12:03 AM
@J.M. Since there are no private chatrooms on SE, do you know how we could have (beside mail) a conversation in private?

@Mauris Here's another way:
Plot3D[-4 x^5 - x^3 y^4, {x, -3, 3}, {y, -3, 3},
MeshFunctions -> {#1 &, #2 &, #1 &}, Mesh -> {15, 15, {0}},
]

@halirutan Private? Over the internet? Ha!

@belisariushassettled over an NSA proxy

12:22 AM
@halirutan Apart from mail, I've no idea. :(

@J.M. too bad. With mails it is harder to have an actual conversation.

@halirutan Once one realize that Google had stored all wifi passwords on Earth and beyond, one is somewhat less confident on privacy
There is RADIUS and two factor auth, of course. But still
@RajAcharya Welcome to Mathematica.SE! Please, when you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge.

@belisariushassettled Is this an open access database?

Remote Authentication Dial-In User Service (RADIUS) is a networking protocol that provides centralized Authentication, Authorization, and Accounting (AAA or Triple A) management for users who connect and use a network service. RADIUS was developed by Livingston Enterprises, Inc. in 1991 as an access server authentication and accounting protocol and later brought into the Internet Engineering Task Force (IETF) standards. Because of the broad support and the ubiquitous nature of the RADIUS protocol, it is often used by ISPs and enterprises to manage access to the Internet or internal networks, wireless...

12:40 AM
@belisariushassettled It is however sobering that most of those passwords are for routers named linksys.

12:54 AM

@halirutan Nah. But the origin of the collection is interesting. By default, Android back ups your wifi passwords, along with any other piece of "useful" info to your google drive/gmail account. Almost nobody disables it

1 hour later…
2:17 AM
@OleksandrR. Why the harshness?
0

Manipulate[((b-a)/n)*Sum[f[a + i *(b-a)/n],{i,0,n-1}], {n,35,150} ] How to turn the output into a decimal number.

3:05 AM
@belisariushassettled At some point, one has to learn from ones mistakes... 5 questions, 5 closed/off-topic/duplicates... mathematica.stackexchange.com/users/35439/juan-valdiri

@anderstood know but
enter a notebook. Type Number
press F1
... nothing useful
at least not without reading for half an hour
Doesn't matter. I guess everybody gets fed up with September questions :)

:D
I'm happy there are people like you, but I'm also happy some people react like OleksandrR... Nothing that really matters anyway! :)

@anderstood Yup. Probably. Sometimes I remember my first days with Mma and it hurts :)

@belisariushassettled Since you have settle, may I ask you a question?

@anderstood of course
freedom of speech, ya know

3:16 AM
I'm implementing an arclength continuation algorithm, inside I am using FindRoot. It works well, except sometimes FindRoot returns some negative values, which I do not want. I was thinking of using an constrainted optimisation to replace FindRoot but I am not sure if it is as efficient, in terms of computation cost.

@anderstood Let me search for something I believe may be useful

But now I am thinking that if my continuation goes towards negative values, since the solutions is a 1D manifold, there's nothing much I can do against it...

@anderstood Take a look
85

As promised in the comments on my first answer, here is an implementation of an all-compiled-code Nelder-Mead minimizer, which hopefully represents a more useful response to the question. The algorithm used here corresponds to that given by Lagarias et al. in SIAM J. Optim. 9 (1), 112 (1998) (abr...

@belisariushassettled Thank you very much. From what I understand, it cannot work for constrained optimisation because "Unlike for NMinimize, constrained optimization is not supported, because the Nelder-Mead algorithm is fundamentally an unconstrained method."

@anderstood ugh. Didn't remember that. Sorry
So.

What is an arclength continuation ?

Numerical continuation is a method of computing approximate solutions of a system of parameterized nonlinear equations, The parameter is usually a real scalar, and the solution an n-vector. For a fixed parameter value , maps Euclidean n-space into itself. Often the original mapping is from a Banach space into itself, and the Euclidean n-space is a finite-dimensional approximation to the Banach space. A steady state, or fixed point, of a parameterized family of flows or maps are of this form, and by discretizing trajectories of a flow or iterating a map, periodic orbits and heteroclinic orbits...
I am using it to solve an nonlinear system of equation: $G(u)=0$ where $G:\mathbb R^n\rightarrow R^{n-1}$.

Looks useful ...and not trivial

Yeah it seems quite efficient; not so complicated in fact. The solution is (a.e.) a manifold of dimension 1.

So you have something to estimate the next arclength ... and it turns out negative sometimes?

3:30 AM
A starting point can be found with FindRoot. Then, if you know the direction, you can get to the next point by adding a $n$-th condition: that the projection of the difference of two consecutive points on the direction is equal to the parameter.
Absolutely! You can choose the direction such that its inner product with the previous one is positive (equal to one)

then ... how comes you get negative answers if you can chose the direction?

Well, I believe asking the question helped me (as often): some values become negative simply because the 1D manifold (curve) crosses one hyperplane x_i=0; nothing I can do against it, except searching in the other direction I guess. There might be bifurcations too...

ahhh ... you solved it!
remember to acknowledge me in your paper :=)

Because once you have chosen a direction, you must not reverse or you'll find the same points.
If i acknowledge you, you won't be able to change your username. OK? :D
Btw, I could acknowledge you for the other useful answers you provided.
Did you learn MMA on your own?
(mathematica, not mixed martial arts)

@anderstood yep. And here. But not sure if learned too much.
In any case, it's fun
Gotta go. Good luck with your algo!
"if learned too much."--> "If I learned much". language barrier here :)

3:38 AM
Thanks, bye!

5 hours later…
8:40 AM
@halirutan I believe GPG is considered unbroken, still, with a 4096-bit key.

2 hours later…
10:21 AM
@belisariushassettled hm, I didn't intend it to come across as harsh. I just meant that this user should spend more of their own effort before asking for help, because this site is not meant as "teach yourself Mathematica in only 1000 easy questions". This particular question is also ill-posed because a and b are not defined, and there is no attempt at formatting.

10:42 AM
@belisariushassettled @OleksandrR. I would consider the OP's reply comment as a no-go. He is not even thinking about Oleks advice but he's instantly backfiring. I really wonder what happened to our voting system, because usually bad questions should be down-voted. In a long run this downvoting would probably drive him away and he will think, what a shitty site. Pointing him to what's wrong with his questions is the correct way.

@halirutan For what its worth, I didn't read his reply like that. Of course your interpretation could be right, but the comment can also be genuine imo.

11:07 AM
@JacobAkkerboom Well the "I may come back later on when I´am a professional someone" part sounds like being pissed off because he didn't got "da codez" but a more general help instead, which makes his life harder, because now he has to read documentation.
Probably it's only me and my experience with some people. It's like one of their standard-arguments: "Can it be wrong to be interested and ask questions?".. "I always thought asking questions is a good thing..."
@JacobAkkerboom I recently came across this which gave me the same feeling:
111

So I was just put on a Performance Improvement Plan (PIP) after six months at my first job out of college. I know this means I'm going to get fired when the time is up, but until then I would like some advice on how to improve for the future. There were four points to the plan, and I'd like to g...

11:40 AM
@halirutan It could also be that he attempted to make a promise to come back later to answer to answer questions himself. Or he tried to do this and lie so that he would get answers without effort :P. Ah, the good old miscommunication problem when talking in text messages :).

3 hours later…
2:12 PM
@OleksandrR. Maybe it's just me, but I read it as a restrained exasperation rather than a harsh snipe. Had I not been a mod, I might have said even choicer words.
@JacobAkkerboom And yet, the lot of us manage, no?
@anderstood Continuation methods are definitely a nontrivial project. We have NDSolve[] but the niggling details are all in making sure the paths you picked aren't just fumbling about in solution space. Again, nontrivial.

@J.M. Thanks, I feel somewhat less stupid now :)
@halirutan @OleksandrR. All good. I just tried to imagine a poor soul, three days into Mathematica and programming getting stupid fractions instead of solid decimals. Then I stared at a blank notebook and asked myself "How could I ask for help on this with F1?" and I felt the pain. You're very right that the question is bad. But so is Mathematica documentation. At least until you grasp it

@anderstood, anyway, I should hope you've seen at least this.
@belisariushassettled see, people who have never started with the command line are spoiled... :P

@J.M. echo < "Y" > "Chat"
@J.M. Humorous, but probably very related to that. I've managed to survive so many OSs without any kind of "helpy" thing that I believe reading manuals and docs is tattooed in my brain.

2:30 PM
@belisariushassettled and almost always, commands do have some sort of help syntax built into them: cat --help
...and that's also why I find it highly annoying when it turns out the docs were not exactly accurate either. ;)

3:05 PM
@J.M. Oh, but *NIX is modern. Try that with VSE, VTAM, ICCF, ...:)

3:19 PM
@belisariushassettled it's a fair point. My usual approach is only to bother others with questions after I exhaust my own resources, and it seemed that this is very much the opposite (or the user is not very resourceful). I only started to engage with the Mathematica community after I was already quite competent with it. And this was a deliberate choice

@belisariushassettled Point taken. ;)

@OleksandrR. I believe most of us moved along the same lines by choice or by accident. But I believe that also may depend on how much time you have until understanding &/@x@@@y//z for your next exam:)

@OleksandrR. (I'm pretty sure we've talked about this before; déjà vu?) I usually only ask other people questions only after straining hard to answer them myself but not getting anywhere. I think the fact that resources were not easily available back in the day (otherwise considered a handicap) helped develop the self-reliant attitude.
@belisariushassettled or next work deadline, for that matter.

@J.M. work has other standards (at least for me). Once you decided to get paid for something, you should be able to find your own way through

@belisariushassettled That's my thinking too, but we have had people asking about work-related stuff here as well, if memory serves...

3:37 PM
@J.M. Well, there is a thin line between trying to do your job better and asking someone else to do it :)
@J.M. Do you know this syntax:
Graphics@{{FilledCurve[{{{0,2,0},{0,1,0},{0,1,0},{0,1,0}}},{{{29,23},{29,26},{3‌​,26},{3,23},{29,23}}}]}}

@belisariushassettled :D Did you read about the case of the highly-paid programmer who was found to be subcontracting all his work to Indian(?) programmers for much less pay?

@J.M. A real entrepreneur :)

@belisariushassettled Does that even work? You have 3D coordinates in there...

@J.M. Nope. They are somekind of node spec
not sure relly
really
or knots.sometng like that
@J.M. Apparently not docummented

@belisariushassettled Yes, apparently not. You could ask a question, but...

3:44 PM
@J.M. he
Ha!
:)
Found it while trying curve = ImportString[ExportString[Style["-", FontSize -> 72], "PDF"],
"TextMode" -> "Outlines"][[1, 1]];

g = Graphics[curve]

Hmm...

@belisariushassettled GeometricFunctionsDecodeFilledCurve[FilledCurve[{{{0, 2, 0}, {0, 1, 0}, {0, 1, 0}, {0, 1, 0}}}, {{{29, 23}, {29, 26}, {3, 26}, {3, 23}, {29, 23}}}]]

@J.M. Does that work on v9?

@belisariushassettled That thing has been available since version 8, IIRC.

3:52 PM
21

There is a public, but undocumented, function called GeometricFunctionsDecodeFilledCurve which helps to decode this type of undocumented FilledCurve: GeometricFunctionsDecodeFilledCurve[ FilledCurve[{{{0, 2, 0}, {0, 1, 0}, {0, 1, 0}, {0, 1, 0}, {0, 1, 0}}}, {{{12.887695983062486, 5.160000...

:)

@belisariushassettled Aha!

@J.M. But the GeometricFunctionsDecodeFilledCurve doesn't seem to work here

2 hours later…
5:33 PM
heya all
does anyone know python in here that would be willing to help me translate some iterative python code?
I think i need to be using a table but oh my gosh haha this is difficult

5:52 PM
oh god i have to go use those awful imperative functions
the ones where mathematica tries way too hard to make if statements functional and it turns into frankenstein's monster
/me cries

6:11 PM
@J.M. @J.M. No, in fact. For the moment my program works quite well! If I need more, I'll check the reference, thank you.

6:42 PM
@J.M. Hey, can I ask you a silly question?
@MichaelE2 if you're active, would you mind too

6:54 PM
wait
I did it in the wrong variable, that's the issue.
Oops, sorry.