09T05:56:08Z https://community.wolfram.com/groups/-/m/t/326240 Triggered by the recent outbreak of Ebola India Bruckner, a pupil from Aberdeen's [St Margaret's School for Girls][1], and myself worked on a little model this summer to understand the basics of the spreading of diseases in populations and the relationship to transportation networks. The model is very basic, but shows some interesting features and is very straight forward to implement in Mathematica. When I was typing these lines I saw that Arnoud Buzing had posted something, reason enough to interrupt my typing and to check out what he had posted: [Visualizing the Ebola Outbreak][2]. I hope that my post is going to complement Arnoud's to some extent. So, my question is how the global air transport network might lead to a spreading of a disease. I will use a very standard SIR (susceptible-infected-recovered) model, which is certainly far from being ideal for Ebola; [but similar types of models are to too bad either][3]. It rather simulates an outbreak of some generic disease from which you recover. If we assumed that everyone died in an outbreak the SIR model might also be appropriate. I will introduce the equations below. I also need a list of all airports and all flight connections. On the website [Openflights.org][4] you will find all data we need. I saved the file "airports.dat" and the file "routes.dat". So that's the data. I first import the data: airports = Import["~/Desktop/airports.dat", "CSV"]; routes = Import["~/Desktop/routes.dat", "CSV"]; This is a plot of all airports in that database. GeoRegionValuePlot[Table[GeoPosition[airports[[i, {7, 8}]]] -> 1., {i, 1, Length[airports]}], PlotStyle -> PointSize[0.003], PlotRange -> 1, ImageSize -> Full] which gives ![enter image description here][5] Alright, now the routes. First, we create a list of rules for all airport IDs and their coordinates: codecoords = Table[airports[[i, 5]] -> GeoPosition[airports[[i, {7, 8}]]], {i, 1,Length[airports]}]; We then calculate the links: links = Monitor[Table[routes[[j, {3, 5}]] /. codecoords, {j, 1, Length[routes]}], ProgressIndicator[j, {1, Length[routes]}]]; and clean out missing data: linksclean = Select[links, Head[#[[1]]] == GeoPosition && Head[#[[2]]] == GeoPosition &]; Now comes a nice figure: With[{locations = RandomChoice[linksclean, 14000]}, GeoGraphics[{{Green, Opacity[0.3],AbsoluteThickness[0.0001], GeoPath[locations, "Geodesic"]}}, GeoRange -> "World", GeoProjection -> Automatic, GeoBackground -> GeoStyling["ReliefMap"], ImageSize -> {1200, 600}]] which gives ![enter image description here][6] Ok. Interestingly I can only plot 16000 max at a time. Somewhere between 16-17k the Kernel quits. That might be Integer related. Could be a limit in the programming of Geographics. Not sure. I have more than enough memory and can plot the remaining 2-3k airports in a second figure and use Show to display all. (It would be great if someone from WRI could comment.) Anyway, let's go to some modelling. The basic idea of an [SIR model][7] is that a population is modelled in three compartments Susceptibles (S), Infected (I) and Recovered (R). I will use a time-discrete model; there are continuous models around, too, and if anyone is interested I can provide the ODE model as well. Here are the three equations: sus[i_] := sus[i] = sus[i - 1] - [Rho] sus[i - 1] inf[i - 1]; inf[i_] := nf[i] = inf[i - 1] + [Rho] sus[i - 1] inf[i - 1] - [Lambda] inf[i - 1]; rec[i_] := rec[i] = rec[i - 1] + [Lambda] inf[i - 1]; The meaning of sus, inf and rec should be clear by now; they are given as percentages of the total population, their sum is 100%. The variable i represents time. $ ho$ is an infection rate and $lambda$ is a recovery rate. The infections increase with the product of susceptibles and infected. By adding the right hand sides it becomes clear that the population does not change over time. We come up with some values for the parameters and iterate: sus[1] = 0.95; inf[1] = 0.05; rec[1] = 0; [Rho] = 0.2; [Lambda] = 0.1; tcourse = Table[{sus[i], inf[i], rec[i]}, {i, 1, 100}]; The time course looks like this: ListPlot[Transpose[tcourse]] ![enter image description here][8] The monotonously decreasing function show susceptibles, the increasing function recovered, and the remaining curve the infected. We now need to do some cleaning up of the original airport data, before we proceed to a multi-airport/city model. (*Extract the names and GPS coordinates*) rawdata = Sort[Select[airports[[All, {5, 7, 8}]], #[[1]] != "" &]][[81 ;;]]; (*These are just the coordinates*) airportcoords = rawdata[[All, {2, 3}]]; (*These are just the names. *) names = rawdata[[All, 1]]; (*Here are the names to indices*) rules = MapThread[#1 -> #2 &, {names, Range[Length[names]]}]; (*The "population" is initially set to 1 for all airports, this allows us to take different airport sizes into consideration later.*) pop = Table[1., {j, 1, Length[names]}]; routesraw = Import["~/Desktop/routes.dat", "CSV"]; (*There are many links so this takes a while*) links = Select[routesraw[[All, {3, 5}]] /. rules, NumberQ[#[[1]]] && NumberQ[#[[2]]] &]; From that we now construct (a first guess at) the coupling or adjacency matrix: couplingdummy = Table[0, {i, 1, Length[names]}, {j, 1, Length[names]}]; For[k = 1, k <= Length[links], k++, couplingdummy[[links[[k, 1]], links[[k, 2]]]] = 1; couplingdummy[[links[[k, 2]], links[[k, 1]]]] = 1]; I do know about ConstantArray, but for some reason that does not work. The first line constructs a matrix full of zeros and the second adds ones where there are links. The problem is that apparently in that dataset some airports are not linked at all. We can sort them out by: indices = Select[Table[If[Total[couplingdummy[[i]]] > 0, i], {i, 1, Length[couplingdummy]}], NumberQ]; We now delete the columns and rows of the couplingdummy matrix intermed = couplingdummy[[#]] & /@ indices; transintermed = Transpose[intermed]; coupling = transintermed[[#]] & /@ indices; Again I had a much more elegant way of doing this, with the advantage that it did not work. To speed up the following calculations I use that the coupling matrix is sparse, but I like the original too much to throw it away just yet. coulinginterm = coupling; coupling = SparseArray[coulinginterm]; We adapt our "population/airport size" vector: pop = Table[1., {j, 1, Length[indices]}]; and set the following parameters: [Rho] = 0.2; [Lambda] = 0.1; Mairports = Length[indices]; [Mu] = 0.05; $\rho$ and $lambda$ are as before. Mairports is the number of airports that we model and $\mu$ is a "migration rate". It comes from the original model which we built for different cities were it describes the migration between different cities. Here is models the "propensity to fly". We now define an effective coupling matrix. It is the adjacency matrix times the population vector (i.e. people in the catchment area of the airport). In our case the vector has all ones, so it is just the adjacency matrix. It allows us later to model more general situations. meanNN = coupling.pop; When we want to model the outbreak as populations at the positions of all airports, each of which is described by an SIR model, we need to couple lots of populations, because there are lots of airports. The following uses the sparsity of the adjacency matrix to speed up the calculation. sumind = Table[Take[Flatten[ArrayRules[coupling[[k, All]]][[All, 1]]], Length[ArrayRules[coupling[[k, All]]]] - 1], {k, 1, Mairports}]; It generates a list of all airports that are coupled/linked to a given airport. Now we are ready to write down the central equations: sus[i_, j_] := sus[i, j] = (1 - [Mu]) (sus[i - 1, j] - [Rho] sus[i - 1, j] inf[i - 1, j]) + [Mu] Total[Table[sus[i - 1, sumind[[j, u]]]*pop[[sumind[[j, u]]]], {u, 1, Length[sumind[[j]]]} ] ]/meanNN[[j]]; inf[i_, j_] := inf[i, j] = (1 - [Mu]) (inf[i - 1, j] + [Rho] sus[i - 1, j] inf[i - 1, j] - [Lambda] inf[i - 1, j]) + [Mu] Total[Table[inf[i - 1, sumind[[j, u]]]*pop[[sumind[[j, u]]]], {u, 1, Length[sumind[[j]]]} ] ]/meanNN[[j]]; rec[i_, j_] := rec[i, j] = (1 - [Mu]) (rec[i - 1, j] + [Lambda] inf[i - 1, j]) + [Mu] Total[Table[rec[i - 1, sumind[[j, u]]]*pop[[sumind[[j, u]]]], {u, 1,Length[sumind[[j]]]} ] ]/meanNN[[j]]; The terms with the Total are "migration terms" that describe the travelling behaviour of the people in the catchment area of the airports. i is the time index and j labels the airports. Next come the initial conditions: For[i = 1, i <= Mairports, i++, sus[1, i] = 1.; inf[1, i] = 0.0; rec[1, i] = 0.;] sus[1, 1] = 0.95; inf[1, 1] = 0.05; rec[1, 1] = 0.0; In the catchment areas of all airports there are only susceptibles, apart from airport number 1, which will have 5% infected people. Now we can finally iterate the whole thing: tcourse = Monitor[Table[{sus[i, j], inf[i, j], rec[i, j]}, {i, 1, 500}, {j, 1,Mairports}], ProgressIndicator[i, {0, 500}]]; Great. Let's save that just in case your notebook tends to crash at this point, just l like mine did when I was playing with this. Export["~/Desktop/SIR-tcourse.csv", tcourse]; If you wish you can now plot the time course of some of the airport catchment areas: ListPlot[Flatten[Table[{tcourse[[All, j, 1]], tcourse[[All, j, 2]], tcourse[[All, j, 3]]}, {j, 1, 200}], 1], ImageSize -> Large] ![enter image description here][9] Now that is not very helpful yet. To generate nicer plots, i.e. to normalise, we first calculate the maximal number of sick people at any of the airports: maxsick = Max[Flatten[tcourse[[All, All, 2]]]]; We then generate movie frames, and go and get some coffee.... frames = Monitor[Table[GeoRegionValuePlot[Table[GeoPosition[airportcoords[[indices[[i]]]]] -> inf[k, i]/maxsick, {i, 1, Length[indices]}], PlotStyle -> PointSize[0.003], PlotRange -> 1, ImageSize -> Full,ColorFunction -> "Rainbow"], {k, 1, 300}], ProgressIndicator[k, {0, 300}]]; Actually, you might want to get another coffee when you want to export the frames: Export["~/Desktop/SIR-frames-World.gif", frames]; Alright. That gif is a bit large to embed it into this post, but you can download it from [here][10]. All I can do is show you some frames to get an idea of how this looks: ![enter image description here][11] Of course we can look at the network structure and try to understand the pattern of infections. This command is useful: CommunityGraphPlot[AdjacencyGraph[Normal[coupling]]] ![enter image description here][12] You clearly see the communities in North America, Europe and Asia. This one is also pretty: Show[TreePlot[Subgraph[grph, ConnectedComponents[grph][[1]]], Center, PlotStyle -> Directive[Gray, Opacity[0.02]]], TreePlot[Subgraph[grph, ConnectedComponents[grph][[1]]], Center, EdgeRenderingFunction -> None]] ![enter image description here][13] We have several enhancements of this. First we can easily look at different countries individually. What if an ebola patient arrives at some airport in the US? [See simulation here][14]. (Careful 50 MBs!) There is also something we can do if we want to go the the level of cities. The main problem is that the Wolfram database does not yet have data for all streets between cities. In one of the online conferences it was said that that will be introduced in some later version, which I cannot wait to play with. Until then we have to cheat. (or use some online database; I prefer cheating.) We developed a model of the spreading of a disease in Nigeria. So we could go about this like this: Clear["Global`*"] CountryData["Nigeria", "Population"] Graphics[CountryData["Nigeria", "Polygon"]] Then get city names, coords and population: names = CityData[{All, "Nigeria"}]; citypop = Table[CityData[names[[i]], "Population"], {i, 1, Length[names]}]; citycoords = Table[CityData[names[[i]], "Coordinates"], {i, 1, Length[names]}]; Here comes the cheat. Because we don't have the streets we use Delaunay triangulation: Needs["ComputationalGeometry`"] dtri = DelaunayTriangulation[citycoords]; list = {}; Table[ Do[AppendTo[list, {i, dtri[[All, 2]][[i, j]]}], {j, 1, Length[dtri[[All, 2]][[i, All]]]}], {i, 1, Length[dtri]}]; coupling = Table[0, {i, 1, Length[names]}, {j, 1, Length[names]}]; For[i = 1, i < Length[list] + 1, i++, coupling[[list[[i]][[1]], list[[i]][[2]]]] = 1;] coulinginterm = coupling; coupling = SparseArray[coulinginterm]; which gives the following network Graphics[Join[ Table[Circle[citycoords[[i]], 0.02], {i, 1, Length[names]}], DeleteCases[ Flatten[Table[ If[coupling[[i, j]] == 1, Line[{citycoords[[i]], citycoords[[j]]}]] , {i, 1, Length[names]}, {j, 1, i}], 1], Null]]] ![enter image description here][15] The point in the middle corresponds to the airport where it all starts; then come its neighbours and then their neighbours etc. You could now animate this and change the colours to see how the disease spreads through the different layers. It would be nice if someone could implement that. There are obviously some problems, i.e. "streets" leaving the country etc, but the general idea should work. The rest is quite the same as before: (*Paramters*) [Rho] = 0.2; [Lambda] = 0.1; Mcities = Length[names]; [Mu] = 0.05; (*Initiation*) For[i = 1, i <= Mcities, i++, sus[1, i] = 1.; inf[1, i] = 0.0; rec[1, i] = 0.;] (*Starting Outbrake at the following city*) sus[1, 1] = 0.95; inf[1, 1] = 0.05; rec[1, 1] = 0.0; meanNN = coupling.citypop; sumind = Table[ Take[Flatten[ArrayRules[coupling[[k, All]]][[All, 1]]], Length[ArrayRules[coupling[[k, All]]]] - 1], {k, 1, Mcities}]; sus[i_, j_] := sus[i, j] = (1 - [Mu]) (sus[i - 1, j] - [Rho] sus[i - 1, j] inf[i - 1, j]) + [Mu] Total[ Table[sus[i - 1, sumind[[j, u]]]*citypop[[sumind[[j, u]]]], {u, 1, Length[sumind[[j]]]} ] ]/meanNN[[j]]; inf[i_, j_] := inf[i, j] = (1 - [Mu]) (inf[i - 1, j] + [Rho] sus[i - 1, j] inf[i - 1, j] - [Lambda] inf[i - 1, j]) + [Mu] Total[ Table[inf[i - 1, sumind[[j, u]]]*citypop[[sumind[[j, u]]]], {u, 1, Length[sumind[[j]]]} ] ]/meanNN[[j]]; rec[i_, j_] := rec[i, j] = (1 - [Mu]) (rec[i - 1, j] + [Lambda] inf[i - 1, j]) + [Mu] Total[ Table[rec[i - 1, sumind[[j, u]]]*citypop[[sumind[[j, u]]]], {u, 1, Length[sumind[[j]]]} ] ]/meanNN[[j]]; This time we try to work in parallel: LaunchKernels[]; tcourse = ParallelTable[{sus[i, j], inf[i, j], rec[i, j]}, {i, 1, 500}, {j, 1, Mcities}]; // AbsoluteTiming (There is something strange here. This ran in MMA9 in 6.3 seconds- I still have data from the course I taught last year. MMA10 takes ages. After the installation of MMA10 also MMA9 seems to take longer 43 seconds. Is this a bug report?). Note that this time the population sizes of the cities are taken into account and are relevant. If you run Manipulate[ Graphics[{Line[Flatten[CountryData["Nigeria", "Coordinates"], 1]], Join[Table[{ RGBColor[tcourse[[t, i, 2]], tcourse[[t, i, 1]], tcourse[[t, i, 3]]], Disk[citycoords[[i]], 0.1]}, {i, 1, Length[names]}]]}], {t, 1, 500, 1}] or poly = Graphics[Polygon[Flatten[CountryData["Nigeria", "Coordinates"], 1]], ImagePadding -> None]; Animate[ImageSubtract[ Graphics[ListDensityPlot[ Join[{{4, 3.25, 0}, {4, 14, 0}, {14, 3.25, 0}, {14, 14, 0}}, Table[{citycoords[[k]][[1]], citycoords[[k]][[2]], 1. - tcourse[[t, k, 1]]}, {k, 1, Mcities}]], InterpolationOrder -> 3, ColorFunction -> "Rainbow", PlotRange -> All, Frame -> False, PlotRangePadding -> None]], poly], {t, 1, 500, 1}, DefaultDuration -> 20.] or better infmax = Max[tcourse[[All, All, 2]]]; frames = Table[ ImageSubtract[ Graphics[ ListDensityPlot[ Join[{{4, 3.25, 0}, {4, 14, 0}, {14, 3.25, 0}, {14, 14, 0}}, Table[{citycoords[[k]][[1]], citycoords[[k]][[2]], tcourse[[t, k, 2]]/infmax}, {k, 1, Mcities}]], InterpolationOrder -> 3, ColorFunction -> "Rainbow", PlotRange -> All, Frame -> False, PlotRangePadding -> None, ColorFunctionScaling -> False]], poly], {t, 1, 500, 6}]; you get nice animations like this one: ![enter image description here][16] I have noticed that Nigeria needs to be rotated, but I hope that the idea becomes clear. I am also aware that there are many flaws in this. SIR is certainly not the best way forward to model Ebola. Any population dynamicist and/or health expert can certainly come up with an endless list of problems. The network is not perfect. For more serious applications we actually use models for the cities, i.e. street connections among close cities and airport connections among countries etc. The problem is that if we simulate between 200-20000 cities per country plus the airports, a standard laptop runs into trouble. On the bright side, we have a cluster on which this kind of larger simulations work just fine. Hope that you like this anyway, Marco [1]: http://www.st-margaret.aberdeen.sch.uk [2]: http://community.wolfram.com/groups/-/m/t/325956 [3]: http://mtbi.asu.edu/files/Mathematical_Models_to_Study_the_Outbreaks_of_Ebola.pdf [4]: http://openflights.org/data.html [5]: /c/portal/getImageAttachment?filename=Airportsall.jpg&userId=48754 [6]: /c/portal/getImageAttachment?filename=Airports-world.jpg&userId=48754 [7]: http://en.wikipedia.org/wiki/Compartmental_models_in_epidemiology [8]: /c/portal/getImageAttachment?filename=SingleSIR.jpg&userId=48754 [9]: /c/portal/getImageAttachment?filename=SIR-airportstcourse.jpg&userId=48754 [10]: https://www.dropbox.com/s/9y6d16z82qjw261/SIR-frames.gif?dl=0 [11]: /c/portal/getImageAttachment?filename=SIR-airports-frames.jpg&userId=48754 [12]: /c/portal/getImageAttachment?filename=Airports-CommNetwork.jpg&userId=48754 [13]: /c/portal/getImageAttachment?filename=SIR-networkrings.jpg&userId=48754 [14]: https://www.dropbox.com/s/0qwgmi7ks8dpsjh/SIR-USA-frames.gif?dl=0 [15]: /c/portal/getImageAttachment?filename=SIR-Nigerianetwork.jpg&userId=48754 [16]: /c/portal/getImageAttachment?filename=SIR-movie.gif&userId=48754 Marco Thiel 2014-08-22T20:18:58Z https://community.wolfram.com/groups/-/m/t/1070264 [Wolfram Language][1] (WL) is a powerful multi-paradigm programing language. There is a set of common mistakes that repeatedly tend to entrap new users. **This is a call to describe such mistakes building a "black-listing" guide for novice coders.** Please consider contributing. I suggest following simple rules (with gratitude adapted from a [similar effort][2]): - One topic per answer - Focus on non-advanced uses (it is intended to be useful for beginners and as a question closing reference) - Include a self explanatory title in header style (example: "# Basic built-in function syntax"; see [syntax guide][3] ) - Explain the symptoms, the mechanism behind the scenes and all possible causes and solutions you can think of. Be sure to include a beginner's level explanation (and a more advance one too, if you can) *Please, use "**Reply**" to a specific comment for structured clarity of nested comments.* ---------- ## Table of Contents - [Basic syntax of built-in functions][4] - [Learn how to use the Documentation Center effectively][5] - [Sorting numerical data and the behavior of Sort][6] - [What does @#(%=<\[!} et cetera mean?][7] - [Consider Reap/Sow Instead of AppendTo][8] - [Alternatives to Reap/Sow][9] - [Case sensitivity and typos][10] - [Numerical vs Symbolic -- What to do when plots are blank][11] - [Import and "CurrencyTokens"][12] [1]: https://www.wolfram.com/language [2]: https://mathematica.stackexchange.com/q/18393/13 [3]: http://community.wolfram.com/groups/-/m/t/270507 [4]: http://community.wolfram.com/groups/-/m/t/1069885 [5]: http://community.wolfram.com/groups/-/m/t/1070285 [6]: http://community.wolfram.com/groups/-/m/t/1070705 [7]: http://community.wolfram.com/groups/-/m/t/1070946 [8]: http://community.wolfram.com/groups/-/m/t/1084289 [9]: http://community.wolfram.com/groups/-/m/t/1164555 [10]: http://community.wolfram.com/groups/-/m/t/1084920 [11]: http://community.wolfram.com/groups/-/m/t/1086929 [12]: http://community.wolfram.com/groups/-/m/t/1086969 Vitaliy Kaurov 2017-04-23T23:54:23Z https://community.wolfram.com/groups/-/m/t/1034626 I recently watched "Arrival", and thought that some of the dialogue sounded Wolfram-esque. Later, I saw the following blog post: [Quick, How Might the Alien Spacecraft Work?][1] Along with many others, I enjoyed the movie. The underlying artistic concept for the alien language reminded me of decade old memories, a book by Stephen Addiss, [Art of Zen][2]. Asian-influenced symbolism is an interesting place to start building a sci-fi concept, even for western audiences. I also found Cristopher Wolfram's broadcast and the associated files: [Youtube Broadcast][3] [Github Files ( with image files ) ][4] Thanks for sharing! More science fiction, yes! I think the constraint of circular logograms could be loosened. This leads to interesting connections with theory of functions, which I think the Aliens would probably know about. The following code takes an alien logogram as input and outputs a deformation according to do-it-yourself formulation of the Pendulum Elliptic Functions: ![Human Animation][5] ## $m=2$ Inversion Coefficients ## MultiFactorial[n_, nDim_] := Times[n, If[n - nDim > 1, MultiFactorial[n - nDim, nDim], 1]] GeneralT[n_, m_] := Table[(-m)^(-j) MultiFactorial[i + m (j - 1) + 1, m]/ MultiFactorial[i + 1, m], {i, 1, n}, {j, 1, i}] a[n_] := With[{gt = GeneralT[2 n, 2]}, gt[[2 #, Range[#]]] & /@ Range[n] ] ## Pendulum Values : $2(1-\cos(x))$ Expansion Coefficients ## c[n_ /; OddQ[n]] := c[n] = 0; c[n_ /; EvenQ[n]] := c[n] = 2 (n!) (-2)^(n/2)/(n + 2)!; ## Partial Bell Polynomials ## Note: These polynomials are essentially the same as the "**BellY**" ( hilarious naming convention), but recursion optimized. See timing tests below. B2[0, 0] = 1; B2[n_ /; n > 0, 0] := 0; B2[0, k_ /; k > 0] := 0; B2[n_ /; n > 0, k_ /; k > 0] := B2[n, k] = Total[ Binomial[n - 1, # - 1] c[#] B2[n - #, k - 1] & /@ Range[1, n - k + 1] ]; ## Function Construction ## BasisT[n_] := Table[B2[i, j]/(i!) Q^(i + 2 j), {i, 2, 2 n, 2}, {j, 1, i/2}] PhaseSpaceExpansion[n_] := Times[Sqrt[2 \[Alpha]], 1 + Dot[MapThread[Dot, {BasisT[n], a[n]}], (2 \[Alpha])^Range[n]]]; AbsoluteTiming[CES50 = PhaseSpaceExpansion[50];] (* faster than 2(s) *) Fast50 = Compile[{{\[Alpha], _Real}, {Q, _Real}}, Evaluate@CES50]; ## Image Processing ## note: This method is a hack from ".jpg" to sort-of vector drawing. I haven't tested V11.1 vectorization functionality, but it seems like this could be a means to process all jpg's and output a file of vector polygons. Anyone ? LogogramData = Import["Human1.jpg"]; Logogram01 = ImageData[ColorNegate@Binarize[LogogramData, .9]]; ArrayPlot@Logogram01; Positions1 = Position[Logogram01[[5 Range[3300/5], 5 Range[3300/5]]], 1]; Graphics[{Disk[#, 1.5] & /@ Positions1, Red, Disk[{3300/5/2, 3300/5/2}, 10]}]; onePosCentered = N[With[{cent = {3300/5/2, 3300/5/2} }, # - cent & /@ Positions1]]; radii = Norm /@ onePosCentered; maxR = Max@radii; normRadii = radii/maxR; angles = ArcTan[#[[2]], #[[1]]] & /@ onePosCentered; Qs = Cos /@ angles; ## Constructing and Printing Image Frames ## AlienWavefunction[R_, pixel_, normRad_, Qs_, angles_] := Module[{ deformedRadii = MapThread[Fast50, {R normRad, Qs}], deformedVectors = Map[N[{Cos[#], Sin[#]}] &, angles], deformedCoords }, deformedCoords = MapThread[Times, {deformedRadii, deformedVectors}]; Show[ PolarPlot[ Evaluate[ CES50 /. {Q -> Cos[\[Phi]], \[Alpha] -> #/10} & /@ Range[9]], {\[Phi], 0, 2 Pi}, Axes -> False, PlotStyle -> Gray], Graphics[Disk[#, pixel] & /@ deformedCoords], ImageSize -> 500]] AbsoluteTiming[ OneFrame = AlienWavefunction[1, (1 + 1)* 1.5/maxR, normRadii, Qs, angles] ](* about 2.5 (s)*) ![Alien Pendulum][6] ## Validation and Timing ## In this code, we're using the magic algorithm to get up to about $100$ orders of magnitude in the half energy, $50$ in the energy. I did prove $m=1$ is equivalent to other published forms, but haven't found anything in the literature about $m=2$, and think that the proving will take more time, effort, and insight (?). For applications, we just race ahead without worrying too much, but do check with standard, known expansions: EK50 = Normal@ Series[D[ Expand[CES50^2/2] /. Q^n_ :> (1/2)^n Binomial[n, n/2], \[Alpha]], {\[Alpha], 0, 50}]; SameQ[Normal@ Series[(2/Pi) EllipticK[\[Alpha]], {\[Alpha], 0, 50}], EK50] Plot[{(2/Pi) EllipticK[\[Alpha]], EK50}, {\[Alpha], .9, 1}, ImageSize -> 500] Out[]:= True ![Approximation Validity][7] This plot gives an idea of approximation validity via the time integral over $2\pi$ radians in phase space. Essentially, even the time converges up to, say, $\alpha = 0.92$. Most of the divergence is tied up in the critical point, which is difficult to notice in the phase space drawings above. Also compare the time of function evaluation: tDIY = Mean[ AbsoluteTiming[Fast50[.9, RandomReal[{0, 1}]] ][[1]] & /@ Range[10000]]; tMma = Mean[AbsoluteTiming[JacobiSN[.9, RandomReal[{0, 1}]] ][[1]] & /@ Range[10000]]; tMma/tDIY In the region of sufficient convergence, Mathematica function **JacobiSN** is almost 20 times slower. The CES radius also requires a function call to **JacobiCN**, so an output-equivalent **AlienWavefunction** algorithm using built-in Mathematica functions would probably take at least 20 times as long to produce. When computing hundreds of images this is a noticeable slow down, something to avoid ! ! Also compare time to evaluate the functional basis via the Bell Polynomials: BasisT2[n_] := Table[BellY[i, j, c /@ Range[2 n]]/(i!) Q^(i + 2 j), {i, 2, 2 n, 2}, {j, 1, i/2}]; SameQ[BasisT2[20], BasisT[20]] t1 = AbsoluteTiming[BasisT[#];][[1]] & /@ Range[100]; t2 = AbsoluteTiming[BasisT2[#];][[1]] & /@ Range[25]; ListLinePlot[{t1, t2}, ImageSize -> 500] ![Series Inverse][8] The graph shows quite clearly that careful evaluation via the recursion relations changes the complexity of the inversion algorithm to polynomial time, $(n^2)$, in one special example where the forward series expansions coefficients have known, numeric values. ## Conclusion ## We show proof-of-concept that alien logograms admit deformations that preserve the cycle topology. Furthermore we provide an example calculation where the "human" logogram couples to a surface. Deformation corresponds to scale transformation of the logogram along the surface. Each deformation associates with an energy. Invoking the pendulum analogy gives the energy a physical meaning in terms of gravity, but we are not limited to classical examples alone. The idea extends to arbitrary surfaces in two, three or four dimensions, as long as the surfaces have local extrema. Around the extrema, there will exist cycle contours, which we can inscript with the Alien logograms. This procedure leads readily to large form compositions, especially if the surface has many extrema. Beyond Fourier methods, we might also apply spherical harmonics, and hyperspherical harmonics to get around the limitation of planarity. The missing proof... Maybe later. LOL! ~ ~ ~ ~ Brad And in the Fanfiction Voice: Physicist : "It should be no surprise that heptapod speech mechanism involves an arbitrary deformation of the spacetime manifold." Linguist : "Space-traveling aliens, yes, of course they know math and physics, but Buddhist symbology, where'd they learn that?" [1]: http://blog.stephenwolfram.com/2016/11/quick-how-might-the-alien-spacecraft-work/ [2]: https://books.google.com/books/about/Art_of_Zen.html?id=4jGEQgAACAAJ [3]: https://www.youtube.com/watch?v=8N6HT8hzUCA&t=4992s [4]: https://github.com/WolframResearch/Arrival-Movie-Live-Coding [5]: http://community.wolfram.com//c/portal/getImageAttachment?filename=Deformation.gif&userId=234448 [6]: http://community.wolfram.com//c/portal/getImageAttachment?filename=AlienPendulum.png&userId=234448 [7]: http://community.wolfram.com//c/portal/getImageAttachment?filename=EllipticK.png&userId=234448 [8]: http://community.wolfram.com//c/portal/getImageAttachment?filename=BellPolynomial.png&userId=234448 Brad Klee 2017-03-18T20:23:59Z https://community.wolfram.com/groups/-/m/t/491285 I want to get the result of a multiplication between a matrix and a vector, basically I want to do a change of reference frame of a vector, so I cannot perform this operation in mathematica. A = {{2, 4}, {2, 1}} // MatrixForm C2 = {{6}, {5}} // MatrixForm A.C2 (*This does not work*) I want to get something like this: {{2, 4}, {2, 1}}.{{6}, {5}} // MatrixForm Actually I have something more complex, but this is enough to show my problem. Do you have any suggestion? Alberto de la Torre 2015-05-04T18:17:40Z https://community.wolfram.com/groups/-/m/t/418720 ## This is it. The perfectly centered billiards break. Behold: ![enter image description here][2] <h2>Setup</h2> This break was computed in *Mathematica* using a numerical differential equations model. Here are a few details of the model: * All balls are assumed to be perfectly [elastic][3] and almost perfectly rigid. * Each ball has a mass of 1 unit and a radius of 1 unit. * The cue ball has a initial speed of 10 units/sec. * The force between two balls is given by the formula $$F \;=\; \begin{cases}0 & \text{if }d \geq 2, \\ 10^{11}(2-d)^{3/2} & \text{if }d < 2, \end{cases}$$ where $d$ is the distance between the centers of the balls. Note that the balls overlap if and only if $d < 2$. The power of $3/2$ was [suggested by Yoav Kallus][4] on Math Overflow, because it follows [Hertz's theory of non-adhesive elastic contact](https://en.wikipedia.org/wiki/Contact_mechanics#Hertzian_theory_of_non-adhesive_elastic_contact). The initial speed of the cue ball is immaterial -- slowing down the cue ball is the same as slowing down time. The force constant $10^{11}$ has no real effect as long as it's large enough, although it does change the speed at which the initial collision takes place. <h2>The Collision</h2> For this model, the entire collision takes place in the first 0.2 milliseconds, and none of the balls overlap by more than 0.025% of their radius during the collision. (These figures are model dependent -- real billiard balls may collide faster or slower than this.) The following animation shows the forces between the balls during the collision, with the force proportional to the area of each yellow circle. Note that the balls themselves hardly move at all *during* the collision, although they do accelerate quite a bit. ![enter image description here][5] <h2>The Trajectories</h2> The following picture shows the trajectories of the billiard balls after the collision. ![enter image description here][6] After the collision, some of the balls are travelling considerably faster than others. The following table shows the magnitude and direction of the velocity of each ball, where $0^\circ$ indicates straight up. $\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|} \hline \text{ball} & \text{cue} & 1 & 2,3 & 4,6 & 5 & 7,10 & 8,9 & 11,15 & 12,14 & 13 \\ \hline \text{angle} & 0^\circ & 0^\circ & 40.1^\circ & 43.9^\circ & 0^\circ & 82.1^\circ & 161.8^\circ & 150^\circ & 178.2^\circ & 180^\circ \\ \hline \text{speed} & 1.79 & 1.20 & 1.57 & 1.42 & 0.12 & 1.31 & 0.25 & 5.60 & 2.57 & 2.63 \\ \hline \end{array} $ For comparison, remember that the initial speed of the cue ball was 10 units/sec. Thus, balls 11 and 15 (the back corner balls) shoot out at more than half the speed of the original cue ball, whereas ball 5 slowly rolls upwards at less than 2% of the speed of the original cue ball. By the way, if you add up the sum of the squares of the speeds of the balls, you get 100, since kinetic energy is conserved. <h2>Linear and Quadratic Responses</h2> The results of this model are dependent on the power of $3/2$ in the force law -- other force laws give other breaks. For example, we could try making the force a linear function of the overlap distance (in analogy with springs and [Hooke's law][7]), or we could try making the force proportional to the *square* of the overlap distance. The results are noticeably different ![enter image description here][8] ![enter image description here][9] <h2>Stiff Response</h2> Glenn the Udderboat points out that "stiff" balls might be best approximated by a force response involving a higher power of the distance (although this isn't the [usual definition][10] of "stiffness"). Unfortunately, the calculation time in *Mathematica* becomes longer when the power is increased, presumably because it needs to use a smaller time step to be sufficiently accurate. Here is a simulation involving a reasonably "stiff" force law $$F \;=\; \begin{cases}0 & \text{if }d \geq 2, \\ 10^{54}(2-d)^{10} & \text{if }d<2. \end{cases}$$ ![enter image description here][11] As you can see, the result is very similar to my first thought: > The two balls in the back corners shoot away along rays parallel to the two sides of the triangle. Here is a picture showing the forces, with each force vector emanating from the point of contact. > ![enter image description here][12] This seems like good evidence that above 1st-thought behavior is indeed the limiting behavior in the case where the stiffness goes to infinity. As you might expect, most of the energy in this case is transferred very quickly at the beginning of the collision. Almost all of the energy has moves to the back corner balls in the first 0.02 milliseconds. Here is an animation of the forces: ![enter image description here][13] After that, the corner balls and the cue ball shoot out, and the remaining balls continue to collide gently for the next millisecond or so. While the simplicity of this behavior is appealing, I would guess that "real" billard balls do not have such a stiff force response. Of the models listed here, the intial Hertz-based model is probably the most accurate. Qualitatively, it certainly seems the closest to an "actual" break. <h2> Full Code </h2> &[Wolfram Notebook][14] ---------- I wrote this post originally for [Math Stack Exchange][15]. [1]: http://math.bard.edu/belk/code.htm [2]: http://i.stack.imgur.com/Y9ixR.gif [3]: https://en.wikipedia.org/wiki/Elasticity_%28physics%29 [4]: http://mathoverflow.net/questions/156263/perfectly-centered-break-of-a-perfectly-aligned-pool-ball-rack/156407?noredirect=1#comment400402_156407 [5]: http://i.stack.imgur.com/WY37i.gif [6]: http://i.stack.imgur.com/wHVJA.png [7]: https://en.wikipedia.org/wiki/Hooke%27s_law [8]: http://i.stack.imgur.com/a1l3b.gif [9]: http://i.stack.imgur.com/xM76n.gif [10]: https://en.wikipedia.org/wiki/Stiffness [11]: http://i.stack.imgur.com/nMJyT.gif [12]: http://i.stack.imgur.com/GKGT9.png [13]: http://i.stack.imgur.com/VuUWT.gif [14]: https://www.wolframcloud.com/obj/8c6b7e81-4a5c-4e3a-bb13-a3d47e728e64 [15]: http://math.stackexchange.com/a/659318/28293 Jim Belk 2015-01-08T18:02:10Z https://community.wolfram.com/groups/-/m/t/2458169 A new study group devoted to Cryptography begins next Monday! I will be leading this group daily, Monday to Friday, over the next three weeks, until February 18. The curriculum follows the upcoming Wolfram U course "[Introduction to Cryptography][1]". A list of daily topics can be found on the [Daily Study Groups][2] page. The study group sessions include videos and reading materials for the course and time for discussion and Q&A. After the study group will help you may complete the course quizzes and achieve the "Course Completion" certificate for the "[Introduction to Cryptography][3]" . > **REGISTER HERE:** > https://www.bigmarker.com/series/daily-study-group-intro-to-cryptography/series_details > **About This Study Group:** Cryptography has existed for thousands of years and currently is a fascinating topic to study because of the close ties it forges between theory and practice. It makes use of bitwise computations, advanced algebra, string operations and everything in between. The Wolfram Language, with functions that provide basic building blocks for modern cryptography, makes a great tool for hands-on understanding of various concepts. In this Study Group, you will follow video lessons from the new Wolfram U course Introduction to Cryptography and learn about the concepts, underlying math and connections to real-world scenarios. ![enter image description here][4] P.S: Don't worry if you are unable to attend every session live: the recordings will be sent out to those who registered! [1]: https://www.wolfram.com/wolfram-u/introduction-to-cryptography/ [2]: https://www.bigmarker.com/series/daily-study-group-intro-to-cryptography/series_details [3]: https://www.wolfram.com/wolfram-u/introduction-to-cryptography/ [4]: https://community.wolfram.com//c/portal/getImageAttachment?filename=WolframUbanner.jpg&userId=20103 Dariia Porechna 2022-01-28T13:31:36Z https://community.wolfram.com/groups/-/m/t/139463 Hello all, I've been trying to create a 3 D plot from my data. I have my X,Y, Z matrices, where Z = f(X,Y). I have my matrices in excel, which I've imported to Mathematica. A little bit of background here - I've not used Mathematica before. But fairly comfortable with Matlab. I'm trying to use Mathematica because it lets me create plots where I can create 3D models with tubes as opposed to surfaces on Matlab. My final goal is to create a 3D surface and export it as an STL file so that it can be printed using a 3D printer. Here's an image of what I'm trying to make: [img=width: 291px; height: 451px;]/c/portal/getImageAttachment?filename=aO24IZc.jpg&userId=11733[/img] I'd appreciate it if someone can help me with this. Thanks. Phalgun Lolur 2013-10-16T00:43:33Z https://community.wolfram.com/groups/-/m/t/209188 Hello everybody.Can Wolfram Mathematica software show steps and not only solution, for ex of this problem: http://s4.postimg.org/vsp9vu9m5/Untitled.png Greetings from Montenegro :) Lukas 007 2014-02-27T07:29:56Z https://community.wolfram.com/groups/-/m/t/114911 How small can a description of a large prime number be? There are the Fermat primes 2^n-1 for certain n, and in base 2 these are a sequence of ones.  In base 10, if you have just zeros and two ones, then the only primes of that form are 11 and 101.  If there are three ones then it is divisible by three.  But what about four ones?  It seems wrong to me that there might be an unbounded number of such primes. That's what some brief experiments suggest though. [mcode]1+10^4+10^18+10^201 == 1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001000000000000010001[/mcode]is the largest one I found.  Also I don't notice any patterns, e.g. here in the first 200 such primes [img=width: 68px; height: 432px;]/c/portal/getImageAttachment?filename=primes-1111.jpg&userId=23275[/img] and here are the first 254 primes with nonzero digits {1,2,1} [img=width: 347px; height: 432px;]/c/portal/getImageAttachment?filename=primes-121.jpg&userId=23275[/img] the largest found is [mcode]1+2*10^14+10^201 == 1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000200000000000001[/mcode]Does anyone else have any primes which don't seem like they should be prime?  The more extreme the better. Todd Rowland 2013-09-03T04:50:53Z https://community.wolfram.com/groups/-/m/t/163848 Hi everyone, I would like to briefly present the [url=http://www.wolframalpha.com/problem-generator/][b]Wolfram Problem Generator[/b][/url], which I have been working on for the past months. There are two features that I think make it special: the ability to generate unlimited problems and free-form input for answers. [url=http://www.wolframalpha.com/problem-generator/][img=width: 505px; height: 360px;]http://blog.wolframalpha.com/data/uploads/2013/10/Topic-List.png[/img][/url] Some problem generators do not make random problems on the fly: in fact, they take problems from a repository (which, even if big, is still finite). By contrast, our system generates a totally new problem just for you. Also, we double-check that we haven't shown it already--in this case, we just show you a new one! Our free-form input takes full advantage of all the technology created by Wolfram|Alpha. Given a problem to try, you can write your answer in whatever way makes sense to you and we will be able to recognize it each time. For example, for [img=float: left;]/c/portal/getImageAttachment?filename=ScreenShot2013-12-02at12.18.49PM.png&userId=50011[/img] you can type "2 sqrt(3)", "two times square root of three", "two radical 3" or "2 * 3^(1/2)". Wolfram Problem Generator was not without its challenges in development. The most challenging aspect of this project was implementing a way to distinguish right answers from incorrect attempts. For example, if your problem is "6 x = 10", Wolfram Problem Generator asks you to simplify your result to "5/3" (or "x = 5/3", or "five thirds", etc.), and not leave the answer as "10/6". Because "10/6" is a mathematically correct answer, we relied on Mathematica's powerful pattern matcher to ensure that the answer really was simplified. You can try Wolfram Problem Generator if you're a member of Wolfram|Alpha Pro. Right now, we have coverage for six core subjects, with a lot more in the works. Let us know what subjects and what types of problems you'd like to see! Enjoy! Luca Luca Belli 2013-12-02T17:46:23Z https://community.wolfram.com/groups/-/m/t/958163 I am very new to Mathematica. As a school project, we have to graph images using basically a very large piecewise function. The image I am working on is a Cardinal. I cannot for the life of me figure out how to use the "Show" function to graph multiple functions of various domains and ranges on the same plot. I've read the help guide multiple times and simply can't make sense of it... Any suggestions? Thank you!! Clarissa Hood 2016-11-06T18:31:56Z