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10:01 AM
~$unlearn rickroll
 
@BalarkaSen Forgot command: rickroll
 
@BalarkaSen Learned command: realmusic
 
~$realmusic
 
10:01 AM
Good, good.
@Pseudohuman Have a nice day
 
Yes.
(thanks for asking)
 
Ha! Rickroll got'eM!
 
kektroll'd bruh
 
Alex: Actually, ee=1 is a bad idea, since as soon we have 1+1=0, then x+x=0 for all x, making the structure quite trivial
 
10:13 AM
@BalarkaSen Learned command: tetrathonk
 
We should just teach this bot how to meme
 
~$learn love This is love
 
@BalarkaSen Learned command: love
 
~$learn guessIlldie guessIlldie
 
10:23 AM
@Secret Learned command: guessIlldie
 
~$guessIlldie
 
~$help
 
Available commands:
Builtin: asm, coinflip, cointoss, eval, fibonacci, help, joinroom, learn, leaveroom, list, listcommands, rand, rolldice, room, unlearn, wotd
Learned: /o_o, /why, addition, calloakbot, cat, conv, downgoat, galactic_transform, greet, guessilldie, gypsytime, lenny, look_of_disapproval, love, matrix, mithrandy, o_o, poem, realmusic, salute, scary, se, tetrathonk, tif, wikilookup, xyz
 
Look at all shitty commands we put in there lmao
 
10:27 AM
~$matrix
 
ok that was unexpected
 
I guessed it'd be related to the movie
 
How do we write utility commands like the wikilookup that take user input arguments?
 
I don't think we can
Not sure
 
10:28 AM
~$list
 
Available commands:
Builtin: asm, coinflip, cointoss, eval, fibonacci, help, joinroom, learn, leaveroom, list, listcommands, rand, rolldice, room, unlearn, wotd
Learned: /o_o, /why, addition, calloakbot, cat, conv, downgoat, galactic_transform, greet, guessilldie, gypsytime, lenny, look_of_disapproval, love, matrix, mithrandy, o_o, poem, realmusic, salute, scary, se, tetrathonk, tif, wikilookup, xyz
 
3 commands for the same thing -_-
~$wikilookup manifolds
 
In mathematics, a manifold is a topological space that locally resembles Euclidean space near each point. More precisely, each point of an n-dimensional manifold has a neighbourhood that is homeomorphic to the Euclidean space of dimension n. In this more precise terminology, a manifold is referred to as an n-manifold. One-dimensional manifolds include lines and circles, but not figure eights (because they have crossing points that are not locally homeomorphic to Euclidean 1-space). Two-dimensional manifolds are also called surfaces. Examples include the plane, the sphere, and the torus, which can...
 
hmm...
 
The dollar sign ($ or ) is a symbol primarily used to indicate the various units of currency around the world. The symbol can interchangeably have one or two vertical strokes. In common usage, the sign appears to the left of the amount specified, as in $1 (read: one dollar). == Origin == === Pillars of Hercules === A common hypothesis holds that the sign derives from the symbolic representation of the Pillars of Hercules – this representation can either have a banner separately around each pillar, or as in the Spanish coat of arms, a banner curling between them. In 1492, Ferdinand II of...
 
10:31 AM
~$learn test $1 $1^2
 
@Secret Learned command: test
 
~$test 5
 
5 5^2
 
noice
so thats how you do it
 
It seems to be bash like...
So I am guessing...
~$learn add2numbers $1+$2
 
10:33 AM
@Secret Learned command: add2numbers
 
~$add2numbers 6 4
 
6+4
 
@Pseudohuman 6+4
 
@BalarkaSen 10
 
It can do math
It's built in
 
10:34 AM
~$learn add2numbers eval($1+$2)
 
@Secret Command already exists.
 
u wot m8
 
~$unlearn add2numbers
 
@Secret Forgot command: add2numbers
 
~$learn add2numbers eval($1+$2)
 
10:35 AM
@Secret Learned command: add2numbers
 
~$add2numbers 6 4
 
eval(6+4)
 
@Pseudohuman sum n = 1 to infinity 1/n^2
 
Did someone program a bot for this chat?
 
@Perturbative It's an SE-wide bot
 
10:35 AM
@BalarkaSen sum_(n=1)^∞ 1/n^2 = π^2/6
 
Dank
 
I'm guessing it has like built in Mathematica
 
~$unlearn add2numbers
 
@Secret Forgot command: add2numbers
 
ok, let's try plotting
 
10:36 AM
~$unlearn test
 
@BalarkaSen Forgot command: test
 
@pseudohuman plot(x^2+y^2)
 
@Perturbative I think it's built in wolframalpha tbh
 
@BalarkaSen I do not understand.
 
That wasn't for you
@Pseudohuman plot x^2 + y^2 = 1
 
10:38 AM
 
@pseudohuman plot3d x^2+y^2 = z
 
If only Pseudohuman could write emails to my profs for me
Spent 17 minutes typing up the last one
Probably will get a one line response
 
Well in australia, we only wrote one long email, after that its mostly one or two liner unless it is an announcement thing
 
@Pseudohuman plot x^4 - x^2 + y^2 = 0
 
10:41 AM
This bot is handy
 
@Pseudohuman plot x^2+y^2+z^2 = 1
 
YES!
@Pseudohuman plot x^2 + y^2 - z = 0
 
@Pseudohuman Solve dx/dt = ax - bxy, dy/dt = cxy - dy
 
10:42 AM
this will be very useful in our discussions
 
@BalarkaSen I do not understand.
 
Oh well
 
@Pseudohuman diff(x^2+1)
 
@Pseudohuman differentiate x^2 + 1
 
10:43 AM
uh....?
 
@BalarkaSen d/dx(x^2 + 1) = 2 x
 
It's like Wolfram Alpha
 
@Pseudohuman Solve dy/dx = 1
 
@Secret I do not understand.
@Secret I do not understand.
@Secret y'(x) = 1
 
ok it works
we can solve ODE with this
 
10:44 AM
That's not a solution
It's just writing down the same thing
 
ooops...
 
@Pseudohuman Solve y'(x) = 1
 
@BalarkaSen y'(x) = 1
 
@Pseudohuman Solve y'(x) = 1 for y
 
@Secret y'(x) y(x) = 1
@Secret y'(x) = 1
@Secret y'(1) = 1
@Secret y'(x) = 1
 
10:46 AM
grrrrr...
 
@Pseudohuman Solve y'(x) = 1 for y(0) = 0
 
@BalarkaSen {y'(x) = y(0), y(0) = 0}
 
Not very good at math are you
Ok I think that's enough meddling with this thing today
~$leaveroom
 
agreed
 
@Pseudohuman ~$leaveroom
 
10:49 AM
@BalarkaSen I do not understand.
 
@Pseudohuman leaveroom
 
@Secret I do not understand.
 
~$help
 
Available commands:
Builtin: asm, coinflip, cointoss, eval, fibonacci, help, joinroom, learn, leaveroom, list, listcommands, rand, rolldice, room, unlearn, wotd
Learned: /o_o, /why, addition, calloakbot, cat, conv, downgoat, galactic_transform, greet, guessilldie, gypsytime, lenny, look_of_disapproval, love, matrix, mithrandy, o_o, poem, realmusic, salute, scary, se, tetrathonk, tif, wikilookup, xyz
 
10:49 AM
~$joinroom
 
~$leaveroom
 
That doesn't do anything
 
~$joinroom
nah....
 
@Pseudohuman Goodbye
 
I don't know why you say goodbye; I say hello!
(according to The Beatles song Hello, Goodbye released as a single in November 1967)
 
10:51 AM
I really wish there's a source code somewhere so we don't need to trial and error
 
~$leaveroom 36
Aha
Nailed it
 
So what we knew so far:
Pseudohuman can do basic arithmetic and calculus, infinite sums and plot 2D, 3D plots
and we can call her in any chat rooms via joinroom
the learn command also accept bash like arguments
 
11:15 AM
@Secret Hey bro
 
11:26 AM
?
 
I have posted a question on the main site but no replies after 2-3 hours as of now
0
Q: Limit with double derivative and modulus

Tanuj Inspite of knowing a thing or two about limits , this particular question just completely stumps me . What does it want me to do and how do I operate with the modulus ? Taking double derivative of $x^p \sin(\frac{1}{x})+x|x^3|$ also seems too complicated.How do I approach this ?

would you kindly have a look if you're free ?
 
ok, so it is a piecemeal function, which means the x=0 ase is taken cared of. All you need to do is to show that the limit as you approach zero from the LHS and RHS are the same. This thus split the f into 2 regions, one where x>0 and one where x < 0. Now you should be able to remove the absolute value.
As for that sine thing, use squeeze theorem
 
11:48 AM
@BalarkaSen I have :')
 
Hi! Could you please tell me how to solve integral $$\int^{ \infty }_0\frac{x^4}{e^{x^2}+1}dx$$ and similiar ones?
 
@Secret How are you defining a cyclic order?
o/
Where did the bot go?
~$wikilookup test
 
~$joinroom 36
@Pseudohuman
 
@Pseudohuman
 
@Alex cyclic order should normally be like e.g. {a,b,c,d}, then a<b<c<d<a
 
11:53 AM
@Pseudohuman ~joinroom 36
 
@Secret How are you defining one rigorously I mean
 
ah, now I get why you ask me about ternery relation...
 
@PrzemysławP Hey , you can start by taking $x^2$ as another variable $t$
 
@Secret Were you avoiding them somehow?
 
no, it's just intitially I was too focused on the algebraic content and forgot that the ordering also need a relation to define it
 
11:55 AM
I can definitely agree that you did that :P
 
and in particular, while I know intuitively what cyclic order means, I am not aware that I need a ternery relation to do it rigorously
 
Well you probably need to know what it means rigorously to ensure people will agree with you
The bot has returned!
 
You have to call her from Sandbox
 
~$wikilookup test
 
Test(s) or TEST may refer to: Test (assessment), an assessment intended to measure the respondents' knowledge or other abilities List of tests == Science and technology == Experiment, a part of the scientific method Test method, a definitive procedure that produces a test result Statistical hypothesis testing, a method of making statistical decisions using experimental data === Biology, medicine, and psychology === Test (biology), the shell of sea urchins and certain microorganisms Medical test, to detect, diagnose, or monitor diseases Psychological testing, the use of behavioral samples to make...
 
11:57 AM
~$commands
~$help
 
Available commands:
Builtin: asm, coinflip, cointoss, eval, fibonacci, help, joinroom, learn, leaveroom, list, listcommands, rand, rolldice, room, unlearn, wotd
Learned: /o_o, /why, addition, calloakbot, cat, conv, downgoat, galactic_transform, greet, guessilldie, gypsytime, lenny, look_of_disapproval, love, matrix, mithrandy, o_o, poem, realmusic, salute, scary, se, tetrathonk, tif, wikilookup, xyz
 
~$tetrathonk
 
Excellent
 
@Alex Well, agree is too strong, they need to be able to comprehend first, which I agree with what you said about my lack of rigor
being one reason why people don't understand
 
11:58 AM
'Agree with you that you've given a cyclic order'*
 
ok
 
I wonder if sticking in a command anywhere works for the bot
I'm in ~$love with the damn train
Nope
 
~$learn test YOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO
 
@Alex Learned command: test
 
~$test
 
12:01 PM
YOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO
 
~$exec print('a')
~calloakboat
~$calloakboat
oh
~$calloakbot
 
/join 138769
 
~$downgoat
 
~$gypsytime
 
12:03 PM
IT'S TIME FOR A FORTUNE!!!
 
~$poem
 
Roses are red, Violets are blue, Haven't you got anything better to do?
 
btw, I found a better intuition to describe what I am constructing: Basically, I want to construct a finite field that is like a discrete analogue of the real projective line
 
@Secret What does that mean?
 
Recall the real projective line is what happens when you join both ends of the real number line with a point at infinity
I want to do something similar, but only with a finite set, so one of the elements will effectively be like a point at infinity
 
12:05 PM
Well that's one way to think of it, one point compactification
 
@BalarkaSen Command already exists.
 
oops freudian error
 
@BalarkaSen Learned command: upboat
 
@Secret How do you plan to construct this? Are you just doing $\Bbb P^1(\Bbb F_p)$?
Or you want a one point compactification for something?
Oh I'm not paying full attention, you just mean by analogy
You don't actually care about constructing something from the analogy :P
Probably a bad analogy though, but I think I see what you mean, but you want an order on the points you started with then?
 
12:10 PM
@PrzemysławP did you get it ?
 
@Tanuj Thank you. After some additional steps I now need to solve $\int^{ \infty }_1\frac{(lnb)^{5/2}}{(b+1)^2}db$
 
3 hours ago, by Secret
@AlessandroCodenotti Ah right, as finite fields cannot be linear ordered. I however don't know if they must be cyclically ordered like the Z/primes
Which makes me think you are missing your own objective ^
 
@PrzemysławP Okay
 
But I may have made some mistake on the way there. Am I moving in the right way even?
 
12:12 PM
Oh lord no
 
@PrzemysławP Can't say , every answer is different in integration , right ? But the final value after putting the limits should be same
 
And how will you make cyclicity work?
 
Intuitively it seems to make sense, since the function that maps the ordering of the elements to some positive real number adjoined with infinity will be continuous from the left, and the whole thing wrap around in a circle.

Rigorously, still working on it, for I don't know what happens to the ternery relation when you introduce an infinite element
 
It looks like you are now only dealing with putting a cyclic order on a finite set?
I.e. you made the problem super easy?
 
yeah, I think that's actually what happen, and one element being marked infinite, meaning that the sum of all other elements can never reach it. but I also want as much field structure as possible, which I am not sure how to do it yet
 
12:19 PM
No point calling it infinity since it doesn't serve any role that motivates that name now
Hell you have a finite set, you don't even need to adjoin an extra element
What sum?
 
@Alex You still interested in an answer to this?
 
Yep, haven't got around to looking at it again
They said something about measuring the failure of something from being a chain complex
 
That's true
So I can quickly tell you the formal story
 
uh... let the set be $S$, if I have some element $e \in S$ in the set such that for all $x,y \in S$ and $x,y \neq e$, $x+y < e$ then $e$ will be like an infinite element?

O wait, the whole thing is cyclic, so $e < x+y$. Ok nvm then, looks like I have been contradicting myself the whole time
 
It'd have to be this guy $\mathsf{C}^\bullet =(\Gamma(M,E\otimes\bigwedge^i T^*M),D)_{i\in \Bbb N}$ I guess
For $E\to M$ the bundle
 
12:24 PM
I thought I can make something like a non achimedian field with finite elements and then wrap the whole thing into a circle, but since the ordering is cyclic, there is no such thing as an element larger than all other element hence a notion of infinite element is undefined
 
@BalarkaSen Tell me in Grothendiecks dining room
/brb 2 min, need dat coffee
 
@Alex Right, if $E$ is a vector bundle over $M$, $\Gamma(E) = \{s \in C^\infty(M, E) : p \circ s = \text{id}\}$ the $\Bbb R$-vector space of sections of $E$, $D$ is a linear map $D : \Gamma(E \otimes \Lambda^k T^*M) \to \Gamma(E \otimes \Lambda^{k+1} T^* M)$ satisfying $D(\omega \otimes s) = d\omega \otimes s + (-1)^k \omega \wedge Ds$
Ok going there
I should have said $s \otimes \omega$. I got the order wrong there.
 
1:05 PM
@Secret cool--glad it worked!
 
1:49 PM
Hey my dudes
If I have a $C^0$ piecewise smooth manifold, is it possible to define the tangent bundle in some way
Using weak derivatives or something
 
You can define a notion of tangent bundle just for $C^0$ manifolds. But I do not think you can define classical tangent spaces on piecewise smooth manifolds on the non-smooth locus
Anyhow look up tangent microbundles
The construction is basically that if $M$ is a $C^0$-manifold with an atlas $U_\alpha$ and transition functions $\varphi_{\alpha\beta} : U_{\alpha} \cap U_\beta \to U_\alpha \cap U_\beta$, you can construct $M$ out of it as $\bigsqcup U_\alpha/x \sim \varphi_{\alpha\beta}(x)$
So now think of $\bigsqcup U_\alpha \times \text{TOP}(n)$ with $U_\alpha \times \text{TOP}(n)$ glued to $U_\beta \times \text{TOP}(n)$ glued by $\psi_{\alpha \beta} : (U_\alpha \cap U_\beta) \times \text{TOP}(n) \to (U_\alpha \cap U_\beta) \times \text{TOP}(n)$ defined as $\psi_{\alpha\beta}(x) = (\varphi_{\alpha \beta}(x), \varphi_{\alpha \beta}'(x))$
 
"introduced by the American mathematician John Milnor"
 
Where $\text{TOP}(n)$ is the germ of homeomorphisms of $\Bbb R^n$ fixing the origin, at the origin. And $\varphi'$ means the germ of $\varphi$ at the origin
 
Milnor again!
 
2:04 PM
This gives a $\text{TOP}(n)$-bundle on $M$ defined analogously as the tangent bundle of a smooth manifold, except fibers are now not tangent spaces, but space of topological germs
That's your tangent microbundle
Although you'll usually see the definition using the diagonal in $M \times M$ etc
@Slereah "Milnor, we meet again!"
 
What is TOP(n)
 
Germ of homeomorphisms $f : \Bbb R^n \to \Bbb R^n$ with $f(0) = 0$ at the origin.
The right analogue of the first derivative.
@Pseudohuman Google tangent microbundle theory
 
@BalarkaSen I do not understand.
 
bad bot
~$wikilookup microbundle
 
In mathematics, a microbundle is a generalization of the concept of vector bundle, introduced by the American mathematician John Milnor in 1964. It allows the creation of bundle-like objects in situations where they would not ordinarily be thought to exist. For example, the tangent bundle is defined for a smooth manifold but not a topological manifold. Use of microbundles allows the definition of a topological tangent bundle. == Definition == The precise definition of a microbundle follows. Let B be a topological space. Then a n-microbundle consists of a triple (E, i, p) where E is a topological...
 
@BalarkaSen Learned command: microbundles
 
Good.
~$learn longliveUSSR This is not a baboon
 
@BalarkaSen Learned command: longliveUSSR
 
morning
 
@MatheinBoulomenos Here's a strange one; my textbook says the ideal class group of $\Bbb Q(\sqrt{65})$ is isomorphic to $\Bbb Z/(2)$, generated by the class of $(5, \sqrt{65})$, even though the minkowski bound is $\approx 4$. Is there any special reason for this ($65$ being $1 \bmod 4$ probably complicates things, but I'm not sure how?)
 
2:20 PM
Actually I don't want to associate USSR with this
~$unlearn longliveUSSR
 
@BalarkaSen Forgot command: longliveUSSR
 
~$learn longliveRussia absolute madlad
 
@BalarkaSen Learned command: longliveRussia
 
@BalarkaSen Learned command: literotica
 
2:22 PM
~$help
 
Available commands:
Builtin: asm, coinflip, cointoss, eval, fibonacci, help, joinroom, learn, leaveroom, list, listcommands, rand, rolldice, room, unlearn, wotd
Learned: /o_o, /why, addition, calloakbot, cat, conv, downgoat, galactic_transform, greet, guessilldie, gypsytime, lenny, literotica, longliverussia, look_of_disapproval, love, matrix, microbundles, mithrandy, o_o, poem, realmusic, salute, scary, se, test, tetrathonk, tif, upboat, wikilookup, xyz
 
hmm...
you are making her russian...
 
I'm done here. Keep teaching this bot about meme and mathematics. Ideally test it out more in the Sandbox room, but you can call it from there by writing "~$joinroom 36" directly in the sandbox. Imma make it leave now
~$leaveroom 36
and im going to flee too
 
afternoon
f: X → Y means the function f maps the set X into the set Y.
what means these in layman terms, please?
Trying to explain this to my son in law
 
Do you know what a set is?
 
2:29 PM
like A B
 
Ahh
 
with interesection, union etc right ?
 
It's a collection {x_1,x_2,x_3,...} right?
 
yep
 
Could be an uncountable collection
 
2:30 PM
got it Alex
 
Like R
Real numbers
 
yes
@Alex following you
 
Okay so you know sets
Do you know the Cartesian product?
 
@Alex yes
 
So $X\times Y$ is a new set consisting of all ordered pairs (x,y) for x\in X and y\in Y
A function f:X\to Y is a subset of this
 
2:32 PM
@Alex got it.
 
Consisting of the pairs (x,f(x))
 
my man
but soz, please carry on
It will be useful in my explanation to my son in law
 
Well that's pretty much it
You have X\times Y
And the function can be seen as a subset of X\times Y, where we have all pairs (x,f(x)) and for each x\in X, there is precisely one associated f(x) for our pair (x,f(x))
So we don't have f(x)=y and f(x)=y' for y\ne y'
 
Thanks Alex
btw
when you are using LaTex here, how do you compile the formula?
you use LaTex the software
 
@MatheinBoulomenos Ahhhh I was using the wrong minimal polynomial, that's something
 
2:45 PM
No, the calculation I just posted was wrong
 
@MatheinBoulomenos Fair, I was using the wrong polynomial anyway though
 
Go there, and use that bookmarklet
 
If I send $x$ to $\frac{1+\sqrt{65}}{2}$, then $\sqrt{65}$ is the image of $2x-1$, not $x$
 
@Alex awesome
 
2:47 PM
I wasn't actually latexing it though, just habit (since I knew you weren't rendering it)
 
Ok guys, I now have a very general result about division by zero algebras that does not even care if the distributive structure is associative:
 
division by zero algebras?
What the sheeet does that mean
afk coffee
$\Bbb Q/\Bbb Z$
 
So $\Bbb Z[\frac{1+\sqrt{65}}{2}]/(5,\sqrt{65}) \cong \Bbb Z[x]/(5,x^2-x-16,2x-1) \cong \Bbb F_5[x]/(x^2-x-16,2x-1)$ $= \Bbb F_5[x]/(x^2+4x+4,2x+4) =\Bbb F_5[x]/((x+2)^2,2(x+2)) = \Bbb F_5[x]/(x+2) \cong \Bbb F_5$
So the ideal norm is $5$
 
that still contradicts the Minkowski bound
so there's something wrong with the statement
 
2:50 PM
Indeed, it's confusing
because
The book defines the minkowski bound to be $M_K = \left(\frac{2}{\pi}\right)^s \sqrt{\lvert d_K\rvert}$
Without the factor of $\frac{n!}{n^n}$ (degree of $K$)
 
that's odd
 
Indeed
 
Division by zero algebra Theorem 1: Given any algebraic structure $(S,\cdot,+)$ such that $\cdot$ left distribute over $+$ and the underlying set $S$ arbitrary, and that there exists a left or right multiplicative inverse to a left or right additive identity, the addition structure must have the following form: Proof to be typed shortly
 
So while I'm looking for primes lying over $2$ and $3$, they're giving the generator as the class of a prime over $5$
 
If you use the weaker Minkowski bound, then there's no contradiction, but that doesn't really help, since the stronger Minkowski bound holds
 
2:52 PM
yeah
 
@Secret math.stackexchange.com/questions/2664696/… check it out if you got time
 
Proof:
 
@Secret Why don't you just read a nice noncommutative algebra text?
 
you don't have to look at the primes above $3$, since $\left(\frac{65}{3}\right)=-1$, $3$ is inert in $\mathcal O_{\Bbb Q(\sqrt{65})}$, so the unique prime above $3$ has ideal norm $9$
 
@Secret Dennis and Farb, or Lam
 
2:55 PM
Yeah, I mean.. I usually just write down the primes less than the minkowski bound and then check the legendre/kronecker symbol, I just meant that I did that for both $2$ and for $3$
not that I'm looking for primes above $3$
 
@Alex I have not get to read yet. Besides, the structure here is so weak that it is nonassociative and noncommutative. Pretty much the only thing holds is distributivity
 
Oh okay
 
@Secret I know, but if you study this stuff properly, you'll be able to do your stuff here really easily (and probably won't find this sort of stuff nearly as interesting - in comparison to the stuff people have spent decades working out)
 
@ÍgjøgnumMeg I think the class group is generated by the ideal above $2$
 
Yeah, I agree
I'm not sure what is meant in the book though
Although, actually
 
2:57 PM
or one of the ideals above $2$, there should be two
 
yeah
$X^2 - X - 16 \equiv X(X+ 1) \bmod 2$
 
but probably both are equal in the class group and generate it
So $(2,\frac{1+\sqrt{65}}{2})$ and $(2,\frac{3+\sqrt{65}}{2})$ are the ideals above $2$
 

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