Mathematics - 2017-06-08 (page 3 of 3)

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Semiclassical

18:00

I may be able to turn it inside out to fix it, but if I'd have to re-tape it I won't bother

Fargle

Or just pretend you meant to do -log from the beginning.

Semiclassical

Lol

Balarka Sen

post a snapshot

Semiclassical

I will, after I redo it. Messed it up while trying to fix that

Balarka Sen

ping me with it when you're done

Semiclassical

18:04

It's for the Lambert W function btw

Balarka Sen

ah. don't think i remember how it looks, all the more reasons i need a snapshot

Semiclassical

Yeah

To see why it's weird, see the presentation of branch points here: functions.wolfram.com/ElementaryFunctions/ProductLog2/04/04/01

You can also find the discussion of branch cuts nearby on that site

N3buchadnezzar

Oh hi

Abcd

How do I solve this: (x^2 - 9)/(3x - x^2 - 24) < 0 ?

N3buchadnezzar

Cry

Alternatively factor both numerator and denominator and draw where the numerator and denominator are positive and negative

Balarka Sen

18:20

lol

Semiclassical

Ok, here's my attempt to take a pcture

i.sstatic.net/lpG2N.jpg @BalarkaSen

That's intended as branches -1,0,1 from bottom to top

Balarka Sen

Cool model

Semiclassical

If I wind around the edge of the three surfaces, I'll go -1 to 0 to 1

Balarka Sen

yea

Semiclassical

But I can also go directly from branch -1 to 1

That's the piece of paper which lies between the pair of double arrow

Balarka Sen

18:31

Semiclassical

Both of which are on the 0 branch. (The arrows are to remind me that said piece of paper doesn't stop me from moving from one side of the 0 branch to the other (the branch cut on this surface starts at -1/e.)

Balarka Sen

Right, I understood that. It lives as an embedded surface in C^2

Semiclassical

That means that the branch point at -1/e is of order 2.

so if I start to the right of -1/e on the 0 branch and make two revolutions around that point, I'll go 0 to -1 to 1 to 0

Which is kinda goofy!

Fargle

The Lambert W function has caused me nothing but confusion. This is no different.

Semiclassical

that's about how I feel tbh

But this may be enough to do what I want.

Namely, there's a difference between this and the log case. There, the difference between log(z) above/below the cut is $2\pi i n$ depending on which branches of log i compare

So if I start at a point on one branch and analytically continue to the same point on another branch, I only have to worry about the winding number

That's not true in the Lambert W case, since I can wind around the -1/e branch either once or zero times

Hence there's an additional ambiguity for analytic continuation, and that's exactly what I want

( I guess I'm going from $\pi_1=\mathbb{Z}$ to $\mathbb{Z}\times \mathbb{Z}_2$)

pjs36

19:19

Shot in the dark here, but: Anyone on that happens to be familiar with writing things to a CSV file, in Python? I have a list of 6,240 "objects" that took a couple hours to generate, and are going to vanish unless I figure out how to record them somewhere...

pjs36

19:49

Downgrade to threat level orange: data appears safely recorded.

Dodsy

20:30

haha

glad to hear you figured it out man

Akiva Weinberger

21:09

Is there an official mathematical definition of "spiral"

that includes logarithmic spirals and helices

Tanner Swett

Do you want the definition to include all kinds of spirals, or is it fine if it mostly covers just those two?

I came up with a sort of "joke" the other day...

Lozansky

What does $\int d^n x$ mean?

Tanner Swett

Mathematics is purely a figment of our imagination; it is not a part of reality. It is, however, a conservative extension of reality.

Lozansky

Does it mean integrating a differential of $n$ components?

Akiva Weinberger

Meaning, whenever mathematics proves something about reality, it's true in reality?

Lucas Henrique

21:20

Hello!

Tanner Swett

@AkivaWeinberger Yeah.

Actually, I have a better phrasing. Mathematics is all made-up; it's not really true. It is, however, a conservative extension of the truth.

zed111

21:36

Anyone here knows the meaning of "normal trajectory of a surface"? (from Differential Geometry)

Hippalectryon

22:33

@TedShifrin @Astyx @LeGrandDODOM A propos de l'endroit exact où se retrouver, je propose le parvis de l'église Notre-Dame-des-Champs

Astyx

22:50

Ça me va

Hippalectryon

@Astyx On se fixe un horaire ? 12h30 par exemple ?

Daminark

Hey everyone!

@TedShifrin you're back! If only for a bit

Semiclassical

hi chat

Fargle

Heya @Semi

Lozansky

Why is it that $\mathcal{B}^{2n}_r(P)$ approaches a square as $n$ increases? This is in $\mathbb{R}^2$

I.e., why does the set of points $\{(x,y) : (x^{2n}+y^{2n})^{1/2n} < r \}$ describe an open "square" as $n \to \infty$?

Thanks!

Do you see why, @Semiclassical?

Semiclassical

23:10

1) you can focus on the case of $r=1$ since you could change coordinates to $u=x/r$, $v=y/r$ and get $u^{2n}+v^{2n}=1$.

Lozansky

Sure

Semiclassical

2) The curve has 8-fold symmetry (symmetric across the axes and upon exchange of y=x), so it's enough to focus on arc where $0<x<y<1$.

Lozansky

Hmmm

8-fold symmetry?

Semiclassical

Sure. It's symmetric w/r/t $x\to -x$ so it's got a reflection symmetry across the y-axis.

Similarly w/r/t $y\to -y$.

Finally, it's also symmetric if we swap $x,y$ (i.e. reflection across the line $y=x$.)

(Those in turn generate a further symmetry across the line $y=-x$.)

Lozansky

Ah, yes that makes sense

Semiclassical

23:16

mmkay

Now, note that $x<y$. Hence as $n\to \infty$ it'll be the case that $x^{2n}\ll y^{2n}$.

So for large $n$ it should be the case that $1=x^{2n}+y^{2n}\sim y^{2n}$.

So $y\to 1$ and $x\to 0$.

Lozansky

Ahh so we get a line

Semiclassical

Right.

Now, that's possibly a bit handwavey.

To do it right, one should probably solve for $y$ in terms of $x$, subject to $0<x<y<1$ since we're in the first octant as I indicated earlier, and show that $y\to 1$ as $n\to \infty$.

Incidentally, this idea shows up outside of geometry: In the realm of analysis, it's the basis of the uniform norm aka sup norm aka $\infty$-norm.

It means that the sup norm is a limiting case of a p-norm.

(Similarly for the inf norm)

Lozansky

Is it obvious where the line(segment) has its start- and endpoint?

Semiclassical

Well, when $(x,y)=(1,0)$ one always has $x^{2n}+y^{2n}=1$.

So that's always a point on the curve.

Similarly, when $x=y$ you've got $x^{2n}+y^{2n}=2x^{2n}=1\implies x=y=2^{-1/2n}\to 1$ as $n\to \infty$.

So the curve will have $(x,y)=(1,1)$ as a point as well.

From there, symmetry does the rest.

Okay, out for now.