Wolfram Mathematica

2:47 AM

@J.M. Yeah, I was very surprised by that as well...

8 hours later…

Sjoerd C. de Vries

10:54 AM

@J.M. The question does not really show much effort, but I guess that people find it an intriguing math question. If its intention was to get a pure numerical solution, it would be easy and an answer indeed exist already here. Finding a symbolical solution (such as a cylinder as a result of a coil) may be quite a challenge. I wonder whether this can be solved in general.

Perhaps finding such a solution would be more ontopic on MathSE.

6 hours later…

4:34 PM

Happy weekend everyone.

acl

5:03 PM

@SjoerdC.deVries I would bet it can't be solved analytically in general. Don't really know though.

3 hours later…

8:02 PM

Graaurr

8:15 PM

@Rojo You're no longer Rojo!

How will I now be able to tell you apart from Verde?

Aye, graaurr indeed. An exclamatory one when you've finally found a fix the hang-up issue in Mathematica's own Documentation Center. Yes...

haha

8:37 PM

@rm-rf It just went away when I edited the profile hidden email challenge of the month

Now I'm just lazy to edit the gravatar back

Btw, I have a list and I want to split it by some separator symbol, just like one would split the lines if the separator symbol was \n. Am I missing sone very straightforward function? (other than SplitBy[list, #==separator&] and then post processing

)

@Rojo Now you can upload an image directly on SE... no need to create account at gravatar.com

That's more practical

@Rojo Nothing comes to mind... I would've probably written a custom function

    ClearAll@f
    f[Shortest[h___],  a ..,  Shortest[t___]] := {f[h]} ~Join~ f[t]
    f[h__] := {h}
    f[] := Sequence[]

    f @@ {1, 2, 3, a, a, 4, 5, 6, a, 7, 8, 9, a}

9:10 PM

@rm-rf Thanks

2 hours later…

11:09 PM

@Ghersic Are you around?

@halirutan I am, sorry for the delay there.

@Ghersic Ok, maybe your question is not really clear to me but what do you try to achieve?

here in this question

I noticed your update to my code on a question, sorry for the double `.

@Ghersic No problem.

I mean, you could apply FullSimplify only to both sides and not to the complete equation.

Like this:

FullSimplify /@ (z1'[t] == -a kf + b kr)

and it would leave the z1 alone.

Ohh, I see.

11:21 PM

@Ghersic And, as you already found out z1' is not really useful because the depended variable is missing.

This will give you trouble inside NDSolve

Because later I'm hoping to use the shorthand of eqns instead of inserting each manually, I was hoping to construct from my eqns = {z1'== -a*kf + b*kr, eqn2, ..., eqn3} a simplified set of equations, seqns` all simplified but still in terms of zn' still.

@Ghersic Yes, I thought you might want to do that and it is surely possible.

Using the same equations as in my example there (autonomous, so like in the Wiki article I linked at the start of my question, it has to be solved numerically because they don't explicitly reference (t) - unfortunately.

@halirutan phew, that is great to hear!

@Ghersic But they do reference t, they just don't write it.

Do you mean the example?

Ah yes, that's what I meant sorry.

11:24 PM

@Ghersic Yes, just replace it:

y'[t]==(2-y[t])*y[t]

this leaves you with

NDSolve[{y'[t] == (2 - y[t])*y[t], y[0] == 2}, y, {t, 0, 10}]

for example

Ah, alright. I knew that appending a [t] to my actual system would be necessary to incorporate what I'm trying to do later in NDSolve.

Just because the t is not explicitly used (I mean, it's autonomous) doesn't mean you don't have to write the dependence on t ;-)

Ahh ok. Do you see anything fishy going on here? (If you don't mind my showing you the dummy equations' Maniuplate[...Plot[...]] function?)

That makes me think I might be putting something unnecessary in my code for it??

eqns[k1f_, k1r_, a0_, k2f_] = {
a'[t] == -k1f*a[t] + k1r*b[t],
b'[t] == k1f*a[t] - k1r*b[t] - k2f*b[t],
c'[t] == k2f*b[t],
a[0] == a0,
b[0] == 0.0,
c[0] == 0.0};
Manipulate[
Plot[
Evaluate[{a[t], b[t], c[t]} /.
NDSolve[
eqns[k1f, k1r, a0, k2f],
{a, b, c}, {t, 0, 1500}, MaxSteps -> 1000, AccuracyGoal -> 10]],
{t, 0, 1500}, PlotRange -> {{0, 1500}, {0, 0.11}},
AxesLabel -> {"Time (Seconds)", "Concentration"},
PlotStyle -> {{Red, Thickness[0.008]}, {Green,
Thickness[0.008]}, {Blue, Thickness[0.008]}}],

@Ghersic And now you want to apply FullSimplify on all RHS in eqns?

(I mean in you real code)

Yessir

preferably just in between my eqns[k1f_, k1r_, a0_, k2f_] and the Maniuplate, placeing the simplified RHS seqns[k1f_, k1r_, a0_, k2f_] into the Manipulate instead.

placing*

11:32 PM

`CUDAImageConvolve[
ExampleData[{"TestImage", "Lena"}]~ColorConvert~"Grayscale",
BoxMatrix[20]/((2 20 + 1)^2) // N]` looks suspiciously wrong, and suspiciously different from the suspiciously right looking version with `ImageConvolve`

@Ghersic OK, what about this really verbose way:

eqns[k1f, k1r, a0, k2f] /. (lhs_ == rhs_) :> (lhs == FullSimplify[rhs])

Ohhh man, that looks like it works.

@Ghersic If you only want to replace the equations where a derivative is on the left side, you could put another specifying pattern on the lhs.

@Ghersic For instance smth like this

eqns[k1f, k1r, a0, k2f] /. (lhs : Derivative[1][_][t] == rhs_) :> (lhs == FullSimplify[rhs])

Yes, and because seqns is what is being placed into the Manipulate I can even name eqns without the bracketed phrase after it, using seqns[k1f_,...] instead!

Ahh ok, that will be valuable in not changing th initial conditions housed with eqns and seqns.

@halirutan Whenever I interact with people on this forum, I think... "If only I could come up with so sexy / brief a code as them..." Thank you, sir.

@Ghersic OK, I have to run now. Hope this gives you a start.

@Ghersic And don't call me sir.. ;-)