wolfram-mathematica

问题 In any Mathematica chart or plot how can I show % values on the y axis? I may have data like this: data = {{{2010, 8, 3}, 0.}, {{2010, 8, 31}, -0.052208}, {{2010, 9, 30}, 0.008221}, {{2010, 10, 29}, 0.133203}, {{2010, 11, 30}, 0.044557}, {{2010, 12, 31}, 0.164891}, {{2011, 1, 31}, 0.055141}, {{2011, 2, 28}, 0.114801}, {{2011, 3, 31}, 0.170501}, {{2011, 4, 29}, 0.347566}, {{2011, 5, 31}, 0.461358}, {{2011, 6, 30}, 0.244649}, {{2011, 7, 29}, 0.41939}, {{2011, 8, 31}, 0.589874}, {{2011, 9, 30},

问题 I would like to numerically find the solution to u_t - u_xx - u_yy = f on the square y ∈ [0,1], x ∈ [0,1] where f=1 if in the unit circle and f=0 otherwise. The boundary conditions are u=0 on all four edges. I have been trying to use ND-solve for ages but I keep getting error messages. I'm new to Mathematica so I don't know how to define f before ND-Solve. Thank you in advance. 回答1: The Mathematica help files are incredibly complete for stuff like this. I'm going to reference the online

问题 Let us interpolate the same data using Mathematica 5.2, 7.0.1 and 8.0.1: Interpolation[{{0, 0}, {1, 1}, {2, 3}, {3, 4}, {4, 3}, {5, 0}}] // InputForm The outputs are: Mathematica 5.2: InterpolatingFunction[{{0, 5}}, {2, 0, True, False, {3}, {0}}, {{0, 1, 2, 3, 4, 5}}, {{0}, {1}, {3}, {4}, {3}, {0}}, {Automatic}] Mathematica 7.0.1: InterpolatingFunction[{{0, 5}}, {3, 1, 0, {6}, {4}, 0, 0, 0, 0}, {{0, 1, 2, 3, 4, 5}}, {{0}, {1}, {3}, {4}, {3}, {0}}, {Automatic}] Mathematica 8.0.1:

问题 As a continuation of my previous question, Simon's method to find the list product of a PackedArray is fast, but it does not work with negative values. This can be "fixed" by Abs with minimal time penalty, but the sign is lost, so I will need to find the product sign separately. The fastest method that I tried is EvenQ @ Total @ UnitStep[-lst] lst = RandomReal[{-2, 2}, 5000000]; Do[ EvenQ@Total@UnitStep[-lst], {30} ] // Timing Out[]= {3.062, Null} Is there a faster way? 回答1: This is a little

问题 I'm puzzled by Mathematica's responses to the following: ClearAll[n] #^2 & /@ Range[n] #^2 & /@ Range[n] // StandardForm It seems that even Mathematica (8.0) doesn't believe what it has just said: #^2 & /@ Range[5] Range[5^2] Any thoughts about what is happening? Edit: The original context for this question was the following. I had written PrimeOmega[Range[n]] - PrimeNu[Range[n]] and since n was going to be very large (2^50), I thought I might save time by rewriting it as: PrimeOmega[#] -

问题 Consider the following list : dalist = {{47.9913, 11.127, 208}, {47.5212, 10.3002, 208}, {49.7695, 9.96838, 160}, {48.625, 12.7042, 436}} Those are coordinatees of Eye fixations on a screen where, within each sublist, #1 is the X coordinate, #2 the Y coordinate and #3 , the duration spent at that particular location I then use the following : Disk[{#[[1]], #[[2]]}, 3N[#[[3]]/Total[dalist[[All, 3]]]]] & /@ dalist to draw disk with duration weighted diameter. I would like to draw cross instead

问题 a=2^Power[10^6, 10^9] 3^Power[4^9, 7^5] TwoTower[n_] := Nest[2^# &, 1, n] What's the smallest n such that TwoTower[n]>a ? This question had a pen-and-paper answer on Quora, is there a way to use Mathematica here? 回答1: Just some thoughts (did not carefully check). If we follow the suggestion in that link and start taking logs (base 2), first thing which seems obvious is that we can safely forget the prefactor (the power of 3), since Log[Log[a*b]] = Log[Log[a]+Log[b]] = Log[Log[a]]+Log[1+Log[b]

问题 Verbeia opened up a rather interesting discussion on the performance of the functional programming style in Mathematica. It can be found here: What is the most efficient way to construct large block matrices in Mathematica? ) I'm working on a problem, and after doing some timing of my code one particularly time consuming portion is where I calculate entries of a matrix through a recurrence relation: c = Table[0, {2L+1}, {2L+1}]; c[[1, 1]] = 1; Do[c[[i, i]] = e[[i - 1]] c[[i - 1, i - 1]], {i,

问题 If I run Mathematica as "math foo bar", what variable(s) hold foo and bar? I'm guessing it's $Something, but haven't found it. Googling told me Mathematica accepts command-line options like -pwpath, -pwfile, but I couldn't find the right search phrase for actual command line arguments (not options). 回答1: The $CommandLine variable holds the command line arguments. $ math foo bar In[1]:= $CommandLine Out[1]= {math, foo, bar} 来源： https://stackoverflow.com/questions/7149252/if-doing-math-foo-bar

问题 It is known that output expressions are passed through MakeBoxes to turn the graphics expressions into the box language which the front end uses to represent graphics (when $Output has default option FormatType->StandardForm ). For example, if we evaluate: HoldComplete[Graphics[Disk[]]] we get a disk wrapped by HoldComplete : This is because HoldComplete does not stop MakeBoxes from converting its contents to typeset expression: In[4]:= MakeBoxes@HoldComplete[Graphics[Disk[]]] Out[4]= RowBox[