@robjohn No, I don't think so. I guess the takeaway is that the chat works reliably and as expected. The main site, otoh...

Yes again.

I don't get the second step in your unimaginative solution, what's the message number @Srivatsan?

It's the number following the # in the permalink. I am not sure that's the official terminology but that seems to be a logical name.

@Gigili Your message has the permalink http://chat.stackexchange.com/transcript/message/2546179#2546179. I mean the number 2546179. So I would write :2546179 Hi Gigili.

@Gigili Hi

I just noticed that Ninefingers is a mod.

Did you see this: meta.crypto.stackexchange.com/questions/113?

@Srivatsan How exciting, thank you.

They copied Asaf's idea... =)

@Srivatsan Are you talking about the online lecture, or did Asaf suggest a meet the moderators event?

Online lecture idea.

Okay, this sounds more like a chat when the mods will be present. I imagine they could have gottent the idea from Asaf, however.

Now sure how we could find out.

I forgot to answer the question that was actually asked. I just fixed that. :-)

By the way, your complex conjugate answer is excellent.

@Srivatsan ask a moderator?

@Srivatsan I actually just added the answer to the question :-)

Where to learn more of such goodness?

I don't know... I actually just figured that info out for myself over the last couple of days.

@robjohn Cool... :)

We were very busy running around town, so I had some time to cogitate.

@Srivatsan Useless to me, but cool :-) [unless I ask a question, of course] I was hoping there was some text of some sort...

@Srivatsan If I find something, I will let you know. I myself have been unsettled about the notation, so when I saw the question, I felt I had to figure it out, for myself as well as the OP.

Thanks.

I have never seen a decomposition of a general function from R^2 to R^2 into the conformal and conjugate conformal before.

There must be some literature on it.

@robjohn: That answer is excellent. It has made completely clear in my mind what d/d\bar{z} is.

@ZhenLin Thanks :-) that is why I wrote it, because I was unclear on the exact relationship before.

@robjohn Can you do such a decomposition globally? Your answer explains that it can be done at a single point, but it's not clear to me that you can make it work for all points at the same time.

@HenningMakholm That is something I will have to investigate. That was beyond the scope of the question, but it is something that I was wondering myself.

I think I once saw a question that asked how to prove that there is a global decomposition.

I thought about it for some time without finding an argument that it would have to exist.

@robjohn I am slightly slow in such things. What conditions do we need the function to satisfy, even for decomposition at a point?

All partial derivatives must exist and be continuous in a neighborhood, or something like that.

That the four partials exist. The rest is pretty straightforward linear algebra.

@HenningMakholm for a single point, I don't think they need to be continuous.

Would |z|^2 qualify? Seems like it would.

It would indeed.

If it ain't too much trouble, can you do this example in a bit more detail? What does the decomposition look like?

u=x^2+y^2, v=0

so...

@robjohn The partials themselves only tell us how the function behaves on the axes. We'd need something to guarantee that it is not completely wild off the axes.

I don't know why I picked that example. =) Unsurprisingly, it's quite mundane.

my understanding is that to get the conformal parts is that you do a change of variables (x,y) <-> (x+iy, x-iy)

so instead of looking at d/d(x,y), you look at d/d(x+iy,x-iy)

@HenningMakholm That is true if we want to expand, but for the decomposition of the partials at a point, I don't think it is necessary. I could be wrong.

and you go from one to the other by using the matrix d(x,y)/d(x+iy,x-iy)

for |z|², you should get d|z|²/dz = \bar z, and d|z|²/d\bar z = z

@robjohn Yes, I was assuming that we wanted to expand f(z+h) = f(z)+Ah+Bh*+o(h) for h small in general.

If we're satisfied with decomposing the Jacobian, then that is of course purely a matter of linear algebra.

@HenningMakholm Yes. I was thinking just about the decomposition into conformal and conjugate conformal once you had the partials, but if we want the conformal etc to really be conformal, then yes, we need continuity.

@HenningMakholm That is exactly what I was thinking about :-)

Does anyone know how to rank answers according to length. I wanna check where mine stands and how much more I should write to get to the top [just in terms of length]... =)

Can someone proof-read my answer here please: math.stackexchange.com/questions/86148/…

@Srivatsan you mean you want to write a book here ?

New library this time. It's 5 stories of books but I checked and the math section is literally one row and two feet wide. Yeesh.

I just fancied doing some Stokes...

I should know it from Analysis II but that's one of the things I didn't have time to look at.

So I thought I'd learn it today : )

Also, the only thing higher than high school level was MacLane and Birkhoff's Algebra, which I checked out. I strongly suspect the library buyers simply mistook it for an easy book.

@Matt: You can make matrices look good with \begin{pmatrix}

Ah, thanks, @anon!

That is for |z|^2

@Matt, I answered to your question.

@AsafKaragila: Aces! Thank you!

@Matt: you forced him to answer :-)

: D

Can someone explain Phira's comment here: meta.math.stackexchange.com/questions/3256? I am not sure I follow the reference (is that a proverb?). Checking the dictionary now..

I guess Phira is simply saying that you will lose the points anyway, it is just your fault if you do a recalc and lose the points sooner.

I think I'll have to look at it tomorrow though. It's late here and I have to be at uni at 9 tomorrow because the buggers concocted a series of "short tests" that you must attend in order to be admitted to the exam. It totally wrecks every other of my Mondays because I have to get up so early : /

Ah, I surmised correctly, and also replied: meta.math.stackexchange.com/questions/3256/…

Are you asking about Fool me once, shame on you; fool me twice, shame on me?

Amusing that we used very similar words, @robjohn.

@robjohn I think I understood now. It's not a very natural thing to say, where I come from. I guess it is more common to regard it shameful to fool a person repeatedly, whereas the fooled is usually not chided for getting fooled. =)

@Srivatsan It is sort of a modern day proverb at least in the US from what I have heard.

Good night folks, it has been a pleasure talking to you (as always) ; )

@Matt Good night!

I think he's tacitly assuming the "you" instictively takes the ostensible rep count for granted, hence is being fooled once by autorecalc and he learns twice after a manual recalc

@Srivatsan I guess it is the idea that we should learn and be wary the next time.

A very cynical pov I guess.

@anon Aw, that is a nice interpretation. I am not attaching significance to rep, but the rep got added and taken away for a reason; I want to know that reason, that's all.

@robjohn OTOH the other pov is all too forgiving. There are pros and cons for both.

@Srivatsan: Phira once posted a request not to have mod tools at 10k, which is quite the silly request I'd think.

@AsafKaragila When would Phira like to have mod tools? never? earlier? later?

learning to hate internet explorer ... redux

@robjohn Never, of course. Otherwise the request is just plain silly.

@AsafKaragila @meta?

Indeed. Look at her user profile, it's there somewhere.

Or just wait for me to do that for you...

This one: meta.math.stackexchange.com/questions/2239. Thanks, but I found it myself too. =)

@AsafKaragila Well, I am silly, so I had to ask :-)

Sure. :-)

@AsafKaragila Is not that related, but why "her"?

Phira has stated somewhere on the meta that she is in fact a female.

Myself and J.M. talked about it yesterday, you can find it somewhere in the log.

Uhum, I will.

Anyway, I am going to bed now. Be seeing you folks.

@AsafKaragila Sleep well.

Wow. My answer about d/dz and \bar{z} is almost a "nice answer" already.

Am I the one who is reading the question wrongly?

See R.'s comment under it.

@robjohn It deserves it.

Not sure why, but I liked many of your answers this week in particular.

@Srivatsan How is it you are reading the question?

That's kind of summarised in my second comment.

@Srivatsan Thanks. Of course, now my head may not be able to get out of the door to go to dinner :-)

reading the second comment

The question is clear: give an interpretation that justifies (1), (2), (4), but still (3) is incorrect. My stand is that one needs to look no further than the set-of-functions interpretation, provided one interprets equals to mean set containment (either \in or \subseteq -- whatever is appropriate).

but does = mean only one direction of inclusion?

That's not to say that it is simple to use. We all have struggled, we have seen our children struggle. That's almost a rite of passage.

in other words, if a=b, does b=a?

@robjohn Yes, only \in or \subseteq. Never \ni or \supseteq. Just like O(n) = O(n^2), but the other way round.

The human centipede 2 must have been the worst film I have seen in ages.

Okay. Then this is only for O and not \Theta.

@JonasTeuwen It must have been, if you say so :-)

It's just that someone decided to say apples=fruits instead of apples \subseteq fruits.

@Srivatsan FWIW, I think your answer answers the question as asked. I don't really like that interpretation intuitively (it seems to me to give too short shrift to the usual intuitions about equality and equations), but it appears to be technically consistent.

@robjohn Shh... Keep your voices down. Henning might hear us. =)

Your solution can handle nested big-O's, which mine can't. Whether or not that is an advantage is somewhat debatable.

@Srivatsan I'll keep the ones down whose names I know... they keep changing.

@HenningMakholm I think that nested big-O's might be a good thing :-D

@HenningMakholm Um, I see that point as well. Some time in the past, I took some effort to understand the notation and get it straight. It grew on me.

@robjohn I intended the question to cover Theta as well as O, Omega, o .. the works.

@HenningMakholm I know that was your intent, but the one-way inclusion doesn't work for transitive relations such as \Theta.

Wait, I take back my meek stance (huh) w.r.t. Theta, @robjohn. I suspect it might work for \Theta as well.

@robjohn Can you elaborate on that?

@Srivatsan okay, I will have to read again.

@HenningMakholm I have to read Srivatsan's post again. I may have missed something.

@robjohn No, my answer does not talk about Theta. I am just saying that for Theta it might not matter whether you take = to mean equality or set containment (inclusion/membership).

@Srivatsan You need inclusion for "5x² = Theta(x²)", I think.

Let's see. Say Theta(f(n)) = Theta(g(n)). So, under my definition, f(n) \in \Theta(g(n)).

@Srivatsan Okay. That is reasonable, I think. Since O(f) represents a class of functions, g \in O(f) has to mean inclusion.

@HenningMakholm I will conflate \in with \subseteq and call it containment. It should be clear from the context.

Further, f \in O(g) is the same as { f } \subseteq O(g), so that's not too much to ask, I hope.

Even more: x³ + Theta(x²) = Theta(x³)

Yes, the Theta(x^2) on the left means an element of Theta(x^2)

@Srivatsan My point was that equality wouldn't work there. Not distinguishing strictly between inclusion and containment here.

@HenningMakholm Ok. You're right. I'm getting ahead of myself. Let me refine the claim: containment is a right way to interpret. It covers all cases and works as you would expect...

It tells us that x^3 + Theta(x^2) = Theta(x^3), and it tell us that the other way isn't true. Check, Check.

Another test is this: we would expect Theta(f) = Theta(g) to imply that Theta(g) = Theta(f). I'm claiming that this also work, thanks to our dear old Theta.

@Srivatsan I think that is a Theta-specific fact, namely that any two Theta classes are either disjoint or identical.

@HenningMakholm Yes, that is more to satisfy @robjohn. He raised this point implicitly here: chat.stackexchange.com/transcript/message/2546637#2546637

I am wondering if he is convinced now.

@Srivatsan but I thing I rescinded in chat.stackexchange.com/transcript/message/2546685#2546685

I read f=O(g) to mean that f \in the class of functions that grow no greater than a constant multiple of g

@robjohn Sorry, I didn't know how to read that comment.

Plus, this is all happening a bit too fast for me. For my typing to cope, I mean... =)

and f+O(h) = O(g) I see as f + any element of the class that doesn't grow any faster than a constant multiple of h doesn't grow any faster than a constant multiple of g

Yep, precisely.

so the = is non-symmetric and essentially means \in

@robjohn Oh no, it is transitive. That's perhaps the most important property of =. Not symmetry, it seems ;)

and O on the left means different than O on the right

Good night guys.

Yes, Henning's answer makes that clear--you will need to use universal and existential quantifiers respectively.

@Srivatsan I mis-typed

@JonasTeuwen Sleep well, @Jonas.

@JonasTeuwen nighty-night :-)

Bye :-).