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12:04 AM
Hello!! Which is the degree of the curve $f(x, y)=1+2x-x^2$ at the point $(0,0)$?? Is it $0$??
 
Prof. @Ted - surprisingly I got that their covariance is 0, rather quickly - And I don't think I made any mistake. It's surprising, though
 
12:26 AM
Yup, @Studentmath, it is. Another example of dependence with covariance $0$.
@MaryStar: First, the point isn't in the curve. Second, what is degree of a curve at a point?
 
@TedShifrin Look here the question has changed.
 
@TedShifrin "Let $F$ be any curve, $P = (0, 0)$. Write $F = F_m +F_{m+1}+···+F_n$, where $F_i$
is a form in $k[X,Y ]$ of degree $i$, $Fm \neq 0$. We define $m$ to be the multiplicity of $F$ at $P = (0, 0)$, write $m = m_P (F)$. " Which is this $m$ in this case??
 
Ah, multiiplicity, not degree. If the point isn't on the curve, it's $0$. Silly question ...
 
12:52 AM
It's strange that the mathematical notation is not global.
It's late. Good night.
 
1:23 AM
@Ted It's frustrating to see PSQs get highly upvoted...
 
PSQ?
@Mike: You enjoying your Balarka sabbatical?
 
Problem-statement question... aka, someone posting their homework with no thought of their own.
I'm too busy doing commutative algebra to enjoy anything :)
 
Good to hear :)
 
1:55 AM
I just want to say thanks for closing my art post the other day. I reposted it on Facebook where it belongs according to you guys. You're all wrong though and are unaware. Good luck. I'll be back when I've put some good hours in at work and have time to study the unbroken symmetries that are in Math lala land.
You're all tripping pretty badly and are also unaware of that.
 
2:40 AM
@EnjoysMath where is your art post? i couldnt find it
where did chris'ssis tell me to make 1 into n-n+1, what does that do?
i meant why did she tell me to do it
i got so far:
$$\sum \limits_{n=1}^\infty \frac{1}{n(n+1)(n+1)!}=\sum \limits_{n=1}^\infty \frac{(n-1)!}{(n+1)!^2}$$
@kaj are you good at big sums?
finals this week @kaj
?
 
3:02 AM
i cant get rid of that square on the factorial noooo
 
3:14 AM
@Integrator how do i get rid of the square in the bottom factorial?
professor!
@zach everyone logs in and noone talks hehe. will you tell me how to get rid of the square on my bottom factorial?
 
I'm likely not good enough to do so :P
 
noooo zachhhhh it is just a big sum thingy
 
@Ted have any insightful comments for my problem?
24 hours ago, by Zach Saucier
What trick am I missing? This problem is too nasty to solve the normal way

"Let $\textbf{n}$ be the outer unit normal of the elliptical half-shell $S: 4x^2 + 9y^2 + 36 z^2 = 36, z \geq 0$, and let
$\textbf{F} = 5 y \textbf{i} + 5 x^2 \textbf{j} + 5 (x^2 + y^4)^{3/2} \sin e^{\sqrt{xyz}}\textbf{k}$.
Find the value of $\mathop{\iint}_S \big(\nabla \times F) \cdot \textbf{n}~\textrm{d}\sigma$

Hint: One parametrization of the ellipse at the base of the shell is $x = 3 \cos t, y = 2 \sin t, 0 \leq t \leq 2 \pi$. But you can avoid a line integral altogether."
 
hi @emrakul
 
@beginner His question was deleted, I think, so you won't be able to see it.
 
3:24 AM
@MikeMiller he seems really upset about it, was it a big question with lots of effort?
 
It was originally closed as not being math within the scope of the help center, if I recall. I'm not him, so I won't comment on the amount of effort involved.
 
maybe he shouldnt complain so much when it doesnt achieve anything and we arent the people who did it
@MikeMiller help with getting that square gone?
$$\sum \limits_{n=1}^\infty \frac{1}{n(n+1)(n+1)!}=\sum \limits_{n=1}^\infty \frac{(n-1)!}{(n+1)!^2}$$
 
@beginner Try thinking about the definition of $(n+1)!$.
 
oh wow that is it!!!
thank you
 
Hi @zach @beginner
 
3:29 AM
hi @ted how are you?
$(abcd)^2=a^2b^2c^2d^2$ right?
 
how can you avoid a line integral (although it's not a bad one)? @zach
 
but $(a+b)^2=a^2+2ab+b^2$
$$\sum \limits_{n=1}^\infty \frac{1}{n(n+1)(n+1)!}=\sum \limits_{n=1}^\infty \frac{(n-1)!}{(n+1)!^2}=\sum \limits_{n=1}^\infty \frac{1}{(n+1)^2n^2(n-1)!}$$
 
She wants you to play with the numerator @beginner
 
the 1+n-n thing. Is that so i can make it into three sums?
 
Two sums, one with a $-$, maybe :)
 
3:33 AM
ohhhhhhh
thank you!
was that a clue to me that you deleted before i saw integrator?
 
user61230
@beginner Hello!
 
@Emrakul how are you? why is your name blue?
 
user61230
I'm a moderator on another Stack Exchange site, and moderators' names always appear in blue.
 
user61230
Also, I'm doing rather well!
 
glad you're sure of that, @Emrakul
 
3:41 AM
@Emrakul do you ban people who are too young to be on here, im not cause i am 13, just wondering
 
user61230
Moderators typically don't handle those cases ourselves, but yeah, it's against the ToS.
 
@Emrakul oh cool, who handles it? just cause my younger brother friend wanted to join and i told him he couldnt cause he would get caught
 
usually, you don't get caught if you don't tell anyone... :)
 
user61230
Stack Exchange staff. Once stuff starts getting into legal territory, it's (almost always) out of moderator jurisdiction.
 
whispers to @beginner You act more mature than many much older than you.
 
3:44 AM
@TedShifrin yay :) does that include balarka :P
@Emrakul thank you very much! do you enjoy math?
 
Who? :) You're both younger than most of my clothing ;)
3
 
@TedShifrin hahahaha, you need to go shopping maybe :)
 
@Zach, no reply?
 
I was catching up on the transcript
G'day @Ted :)
 
There's some exciting stuff to catch up on for sure :P
 
3:48 AM
snores
 
user61230
@beginner Yeah! I probably barely qualify as "enthusiast," though. Considering switching to a math major.
 
@Emrakul really?? that is great, i think it is really eye opening
 
user61230
Hopefully at some point I'll move from "enthusiast" to "somewhat knowledgeable."
 
@TedShifrin haha
did you see the problem I'm stuck on?
 
I vote you should, @Emrakul!
 
3:50 AM
Watch out, @Emrakul: The world is full of mean ol' math profs :D
 
I'm terrible at finding tricks to do things
 
@Emrakul i hope to make that move aswell
@MikeMiller yes we should have a vote on the unbiased math chat room
 
this is an important point, @Zach ... What surface to pick ...?
 
user61230
@Ted [grins maniacally]
 
user61230
Physics (where I am) and math luckily are closely dependent, so I have some time to decide.
 
3:51 AM
doesn't it give the surface to use?
 
Anonymous
@Emrakul You should follow Freeman Dyson
 
I've lured my share from physics to math @Emrakul
 
@TedShifrin am i meant to keep getting smaller sums spat out?

$$=\sum \limits_{n=1}^\infty \frac{1}{(n+1)^2n^2(n-1)!}=\sum \limits_{n=1}^\infty \frac{1}{(n+1)^2 n^2(n-2)!} - \sum \limits_{n=1}^\infty \frac{1}{(n+1)^2 n (n-1)!}$$
 
Do you always follow orders, @Zach? Break the rules!
 
haha, that's not what we're taught to do in math class, now is it?
 
Anonymous
3:53 AM
@TedShifrin Not doing physics,but only math is similar to eating and not learning to cook.
 
I didn't pay close attention, @beginner. What does the sum become with that algebra?
 
if i keep doing it, i will get rid of the bottom factorial i guess
 
So what is learning to cook but not eating?
 
i guess there will be an $e$ in there?
 
user61230
@Ted Oh? (Wait, are you the one I wanted to talk with about how calculus is taught horribly incorrectly?)
 
3:54 AM
i love cooking ... :)
 
Anonymous
Killing yourselves :D You won't survive not eating 0_0
 
I dunno :D a few of my students who come here can complain about me :)
 
my internet is lagging and the messages come through in big blocks :(
 
Yup @beginner
 
any more hints, @Ted?
 
3:57 AM
@TedShifrin it goes to $$=\sum \limits_{n=1}^\infty \frac{1-n(n-2)}{(n+1)^2n^2(n-2)!}$$ and i have to get an $e$ out somewhere hmm
 
Change surfaces @Zach ... Only the boundary curve must stay the same.
Way worse @beginner
 
@TedShifrin oh woops, i thought that might be the what becomes with algebra hehe, but the $e$ guess was right
 
Simplify without fretting about the factorial, first.
 
oh wow i cant believe i missed the thing on the right denumenator!
 
4:01 AM
I mean ... Go back to the original problem, right? @beginner
 
@TedShifrin the very start?
 
@TedShifrin can I ask you for some advice?
 
i keep getting kicked and locked out of the chat
 
@Ted Thanks for the help, I didn't remember we could do that :P
 
@TedShifrin did you want me to go all the way back to the left here, or was my right simplification the right direction?$$\sum \limits_{n=1}^\infty \frac{1}{n(n+1)(n+1)!}=\sum \limits_{n=1}^\infty \frac{(n-1)!}{(n+1)!^2}$$
 
4:11 AM
Use @chris'ssis idea on the original, left. @beginner
@Twink Sure, but it's my bedtime.
 
@TedShifrin it's quick
 
Ok
Maybe it knows you're too young now, @beginner :D
 
if this is wrong i wont bother you with it anymore because i clearly need to read more about these things
$$e\sum \limits_{n=1}^\infty \frac{1}{n(n+1)^2}$$
@TedShifrin noooooo i am 12-05-2001(age 13)
@Twink 6 of what? 10?
 
yes
 
if you retake you would get the 5 and whatever you get again, which averages to at best a 7.5 though worth twice the value?
 
4:16 AM
To be blunt, @Twink: Whether a 5 or a 6, this may be an indicator that grad school may not be the best idea.
 
@TedShifrin I can do better if I try hard :(
:'(
 
and it fills up another class slot, which you could just do well in next time?
so you are only lowering your gpa
and you can study it in your break
 
@TedShifrin I haven't been dedicating the time I should to school
 
my brothers when they were in uni spent every saturday working all day as a rule(and they worked in the week by going to all lectures and tutorials), and sunday was rest day
are you doing something like that?
 
But maybe that means you don't live and breathe math, which is great, but not right for being a grad student.
 
4:20 AM
do you think a student with a 6 won't be accepted in a good university?
 
i am starting to live and breathe math, and my parents said they will ban me from doing more than two hours of math a day, unless i do all my other classes for atleast as long lol
 
I can't speak for your country.
 
@Twink you need to have an upward trend and then it doesnt matter
 
@TedShifrin Hello Mr.
 
@Twink if you were a 5 average but your last year you got all 10s
you are fine(assuming they were actually last year courses, you can't do first year electives to fill it)
 
4:21 AM
hi @Pedro
 
@TedShifrin what about for USA?
 
my parents are both professors, but not in USA or math
 
@beginner: I suspect you still need some calculus, even after the trick. You're not yet ready for all of chris'ssis's gymnastics.
 
thank you.
 
So was this wrong? i will go and do calculus if it is wrong
$$\sum \limits_{n=1}^\infty \frac{1}{n(n+1)(n+1)!}=e\sum \limits_{n=1}^\infty \frac{1}{n(n+1)^2}$$
balarka gave me hammocks book of proofs(set theory) to do first
 
4:24 AM
@beginner: Be careful giving advice about things where your parents may be knowledgeable but you're not :)
 
@TedShifrin ok hehe, you are right
i think what i was doing is called prestige bias
 
you lost the factorial, @beginner. You need to simplify the $1/(n(n+1))$ part, but the factorial stays.
ok, I have two lectures to give tomorrow I've never given before. G'night!
 
@TedShifrin okay thank you for all of your help ted, i will go learn more calculus :)
good luck with the lectures hehe
 
5:07 AM
is this correct math notation?

$$n=\frac{x}{1+x}, \lim \limits_{x\to\infty} \frac{x}{x+1}=1$$

but since $n$ never truly reaches $1$, i can say that $\operatorname{sup} n = 1$
 
@beginner What is $\sup [0,1)$?
 
@pedro 1
 
hello karl.
 
so my way of saying $n=$ stuff and limit thing is all alright in math language
 
hi @MikeMiller and @pedro
 
5:13 AM
@beginner OK. What is $\sup\{x(1+x)^{-1}:x>0\}$?
 
@Karl I have a question for you
 
@KarlKronenfeld You gave me a heart attack.
 
what... is your favorite color?
 
I submitted those notes to my prof.
 
@PedroTamaroff 1?
 
5:14 AM
@beginner Yes.
 
@PedroTamaroff ok so my thingy was right
 
@PedroTamaroff I still don't get it.
 
how can I ping someone that doesn't appear when write @
 
@MikeMiller Haven't given enough of a shit to decide.
 
also, if $J=(a_1, \dots, a_n)$ is a regular sequence in a Noetherian local ring $A$, why is $J^k/J^{k+1}$ isomorphic as an $A$-mod to a sum of copies of $A/J$?
 
5:14 AM
@KarlKronenfeld You said one of the theorems was wrong.
 
?
 
@PedroTamaroff oh lol
 
@KarlKronenfeld I was like "OHFU--"
 
@MikeMiller I know regular sequences are important. I haven't learned about them, though.
 
@MikeMiller I think I got something.
Suppose $I=(a,b)$.
Wait, let me think about it myself a little bit more.
In a Noetherian local ring, what do you know about flatness @MikeMiller ?
 
5:19 AM
I don't know anything about anything, @PedroTamaroff
 
@beginner it is not correct:
$$
\begin{align}
\sum_{n=1}^\infty\frac1{n(n+1)(n+1)!}
&=\sum_{n=1}^\infty\left(\frac1n-\frac1{n+1}\right)\frac1{(n+1)!}\\
&=\sum_{n=1}^\infty\left(\frac1n-\frac1{n+1}-\frac1{(n+1)^2}\right)\frac1{n!}\\
&=\sum_{n=1}^\infty\left(\frac1{nn!}-\frac1{(n+1)(n+1)!}\right)-\sum_{n=1}^\infty\frac1{(n+1)!}\\[3pt]
&=1-(e-2)\\[12pt]
&=3-e
\end{align}
$$
 
@MikeMiller Dude, I don't think what Matsumura says it too hard to see.
I mean.
But, I think I'm not off here.
It is hard.
 
say you after an hour :P
 
$A/(a,b)$ is $\overline A/\overline{(a)}$ where we mod out $(b)$.
Suppose that $A/(a)\to A/(a)$ where $\overline x\to\overline{bx}$ is injective.
 
mhm
 
5:25 AM
I keep losing my train of thought.
I'm cracking up on that.
 
It's true.
There are also more stars in the universe than Adams.
 
5:43 AM
Hey is $$\lim_{n \to \infty} \sin(nx)$$ conintuous?
 
The set of $x$ for which that limits exists has measure cero.
So your function is not defined almost everywhere.
 
the dual notion of nintuity.
 
Bummer.
 
Yeah, that's what I am thinking
 
@KarlKronenfeld LOL.
@KarlKronenfeld
I have given Mike's problem some thought.
I am looking at the case of two generators $(a,b)$.
 
5:46 AM
So then how can I come up with a continuous function on $[0,1]$ that does not have bounded variation on any closed sub-interval?
 
So $x\mapsto ax$ and $\bar x\mapsto \overline{bx}$ are both injective.
I am trying to use that $A/(a)\otimes A/(b)= A/(a,b)$.
 
Every function I keep thinking is discontinuous a.e.
 
Also that the second map gives an isomorphism of $A/(a)$ onto $(b,a)/(a)$ and $A/(a,b)$ is $A/(a)$ over $(a,b)/(a)$.
One already knows that $A/(a)$ is isomorphic to $(a)^k/(a)^{k+1}$ for any $k$.
 
6:02 AM
Indeed, I have finals Thursday and Friday @beginner. What sums are you looking at?
 
@KajHansen hey
 
Hey there
 
You should be asleep at this moment.
 
It's not that late, and I have some homework to do.
 
1AM is primetime for homework.
 
6:05 AM
Ah, ODEs are precisely something which can keep you awake @Kaj
@MikeMiller We're nocturnal.
 
Indeed it is @Mike.
 
Rather, conocturnal.
 
I swear, the vast majority of my undergraduate work has been completed between 12 AM and 5 AM.
 
Quantum Field Theory, @Karl? At this time?
 
6:07 AM
If I didn't have classes at 9 and teaching at 8 I'd probably swap my sleep cycle, be awake until 5 and sleep until 11-12. But I do and so don't.
 
What the hell are physicists really doing? Their topic names are getting bigger and bigger to fit in the margin of their PhD paper.
Twistor theoretic topological quantum field theory
SHEESH.
 
6:24 AM
@KajHansen i am doing sums that chris'ssis gives me
@KajHansen finals are so soon, i am sure you will do well
 
you should really do set theory @beginner
 
is $$\sum \limits_{n=1}^\infty \frac{1}{n(n+1)}= \sum \limits_{n=1}^\infty \frac{1}{n}-\sum \limits_{n=1}^\infty \frac{1}{(n+1)}$$ well known?
i can see how to prove it easily enough, even with the clue you gave me balarka, but i dont know how someone initially saw it
@BalarkaSen i am doing it as well i promise
 
@beginner what do you mean well known?
 
@BalarkaSen robjohn, you and chris'ssis all knew it instantly on seeing the LHS
 
sure, it's called partial fractions.
pretty important while doing these stuff.
 
6:27 AM
@BalarkaSen partial fractions is a general thing, or just that specifically?
 
@beginner just a bit of practice.
@beginner a bit general. google it.
 
@BalarkaSen thanks balarka
@robjohn thanks robjohn that is really clever
hi @hat
 
Hello
 
how much have you learned set theory, @beginner?
 
why are you the hat man?
heaps
 
6:35 AM
Does there exist hat men? I don't know
 
Sure. Existence is easy. Uniqueness, on the other hand...
 
@beginner countability?
@KajHansen LOL
 
@beginner The important thing is that we cannot factor terms as it looks you were trying to do. (factoring out the $e$)
 
@robjohn i was trying to say that since $\sum \limits_{n=0}^\infty \frac{1}{n!}=e$ then, $\sum \limits_{n=0}^\infty \frac{1}{an!}=e\sum \limits_{n=0}^\infty \frac{1}{a}$
is this false?
 
@beginner Yes. This is a telescoping series
 
6:37 AM
woops
 
@KajHansen I got to know about Kummer's proof of FLT. Or at least a sketch of it.
 
@beginner that is true, as long as you are factoring out a constant, but you were factoring out non-constants
 
That's pretty cool
 
It came up in a number theory lecture explaining the connection between the representation theory of Gal(\bar Q/Q) and modern algebraic number theory
 
@robjohn so it is because $\sum \limits_{n=0}^\infty \frac{1}{n!}\sum \limits_{n=0}^\infty \frac{1}{n!}\ne \sum \limits_{n=0}^\infty \frac{1}{(n!)^2}$?
 
6:39 AM
A real-life lecture, or a Youtube video?
 
real-life lecture.
 
@beginner That is definitely a case of what I am saying
 
really?? shouldn't that be $e^2$??
 
That's never true for general sums @beginner. Consider $(1 + 3)(2 + 4) \neq (1 \cdot 2 + 3 \cdot 4)$.
 
@beginner the left side is $e^2$, the right side is something less
 
6:41 AM
@KajHansen the two dimensional representation theory was fun. you know about elliptic curves?
 
ohhhh i see, $\left(\sum \limits_{n=0}^\infty \frac{1}{n!}\right)\left(\sum \limits_{n=0}^\infty \frac{1}{n!}\right)\ne \sum \limits_{n=0}^\infty \frac{1}{n!}\sum \limits_{n=0}^\infty \frac{1}{n!}$
 
Not really @BalarkaSen
 
that was my confusion, thank you!
 
@beginner no, those are equal
 
OK, @KajHansen. You know how the plot of y^2 = x^3 + ax + b in R^2 looks like in general?
 
6:42 AM
Yeah
 
but... kajs thing made me think that we added each part of the sum..
 
OK, cool. @KajHansen now imagine a torus in C^2.
 
Okay
 
Disect it right from the middle so you get two handles.
 
Sure
 
6:43 AM
Now if you can press one of your handle onto a copy of C, you'll get two disjoint circles as a result right?
 
ok ok i probably dont get it, so i will have to go read some stuff. so sums are treated with parentheses, i just cant multiply the inside of the sum, only note that the sum of two series is the series of their sum, but not multiplication
 
What I'm saying is $\sum a_n \sum b_n \neq \sum a_nb_n$
 
but $\sum a_n + \sum b_n = \sum (a_n + b_n)$ right?
 
Whatever you mean by "press" @BalarkaSen
Oh I think I see
Yes @beginner
 
@KajHansen ok thank you :)
 
6:45 AM
@KajHansen it's like getting a trace onto the copy of C by which you're disecting
the edge of one of the handles, two disjoint circles
 
[Heat Equation in Polar Co-ordinates with additional terms][math.stackexchange.com/questions/1056956/…
Hello!
 
@mod0 [text](link) that is the right way.
 
if you're seeing it, @Kaj, then you have two circles and in the projective sense if you delete a point in one of the circle and replace it by the point at infinity, then it precisely looks like the plot of an elliptic curve, no?
 
Thanks Hat Man! I will ensure I post it like that next time. I guess I missed the fact that it is a parenthesis. Sorry!
 
Yeah, I think I see it, but this is fairly hand-wavy lol
 
6:49 AM
Fact : Riemann surface of an elliptic curve over C is a torus.
What I gave you is a visual represetation of this fact :P
 
@KajHansen Go outside and look at the moon
 
how do i write the the set $\{2,4,8,16,32,64,\dots \}$ in set builder notation? i see that it is $x=2^k$ do i write $\{x\in\Bbb{Z}:x=2^k, k\in\Bbb{Z}\}$?
 
I saw your status @Zach. I will soon
 
I'd say $\{2^k : k \in \mathbb{N} \}$.
 
6:50 AM
sorry i mean
 
Did any of have some glasses moonshine?
 
okay yeah that is right
thanks
 
I don't see the point of moon then :P
 
There are all sorts of conventions for that though @beginner. There isn't really one "right" way.
 
@KajHansen yeah that is true, but yours was short and sweet :). i meant to write $k\in\Bbb N$ but even still mine was longer
 
6:52 AM
OK, @Kaj. Now so the ellcurve y^2 = x^3 + ax + b over C is just a torus, which is Z \times Z. So the m-torsion points (i.e., the points (x, y) in the ellcurve such that (x, y) + (x, y) + ... + (x, y) is (0, 0) after m times summing, where plus is the elliptic curve group operation) form the group Z/m \times Z/m
does that make sense?
 
Sure. Again, I don't have proof for any of this, but I follow what you're saying at face value.
Also, I don't know what the "elliptic curve group operation" is.
 
@KajHansen ah just take two points in the graph of the elliptic curve and join it by a line.
the line intersects at another point on the curve
so you have a sense of binary operation.
 
point + point = point of intersection?
 
mmhmm
 
And that will be only one point..?
 
6:55 AM
yep
it can all be proved
 
Not convinced, but sure.
 
in other words elliptic curves are godly
 
@Kaj look at a graph, convince yourself.
LEL @Karl
 
they're alright
 
do you take the shell of this torus, or the torus is solid?
 
6:56 AM
In some cases that's not true. Some points you can join and the line won't intersect anywhere else.
 
the real cool stuff is the behavior of rational or integral points of an elliptic curve $E/\Bbb Q$ but that's too hard
 
@KajHansen it will. at the point at infinity.
 
I guess it's true where it matters, at the torsion points?
 
maybe those points arent closed
 
@MikeMiller Still thinking about the regular sequence problem?
 
6:57 AM
Oh ok.
 
you should be all projective when thinking about ellcurves @Kaj
 
@KarlKronenfeld not really; in the solution I wrote down I just said "Matsumura says this is true but I don't know why" and moved on :P
if you've ideas I'm glad to listen
 
no, I'm stuck
 
@Balarka That makes me feel naughty.
4
 
he seems to think it's a triviality, but he does that for a lot of things
 
6:58 AM
I know it's a generalization of a much easier result matsumura stated earlier on.
 
what result are you referring to?
 
something he said after the artinian implies noetherian theorem
 
@KajHansen so the m-torsion points forms the group Z/m \times Z/m.
 
I don't remember the exact result
 
6:59 AM
For what $m$ do $m$-torsion points exist?
 
don't see it, but meh
 

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