12:03 AM
@acl There's one way to find who it was and force their hand :P

12:13 AM
@rm-rf Is it our dear random downvoter?

12:49 AM
@rm-rf pray tell!

1:00 AM
(unless your comment there was The Way To Find Out Who It Was, in which case it flew well over my head)

@rm-rf JMHO but I'm not enthusiastic about cutesy chatroom names. I like it well enough as it is.

@OleksandrR. I thought you were off to write the thesis...

@acl it's 2am... I came home and am about to go to bed.
Bye!

@OleksandrR. excuses are cheap :)
night

@acl No, I'd have to close and reopen which will then show who it was :)

1:13 AM
@rm-rf show how? is this (viewing those who voted to close) some kind of Yodaesque power that I don't have due to my puny rep and pathetic non-mod status?

@acl No, if a question is closed, then the names are made public
You can always find who the closers were in the history (after it has been closed)

@rm-rf I see, much simpler than I thought. But what would be the reasoning behind closing it?
or voting to close

I don't know, and I wouldn't have voted to close. I guess whoever it was, probably thought it was an excel question even though the very first line says otherwise
Meh, not worth getting hung up over stray votes... if 5 of them voted to close, then we have a problem :)

@rm-rf maybe it's an Eloi vs Morlock thing
@rm-rf right, I was just wondering whether I missed something
I fairly recently discovered two things: 1) I am not all that unique in enjoying Neal Stephenson's books (I thought he was a fringe writer and I the discoverer of some hidden treasure, until I googled him...) 2) he's a Mathematica programmer

2:02 AM
@rm-rf Maybe the FullForm[] for verbosity's sake: Get["Chat"]
@acl I read his stuff, but I still think he's relatively fringe. He's not Dan Brown or even Neil Gaiman-level of being famous, I'd say.

2 hours later…
4:20 AM
0

I wanted to start out by saying that I love the new MMA SE and I hope the meta is the correct place for this post. The MMA SE is turning out to be great and the community is really helpful and I much prefer the interface to the older listservs and usenet newsgroups. However, listservs are easily ...

1 hour later…
5:47 AM
@sjoerdc.devries I don't think this question is good enough to merit migration.

6 hours later…
11:28 AM
hello; quick question: can mathematica present a set of {theta,phi} points as a Mollweide representation? Can we do the reverse of mathematica.stackexchange.com/questions/11834/… with a set of coordinates?
it seems the function Mollweide exists and is being used by the WorldPlot/ref/Mollweide package but I did not understand how... May be I should look at the code?

11:56 AM
I add coded something similar in 2001 (!) which extracted level curves from a given contour and then built a country "astro". But it seems my code is not mm8 friendly...

@chris, what exactly do you want to do?
I mean if you only want to transform from {theta,phi} to {x,y} you could just use the formula from wiki.

@halirutan I would like to do a 2D plot which shows the whole sphere at once with a line corresponding to the trajectory of a point on the surface of the sphere.
ie I would like to do !Mathematica graphics
but in a Mollweide representation
as I mentionned I wrote a code which used to convert
but i) this code is for contours and ii) it doesn't work anymore ;-(
Do you think I should set this up as a question?
my code was doing something like 'Do[WorldData /: WorldData["astro" <> ToString[i]] =
Join[#1, #2] &, {LevelCurve[cp, cts[[i]]][[2]] /. Line[a_] -> a,
LevelCurve[cp, cts[[i + 1]]][[2]] /. Line[a_] -> a}]*10000/
Pi /. {a_, b_} -> {a/4, b};, {i, 1, Length[cts] - 2}];
WorldPlot[{Table["astro" <> ToString[i], {i, 1, Length[cts] - 2}],
RandomGrays}, WorldProjection -> Mollweide]'
where LevelCurve extracts from a given contour plot cp the contours cts[[i]]

@chris Yes, this seems worth a question.

12:52 PM
@chris Still here?

1:07 PM
@halirutan yes... (had to attend a phd defense)

(Maybe head start is the wrong word here..)

@sure ! Well I am not sure what you mean but...

Starting with the first equations from the wiki page of the Mollweide projection:
cart[{lambda_, phi_}] := With[{theta = fc[phi]},
{2 Sqrt[2]/Pi*lambda Cos[theta], Sqrt[2]*Sin[theta]}
]
Now we need an solver for theta which is used inside and has no analytical solution:
fc[phi_] := Block[{theta},
If[Abs[phi] == Pi/2,
phi,
theta /.
FindRoot[2 theta + Sin[2 theta] == Pi Sin[phi], {theta, phi}]
]
];
Create a grid to test the stuff:
grid = With[{delta = .1},
Table[{lambda, phi}, {phi, -Pi/2, Pi/2, delta}, {lambda, -Pi/2,
Pi/2, delta}]];
gr = Graphics[{Line /@ grid, Line /@ Transpose[grid]},
AspectRatio -> Automatic];
And now we transform all points inside the graphic with our transformation cart
gr /. Line[pts_] :> Line[cart /@ pts]
Maybe this helps, but remember, I have no idea what I'm doing here. First time I looked at the formulas of the Mollweide transformation.
With your graphics, which is a set of points or lines too, the approach should be similar.

@halirutan thanks... I ll give it a shot... I was just thinking of translating my problem to the existing WorldPlot formulation as it has all sorts of projections already built-it. But if it fails I ll try your method.

2:10 PM
@halirutan The equation being solved there looks awfully like the Kepler equation...

1 hour later…
3:22 PM
@J.M. hi! Do you remember if there was a q about using the conditions resultant from Resolve[] to define Piecewise functions?

4:12 PM
@belisarius If there was, I do not remember it...

@J.M. mmm... perhaps it's time to post it

5

I have Solve[] return a list like {{F->ConditionalExpression[EXPR1, 0<z<a1]},{F->ConditionalExpression[EXPR2, a1<z<a2]},...} and so on -- EXPR* and a1,a2,... are somewhat complicated expressions, containing things like Root[]: EQ = F^4 (138880 z + 318688 F z + 98415 F^10 z + 6...

Hehe ... but that one doesn't have generated parameters ...C[1] \[Element]
Integers && ((\[Pi] + x > 2 \[Pi] C[1] && \[Pi] + 2 x <
4 \[Pi] C[1]) || \[Pi]/2 < x - 2 \[Pi] C[1] < \[Pi])

$c_1\in \mathbb{Z}\land \left(\left(x+\pi >2 \pi c_1\land 2 x+\pi <4 \pi c_1\right)\lor \frac{\pi }{2}<x-2 \pi c_1<\pi \right)$
@rm-rf O my
@rm-rf Example seems a forbidden word
@rm-rf @rm-rf Using pastebin as an external repository doesn't ensure data survival

4:35 PM
@belisarius No, but in case it goes down (even though I set it to "never expire") or goes out of business, the original data is available in the history here. So as long as paste bin is alive, we can enjoy a short code block. — rm -rf 1 min ago

@rm-rf @rm-rf You're right. I guess in these cases it's better to post a warning notice about the data being in the posts' history upon modificat :DDD

@belisarius Warning for the OP?

@rm-rf No, for future viewers
If pastebin isn't accessible ... etc
Anyway, the question in question seems nonsensical

@belisarius Ok, see the edit now

@rm-rf Yep. Looks very nice
@rm-rf Congrats! ;P

4:42 PM
I doubt if we'd really want to archive the data in this one :P Looks like it's too localized :D

@rm-rf I'm trying to understand the integral. Soooo messed up

@belisarius Maybe Artes' answer here might help: mathematica.stackexchange.com/a/1752/5
Might even close it as a dupe...

Calculating now
@rm-rf he used Artes' code

I don't see what the question is actually
why are they irrelevant, for instance?
It looks like NARQ to me

{epsilon1, w} =
ToExpression@Import["http://pastebin.com/raw.php?i=Z08GwMa8"];
f = Interpolation[Transpose[{Flatten[w], Flatten[epsilon1]}]]
epsilon2[w_] :=
epsilon2[w] =
2/Pi NIntegrate[a f[a]/(a^2 - w^2), {a, 0, 5.5793},
Method -> "PrincipalValue", Exclusions -> {(a^2 - w^2) == 0}] //
Quiet
ListLinePlot@Flatten@epsilon1
Plot[epsilon2[w], {w, 0, 5.57932}, AxesOrigin -> {0, 0}]

4:59 PM
@belisarius it's taking too long here

@acl here too
I just wanted to show you "working" code
not the one the OP posted

But I guess my problem with this sort of question is that it looks as if they wrote down literally the definition, it didn't work, and they hurled this at us

@acl "it looks as if" == "We're all certain that"

@belisarius right, I'd be surprised if they did try anything else

@acl He copied Artes' code for a different problem, and it (of course) didn't work
so ...
here we are

5:04 PM
aha, yes I just saw it!
ha

5:48 PM
Sup

5:59 PM
@Rojo Inf

@Rojo At least I'm frogs now, and not a RGB component

Hehe
How's everything going

6:16 PM
@Rojo Nice. Still alive.

6:30 PM
Alex Miller on October 15, 2012

It’s Back! Welcome to episode #33 of the Stack Exchange podcast.  We’ve got a brand new co-host (Jay Hanlon, our new VP of Community Growth) plus our guest this week is David Fullerton, VP of Engineering at Stack Exchange.

So what’s new in the seven months since our last podcast? Check out the new and improved review queue! If you’ve got enough reputation, you can see the review button at the top of any Stack Exchange site. The new system is clearer to use and it’s fast thanks to a ton of AJAX goodness. …

2 hours later…
8:20 PM
@Mr.Wizard Here is the version 7-save version (I hope) of the white space problem demonstrator.
Show[
GraphPlot[GraphData["DodecahedralGraph", "EdgeRules"],
EdgeRenderingFunction -> (Function[{e, v},
Inset[Graphics[Rectangle[{0, 0}, {10, 1}]], e[[1]], {0, 0},
EuclideanDistance[e[[1]], e[[2]]], e[[2]] - e[[1]]]]),
VertexCoordinateRules ->
MapIndexed[First[#2] -> #1 &,
GraphData["DodecahedralGraph", "VertexCoordinates"]]],
Method -> {"ShrinkWrap" -> #}
] & /@ {False, True}

8:47 PM
(I thought some here would appreciate that)

9:10 PM
this gentleman found the default mma IDE lacking, and asked for help:
-6

I find the default Mathematica IDE lacking. I am missing a simple project file tree on the left (as in eclipse, visual studio etc.) The following would be very cool: All your files (in a given directory) would be displayed in a Mathematica Palette. Clicking them changes to the clicked file or ...

is any kind soul going to help me get a gold badge by reaching +20 votes?
2
(yes I am soliciting upvotes :-| )

9:25 PM
@acl you're right behind me in rep. After the good old rules of tactical downvoting I should not have upvoted.

@halirutan but we're not doing this for rep, but for altruistic reasons, right? :)
thanks though, I'd like to have a gold badge for something other than coming here every day

@acl yep. That's why you need only 4 more upvotes now ;-)
Apropos we doing it for...
My stackexchange t-shirt, sticker, .. arrived today ;-)))

@halirutan and mine a few days ago. although come to think of it I forgot to even open the package
way too much work, leaving me with a half-asleep brain every night

And the best part is, when I bought "Joel on software" I didn't really know who this "Joel Spolsky" was and now I have a letter almost directly from him.
@acl I use the nice trick of sleeping often, not long..

@halirutan doesn't work for me. I go for "work 12h, interact with people and drink beer for 4, sleep for 8"
this in times of high workload of course, I don't always work 12h a day!
nice t-shirt, I just opened the package

9:50 PM
@SjoerdC.deVries Thanks, that clearly shows the problem.

@Mr.Wizard Reminds me of what SphericalRegion does

I don't think the tags here are the right ones
any ideas?
I'm guessing that and ended up there because he/she thinks of their problem as a "complex system", but here they mean something else

@acl let me look
What tag(s) do you propose? There is no "system-modeling" but there is rather confusing hyphenated "system-modeler" tag.

10:05 PM
@Mr.Wizard I don't know. I thought the system-modeler one was for the system modeler, while this question isn't about that

@acl Yes, exactly. I was about to creat a "system-modeling" tag but I thought it would be confusing alongside that one. I guess I see why system-modeler was hyphenated but maybe it would be better if it wasn't, for distinctiveness?

@Mr.Wizard my reasoning would be: suppose I was looking for an answer to this. what tag would I be looking for?

@acl Not inappropriate, but sufficient? Clearly the OP was focusing on "complex" and "system" in his own mind...

and I guess I'd be looking for the two I mentioned, and I can't think of a tag which would refer to the perturbation
@Mr.Wizard and we don't have that many questions on delay equations so I don't know if it makes sense to have a tag about them
@Mr.Wizard that's too domain specific to be useful here, I think. what's complex to me may be simple to you and vice versa
anyway I don't know. that's why I asked here

@acl Right, I'm not for that. ;-)
Do you not see value in a new tag "system-modeling" ?

10:12 PM
@Mr.Wizard ouch
@Mr.Wizard well, what does it mean? or rather what does it exclude? it seems to me to be describing the motivation behind solving a problem (namely, to model some system), not any property of the set of equations you're solving

I see your point. It might be informative to look for similar tags on other SE sites to see how well they appear to be working. Thanks for raising this question, BTW. Perhaps it is worth a Meta post.

@Mr.Wizard probably. we can't decide what to do, so let's ask someone :)

10:57 PM
@acl are you still here?

yep

I'm not understanding this question:
3

This probably has a one line answer, but I'm totally stuck. I have two tensors, i.e. two objects that depend on NN abstract indices which i've labelled i[m] (m=1,...,NN). I want to keep NN general for now. Each index i[m] ranges over the values {1,2}. I want to sum over all the i[m]s, i.e. Sum[...

Could you try explaining it to me, please? :)

I think he literally wants to do Sum[a[i[1],...,i[NN]] * b[i[1],...,i[NN]], {i[1],1,2},{i[2],1,2},...,{i[NN],1,2}]
so, he has two N-dimensional tensors a and b
that is, $a$ and $b$ are like matrices but, instead of having 2 indices, they have $N$; so the elements of $a$ may be specified as $a_{i_1,i_2,...,i_N}$
so I think he wants to do this, with the sums ranging from 1 to 2
is that clearer?

Better, yes.

this looks like it should be solvable by some sort of awe-inspiring one-liner, fit to scare small children with, but I've always used the same approach jVincent describes
(by the way did I explain whatever it was that was puzzling you? or did I miss the problem?)

11:11 PM
I think so, but I'm still looking at at. If SeedRandom[1]; {a, b} = RandomInteger[9, {2, 2, 3, 4}] then ...?

No I think Dimensions[a] should be {2,2,2,2,...,2} (and there should be NN of them, ie, Length@Dimensions[a] should be NN

Damn, I'm just dumb as a stump today. <:-/

so for NN=3 it should be something like
SeedRandom[1];
{a, b} = RandomInteger[9, {2, 2, 2, 2}]
Total@Flatten@
Table[a[[i, j, k]] b[[i, j, k]], {i, 1, 2}, {j, 1, 2}, {k, 1, 2}]
or anyway I think so, it's pretty late here and I'm tired

Thanks for humoring my stupidity.

(and I'm also not good at thinking in front of computers so take all this with a grain of salt)
no problem. someday, and that day may never come, I will ask a favor of you.

11:17 PM
lol

ah found the actual quote: "Some day, and that day may never come, I will call upon you to do a service for me." better :)

11:31 PM
@acl After staring at this for a while trying to see the hidden complexity, it looks like nothing more than Total[a b, -1] -- am I still missing something?

@Mr.Wizard awesome

Seriously? That's all?

apparently

LOL -- damn, talk about making a mountain out of a mole hill. I was sure it wasn't that simple.
I'm gonna get some food. See ya later.

bon appetit
I guess this would be more involved if it was something like this
well here you could transpose and do what you did, but with more indices, contracted with non-corresponding indices, it would probably be too much to keep in one's head at once

11:36 PM
This may seem kinda sophomoric but how do you find the Min value of an interpolating function? I have an interpolating function in x y and t. With FindMinValue what values t` must I use (the inter. func. doesnt just have 3 rows so I am dumbfounded)?