Mathematics - 2018-11-28 (page 4 of 4)

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loch

23:00

urop?

MatheinBoulomenos

@LeakyNun maybe something related to Langlands? that's a huge field and you have experience with that

user330477

Where is the complex square root function analytic?

Rithaniel

Hmmm, I think this is correct, but electronic verification implies that I've screwed up a step somewhere.
$\frac{z+1}{(z^2-2z)}=\frac{1}{z}+\frac{3}{(z^2-2z)}=\frac{1}{1+(z-1)}-\frac{3}{2z}(\frac{1}{1-z/2})=(\sum (-1)^n (z-1)^n )-\frac{3}{2z}(\sum(\frac{z}{2})^n)$

Leaky Nun

23:18

@loch m3r

Rithaniel

Which seems to imply that the residue at 0 is 0, and the residue at 2 is -3/2

Semiclassical

$\frac{1}{z}+\frac{3}{z^2-2z}=\frac{z^2-2z+3z}{z^2-2z}=\frac{z^2+z}{z^2-2z}$

when it should be $1+z$ in the numerator

loch

@LeakyNun you can do m3r in y2 ? or are you just planning ahead lol

Semiclassical

You've got $\frac{z+1}{z(z-2)}$

Rithaniel

Hmmm, my steps were $\frac{z+1}{z^2-2z}=\frac{(z-2)+3}{z(z-2)}=\frac{z-2}{z(z-2)}+\frac{3}{z(z-2)}$

Semiclassical

23:21

oh, doh

your algebra is right there, mine was careless

@Rithaniel in what domain are you expanding?

you've got poles at z=0 and z=2, so that will matter

Leaky Nun

@loch planning ahead

loch

Rithaniel

I'm trying to find the integral along the simple contour of radius 3 centered at the origin.

So, I'm trying to find the residues of the two poles, sum them, and multiply by $2\pi i$

Is my approach not good for that?

I have a suspicion that might be the case, to be honest.

MatheinBoulomenos

your function has simple poles, finding the residue at a simple pole is rather easy

Balarka Sen

Hi @Sawarnik

Sawarnik

23:30

Hi @BalarkaSen

MatheinBoulomenos

so if $f$ has simple pole at $z_0$, then you get that $\mathrm{Res}(f;z_0)=\lim\limits_{z \to z_0}(z-z_0)f(z)$

Balarka Sen

How's it going

Sawarnik

Your semester exams over?

Balarka Sen

yeah

Sawarnik

So how's ISI?

I think IIT was a better choice for me :)

MatheinBoulomenos

23:33

@Rithaniel so for example the residue at $0$ is just $\lim\limits_{z \to 0} z \frac{z+1}{z^2-2z}=\lim\limits_{z \to 0}\frac{z+1}{z-2}=-1$

Balarka Sen

@Sawarnik ISI is bullshit but I think that's minority opinion

Glad you're happy with where you are

Rithaniel

Really? I plugged things into wolframalpha (which can be hit or miss, admittedly) and it claimed that the residue at $0$ was $-\frac{1}{2}$

MatheinBoulomenos

lol

I can't plug in $0$ into things

Rithaniel

Ah, yeah, I see the mistake.

MatheinBoulomenos

but at least the approach is right

Sawarnik

23:36

@BalarkaSen Interesting. Are you satisfied with your profs?

Rithaniel

Okay, but I was trying to find the expansions for the functions, but was unable to arrive at the correct result. So, clearly I need more practice taking series expansions.

Sawarnik

That's one thing I found lacking here. They are worse than school teachers even (as teachers).

Balarka Sen

@Sawarnik Faculty is extremely good here. But students are terribad

Sawarnik

@BalarkaSen totally opposite case here lol

Balarka Sen

Weird!

user193319

23:39

If $f$ and $g$ are paths in $X$ from $x$ and $y$ such that $\overline{f} \star g$ is path homotopic to the constant path at $x$, how does one show that $f$ and $g$ are path homotopic?

Balarka Sen

@Sawarnik I heard you had a chat with Nikhil. He's a good friend

MatheinBoulomenos

@user193319 is $\overline{f}$ the inverse parth of $f$?

Sawarnik

@BalarkaSen Yup, we are from the same state.

user193319

Yes, that's right.

Balarka Sen

Mhm

Sawarnik

23:44

He had contacted me when the results came out. Been in touch since..

moneydhaze

Hey there, is here anyone who may help me with some basics of neural networks?

Sawarnik

Going to sleep now, good night :) @BalarkaSen

Balarka Sen

Night, talk later

MatheinBoulomenos

okay, then you have an "equation" $\overline{f} * g \simeq \text{const}$, then you concatenate both sides with $f$ from the left and use that concatenation behaves nicely with homotopies

this uses the facts that concatenation is associative up to homotopy, respects homotopy and that the constant path is a neutral element up to homotopy and that $f*\overline{f}$ is homotopic to the constant path

user193319

Ah, I see. I guess what I'm trying to show is that concatenating both sides with $f$ is allowed and makes sense, but that's probably a very involved proof. Perhaps I should consult a book rather than try proving something like this on my own.

In fact, I just looked at Munkres' book and he uses several pages to prove these things.