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00:00 - 22:0022:00 - 00:00

00:14
@anon Of course the hemisphere is isometric with the disc! Where have you been?
@user60887 No, the sequence is not necessarily decreasing.
But if $y_n=x_{2n}$ and $z_n=x_{2n+1}$ then they are both subsequence that converge, yet $y_n\geqslant 0$ and $z_n\leqslant 0$.
Thus $0\leqslant \lim \limits_{n\to\infty} x_n\leqslant 0$.
Stumbles in on sardonic sentences
 
1 hour later…
leo
leo
01:35
-1
Q: Find the norm of $f (x) = 3x^3 −2x^2 +x −1$ on $C [−1,1]$?

Terry MobleyThis is from the book Linear algebra demystified.

And that's all
Ridiculous.
 
1 hour later…
02:37
Hello all
@T.Bongers howdy
Hey, @robjohn. I have a question for you - what should I do if I see a user that I think is misusing the site in a completely non-mathematical way? There's a particular user I've seen with about a half-dozen instances in the last week of serial voting reversed, and who I think is misusing the review queues.
Should I just flag a random question of this user's and explain my concerns?
03:04
Hey guys, is anyone here familiar with pseudodifferential operators? I want to show that if two $T_\alpha$ and $T_\beta$ are two elliptic pseudo-differential equations, then $T_\alpha T_\beta$ is also an elliptic pseudo-differential operator.
03:42
I need help with a Laplace Transformation proof. :/
03:53
@T.Bongers That would be the best way to contact the moderators.
@robjohn Done, thank you.
04:08
0
Q: Laplace Transformation Proof - Show that $L\{f_a(x) \} = \hat f(s+a)$

usukidollFor any number $a \gt 0$, let $f_a$ be defined as follows: $f_a(x)=\begin{cases}f(x-a)&x\geq a\\0&0\leq x<a\end{cases}$ Show that $L\{f_a(x) \} = \hat f(s+a)$ This is what I did so far $f_a(x)=\begin{cases}f(x-a)&x\geq a\\0&0\leq x<a\end{cases}\\\qquad=f(x-a)h(x-a)$ $\mathcal L\{f_a(...

04:24
why is the 0 ring not a final object in the category of unital rings?
04:40
0 would be terminal, so presumably it's not considered unital in whatever context you're in
in fact the wikipedia page on initial and terminal states that 0 is terminal in the category of rings with unit!
 
2 hours later…
06:30
Hello :) I have a question on group theory
I understand what this question means, but I don't know how to prove it
I need to show that two reflections on the complex plane are a rotation if the lines of reflection are intersecting, else they form a translation
Also, what are non-trivial subgroups? Subgroups without zero?
hi. I can answer your second question. A nontrivial subgroup is a subgroup other than {e} where e is your identity. A subgroup that has the identity element alone is a trivial subgroup which doesn't really do anything.
Ah. Does every group have {e} subgroup?
yes because every group by definition must have the identity "e" or whatever you denote it by.
Ok :)
Thank you :)
and as for your other question are you talking about a complex plane like where the x axis is the real numbers and the y axis are the complex i?
06:40
Yes
hmm seems like an interesting problem. id try it out tomorrow.
ill see you later peace.
haha sure :)
cheers
 
1 hour later…
08:11
hey hey, anyone in here?
suppose I have a string of size $n$. How many possible permutations does it have? for example the string "NQEL" has 12 substrings: N, Q, E, L, NQ, QE, EL, LN, NQE, QEL, ELN, and LNQ
I just want the count
so are you ordering the letters cyclically and counting substrings?
you can't expect others to magically read your mind
do we assume each letter has multiplicity one in the string?
08:20
for context, its a biology problem. "NQEL" is a peptide, and I am looking for all possible sub-peptides.
I am not sure what you mean, but if it means repetition then no. For example the string "ELEL" also has 12 substrings.
again, the context being that in "ELEL", the first 'EL' and second 'EL' could be generated from different DNA strands.
Sample Input:
31315

Sample Output:
980597910

is what is given to me.
and by "substring" you mean some kind of contiguous block of letters in the original string? (for example, QL does not seem to be listed as a substring of NQEL)
yes, sorry I shouldve mentioned that.
it is cyclic though so we do have LN as a "substring" of NQEL
i've been playing around with combination formulas but can't get exactly what I need.
and you do not count the original string as a substring, correct?
I do not think there will be any formula for this. it depends on more than just the number of letters, or even the multiplicity profile of all of the letters: it requires their exact arrangement (up to cycling). in general, you will have to settle for an algorithm to compute the number of contiguous substrings of a cyclically ordered string.
Yeah I was hoping to avoid programming this.
if one assumes every letter is distinct though the answer is pretty clear
08:28
Its part of my online course on coursera on Bioinformatics (my field of interest) but I am a grad student in Applied math
and courses like functional and PDEs take up all my time
with multiplicities, I think programming or computer work is unavoidable
Hello All, I would like to bring your attention to this problem math.stackexchange.com/questions/575127/… I have been able to provide a partial answer. If anyone can complete the solution or provide another solution it would be nice.
thanks @anon i'll see if I can program it in
1
A: Laplace Transformation Proof - Show that $L\{f_a(x) \} = \hat f(s+a)$

copper.hatThe statement is incorrect. It should be that $({\cal L} f_\alpha)(s) =e^{-sa}({\cal L} f)(s)$. You finished right before you wrote "Everything ....". There you showed that $({\cal L} f_\alpha)(s) = \int_0^\infty f(t) e^{-s(t+a) } dt$, which is almost finished. Continuing, $\int_0^\infty f(t) ...

I'm so close...what am I doing wrong D:
it's weird that you use a hat over f and the L symbol for the laplace transform
I've only ever seen the hat used for fourier transforms
08:39
I'm using a very old differential equations book from the 1960s T_T
it's bad I know
what *is* true is that:
(a) applying L to f(t-a)h(t-a) yields exp(-as)F(s)
(b) applying L to exp(-at)f(t) yields F(s+a)
:S this is the only problem that driving me bananas.
let me ask you a question
why are you trying to prove an incorrect statement?
either you have misunderstood the question (and therefore miscommunicated it to us), or the problem is simply wrong, and if your instructor has any sense they should accept e.g. a proof of the (a) & (b) I gave above as remedy
because that's what it said in the ebook. I'll gladly give you a link to it so you can see that I'm not making this up at all.
http://scholarspace.manoa.hawaii.edu/handle/10125/21735
page 139 on the pdf file and I really am not a fan of this book.
there are so many good diff equ books, it'd be a shame to have to use a bad one
I'll take a look
08:47
yeah T_T
but I'm passing the course... so after that I won't be using this thing
unless it's going to show up in the second part of differential equations, then...NOOOO!!!!!!
yeah, the problem is incorrect
so just prove (a) and (b) I gave above as remedy
oh man D:
you're serious?!
you already did (a)
yes
told ya the book is coo coo @_@
where did I do a? for #9??
12 mins ago, by anon
what *is* true is that:
(a) applying L to f(t-a)h(t-a) yields exp(-as)F(s)
(b) applying L to exp(-at)f(t) yields F(s+a)
your notation is sloppy but you've shown (a)
08:54
what should be the correct notation?
for example, you get the letter "t" out of nowhere in the last line
so it should be x?
and you should say L{f}(s) rather than L{f(u)}, for example, as copper.hat mentions
@usukidoll no, x is a dummy variable, it can't exist outside of the integrals it appears in
go back and read what I wrote for (a) and you'll see which letter it should be! (hint: you had it correct the very line before it, but you changed the letter inexplicably)
08:57
$\int_0^\infty f(t) e^{-s(t+a) } dt = e^{-sa} \int_0^\infty f(t) e^{-st } dt = e^{-sa}({\cal L} f)(s)$
right
hmm where do I go from here? to prove b...
you can start by writing out the definitions of things
$\cal L( f(s)) = \int_0^{\infty} e^{-st} f(t) dt \\
\cal L( f(s+a))=\int_0^{\infty} e^{-(s+a)t} f(t) dt$
the function is shifting...
whoah there
it's ${\cal L}\{f\}(s)$, not ${\cal L}(f(s))$
and besides, (b) is about the laplace transform of exp(-at)f(t)
you get F(s+a) as the result of that transform.
09:03
so we have to do this one $\cal L( [f](s)) = \int_0^{\infty} e^{-st} f(t) dt $
so do I use the $e^{-st}$ laplace transform?
the equation you have written after "we have to do this one" is just the definition of the laplace transform
you do not have to prove the definition of L. one doesn't prove definitions.
arghh
this is hard T_T
one needs to observe that the laplace transform of $e^{-at}f(t)$ is $$\int_0^\infty e^{-at}f(t)e^{-st}dt=\int_0^\infty f(t)e^{-(s+a)t}dt=F(s+a)$$ where $F(s)=\int_0^\infty f(t)e^{-st}dt$. not hard at all.
unless you're a beginner at proof writing then it is
I'm still trying to get the hang of it. It's not as easy as matrices
but you hadn't even got to the proof. you were having trouble writing down the problem itself. writing down the problem is a matter of following directions. when part (b) says to apply the laplace transform to exp(-at)f(t), then that's what you do!
15 mins ago, by anon
12 mins ago, by anon
what *is* true is that:
(a) applying L to f(t-a)h(t-a) yields exp(-as)F(s)
(b) applying L to exp(-at)f(t) yields F(s+a)
09:08
so I do have to apply the laplace transformation for the exponential $e^{-at}f(t)$
yes, which is what I just did above. rewriting it: $\int_0^\infty \color{Blue}{e^{-at}f(t)}e^{-st}dt$
that's $/frac{1}/{s+1}$
blah need a frac latex
the laplace transform of exp(-at) is 1/(s+1), but not the laplace transform of exp(-at)f(t)
notice how 1/(s+1) does not depend on the function f(t) at all, so it should immediately ring alarm bells that 1/(s+1) is not the transform (not to mention it has nothing to do with the statement of (b))
partial fractions ^^
$\frac{1}{s+a} $??
huh?
09:13
there must be another laplace transformation somewhere that will satisfy b
I have no idea what you're talking about
there is only one laplace transform, and (b) is true for all nice functions f(t)
so I have to use the laplace transform of b,...?
> (b) applying L to exp(-at)f(t) yields F(s+a)
apply L to exp(-at)f(t), get F(s+a). that is the task in part (b).
I actually did the whole problem in one line above, if you were reading!
le sigh
:(
that was it just one tiny line? O_O
yes
09:18
do I have to expand it to show some work behind this? Proofs can't be that short right
oh, but they can
X(
whew got that under control
now to finish up my portfolio for English class
09:34
Greetings
09:54
@robjohn that result I said yesterday is the correct one. By the way, I lost my favourite puppy :-(
 
3 hours later…
12:30
@Chris'ssis I am very sorry. I lost my dog that I had for 12 years earlier this year. I know it is very hard :-(
:-(
@Chris'ssis The sum is $0.48864370569403250011$ yet the answer you cited is $0.48896786129668636985$. I don't see how they can be the same.
@robjohn Maybe Mathematica is wrong.
@Chris'ssis The sum converges quite quickly. Mathematica should be able to handle it. I will do it by hand...
sage: k = var('k')
sage: float(sum( 1 / (3**k) * ((pi / (pi^2 - 3*pi + 3))**(1/(3**(k)))), k, 1, 100))
0.48864370569403254
@Chris'ssis I had looked at it yesterday too.
12:45
@PabloRotondo did you compute it?
see above
@PabloRotondo I was referring to a proof.
@robjohn hmmm, if so, then something is simply wrong there.
@Chris'ssis Do you have a proof actually? Because it doesn't seem to fit with the numerical value (see the sage output above)
what are the benefits of sage over maxima?
@PabloRotondo No, not at the moment. I hardly worked in the last hours on some different questions. I didn't manage to attend this one yet.
12:49
@MickLH I don't know much about maxima
AFAIK it's the engine that does the symbolic math inside sage
the open source fork of MACSYMA
@Chris'ssis I see, did you get the limit from the other day (the integral / N)?
Yes, in 2 different elementary ways, without pen and paper.
@PabloRotondo 1) horizontal asymptotes; 2) by a special theorem of L'Hopital where it's enough to have infinity in denominator.
@PabloRotondo I wrote above the 2 ways.
The integral looks like one given by Ramanujan (the first solution is as clever as his solutions) :-)
13:00
@Chris'ssis Is this the integral that evaluates to $1$?
@robjohn Yes.
Did any one see my question about a Dirichlet character sum for the von Mangoldt function? I would like generalize it to different values of $j$ like in this Mathematica code: Table[DirichletCharacter[7, j, n], {j, 1, EulerPhi[7]}, {n, 0, 3}] // Grid
Gotta go for a while. BBL
@robjohn OK. I'm around for some hours.
@Chris'ssis How can it look like one given by Ramanujan?
13:13
@PabloRotondo aesthetically speaking ...
Ah, hahaha
@PabloRotondo no one of the students I talk to managed to compute it. :-)
@Chris'ssis Ha
@Chris'ssis Your students? or your fellow students?
@PabloRotondo some friends (students) of mine
@Chris'ssis What are you working on now?
13:25
@PabloRotondo I'm trying to find a way to elementarily compute $$\lim_{n\to\infty}\frac{1}{\sqrt{n}}\sum_{k=0}^{n}\frac{n^k \Gamma(n + 1)}{\Gamma(n + k + 1)}$$
Elementary = Beta not allowed ?
@PabloRotondo Yes, it's allowed.
@Chris'ssis Are you sure it is not $\infty$?
@PabloRotondo what do you mean? The limit is $\sqrt{\pi/2}$
Look at it this way
Assume $k$ were a positive integer first
Oh wait
nevermind
13:47
@PabloRotondo thanks for the information about the Mobius inversion.
Congratulations to @amWhy for the 100k.
14:04
@Chris'ssis ow, Chris... hugs
@Chris'ssis Help with this limit?
$$\lim_{x\to0}\frac{\log^{n-k}x}{x^{n-1}}$$ $n, k$ are constants.
@Charlie hugs :D. How are you doing?
@Alizter this is a weird limit. It's created by you?
brb in 20 min
@Chris'ssis I'm fine thanks :) and you?
@Charlie This is bad day! I lost my puppy! :-(
I go out again to search for him.
@Chris'ssis :((((
14:12
The odds to find him are really poor.
@Chris'ssis ask for help
@Charlie Yeah, I asked for that but I think some kids took it. It was a great puppy.
@Chris'ssis oh noes!
@Chris'ssis The limit is very weird. It came out form my tediously long attempt at $$\int_0^e x^{1/x}\;dx$$
@Alizter interesting integral.
14:17
@Chris'ssis My progress
@Alizter I upvoted. I'll look at the details when I'm back.
Out for some minutes.
14:31
yay $$|\mathscr{P}(\Bbb N)|=|\Bbb R|=\aleph_1$$
leo
leo
@Alizter undeed
14:55
@Alyosha You are welcome.
15:29
nn = 32;
b = Table[
Exp[MangoldtLambda[Divisors[n]]]^-MoebiusMu[Divisors[n]], {n, 1,
nn}];
j = 1;
MatrixForm[
Table[Table[
Product[(b[[n]][[m]]*
DirichletCharacter[b[[n]][[m]] + 2, j + 1, k] - (b[[n]][[m]] -
1)), {m, 1, Length[Divisors[n]]}], {n, 1, 32}], {k, 1, 32}]]

Above is as a Mathematica program a generalization of the Dirichlet inverse of the Euler totient function to Dirichlet characters.
In a Greatest Common Divisor matrix.
32 in the last row of the code should be set to nn, in case you want to change the variable nn.
huhu
@robjohn long time didn't read :D
15:46
Hi, everyone.
@FrankScience Hello.
Suppose $\gamma$ is a closed convex curve, and $u(x)=d(x,\gamma)$ is the distance function defined outside $\gamma$. Is it obvious that $\lvert\nabla u\rvert=1$?
Didn't catch that, sorry.
@FrankScience $\gamma$ is a curve in $\Bbb R^n$?
plane curve.
16:05
Oh, in $\Bbb R^2$?
Hm. I don't know if I can help. Can you give me some background?
V.I.Arnold gave $u$, then said that it's obvious when we know that the geometrical interpretation for $\lvert\nabla u\rvert$ is the maximal rate of increase.
It's no obvious to me.
@FrankScience You probably know that $\lvert u(x) - u(y)\rvert \leqslant \lVert x-y\rVert$, so from that $\lVert \nabla u\rVert \leqslant 1$. Now fix $x$ and consider the projection $Px$ of $x$ on the curve. Since the curve is convex, it is entirely on one side of the line orthogonal to $x-Px$ through $Px$, on the other side than $x$. So $u(x+t\cdot (x-Px)) = u(x) + t\lVert x-Px\rVert$ for $t \geqslant 0$.
That means the maximal rate of change is $1$, hence $\lVert \nabla u\rVert = 1$.
@DanielFischer Projection?
16:18
Projection onto a closed convex set: the point in the convex set closest to $x$. The convex set here is the curve with its interior.
I see. The velocity along the normal line is exactly 1.
@Chris'ssis
16:42
Hi all!
leo
leo
@nullgeppetto hello
What are Dirichlet characters anyways? math.stackexchange.com/questions/228751/…
@MatsGranvik A Dirichlet character is a function $\chi:\Bbb Z/(k)^\times \to S^1$ such that $\chi(ab)=\chi(a)\chi(b)$, $\chi(1)\neq 0$ (you can extend to all of $\Bbb Z/(k)$ by $\chi(a)=0$ whenever $(a,k)>1$.)
16:58
@PedroTamaroff I have seen this before. I should try to figure out how to program them from scratch, in Mathematica.
But I am not good at figuring things out. That requires understanding. Understanding things is hard.
I think it's wierd! No one is interested in a question like this(math.stackexchange.com/questions/556977/…)?
17:22
@nullgeppetto Why?
@PedroTamaroff, Why it's wierd, or why is that question interesting? :)
@nullgeppetto Both.
1. I don't know. 2. I think it's challenging to know what is the probability for an $n$-dimensional point to lie on an $n$-dimensional half-space, while it follows the $n$-dimensional Gaussian distribution... No? :/
@PedroTamaroff, I forgot to mension you in my last message... Really, does it seem to you like a stupid question?
 
2 hours later…
19:07
The geometric answer should recieve more attention here math.stackexchange.com/questions/574608/…
It is a very nice answer.
@N3buchadnezzar, nice answer, indeed!
@robjohn are you around?
@Chris'ssis
@Alizter
:-)
19:17
@Alizter I appreciate that with an upvote. :-)
@Alizter, one "up" from me too, I can't help, but I would like to see the answer(S)!
@nullgeppetto So far $$-e^{1/e}+e^{1+1/e}+???$$
Hey people - am I right in thinking that this expression is right? mathbin.heroku.com/hPJkDUU (I'd type it into chat, but MathJax doesn't display for me in it and typing that out would be horrendous)
This is what he wrote:
$$\frac{\partial}{\partial t} \Psi( \hat{x} - \hat{u}t, t) = \frac{\partial}{\partial t} \Psi(\hat{x},t) - \hat{u} \bullet \nabla( \Psi(\hat{x} - \hat{u}t, t))$$
@Alizter Thanks
19:36
>[I am text](I am link)
@Alizter, Thanks!
@Alizter, all, sorry for being such a newbie...
@DanielFischer, you made a comment here talking about rotation. Could you kindly tell me more about? Thanks in advance!
19:54
if anyone has a moment, I would appreciate some feedback from the types who hang out here on my project, it's a latex calculator szifler.com/calc/calculator.html
it's only a few days old so I don't expect it to be rock solid, and it would have no idea how to solve a polynomial above order 3
@MickLH Nice job. It had trouble solving quartics though ;)
thanks :) if you give me the exact equations you entered it would greatly help the development process
$ax^4+bx^3+cx^2+dx+c$
;)
@MickLH, very nice! Would you like to test something specific?
heh, sorry, it's not smart enough to actually solve something like that, it just has the quadratic and cubic equations built-in
19:58
I do not always stop my sentences midway but
da dum tss
@nullgeppetto well the thing is being a high school dropout, I don't know how educated people actually use calculators :P
so I was hoping you guys would tell me, "it fails on f(X)"
@MickLH, I see!
oops, also specifying bases / powers of functions is broken right now
20:10
ok well im off to the shower, I'll add quartics as soon as I get back :)
I plan to stop at decic though, if someone needs more they should probably use a whole computer algebra suite
20:28
@MickLH There is no formula past degree 4
It is impossible
The program has to be able to solve it using known algorithms
@Alizter Hah. gaaaaaaaay
@N3buchadnezzar I'm sorry, what?
Oh it was a joke, watched community?
@N3buchadnezzar What is that?
20:32
there are ofcourse better tv shows out there
Season 1 + 2 are actually some of the best newer series I have seen. Alas it can not compare to older series such as Fawlty Towers, Black adder usw.
@N3buchadnezzar: Isn't there a more serious show you like?
something sci-fi, perhaps?
Oh, I compared generes. I have watched a ton of Sci-fi series
and your favorite please
Babylon 5, Battlestar Galactica, Kyle xy, Dark Angel, Black Orhphan (ish), Doctor Who, Firefly, probably a few more
20:38
and which one is the most important right now?
It really deppends on my mood, and they are hard to compare. I really liked Dark Angel when I was younger. It is not a particularly great one. Recently I really liked Black Orphan. And at the time I saw Firefly it really made an impression one me.
You?
yes, that happens.
It's one (actually more) of those shows you mentioned
especially today
Doctor Who ;)
can you guess which one?
50th anniversary
20:46
absolutely right.
I love it because it's all timey-wimey.
@Chris'ssis just got back.
@robjohn Welcome back.
@robjohn I wanted to show you something.
@Alizter completing the square?
@Chris'ssis okay
@robjohn Sent it.
20:57
@Chris'ssis got it
@robjohn ok :-)
@Chris'ssis Oh, I thought this was the sum. It is the 1 integral
@robjohn No, it's the 1 integral.
Ok, here's a fun little question to all of you mathsy-patsy folks. What is, in your opinion, the scariest looking number?
@robjohn Indeed, maybe that integral limit is too easy. I personally like it pretty much.
21:03
@Nick 7
lol creepy ass 7, I was gonna say 7 also
@Chris'ssis It requires some work, but it is not too hard.
@robjohn Agree.
@N3buchadnezzar: cuz 7 ate 9? lol
no because it's the punk that started the feud between the fibonacci sequence and the primes
the universe could be so simple...
21:06
what? what happened there
how did it end the war?
the war goes on forever
did it have a sonic screwdriver?
@robjohn That sum I posted yesterday seems from other world. I don't know to compute it yet.
@Chris'ssis Yeah. I can compute it pretty easily, but not a closed form. It converges pretty quickly
Did "the seven" regenerate 6 times and answer the question to life the universe and everything?
21:10
@robjohn Really? That's great!
@Chris'ssis well, it converges at almost 1/2 digit per term
....yeah, I know. That was a bit tooo much but you can't blame me for being excited!
hi hi
@robjohn that's fast.
I think the scariest number has to be 1,000,000,000,000,066,600,000,000,000,001. It's called the Belphegor's prime. Belphegor, fyi, is a price of hell.
21:16
@Chris'ssis each term is smaller than $1/3^n$
@robjohn true
@Chris'ssis I would be surprised if there were really a closed form, but pleasantly surprised.
@Chris'ssis hi, chris, any news?
@Nick hi nikhilly
@Charlie hi. I was preparing to call another colleague to clarify some things. Welcome back! :-)
@Chris'ssis ah :)
21:24
grumpy cat! hugs
@usukidoll I hugged once, it was awful. Kidding :P
I have a cat that looks similar to grumpy cat only she's black and white persian cat
@usukidoll I had a cat when I was a kid, now I have two dogs
cats are cute hugs
all I gotta do for my math is to write this down on my paper and I"m done :D. THank goodness that my calc iv homework is only 4 problems and due on Wednesday. But, I like to get things done
Hehehe :)
21:27
because the sooner the better
and I am mad at my differential equations book
it's garbage
with siracha sauce
Mine is worse
what's it called
@Chris'ssis: I thought this was interesting. It took me a while to come up with the slick answer.
E. Süli and D. Mayers, An introduction to Numerical Analysis, Cambridge University Press (2003).
prefer applied Mathematics T_T
21:30
@robjohn I didn't look at your answer yet, but the first thing that come to mind was mean value theorem. Did you use that?
@robjohn Oh, this is a tricky.
@Chris'ssis Yes, but the best way is a tricky use of the mean value and intermediate value theorems
@robjohn Very clever your approach.
@Chris'ssis reminiscent of integrating factors
@robjohn This is good to be kept in mind. I add a star there.
00:00 - 22:0022:00 - 00:00

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