polynomials are interesting because of their properties. :-)

What happens if you plot the derivative of a lambert w function.

Am I asking stupid questions? Probably.

Well, we know that $W_0(x)e^{W_0(x)}=x$ for $x>-1$.

Hi @TedShifrin. Where are you now?

Hi @Ted, @Mike, everyone else

So if I differentiate both sides w/r/t $x$ I get $W_0'(x)e^{W_0(x)}+W_0(x)W_0'(x)e^{W_0(x)}=1$.

Hi everyone

Hi Astyx

Hi all. Genoa, Italy for the night.

Astyx :)

@Semiclassical oh that's kinda interesting

Ted I miss your jumbles.

LOL ... BACK

I can simplify the second term there to $xW'_0(x)$ since $We^W=x$.

How was your travel from France to Italy ?

oops .... In two weeks.

Ah, so you already passed through France?

Left Nice and it was only a couple hours, Astyx. I had fun meeting you guys :)

So that gives $W_0'(x) = \dfrac{1}{(1+W_0(x))e^{W_0(x)}}$

Yup, Mike. 8 days in France.

(Ignore the second simplification I made, I'm not actually using it yet)

Hey everyone!

Where are you visiting next?

Macron hasn't demicated labor laws yet :)

If I multiply top and bottom by $W_0(x)$ and use $W_0 e^{W_0}=x$, that gives $W_0'(x) = \frac{W_0}{x(1+W_0)}$

Hi Demonark.

So in principle I can compute $W_0'(x)$ from knowledge of $W_0(x)$.

And that's enough for a plot if nothing else.

It all gets a bit complicated fast, though.

@Semiclassical I see.

Oh that's not too long then ! It was nice to meet you too

Now, I talked about the k=0,k=-1 branches

How've your travels been?

But in fact there's an infinite number of branches $W_k(x)$ with $k=0,\pm 1,\pm 2,\cdots$

@Daminark Hello.

That's not all that strange if you know a bit about how the log function works in the complex plane, though.

Namely, from Euler's formula one has $1=e^{0i}=e^{\pm 2\pi i}=e^{\pm 4\pi i}=\cdots$

So there's an infinite number of solutions to the equation $1=e^x$, all labelled by a particular integer.

Fun, Demonark, but tiring (still 7 more stays). I'll be glad to get home, too.

So even just inverting $y=e^x$ to get $\log x$ presents some complex analysis weirdness.

It's not surprising that things get even weirder for $y=xe^x$.

that's pretty cool.

Yeah.

Branch cuts...

In particular, with the log function the differences between the branches are just multiples of $2\pi i$.

And with inverse sine, for instance, the differences are just multiples of $2\pi$.

With $y=xe^x$, though, there's no simple relationship that I know between the branches.

And this article I was looking at supports that claim, and has some useful-looking remarks in its abstract to that effect...

...which it entirely fails to follow up on in the note itself >:(

@Semiclassical so if I do graph $y=-xe^x$ it does reflect it in the x-axis, but it doesn't change things much in the sense that it doesn't achieve what we're trying to do?

It's just not very well written at all.

@Dodsy Yeah. It's like doing $y=-e^x$ instead of $y=e^x$.

right.

The inverse function for the first one is $\log(-x)$, and the inverse function for the second is $\log(x)$.

Not much different at all.

Did you read the wikipedia page?

:P

I probably have? Lemme see if it has details on this particular point.

it's a pretty hefty read

my girlfriend would slap me if she knew I recommended the wiki page.

lol

For math pages I think it's pretty defensible.

Hey!

Question: Why is sin(x) = y coordinate?

It doesn't seem to cover the issue I'm interested in, though.

cs.uwaterloo.ca/research/tr/1993/03/W.pdf

PS: i have the unit circle diagram

have you read this Canadian article?

@Dodsy heh, I already had that open in another tab :)

You'll have to read a lot of "eh's"

There's at least one name among the authors there you should take note of, though

namely, D.E. Knuth :)

Can someone please answer?

aka Donald Knuth, the person who created TeX.

Hm interesting...

How's it going @Dodsy?

I wonder why he is listed on a paper from Uwaterloo

Well, presumably because he collaborated on it?

And lol @Ted, for sure. Enjoy!

@Daminark It's going well! Semiclassical helped me study for my functions test, so I am pretty confident that I did well!

I mean, take a look at the locations the authors are associated with

@Semiclassical hahahaha that would make sense!

Nevermind, I understood.

@Semiclassical OH I DIDNT EVEN NOTICE THATTTTTTT

The first school listed, I applied to.

hah, nice

@Abcd: By definition! Draw the right triangle.

I find out next week.

Ted might be back by then

It is a bit random.

I wonder what the story behind it is.

Behind the paper?

Like, why there are so many collaborators at such far distances?

Yeah, behind their collaboration

it's interesting.

also, for a more accessible introduction, see here: americanscientist.org/libraries/documents/2005216151419_306.pdf

@TedShifrin Thanks. I understood.

The stuff I'm interested in at the moment has to do with Figure 4 in that article

Does it say the article is from 1993?

@TedShifrin DOes this work for non-unit circles too?

$\mu$

This is interesting stuff, though over my head quite a bit.

Right.

There's a bit of the history in the SciAm article, though not much.

i had an exam 2 days ago, i wanted to say thanks to everyone that helped me.

$\LaTeX$

How'd it go? @AbdullahUYU

how do i get latex to work

Use the 'Latex in chat' link in the room desc.

@Abcd: you need the radius, but otherwise yes.

passable, i give feedback when results are announced.

mmkay

lmfao.

I wish I had have studied trigonometry a bit more before my test.

But @Semiclassical I found out that my calculator can do compound angle questions.

nice

and can do unit circle questions.

My book says: $cosec (x) = \frac1{sinx}, x \ne n \pi $

actually it is a university admission test

$\alpha$

@Semiclassical I did something really weird though, I converted to an angle theta, then converted back to radians, then subtracted the two.

Why does it include; x is not eual to n pi

?

n is any integer

its not working

I believe the question was $cos(x+y) = cos(\frac{15\pi}{4})-cos(\frac{10\pi}{4}$

Because they don't want to divide by 0, @Abcd.

Not sure what that equation is supposed to mean.

Oh sorry.

It's not important. :)

I mean, $\cos(15\pi/4)=\cos(3\pi+\pi/4)=-\cos(\pi/4)=-1/\sqrt{2}$ and $\cos(10\pi/4)=\cos(2\pi+\pi/2)=\cos(\pi/2)=0$.

So that gives $\cos(x+y)=-1/\sqrt{2}$.

But that only constrains $x+y$, not $x$ or $y$ by themself.

hm.

the angles were 60 and 30 I believe. I don't even remember.

Weird.

@Abcd you could have pinged me next time

@TedShifrin How does it lead to division by zero? Sorry, I am new to these concepts

@LeakyNun TYSM :)

@Abcd try to substitute $x=\pi$ to the equation

@Semiclassical maybe I got that question wrong too.

@LeakyNun Done. What next?

:/

:D

@Abcd what do you get?

I'm just kidding, I nailed it.

My memory is just terrible.

mmkay

@LeakyNun cosec n pi = 1/ sin n pi

Damn smoke, Nate ...

@Abcd can you evaluate both sides?

and I said $x=\pi$, not $x=n\pi$.

I don't think smoking has much to do with memory.

@TedShifrin The smoke and the bubbly.

Who knows ...

@LeakyNun How?

I've hit my head a few times too.

@LeakyNun Ok

Not cool, Nate.

@Abcd what is $\sin \pi$?

It'd likely take a while for that to set in but shrugs

Head hitting, not fun

You need brain cells for math(s).

@MikeMiller You have terrible memory though. (So do I, but)

I suspect I have a selectively bad memory :/

@LeakyNun Oh. sin 180 deg = ? I am new to radians. But what's sin 180 deg?

$\mu$

I have an amazing memory for everything I have no need to remember.

^

@Abcd never mind, just substitute $n=0$ then

@BalarkaSen At least, that's not from tobacco.

I can remember licence plates very well.

Hehe, indeed

@LeakyNun undefined

@Abcd exactly. Does that answer your question?

My memory is great if I focus. For example, the degree of a map between manifolds is its derivative at zero or something, right?

I'd insert some jocular comment re: smoking weed, but uh

I've never actually smoked it.

@LeakyNun whats sin pi?

$\alpha$

@Abcd 0

Right, Semi, we all believe that.

You're like weedy harrelson.

I don't know what that means :/

@LeakyNun what's sin 2pi or sin 360 deg?

I read that as weedy.

The actor.

@Abcd 0

@BalarkaSen fixed

@LeakyNun How is sin 360 deg possible when we dont even get a triangle with 360 deg?

$\alpha$

@Abcd ah... so this is where your problem is.

oh

@Semiclassical I'm just kidding (my jokes aren't very good today)

hah

@Abcd are you following a textbook or a course or what?

I used to smoke weed in high school.

Some people can be highly functioning on that stuff.

@LeakyNun Grade 11 textbook.

blargh. My advisor hasn't been here in days, and none of his other grad students know where he is either :/

Canada is starting to take a more liberal view on it.

@Abcd does your textbook talk about values of $\sin(x)$ where $x$ is not an acute/obtuse angle?

Very specific parts of the US are starting to do so as well.

@LeakyNun NO

Though the leading edge of that is medicinal use rather than recreational.

The fact that weed is criminalized has always struck me as deeply absurd.

Lol so I smelled weed for the first time recently and was just like ugh, this is why it's illegal, it smells worse than tobacco

haha

@Abcd do you have 7 minutes?

@LeakyNun Yes

oh it was $cos\frac{4\pi}{12}-cos\frac{2\pi}{12}$

hmm, so $\pi/3$ and $\pi/6$

right.

I am not used to Windows 10... this has too much stuff for me to handle

it's better than windows 8.

They should probably take the NY smoking regulations and just apply that to weed. It basically seems to be hallucinatory tobacco or smth

that was garbage.

I am going to disable and remove all the extra stuff till it starts looking like XP or something lol

@Daminark why can you basically smoke it anywhere?

@Abcd I was going to link you to a 7-minute video, but then I decided that the video isn't really about this.

So I'll just teach you here.

ok

are you ok with that?

NY puts heavy restrictions on where you can smoke, since the second hand fumes are also very dangerous

Of cigarettes?

@LeakyNun yes.

Except does second-hand marijuana smoke actually cause cancer?

I don't know about that but it smells p bad so...

One thing I really hate, is seeing people have those tobacco vapes in public.

And yeah cigarettes @Dodsy

Iam stuck on something, probably simple, it is in a proof: Suppose there is a parameter $\psi$ of interest that can be expressed linearly in the canonical parameter $\theta$, so we may represent $\theta$ as $(\lambda,\psi)$.. suppose $\theta = (\frac{\mu}{\sigma^2},-\frac{1}{2\sigma^2} )$ can we express $\sigma^2$ linearly in $\theta$? can we express $\mu$ linearly in $\theta$ , how? can some one showme

They walk around acting so cool with their vapes

@Abcd consider a unit circle centered at origin.

tucked into their hand, taking puffs every 5 seconds.

k

Marijuana smells kinda terrible

@Abcd can you see the angle θ?

and the point C = (-0.8,0.6)

@LeakyNun yes

the angle θ is defined to be how much you must rotate anticlockwisely from the positive x-axis to get to that point

@Daminark Right, Canada is very regulated about that stuff. You cannot smoke in any public building, on any patio, or within a certain meter of a building. (cigs) But if you go to places in the states, people are smoking in Casinos, in bars, etc.

hence why I drew an arrow @Abcd

@LeakyNun Ikr

if the point C were in the third quadrant, the angle would be larger than 180 deg

yes

question: why is C on the unit circle?

I think it's like that everywhere, but NY is more severe, you can't be within 100 feet, I think

@LeakyNun 0.8^2 + 0.6^2 = 1

guys.

I'm teaching @Abcd.

Or at least more severe than other states, not sure about Canada

@TheGreatDuck please delete.

@LeakyNun Idk

@LeakyNun. You might want to create a new room, then. Otherwise we're likely to walk all over each other.

@LeakyNun I didnt understand great's stuff

@Semiclassical do we agree that this isn't the first time when @TheGreatDuck is trolling?

what?

you asked why it was on the unit circle

I'm not going to touch that.

He didn't ask you

@Semiclassical Me either.

meh

@Astyx and that means I'm not allowed to answer?

Not this again.

@Daminark I see.

nvm

ffs I can't answer a fricken question?

To be fair, if you don't want other people to get involved in your conversation, you can create a room easily enough.

^^

Getting annoyed at people interjecting on the main room just strikes me as silly.

I haven't seen Sha in a while

I saw her here yesterday.

She's still around

Is all well with her?

@LeakyNun I'd prefer you delete this. That's just rude.

She might be studying or taking an exam?

I see.

I know she was making a lot of prep for optics

@Semiclassical you don't think I was trolling do you? I mean... it seems like that's what she was asking?

Nah. Trolling implies malice. I don't see any of that.

@LeakyNun I hope you don't mind if I just sit in on the room, I just want to listen in, I will not interject.

I do think you could be a touch more restrained about what conversations you interject in :)

wat.

(removed)

I was sitting in chat and someone posted a random comment asking why something was on the unit circle

Well, that was part of an ongoing conversation.