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12:11 AM
is it viable to see wether a limit exists as x approaches a certain value c (not infinity) to plug in (c+$\epsilon$) and (c-$\epsilon$) and see wether those are the same
where epsilon is a arbitary small value above 0
 
@KevinDriscoll That's pretty much dead on. It's the eigenspectrum of a particular 4-by-4 Hamiltonian as a function of a control parameter.
(If it were a dispersion relation, that control parameter would be (pseudo)-momentum. In this case, it's actually time i.e. this is the (adiabatic) spectrum of H(t).)
 
12:37 AM
hello
I have a differential equation and I have to show that it is a bessel equation in disguise. can anybody help me?
as
 
Fermat's last theorem states that $a^n + b^n = c^n$. Can a and b be the same integer?
 
It states that there are no positive integer solutions when $n\ge 3$.
 
12:57 AM
Quick question, is there any set-theoretic identity that allows us to write the product of union of sets as the union of products of sets?
Which holds for arbitrary products and unions?
 
Sure. Draw a picture and figure it out.
 
well it's also easy to see it's impossible for $a = b$. that would imply that $2a^{n} = c^{n}$, which means that $a = \frac{c}{2^{\frac{1}{n}}}$, which can only be an integer for $n = 1$.
 
@user32720 You presumably mean that you want to show that the second equation is actually an instance of the first.
The only thing that comes immediately to mind is a substitution $z=x^b$.
 
That doesn't look like it works, @Semiclassic.
 
Well, I did say 'immediately'. Didn't say it actually worked :P
And, in fact, if I plug that into Mathematica it gives a solution in terms of $J_{\pm p}(\sqrt{x^{2b}})$
 
1:01 AM
Hmm, interesting ... how does the $(2c+1)$ coefficient arise? Are $b$ and $c$ related?
 
So I think you pretty much have to do that change of variables (there's also a prefactor, though, so that's not the only thing one has to do)
One should find that $w=yz^d$ for appropriate $d$.
There's not a relationship between b and c.
back in a bit
 
@Semiclassical I understand, but I could not develop this idea. Could you give me a hand?
 
how are you today @Ted
 
hi @Ali ...
 
@Ted I heard about this interesting geometric construction
Take a square
Divide it into quadrants
Fill each quadrant with a circle
then fill the middle of the square with a circle so it touches the other circles
 
1:13 AM
Hello chat!
 
The same can be done for a cube by considering octants and spheres
 
@Fargle!
OK, Ali.
 
@Ted!...?
 
Infact you can carry on doing this for hypercubes and hyper spheres
However at dimension 9 the inside hypersphere touches the edges of the hypercube
and above that it extends past it
 
Yeah, the diameter of the inner hypersphere goes to infinity
 
1:14 AM
dont use the word hyper unless its an absolute necessity
 
This is one of those funny things with volume of the unit ball dying off as dimensions go up.
 
Oh yeah, @Ali, someone mentioned that recently on here. That result still hurts my brain.
 
@KasmirKhaan why is it not necessary?
 
@KasmirKhaan I mean, it's an $n$-dimensional sphere. That's the definition of a hypersphere.
 
This is somehow an instance of geometry and algebra contradicting one another?
No, Kasmir is right.
We say hyperplane for a linear space of codimension 1. But spheres are already hypersurfaces by definition.
 
1:15 AM
I am of the impression hyper is for dim > 3
 
No, that's a misuse of language.
hyper should mean codimension 1.
 
Hyper is reserved for spaces one dimension below the ambient space, I believe.
 
Maybe they'll listen to you, @Fargle. They're not listening to me.
 
They still are, though
 
I am listening :)
 
1:17 AM
I would interpret a hypersphere to be one dimension lower than top allowed, DogAteMy.
 
At dimension 9, when it touches the hypercube's edges, it's an 8-sphere
 
What should I say? 9-cube?
 
So if you insist on this, I'll say that $S^n\subset\Bbb R^{n+1}\subset\Bbb R^{n+2}$ is a hypersphere ... :P
 
Ok so the 8-sphere inside the 9-cube touches
 
Most mathematicians with whom I'm acquainted will say $S^k$ to specify it's a $k$-dimensional manifold.
 
1:18 AM
They still are of codimension one so they satisfy it by your definition.
 
I'm saying a sphere is assumed to be a hypersurface.
So hyper- is a redundant prefix.
 
I will no longer use hyper then
 
It's making me hyper, @Ali.
 
From what I read in your comments i think its all depends on your start point , but the wayI know it agrees with Ted =p
 
If we take the hyposphere inside the ultracube
 
1:19 AM
For the record, I wasn't going to say anything, but I wanted to defend @Kasmir (for once).
 
Hyposphere! :D
 
I need my blood pressure meds for that, @Ali.
 
@TedShifrin Thank you Ted ! I like you more each day :D
 
No need for defending, I wasn't attacking anybody :P
 
I still can't say I believe what you guys are saying. I guess I need to write out the algebra.
 
1:20 AM
Lets do it together
 
I have a half hour and then I'm off to dinner and bridge.
 
Weirdly the n-cube has diagonal of length $\sqrt n$
 
That's not weird.
 
Radius of inner sphere + radius of corner sphere + distance from corner sphere's center to corner of cube is $\sqrt n/2$
 
Or maybe not weird
 
1:22 AM
Slow down, DogAteMy.
 
Radius of corner sphere is a constant, $1/4$.
The latter distance is $\sqrt n/4$.
Thus, the radius of the inner sphere is $\frac{\sqrt n-1}4$, if I haven't made any mistakes.
 
Slow down.
OK. I'm at your first equality.
 
At dimension $9$, this is $\frac12$, exactly enough for it to touch the cube's edge.
 
However it will never engulf the n-cube
 
Right. You need a sphere of radius $\sqrt n/2$ for that.
 
1:25 AM
Even for me this is totally unbelievable.
 
I'm assuming side lengths are $1$, of course.
 
Counterintuitive, for sure.
 
It contradicts betweenness on line segments.
 
@TedShifrin That's because you're not correctly visualizing a cube with 512 vertices.
 
Then are we sure our computation is correct?
We're drawing a cross-section and doing 2-D geometry. Are we correct?
Your sum is based on betweenness, which is obviously violated.
 
1:26 AM
I mean, you could surely do it with more rigor. It's going to involve lots of coordinate geometry and set notation, probably.
But it would be the same argument.
 
Lets try rigor?
 
@TedShifrin ¿Qué?
 
But your answer contradicts the picture that gave you your sum.
LOL @Ali
Your equality had a sum based on betweenness, which your answer then contradicts.
So I do not accept this as valid ...
 
Well it works in 3 dimensions so it must be correct
 
How does the answer contradict it?
 
1:28 AM
Because it has the point of contact of the inner sphere and outer sphere being way past the middle of the line segment.
 
What do you mean?
 
Someone whip out ms paint
 
The latter is further away from the origin than the former, surely, but that's just because it's near the corner
 
You drew the same picture I did. You went from the origin to the point of contact of the two spheres to the center of the outer sphere to the outermost vertex of the cube, and you said the sum of the three lengths was the diagonal of the small cube.
That assumes betweenness.
 
1:29 AM
But then your number said the point of contact was way further out and couldn't be in between.
 
The answer doesn't contradict it.
The point of contact is $(\sqrt n-1)/4$ away from the origin, and the middle of the edge is $\frac12$ away from the origin, yes
but that's not a contradiction.
Even in 2D and 3D, things near the corner are further than $\frac12$ away from the origin.
 
Oh, OK. So for large $n$, the center of the outer sphere is far outside the sphere of radius 1/2 centered at the origin.
 
Yes.
That point is $\sqrt n/4$ away from the origin, as it's a quarter of the way up the diagonal.
 
Yeah, OK. But how can the entire outer sphere be outside that sphere of radius 1/2?
I guess that's the truth, but unintuitive part.
 
What sphere of radius 1/2?
 
1:32 AM
The one centered at the origin.
 
But that doesn't have radius 1/2?
 
What's the ratio of the volume of the sphere of radius 1/2 and the volume of the sphere of radius 1
@AliCaglayan We drew a new one
 
oh whoops
 
No, I mean the sphere of radius 1/2 inscribed in the big cube.
 
at dimension 9
 
1:33 AM
Right, DogAteMy. This comes back to the bizarre behavior of volume going to 0.
Which is what I knew, but I still find this challenging :P
 
I guess I'm ignoring just enough of what I know about the world, such that it stops being unintuitive.
 
What is the volume of the parts of the cube outside the sphere
 
Later I'm going to remember all that stuff, and it's not going to make sense anymore.
 
As our tweeter-elect would say, @Ali, Yuuuge.
The sphere is an infinitesimal portion of the volume of the cube :)
(for large $n$).
 
hmm you are right
 
1:35 AM
I should have discussed this in my multivariable course. Computing the volume of the n-ball is an exercise, but I didn't think to put this paradoxical stuff in as a postscript.
Well done, DogAteMy et al.
 
Well done Akiva
 
Hi chat
 
rehi Semiclassic
 
@TedShifrin One place the formula for the n-ball comes up, interestingly enough, is QFT, but not for the reason you might expect.
 
I wouldn't expect anything.
 
1:38 AM
Heh.
 
@TedShifrin You mean, Well done, Ali
et al
 
Well, you start with three dimensions of space and one dimension of time.
 
since he brought it up
 
I only cause problems not solve them
 
That he did.
LOL
I rarely solve any.
I just ask questions.
 
1:39 AM
But that leads to "weird things" in QFT, so you do analytic continuation to 'imaginary time' and therefore have four spatial dimensions (Euclidean instead of Minkowski).
 
Did you just say that imaginary time is a space dimension
 
How does this get us to 9 and larger?
 
@AkivaWeinberger Yuuuup
 
@Ted have you seen 3blue1browns video on analytic continuation of RZ?
 
I have no idea what you just asked, Ali
 
There's a YouTube channel called 3Blue1Brown.
 
hi
yeah @AkivaWeinberger
very good channel.
 
I like his explanation alot
 
I agree.
 
Oh, I didn't know what RZ was. I don't think about that stuff.
 
1:41 AM
hi everyone
 
hi Karim
 
Actually, it's not so absurd. Keep in mind that in special relativity, an observer in motion will see different distances than observers at rest.
 
I'm fine with imaginary time being distance. How do we get to $\ge 9$?
 
times by 3 and add 1
 
Hence one doesn't talk about the distance via $r^2=x^2+y^2+z^2$, but rather the 'space-time interval' $s^2=t^2-x^2-y^2-z^2$.
 
1:42 AM
glares @Ali
 
And if substitute $t=i\tau$, then that fourth dimension of time acts like the other three spatial dimensions.
 
I am fairly sure special relativity is robust enough to have all sorts of crazy metrics
like -, -, +, +, +
 
No, then you're not doing relativity, Ali.
 
So at least at the level of "I can make time look like space in my metric if I make time imaginary" it's consistent
 
eh who is
 
1:44 AM
Why one should do that, that's harder to answer and I don't want to
It's either +++- or ---+.
+++- is nice from the point of view of QFT, since then you get $s^2=x^2+y^2+z^2+\tau^2$ as your metric.
 
I need to leave. So how do I get to large dimensions for balls?
 
@TedShifrin grow a pair
 
Not -large- dimensions. But when you do integrals in QFT, you often times need to do regularization to make them converge.
 
smacks @Ali
 
Have a good day @Ted
 
1:46 AM
What does that have to do with dimension large, @Semiclassic?
 
And one way of doing that is so-called dimensional regularization. Which basically amounts to saying "You want n=4? Okay, let's take $n=4-\epsilon$ and analytically continue to $\epsilon=0$."
 
GAAAAH.
OK, carry on.
 
LOL
 
Yuuup.
 
I think the answer to my question is that the volume of n-balls for n>>0 is not relevant to QFT.
 
1:47 AM
Basically, you treat the dimensionality of your space as though it's a continuous variable.
 
oh thats pretty cool
 
But in order to do that, you need to know how the volume of an n-ball behaves for arbitrary n.
 
The formula is derived only for natural numbers, but you're going to use the gamma function to make it make sense for real numbers. Pfeh.
 
Yuuup.
 
Typical physicists' bull.
 
1:48 AM
FWIW, the "much larger than" symbol is \gg $\gg$
 
I wasn't in math mode, DogAteMy.
I usually am, but I'm giving in to the slack behavior of people in this room.
 
Yeah, but it's something you might not have known
 
I knew :P
 
(Tom Scott voice)
 
But I'm leaving. Y'all can have fun sans moi.
 
1:49 AM
We can all misbehave without you.
 
For details on the horror of dimensional regularization, Wikipedia's got an article here: en.wikipedia.org/wiki/Dimensional_regularization
Bye @Ted
 
I'm quite confident you will, DogAteMy.
 
Quick everybody stand on the tables
 
@AkivaWeinberger $\gg$ no re
 
1:50 AM
I've also seen it used in condensed matter theory for understanding two-dimensional physics.
 
That's a reference Ted probably wouldn't get, were he still here @Fargle
 
You never know. Maybe he plays Battlefield.
 
How many unit n-spheres can I fit inside an n-sphere of radius 2?
 
Hm. $\gl \hf$
Nope.
@AliCaglayan I would assume $2^n$, no? EDIT: OH WHOA
WE CAN PUT ANOTHER ONE IN THE CENTER, BECAUSE WE JUST PROVED THAT
I am way too tired.
Yeah so I have no idea.
 
Well in 2 dimensions
 
1:55 AM
@AliCaglayan Also, I misread what you wrote completely.
 
I am getting stuck in something stupid.
 
I count 3
 
I thought you said fitting it in an n-cube of sidelength 2.
Which is very different.
...Three?
 
no 2
 
1:56 AM
my bad my geometry uncuclideanised
English was not my strong point
uneuclidianised
there we go
 
$\rm(e)-(c)=(-)$
typographically
 
how many unit spheres can you fit in a radius 2 sphere
 
hm, im working on a problem right now and it is utterly confounding me
i kinda like it
but mostly hate it
 
I count 2 for that as well
 
2:00 AM
What about 3-spheres
 
Because they all contain the origin (either in the interior or in the boundary).
And only two can have it in their boundary at once.
So it's going to be 2.
 
hmm correct
But If I spice things up and say an n-sphere of radius n
 
Despite the fact that more and more space is wasted.
 
how many unit n-spheres fit then?
 
I think that might be unsolved even in 2D
 
2:02 AM
In 2d it is 2
 
Oh, sorry. I thought we had n 2-spheres.
So the radius is equal to the dimension.
 
A radius 3 3-sphere
How many unit 3 spheres fit
Well at least 3
bump that upto 5
and I guess 7 why not
 
At least 9 I think?
Octahedral symmetry
 
yup
I can't tell if there are more
Does an optimal packing have to have a center sphere?
 
alright switching to a different technique
 
2:10 AM
4-sphere has maybe 14?
I am too tired to carry on however
Good night all!
Especially @AkivaWeinberger
 
G'night
 
Hello everyone
I want to know that what should I know before go on to inner product space
 
2:34 AM
0
Q: Compute $E(Y\,|\,X)$ as a function of $X$

Simple Let $Y$ be a Poisson $\lambda$ random variable, and define $X=I_{[Y>0]}$. Compute $E(Y\,|\,X)$ as a function of $X$ and find $E(|Y-X|)$. So far, I found the mass function of $X$, $$f_X(x)=\begin{cases} P(X=0)&=e^{-\lambda}\\ P(X=1)&= 1-e^{-\lambda} \end{cases}$$ I don't see how to f...

 
hmm I solved this problem in a very long way.
there should be an easier solution.
 
I just learn that one of a property of $\vec A \cdot \vec B$ is $(\vec u + \vec v) \cdot \vec w= \vec u \cdot \vec w + \vec v \cdot \vec w$
 
brb
 
I am struggling to do that problem, can someone give me a little help
 
How to write MaxJax here?...
 
3:02 AM
Don't bookmark that link, click on it and bookmark the link that says ChatJax
and follow the instructions
 
^^What Akiva said
 
4:00 AM
@Perturbative@AkivaWeinberger thanks
 
This paper ( arxiv.org/abs/hep-th/0401052 ) dated 2004 claims it is proven there is no odd perfect numbers, but in 2013 (at least) they are still looking. What happened?
 
@Perturbative I'm using my phone to chat here, is that still work for me?
@Perturbative and that tool is to make only myself to see or others also?
 
4:22 AM
testing: $\vec A$
 
4:33 AM
\vec A
@Perturbative It works perfectly on my laptop though...
The complete question is here. Take a look if you know the others property like this one...
 
4:52 AM
Ack. I hate coming in here when there's nobody around.
Anyway, I need help!
1
Q: $|G|=pq$, where $p$ and $q$ are primes, is the semidirect product of subgroups of orders $p$ and $q$

JessyunBourneI am currently working on the following exercise: Let $G$ be a group of order $pq$, where $p$ and $q$ are primes and $p > q$. Prove that $G = N \rtimes H$ for some subgroups $N$ and $H$ of orders $p$ and $q$, respectively. Now, I just proved before this that for $G$ where $|G|=pq$ for $p$, ...

Cat Damon. Meow.
 
5:10 AM
@Joshua A claim to have a proof is not the same as actually having a proof.
 
Who's Joshua?
oo.
 
5:44 AM
@BalarkaSen the velocity of the object thats moving, basically I'm trying to figure out the equation apple's using for its bouncy scroll effect
 
6:17 AM
Hello! I am confused little bit that : if q(n-1)+n.s<=t, then q(n-1)>=t-n.s is correct? or q(n-1)<=t-n.s correct?

That is when we subtract `n.s` on both sides then sign `<=` changed to `>=` or still we same? Please, clarify.
 
7:10 AM
I have certain deceleration, how do I get the distance the body will move with that deceleration
 
@SemiC especially when it's tagged hep-th ;)
 
7:38 AM
@MikeMiller I am somewhat surprised that one got through even arXiv's minimal screening, as I thought one of the things they screened for were papers in blatantly wrong categories.
I mean, this is the sort of thing one would expect to find on vixra
 
8:05 AM
@Bhargav The velocity is ever changing, because there is always acceleration. At which position are you measuring it?
 
yes there is decelration
I have that deceleration value
i have the initial velocity and the velocity at position from which i want to calculate displacement, and I have the dt too, so I have dv/dt
I need to know the distance that it will further travel
and also I need to apply a springs equation on that specific movement
basically imagine a car moving with deceleration and one point it crashes against an elastic Wall
how far will the car move
and describe the movement mathetmaticall
so I know the initial velocity when the car started decelerating
I know the velocity when the car hits the elastic wall
is this the right place to ask this?
 
8:25 AM
@Bhargav If it digs x distance into the elastic wall, then F = kx is the force exerted on the car by Hooke's law. Suppose initial velocity of the car is v right before it hits the wall. Then 0^2 = v^2 - 2 (F/m) x.
There isn't anything more to it. Anyway, no, this is the mathematics chat. You'd get more help in the physics chat (h-bar)
Actually scrap that; the force exerted on the car is not constant.
 
How does one prove the well-orderness of (N,<) using Peano's formulation of N without resorting to the ZF-formulation?
 
It should be 1/2mv^2 = 1/2kx^2 by conservation of energy
 
@DHMO Can the statement of being well-ordered even be formulated in PA?
 
Indeed, in 0^2 = v^2 - 2(F/m)x if I replace F by (0 + F)/2 that's exactly what you get. The correction comes from the fact that you average the force exerted on the car to model it with a constant acceleration.
 
Hello! I am confused little bit that : if q(n-1)+n.s<=t, then q(n-1)>=t-n.s is correct? or q(n-1)<=t-n.s correct?

That is when we subtract `n.s` on both sides then sign `<=` changed to `>=` or still we same? Please, clarify.
 
8:32 AM
@MithleshUpadhyay addition and subtraction does not change the inequality
 
@TobiasKildetoft forall S subset N: exists n in S: forall s in S-{n}: n<s.
 
@DHMO Ahh, I didn't know PA was allowed to take all subsets
 
eh, i know nothing about PA
 
In that case, take a set with no minimal element and consider its complement, just like you would do in ZF
 
PA cannot take all subsets?
and then?
 
8:36 AM
I have no idea actually
 
@Alessandro I'm up if you want to talk about forms.
 
@BalarkaSen Modular?
 
@BalarkaSen can you help me?
 
Differential
@DHMO Depends
 
ahh, you'e no fun anymore
 
8:42 AM
Modular forms are actually some sort of differential forms
 
@BalarkaSen prove that all subsets of N (Peano formulation) have a minimum
except, you know, the empty set
 
@DHMO Sorry, I can't help you with that
 
ok thanks
 
@TobiasKildetoft I think if $f$ is a modular form on $\Bbb H^2$ then $f(z)dz$ is a PSL_2(Z)-invariant differential form. So quotienting gives you a differential form on a modular surface downstairs.
 
@BalarkaSen They are? I did not even know the term "form" meant the same in the two cases
 
8:46 AM
Actually that's true if it's weight 2 (is that the right word?)
That is $f(z') = (cz + d)^2 f(z)$ where $z' = (az + b)/(cz + d)$
then $f(z') dz'$ is det(a, b, c, d) times $f(z) dz$.
So you get a well-defined holomorphic 1-form on $\Bbb H^2/\text{SL}_2(\Bbb Z)$, or something
 
@BalarkaSen I do, but I really should be studying numerical analysis now :(
 
Oh, sure, definitely do that
 
@BalarkaSen Yeah, that is the weight
 
@AlessandroCodenotti puedes ayudarme?
 
gotcha
 
8:53 AM
I've seen the well ordering principle proved from the induction one before @DHMO but I'm not sure if the proof can be carried out in PA and I don't have time to think about it now, sorry
 
ok thanks
 
@DHMO I've only seen the reverse proof, doing induction out of well-ordering
But I can try the reverse direction
So let's say you have some $B\subset\mathbb{N}$ such that $B$ does not have a least element
Well finite sets could probably be done by inducting on cardinality
But well, we know that $1\notin B$ since it'd be a least element
I guess you could induct by taking $\mathbb{N}\setminus B$
It'd need to be strong induction, I guess
 
@TobiasKildetoft , thanks.
 
If $1,\ldots,n \notin B$, then if $n+1\in B$, it would be a least element
That should probably do it
Like it's tricky to rattle off about PA exactly
I don't believe it carries with it a formally defined notion of a set
 
9:11 AM
Yeah, PA is not very useful for most purposes.
 
I mean I guess it depends on how you state induction there
 
I wrote a blog post about induction some time ago math.blogoverflow.com/2015/03/10/when-can-we-do-induction
though that is very specifically not about PA
 
(Also @DHMO sorry for going off on the proof, I had only read the most recent messages and I thought you were just looking for a proof by induction)
I usually only see Peano when a class is trying to give an intro to proofs, and want to do some axiomatic buildup but don't feel like going into /too/ much detail on ZFC
And I guess in the incompleteness theorems you're trying to formulate stuff in any system that can contain Peano, so you have to appeal there
But beyond that, you usually couple it with some notion of subsets
Well anyway, good luck, I will need to go to bed because classes and stuff
 
9:32 AM
alright
 
Could someone please help me with the integral of sqrt(x^2-x-1)/(x+1)?
 
@user379685 Hint: x+1=0 when x is -1. Now is -1 a root of x^2-x-1?
 
No it isn't.
Could you elaborate more?
 
yes you're right! Sorry I didn't pay attention, I'll have to think about it
 
9:52 AM
@user379685 maybe u=x-0.5?
 
@user379685 it seems to me that there will be a lot of substitutions involved I would try to rewrite the numerator as sqrt(u^2-1) or something. Sorry this is the best I can think of. May be try posting it as a question
 
@DMHO didn't help much :/
 

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