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13780
How to set Math text size & line spacing
I have chosen different font for text. Unfortunately this font is larger than the standard Times font. So all math symbols (e.g. greek) are now a little too small. If I could just set the FontSize for the math font to 13 it would probably look good. How do I accomplish that? If it is not that simple: In defining a custom Style via Stylesheet (or private Style Definitions), what is the correct way to define a custom size for the math font. There are Style definitions for "Display Formula" and "Inline Formula", but if I type just an inline cell, the latter does not apply.
34527
What's wrong with my palindrome finding function?
I'm trying to find the palindromes in the dictionary, but my output isn't giving me all of them. I have this so far Palindrome[] := DictionaryLookup[y_ /; (y == StringReverse[y])] Palindrome[] And my the output just it gives me `{a,I}`. What's wrong with my input? I don't see it. * * * Thanks.
59351
Switching parameter lines on/off in a 3D parametric plot
In a classroom demonstration for ParametricPlot3D [{f[u, v], g[u, v], h[u, v]}, {u, ... }, {v, ... }] I want to show * u lines only without v, * v lines only without u, * no lines, only surface, * no surface, only u, v lines. How can it be done?
34733
How do I assign specific data points in an array to a new array?
I am trying to learn how to script/program in mathematica and I am retyping some simple C and FORTRAN programs into Mathematica as a way to learn. I am writing a Monte Carlo code to calculate the value of pi. In the method attached a 10000x2 dimensional array is created using table with values ranging from 0 to 1. The magnitude of each row of the array is determined inside a For loop and if it is greater than 0 it is rejected and if it is less it is used in the running tally that is divided by the total number of points and then multiplied by 4 to give an estimate of pi. On top of calculating pi, I would like to create two more arrays to use in a list plot. In a second array I would like to store values of the first array that met the rejection criteria and in third array I would like to store values that were rejected, so I can plot the rejected points as red and the accepted points as blue. However, no matter what I try Mathematica will not seem to let me transfer points from the first array titled "vector". I am attaching the current code to this question that effectively calculates the estimate of pi. If anyone can show me how to transfer the rejected points to another array as determined inside the For loop I would be very grateful. `[counter, samples, a, vector, tally, pi, tally2] counter = 1; samples = 10000; tally = 1; a := Random[Real, {0, 1}]; Timing[vector = Table[a, {i, samples}, {j, 2}]; For[i = 1, i <= samples, i++, {If[Norm[vector[[i]]] < 1, tally++, ""]}]; pi = (tally/samples)*4 // N]` `
36847
How can I implement the method of Lagrange multipliers to find constrained extrema?
I want to form the function $h=f-\lambda_{1}g_{1}-\lambda_{2}g_{2}$ where $f$ is the function to optimize subject to the constraints $g_{1}=0$ and $g_{2}=0$ so that I can solve the first partial derivatives with respect to $\lambda_{1}$ and $\lambda_{2}$. Can someone get me started using $f(x,y,z)=xy+yz$ subject to the constraints $x^2+y^2-2=0$ and $x^2+z^2-2=0$?
43566
Adding Elements to a List of Lists
I'm trying to do the following: Given a function `f[x,y]`, I want to turn the list `{{x1,y1},{x2,y2},...,{xn,yn}}` into `{{x1,y1,f[x1,y1]},{x2,y2,f[x2,y2]},...,{xn,yn,f[xn,yn]}}`. In other words, I want to replace each 2-tuple in the original list of lists with a 3-tuple, where the 3rd element is a function evaluated using the 2-tuple. For example, if `f[x,y]` is `x+y`, then I'd want `{{1,2},{5,6}}` to be turned into `{{1,2,3},{5,6,11}}`. I've seen syntax for many similar tasks, but not one quite like this. I would tremendously appreciate any help -- I've been stumped on this one for a while!
36842
NDSolve - 2nd order ODE - how to solve for y and y'
i am trying to workout how to solve a simple 2nd order ODE for y and y'. I can solve for y[t] as follows but how can i solve for y[t] and y'[t] ? m = 1; g = 9.82; sol = Flatten @ NDSolve[{y''[t] == -m g - 0.3 y'[t], y'[0] == 10, y[0] == 0, WhenEvent[y[t] < 0, tmax = t; "StopIntegration"]}, y[t], {t, 0, 3}]; Plot[y[t] /. sol, {t, 0, tmax}] Also the output from NDSolve is an interpolating function but how do i turn this into a function ? i.e i thought it would be something like this but that doesn't seem to work ? g[t_]:= y[t]/. sol Thanks David.
36843
using a Mathematica function to define a new function
I'd like to define a function of three variables which produces a new, named function of a single variable, where this final variable is not a member of the first three. So I'd like something where I have `f0[x_ , y_ , z_] := << complicated function >>` which gives `f1[p_]` Furthermore, I'd like to have the functions be defined in such a way that when I write `IN:= f0[x0,y0,z0]` I get `OUT:= fx0y0z0[p_]` so that I can easily identify which inputs produced the function. So, basically, I'd like to wind up with a function produced via Interpolation and I'd like that function to have a specific name defined by my inputs. For instance, let's say I'm creating a numerical function by writing down a table in two variables (say, p and q) that requires x, y, and z as inputs. Then I'd like to integrate over q and interpolate over p so that my final function is just a function over q only. So, let's say that my first function is `g[x_,y_,z_,p_,q_] := (x + y + z) p / q` Now I'd like to tabulate this over p and q for given values of x, y, and z. Then I'd like to integrate over p and have a function of q only. I can for instance do the following `f1[x_,y_,z_] := NIntegrate[ Interpolation[ Flatten[Table[{q, p, g[x,y,z,p,q]}, {p,p0,p1}, {q,q0,q1}],1]][#, p], {p, p0, p1}] &` and then this is a well-behaved function of q which I can make tables of, which I can then plot, integrate, etc. just by writing f1[x0,y0,x0][q]. But this is not very convenient for me since it requires me to write out a new function name every time I want to examine the behavior as a function of different values of x, y, and z, and I will ultimately need many values of x, y, and z. Is there any way to write a meta-function that is capable of producing a brand new interpolating function of q only, with the name including the input values of x, y, and z? Thanks for your help, Sam
51823
Chart with its ticks in geometric progression
I have to render a chart with an axis that has ticks with the following values: 250, 500, 1000, 2000, 4000, 8000 I first thought it was a logarithmic scale. But then I realized the pattern here is simpler (a tick has the value of the previous tick * 2). ![custom scale](http://i.stack.imgur.com/7IUct.png) Does this type of scale have its own name, or is it simply a logarithmic scale with a "special" base?
14626
How do I stack a RectangleChart?
What I would like to do is represent the state of a machine over a period of time using a RectangleChart (or anything else that can do it) . The chart will need to be made up of equally sized blocks that change colour based on the state of the machine (there are 7 states 1-7) at that time. I've created this chart which is similar to what I want; data = Table[{RandomInteger[{1, 7}], 5}, {i, 1, 20}]; RectangleChart[data, ChartLayout -> "Stacked", BarOrigin -> Left] ![Almost what I'm after](http://i.stack.imgur.com/7YwOj.jpg) The problem with this chart is my state data is being represented by the height of the bar, not the colour. What is the best way to get the results I'm after?
14623
How to do conditional matrix division when elements are a combination of zero and real numbers?
I tried to divide two large matrices DLand PL of size 100,000 x 5 each and they look like this, DL=![enter image description here](http://i.stack.imgur.com/WpATf.png) and PL=![enter image description here](http://i.stack.imgur.com/vwcHi.png) By using conditional function: Spread1[DL_, PL_] := SparseArray[ {i_, j_} /; PL[[i, j]] != 0 :> 100000 DL[[i, j]]/PL[[i, j]], {1000, 5}] test = Spread1[DL, PL]; MatrixForm[test] and then I get result as shown, but why do I get those messages and some of the resulting test matrix elements say ComplexInfinity? Not sure if I should ignore those messages as all zero elements are aligned? ![enter image description here](http://i.stack.imgur.com/wol1z.png)
55721
Finding linearly independent combination of vectors with a certain symmetry property
I have two linearly independent vectors $v_1$ and $v_2$. Is there a way for me using Mathematica to find two linear combinations $w_1$ and $w_2$ such that: 1. $w_1$ and $w_2$ are linearly independent. 2. If I cut $w_2$ in half and re-order the halves to get $w_2'$, i.e. $w_2=\\{1,2,3,4,5,6\\}\rightarrow w_2'=\\{4,5,6,1,2,3\\}$, then $w_2'=w_1$? $v_1$ and $v_2$ are chosen such that I am guaranteed $w_1$, $w_2$ exist.
52290
Specifying shortcut without modifying KeyEventTranslations.tr
Until recently, I followed the suggestions(e.g., http://stackoverflow.com/questions/7327411/is-there-a-way-around-using-and- for-part-in-mathematica) to introduce some new shortcuts. However, I moved to another place recently, and here I am not allowed to modify `KeyEventTranslations.tr` personally. (I am now a user of linux system) To find an alternative solution, I encountered this post. Best way to add KeyEvents and faster Quit In the selected answer, the author introduced a seemingly alternative way but honestly I don't quite understand what it does and how I revise it to meet my needs. Can you show me some simple examples in this direction? Or, it would be great if you suggest another solution.
14629
How can I use several data fields drawn from curated data at once?
How can I use `"MinTemperature"`, `"MaxTemperature"`, and `"Temperature"` properties of `WeatherData` for one or more cities in one line of code?
98
How can I improve the speed of eigenvalue decompositions for large matrices?
I often need to compute the eigenvalues of large matrices, and I invariably resort to MATLAB for these, simply because it is much faster. I'd like to change that, so that I can work entirely inside my notebook. Here's a plot comparing the timings between the two for eigenvalue decompositions for matrices of varying sizes (left). The y-axis shows the time in seconds. As you can see, there's about a factor 3 difference between the two (right). ![enter image description here](http://i.stack.imgur.com/w8cAnm.jpg) ![enter image description here](http://i.stack.imgur.com/3i8V5m.jpg) Here's a sample code in Mathematica to generate timings: timings = With[{x = RandomReal[NormalDistribution[], {#, #}]}, Eigenvalues[x]; // Timing // First] & /@ Range[500,5000,500] and its equivalent in MATLAB: s = 500:500:5000; t = zeros(numel(s),1); for i = 1:numel(s) x=randn(s(i)); t1=tic;eig(x);t(i)=toc(t1); end I do not think that Mathematica's algorithms are inefficient, as the fastest algorithms for eigenvalue decompositions (in the general case, not exploiting symmetry and such) are $\mathcal{O}(N^{2.376})$ and the timings both MATLAB's and Mathematica's implementations have the same correct slope on a log-log plot. I suspected unpacking in the background during the call to `Eigenvalues` and turning `On["Packing"]` confirms this. However, I don't think this alone could be the cause for a 3 fold speed reduction. I'm not expecting the timings to be exact either, as I understand that arrays and matrices are baked into the core of one and not the other, which can lead to performance differences. However, I'm interested in knowing if 1. there are reasons other than the simplified one I gave above for the difference in timings and 2. there ways in which I can improve the speeds or at least, reduce the difference by _some_ amount. Or is this something that one has to accept as a fact of life?
38646
Cleaning up InverseLaplaceTransforms in Mathematica
I am attempting to use _Mathematica_ to model an electrical circuit in the s domain. The circuit contains capacitances and and inductances. I've written a series of equations using Kirchoff's laws to solve for various currents in the circuit; however, this output is in the s domain. I would like to transform the output back into the time domain. The outputs that I obtain using `ExpToTrig[InverseLaplaceTransform[expr,s,t]]` are large and complicated. They contain several complex sin and hyperbolic sin terms. I know, from my knowledge of the circuit that the time domain solution should consist of three or four exponentially decaying terms and a single steady state sinusoid. How can I convert this complicated output into one that consists of a few decaying exponentials and a single, phase shifted sinusoid? The s domain expression for ig that I want to transform into the time domain is: ig=(3.25269*10^7 s)/((424000. + 923. s) (142122. + s^2)) Below, I have copied some of the code that I've tried. In[168]:= ig = ( 3.2526911934581187`*^7 s)/((424000.` + 923.` s) (142122.30337568675` + s^2)) Out[168]= (3.25269*10^7 s)/((424000. + 923. s) (142122. + s^2)) In[169]:= ExpToTrig[InverseLaplaceTransform[ig, s, t]] Out[169]= 3.25269*10^7 (-1.40932*10^-6 Cosh[ 459.372 t] + (Cos[(376.991 + 0. I) t] - I Sin[(376.991 + 0. I) t]) ((7.0466*10^-7 + 5.78292*10^-7 I) + (7.0466*10^-7 - 5.78292*10^-7 I) Cos[(753.982 + 0. I) t] + (5.78292*10^-7 + 7.0466*10^-7 I) Sin[(753.982 + 0. I) t]) + 1.40932*10^-6 Sinh[459.372 t]) In[170]:= Simplify[%] Out[170]= 3.25269*10^7 (-1.40932*10^-6 Cosh[ 459.372 t] + (Cos[(376.991 + 0. I) t] - I Sin[(376.991 + 0. I) t]) ((7.0466*10^-7 + 5.78292*10^-7 I) + (7.0466*10^-7 - 5.78292*10^-7 I) Cos[(753.982 + 0. I) t] + (5.78292*10^-7 + 7.0466*10^-7 I) Sin[(753.982 + 0. I) t]) + 1.40932*10^-6 Sinh[459.372 t]) In[171]:= FullSimplify[%] Out[171]= 0. - 45.8409 E^(-459.372 t) + (22.9204 + 18.81 I) E^((0. - 376.991 I) t) + (22.9204 - 18.81 I) E^((0. + 376.991 I) t) In[172]:= ExpToTrig[%] Out[172]= 0. + (45.8409 + 0. I) Cos[(376.991 + 0. I) t] - 45.8409 Cosh[459.372 t] + (37.6201 + 0. I) Sin[(376.991 + 0. I) t] + 45.8409 Sinh[459.372 t]
38640
InputField Insertion Point Movement
I am trying to make the Return Key move the insertion point to the end of the input field's text. Please see the example below for the problem I am facing. Namely, when the return key is pressed, the input field's text is updated but the insertion point is not moved. DynamicModule[{thing = "foo"}, nb = EvaluationNotebook[];EventHandler[InputField[Dynamic@thing,String], {"ReturnKeyDown" :> (thing = thing <> "bar";SelectionMove[nb, Next, Word])}]]
38642
Why cannot import 3ds file into 3dsmax
![enter image description here](http://i.stack.imgur.com/jdjBI.jpg) ![enter image description here](http://i.stack.imgur.com/FB7Od.jpg) The obj file can be imported into max rightly, but when I import the .3ds file into max I get some camera objects. Some 3ds files can be imported into max. What's wrong with the file? Is it possible to change something in Mathematica and make the .3ds file could be imported into 3dsMax? my file is here:
58211
Ordering the Boundary points of a Polygon
I feel like this should be simple, but I keep running into walls. Say someone gives you the coordinates of the vertices of a pentagon and the center point of the hexagon. Is there any way to get an "ordered" list of the boundary points? Let me try to put this in a picture. For instance, if someone gave me a list `{+,m,\pi,[],2}` plus the center point C, and I plot it and find 2 m + C [] \pi Is there anyway I can extract the list: `{2,+,\pi,[],m}` up to cyclic permutation? Ideally I would like to be able to differentiate orientations, clockwise vs counterclockwise, but I don't really care about cyclic permutation. Thanks for any help in advance! Big CRUCIAL edit: Thanks everyone for your help, I have been trying lots of suggestions, but I let out a crucial element in all of this. **My points are in 3D!** It seems to me that all of these routines, ConvexHullMesh, FindCurvePath, etc. all work with **2D** coordinates. I was thinking about trying to project onto a plane perpendicular to the central point, since I know the center, but that might be too lengthy. * * * Here are the points: Center: {0.951057,-0.309017,0.} Neighbors: {{0.723607, -0.525731, -0.447214}, {0.850651, 0., -0.525731}, {0.894427, 0., 0.447214}, {0.951057, 0.309017, 0.}, {0.587785, -0.809017, 0.}, {0.688191, -0.5, 0.525731}}
58210
Interpolation of 4D data
To create a table and an interpolation function of 3D data T=f(X,Y) I have the answer here: interpolation of 3D data How to create a table/interpolation function of 4D data T=f(X,Y,R)? If I do dada = Flatten[Table[{x, r, t, func[x, r, t]}, {x, 0, 1}, {r, 0, 0.12}, {t, 0, 1}], 1]; intfunc=Interpolation[data]; doesn't work. func[x,r,t] is a valid function for this domain. I think it has to do with Flatten.
58213
Recursive function with array of variables
I am trying to create a recursive function which works with an increasing array of variables. Since the formula is quite complex, i will try to simplify my example to the core of my problem. The initial value of the array should be $f[\\{1\\}]=2$ and all other arrays with a single element should have the value $0$. When an array of size $n$ is given to the function, it will reduce the size of the array to $n-1$ by dropping the first element. So if an array with the values $\\{0,0,1,0,2\\}$ is given, it will reduce it to $\\{0,1,0,2\\}$, or more generally the array $\\{r_5 ,r_4 ,r_3 ,r_2 ,r_1\\}$ will be reduced to $\\{r_4 ,r_3 ,r_2 ,r_1\\}$. The computation in this case should be: $$f[\\{r_5 ,r_4 ,r_3 ,r_2 ,r_1\\}]=(r_4 -1)*f[\\{r_4 -1,r_3 ,r_2 ,r_1\\}]+(r_3 -1)*f[\\{r_4 ,r_3 -1,r_2 ,r_1\\}]+(r_2 -1)*f[\\{r_4 ,r_3 ,r_2 -1,r_1\\}]+(r_1 -1)*f[\\{r_4 ,r_3 ,r_2 ,r_1-1\\}]$$ To speed up the computation, the function should memorize results and arrays containing a negative element ($-1$) should not be computed and instead given the value 0. My attempt so far starts with: f[r_] := f[r] = With[{n = Length[r], k = Rest[r]} ... To change the values of the $r_i$, I generated a For-Loop (from $1$ to $n$) containing: f[ReplacePart[k, -i -> Part[k, -i] - 1]] Any help is highly appreciated.
92
How can I ensure graphics exported in WMF format don't have text-spacing problems?
A lot of my work involves manipulating data, drawing a pretty plot of it and then inserting that graphic into a report written in Microsoft Word, under Windows.1 The fun comes with the exporting part. If the audience don't have the Mathematica fonts installed, negative signs, parentheses etc will come up as missing-characters in the WMF graphic and in Word. You can fix this using the `PrivateFontOptions` option, either in the notebook or in the Options Inspector. SetOptions[$FrontEnd, PrivateFontOptions -> {"OperatorSubstitution" -> False}] But I still find the output a bit disappointing, in that the spacing around parentheses and other characters in the WMF file looks wrong (sorry, I can't show the whole graphic because the data are confidential). ![enter image description here](http://i.stack.imgur.com/DKmZb.png) The code to produce the graph above rests partly on a package, so here is a minimal example: test = Grid[{{DisplayForm[ AdjustmentBox[ Style["Source: REDACTED (confidential)", 13, FontFamily -> "Arial", Black], BoxMargins -> {{3, 0}, {0, 0}}]]}}] ![enter image description here](http://i.stack.imgur.com/cF2wT.png) This is how it looks in Win7 and Office 2010: the first two are with operator substitution ON, the second pair are with operator substitution OFF. There is still a bit too much space but it is better than in Word 2007. ![enter image description here](http://i.stack.imgur.com/A6Heo.png) As highlighted, there is just too much space between the text and the closing parenthesis. This is not apparent in the Mathematica notebook, where it looks just fine. It must be something to do with Mathematica's WMF export routine: other applications with WMF export don't do this. Is there any way of automatically ensuring that the spacing around letters in the resulting WMF file is a bit more acceptable? Some kind of auto-kerning option? Ideally it should be something I can set in a package so my colleagues don't have to know the internals of how to do it. 1Actually I have people to do that now, but you get the idea.
58215
UnFlatten an array into a ragged matrix
Trying to unflatten an array that was part of ragged array of sub matrices. As input we have flattened array: myflatarray={7.7056, 4.20225, 3.02775, 7.60807, 9.77169, 6.18476, 4.80993, 6.32965, 2.69882, 0.268505, 1.78048, 9.67702, 0.875699, 7.18504, 1.62811, 0.0313174, 5.66048, 1.61613, 5.71987, 0.971484, 3.87503, 2.8537, 2.68776, 3.47433, 2.81483, 4.0387, 1.74939, 3.12037, 4.18016, 2.35264, 0.853583, 1.22412, 0.382633, 3.67874, 8.68059, 8.02419, 6.9547, 7.11112}; and the original dimensions of the submatrices in the ragged array. mydimensions={{4, 5}, {3, 4}, {2, 3}}; In other words the original object looked like original={ {{7.7056, 4.20225, 3.02775, 7.60807, 9.77169}, {6.18476, 4.80993, 6.32965, 2.69882, 0.268505}, {1.78048, 9.67702, 0.875699, 7.18504, 1.62811}, {0.0313174, 5.66048, 1.61613, 5.71987, 0.971484}}, {{3.87503, 2.8537, 2.68776, 3.47433}, {2.81483, 4.0387,1.74939, 3.12037}, {4.18016, 2.35264, 0.853583, 1.22412}}, {{0.382633, 3.67874, 8.68059}, {8.02419, 6.9547, 7.11112}} }; I tried using the undocumented function Internal`Deflatten, and it didn't work, but I was happy that it didn't crash my kernel! I'm able to do it rather inefficiently this way: arraytoweights[myflatarray_,mydimensions_]:= MapIndexed[Partition[#1, mydimensions[[Last@#2, 1]]] &, Take[myflatarray, #] & /@ ((# - {0, 1}) & /@ (Partition[ FoldList[Plus, 1, (Times @@@ mydimensions)], 2, 1]))]; My main problem is that this is part of an optimization function, that needs to be called many times during the optimization, and this is atm the slowest part. It also can't seem to be compiled. I suppose with Compile I have to make sure there are no ragged arrays generated in-between neither. Any suggestions are appreciated! **Edit** : So it turns out mine is not as bad as I thought. Though I like the extensions to more structure as @Leonid did. I have no need for it now, but I see how I can use it in the future. Here is the current tally of the answers and their timing, using a random larger result, the typical size I need. (only with submatrices atm) mydimensions = {{30, 4}, {10, 5}, {3, 4}, {2, 1}}; SeedRandom[1345] original = ((RandomReal[{0, 10}, ##] & /@ mydimensions)); myflatarray = Flatten[original]; For the timing I ran the result from each with $10^3$ runs. For example, here the set for my results. lalmeiresults = Table[First@ Timing[lalmei = arraytoweights[myflatarray, mydimensions];], {i,Range@1000]}] lalmei === original > > True Here is the BoxWhiskerChart, with only the 95% quantile, no outliers. (Not sure if this the best way to do timing, but here it goes ) ![enter image description here](http://i.stack.imgur.com/5aUql.jpg) Couple of notes about the image: the distribution of mine and Leonid's, and seismatica's is about the same. Picket's is only a bit larger 25% only. Mr.Wizard's, kguler's and WReach's results only shoot up as the size of the sub-matrices gets larger. With a small examples, Mr. Wizard's is the fastest. mydimensions = {{3, 4}, {2, 5}, {3, 4}, {2, 1}}; SeedRandom[1345] original = ((RandomReal[{0, 10}, ##] & /@ mydimensions)); Here the distributions in that case: ![enter image description here](http://i.stack.imgur.com/NElRx.jpg)
38649
Tooltip in Listplot - Export to PDF
I would like to export a listplot as a PDF document without loosing the tooltip information on the single points. It kind of works with hyperlinks. Nevertheless this workaround is pretty limited and does e.g. not offer graphics as tooltips. Looking forward any ideas! Thanks and best regards Patrick data = {{1, 2}, {2, 3}, {3, 5}, {4, 7}}; hlinks = Text[ StringForm["``", Hyperlink[" ", "http://" <> ToString[#[[2]]]]], #] & /@ data; chart = Show[ListPlot[data, AxesOrigin -> {0, 0}], Graphics[hlinks]]; Export["C:\\Users\\myname\\Desktop\\scatterplot.pdf", chart]
97
What causes Manipulate to stop evaluating expressions?
I'm working on a N-body simulation within Mathematica, and currently I have something along these lines: Manipulate[ {r, v, a} = Step[G, m, r, v, a, ts dt]; Show[Visualize[names, colors, radius1, r1], PlotRange -> {{-d, +d}, {-d, +d}}, ImageSize -> 1200], Dynamic[MatrixForm[r]], {{dt, 0.02}, 0.0001, 1, 0.0001}] `Step[...]` simply steps the simulation forward in time. This works fine and I can confirm it works perfectly fine independent of anything else. However, occasionally the `Dynamic[MatrixForm[r]]` expression stops evaluating. I can't really figure out why it does this. Furthermore, sometime after the `Dynamic` stops evaluating, the `Show[Visualize[...], ..]` expression stops evaluating. At some point earlier when I had something much simpler along the lines of Manipulate[f = Step[f]; Show[SimpleVisualize[f]], {...}] The code within `Manipulate` would run indefinitely. Now it seems to randomly stop. It never stops at the same time, but it always does. Any ideas why? **An example of where this occurs is available here:http://dl.dropbox.com/u/9450720/minimal.nb** **Environment Information: Mathematica 8.0.4.0 on Windows 7 64-bit** * * * Here is a basic example of what I'm doing: Step[r_, v_, a_, dt_] := Block[{nr, nv, na}, nv = v + dt/2 a; nr = r + dt nv; na = -nr; nv = nv + dt/2 a; {nr, nv, na}] r = {1, 0}; v = {0, 1}; a = {-1, 0}; Manipulate[{r, v, a} = Step[r, v, a, dt]; Dynamic[Show[{Graphics@Circle[{0, 0}, 1], Graphics[{PointSize[Large], Red, Point[r]}]}]], Dynamic[{r, Norm@r}], {{dt, 0.005}, 0.00001, 1, 0.00001}] This evaluates fine and runs indefinitely.
28703
what is the difference between these definitions?
I define 2 functions: g[s_][x_,y_] := the gaussian (1) g1[s_,x_,y_] := the gaussian (2) I get the same answer using `g[s][ x, y]` and `g1[ s, x, y]`. So, what is the difference? Is there some advantage to be gained in definition `1` over `2`?
7999
Define parameterized function
I would like to be able to define the gain function of a system from its parameters. Specifically, I'd like to define a function that accepts two inputs, call them $b$ and $w$, and returns a function that accepts a single parameter, $v$ and returns $1/\sqrt{(w^2 - v^2)^2 + b^2 v^2}$ . What's the right way to do this? Presumably I'd like to be able to write gain[b,w][v] to evaluate the function, with parameters b and w, at v. But I'd also like to be able to write f[v_] = gain[b,w] and have `f[v]` be the same as `gain[b,w][v]`.
30411
Key Mapping in Command Line Mode
I would like to map Ctrl-U to "kill line" when using Mathematica 9.0 in command line mode (i.e. not within the front end, but when called from a Unix command line by "math"). I am aware that (on a Mac) the directory /Applications/Mathematica.app/SystemFiles/FrontEnd/TextResources/Macintosh contains the file "KeyEventTranslations.tr", however this file seems to only affect the front end, not the behavior of the kernel when called from the command line. I also know that Ctrl-C and Ctrl-G are already mapped to "Kill line" (although they don't seem to have corresponding entries in "KeyEventTranslations.tr"). But I would prefer to use Ctrl-U since this is what I am using in all other applications. (Note: Ctrl-U used to work in Math 8.0).
57677
How to set the boundary fixed and apply the Young's modulus in Mathematica's FEM?
Provided the drawing .stl is given, and the Young's modulus is know, how to start a FEM vibration mode test in Mathematica? The boundary is required to be fixed and the interested focus is on the different responses over various Young's modulus. ![enter image description here](http://i.stack.imgur.com/yGR80.jpg)
30413
Manipulating Frame labels when legends are also there
I am plotting a few curves inside a frame with labels on three of its sides; left, right and bottom. I want legends to be places at the top. The problem which I am facing is that the top label occupies white space and I want it not to do so. In the code it is labeled as "I want this removed". Please guide me how can I remove the top label completely leaving no gap between frame and legends. Plot[{Sin[h],Cos[h],Tan[h]},{h,5,10}, PlotStyle->{{GrayLevel[0],Dashing[None],Thickness[0.005]}, {GrayLevel[0],Dashing[Tiny],Thickness[0.005]}, {GrayLevel[0],Dashing[Small],Thickness[0.005]}, {GrayLevel[0], Dashing[Medium],Thickness[0.005]}}, Axes -> False, Frame -> True, FrameLabel->{{"Left", Rotate["Right",180 Degree]},{"Bottom","I want this removed"}}, PlotLegends -> Placed[LineLegend[Automatic, {"sin", "cos", "tan"}, LegendMarkerSize->{{45, 15}}],Top]]
50641
Segment parts of a picture
I remember that there was some thread of this sort, but I could not find it, so I hope it's ok to open a new thread. I want to segment pictures of the brain, i.e. I only want the area within the skull without the bones (all the grey matter/pixels. White parts are bones of the skull). I have the following two pictures. In the first picture I also want all the grey parts, but you see that there are some white parts intermingled between. So basically the algorithm should segment four different images. In the second pictures it seems to be easier. Can you guys give me a hint to a function or some examples or threads? ![enter image description here](http://i.stack.imgur.com/U6adl.jpg) ![enter image description here](http://i.stack.imgur.com/zCxeQ.jpg)
30415
How can I save the current evaluation state of a notebook and start from it later?
I currently have a nested `Table` command running (in Mathematica 9), with two progress bars to track the variables. I have to shut down or hibernate the computer in a couple of minutes, and the calculation is only halfway done. What can I do in such a situation. I'm pretty sure simply hibernating won't work, as I recall it not working in Mathematica 8. Will entering a subsession and _then_ hibernating work? Or entering a subsession and dumping the kernel? While I don't have any preference towards shutting down or hibernating in my current situation, a solution that allows me to close Mathematica and resume evaluation at some later time (after shutting down, etc) would be ideal. Note: This question is about a cell which is already running (so I can't modify it), however if it's not possible to do this here, an answer that shows how to run `Table` in an interrupt-and-saveable manner would be OK1 1\. I have a pretty good idea of how to do this (dynamically append data to a list with `For` instead f using `Table`), but there may be better ways of doing this.
17734
Optimising 2D binning code
I have a set of (x,y,z) data, 45,000 to be precise and I want to bin the z values in 256 equidistant bins based on their (x,y) values. The final array should be a set of 256x256 array with each slot containing an average of binned z values. Being new to mathematica, I came up with the following code: data = RandomReal[{12000, 35000}, {45000, 3}]; data1 = data[[All, {1, 2}]];(*strip the zvalues from the set*) xValues = data[[All, 1]]; yValues = data[[All, 2]]; zValues = data[[All, 3]]; (*Compute maximum/minimum of x values*) maxXvalue = Max[xValues]; minXvalue = Min[xValues]; (*Compute maximum/minimum of y values*) maxYvalue = Max[yValues]; minYvalue = Min [yValues]; (*Compute maximum/minimum of z values*) maxZvalue = Max[zValues]; minZValue = Min[zValues]; bbx = {Floor[minXvalue], Floor[maxXvalue], Floor[((maxXvalue - minXvalue)/256)]}; (* equidistant x bins*) bby = {Floor[minYvalue], Floor[maxYvalue], Floor[((maxYvalue - minYvalue)/256)]};(* equidistant y bins*) bList = BinLists[data1, {bbx}, {bby}]; bCount = BinCounts[data1, {bbx}, {bby}];(*Gives a count of the number of items in \ each bins*) (*Defining array to contain final z average values*) meanZValues = Table[0, {Length[bList]}, {Length[bList]}]; i = 0; (*initialising loop variables*) j = 0; k = 0; f[x_] := zValues[[x]];(*Defining function to get z values back*) For[i = 1, i <= Length[bList], i++, For [j = 1, j <= Length[bList], j++, m1 = {}; (*Re-empty m1 list*) For [k = 1, k <= Length[bList[[i, j]]], k++, AppendTo[m1, Position[data1, bList[[i, j]][[k]]] (*accessing only the x- coordinate index of the position on original matrix*) ]; (*Getting the indices of the binned values*) indices = Flatten[DeleteDuplicates[Take[m1, All]]]; (*Position command above gives multiple indices if these values occur more than once, hence deleting the duplicate ones*) meanZValues[[i, j]] = Mean[Map[f,indices]]; (*Compute average values of Z by accessing the original array, getting the z values *) ] ] ] meanZValues It gives an output in a reasonable amount of time for up to couple of thousand values, however, it lags and maybe crashes without any output for 45,000 set of data. How do I make this code more efficient? Thank you
29540
Partitioning a data set of two-dimensional coordinates into subsets of coordinates underlying non-overlapping tiles of a bounding box
Image I have a list of coordinates of the form: CoordinateList = {{{107.187, 130.25}, {160.742, 131.872}, {22.1312, 86.4814}}, {{251.357, 47.4634}, {177.773, 223.6}, {251.487, 126.181}}, {{51.1981, 15.1508}, {10.17, 44.4388}, {75.039, 4.44131}}, {{21.6951, 47.7368}, {73.1692, 130.974}, {93.9724, 139.979}}, {{107.914, 237.77}, {139.735, 199.392}, {20.8538, 88.0626}}, {{3.47162, 71.022}, {150.473, 45.0054}, {83.2402, 228.76}}, {{244.43, 156.011}, {126.716, 77.7305}, {247.013, 87.2547}}, {{85.3221, 108.568}, {16.3881, 167.449}, {147.638, 107.363}}, {{135.401, 97.1536}, {210.616, 104.893}, {195.791, 210.578}}, {{95.4441, 115.686}, {236.267, 69.9433}, {160.552, 30.2893}}, {{92.4965, 199.097}, {201.669, 56.4691}, {186.667, 113.938}}, {{98.7167, 45.6133}, {151.794, 88.8944}, {245.539, 177.681}}, {{103.834, 250.27}, {56.5259, 176.679}, {90.3859, 164.626}}, {{64.3762, 197.027}, {78.5971, 216.795}, {52.4567, 200.263}}, {{31.2684, 73.8046}, {12.6404, 246.689}, {3.62497, 177.448}}, {{210.712, 104.823}, {151.347, 13.8795}, {63.3472, 197.042}}, {{92.1903, 12.4941}, {141.632, 112.287}, {221.041, 189.748}}, {{237.627, 76.7537}, {20.2009, 117.81}, {150.973, 110.379}}, {{24.1639, 218.049}, {251.751, 185.44}, {161.526, 231.584}}, {{125.477, 237.408}, {38.6014, 73.5206}, {4.64447, 49.3799}}, {{80.0287, 123.736}, {153.407, 131.396}, {11.0003, 202.104}}, {{250.583, 51.9734}, {69.2102, 89.4853}, {10.0789, 163.632}}, {{96.328, 67.6535}, {210.743, 104.839}, {175.073, 114.937}}, {{136.214, 115.21}, {58.048, 144.602}, {244.051, 57.6266}}, {{60.0216, 18.7106}, {173.178, 177.361}, {87.0396, 112.919}}, {{4.67427, 49.3867}, {60.0381, 18.7024}, {77.1039, 162.763}}, {{134.911, 10.2993}, {79.7796, 61.8847}, {168.708, 104.301}}, {{200.808, 221.631}, {84.5544, 137.876}, {136.434, 150.225}}, {{110.313, 30.4609}, {37.538, 199.435}, {10.6894, 230.592}}, {{132.621, 98.0955}, {213.781, 124.4}, {21.3729, 31.0911}}}; Note that this is a list (length = 30) of subsets of coordinates. These points fit within a bounding box of dimensions: BoundingBox = {Min[CoordinateList[[All,1]]], Max[CoordinateList[[All,1]]], Min[CoordinateList[[All,2]]], Max[CoordinateList[[All,2]]]}; I would like to partition this box into the largest possible set of $A \times B$ rectangular tiles - the bounding box for the points can be trimmed as necessary. Call this number of tiles $k$. Then, without disturbing the ordering of the coordinate subsets, I would like to generate a list of $k$ subsets of the CoordinateList consisting of the points that fall in each rectangular region. Here's a brief illustration (by mouse, so it's inexact!) of how I'd like to partition the points: ![enter image description here](http://i.stack.imgur.com/CajWf.png) Here, we do the best we can to tile a bounding box for the list of points with tiles of some chosen dimension (here - roughly $A=B=50$), and then if a point falls underneath a certain tile, it is assigned to a subset for that tile. The subset must respect the primordial subsets in the original CoordinateList s.t. in each subset we can see while element of CoordinateList a point corresponds to. Is there a nicer way to proceed than defining a coordinate list for each box, and checking for membership by a few inequality checks, or more compactly, computing a winding number using: Graphics`Mesh`InPolygonQ[poly, pt] Perhaps this could be done via the use of Pick[]? The issue I have is that my CoordinateList is actually quite a bit larger than the example provided, and I need to do this partitioning quickly.
27308
Coarse-graining and/or sorting of a list of arbitrary triades {x,y,z} in the (x,y) plane
In general my question is related to an approach to make an interpolation of a data which is not on a regular mesh. Another aspect of this question has already been discussed after my question "Numerical integration of a numeric data available as a nested list". I have data in the form of a list of triads `{x, y, z}` that came from an external simulation. It may be and may be not ordered. Further, x and y lie within a rectangle R, and are not regularly spaced. I would like to coarse- grain and simultaneously sort this data. The coarse-graining consists in partitioning the rectangle R into multiple small rectangles and constructing a list `{xij, yij, zij}` where xij and yij are the coordinates of the centers of the rectangles, while zij is the average of all z values for which x and y belong to the rectangle specified by i and j. To be specific, let us build a list mimicking the real one, but small enough: a = RandomReal[{0, 10}, {10000, 2}]; b = RandomReal[{-1, 1}, {10000}]; lst = MapThread[Insert[#1, #2, 3] &, {a, b}] /. {x_, y_, z_} -> {x, y, 10 Exp[-((x - 5)^2 + (y - 5)^2)/4] + z}; The list lst has 10000 elements with x and y located between 0 and 10. The coarse-grained list may be obtained by the averaging over squares with the sizes 0.5*0.5 like this: lstCoarseGrained = Flatten[Table[ (s = Select[ lst, ((i <= #[[1]] <= i + 0.5) && (j <= #[[2]] <= j + 0.5) &)]; {{i + 0.25, j + 0.25}, Mean[Transpose[s][[3]]]}), {i, 0, 9.75, 0.5}, {j, 0, 9.75, 0.5}], 1]; It has Length[lstCoarseGrained] > 400 elements and can be straightforwardly interpolated: f = Interpolation[lstCoarseGrained, InterpolationOrder -> 3, Method -> "Spline"] plotted: Row[{ ListPlot3D[lst, PlotRange -> All, ImageSize -> 250], Plot3D[f[x, y], {x, 0, 10}, {y, 0, 10}, PlotRange -> All, ImageSize -> 250] }] ![enter image description here](http://i.stack.imgur.com/0bgtC.jpg) and then further post-processed. On the figure above the left image shows the data before, and the right - after the coarse-graining and interpolation. However, the coarse-graining by this approach requires Timing[ Flatten[ Table[ ( s = Select[lst, ((i <= #[[1]] <= i + 0.5) && (j <= #[[2]] <= j + 0.5) &)]; {{i + 0.25, j + 0.25}, Mean[Transpose[s][[3]]]} ), {i, 0, 9.75, 0.5}, {j, 0, 9.75, 0.5}], 1];] > {13.468750, Null} 13 seconds. My realistic lists have about 10^6 triads and will require about 21 min. My question is, if you can see a faster way to do this? In principle, if I could, say, use something like Partition instead of a Table, it would go faster. However, to do this, the list should be first sorted in 2D, and I do not see, how. So the second question is if you see the way to sort the list with elements {x,y,z} with arbitrary order in the (x,y) plane such that it is organized as like, say, the list given by the table: Table[{i, j}, {i, 1, 5}, {j, 1, 5}]
38597
Calculating Second-Order Tikhonov Regularization Parameter in Mathematica
I am trying to map slowness of underwater sound velocity in a river using some tomographic device. The location of each acoustic receiver/transmitter is shown in the picture below ![enter image description here](http://i.stack.imgur.com/dM7wr.png) To find the slowness m={S11,S12,S13, ...S112,...,S71,S72,S73, ... , S712} from the observations d={t_R1L1,t_R1L2,...,t_R4L4} I need to solve the regularized least-square equation min||Gm-d||^2+a^2||Lm||. I, therefore, established matrices G={{0, 12.1524, 2.7609, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 15.6297, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 15.6297, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 5.65726, 9.97241, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 15.6297, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 15.6297, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 14.3434, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 14.912, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 15.6283, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 15.6283, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4.50649, 11.1218, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 15.6283, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 15.6283, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 11.0604, 2.88671, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 19.1877, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2.16882, 17.9406, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4.58072, 15.5287, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 6.99261, 13.1168, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 9.40451, 10.7049, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 11.8164, 8.293, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 14.2283, 0.877225, 0, 0, 0}, {0, 0, 17.3355, 7.70369, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 10.5772, 15.6647, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2.61622, 18.2809, 5.34477, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 12.9361, 13.3058, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4.97514, 18.2809, 2.98585, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 15.295, 10.9468, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 7.33406, 14.2199}, {0, 0, 0, 0, 14.4946, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 9.24233, 9.34103, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2.41548, 16.1679, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 18.5834, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 14.172, 4.41137, 0, 0, 0, 0, 0, 0, 0, 0, 0, 7.34514, 11.2382, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 17.054, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 11.8019, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 15.1311, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 15.1311, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 15.1311, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 15.1311, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 15.1311, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 13.5034, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 2.1671, 10.7843, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 16.6048, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 7.58393, 9.02091, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 16.6048, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 9.34735, 7.25749, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 16.6048, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 11.1108, 1.36223, 0, 0, 0}, {0, 0, 0, 0, 1.29176, 15.5556, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 5.29103, 16.3088, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4.53791, 17.0619, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3.78479, 17.815, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3.03167, 18.5681, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2.27856, 19.3212, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1.52544, 16.2157}, {0, 0, 0, 0, 0, 0, 18.5392, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 19.4035, 2.06375, 0, 0, 0, 0, 0, 0, 0, 0, 0, 19.9029, 1.56439, 0, 0, 0, 0, 0, 0, 0, 0, 0, 20.4023, 1.06503, 0, 0, 0, 0, 0, 0, 0, 0, 0, 20.9016, 0.565678, 0, 0, 0, 0, 0, 0, 0, 0, 0.433035, 20.9679, 0.0663213, 0, 0, 0, 0, 0, 0, 0, 0, 0, 19.7006, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 14.1187, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 16.3486, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 9.75723, 6.5914, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 16.3486, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4.73928, 11.6094, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 16.3486, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 14.59, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 1.65553, 11.4627, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 15.1901, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 15.1901, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 15.1901, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 15.1901, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 15.1901, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 7.70533, 3.70495, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0.447913, 15.4943, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 10.2405, 8.21959, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 17.5151, 0.944913, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 18.4601, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 6.32976, 12.1303, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 13.6044, 4.85561, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 15.1623}, {0, 0, 0, 0, 0, 0, 0, 0, 6.77971, 17.4671, 0, 0, 0, 0, 0, 0, 0, 0, 0, 14.9997, 11.4365, 0, 0, 0, 0, 0, 0, 0, 0, 5.00342, 18.2163, 3.21656, 0, 0, 0, 0, 0, 0, 0, 0, 13.2234, 13.2128, 0, 0, 0, 0, 0, 0, 0, 0, 3.22712, 18.2163, 4.99286, 0, 0, 0, 0, 0, 0, 0, 0, 11.4471, 14.9891, 0, 0, 0, 0, 0, 0, 0, 0, 0, 17.4914, 6.76916, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 18.2098, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 16.1093, 3.74475, 0, 0, 0, 0, 0, 0, 0, 0, 0, 13.0672, 6.78682, 0, 0, 0, 0, 0, 0, 0, 0, 0, 10.0251, 9.8289, 0, 0, 0, 0, 0, 0, 0, 0, 0, 6.98307, 12.871, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3.941, 15.913, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 17.7183, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 14.3166, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 15.6093, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 15.6093, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 9.15931, 6.45003, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 15.6093, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 15.6093, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 11.7252, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 2.36708, 11.8831, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 15.5369, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 15.5369, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 14.6006, 0.936286, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 15.5369, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 15.5369, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 12.7613}} and L=Flatten[    Table[ M = ConstantArray[0, {7, 12}];, M[[i, j]] = -4; M[[i, j + 1]] = 1; M[[i, j - 1]] = 1; M[[i + 1, j]] = 1; M[[i - 1, j]] = 1; Flatten[M], {j, 2, 11}, {i, 2, 6}], 1]; Given that `{{U, V}, {Lamb,M}, X} = SingularValueDecomposition[{G, L}]` , where `G=U.Lamb.Conjugate[Transpose[X]]` and `L=V.M.Conjugate[Transpose[X]]`, **Why the generalized singular values of G and L, gama, become indeterminate or zero? And, how can I fix this issue?** gama=lambda/mu where lambda=Sqrt[Diagonal[Transpose[Lamb].Lamb]] ={0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.} and mu=Sqrt[Diagonal[Transpose[M].M]] ={0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1.}
30418
Why cant I open CDF's in the Chrome browser on Linux?
I have google chrome on ubuntu 13.04. Every time I try to run a CDF on wolframs website, I'm required to download the CDF first, and then open it in my computer, but when I was using windows it was ready to use on the browser itself, any ideas on how to fix it? Is it a better topic for the ubuntu stackexchange?
43717
Launching external program in background process with ParallelSubmit
I work on the implementation of an image processing pipeline in _Mathematica_ in which I also would like to use external programs to do certain processing steps. As an example I have macros for the program FIJI that are written, parameterized and then executed with the `Run` command in _Mathematica_. Some of these processing steps just need to be executed and I do not necessarily need the results in the pipeline later on. For those computations I thought it would be nice to let the computation be run in a background process and continue with other steps while the background process is still running. I've read about pushing computations to background kernels (Computing Many Slow I/O Operations, Concurrency: Managing Parallel Processes) and it works fine. What I've tried is the following: I launch a bunch of kernels, generate some processes and queue those processes using `ParallelSubmit`. I then use `QueueRun` to start the processes in the background and collect the results with `WaitAll` after a while. CloseKernels[]; Needs["Parallel`Developer`"]; ResetQueues[]; LaunchKernels[5]; ids = Function[i, ParallelSubmit[(Pause[i^2]; {i^2, DateString[]}), Scheduling -> i]] /@ {1, 2, 3, 4, 5} ![enter image description here](http://i.stack.imgur.com/oHcB1.png) {QueueRun[], $QueueLength} > {True, 0} ![enter image description here](http://i.stack.imgur.com/wHpGB.png) WaitAll[ids] > {{1, "Mon 10 Mar 2014 09:37:53"}, {4, "Mon 10 Mar 2014 09:37:56"}, {9, "Mon > 10 Mar 2014 09:38:01"}, {16, "Mon 10 Mar 2014 09:38:08"}, {25, > "Mon 10 Mar 2014 09:38:17"}} ![enter image description here](http://i.stack.imgur.com/RudKp.png) DateString[] > Mon 10 Mar 2014 09:41:23 I included `DateString` to see, when the tasks have been processed in the background kernels. As can be seen from the timings it worked as expected and I also was able to do computations in the front end while the tasks were running. Now I'd like to do the same thing with calling external programs like FIJI but for some reasons the tasks are processed not before `WaitAll` is evaluated. The kernels receive the jobs but they do not start until I evaluate `WaitAll`. CloseKernels[]; Needs["Parallel`Developer`"]; ResetQueues[]; LaunchKernels[3]; ids = { ParallelSubmit[runFijiMacro["command line code"]], ParallelSubmit[runFijiMacro["command line code"]], ParallelSubmit[runFijiMacro["command line code"]] } ![enter image description here](http://i.stack.imgur.com/9Kepv.png) {QueueRun[], $QueueLength} > {True, 0} ![enter image description here](http://i.stack.imgur.com/dzNo2.png) ( _The returned EvaluationObjects seem to be running but in fact they don't_ ) WaitAll[ids] ( _now FIJI is started and the tasks are processed. But, also the main kernel is blocked._ ) **Questions** 1. Why does this construct not work with `Run`? 2. Is there a possibility to force the evaluation on the background kernels? 3. Is there a workaround to do this? Thanks for any help or comments!
59349
Windows Insert File Path
Suppose I want to Save["filename", foo] under Windows. Ordinarily, the way to deal with file names under windows is to open the Insert dialogue and find a file path, but in this use case, the file is new, so it does not exist (yet). Unlike every other windows program I am familiar with, Mathematica is unwilling to create a new file (that is, once you locate the relevant folder, you type in "mynewfile.m", Mma balks. This seems like a bug (sorry, Dan Lichtblau, a misfeature :)), but is there any reasonable workaround? Am I missing something obvious?
13052
Excel-like UI in Mathematica
This is NOT about importing/exporting Excel data to/from Mathematica. This is purely a UI question. The Mathematica notebook interface is very "linear" — I type in an input cell and _Mathematica_ produces an output cell. Is it possible to implement something like an Excel interface for _Mathematica_? For example, if I have: * cell `a1 = "x+y"` * cell `b1 = "x-y"` * cell `c1 = "= a1^2 + b1^2"` Then, _Mathematica_ puts "2(x^2+y^2)" into `c1`. If I modify `a1` to say `"3"`, _Mathematica_ puts `"(x-y)^2 + 9"` into `c1`.
13050
How to make the docked cell and the navigation toolbar in the Slide Show?
I have a notebook with a docked cell displaying a banner with some graphics. When I convert this notebook to the Slide Show the banner is nicely displayed on each slide. But the Navigation Toolbar is not visible, apparently covered by my docked cell. How to display both, the Navigation Toolbar and my graphical banner in the Slide Show?
13056
How to model Macroeconomic dynamics?
I am rather new to _Mathematica_ and I wanted to see if I could get some help with the following. I am trying to generate a model of macroeconomic indicators defined by the following functions, but I do not know how to input these in _Mathematica_ or be able to manipulate them a bit more. Y(T+2) = C(T+2) + I(T+2) + Govlev(T+2) + X(T+2) ; YD(T+2) = (1 - TX(T+2)) * Y(T+2) ; C(T+2) = a + b * YD(T+2) ; I(T+2) = e - d * R(T+2) ; Mlev(T+2) / plev(T+2) = k * Y(T+2) - h * R(T+2); Pie(T+2) = alpha * Pi(T+1) + beta * Pi(T) ; Pi(T+2) = Pie(T+2)+ f * (Y(T+1) - YNlev(T+2)) / YNlev(T+2) ; plev(T+2) = plev(T+1) * (1 + Pi(T+2)) ; Ex(T+2) * plev(T+2) / plevw(T+2) = E = q + v * R(T+2) ; X(T+2) = gamma - mu * Y(T+2) - n * ( Ex(T+2) * plev(T+2) / plevw(T+2)) ; GD(T+2) = Govlev(T+2) - TX(T+2) * Y(T+2) ; U(T+2) = UN(T+2) - s * (Y(T+2) - YNlev(T+2)) / YNlev(T+2) ;
28585
How can I account for assumed X and Y errors when using findfit?
I'm using very noisy biological data, and I need to assume, when doing linear regression, that there are errors in both `X` and `Y`. To my understanding normal regression only assumes errors in `Y`. Is there a way to specify error values for `X` and `Y` in _Mathematica_ , and take that into account when finding a linear relationship between the two?
48344
Data Fitting with xerrors and yerrors using findfit
I'm trying to fit data with xerror and yerror using "Findfit", so i'm trying the same form than in tuto : http://reference.wolfram.com/applications/eda/FittingDataToLinearModelsByLeast- SquaresTechniques.html which is {{x,errx},{y,erry}} So my code is : data2201vol = ReadList["data.dat", {{Number, Number}, {Number, Number}, {Number, Number},{Number,Number}}][[All, {1, 3}]] data1 = data2201vol res1 = FindFit[data1, model, t, eta]; But when i evaluate i get that : FindFit::fitd: First argument {{{7.98187,2.92332},{65326,255.59}},{{13.0174,2.91387},{61510,248.012}},{{23.0829,2.98836},{55135,234.808}},{{43.1855,3.0372},{45142,212.466}},{{53.2883,3.11867},{41257,203.118}},{{63.3186,3.16282},{38048,195.059}},{{83.3984,3.23615},{32628,180.632}},{{103.479,3.26169},{28339,168.342}},{{123.585,3.36999},{24976,158.038}},{{203.785,3.47143},{16381,127.988}},{{243.893,3.62374},{13685,116.983}},{{303.901,3.54541},{10845,104.139}},{{363.98,3.57129},{8753,93.5575}}} in FindFit is not a list or a rectangular array. >> Of course "model" is well defined, it works when i use a more classical input without yerror. I didn't find so much on that, if someone have an idea !
38888
Scheduling Mathematica scripts to run from a command line
I want to run _Mathematica_ "jobs" on a daily/weekly basis to do system maintenance and as part of daily processing operations. What is the best way to do this?
18703
Determining the partitioning of Do loop iterations between kernels?
Say I have some `Do` loop that I'd like to evaluate in parallel: `Do[Something, {i, 1, M}]`. I'd like to split this list up into chunks which individual kernels evaluate. However, I would like to explicitly specify these chunks, which is something `ParallelDo` does not seem to allow for. How does `ParallelDo` actually partition values of $i$ to be evaluated during a `Do` loop? If I have a list $L$ where I need to access $L[i]$ at the $i$th iteration of the `Do` loop, how might I avoid slowdowns associated with using some large global list?
58266
Specifying initial conditions for a PDE
I have simple diffusion equation with point source at `c(0, t) = 1` and initial condition `c(elsewhere, 0) = 0`. How should I apply `DSolve` or `NDSolve` to solve the equation ? I have tried specifying my initial conditions like this: ic = {c[x > 0, 0] == 0, c[x < 0, 0] == 0, c[0, t] == 1, Derivative[1, 0][c][1, t] == 0} but it fails. Any suggestions? Diff = D[c[x, t], t] - D[c[x, t], {x, 2}] == 0; ic = {c[x > 0, 0] == 0, c[x < 0, 0] == 0, c[0, t] == 1, Derivative[1, 0][c][1,t] == 0}; s1 = NDSolve[{Diff, ic}, {c[x, t]}, {x, 0, 1}, {t, 0, 1}]; Plot3D[c[x, t] /. s1, {x, 0, 1}, {t, 0, 1}]
33853
How do I incorporate metadata using Export?
I have Export["file.png",codeThatMakesAGraphicsObject[],"PNG"]; and want to embed some specific metadata information. Is there a simple way to do that? * * * Note that this code already (automatically, apparently) includes metadata information for "Horizontal resolution", "Vertical resolution", "Software used", and "Date and time of digitizing", and I don't want to disturb that with any fields I add.
18
Where can I find examples of good Mathematica programming practice?
I consider myself a pretty good _Mathematica_ programmer, but I'm always looking out for ways to either improve my way of doing things in _Mathematica_ , or to see if there's something nifty that I haven't encountered yet. Where (books, websites, etc.) do I look for examples of good (best?) practices of _Mathematica_ programming?
25072
Where to begin?
I was looking at the Mathematica website today and just decided to buy the software. I've spent too much time juggling annoying symbolic expressions by hand that actually learning to do things with a computer should be a worthwhile investment. Similarly for making conjectures about formulas etc. a computer would seem like a useful tool. My question: Where should I begin? I have a Ph.D. in math and would consider myself an expert level programmer in C, C++, x86 assembler, OCaml and Perl. I understand the functional programming paradigm well, so I don't need any introductions. To begin with, I would like to be able to work efficiently with linear algebra, number theory and random sampling. Oh, and plotting would be nice too and how to embed formulas written in LaTeX into those plots. Any advice?
26625
mathematica for learning?
_Mathematica_ looks like an extremely powerful tool. I would love to have it in my toolbox. Unfortunately, I can hardly add two numbers together. I think the application of math through programming is the best way to learn. Is there any book that takes you from _no math skills_ to advanced levels through the lens of _Mathematica_?
5059
What Mathematica book to buy?
I have used Mathematica for several years but at a pretty low level - piecing together built-in function inefficiently and fearing the sight of # and &'s when I see others use them (I never do). I would like to improve my skills. Which book would be best to read for someone familiar with Mathematica basics but would like to learn more sophisticated uses of Mathematica?
11530
Good Introduction / Tutorial to Mathematica **Books**
> **Possible Duplicate:** > Where can I find examples of good Mathematica programming practice? I'm a high school senior and am using Mathematica for a research project. However, I find it a bit slow to learn Mathematica by watching videos. I know that there are many helpful videos of Mathematica topics online, but I was wondering if there are any books or reference manuals I can read offline. I've looked at the question at What is the best Mathematica tutorial for young people? But again, it only lists online resources. So are there are any books out there detailing an introduction to Mathematica?
51990
Manipulate Tutorials?
I am wanting to combine equations with sets of static data points in a manipulate and do the curve fitting by moving the slider bars on my independent variables. The best examples of this art may be found at: http://demonstrations.wolfram.com/search.html?query=Normand Could someone please point me to one or more detailed tutorials that would make me as good a Mathematica programmer as Mark D. Normand?
33500
Plotting a function that uses the solutions to a system of differential equations
I am trying to draw a plot by using the solution of differential equations z = 0; Ω = 2.2758; τ = 13.8; T2 = 200; ω0 = 1; r = 0.7071; Δ = 1.7758; ΩR = 2.2758; ω == 0.5; system = {u'[t] == Ω*v[t], v'[t] == Ω*u[t] - 2*ΩR*w[t] - v[t]/T2, w'[t] == 2*ΩR*v[t]}; initialvalues = {u[0] == 0, v[0] == 0, w[0] == -1}; sol = DSolve[Join[system, initialvalues], {u[t], v[t], w[t]}, t] FT = (-2*2.2758*r*u[t])/ω0^2 E^(-(r^2/ω0^2) - (t^2/τ^2))* Cos[kz - ωt]; p1 = Plot[FT, {t, -20, 20}, FrameLabel -> {"t", "FT"}, Frame -> True, RotateLabel -> True, PlotRange -> {-0.1, 0}, PlotStyle -> {Thickness[0.002]}] ![my code](http://i.stack.imgur.com/9kWLE.jpg) right now I went to plot "FT" by using the result of "u" , How I can do that ?
18704
how to print stack trace when TimeConstrained times out
According to the documentation, `TimeConstrained` generates an interrupt which interrupts the computation. This interrupt is treated just like an abort, at least in the sense that it respects `AbortProtect`. I want to see the contents of the stack at the time the abort is generated, before the aborted evaluation is removed from the stack. I tried this: changeAbort[] := ( Unprotect[Abort]; testAbort = True; Abort[args___] := ( Print[Stack[_]]; (* in real life, to a log file *) Block[{testAbort = False}, Abort[args]] ) /; testAbort Protect[Abort]; ) This works for aborts that I generate myself. For example, While[True, Abort[]] aborts and prints the following stack trace: (* {While[True,Abort[]],Print[Stack[_]];Block[{testAbort=False},Abort[]],Print[Stack[_]]} *) Along (spiritually) similar lines, I can print the the stack at the time a message is displayed with this code: Internal`AddHandler["Message", Print[Stack[_]]&]; Getting back to `TimeConstrained`, neither of the following give me a stack trace: changeAbort[];TimeConstrained[While[True, Null], 1] TimeConstrained[changeAbort[]; While[True, Null], 1] The last line was written in case `TimeConstrained` was implemented by temporarily redefining `Abort`. Unfortunately, it doesn't work. Is there some way to intercept the interrupt generated by `TimeConstrained` so that I can get a stack trace of the aborted computation, before it is removed from the stack?
33858
Judging the quality of fitting an EventData object to a distribution using LogRankTest
I have $389$ data points defined in an `EventData` object and want to judge the quality of fit of this data to a fitted $3$-parameter Weibull distribution. This particular data set contains no censoring but in general there will be censoring in my future data sets. My approach is to find the three parameters using `EstimatedDistribution` and then to represent the found distribution by generating a number of random variates from it so that they can be compared to the original `EventData` object. I’m using LogRankTest to produce the $p$-value to judge the fit. My fundamental question is, is there a more elegant way of doing this in _Mathematica_? My data and code are below. Here are the raw data: data1=Table[40.8,{10}]; data2=Table[42.075,{23}]; data3=Table[43.35,{48}]; data4=Table[44.625,{80}]; data5=Table[45.9,{63}]; data6=Table[47.175,{65}]; data7=Table[48.45,{47}]; data8=Table[49.725,{33}]; data9=Table[51,{14}]; data10=Table[53.55,{6}]; data=Flatten[{data1,data2,data3,data4,data5,data6,data7,data8,data9,data10}]; Here is where I define the `EventData` object and determine the fitted distribution, distribution. censoring=Table[0,{Length[data]}]; eventdata=EventData[data,censoring] dist=EstimatedDistribution[eventdata,WeibullDistribution[a,b,c]] EventData[ ] WeibullDistribution[2.62045,7.14193,39.7656] And here is where I non-parametrically determine the $p$-values via `LogRankTest`. n=200000; i=0; Do[ {i=i+1, If[i>4,Break[]], samplevalues=RandomVariate[dist,{n}], logrankpvalue=LogRankTest[{eventdata,samplevalues}] , Print["i=",i," Log Rank p Value=",logrankpvalue]}, {5}] i=1 Log Rank p Value=0.999826 i=2 Log Rank p Value=0.973828 i=3 Log Rank p Value=0.948566 i=4 Log Rank p Value=0.976803 Note that with $n$ set at a relatively large value ($200\,000$), $p$-values of around $0.97$ are produced indicating that the hazard rates of the two data sets are in pretty good agreement. I think this tells me that `EstimatedDistribution` did a good job in the fitting. Conversely, in general, small values of $n$ (e.g., $10$) produce a much poorer agreement as expected and as evidenced by smaller $p$-values. I generate four $p$-values because I’ve observed that depending upon the particular random values chosen that it is possible to produce a large $p$-value with small values of $n$. My questions are 1. Is there a more elegant way of determining the degree of similarity between the `EventData` object and the found distribution? Perhaps comparing to the theoretical distribution, dist directly instead of sampling from dist? 2. If this approach (or any improved suggested approach) is robust for the _no-censoring_ case is it also robust if the `EventData` object contains _censored values_?
18709
Sinusoid modelling
There are lots of great demonstrations that show how to get periodic functions in the unit circle. I'm wondering how hard it would be to simulate a "moving grid" similar to this video on youtube (at about 1 minute) The circle and moving points are easy , but I don't know how to simulate that grid and sinusoid moving under the moving points. Any help would be appreciated. Here is my start, lots of possibilities to make this look better cosmetically, but the moving "paper" is something that would really help to illustrate the idea. Manipulate[Graphics[{ Circle[], PointSize[0.012], Point[{Cos[t], Sin[t]}], Point[{0, Sin[t]}] }, PlotRange -> {{-1.1, 11}, {-1.1, 1.1}}, ImageSize -> {500, 100}], {t, 0, 10}]
18708
How can I deal with an NDSolve::ivone message regarding boundary values?
I have a question regarding the error message: > > NDSolve::ivone: Boundary values may only be specified for one > independent variable. > Initial values may only be specified at one value of the other > independent variable. >> > I'm curious as to when this error shows up. If I run the heat equation example from the `NDSolve` documentation, I get a valid solution. However, consider an edit to the p.d.e to NDSolve[{D[u[t, x], {t, 2}] == D[u[t, x], {x, 2}], u[0, x] == 0, u[t, 0] == 0, u[t, 5] == 0}, u, {t, 0, 10}, {x, 0, 5}] That is, I changed $u_t = u_{xx}$ to $u_{tt} = u_{xx}$ and also changed an initial condition. (A solution of course is $u(x, t) = 0$.) Now, if I run this, I will get the error quoted above. I believe `NDSolve` doesn't even try to run. How do I need to edit the boundary conditions in order to get a run attempt? Does `NDSolve` only accept certain types of boundary conditions dependent on the order of the p.d.e? How do I find out what boundary conditions are acceptable? I'm asking about this because I'm trying to solve a very complicated third- order, nonlinear, p.d.e resulting from an optic equation and every set of initial conditions that I have given `NDSolve` for this 3rd-order equation has resulted in the above error.
33509
Curve Fitting Errors
I am trying to fit a part of a dataset to the function $$ae^{-b\sqrt{x-c}}$$. My friend was using GNUPlot to this and got reasonably good results. However when I try to this with _Mathematica_ : NonlinearModelFit[Tmean, a*Exp[-b*Sqrt[x - c]], {{a, 45}, {b, 2}, {c, 1.8}}, x]). > > -188.006 E^{1.17551\sqrt{-<<18>> + x}} > > > Data= {{2.76134, 6.134}, {2.7552, 6.081}, {2.7491, 6.224}, {2.74301, 6.288}, > {2.73696, 6.629}, {2.73093, 6.486}, {2.72493, 6.719}, {2.71895, 6.989}, > {2.713, 7.076}, {2.70708, 7.293}, {2.70118, 7.152}, {2.69531, 7.37}, > {2.68946, 7.626}, {2.68364, 7.59}, {2.67784, 7.74}, {2.67207, 7.723}, > {2.66633, 7.606}, {2.66061, 7.855}, {2.65491, 7.668}, {2.64923, 8.03}, > {2.64359, 8.16}, {2.63796, 7.932}, {2.63236, 8.174}, {2.62678, 8.335}, > {2.62123, 8.171}, {2.6157, 8.292}, {2.61019, 8.247}, {2.60471, 8.363}, > {2.59925, 8.541}, {2.59381, 8.357}, {2.5884, 8.347}, {2.583, 8.404}, > {2.57763, 8.414}, {2.57229, 8.363}, {2.56696, 8.509}, {2.56166, 8.556}, > {2.55638, 8.297}, {2.55112, 8.562}, {2.54588, 8.335}, {2.54066, 8.359}, > {2.53546, 8.581}, {2.53029, 8.311}, {2.52514, 8.308}, {2.52, 8.276}, > {2.51489, 8.337}, {2.5098, 8.075}, {2.50473, 8.098}, {2.49968, 8.119}, > {2.49465, 7.891}, {2.48964, 7.682}, {2.48465, 7.519}, {2.47968, 7.593}, > {2.47473, 7.482}, {2.4698, 7.567}, {2.46489, 7.053}, {2.46, 7.213}, > {2.45513, 7.077}, {2.45028, 6.784}, {2.44545, 6.81}, {2.44063, 6.796}, > {2.43584, 6.643}, {2.43106, 6.633}, {2.42631, 6.323}, {2.42157, 5.994}, > {2.41685, 6.198}, {2.41214, 6.063}, {2.40746, 5.797}, {2.40279, 5.808}, > {2.39815, 5.684}, {2.39352, 5.506}, {2.38891, 5.307}, {2.38431, 5.531}, > {2.37974, 5.207}, {2.37518, 5.193}, {2.37063, 5.247}, {2.36611, 4.985}, > {2.3616, 4.882}, {2.35711, 4.696}, {2.35264, 4.501}, {2.34819, 3.966}, > {2.34375, 4.399}, {2.33932, 4.297}, {2.33492, 4.154}, {2.33053, 4.011}, > {2.32616, 3.853}, {2.3218, 3.654}, {2.31746, 3.521}, {2.31314, 3.505}, > {2.30883, 3.519}, {2.30454, 3.627}, {2.30026, 3.49}, {2.296, 3.412}, > {2.29176, 3.624}, {2.28753, 3.43}, {2.28332, 3.345}, {2.27912, 3.503}, > {2.27494, 3.2}, {2.27077, 3.169}, {2.26662, 3.015}, {2.26249, 2.968}, > {2.25836, 2.76}, {2.25426, 2.523}, {2.25017, 2.624}, {2.24609, 2.567}, > {2.24203, 2.654}, {2.23798, 2.525}, {2.23395, 2.413}, {2.22993, 2.279}, > {2.22593, 2.262}, {2.22194, 2.47}, {2.21796, 2.288}, {2.214, 2.324}, > {2.21006, 2.204}, {2.20612, 2.224}, {2.20221, 2.316}, {2.1983, 2.183}, > {2.19441, 2.323}, {2.19053, 2.085}, {2.18667, 1.981}, {2.18282, 2.047}, > {2.17898, 2.096}, {2.17516, 2.062}, {2.17135, 2.091}, {2.16756, 2.057}, > {2.16377, 2.06}, {2.16, 2.048}, {2.15625, 2.231}, {2.1525, 1.956}, {2.14877, > 2.066}, {2.14506, 2.082}, {2.14135, 2.019}, {2.13766, 2.184}, {2.13398, > 2.043}, {2.13031, 1.94}, {2.12666, 1.985}, {2.12302, 2.172}, {2.11939, > 2.175}, {2.11577, 2.161}, {2.11217, 2.184}, {2.10857, 2.082}, {2.10499, > 1.994}, {2.10143, 2.081}, {2.09787, 2.066}, {2.09433, 2.239}, {2.0908, > 2.167}, {2.08728, 2.192}, {2.08377, 2.207}, {2.08027, 2.3}, {2.07679, > 2.342}, {2.07331, 2.346}, {2.06985, 2.37}, {2.0664, 2.391}, {2.06297, > 2.436}, {2.05954, 2.321}, {2.05612, 2.433}, {2.05272, 2.528}, {2.04933, > 2.714}, {2.04594, 2.569}, {2.04257, 2.682}, {2.03921, 2.621}, {2.03587, > 2.899}, {2.03253, 3.089}, {2.0292, 3.136}, {2.02589, 3.259}, {2.02258, > 3.158}, {2.01929, 3.147}, {2.016, 3.295}, {2.01273, 3.279}, {2.00947, > 3.294}, {2.00622, 3.47}, {2.00298, 3.481}, {1.99975, 3.544}, {1.99652, > 3.882}, {1.99332, 4.078}, {1.99012, 3.939}, {1.98693, 4.153}, {1.98375, > 4.395}, {1.98058, 4.377}, {1.97742, 4.471}, {1.97427, 5.051}, {1.97113, > 5.072}, {1.968, 5.454}, {1.96488, 5.711}, {1.96178, 6.012}, {1.95868, > 6.169}, {1.95559, 6.691}, {1.95251, 7.021}, {1.94944, 7.428}, {1.94638, > 7.902}, {1.94333, 8.348}, {1.94028, 9.005}, {1.93725, 9.836}, {1.93423, > 10.504}, {1.93122, 11.491}, {1.92821, 12.563}, {1.92522, 13.621}, {1.92224, > 14.784}, {1.91926, 16.286}, {1.91629, 17.614}, {1.91334, 19.183}, {1.91039, > 20.854}} when `a` should be around $50$, `b` around $5$-$7$ and `c` around $1.88$. We both have the same initial guesses. What does the `<< >>` mean and why am I getting such bad results? Data: _Mathematica_ : EDIT: Added Data
41564
For-loop not working as I expect
I want to do this: * When `10 <= t < 130`, give `t -> 1`; * When `130 <= t<300`, give `t->3`; * When `300 <= t< 1000`, give `t -> 4`; * When `1000 <= t < 2000`, give `t -> 5`; My code is this: data = {10, 130, 300, 1000, 2000}; fun[t_] := Part[data, t]; comf[s_] := For[t = 1, t <= 4, t++, If[fun[t] <= s < fun[t+1], Print[t]]]; I could get this: comf[110] > > 1 > comf[1300] > > 4 > But when I evaluate this comf[110] + 100 I didn’t get the right result, rather I got this: > > 100 + Null > I would like to know how I could fix this problem.
41561
Integration and plot of an interpolating function
FrenetFucntion = { t'[s] == s* n[s], n'[s] == -s* t[s] + 0.3* b[s], b'[s] == -0.3* n[s], t[0] == {{1, 0, 0}}, n[0] == {{0, 1, 0}}, b[0] == {{0, 0, 1}} }; sol = NDSolve[FrenetFucntion, {t, n, b}, {s, 0, 50}]; How to integrate the solution of `NDSolve`. Also is there any way of plotting the integration function of interpolating polynomial? I used `NIntegrateInterpolatingPolynomial` but then it will give me definite integral only because I have to specify the limits of s. Is there any way of defining s limits like from 0 to s so that I may plot the result of integration as well.
24360
Plotting derivatives of B-splines
I am trying to solve an electrodynamics problem numerically by using B-splines a basis function. Please don't ask me why, but if you really wanna know, here is the assignment. Using the following code, I have successfully plotted the B-spline basis functions in two dimensions, given the specific knot sequence and the respective x and y intervals. knots = {0, 0, 0, 0, 0.2, 1, 1, 1, 1} Plot3D[Evaluate[Table[BSplineBasis[{4, knots}, i, x], {i, 0, 3}]] Evaluate[ Table[BSplineBasis[{4, knots}, i, y], {i, 0, 3}]], {x, 0, 1}, {y, 0, 1}] The corresponding plot looks like this: ![enter image description here](http://i.stack.imgur.com/Kwtb5.png) Now I am trying to do the same for the second derivatives of the B-splines, using the following code: knots = {0, 0, 0, 0, 0.2, 1, 1, 1, 1} Plot3D[Evaluate[Table[D[BSplineBasis[{4, knots}, i, x], {x, 2}], {i, 0, 3}]] Evaluate[Table[D[BSplineBasis[{4, knots}, i, y], {y, 2}], {i, 0, 3}]], {x, 0, 1}, {y, 0, 1}] However, it does not work and produces an empty graph and lots of red error messages. I have tried several combinations of the D, Table and Evaluate operators, but nothing seems to work. What am I doing wrong?
41563
Extracting angles from a simple picture
I'm not sure if this is even possible with _Mathematica_ , but I want to extract the angles between all lines in this picture: ![enter image description here](http://i.stack.imgur.com/OiTnw.png) It consists of $12$ lines, so I want a matrix with $144$ entries. I really have no idea even how to begin. Any ideas?
17828
Problem with inequality using refine
I am using inequalities to follow some line of reasoning in Mathematica and I came across a problem I can not explain to myself. Does from a>=0 and b>=0 not follow a+b >= 0? How does Mathematica evaluate these statements? Thanks! In[225]:= Refine[(A0 c0 (c - c1) >= 0), A0 >= 0 && A >= A0 && A1 >= A && c1 >= 0 && c > c1 && c0 >= c] Out[225]= True In[226]:= Refine[(A (c0 - c1) c1 >= 0), A0 >= 0 && A >= A0 && A1 >= A && c1 >= 0 && c > c1 && c0 >= c] Out[226]= True In[227]:= Refine[(A0 c0 (c - c1) + A (c0 - c1) c1 >= 0), A0 >= 0 && A >= A0 && A1 >= A && c1 >= 0 && c > c1 && c0 >= c] Out[227]= A0 c0 (c - c1) + A (c0 - c1) c1 >= 0
45028
I am having a Plot and i need inside a Do loop to export 100 frames (changing the bounds of the plot)
for some reason i was thinking of doing this Do[ Plot[F[x],{x,-i,i}]; Export["name_0" + i + ".jpg", %, "JPEG", ImageSize->{250,250}], {i,5}] but i get an error Any ideas or help is welcome ps. i am a terribly new at this :-) Regards, J.
37143
Exporting a Table of Graphics into Individual Files
The command `Table[Polt[Sin[k x],{x,1,10}],{k,1,30}]` generates a list of graphics. I want to export each list item as an individual jpg file, named according to the order of appearance in the list. How can I do this? Thanks in advance.
50937
how to export multiple graphics from Do loop?
I would like to export graphics generated from do-loop as jpg. There will be like more than 100 graphs. I want to combine it later in latex beamer as animation. When I put Export command inside the loop, I could not manage to change the name of the files since the name is inside "". Is there any way to solve this problem? Thank you very much.
17825
Simple Animate[] question
I am trying to make a visualization of random process: coin flip. What I want to visualize is the result of i-th try: 1 or 0 value (the dot) and the mean (line) at i-th step. My code is as follows: m = {}; c = {}; k = {}; Animate[ AppendTo[c, RandomInteger[]]; AppendTo[m, Mean[c]]; AppendTo[k, i]; Show[ ListPlot[Transpose[{k, c}]], ListPlot[Transpose[{k, m}], Joined -> True, PlotRange -> {{1, 100}, {0, 1}}, AxesOrigin -> {1, 0}] ] , {i, 1, 100, 1}, AnimationRepetitions -> 1,AnimationRunning -> False ] What I can't figure out is: Why at the very beginnig I get such a plot: ![I get such a plot:](http://i.stack.imgur.com/MJzVD.png) The line is not a mean value as should be expected. The second thing: Why there are two points at the moment, despite the fact that I have AnimationRunning->False? And the third: why k list is filled with ones only? Shouldn't it contain 1 to 100 values?
19843
Dynamically splitting Disks with Mouseover
I've recently stumbled across this site: Koalas to the Max, and the first thought that came to my mind was "I want to recreate this with Mathematica". As a first step I tried to create a `Disk` that splits itself up into four disks, when the mouse cursor touches it. Those four disks should in turn split themselves up. Here's what I have so far: makeDisks[pos_] := makeDisks[1, pos] makeDisks[level_, pos_] := Module[{ disk = Disk[pos, 1/level]}, Mouseover[Dynamic@disk, Dynamic[disk = makeDisks[level + 1, #] & /@ (pos + # & /@ (1/(2 level)) Tuples[{-1, 1}, 2])]] ] (* test with the following *) Graphics[{makeDisks[1]}, PlotRange -> {{-1, 1}, {-1, 1}}] For the first disk this seems to work just as I want it. For later steps, however, I got the positioning wrong and some of the disks go back to larger ones, when another one is touched. How can these issues be fixed? As a next step I want to add colors to the disks. Maybe this can be done by using `MapIndexed` with `makeDisks` on the `ImageData` of some image. The colors in each step should probably be the mean values of appropriate subsets of the image data. Any help to move this little program forward is highly appreciated!
49246
Can I change Legend text in a stylesheet?
I prefer a sans serif font in graphics, which I do by creating a style for "Graphics" in a stylesheet. However, this does not affect the style used by legend labels: Plot[{x, 2x}, {x, 0, 1}, PlotLegends -> {"Times", "New Roman"}] ![enter image description here](http://i.stack.imgur.com/JT9NB.png) Is there any way to change the legend text style in the stylesheet, or can it only be done through the `LabelStyle` option?
4643
How to use Mathematica functions in Python programs?
I'd like to know how can I call Mathematica functions from Python. I appreciate a example, for example, using the Mathematica function **Prime**. I had search about MathLink but how to use it in Python is a little obscure to me. I tried to use a Mathematica-Python library called _pyml_ but I hadn't no sucess, maybe because this lib looks very old (in tutorial says Mathematica 2 or 3). So, someone knows a good way to write python programs who uses Mathematica functions and can give me an example? **Old Edit:** _Maybe this edit can help someone who wants to use mathlinks directly._ _To another solution, please see the answer accepted._ Using the _Wolfram/Mathematica/8.0/SystemFiles/Links/Python_ I could had sucess in compiling the module changing some things in setup.py. _My architechture is`x86-64`._ 1-Change the `mathematicaversion` to `8.0`. 2-Changing the lib name `ML32i3` to `ML64i3`. 3-Copying the file `Wolfram/Mathematica/8.0/SystemFiles/Libraries/Linux-x86-64/libML64i3.so` to the path pointed in setup.py `library_dirs = ["/usr/local/Wolfram/Mathematica/" + mathematicaversion + "/SystemFiles/Links/MathLink/DeveloperKit/Linux/CompilerAdditions"]`. 5-Compiling the source with `sudo python setup.py build`. 6-Installing the lib with `sudo python setup.py install` 4-Editing the file `/etc/ld.so.conf` and putting the line ` include /usr/local/lib`. 5-Creating a directory in `/usr/local/lib/python2.6/dist-packages/mathlink` with the lib `libML64i3.so`. 6-Running `sudo ldconfig` I had tested the scripts `guifrontend.py` with `python guifrontend.py -linkname "math -mathlink" -linkmode launch` and `textfrontend.py` with `python textfrontend.py -linkname "math -mathlink" -linkmode launch` and worked fine. Looks like I almost. But the script >>> from mathlink import * >>> import exceptions,sys, re, os >>> from types import ListType >>> mathematicaversion = "8.0" >>> os.environ["PATH"] = "/usr/local/Wolfram/Mathematica/" + mathematicaversion + ":/usr/local/bin:/usr/bin:/bin" >>> e = env() >>> sys.argv=['textfrontend.py', '-linkname', 'math -mathlink', '-linkmode', 'launch'] >>> kernel = e.openargv(sys.argv) >>> kernel.connect() >>> kernel.ready() 0 >>> kernel.putfunction("Prime",1) >>> kernel.putinteger(10) >>> kernel.flush() >>> kernel.ready() 0 >>> kernel.nextpacket() 8 >>> packetdescriptiondictionary[3] 'ReturnPacket' >>> kernel.getinteger() Traceback (most recent call last): File "<stdin>", line 1, in <module> mathlink.error: MLGet out of sequence. breaks in the last command and I don't know why. How can I fix this?
35569
calling mathematica code in Python
I am wondering if there is any working example of calling mathematica 9 outputs in python? or plotting mathematica output in python?
20259
How to change the density range in DensityPlot?
This is an example from Mathematic Help, DensityPlot[Sin[x] Sin[y], {x, -4, 4}, {y, -3, 3}, ColorFunction -> "SunsetColors", PlotLegends -> Automatic] We can notice that the density values are from -1.0 to 1.0. However, I want to change this automatic density range to other values, for example, 0 to 0.5. Of course, the density plot should be different. This should be simple, however, I have no idea about it. Can you tell me how to change it? Thanks!
20252
Is it possible to select outliers on a graph and determine the index of the outliers?
I have a large data set of experimental data for which I have determined a theoretical fit. However, for some of the experimental data I took, there was systematic error, leading to the data deviating from the expected fit as shown below (experimental values are plotted on x-axis and theoretical values are plotted on y-axis). ![graph of data and outliers](http://i.stack.imgur.com/4K3Wo.jpg) Is it possible to write some code to allow one to simply click on the outliers and determine their index, or is there any way to do so natively in Mathematica? It would also be possible for me to divide the values plotted on the y-axis by those plotted on the x-axis, but I'm wondering whether there's a general method allowing you to click on to identify outliers on any graph. My code for the graph is as follows (I'm only plotting the first 500 values): Show[ListPlot[Prepend[Transpose[{experimentalCR[[1 ;; 500]], theoreticalCR[[1 ;; 500]]}], {0, 0}]], ImageSize -> {500, 500}, AspectRatio -> Full] A sample of the data can be found here: http://pastebin.com/mA6VxPfw
31184
Trouble carrying out symbolic operations on matrixes
I have some problems. First of all, I declare matrices like below $Assumptions = { Element[X, Matrices[{m, n}]], Element[T, Matrices[{n, k}]], Element[Y, Matrices[{m, k}]]}; and I want to apply transpose(X T - Y).(X T - Y) = transpose[T].transpose[X].X.T - transpose[T].transpose[X].Y - transpose[Y].X.T + transpose[Y].Y so I tried 1. `Inner[Times, {X \[Transpose]}, {T \[Transpose]}, Plus] - \[Transpose][Y]` 2. `Inner[NonCommutativeMultiply, {X}, {T}] - {Y} = {-Y + X ** T} ...` But I can not multiply the two above expressions. Inner[NonCommutativeMultiply, Inner[Times, {X\[Transpose]}, {T\[Transpose]}, Plus] - Transpose[Y], Inner[NonCommutativeMultiply, {X}, {T}] - {Y}] because I get errors. Could you help me?
27576
Implementing new algebra for list-like objects
I would like to implement an algebra (addition, multiplication, etc.) for what is called "truncated power series" (TPSA) (Differential Algebraic description of beam dynamics). . These 'objects' (sets of complex numbers) have special rules for addition and multiplication. Of course the easy way to go is just to use plain lists and redefine functions like myAdd[], myMult[], etc. acting on these lists. However in most implementations in other languages these objects are represented as real objects (OO) of a 'TPSA' class, and operator overloading is used to manipulate them. My question is: I know the last solution is not the way to go with Mathematica, but the first easy way doesn't seem the best one either. How would you advice to implement that with Mathematica ? **Edit:** The root of my question was apparently "what is an up-value", the answer below about dual number also points in that direction.
20257
How to change the output format of a built-in function
Is it possible to change the output of build-in functions? I would like to change the output for IdentityMatrix[n] to "I", so I tried Format[IdentityMatrix[n]] := "I" For comparison I also tried Format[Sin[x_]] := "Sine(" <> ToString[x] <> ")" Both result into an error: "Tag IdentityMatrix in IdentityMatrix[n] is Protected." I guess these functions are protected because they are build-in functions. Is there a way to change the output format of a build-in function?
27575
DictionaryLookup very slow
I'm trying to plot, given a list of words, how many words belongs to each language. I'm using the following code: Total /@ Transpose@ ParallelMap[ Function[word, Length@DictionaryLookup[{#, word}] & /@ DictionaryLookup[All]] , wordList] I am finding it way too slow, the lengths of the lists of words which I am using are about 45000 (sample file: https://dl.dropboxusercontent.com/u/35192406/words.txt) Is DictionaryLookup really slow? Does it use Internet? How can I speed up my function?
44427
Why there is no SlotSequence (##) form analogical to #[[ ;; n]]
I'm aware that this may be closed as offtopic but there are people around with "offtopic" knowledge :) So, in `List` context we can: `#[[n ;; ]]` and `#[[ ;; n]]`, but we can't do this in context of `Sequence`. Only `##n` is available. One may do f[n_] := Unevaluated[##][[;; n]] & f[2][1, 2, 3, 4] > > Sequence[1, 2] > but it is not what I'm looking for. It is very often when I would like to use it. Of course thare are no problems with workarounds. I'm just curious. p.s. Syntax is an issue for sure, I thought something like `###2` could stand for this but `###` is automatically interpreted as `##1 #1`.
40743
Select repeating cluster of numbers
I have a list with measured data. There are multiple measurements in the list, but each measurement requires some time to find its equilibrium. I'd like to know the positions of the equilibria, but because of errors (during the measurement and in the behaviour of the measured quantity), I'd like to get five consecutive values from each equilibrium, so I can take the average (or at least compare them). Here is the graph of the data: ![picture of data](http://i.stack.imgur.com/CVmja.png) The useful data can be extracted at the positions used by data[[450;; ;; 500]] I know the following will work Tranpose@Table[data[[445+i;; ;;500]],{i,0,10}] But I expected the following to work (which it does not sadly) data[[Range[445,455];; ;;500]] Can this be done without using `Table` (or similar constructs)?
40740
Law of Cosines to determine angle change along spiral
# The Spiral I'm trying to place circles along the outside of an Archimedean spiral. Shown below: ![Archimedian Spiral from 0 to 6 Pi](http://i.stack.imgur.com/nbgyo.png) ParametricPlot[{θ/(2*π)*Cos[θ], θ/(2*π)*Sin[θ]}, {θ, 0, 6π}, PlotStyle -> Red] # The Circles ![Circle placement](http://i.stack.imgur.com/k60VS.png) To do this effectively I have been trying to use the Law of Cosines to determine the change in the angle. I found the path that would contain the circle centers: f[θ_] := {((π+θ) Cos[θ])/(2π), ((π+θ) Sin[θ])/(2π)} Using the Law of Cosines for the following: ![Law of Cosines for this example](http://i.stack.imgur.com/bTZ7J.png) Seems pretty straight forward, right? Well, when I went to solve for the angle alpha Mathematica seems to have trouble. ![code](http://i.stack.imgur.com/pUoAQ.png) seems to work "fine", but any subsequent evaluations don't seem to go smoothly. In fact, any evaluation that does not contain Pi seems to fail, and an evaluation of the distance between the points shows that the angle is off (!=1). ![Distance](http://i.stack.imgur.com/hK9FI.png) # The Question What am I doing wrong to be more than 5% off from what I expect. Also, is there a more efficient and correct method to solve for the angle? # An End Product Thanks to KennyColnago for his response. ![Final Product](http://i.stack.imgur.com/YVpvo.png)
48029
Question about plotting
I have to plot this graph http://i.imgur.com/hf54aiT.jpg I tried `Graphics[{Circle[{0, 0}, 3, {-2Pi, 2Pi}]},PointSize[.025]]` I don't know how to proceed from here. Any help would be appreciated!
18393
What are the most common pitfalls awaiting new users?
As you may already know, _Mathematica_ is a wonderful piece of software. However, it has a few characteristics that tend to confuse new (and sometimes not-so-new) users. That can be clearly seen from the the fact that the same questions keep being posted at this site over and over. Please help me to identify and explain those pitfalls. Suggestions for posting answers: * One topic per answer * Focus on non-advanced uses (it's intended to be useful for beginners and as a question closing reference) * Include a self explanatory title in h2 style * 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're in the mood) * Include a link to your answer by editing the **Index** below (for quick reference) ## Index * ### Basic syntax issues * ### Avoiding procedural loops * ### Understand what `Set` (=) really does * ### Understand the difference between exact and approximate (Real) numbers * ### Understand the difference between `Set` (or `=`) and `SetDelayed` (or `:=`) * ### Lingering Definitions: when calculations go bad * ### Learn how to use the Documentation Center effectively * ### Understand that semicolon (;) is not a delimiter * ### Using the result of functions that return replacement rules * ### Attempting to make an assignment to the argument of a function * ### Use Consistent Naming Conventions * ### Assuming commands will have side effects which they don't: * ### Undo is not available in general * ### What the @#%^&*?! do all those funny signs mean? * ### User-defined functions, numerical approximation, and NumericQ * ### The displayed form may substantially differ from the internal form * ### Why is my picture upside-down? * ### Mathematica can be much more than a scratchpad * ### Mathematica's own programming model: functions and expressions
29020
How to define a function based on the output of DSolve?
I would like to define a function based on the output of the following DSolve call: `DSolve[{b Varx^2 - 2 b Varx x'[t] + b x'[t]^2 + m x''[t] == 0, x[0] == x0, x'[0] == V0x}, x[t], t]` I've tried to do the following: `x[t_] := DSolve[{b Varx^2 - 2 b Varx x'[t] + b x'[t]^2 + m x''[t] == 0, x[0] == x0, x'[0] == V0x}, x[t], t]` It showed the message: `$RecursionLimit::reclim: Recursion depth of 256 exceeded.`, so I believe I'm not doing it the right. Any ideas?
47516
Find minimum value of function which calls NDsolve
I have defined a function which takes a coefficient for a differential equation as an argument, solves the differential equation with NDSolve, and returns the value of the solution at time t=10. I can plot this, but I can't seem to find the minimum value. This code demonstrates the problem i'm having: f[k_] := y[10] /. NDSolve[ {y''[t] == -k*y[t], y[1] == 2 , y'[2] == 1}, y, {t, 0, 10} ]; Plot[f[k], {k, 0, 2}] (* works great *) NMinimize[{f[k], k > 0, k < .75}, k] (* fails *) > The NMinimize fails with the error NDSolve::ndnum: Encountered non-numerical > value for a derivative at t == 1.'. So how do I find the minimum of my function f?
27634
Converting the output for variable assignments to a usable array form
Output from built-in _Mathematica_ functions often has this form (e.g. from `Solve`): > {1 -> {166., 254.214}, 2 -> {237.214, 254.}, 3 -> {121., 253.5}, 4 -> > {221.318, 251.682}, 5 -> {45., 246.}, 6 -> {125., 245.214}, 7 -> {166.5, > 244.7},...} How would I efficiently (without converting back and forth from a string) convert this to array of the form: > {{166., 254.214}, {237.214, 254.}, {121., 253.5}, {221.318, 251.682}, {45., > 246.}, {125., 245.214}, {166.5, 244.7},...}
59195
How do I assign a formula to a variab
I want to compute equations and assign them to variable. This is a wxMaxima program that calculated symbolic PID gains for an integrating first order system remvalue(all)$ /* Undefine */ Gp: K*%alpha/(s*(s+%alpha)); /* System transfer function */ Gc: Ki/s+Kp+Kd*s; /* Simple PID controller */ CLTF: ratsimp(Gc*Gp/(1+Gc*Gp)); /* Closed loop transfer function */ CE: expand(denom(ratsimp(CLTF))); /* Characteristic equation */ DIFCE: expand(CE-(s+%lambda)^3); Ki: solve(coeff(DIFCE,s,0),Ki); Kp: solve(coeff(DIFCE,s,1),Kp); Kd: solve(coeff(DIFCE,s,2),Kd); grind(Ki,Kp,Kd); The grind outputs text that I can paste here [Ki = %lambda^3/(%alpha*K)]$ [Kp = 3*%lambda^2/(%alpha*K)]$ [Kd = (3*%lambda-%alpha)/(%alpha*K)]$ Notice that Ki, Kp and Kd represent equations. This is what I want because I can change the values and compute new gains. This is my Mathematica equivelent ClearAll["Global`*"]; Subscript[G, p][s] = (K*\[Alpha])/(s*(s + \[Alpha])) Subscript[G, c][s] = Ki/s + Kp + Kd*s T[s] = Subscript[G, c][s]* Subscript[G, p][s]/(1 + Subscript[G, c][s]*Subscript[G, p][s]) // Simplify Solve[{CoefficientList[Denominator[T[s]], s] == CoefficientList[(s + \[Lambda])^3, s]}, {Ki, Kp, Kd}] This looks ugly because I haven't found the Mathematica version of grind() yet. Notice that the results are in the form Ki->Lamba^3/(K*Alpha). The -> is not an assignment so I can't treat Ki as a formula. 3 2 \[Lambda] 3 \[Lambda] -\[Alpha] + 3 \[Lambda] {{Ki -> ---, Kp -> ----, Kd -> --------}} K \[Alpha] K \[Alpha] K \[Alpha] How do I assign Lambda^3/(K*Alpha) using equations instead of by hand? I give up. formatting what get much better. Edit Solve[{CoefficientList[Denominator[T[s]], s] == CoefficientList[(s + [Lambda])^3, s]}, {Ki, Kp, Kd}] /. Rule-> Equal Does not work. When I type Ki for I get Ki in return and not the equal for Ki. This looks like it works s = Solve[{CoefficientList[Denominator[T[s]], s] == CoefficientList[(s + [Lambda])^3, s]}, {Ki, Kp, Kd}] {{Ki, Kp, Kd}} = { Ki, Kp, Kd } /. s
45223
Storing solution to x and y
If I do sol=Solve[{2x+2==y,3x+9==y},{x,y}] then Mathematica will tell me {{x -> -7, y -> -12}} But I would like to store values to x and y even if I don't write the following thing explicitly. x = -7 y = -12 At least I know this is possible, because I found how to do it by myself. But now I forgot. But I am sure it is possible.
17767
How to modify function argument?
Usually in programming languages, function arguments are normal local variables, which can be modified. Is this not true in Mathematica? In[94]:= TTT[x_] := Block[{}, x += 2 ]; In[95]:= TTT[3] > During evaluation of In[95]:= AddTo::rvalue: 3 is not a variable with a > value, so its value cannot be changed. >> Out[95]= 3 += 2 Is it possible to simulate normal argument behavior somehow?
43658
How to get a number from Solve
It's often useful to get just one number from `Solve`. I use the following construction for acquiring one solution from the expression returned by `Solve`. q = t /. Solve[p == 2 t + 1, t][[1]] Mostly I don't like to use the different variable name within `Solve`. And the rule-applying seems very unnatural. I think a solution that is more elegant must exist. Can anyone show me this?
14803
Assign the result of Solve in a list
> **Possible Duplicate:** > Assign the results from a Solve to variable(s) I want to assign the results of a Solve command in a list. For example Solve[-((287 s^3)/50) == 0, s] returns {{s -> 0}, {s -> 0}, {s -> 0}} I want it to return {0,0,0}. Can this be done automatically and not manually? Thank you very much
23153
Error in extracting a column
I imported excel data to Mathematica using this command. kuruzD = Import[ "C:\\Users\\TMH\\Documents\\Problem Set 4\\kuruzData.xlsx"][[1]] //TableForm I want to extract the 5th column. So, I used the command, kuruzD[[All, 5]] but, it results the data in 5th row? Could anyone please tell me what is my mistake?
48054
Table of values for functions with random variables
I'm trying to get a table of values for functions with random variables. First of all, i have a function like that: t = 100.0; LimMatrix3x3[a11_, a12_, a13_, a21_, a22_, a23_] := SetPrecision[ TableForm[{ MatrixForm[ MatrixPower[{ {a11, a12, a13}, {a21, a22, a23}, {1 - a11 - 21, 1 - a12 - a22, 1 - a13 - a23} }, t], TableAlignments -> Left]}, TableHeadings -> {{"m", "Value"}, Null}, TableSpacing -> {4, 2}], 3] And two matrices: MatrixIn := SetPrecision[ MatrixForm[{ {p11, p12, p13}, {p21, p22, p23}, {1 - p11 - p21, 1 - p12 - p22, 1 - p13 - p23} }, TableAlignments -> Left], 3]; MatrixOut := LimMatrix3x3[p11, p12, p13, p21, p22, p23]; where variables pij are: p11 = RandomReal[]; p12 = RandomReal[]; p13 = RandomReal[]; p21 = RandomReal[{0, 1 - 11}]; p22 = RandomReal[{0, 1 - p12}]; p23 = RandomReal[{0, 1 - p13}]; Now i want to get the table with values for matrices above and use this function: k = 5; TableForm[ Table[{MatrixIn , MatrixOut} , {, 0, k}], TableHeadings -> {Table[a, {a, 0, k}], {"Basic Matrix", "Matrix infinity"}}, TableSpacing -> {3, 2}, TableAlignments -> Center] As a result i get the table with 5 lines, which are identical. It happens because pij are evaluated only once. But i need to get random values for each pair of matrices. Is there a way to create from `MatrixIn`, `MatrixOut` functions? Or, i heard, it's possible to use `&` in some cases. Any other idea?
34470
How to manipulate lists to get only part of the value
This is really a question about manipulating the output of `Maximize`. I have a function of 2 variables, defined thus: Pqpcum[r_, b_] = Integrate[2 Pi*x*Pqp[x, b], {x, 0, r}, Assumptions -> {b > 0}] I want to find the point where the maximum of a certain function, defined below, occurs: Abs[Pqpcum[x1, b] - Pqpcum[x1, 0]], x1] The function `Maximize` returns an array, like this: In[24]:= Maximize[Abs[Pqpcum[x1, 0.1] - Pqpcum[x1, 0]], x1] Out[24]= {0.0204023, {x1 -> 1.77787}} Now I want to plot the location of the maximum, which is 1.77787 in this case, as a function of b. Attempting to get the second element of the array returns an element like this: {x1 -> 1.777869648006448`} The above cannot be plotted. I need to do something like this: Plot[Maximize[Abs[Pqpcum[x1, b] - Pqpcum[x1, 0]], x1][[2]],{b,0,0.2}] except that the above doesn't return anything, since I have plotted the list. How can I go about doing it? I could discretize the data and manually copy- paste the location of maximum values to do a `ListLinePlot`, but it would be a pain to do it. **EDIT:** I found my answer- I can use ArgMax instead. It directly returns the location of the maximum, so the above can be done. But the question still remains- how in general can one manipulate a list that reads like `{x1 -> 1.777869648006448}` to get only the numeric value `1.777869648006448`?
33546
Do loop - variable does not change
I got the follwing code which should add a '1' 100 times to an empty list. plotVals = {} Do[Append[plotVals, 1], {l, 1, Nmax}]; The output is again an empty list, why is this the case?
21377
Plot not displaying
I'm trying to perform a basic plot to show that two equations align, but I'm not getting any results. Plot[{[Pi]*Cos (3 x - [Pi]), [Pi] * Sin (3*[Pi] - 3 x)}, {x, -3 [Pi], 3 [Pi]}] To me this seems fairly straightforward, but I'm getting a graph without the functions.
20459
Mathematica, nested optmization
I want to use the solution of `Maximize`, defined as a function, in another function. Here's an example: f1[y_] := x /. Last[Maximize[{Sin[x y], Abs[x] <= y}, x]] (* or any other function *) This definition is fine, for example if I give `f1[4]`, I get the answer `-((3 \[Pi])/8)`. The problem is that when I want to use it in another function I get an error. For example: FindRoot[f1[y] == Pi/4, {y, 1}] Gives me the following error: > ReplaceAll::reps: {x} is neither a list of replacement rules nor a valid > dispatch table, and so cannot be used for replacing. >> > > FindRoot::nlnum: The function value {-0.785398+(x/.x)} is not a list of > numbers with dimensions {1} at {y} = {1.}. >> I've been struggling with this for several days now! Any comment, idea, help, ... is deeply appreciated!
27372
On reimplementing the Select function
I am working through the course _Programming Paradigms via Mathematica_. One of the exercises asks that you construct a function that mimics `Select`,even with three arguments. It also includes the following direction: > Do not use the `Select[]` function in your definition. Here is an example of how `Select` works: Select[{1, 2, 3, 4, 6, 8, 10}, EvenQ, 4] (* {2, 4, 6, 8} *) First I tested the following code and got the same result as `Select`: l1 = {1, 2, 3, 4, 6, 8, 10}; l2 = {}; crit = EvenQ; n = 4; While[Length[l1] > 0 && Length[l2] < n, If[crit[First[l1]], l2 = Append[l2, First[l1]]]; l1 = Rest[l1]; ] l2 (* {2, 4, 6, 8} *) I then tried to write a function that would do the same thing (caution causes infinite loop): Clear[whileSelect] whileSelect[l1_List, crit_, n_] := Module[{l2}, While[Length[l1] > 0 && Length[l2] < n, If[crit[First[l1]], l2 = Append[l2, First[l1]]]; l1 = Rest[l1]; ] l2 ] However I got a series of `Set::shape: Lists {1,2,3,4,6,8,10} and {2,3,4,6,8,10} are not the same shape. >>` error messages and the code put the computer into an infinite loop. I have looked at documentation on `Module`,`Rest`,`Set` and `If` but have not been able to figure out what I am doing wrong. I would greatly appreciate a point in the right direction.
333
Combined numerical minimization and maximization
I want to numerically calculate the maximum of a function defined by the minimization of another function, like the following: NMaximize[b/.NMinimize[(a^2+b)^2,{b}][[2]],{a}] Obviously the intended result (for this simple test function) would be: {0., {a->0.}} because the inner minimization would give b=-a^2 for any value of a, and maximizing that gives 0 for a=0. However I get the error message `NMinimize::nnum: "The function value (-0.829053+a^2)^2 is not a number at {b} = {-0.829053}."` and the result `NMaximize[b/.{b},{a}]`. I figured that this is due to premature evaluation of the argument (i.e. before `a` got anumerical value), therefore I tried to wrap either the whole first argument of just the `NMinimize` call in `Unevaluated`, but neither helped. So my question is: How can I do this combined numerical optimization?