Wolfram Mathematica

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b3m2a1

1:12 AM

@halirutan Simplicity ;___;

And use existing technology like the Jupyter layout

halirutan

1:42 AM

@b3m2a1 Well, I have several more hypothetical questions and scenarios that I go through. Often, they are driven by seeing that some of Mathematica's functionality/technology doesn't scale well. Especially in the light of bigger and bigger data.

b3m2a1

1:56 AM

@halirutan I think that's big

Clearly (from hearing JF on the live streams) there aren't enough resources to give the FE the love it would need to really be a quality object

halirutan

@b3m2a1 One of the many problems is for instance (!) that we work with MRI data. Usually, it isn't even that large, like 400x400x400, but even at this moderate size, I found the Image3D tools unusable.

Like adjusting the color-function. It's more than buggy every time I try it so I don't even start to rely on it.

One of my more recent investigations concerned the performance of all these nice graphical representations for things like colors, interpolating functions, etc. For instance, just imaging a list of 1000 colors and represent them as simple rgb-triples:

Table[List @@ ColorData["SunsetColors", x], {x, 0, 1, 1/1000.}]

Try this out, resize the notebook, etc.. I works okayish but it's all text. Now remove the List part and do the same. I can tell you, in 12.0 and Linux the front end freezes for seconds.

So one important question is, how much performance can you expect from a notebook that pushes the limits, when everything in it is iconized like it is nowadays?

Even if you consider a new front end that was rewritten using the latest technology.

b3m2a1

2:54 AM

@halirutan I'm a fan of the iconization, but I do wish Mathematica would default to showing me less

I think the FE could be smart about that

Like give a max of one second for output rendering for anything non-Graphics

At which point just provide a ... for anything non-displayable in that time constaint

5 hours later…

CA Trevillian

8:14 AM

Just to add to these comments questions & etcetera @halirutan do you have trouble when you save those notebooks? If you get to that point with such “large” 3D files? I kept having my system hang to “not responding” while saving so much I went for a walk to save on my last run. Got tons of files though! Awesome 3DFFTs with phase plotting & everything. Learned a ton & made the whole class jealous haha!

6 hours later…

kirkus

2:33 PM

@halirutan do you report all those as performance bugs? I always do when I run into something that's just shockingly slow.

1 hour later…

1010011010

4:01 PM

Suppose I have a 3-dimensional array and a 2-dimensional array

and I say "Total[ArrayOne * ArrayTwo]", then what do I get? Am I summing over all of the first positions?

So I would basically get "ArrayOne[[1]] * ArrayTwo[[1]] + ArrayOne[[2]]ArrayTwo[[2] + ... + ArrayOne[[ArrayLength]] ArrayTwo [[ ArrayLength]]"?

2 hours later…

C. E.

5:52 PM

I want to build a graphics library for Python inspired by Mathematica. I can't take matplotlib anymore.

b3m2a1

@C.E. I've actually been doing exactly that kind of thing for my work :)

It's not very good at this point

And it uses matplotlib

But I've tried to make it very OO and nice to use with things like:

Plot(f, [0, 1], axes_labels=["x", "y"])

Lukas Lang

@1010011010 You are essentially using the Listable attrbiute of Times - that is, the first two dimensions (assuming the arrays have the same length) are multiplied element-wise. For the third dimension, you are effectively doing a * {b, c, d}. This means that you are left with a 3D array after this, and then total will sum the elements on the first level together. But the easiest is probably to just try it, no?

Total[Array[a, {3, 3}]*Array[b, {3, 3, 3}]]
(* {{a[1, 1] b[1, 1, 1] + a[2, 1] b[2, 1, 1] +
   a[3, 1] b[3, 1, 1],
  a[1, 1] b[1, 1, 2] + a[2, 1] b[2, 1, 2] + a[3, 1] b[3, 1, 2],
  a[1, 1] b[1, 1, 3] + a[2, 1] b[2, 1, 3] +
   a[3, 1] b[3, 1, 3]}, {a[1, 2] b[1, 2, 1] + a[2, 2] b[2, 2, 1] +
   a[3, 2] b[3, 2, 1],
  a[1, 2] b[1, 2, 2] + a[2, 2] b[2, 2, 2] + a[3, 2] b[3, 2, 2],
  a[1, 2] b[1, 2, 3] + a[2, 2] b[2, 2, 3] +
   a[3, 2] b[3, 2, 3]}, {a[1, 3] b[1, 3, 1] + a[2, 3] b[2, 3, 1] +
   a[3, 3] b[3, 3, 1],

C. E.

@b3m2a1 :O is it published anywhere?

b3m2a1

I sent you a link on the Slack

Caveat emptor: it's not battle tested. I just use it for my own work because MPL was driving me mad, too.

C. E.

@b3m2a1 What is the link to the Slack, I can't even remember how I used to access it.

b3m2a1

5:59 PM

wolframlanguage.slack.com

ded7

6:23 PM

I have a beginner mathematica question. If x(t) = ct + 2 and x(1) = 1, then there is a unique solution c = -1 and so x(t) = -t + 2. How do I express that in Mathematica? I tried

```
DSolve[{x[t] == c*t + 2, x[1] == 1}, x[t], t]
```

But this gives ```DSolve::bvnul: For some branches of the general solution, the given boundary conditions lead to an empty solution.```

Lukas Lang

6:52 PM

@ded7 The problem is that you want to solve for c, not x[t] - Mathematica assumes that c is a fixed parameter in the DSolve command you have shown, which is why it is complaining. Here are two options:

DSolve[{x''[t] == 0, x[1] == 1, x[0] == 2}, x[t], t]
(* {{x[t] -> 2 - t}} *)

x[t_] := c t + 2
Solve[x[1] == 1, c]
(* {{c -> -1}} *)

ded7

@lu

@LukasLang Thanks! If I follow your second solution, how do I tell Mathematica to construct (and simplify) all the possible symbolic expressions for the function x[t], given the values of the coefficients that it has determined? My real example involves two coefficients:

x[t_] := 1 / (c3 t + c4)^2;
Solve[{x[0] == 1, x[2] == 4}, {c3, c4}]

(* {{c3 -> -3/4, c4 -> 1}, {c3 -> -1/4, c4 -> 1}, {c3 -> 1/4, c4 -> -1}, {c3 -> 3/4, c4 -> -1}} *)

I'd like it to give me a list of symbolic expressions for distinct x[t] functions obtained by plugging in the 4 solutions for the coefficients.

1010011010

7:15 PM

@LukasLang Excellent, that is precisely the expected output indeed :)

2 hours later…

1010011010

9:20 PM

@ded7 You could map it to an Array, then loop over it during a replacement operation

Rohit Namjoshi

@b3m2a1 Did not know about the Slack channel. Can I get an invite?

1010011010

9:39 PM

@ded7 Is this what you mean?

x[t_] := 1/(c3 t + c4)^2;
x[t] /. Solve[{x[0] == 1, x[2] == 4}, {c3, c4}]

Rohit Namjoshi

10:10 PM

@1010011010 That is an interesting 3d plot. How was it generated?

1010011010

@RohitNamjoshi It's a bright soliton in the one dimensional repulsive Bose gas at zero temperature

Basically what you're seeing is a case of wave-particle duality for matter; the "depth" axis in the above graphic is time, the width axis represents the spatial parameter and the height is the density profile of the gas

It's not very clear from the picture, but I'm assuming a closed line, i.e. periodic boundary conditions, so the left and the right of the graph represent the same point in space, the wave packet is moving to the left here

halirutan

10:54 PM

@kirkus If I can pin a problem down or if it is severe, I report it.

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