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12:07 AM
@Ted I asked a question on MO
No idea if you are interested.
0
Q: Are immersed curves really richer than the knot theory?

PVALIn his 1991 book Topological Invariants of Plane Curves and Caustics, Arnold asks the following question (which I paraphrase).. Does every topological knot type in $ST^* \Bbb R^2$ arise as a Legendrian curve of an immersion $S^1 \to \Bbb R^2$ (in other words is every knot type parameterized b...

 
12:25 AM
Hey there
 
Hi @Daminark
 
What's up?
 
someone already downvoted my question :/
 
presses F to pay respects
 
You still coherent, Demonark? (Well, as coherent as you ever were ...)
 
12:45 AM
Heya @Eric.
 
hello
 
Yeah, I'd say so @Ted
The pressing F thing is a meme
 
oop i did pretty well on my geometry midterm it looks like :)
 
Great, @Eric. :)
 
Nice
Analysis went well on my side even with the botched question
 
12:48 AM
from the points off it looks like i only missed what i knew i missed
 
He grades pretty fast. That's a bonus point in my book.
 
yeah our course grader is good too
we get our grades back like clockwork
 
Oh yeah
 
@Daminark did you talk to marianna about whatever the issue was you told me about
the integrable convention thing or whatever
 
We were surprised, Neves gave us our grades at around 5-5:30, and he had a class immediately after ours so in reality that was just 5 hours of grading
@Eric No, turns out I was doing something else stupid
 
12:50 AM
He probably graded during Eric's class.
 
I forgot that avoiding $\infty - \infty$ doesn't require that only finitely many terms were of a certain sign, convergent series are a thing which exist
 
lol
 
Though as Ted puts it, thanks to poor offense it still came out well
 
@Ted is there an easy mental picture that explains what Ricci curvature means? we never discussed it in class, only proved relations it satisfies on the psets
 
Oh did Neves give you a pset this week?
 
12:54 AM
no
 
Just average sectional curvature among all planes containing $v$.
 
Yeah same
 
hmm ok, i guess ill need to see it crop up in more places to get a feel for why this matters
 
Marianna shifted the time schedules, so psets are now due on Wednesdays
 
oh that's what life was like for me when i took her class too
 
12:57 AM
Lol imagine if Marianna teaches first quarter of grad analysis next year
Jk she's gonna be at Berkeley that quarter
 
does it make any sense to pass as input to trigonometric functions degrees per second (or radian per second), where the point is to highlight the "per second"?
 
@nbro Not really, trigonometric functions don't really interact with units at all
 
oh yeah she's going to that msri thing
 
@Daminark indeed, it also seemed to be that this doesn't make much sense...
 
(Is it weird that I secretly want Soug to teach functional?)
(Not secretly anymore)
 
1:02 AM
yes @Daminark.
 
I have a problem where the inputs are angular velocities, given in deg/sec
I also have a numerical algorithm whose input should be these angular velocities...
 
I guess Stockholm syndrome of a sort. But I dunno, I appreciated his style in some weird way. How'd you like Smart for functional?
 
the problem is, does it make sense to make these calculations in this numerical algorithm?!
 
@Daminark I like him but he didn't give our midterms back until the final which made me really upset.
 
Eek, well how was Silvestre first quarter?
 
1:06 AM
silvestre is fantastic
easily the best lecturer ive had here
 
Moreso even than Schlag?
 
Yeah
 
Holy crap
The only person I've seen whose lectures I prefer is Laci
 
 
2 hours later…
3:20 AM
Hi guys!
 
3:47 AM
Hey @Baymax!
 
hi @daminark
 
How's it going?
 
yeah nice,how about you?
 
Doing alright, thanks!
 
PBS infinite series is a nice one!
nice set of videos
 
3:50 AM
Nice series of videos?
 
yeah series
well in mathematical sense it is a set of videos!
like we are not adding them so how can it be series!
:)
 
:( You're killing my joke
 
oh
 
4:29 AM
A piecewise linear function may or may not be a linear function.Is this correct?
 
@BAYMAX correct
 
thanks
 
 
1 hour later…
5:58 AM
Hi all, i have a question about proof writing,

Suppose $a$ and $b$ are real numbers, then LHS = $8a^3+12a^2b+6ab^2+10b^3$ and RHS = $9b^3+9b^2a+9ba^2+9a^3$,

To properly show that LHS $\neq$ RHS, what should i do? Or is it suffice to conclude LHS $\neq$ RHS by observation?
 
I am not sure but I think we should assume suppose they are equal and proceed
@LittleRookie
like we may get a kind of contradiction if they are not equal.
 
What sort of contradiction should i obtain?
 
May not be the correct approach but let us see ,after equating we get $-a^3 + 3a^2b - 3ab^2 + b^3 = 0$
I think it is $(b-a)^3 = 0$
Implies $b = a$
So LHS = RHS if $b =a$
so contrapositive will be
If $b \neq a$ then LHS $\neq$ RHS
 
I see.
$a$ and $b$ are actually the lower and upper limit of a quadrature formula
i was doing some questions on degree of precision
 
Gauss Quadrature formula
?
 
6:11 AM
yea
 
Integration rules or Quadrature formulas numerical methods!
 
The lower and upper limit $a$, $b$ should work for any $a,b \in$ domain of function right?
 
Let us see this
$\int_{-1}^{1} \frac{1}{x^2}dx$
I think we would get 2?
but now $\frac{1}{x^2}$ is an even function
so $\int_{-1}^{1}\frac{1}{x^2} dx = 2.\int_{0}^{1} \frac{1}{x^2}dx$
Now we see that $\int_{0}^{1} \frac{1}{x^2} dx = 1$
But we observe that $1 \in$ domain of $f$ but $ 0 \notin$ domain of $f$!
@LittleRookie
 
hmm, then why is $a=b$ a contradiction?
Heres the exact question:

Let $h=(b-a)/3, x_1=a+h$ and $x_2=b$. Find the degree of precision of the quadrature formula, $\int_{a}^{b} f(x) dx$ approximately $\frac{9}{4}hf(x_1)+\frac{3}{4}hf(x_2}$
 
$a = b$ is not the contadiction
It says that if LHS = RHS
then $a$ must be equal to $b$
So those two expressions are equal only when $a=b$
 
6:23 AM
is $a$ okay to be equal to $b$ in the question? im confused =/
 
unable to get the lastl ine!
 
wait let me try typing the question again
Let $h=(b-a)/3, x_1=a+h$ and $x_2=b$. Find the degree of precision of the quadrature formula, $\int_{a}^{b} f(x) dx$ approximately $\frac{9}{4}hf(x_1)+\frac{3}{4}hf(x_2)$
 
will be right back , will think about this!
 
okay thanks!
 
6:45 AM
@Akiva @Balarka The generalisation to $n$ sided polygons follows from the elementary case of the triangle does it not ?
Since a polygon isn't anything more than a reunion of adjacent polygons
 
7:00 AM
triangles*
 
7:25 AM
is degree of precision $3$ @LittleRookie
 
 
2 hours later…
9:20 AM
@BAYMAX How is $\int_{0}^1 {1\over x^2}dx$ equal to $1$ ?
 
10:00 AM
as $\frac{1}{x^2}$ is an even function ,$\int _{-1}^{1} \frac{1}{x^2}dx = -2$(edit) so $\int_{0}^{1}\frac{1}{x^2} dx = -1$ ?
@Astyx
 
Then why does $\int_{-1}^1{dx\over x^2} = -2 $?
 
as $\int \frac{1}{x^2} dx = \frac{-1}{x}$
 
Oh that's bold :p
 
is it not!
 
10:25 AM
Hi, is it true to say that if $f'$ has no analytic continuation through $\partial\mathbb{D}$ (where $\mathbb{D}$ is the unit disc) then $f$ has no analytic continuation through $\partial\mathbb{D}$ (where $\mathbb{D}$?
 
10:45 AM
@Astyx I guess that's true.
 
Anyone here know some probability?
 
I know some, but maybe not enough to help you
Just ask your question anyway
 
Here's the problem: I have $N$ samples ($S_k$) from the set $\{1,2,\dots,n\}$. Using these, how can I determine an approximation of $n$? (Note that these can be repeating samples; if they were unique, then this would be equivalent to the German Tank Problem.)
Obviously, we know that $n\geq\max S_k$
 
What's a sample ?
 
As in, I take a single value of the set, uniformly at random
 
10:58 AM
Oh right
Well $\lim_{N\to \infty } \sum_{k=0}^N {S_k\over N} = {n+1\over 2}$
By the law of large numbers
 
...that's a good point, actually
 
?
I don't quite understand what is $n$ in this case either....
 
A natural integer
 
@RanWang It's the maximum value in the sampled set $\{1,2,\dots,n\}$
 
OK~
The maximum likelihood estimator is actually the sample maximum
If I recall correctly
 
11:02 AM
So $n\approx-\frac12+2\sum_{k=1}^N\frac{S_k}N$ using your formula, correct?
 
$-1$, not $-{1\over 2}$
 
Why so?
 
Because arithmetics :)
 
Oh, I see, forgot
 
Hi Steamy
 
11:04 AM
Equivalently, that would be $n\approx-1+\frac2N\sum S_k$ as it is a finite sum, correct?
 
If you know that the distribution of the $S_k$ is uniformly from the range you applied, I think the MLE is the max of your sample
 
Mmm I don't get your argument but the result is true
@RanWang That is only if you have an infinite amount of samples
 
An MLE does not depend on the number of observations you have
 
That's true for any sum, not only finite ones @LegionMammal
 
It is an estimator
 
11:06 AM
Besides, I'll just take samples until $n_\Sigma-\max S_k<\varepsilon$ for a small $\varepsilon$
 
Sorry I don't know what MLE stands for
 
For this estimator to be consistent, it is required (as the definition of consistency) requires
Maximum likelihood estimator
From a statistical point of view, you have have any kind of estimator for any quantities you are interested in
 
Right
 
But some are good and some are bad
In this situation, the usually used estimator is the MLE which is the max of the sample interval
But this assumes that the r.v. is uniformly distributed
(But I am thinking that this also holds if your sample space is finite)
(I think the above conjecture is wrong, never mind)
(Looks like I have killed the conversation.... good luck guys! I am going to get some lunch)
 
Bon appétit !
I'm not sure I got what you meant
 
11:12 AM
0
Q: Length of shortest path

user123733What is the length of shortest path that begins at the point (-1,1) , touches the x axis and then ends at point on the parabola $(x-y)^2 =2(x+y-4)$ . I put $x=x-y$ $y=x+y-4$ then the starting point would be $(-4,-2)$ . can I proceed by this methos of changing the coordinates After that I g...

can anybody explain me how to do with my method
 
yawn
These samples take about a half-second each, and I'll likely need $N$ in the hundreds(/thousands?) for anything accurate
 
That's 50 seconds
Or 500
 
$\times\frac N{100}$
Ima go get some breakfast
 
Sure
Bon appétit
 
4/29/2017
$2^2/2^1/2^0$
:)
 
11:25 AM
$$\lim_{h\to 0^-}\lim_{x\to a^+}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0^-}\lim_{x\to a^-}\frac{f(x+h)-f(x)}{h}$$
Ah, I know what I am trying to do here:
It appears this thought process leads to the equation that is needed to show under what condition that continuity implies continuity of one sided derivatives (assume it exists)
The condition is that one can interchange the limits, and this is possible only when uniform convergence is obeyed:
7
Q: Under what conditions can I interchange the order of limits for a function of two variable?

DJ KSuppose I have $f:\mathbb{R}^2 \to \mathbb{R}$. What conditions do I need to say that $$\lim_{x \to a} \lim_{y \to b} f(x,y) = \lim_{y \to b} \lim_{x \to a} f(x,y)$$ ? What about in a more general case, by taking $X,Y$ and $Z$ topological (Hausdorff) spaces and $f$ from $X \times Y$ to $Z$ ?...

Now for a solution of a one sided ODE to satisfy its corresponding (two sided) ODE, one needs the constraint:
$$\lim_{h\to 0^+}\lim_{x\to a^+}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0^+}\lim_{x\to a^-}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0^-}\lim_{x\to a^+}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0^-}\lim_{x\to a^-}\frac{f(x+h)-f(x)}{h}$$
That means, letting $D(y)=c$ be an ODE and $D_+(y)=c$, $D_-(y)$ be right and left ODEs, if $f$ is continuous within its domain $\textrm{Dom}(f)$ and satisfy e.g. $D_+(y)=c$, then $f$ does not necessary satisfy $D(y)=c$ unless it also satisfy $D_-(y)=c$
 
11:42 AM
Hello what is the Math Jax text for the following situation
i have a partial derivative i need to show one of the variables constant at the bottom
 
Therefore
$\textrm{f solves $D(y)=c$}\iff \textrm{f solves $D_+(y)=c$ and $D_-(y)=c$}$
but
$\textrm{f solves $D_+(y)=c$ or $D_-(y)=c$}=\not\implies \textrm{f solves $D(y)=c$ }$ in general
 
$(\partial\psi/\partial t)q^i does not work for me
 
@Astyx Can't get it any closer than the range $n\in[26,30.77]$. Thus, I have a new theory: Instead of the set $\{1,2,\dots,n\}$, I'm instead sampling from $\{m,m+1,\dots,n\}$ for $m=5$ or so. Not sure how to find the exact bounds...
 
@gansub $(\frac{\partial\psi}{\partial t})_{q^i}$ something like this?
 
@Secret - beautiful ! works fo rme
 
12:26 PM
Quick question on polynomial interpolation: Given a polynomial function $f(x)$ and a point $(a,b)$, the modified function $f^\prime(x)=(f(x)-a)(x-b)+a$ will always cross through $(a,b)$, correct?
 
12:43 PM
"The prison is deep and of stone; it's form, that of a perfect hemisphere, though the floor (also of stone) is somewhat less than a great circle, a fact which aggravates the feelings of oppression and of vastness"
had to read that thrice :P
 
1:05 PM
@BalarkaSen "Somewhat less than a great circle?" So like this?
 
I think so.
however there is also a dividing wall of this prison which cuts it in the center
 
Some ordinal property in a nutshell: Am ordered set. Adding stuff on the right sticks behind, but adding on the left and if it is not big enough, will be eaten up
 
1:33 PM
Hey everyone!
 
Hi @Daminark
 
How's it going?
 
Not bad
 
Nice, same here!
 
1:48 PM
Hi!
Does anyone here have a solid experience with numerical computations? By solid I mean that you haven't just implemented one iterative algorithm in your life and that you actually know, at least, the main problems when implementing these algorithms.
 
What does that mean @nrbo ?
 
I know nothing of the sort :/
 
what exactly you didn't understand?
 
What kind of numerical computation? There's a big difference between, say, a PDE solver or an MCMC sampler or a large integer factorization algorithm, even though all of those could be called "numerical computation".
 
I mean algorithms like Gauss-Newton, conjugate-gradient, etc
optimization algorithms
 
2:01 PM
Hmm, well, can't say I'm an expert on those. I've read some stuff about them, but I've never really done any serious work with them.
 
ok
 
Anyway, if you don't find any help here, you might want to try your luck over at scicomp.stackexchange.com.
 
@IlmariKaronen The reason I ask here it's because here there's usually a lot more active people than in other more specific communities, like the one you're pointing me too... but their chat is completely desertic...
 
True, not a lot of smaller sites have active chat rooms. But if you have a specific question about optimization algorithms, it might on-topic at the main site there.
 
It's really sad, as I wrote in their chat, that not even a moderator is actively or passively in the chat...
it's like, no one cares
 
Anonymous
2:08 PM
@nbro Why not ask your question on the main site rather than the chat room ?
 
Hello!!! I have a question:

We have the function $f(x)=\left\{\begin{matrix}
-1 & ,-1 \leq x <0\\
1 & ,0 \leq x<1
\end{matrix}\right.$.

In this case $L=1$ and so we should have $T=2L=2$.

But $f(x+2) \leq f(x)$. So which is the period in this case?
 
Anonymous
@Evinda What is $L$ and $T$ ?
 
because I once asked a question and they were very reluctant to answer, it was like, why don't you look up in a 600 pages-book? I mean, if I ask a question, sometimes, it's not just because I don't know currently how to solve a problem, but it's also because I do not have enough time to look up in a book, don't you think so?
 
$T$ is the period, $[-L,L]$ is the interval at which we consider the function @blue
 
Anonymous
@nbro You need to ask a question and wait for 1 or 2 days to get an answer on beta sites. On bigger sites you can expect answers in a few hours. Keep in mind that no one here gets paid to answer your question. People answer questions voluntarily to help other people. So don't lose your patience and go ahead and ask your question on the site which you feel is most suitable.
 
2:13 PM
How to find all the positive integers $n$ of 4 digits such that all its digits are perfect squares and $n$ is a multiple of $2, 3, 5$ and $7$?
 
@Evinda Your function is only defined on $[-1,1)$
 
Yes. How can we find the priod? @SteamyRoot
 
Also, what's the point of talking about a period if your function is only defined on an interval?
 
We assume that the function is extended periodically at the external of the original interval @SteamyRoot
 
Well, that changes things.
Then all you have to do is ask yourself "what's the smallest $T$ such that $f(x) = f(x+T)$ for all $x$
 
2:15 PM
@blue it's not just a matter of waiting for the answer for a longer time, it's also a matter of how they first react to a question, it's like, I'm not going to answer this question because "reasons"
maybe they do not even know what I'm talking about
that's my bet
 
But this seems like a useless question. If you're going to take a function defined on an interval and then periodically extend it
 
Hi all :)
 
Anonymous
@nbro No one is obliged to answer to your question. You need to be patient. Questions which show more research efforts receive faster answers. It is quite possible that they don't know the answer. At the same time it is possible that your question doesn't interest them. If you want immediate positive responses to your question, then this is not the correct site for you.
 
I have been struggling with this question. Let a_n be a diverging sequence. Then can we find a sequence {x_n} such that \sum |x_{n}| converges but the series \sum |a_n x_n| diverges ?
 
I've been here on Stack Exchange websites for a sufficient time to say that, except maybe in this community, people often do not even read completely the question or do not even know what the question is about, but they like to start saying "Create a minimal verifiable, etc, example" or, again, "there are plenty of examples around", yes? where? and things like this. This only makes me think more than most of the people are stupid.
 
2:20 PM
yeah and they also downvote your question if you don't have any idea of how to solve a problem
 
Anonymous
@nbro Yes, I agree there are such problems as you describe. It's more like if you are lucky you will get an answer. If you're not, you won't. But such is life. No promises.
 
@S.C. a_n = e^n, x_n = 1/n^2
a_n = 4^n, x_n = 2^-n
 
But then again, you can't expect everyone to read all questions fully, as a LOT of questions either have not been thought about by the poster, or are incomprehensible for anyone except the poster, which leads to people not wanting to "waste time" on questions
 
Anonymous
@Astyx Agreed. Add to that "bad formatting".
 
Oh yeah, and that
 
2:23 PM
@nbro To be honest, I find this remark quite rude. You claim not to have enough time to look something up, but you expect others to take their time to answer a question for you?
Also, if you use the index or table of contents well, it doesn't matter that much if a book has 50 pages or 600 pages.
 
:37050091Will the period of the extension be equal to 2?
 
@nbro In any case what is your question exactly ?
 
@Evinda Yes
 
Because the function will be equal to -1 at the interval [1,2), 1at the interval [2,3) and so on, right? @SteamyRoot
 
Yup
 
2:27 PM
Nice... Thank you @SteamyRoot
 
@SteamyRoot If someone knows the answer, it just loses the time to put thoughts together and write it; OTOH, if someone doesn't know the answer, needs first to find the right resource, read it, which may take a lot of time, read it again to understand it better, etc... I mean, I don't know what you want to demonstrate, but you seem to be the type of person that thinks rules exist in an absolute sense, and that's sad
 
Why would you think someone who knows the answer has read or even seen your question ?
 
never mind guys
I do not have time to keep discussing
I have important things to do
bye!
 
I have no idea how you come to that conclusion of rules existing in the absolute sense, or what that has anythign to do with it...
 
Bye ! Still don't know what your question was though
 
2:35 PM
While I can said a few issues about numerical calculation in general (such as numerical instability, eigenvalues too small, nonconvergence or very slow convergence, well posedness of a problem, curse of dimensionality etc.) I do nto have suffiicient experience on the issues for the 3 specific algorithms in question
 
Anonymous
Sometimes I feel that Stack Exchange should be made a paid service so that people start valuing the time others spend on answering their questions.
 
That would ruin the point of StackExchange unfortunately
 
I think that would defeat the purpose of the site, but I get what you mean
there's a lot of abuse.
 
If you let $C_n$ be the $n$-th cake number, can you cut a cake in $C_n$ equal parts with $n$ slicing steps ?
I'm not sure how to start with this
 
that sounds rather like the ham sandwich theorem
 
2:44 PM
Looks like it, however you have more than $3$ objects and you always remain in dimension $3$
 
@blue people are already impatient now when it's about people answering out of their generosity, if you make it paid then people will more rightfully demand instant and full responses to every homework question ever.
@Semi In office hours I saw my professor's chalkboard had the term "Polynomial ham sandwich theorem"
 
I don't know what the discussion was about but I find this chat is a more helpful community than the SE. And not helpful in the sense that "we'll do your homework" (which is not helpful!).
But of course if 5 million people invaded this room that would break everything.
So maybe that's just because it's a small, stable community.
 
3:02 PM
Yeah probably
They serve different purposes though, sometimes you need people to discuss with, sometimes you just need a complete answer to a specific question
 
True enough.
True enough.
Meh stupid internet
 
@LeakyNun Thanks. Shit i missed a simple one. Thanks :)
 
press F to pay respects for Balarka's internet
 
@LeakyNun But you have just given an example. I am looking for a proof which works for all diverging sequences a_n
 
hello. i have a quick question. If i have a graded ring A=\bigoplus_{i=0}^\infty A_i, then each A_i is an abelian group. But 0 is homogeneous degree 0, and hence shouldn't be in each A_i but for A_0
Do we simply put 0\in A_i?
Can't reconcile how it's a group otherwise
 
3:17 PM
fyi, you can use the 'LateX in chat' link in the room desc to get your symbolic math to render
 
@S.C. Sorry I misinterpreted your question
you might want to ask in main
 
@alggeo So in your case you've got $A=\bigoplus_{i=0}^\infty A_i$, and you want to know where the zero of each abelian group is
Seems like it might be helpful to take a concrete case, e.g. $A=k[t_1,\ldots,t_n]$ is graded by degree (direct sum of homogeneous polynomials of degree $l$).
 
Yep, and $0$ is not in each of them, since it has degree $0$
Or I am wrong for some reason
 
my guess would be that 0 is an exceptional case. if you add two nontrivial homogeneous polynomials of degree n you in general get another homogeneous polynomial of degree n.
but if you add 0 to any polynomial, the degree doesn't change.
I think the way to look at is that 0 is an element of A, not of one of the A_i's
 
That's what I am thinking, but I haven't found any place that actually says that
It has to be one of the $A_i$'s, since they are $A_0$ modules, and $0\cdot A_i \subset A_i$
 
3:26 PM
hm.
I think one has to somehow distinguish, as it were, between $0_i \in A_i$ and $0\in A$.
But I'd like to be able to point to an actual reference...
 
But as you said, $A=K[x,y]=\displaystyle\bigoplus_{i+j=d\geq 0} A_d$, $0\in A=(?)0\in A_d$
 
Yeah. hrm. (I don't actually know enough about this to be more than a sounding board, to be honest)
 
Not a problem, just very strange that I have never thought about this
 
And the wikipedia article specifically says: "$A_0$ is a subring of $A$; in particular, the additive identity 0 and the multiplicative identity 1 are homogeneous elements of degree zero."
Which makes it sound like your statement more than mine.
 
Yep, which breaks my mind
Oh well I will afk, and hopefully someone who knows can come by
 
3:30 PM
Oh, this looks relevant:
1
Q: Is the zero ideal of a graded ring considered homogeneous?

Sheaf KeefI'm working through Vakil's notes, and he defines a $\Bbb{Z}^{\ge0}$-graded ring $$S_{\bullet}=\oplus_{n\ge0} S_n$$ and $X:=\operatorname{Proj}S_{\bullet}$ to be the collection of homogeneous prime ideals of $S_{\bullet}$ not containing $S_+:=\oplus_{n>0}S_n$. My question is: If $S_{\bulle...

particularly the answer.
 
Homogeneous of all degrees, very nice
What a loophole!
 
See also the comments, though.
 
3:49 PM
@BalarkaSen That and the fact that people here a genuinely nice (or at least that's my point of view)
 
does something mean to Astyx
 
Except for Daminark, of course
 
@Astyx Speak for yourself! :P
But yeah, that is true.
 
Ok, ok, I'll leave, stop bossing me :(
 
bosses never stop bossing
 
3:54 PM
Words to live by
 
Heh, now I want to make a game set in an office, where the boss fight is literally that you fight your boss
 
rofl
That's a really good idea man where do you get those
 
Lol, thanks!
 

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