Mathematics

2:01 AM

tuw lol

2:51 AM

Any ideas as to why this might've gotten downvoted? It's entirely possible I made a mistake, but I'm not sure what... math.stackexchange.com/questions/870265/…

3:03 AM

possibly somebody didn't take a liking to the fact that you gave a good answer to a homework question

@lolwut, I'm starting to think that might be it. I try not to give away everything, but sometimes that's difficult. I'm trying to think of how to cut it down.

4:07 AM

@KajHansen You look like Matt Damon.

Why is there a game theory SE in creation?

.
<--

What does that mean?

Are you Bart?

no, I am not that eyeglasses kid

4:12 AM

Ah, he has vanished from SE.

I will tell you this. My real name has appeared somewhere at some time yesterday.

Very mysterious.

Can someone please help me with Linear Algebra Done Wrong?

I can't understand why author has chosen "wrong" in its title.

@Sush There is a text called Linear algebra done right :-) ...

@r9m, there is another text called LADW See this

4:21 AM

@Sush ya I know ... :P but I don't know why the author chose that title :P

We will have to use it as textbook in this semester. So I want to know what is 'wrong' there.@r9m

@r9m, ok!

I read its preface, but nothing is mentioned why is that wrong!@r9m

@Sush there is nothing wrong in that book as far as I can remember :P

@r9m, :D

So, why the title has "wrong"?

idk :| ..

;|

4:27 AM

linear algebra done like a boss by lol wut. look for it in bookstores near you

@lolwut XD ... you use the name lol wut even for your books ? :P

yep!

significantly better sales than when I use my real name

XD !

(which is so unfathomably gross that I shall not say it here)

fair enough :-)

4:46 AM

What happened to Mr Eyeglasses?

@JasperLoy, someone else told me that on here not too long ago, lol

@KajHansen Oh, might have been me as well. You are hot, lol.

Why thank you :P

Contemplates posting a picture of myself just to see who Jasper compares me with..

I shan't post my pic here, because I look fat =)

4:54 AM

That's an easy thing to fix.

I put on 10 kg the past year.

As a consolation I tell myself that is because of the meds I am taking.

yeah, medications for psychological problems can be particularly problematic. how's your sleep?

@lolwut, are you an older user under a new account? Same question @JasperLoy. Usually there aren't too many lower-rep users hanging out in chat.

Well, I sleep and eat irregularly, which is OK since I don't work now. I hope to get well enough soon, but that may never happen.

(No offense of course)

4:58 AM

@KajHansen I deleted all my old accounts. I delete accounts regularly, lol.

Ah, that explains it.

Many people in this chat know me. You are considered a newbie, lol.

@KajHansen i don't participate much on main

@JasperLoy I gained 8 kg in last 2 months .. I ain't joking :|

Oh definitely. I just joined at the beginning of the summer thereabouts.

Something like that.

5:00 AM

prefers trolling a math chat room to answering/asking questions

2

I have deleted my account so many times and changed my username so many times that people no longer know who Jasper Loy is.

Yeah, I can see how answering questions can get tiring.

Before I put on weight, I look like Justin Bieber, haha.

If I regularly came up with very good questions, I would definitely be asking them here. Unfortunately, I only come up with questions that either (1) I can answer in a couple of days if not sooner or (2) are just plain too difficult leading me to lose interest quickly

@lolwut, where are you with your education?

5:07 AM

the equivalent of a small child

hahaha

I finished undergrad a decade ago. Hoping to go grad school when my mental illness is well enough...

That's unfortunate @JasperLoy. I had really bad depression about 3 years ago. Flared up again a year and a half ago. It's finally gotten better recently.

@KajHansen I have OCD. Been suffering over a decade. Life has become meaningless. I just hope that my next life is better, since I believe in rebirth.

Damn. I couldn't imagine what that's like...

5:22 AM

...unfortunately, I can do more than imagine.

But life goes on, as they say.

With or without meaning.

John Smith

5:53 AM

In categories what does hom(A,-) denote? hom(A,B) is the set of all morphisms from object A to B. Is hom(A,-) the set of all morphisms from A (to any object, including itself)?

Helllllllo?

Sawarnik

8:55 AM

@Anthony Hello.

@KajHansen Oh. Why did you go into these depressions?

@Sawarnik Hey.

Could you help me with my complex analysis homework?

$i^2 = -1$.

:D

Does that help?

No D:

9:02 AM

What is it you're having trouble with then ?

Showing the following

Can't you just use the theorem of residues and the standard estimate for line integrals?

I haven't learned about residues yet :/

And I'm kinda hazy on this stuff, what standard estimate?

$$\left| \int_\gamma f \right| \leq L(\gamma) \cdot \max_{z \in \operatorname{im} \gamma} |f(z)|$$

I don't really remember to be honest. :(

Whhhhhhat. What's L?

9:14 AM

The length of the line.

But I think you need to work around the poles somehow first. That's why I was thinking of said theorem.

There's just a pole at zero, right?

Yes.

I don't think that's a pole though, thinking about it.

Because of the logarithm.

9:33 AM

Oh.

Sawarnik

@Anthony I know nothing about it actually :D

D:

@DanielFischer Might I bother you again?

@Anthony I don't know. Might you?

Please

PLEASE

0

Q: Bound in Complex Analysis

Can someone direct me towards the right way to approach this problem? Show $$\displaystyle \left|\int_{|z|=R} \frac{\log{z}}{z^2} dz\right| \leq 2\sqrt{2\pi}\frac{\log{R}}{R},\; \text{ for } R>e^{\pi}.$$ Edit: I'm considering the ML estimate, so L would be $2\pi R$ and I'm still debating M. The ...

This problem. @DanielFischer

9:49 AM

Hm, @Anthony, the ML estimate doesn't give you the desired factor of $2\sqrt{2\pi}$, it only gives you $2\pi$. So we need to look at the integral in more detail. Nothing obvious seen so far, however.

Ugh.

Oh shoot.

I have a small typo.

@DanielFischer: I don't quite remember how to solve these kind of problems. How would I compute the residue of $\log z / z^2$? It's an essential singularity at $0$, right? But the logarithm doesn't have a Laurent series, no?

@DanielFischer I ammended it, it was the Principle Log.

On the left.

@Huy It's a branch-point, there is no residue. Only isolated singularities have residues.

I guess that doesn't really matter.

9:51 AM

Oh right.

@Anthony The choice of the branch-cut matters, the chosen branch for that cut doesn't, since $$\int_{\lvert z\rvert = R} \frac{dz}{z^2} = 0.$$

Oh no, I was saying it didn't matter that I forgot to write Log, as opposed to log.

Ugh I have no idea how to do this problem...

10:09 AM

Well damn, I got really close.

@Anthony You can explicitly evaluate the integral. That's easy in this case. You get a strictly better result, though, and I can't see a natural way to get the given bound.

@Anthony I looked at your .png. You misread the bound. The factor is $2\sqrt{2}\,\pi$. Then the ML estimate plus Julian's answer gives you the result.

Oh.

Crap.

I'm sorry.

That makes so much more sense.

@DanielFischer Thanks for your help.

Karl Kronenfeld

10:33 AM

I can't help but laugh. It appears that someone is downvoting all of my answers in which I am a smartass (e.g. this)

Alizter

@KarlKronenfeld maybe he thought of the same snappy answer and he is jealous.

Karl Kronenfeld

@Alizter I figure someone believes that my answers are bringing down the value of the site (not that those particular questions are any good) and wants to express it through downvotes.

the system will catch'em

Karl Kronenfeld

I don't care if he/she is caught. I just found it funny which answers were targeted.

@skullpatrol At the rate of one downvote per four days?

10:45 AM

@DanielFischer Pardon me, I didn't know the rate.

@KarlKronenfeld my answer was down voted on that answer too >8(

12:21 PM

@lolwut here :-)

Mats Granvik

12:45 PM

Anyone good with Mathematica and q-factorials? I would like to know how to program this sequence with the description given. https://oeis.org/A127926

The Pari program is a bit terse to understand:

(PARI) {a(n)=if(n==0, 1, polcoeff(1-q- sum(k=0, n-1, a(k)*q^k*prod(j=1, k+1, (1-q^j)/(1-q+q*O(q^(n-k))))), n, q))}

1:42 PM

What is the convention of the notion $C(I)$ for some closed interval $I$? Is it $C^0(I)$, i.e. the space of continuous functions on $I$?