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Feb
13
comment Making table and plotting results of NDSolve problem
There are some simple mistakes in your question, but I think the most troublesome part is that under h = 1 it seems to be impossible to find a proper initial guess for shooting method. Using the approach in this post, the best initial guess I can find is sol[h_, L_, i_] := sol[h, L, i] = {i, NDSolve[{f[x]^3 # & /@ (f''[x] == f[x] (h/f[x]^2)^2 - f[x] (1 - f[x]^2)), f[0] == f[L] == 1}, {f}, {x, 0, L}, Method -> {"Shooting", "StartingInitialConditions" -> {f[L] == 1, f'[L] == i}}]}; sol[1, 8, 9000000003/10000000000]
Feb
13
comment VectorScale Explanation
What do you mean by saying "My version (10.0.0) does not support sfun."? (You did set it in your second sample.) Anyway, +1, I didn't know StreamPoints etc. can be used inside VectorPlot!
Feb
13
comment VectorScale Explanation
1. Is "With unitlen = 1 (middle)" a typo? 2. So you think aratio is short for arrowhead ratio rather than aspect ratio?
Feb
13
comment Graphics3D: Opacity limitations
I'm afraid this isn't what OP wants, he wants "the lines further from the eye to be "grayer" than the lines closer to the eye, as if there were a little fog inside the sphere", while this only differs "front" and "behind".
Feb
12
comment Obtaining the number of iterations used in ReplaceRepeated
But you're asking for "a way to obtain the actual number of replacement operation", not a option of ReplaceRepeated that demonstrate the actual number of replacement operation :)
Feb
12
comment Obtaining the number of iterations used in ReplaceRepeated
possible duplicate of How to visualize pattern matching process?
Feb
12
comment Graphics3D: Opacity limitations
@celtschk I got the same disappointing result as OP in v8.0.4 and v9.0.1, Vista 32bit. Can this be a platform or even hardware related issue?
Feb
12
comment NDSolve grid refinement for PDEs
I think it'll be better if you can add the specific equations to your question, maybe someone can find a combination of options that'll circumvent your problem.
Feb
11
comment replacement rules from a pattern and a matching expression
Er……how can I make "Failed!" appear in the result?
Feb
11
comment Alternative to Series
@Hawk Yes. Seems that it can be arbitrary small. Just tried x[1][0] == 10^-160.
Feb
11
comment Alternative to Series
I fixed it for you. NDSolve outputs an InterpolatingFunction, which is already a differentiable approximation. Well, I can post an answer if you like, the graph under your i.c.s isn't interesting though.
Feb
11
comment Alternative to Series
Your code still contains typos, you'd better fix them. The only troublesome part is the x[1][0] == 0 condition, if an approximate nonzero condition like x[1][0] == 10^-10 is acceptable, then the method in the post linked above will handle your equation set.
Feb
11
comment Alternative to Series
If a numeric series solution is desired, then I think this is a possible duplicate of Series expansion of InterpolatingFunction obtained from NDSolve
Feb
11
comment How to visualize pattern matching process?
I tried to improve the expressions of your question. (Wish I didn't make it worse… ) Well, To be honest, personally I feel your Edit a little verbose…
Feb
11
comment the winding number for the circle map (Arnold tongue)
Can you add the specific parameters for the producing of this picture? BTW, the link is broken.
Feb
11
comment How to visualize pattern matching process?
@user15961 You don't need to accept that quick, feel free to wait for 24 hours or more so your question may attract better answers.
Feb
10
comment How to visualize pattern matching process?
@ChipHurst Yeah, Reap and Sow is undoubtedly a better choice. The initial intention of this answer is only to show that trick so I chose the relatively simpler Print to avoid distraction :)
Feb
10
comment Does NRoots own an abstract counterpart? If not, can we write one?
Any plan to implement the other 2 method of NRoots[]? Though I admit this question is motivated by my competitive mood against Matlab, now its target is to get the counterpart of NRoots[] anyway :)
Feb
10
comment Does NRoots own an abstract counterpart? If not, can we write one?
Translation for the comment of @chyaong: According to your new-added code, NRoots in v10.0.2 is more than 4 times as fast as that in v9.0.1. The roots[] in my answer has similar performance as roots() of matlab. Well, that's interesting.
Feb
10
comment Alternative to Series
Do you need a numeric series solution or a symbolic series solution?