The h Bar

user228700

10:01

Wow :-)

user228700

I need some homework-tsy help in Math for now...

user228700

I've been asked to check if the function $f(x) = x^2 + \ln x$ is injective and surjective or not.

user228700

To check if it's injective, I found $f^{'}(x)$, which came out to be $(2x^2+1)/x$

Secret

injective is easy, just assume there is $a, b$ such that $f(a)=f(b)$, plug them into the function, and see if you get $a=b$ in the end. Alternately, as you said, check that the inverse function $f^{-1}$ is such that $(f^{-1} \circ f) (x) =x=(f^{-1}\circ f) (x)$.

user228700

For $ x \epsilon (-\infty, 0]$, $f^{'}(x) \le 0$

Secret

10:06

actually, is $f'(x)$ the inverse function or its derivative?

user228700

@Secret No, no, I'm checking the monotonicity of the function. I typed inverse by mistake because I'm not very fluent in LaTeX.

Secret

Ah I see

Indeed, if a function is monotonic, then it must be injective (i.e. have a left inverse)

user228700

Yeah, so I see that $x \epsilon (-\infty, 0] , f^{'}(x) < 0$.

user228700

And for $x \epsilon [0,\infty), f^{'}(x) > 0$

user228700

Dammit, I didn't mean to include 0. At 0, I guess the function has a vertical asymptote.

Secret

10:11

that's ok, it just means the domain does not include 0

user228700

Okay, so there seems to be no turning point for the graph...which is weird, because for awhile, it decreases, and then it increases. Huh :-|

user228700

Oh, wait.

user228700

Nope, nothing.

user228700

@Sec: Uh, u there..?

Secret

I am thinking a bit, because when I graph it, it seemed to be increasing without turning points

and also, thinking being complicated by wacky cases like as follows:

0

Q: Can the product of an increasing monotonic function and a strictly increasing monotonic function have turning points

Suppose $f:\Re \rightarrow [0,1]$ is an increasing monotonic function and $g:\Re \rightarrow [0,1]$ is a strictly increasing monotonic function. For the product function $h(x) = f(x)g(x)$, will there exists a local maximum? If the gradient of $f$ and $g$ are different, I think there exists a tur...

user228700

10:19

@Secret It first decreases and then increases. I mean, the derivative always increases, but not the graph itself...

Secret

Ok, I guess the better way to say is that, a monotonic function does not have minima nor maxima except for the boundaries

user228700

Wait, what does "monotonic function" really even mean?

Secret

It means given any $x \leq y$, $f(x) \leq f(y)$ or $f(x) \geq f(y)$.

in calculus terms, the function only increases or decreases throughout its domain

user228700

> "A monotonic function is a function which is either entirely non-increasing or non-decreasing. A function is monotonic if its first derivative (which need not be continuous) does not change sign."

Secret

Yes, that's true, if the first derivative change sign at some interval, it means it is decreasing for that interval

user228700

10:23

What? That means that the function $y=x^2$ is not a monotonic function.

user228700

Oh, wait.

Secret

yeah, $x^2$ is not monotonic, because it decreases for negative x and then increases again for positive x, with a stationary point at x=0

user228700

Nvm me, I'm just confused. All cleared up now.

user228700

Ow, I fell off my chair trying to grab a pen that fell down without getting up. And now my back hurts :-(

user228700

Alright, so functions functions, um...

Secret

10:27

@Kaumudi.H Ah now I see why it does not initially seemd to fit the definition of monotonic function: ln (x) is complex for all x<0. Since you are dealing with real functions here, the domain is $x\geq 0$ hence f'(x) > 0 for all x, and hence it is monotonic

user228700

Wait, so since the derivative first decreases and then increases, the function itself is clearly not monotonic so it's not one-one, correct?

user228700

@Secret Oh, yes! You're really smart, I missed that completely! Thanks so much! :-D

user228700

11:00

@Secret: I tried to read that paper and I also sort of read your question and truth be told, I don't really understand what's happening.

David Z

@Kaumudi.H Deletion means an account is anonymized and completely removed from the system permanently (except for some forensic information kept by SE). Suspension (which is what people probably mean when they say "ban") is just temporary, and has no permanent effect on the account.

While suspensions are imposed by mods as a disciplinary measure (more or less), deletion usually happens because the user requested it, because they are a spammer, or because the account is a sockpuppet being used fraudulently.

user228700

Oh, I see. Thanks :-)

David Z

No problem

user228700

If ANYBODY is familiar with the Baker-Nathan effect, I request you to PLEASE help me out _/\ _

user228700

@SirC: I'm assuming you didn't get any sleep last night, as usual?

Jacobadtr

11:26

Hello,

I am having trouble calculating the heat capacity in the Debye model because I am confused about how to take the derivative of

$$U = U_{zero} + 9NK_BT (\frac{T}{\Theta_D})^3 \int^{\frac{\Theta_D}{T}}_0 \frac{x^3}{e^x - 1} dx$$ where I have made the substitution $$x = \frac{\hbar \omega}{K_B T}$$

If anybody could give me a hint, or point me to something that might help that would be great

Thanks!

Is there a rule I need to consider that will help me solve this?

Secret

Basically, if we temporary set aside on why only C-H can hypderconjugate for that reaction (that is something that confuses me and no journal articles nor website sources covers that and hence that ChemSE question).

Then if we use the inductive effect, the halide X will draw electrons from the C that is attached to the benzene ring, which in turn will draw electrons from the para group, thus when you drew all 3 resonance structures, you will get for each structure a formal positive charge at one of the 4 locations where I labelled as $\delta^+$ in black. Meanwhile the carbon at the C-X bon…

So as the article said, both hyperconjugation and inductive effect facilitate the reaction. The trend reversal is due to Me being the best hyperconjugating group and tert is the best electron donating group

Swapnil Das

Wow.

Will I even understand chemistry? I doubt.

Jacobadtr

Ah, the leibniz integral rule is what I want

G. Bergeron

11:41

@Kaumudi.H Who is that?

Secret

that's Sir cumference

Secret

On hyperconjugation. Actually I think it is not very straightforward especially when you realise that a sigma orbital formed by C-C is expected to have a better overlap than its hydrogen counterpart (being larger), yet we observed the opposite trend

user228700

12:09

Wow. Wokay, I don't think that I need to go into such depths in this topic. I'm going to move on. Thanks very much :-)

user228700

@G.Bergeron Yep, that's Sir Circumference.

Secret

Well for your level, I think you only need to remember that adjacent C-H does hyperconjugation, and groups with a lot of carbon are more electorn donating than those with less carbon, plus both hyperconjugation and inductive effect facilitate the baker nathan reaction. The trend reversal is due to Me being the best hyperconjugating group and tert is the best electron donating group

should be more than enough to answer any question about Baker Nathan

Secret

I have a tendency to massively overanalyse a question because of my perfectionism to find the most perfect answer. This explains why I often cannot respond to people in time because I am thinking way too hard

user228700

I see. Thanks :-)

user228700

I come with a homework question again. I'm trying to find the range of the function given by $y = (x^2-2x+5)/(x^2+2x+5)$

user228700

12:24

Writing this equation as a quadratic in terms of $x$, I get $x^2(y-1) + 2x(y+1) + 5(y-1)=0$

user228700

To find the range of the function, I am trying to find a condition on $y$ such that the above quadratic equation has real roots. For this, we know that $D \ge 0$, where $D$ is the discriminant of the quadratic equation given.

user228700

After simplification and everything, this gives me $y^2-3y+1 \le 0$

user228700

I dunno if I'm correct so far...in terms of what to do, I think I am but I'm not sure about what I've done, you know, steps-wise...if it's all free from errors, because I've tried this question bout 5 times already and I'm still not getting the correct answer.

Secret

I think you want instead to find condition for x such that y is real. Also checking the denominator with the discriminant, $D < 0$ thus the denominator has no real roots hence the domain will be all of $\mathbb{R}$

The function $y$ can also be simplified using long division to become as follows:

$$y=1-\frac{4x}{x^2+2x+5}$$

user228700

Huh..? No, to find the range, I'm pretty sure I want a condition on $y$, not $x$. $y$ is the output and $x$ is the input, remember..?

user228700

12:34

And what is it that you're saying about checking the denominator with the discriminant..? I'm unable to fully get what you're trying to say...

user228700

@Secret What to do with this..? To find a condition on $y$, I don't think writing it like this would be very helpful...

Secret

Check the discriminant of the denominator $k=x^2+2x+5$ gives $D=2^2-4*5<0$. Therefore the denominator has no real roots thus there are no vertical asymptotes. Now check whether $k$ is positive or negative (above or below the axis).
Completing the square gives $(x+1)^2+4$, which is $\geq 4$. Therefore the denominator will be any number $\geq 4$. Now for the numerator. $4x$ has the range of $\mathbb{R}$, so the fraction has the range of the reals. Finally 1- the fraction also have the range of all reals (Since 1- any number in reals is going to lie in the reals). Hence y has the range of all…

user228700

That is...almost convincing but the answer is unfortunately wrong :-(

user228700

> "The denominator has no real roots thus there are no vertical asymptotes."

user228700

Um, why would any polynomial have vertical asymptotes? Surely, u are talking about the whole function itself..?

Secret

12:45

yup, I am actually talking about how the demominator affect the whole function y (as for the answer, I am still figuring why somehow I missed out the max and min that bound the function between two numbers)

user4814373

@Kaumudi.H The best way is to draw a graph. The maxima of the function occurs at -sqrt(5) if you differentiate the function y. And the minima of the function occurs at +sqrt(5). BTW there will be no vertical asymptotes.

At either infinites the value will tend to 1

Secret

Hmm, I guess that's a lesson for me to always check for extremas (whether cusps or stationary points or boundary points)

user228700

Well, yes, but the answer still doesn't quite match. The range is given as $[(3- \sqrt{5})/2 , (3+ \sqrt{5})/2]$

user228700

Remember that equation I got..?

user228700

$y^2-3y+1 \le0$

user228700

12:53

The real roots of the quadratic on the right are exactly $(3 \pm \sqrt{5})/2$

Secret

Hmm, I think I can see how that real root argument works above. Because your function is real, for every real y there must be a corresponding real x, thus rearranging into the quadratic involving y like you did above allow you to use the discriminant to pick out the ys that guareentee the x are real hence giving the range of y in the process from the bound set by the discriminant

How does that ensures it gives you the upper and lower bound of the range, I am not very certain

user228700

You put it exactly the way I didn't know how to :-P So, yeah, it works, but that inequality..? It doesn't give the range. I can't factorize that into anything.

Secret

You can find the y s where the discriminant is =0, (i.e. double roots), this is where you can use the quadratic formula to solve for y, which gives you the roots that forms the range

This is because the discriminant is at least = 0 in order for x to lie in the reals

user228700

Yes, it gives me the roots and all, but it doesn't give me the range. Dyou know what I'm saying? I need to get the range using that inequality.

user228700

Ohh, hang on.

user228700

13:00

DANG IT, I knew the answer all along. I just. Ugh. Okay, I just didn't know how to arrive at the last effing step.

Secret

x is real when $D \geq 0$. Therefore the minimum requirement that x is real is when that discriminant $D = 0$. You have found that $y^2-3y+1 \geq 0$. Therefore the largest possible interval of y you can get x to be real (the range) will be given by the interval formed by the solution of y in $y^2-3y+1 = 0$, that is the two roots together form the bound that is your range

user228700

Yeah, yeah, I was just...being dumb (Sigh, @DanielS: I can't help it. I kinda am, okay? Or maybe I'm just tired.)

user228700

Alright, thanks a lot for your help :-) Dyou want to discuss another similar question? :-P

Secret

Don't worry , I intiially forgot about this method despite I learnt that back in my high school grade 5 because it is too hard for me to visualise.

Well, ask

user228700

Grade 5? Man, what kind of a high school did you attend? O_o

Secret

13:03

sorry mistake Grade 10

user228700

Ah, okay :-P

Secret

can't count because in HK we use the Form somethnig system and form x = Grade 6 + x

user228700

Well, if u're still interested to learn to visualize this method, then I suggest that you read yesterday's transcript. Daniel Sank and I had a brief discussion about this, which cleared it up for me...

user228700

@Secret Oh, okay...

user228700

Anyhoo, this is the question:

heather

13:08

hello

user228700

$y = 1/\sqrt{x^2-x}$

heather

ooh, bijective and surjective

cool

user228700

@heather Hey :-) Yeah. How's it going?

user228700

What I did was, I squared both sides, after noting that the domain absolutely cannot include 1 and 0.

Secret

the domain does not include anything that is $x^2-x \leq 0$

user228700

13:10

Oh, hang on...

user228700

@Secret Dang it, yeah.

heather

@Kaumudi.H good, how about for you?

user228700

@heather I wanna say "good", but in all honesty, kinda miserable :-P (Probably should've just said "good" :-P)

heather

@Kaumudi.H oh, I'm sorry! What's making life miserable?

fwiw, thanks for your honesty =P

Secret

I am kinda multitasking: helping kaumudi on the maths, investigating convolutions in the maths chat and still trying to divide by zero

2

user228700

13:13

@heather Nothing that won't go away in awhile :-) Thanks.

user228700

@Secret Lol. I'm sorry I'm taking so long to type...

heather

wait, Doraemon was banned?

why?

user228700

Deleted.

heather

er, deleted.

but why?

user228700

(Just to be clear, this is not me "speculating", because I know (and can prove, if needed) this for a fact) Sockpuppet account. He was S007.

heather

13:19

ah.

is, um, S007 banned?

ACuriousMind

@heather Like suspensions, reasons for deletion are not public unless the user in question chooses to make them public or they are otherwise in plain view. (Some deletions are actually requested by the users themselves)

user228700

@heather Probably.

heather

@ACuriousMind, ah, sorry.

just woke up, not exactly thinking.

@Kaumudi.H, nice hat, btw.

bacon is a good scarf.

user228700

Thanku thanku :-P

heather

@ACuriousMind, oh, while you're around, I think I got serially downvoted.

ACuriousMind

13:21

@heather Oh, I see another mod misunderstood your flag

heather

@ACuriousMind oh, really?

ACuriousMind

I almost made the same mistake; you shouldn't have chosen that exact flag text on a post where there were three other answers which had a single downvote :P

heather

oh, i didn't notice that!

sorry =P

i don't know if mods can see this, but I ended up deleting one of the answers anyway because it was bad, too

hmm, i wonder if the other posts got downvoted across the answer board

DanielSank

@Kaumudi.H XD

heather

@DanielSank, hello

user228700

13:23

@DanielSank :-P Hi.

DanielSank

@heather yo

heather

mmm, bacon

this is a great hat

ACuriousMind

@heather It read as if you were saying someone had "serially downvoted" all the answers to a particular question, which is what the text for declining your flag is about; I left the flag untouched because I could not uncover evidence for serial voting beyond what ticked you off. I'm afraid that unless the perpetrator returns there's little we can currently do about it.

user228700

@Sec: God, this is horrible. Which method do you usually choose to find range?

heather

@ACuriousMind oh, okay

sorry about that (that was really stupid on my part)

it's not that big a deal, i can live with a couple downvotes =P

ACuriousMind

13:27

@heather Just a simple misunderstanding - it's hard to predict how others will interpret what you write because you know what to wanted to say.

Secret

Well, I tend to figure out what the domain is, and then worked from there, as well finding extremas and such. But really after year 2 in uni, I rarely need to worry about functions for some time as I shift towards more abstract algebra pursuits

and not to mention chemistry made up most of my year 2 to honours

so it is not suprising I am kinda rusty on that aspect

user228700

Oh, I see. Good for you :-) I think I'll chuck this and move to other stuff for now. Thanks ever so much for helping me out so much _/\ _ :-)

Secret

But I am pretty sure I will get better once I finish reading munkres, which deal with even more general mappings

user228700

@Secret Say, before I go, dyou ever like, do stuff other than studying and trying to divide by 0? :-P

Secret

Well I do other things in the past, such as analysing time travel stories and games, some quantum mechanics calculations, reading some physics jourhal articles for fun, browsing facebook and so on

user228700

13:31

In the past..?

Secret

not to distant past, like 3 months ago. Currently I am mostly hanging out in the SE chats while continue my divison by zeor pursuit in the backgorund

Aftr all, my phD starts next year, I can do nothign until at least january arrives

user228700

I see. So the short answer would be: No? :-P

Secret

mostly no, other than asking and answering question . Pretty much my role is kinda in parallel to johnrennie nowadays, except I don't drink coffee and I am not actually working on a formal project

user228700

:-P Wow, I see. And u're not even 20 yet, right..?

Secret

I am 21

user228700

13:34

Oh. Sorry :-P I thought you were like, 19.

user228700

Anyhoo, thanks for all your help! :-D

Secret

well, my mental age is actually around 13, and my parents have noticed I have a quantised trend in my maturity

heather

lol

that's fabulous

Secret

that is, I become mature at some intervals

like a quantum jump

user228700

x'D I see. That's...interesting.

Secret

13:35

But, whatever the next jump will be, hopefully it is not one that turn me into a souless working machine, that will be TOO SAD

user228700

:-) Indeed. Good luck! I'll see u tomorrow.

user228700

Bye, everyone.

heather

@Kaumudi.H, hope your day gets better =)

have a good one.

user228700

(I might come back with questions, who knows :-P)

user228700

@heather Thanks! :-) Have a nice day.

heather

13:45

ooh, got the trendsetter secret hat

hats are great

Kenshin

14:19

Sup

heather

nothing much

how are you?

oh, general hbar question: thoughts on having a tag shapiro-delay?

ACuriousMind

@heather I couldn't even tell you what that is beyond mumbling something about GR ;P

Have you seen a sizeable number of questions who would benefit from such a tag?

Kenshin

I'm fantastic ty

If there are enough questions that need the tag then why not

John Rennie

@heather that seems a rather specific tag. I'd be surprised if there were that many questions on it ...

Kenshin

How much rep is required to make a tag

ACuriousMind

14:33

@Kenshin 1k.

John Rennie

27 questions in fact. And given how easy a term it is to search for I'm not sure a tag would be that useful.

ACuriousMind

@JohnRennie Well, don't forget that at least some people filter the site by tags

I'm not saying that Shapiro delay specifically is a tag we need, but there are several other tags (especially the large ones, like quantum-mechanics) who are actually not useful to search for anything specific but nevertheless are useful as an organizing tool

John Rennie

But weren't we discussing whether Shapiro delay specifically is a tag we need?

ACuriousMind

Yeah, just saying that tags don't need to be difficult to search for to be useful

Kenshin

I think with 27 questions a tag can be created

ACuriousMind

14:38

And actually, there are only 5 questions about "Shapiro delay", your other results are answers.

Kenshin

oh

Well that changes my view

ACuriousMind

(add is:question to a search to view only questions)

heather

hmm, true.

thanks for the advice

Kenshin

np but feel free to still create the tag as you have earned the rights to do so

ACuriousMind

@Kenshin Technically true, but it is considered good style to ask for the creation of new tags first, especially on a site as old as ours.

Basically, it's very unlikely that there's a really good tag we haven't yet created

Kenshin

14:43

Yes but I think a 1k rep user would have good enough judgement

heather

not really..i'm 4.5k and i don't; that's why i asked.

ACuriousMind

@heather According to the canonical meta.SE post, it's for changing your hat three days in a row. I just got it, too

wendy.krieger

Sounds like an interesting thing. I have two secret hats.

Kenshin

hi wendy

Heather u have sufficient judgement. We wouldn't create that tag but creating it wouldn't cause harm its a borderline choice

wendy.krieger

hi kenshin

Kenshin

14:55

Wendy r from Brisbane?

wendy.krieger

yes, þe dark side of þe planet.

Where are you from

Kenshin

Same

wendy.krieger

ok

Kenshin

What is ur job

wendy.krieger

retired invalid, mainly.

Kenshin

14:57

Sounds good

wendy.krieger

but i do weights and measures, number bases and physics as a hobby

base 120, that sort of stuff

Kenshin

Do u lift

Wats ur favourite physics topic

wendy.krieger

I cover much of it. At the moment, i'm trying to simplify relativity. I have done a really good job on easter.

Kenshin

Yeah my favourite topic is GR

wendy.krieger

Some of special relativity was pretty easy, the curvature thing.

Kenshin

15:02

Do u mean Christmas or easter

wendy.krieger

GRT

a formula to find easter. It's down to about four lines of code.

Kenshin

That would be handy for planning

wendy.krieger

I solved the schwarzchild metric in my head. You can pull GMm/R^2 as a tension in the fabric.

Kenshin

Wow u must be a genius

wendy.krieger

But generally, i don't believe GRT is the right answer, so i read up on this stuff.

Kenshin

15:06

Na Einstein wouldn't be wrong

heather

hmm, i just earned a bunch of hats

@Kenshin eh, he's been wrong before.

Kenshin

Na

heather

@Kenshin yes he was

Kenshin

Anyone who says Einstein is wrong is wrong

wendy.krieger

Ah, you just pop a bit of arithmetic in base 120 at them, and see what they say.

Kenshin

15:08

U need base 121 tho

wendy.krieger

Oh good lord, no. Base 120 is straight-forward.

Kenshin

How come

wendy.krieger

Cause i've been using it since the seventies.

Kenshin

Do u have 120 fingers and toes?

wendy.krieger

You know, six by twenty is one hundred.

eight fifteens is a hundred

I sometimes translate decimal numbers to it to calculate.

Kenshin

15:11

No six by twenty is ten

wendy.krieger

I'm pretty sure my calculator says six by twenty is a hundred. Twenty is two by ten.

Kenshin

Hmm it depends on how you define hundred

wendy.krieger

No one ever said the base you use has to be called 'ten', or you need digits from the unit to the base. I use just twelve digits.

A hundred is 100. You know 6*20. As it was at the time of christ, and at the time of the crusades.

Kenshin

6*20=120 which in base 120 is 1,0 where a comma separates each order

Y at the time of crusades?

wendy.krieger

You really did not learn arithmetic at school, i take it.

Kenshin

15:15

No

Y r u awake so late

wendy.krieger

Prolly born after 1960 even. Never used real money £sd.

Kenshin

Correct

wendy.krieger

it's only 0116 here

But it's 2016.1230.0116 here, or 16.96.99F.06 using twelfty time

Kenshin

Wow twelfth time dats cool

It's not so late when u put it Lyk dat

wendy.krieger

Most parts of the world, the days start with Y. 'yesterday'. :)

so i must to bed go.

Kenshin

15:21

Ok u gtg sleep now

I gtg*

Didn't mean to type u lol

Laterz

DanielSank

15:50

What is going on in here?

I can barely read the chat log.

Hi, @Stoby.

Stoby

Hi @DanielSank I'm not sure I just got here

DanielSank

@Stoby I don't recognize your username. New around here?

Stoby

Yes

DanielSank

Welcome.

Stoby

Thank you I'm really enjoying this stack exchange, it's a great reasource

DanielSank

15:52

Yes, it is amazing.

This chat room is a lot of tun too.

We have a good rapport.

AccidentalFourierTransform

@Stoby Hi and welcome :-)

DanielSank

Yo, @AccidentalFourierTransform. What's shakin'?

heather

@Stoby, hello, welcome!

DanielSank

'sup @heather?

heather

@AccidentalFourierTransform, hello

@DanielSank, nothing much, how are you?

DanielSank

15:54

Doing well. Getting ready to release a LaTeX package on CTAN.

The system is arcane.

heather

nice!

DanielSank

AFAICT there is no authentication needed to upload a package, so there's no real way to make sure that other people can't modify it.

heather

as for the transcript, it started off with Kenshin and someone else talking and I ignored them both as i didn't really want to read the conversation.

DanielSank

lol ^

heather

@DanielSank that's...strange

DanielSank

15:55

Yeah it's pretty weird.

It's not like PyPI where you need your account credentials to upload a package.

heather

@DanielSank the first bad sign was Kenshin saying Einstein was never wrong and anyone who said he was wrong is wrong (which means I am, therefore, wrong).

DanielSank

o_O

Kenshin = John Duffield

?

heather

@DanielSank seems like there's a not insignificant possibility for spamming then

DanielSank

@heather Yeah I guess we'll see.

::shrugs::

heather

@DanielSank some recent activities have made me unsure in my position as moderator on the new site...on the one hand, I do want to moderate, on the other, Kenshin sometimes takes....difficult positions.

DanielSank

15:57

Mostly I just want TeXLive to pick up my package so I can \usepackage it. Right now I'm using a git submodule which has its own problems.

heather

@DanielSank what's your package?

DanielSank

@heather You could consult the mods here for general mod advice ;)

heather

=)

yeah maybe

DanielSank

Really. They have more experience than you.

They know things.

heather

that does make sense the more i think about it =)

DanielSank

15:57

::spooky music::

rob

I KNOW NOTHING

3

DanielSank

LOLOLOL

Man, what is it with the mods lurking like that?

Y'all just sit silently in front of your keyboards waiting for someone to mention "mods" and then BAM you pounce.

rob

@DanielSank I was writing something in another chat room, and happened to notice your comment on the sidebar.

DanielSank

Wait a minute... do you guys have a chatbot waiting for the word "mod"?

Ah, so you are the chatbot.

heather

you guys really do just pop out of nowhere sometimes

rob

15:59

In fact I was writing in the secret mods-only chatroom where we all discuss our ignorance.

3

heather

ah

your secrets have been revealed!

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