Mathematics - 2022-02-11

geocalc33

00:17

Two fields $F=(\Bbb R, +,\times)$ and $G=(\Bbb R^*,\times, \exp).$ $G$ is not closed under the addition of $F$ but one can still add elements in $G.$

So you can use addition in $G$ but you can't have the closure of the operation I guess?

maybe easier to state with groups, $(\Bbb R,+)$ and $(\Bbb R^*,\times)$ which are isomorphic

$(\Bbb R^*,+)$

that is closed under addition

why is this person saying that the mapping does not commute with addition?

anakhro

01:05

Where is leslie when you want his guessing abilities.

leslie townes

01:17

huh?

anakhro

LESLIE

I no longer need you.

But I can ask you if you wanted, anyway.

leslie townes

i guess you don't

anakhro

slaps their knee

leslie townes

is it interesting or funny? or both?

anakhro

Neither, in fact.

leslie townes

01:21

what a downer

anakhro

I don't have any consolation jokes or facts for you.

leslie townes

as my daughter would say, phbhtbhtth

anakhro

when are you going to start her on operator theory so that it becomes p*hb*htbh*tth

Koro

@TedShifrin 10

But that was not required for my question.

But thanks a lot for catching that. I just miscalculated it at that time.

I discard n=4 as follows: if there were such a union then in that union there will be at most 16-3 =13 elements (distinct) and 3 here is being subtracted due to identities presence in every subgroup that's part of the union.

I did a similar thing to discard n=5.

10 because 4 cyclic and 6 Klein 4.

CroCo

01:39

Hi everyone, I'm trying to use reduced row echelon form in Matlab and Mathematica for an augmented matrix A=[1 -2 x;2 1 y;-3 1 z] where x,y,z are symbolic variables. Both software treat the matrix as 3x3 generating three pivots whereas the system has a most 2 pivots. How can I let Matlab or Mathematica to detect it as augmented matrix?

RowReduce[{{1, -2, x}, {2, 1, y}, {-3, 1, z}}]

yields {{1, 0, 0}, {0, 1, 0}, {0, 0, 1}}

leslie townes

it doesn't sound like they're even treating them like symbolic variables. i dunno. this is a software thing i am unfamiliar with. you could apply the row operations manually and stop when you clear the first two columns.

Koro

So you want the third column to be 'augmented'?

leslie townes

wolfram recognizes the cases. wolframalpha.com/…

CroCo

@leslietownes I've done the manual calculation which is pretty easy but I'm surprised why the software failed to reach to the same result.

@Koro correct

leslie townes

croco: there must be some special syntax for augmented matrices or partitioned matrices where you only want to rref the first x columns, but i don't know what it is.

i never did symbolic calculation in matlab, it wouldn't surprise me if it were bad at that. but i haven't used it seriously in many years.

CroCo

01:44

@leslietownes I think the posted link yields incorrect results.

leslie townes

it sort of makes sense that in general it wouldn't know what to do with a symbolic matrix unless all of the 'variables' are in the last column only. that's really key here. if the matrix is just of data type "any entry is some function of x, y, and z" it's going to choke on that. too many cases in the rref.

CroCo

not sure what's goin' on here.

Actually I've faced this problem a lot, but I was lazy to investigate the problem and tend to do it manually.

leslie townes

croco: it's not treating it like an augmented matrix, but it's recognizing the relevant cases. it suggests that you'd end up with 0, 0, 5x+5y+5z (or with minus signs) in the last row if you do what you want to do.

CroCo

I get (7x-2y)/3, (2x-y)/3, (3z+19x-5y).

hopefully I didn't screw it up.

but nonetheless, it shouldn't generate three pivots.

leslie townes

now try adding multiples of the first and second rows to that one, to clear out the first two entries.

you should get 0, 0, something and it wouldn't surprise me if it's what WA suggests ought to be there. i'm certainly not doing this calculation myself. you would definitely expect there to be three pivots a lot of the time if it's treating it as a 3x3 matrix. 0, 0, nonzero in the final row would, if you treated it as a 3x3 matrix, turn into 0, 0, 1.

CroCo

01:50

I need them as this because I need to check the spanning vectors.

Ted Shifrin

@Koro I do not believe this.

CroCo

@leslietownes but I didn't understand when it says assuming non-zero and zero determinant.

Ted Shifrin

Hi, munchkin’s slave.

leslie townes

croco: if you can get the first two rows to look like 1, 0, [something] and 0, 1, [something], respectively, then you can make the last row look like 0, 0, [something] where something is a linear combination of x, y, and z.

if it's 0, then you have a consistent system. if it's nonzero, you have an inconsistent system. those are the cases.

CroCo

this is exactly what I've done.

leslie townes

01:54

and those happen to be the same cases as the determinant being nonzero and zero

CroCo

i see

leslie townes

this is somewhat unique to this square matrix setting. for general rectangular matrices you could do the same calculation and you'd just have potentially multiple combinations of variables that have to be 0 for the system to be consistent.

in that sense, it's sort of an accident that the determinant has to give you the one condition in this case. but it's interesting that wolfram notices it, where mathematica apparently just assumes whatever expression you get in that last entry is nonzero so it can divide by it.

CroCo

Actually I have two vectors of 3 dim. I'm trying to find their span set which I believe it is a plane

v1=[1 2 -3]^T and v2=[-2 1 1]^T.

Ted Shifrin

Mathematica does the generic case, unless you give it rules that dictate otherwise .

anakhro

@leslietownes I return for guessing abilities: let $\mu$ be an $\mathbb R^m$-valued measure over $\mathbb R^n$. What should the notation $\int_{\mathbb R^n}f(x)\,d\|\mu\|(x)$ mean?

leslie townes

01:57

anak: perhaps the total variation?

or whatever it's called?

Ted Shifrin

Take the norm of the vector and do a real-valued measure.

leslie townes

it'll be a measure in the usual sense if mu is.

anak: en.wikipedia.org/wiki/Vector_measure calls it the "variation". my guess is it's that.

CroCo

@TedShifrin this is the actual augmented matrix
A=
[ 1 -2 | x]
[ 2 1 | y]
[-3 1 | z]

this is system has two pivots.

how to tell Mathematica this info

anakhro

@leslietownes So since x is merely a vector here, it's like Ted suggests and is just the norm, $x\mapsto\|\mu(x)\|$?

Ted Shifrin

You want the constraint equations for consistency. Forget row reduction and use theory.

Where $x$ is a set in your sigma algebra?

anakhro

02:01

I guess yeah it would have to be to make sense

leslie townes

anak: you might need to use a sup formula as in wikipedia in general for countable additivity but yes.

CroCo

@TedShifrin I can do it manually since this is not a big matrix but for matrix 4x4, I keep getting same issue.

this is the example from the book I'm following.

I'm not able to get 11x+5y+2z=0 using software but I've managed to get it manually using rref

Ted Shifrin

Theory (see my book or lectures) tells you to find a basis for the nullspace of the transpose.

CroCo

Unfortunately, I couldn't get your book in my university's library. I will try your videos.

anakhro

Ted, when are you going to release your textbook pdf online for the masses.

CroCo

02:11

@anakhro we had this discussion before and it seems the publishers are the issue here.

anakhro

well let's just say it's out there if you know where to look

coughs something about russian libraries

CroCo

this water is full of dragons

Koro

@TedShifrin why not?

There are 4 elements of order 4.

Nope, let me rethink.

Ted Shifrin

@anakhro Learn undergrad diff geo :)

Koro

There are 12 elements of order 4.

(order from 1st Z4, order for second Z4): (1,4),(4,1),(2,4),(4,2),(4,4)

Ted Shifrin

02:25

So 6 subgroups?

anakhro

@TedShifrin is this a mad diss against me :(((

Koro

So 2+2+2+2+4=12

Cyclic subgroups of order 4 is: 12/2=6

Ted Shifrin

I agree: 6 subgroups and 1 Klein subgroup. Your 4 was nuts.

Oh, right, 6. Even nuttier.

Koro

Order 2 elements are: 2+1=3 so number of K4 is 6.

Total number of subgroups is: 12

Ted Shifrin

You’re crazy.

Give me the Klein subgroups explicitly.

Koro

02:28

{identity, 3 choices for order 2, 2 choices, 1 choice} and we have 3 elements of order 2.

Ted Shifrin

Write them down.

Koro

{(0,0),(0,2),(2,0),(2,2)}

Oh 1 only!

Ted Shifrin

Yup. You need to be careful!

Koro

Thanks a lot. :)

Ted Shifrin

Sure :)

Koro

02:30

@TedShifrin *cyclic

So total number of subgroups of order 4 is 7.

Ted Shifrin

Yes, right.

I believe so.

Koro

:)

geocalc33

there's 16 people in the chat

a few hours ago there were only 5

Koro

participation in the chat was very less yesterday @geocalc33

Ted Shifrin

02:46

I can depart.

“Very less” is an odd construct.

user10478

In this video (youtube.com/watch?v=dNLmvhZWEq8), the professor solves a heat equation with two boundary conditions and one initial condition, $u(x, 0) = 0$ via the similarity solution method. The PDE is used to construct a second order ODE, and two boundary conditions determine the two arbitrary constants from the ODE. However, the initial condition is never used. How is it possible to solve the heat equation without using the initial condition?

Does the similarity solution method only work for that particular initial condition and use it implicitly?

geocalc33

03:11

I expect participation to be high tomorrow

user400188

03:29

I have come across two definitions of what is means for a set to be connected, and I am unsure why they are equivalent.

The first: $A\subset X$ is connected iff there do not exist open sets $U$ & $V$ in $X$ with $A\subset U\cup V$, $A\cap U\neq\emptyset$, $A\cap V\neq\emptyset$, & $A\cap U\cap V=\emptyset$.

The second: $A\subset X$ is connected iff the only subsets of $A$ that are both open and closed in the relative topology of $A$ are $\emptyset$ and $A$.

I like to think of the first as $A\subset U\cup V\rightarrow\left[A\cap U\cap V=\emptyset \rightarrow\left(A\cap U=\emptyset\lor A\cap V=\emptyset\right)\right]$, for every $U$ & $V$ in $X$. This reads as: When $A$ is a subset of the union of $U$ & $V$, if the intersection of $A$, $U$, & $V$ is empty, then either $A$ does not intersect with $U$, or $A$ does not intersect with $V$.

Normal rewriting things like this is sufficient to understand the link between two definitions, but in this case it has not been, because I was unable to write the second definition out in an equally expanded form. So I ask, how exactly are these two definitions equivalent?

anakhro

03:41

@user400188 can you show that the definitions are equivalent if you consider $A=X$?

leslie townes

i don't think that thinking in symbolic sentences is a good way of unraveling one definition into the other. it sometimes is the simplest way (and maybe always is if you are a computer) but maybe not here.

anakhro

That is, can you show "there do not exist non-empty open sets $U,V$ such that $X = U\sqcup V$" and "the only clopen subsets of X are X and the empty set" are equivalent.

From that, it's pretty natural to extend it to the relative topology case.

user400188

03:59

@anakhro I am not sure if I understand the first definition, and how it relates to the original. If there do not exist non-empty open sets $U$, $V$ such that $X=U\cup V$, then isn't that equivalent to saying that all non empty sets $U$,$V$ satisfy $X\neq U\cup V$? If so, that requirement of 'all' non-empty sets would include $X$ itself, which would have to satisfy $X\neq X\cup X$. Hence my confusion.

Ted Shifrin

DISJOINT

user400188

@TedShifrin So it's meant to be "non-empty disjoint open sets..."?

Ted Shifrin

Yes

user400188

Ok, thank you. I'll continue to try and show them equivalent.

Ted Shifrin

Without that, of course you’re totally confused.

user400188

04:07

Exercise: show that topology is inconsistent.

Ted Shifrin

Exercise: Do not trust sloppy crap on the internet.

user400188

That's where most of my learning in abstract mathematics takes place though. Although most of it isn't sloppy. (At least, I don't think it is.)

robjohn

@TedShifrin 💩 . o O ( trust me )

user400188

04:22

I think I have shown that the first definition leads to the second, but not the other way around yet. I'll write out the first to second now just to make sure.

If there do not exist disjoint, non-empty open sets $U,V$ such that $X=U\cup V$.
- All disjoint, non-empty open sets $U,V$ satisfy $X\neq U\cup V$.
- Assume, for the sake of argument, that 'there exists a subset $A\subset X$ which is clopen, nonempty, and not equal to $X$.'
- - $A$ is open and non-empty.
- - $X\neq A\cup V$, for any $V$.
- - If $A$ had elements in common with $X$, but missed out on some, and had no more elements (i.e. $A=X-B$, for some $B$), then we could pick a $V$ to fill in the gaps. This is a contradiction, so it's not the case that for some $B$, $A=X-B$.

user400188

04:47

And I think I have the other direction now, if anyone actually understood what I wrote the first time.

If the only clopen subsets of $X$ are $X$ and the empty set:

- Then assume, for the sake of argument, that there exist disjoint non-empty open sets $U,V$, such that $X=U\cup V$.
- - $X-U=V$
- - $U$ and $V$ are open, so $U$ is clopen.
- - $U\neq\emptyset$, so $U=X$
- - But if $U=X$, then $V=\emptyset$. Although $V\neq\emptyset$.
- - Contraction

- So it must not be the case that there exist disjoint non-empty open sets $U,V$, such that $X=U\cup V$.

So if the only clopen subsets of $X$ are $X$ and the empty set, then there do not exist disjoint non-empty open sets $U,V$ such that $X=U\cup V$.

Did those two proofs(?) (if you can call them that) make sense @anakhro @TedShifrin?

Ted Shifrin

05:20

I didn't read it all. But you need to say that $X-A$ is open (in $X$), which means ...

leslie townes

"then we could pick a $V$ to fill in the gaps" leaves something to be desired, even if it gets at the idea.

Ted Shifrin

Leaves lots to be desired.

Even red leaves in the autumn won't do.

leslie townes

there are a few missing "open"s. it is often the case that U and V represent open sets, the same way that z often represents a complex number and x a real number. but if you're using it you should mention it.

might be possible to untangle some of those negations and contradictions, too, althoug that is more exposition than logic

Alex

Hi, suppose that I'm maximizing the perimeter $P(x,y,z)=x+y+z$ of a triangle in a ellipse $x^{2}/a^{2}+y^{2}/b^{2}=1$. Can I use Lagrange multipliers? Lagrange multipliers says I need to solve the equation $\nabla f=\lambda g$ where $f$ is the objective function and $g$ is the constraint. But here, we have $\nabla P=\lambda g$ with $g=x^{2}/a^{2}+y^{2}/b^{2}-1=0$ and then $(1,1,1)=\lambda(1x,1/2y,0)$ but then $1=\lambda 0$ but is not make sense. How can I use Lagrange multipliers here?

leslie townes

so in P(x,y,z), your symbols x, y, and z are lengths of sides in a triangle in the plane? do i read that right? but in the equation of the ellipse x and y are coordinates of a point in the plane?

Alex

05:32

Oh, well there is a real problem in my approach

Uhm

leslie townes

you might need more letters. certainly possible to parametrize this thing in some way to to get multivariable calculus to apply to it, but not if x and y are different things in different functions.

might help to try the case of a circle first.

user400188

Yeah, I am not sure how I should write it more precisely. $V=B$ does not work because $B$ might not be open, it has to be $V=(B+\text{some other things to make it open if it isn't already})$.

Also, you are right that I forgot to mention a few times that $U,V$ are open. I say so at the start, but I also should have said it whenever I wrote "for 'any' $U,V$.

Alex

I was reading a similar approach here in MSE with a rectangle, but I don't have the link because seems MSE is in maintenance. By the way, "$(1,1,1)=\lambda(1x,1/2y,0)$ should be "$(1,1,1)=\lambda(2/a^{2} x,2/b^2 y,0)$". I'm sorry.

There is the link: math.stackexchange.com/questions/3799368/…

user400188

@TedShifrin Did you happen to mean my $A$ from the proofs I attempted to write, or the $A$ in the original definition?

Alex

@leslietownes Yes, for all the question. I think the problem is that the function $P(x,y,z)$ depends of three variable but $g(x,y)$ the constraint depends of two variables.

leslie townes

05:44

alex: it's a little more than that. the x, y in g(x,y) are coordinates of a single point, and the x, y in P(x,y,z) are lengths of sides between points (not coordinates that satisfy the constraint). you're not maximizing P(x,y,z) subject to g(x,y) but to some more complicated constraint that is not easy to express as a constraint on the lengths of sides.

Alex

Well, I understand that :-(

copper.hat

mse main site seems to be having a lot of issues lately

leslie townes

i think it might help to try the circle case first. you could well assume one of the points is (1,0) and if you parametrize right you can represent the other two points in two variables instead of four and they are not constrained.

user400188

@copper.hat There is a post on meta.se about cyber attacks.

leslie townes

if you want to think of the points in the triangle as pairs subject to the constraint then the perimeter would be a function of six real variables (subject to constraints), or four if you fix one of them at (1,0).

copper.hat

05:47

damn cylons

@user400188 thanks :-)

leslie townes

lots of flexibility in setting these kinds of things up but the choices do matter and the setup that makes the most 'sense' from a 'just write it down' point of view might involve complicated calculations.

pays to be clever.

copper.hat

i am sorry i did not take up a life of crime (apart from driveways).

the opportunities are endless with the interweb

Alex

thank you leslie, I'm going to think about this some more.

Koro

I like how Axler shows that for every operator T in L(V), there is a one dimensional or two dimensional T- invariant subspace of V.

Using complexification :)

Govind75

10:09

@robjohn cheers, so the joint distribution would be $\Pi_{i=1}^n f_{X_i}(x)$, this is only normally distributed when $p = 0$ or $1$, or when the two distributions are the same?

Vrouvrou

10:27

Hello, I have this sequence defined by $u_{n+1}=u_n+\frac{(-1)^n}{n+1}, n\in \mathbb{N}^*$

$u_1=1$

I proved that $(u_{2k})$ and $(u_{2k+1})$ are adjacent

Son they have the same limit

Then $(u_n)$ converge also to the same limit

I want to find an interval for the limit

I say $u_2\leq u_{2k}\leq u_{2k+1}\leq u_3$

Because u_{2k} is increasing and u_{2k+1} is decreasing

My question is : is u_{2k+1} start for k=1 or k=0 ?

That is I must say $u_{2k+1}\leq u_1$ or $u_{2k+1}\leq u_3$ ?

Vrouvrou

10:52

Oh sorry it is less then u_1 I don't know why I don't see this

robjohn

@Govind75 I thought you were interested in the mean of two distributions. The probability of $N(\mu_1,\sigma_1)$ is used is $p$, and the probability of $N(\mu_2,\sigma_2)$ is used is $1-p$.

porridgemathematics

hi, can anyone help me out with this ? imgur.com/a/0HnnnAu, in the highlighted part of the second page, im not sure why $\nabla_{e_i}(e_j)$ can make sense if $\nabla$ is a map from vector fields on the entire manifold squared, but the $e_i$ are only defined on some open subset $U$, im guessing this is because $\nabla$ really only depends on local info but this hasn't been shown yet so

Sergio Basurco

Does it make sense to add a quaternion to a location in 3d space?

I'm programming some physics stuff for games and found some code that does this and I don't quite understand what the purpose is. Code reads :

`const FVector lTargetPositionInWorld = lOwnerTranslation + lOwnerRotation *(
(mInitialRelativeTranslation + mInitialRelativeRotation.Quaternion() * lBoneTranslation));`

Where rotations are quats and translations Vec4

Govind75

11:26

@robjohn ah, I was tryna describe the joint distribution of $X_1,...,X_n$, I thought the joint pdf would be what I said above where each $f_{X_i}(x) = pf_1(x) + (1-p)f_2(x)$

Where $f_1(x)$ is the normal pdf for distribution 1 and likewise for $f_2(x)$

Sergio Basurco

11:50

disregard my question, didn't read the operation well, it's rotating a vector then adding the result -_-

anakhro

14:06

@user400188 Sorry, the point of using \sqcup over \cup is to indicate that the union is disjoint.

geocalc33

14:36

hi

can anyone walk me through what it means for an integral transform to have a symmetry under $f(x) \to f(1/x)$?

anakhro

@geocalc33 do you have a reference for where you have seen this mentioned?

geocalc33

I can try to find the comment

found it

"Indeed, the Mellin transform is a multiplicative-coordinate analogue of Fourier transform, and like the symmetry of Fourier under $f(x) \to f(-x)$ there is a symmetry of Mellin under $f(x) \to f(1/x)$"

Jaakko Seppälä

15:24

Does anyone know how to solve this: Let $X=A_1\cup\cdots\cup A_k$ where $k\geq 2$, sets $A_j$ are Jordan curves and $|A_j\cap A_{j+1}|=1$ (a set that has exactly one element) for every $1\leq j\leq k-1$ and $A_j\cap A_k=\emptyset$ for every $|j-k|\geq 2$. How one can show that $\pi(X)$ is not an Abelian group?

anakhro

@geocalc33 can you link to the comment?

The symmetry for Fourier I thought had have a conjugate, $\hat f(x) = \overline{f(-x)}$.

Sorry, the second f should also have the hat.

geocalc33

15:47

3

A: Mellin transform of $f(1-x)$

No. For one thing, most of the values of $f(x)$ in the integral for $F(s)$ are completely different from the values of $f(1-x)$ in $M(f(1-x))$: the former takes values from $f([0,\infty))$, while the latter uses $f((-\infty,1])$.

it's the comment under this answer

@anakhro

Govind75

Wait can a 1D autonomous equation have infinitely many orbits?

I was thinking yes but my mate keeps saying it’s impossible, $\dot{x} = 1$ surely I can get infinitely many orbits as I can have different paths $t+C$

anakhro

@Govind75 Yes, that's an easy example.

@geocalc33 I don't know the Mellin transform, but I figure it's something like the conjugate symmetry of the Fourier transform.

geocalc33

@anakhro is $\hat f(x)$ the transformed function?

for Fourier

anakhro

Yes.

Ted Shifrin

17:13

@JaakkoSeppälä Have you drawn pictures? How do you show the fundamental group of a figure 8 is not abelian?

@Govind75 You’re misunderstanding. In dim 1 there is precisely one orbit. Look at $\dot x = Ax$ in $\Bbb R^2$.

Govind75

18:20

Why is there only one orbit?

Ted Shifrin

Because one flow line fills up all of $\Bbb R$.

Oh, you said 1D.

Your mate wins.

Adil Mohammed

Not a maths question, but can anyone British, say what's cool/cute about Paddington bear. I got an English assignment on Wojtek the polish and I got no clue what Paddington bear even go to do with it.

Govind75

Is the orbit a range of values?

Ted Shifrin

Orbits are the flow lines — the curves that are the solutions to different IVPs. You need to learn definitions.

Govind75

Yeah, so surely the example I gave gives infinite different orbits?

In 1D

Ted Shifrin

18:35

Surely? There is precisely one orbit.

Govind75

What is it?

Ted Shifrin

Every point is hit by any one of your solutions. It's just at different times. Who cares.

All of $\Bbb R$ is one orbit. The only way you can get more than one orbit is to have singular points (so that each of them will be an orbit by itself).

For example, $\dot x = x(x-1)$. Figure that one out.

Well, maybe start with $\dot x = x$. Then do the other one.

Govind75

$\frac{1}{1-Ae^t}$?

The latter one is $e^t + C$

Ted Shifrin

Nope on both, I believe.

But you're giving $x(t)$, not answering the orbit question.

Govind75

$Ae^t$?

Ted Shifrin

18:40

Yes, that one is correct. So what are the orbits?

Govind75

It's the set of values taken by $Ae^t$ when $A$ is varied

Ted Shifrin

So how many orbits?

Govind75

one

Ted Shifrin

nope

Govind75

Infinite?

yeah

Ted Shifrin

18:44

I gave you a clue when I mentioned singular points. What are the singular points here? (or equilibrium points?)

Govind75

You get equilibrium when $x = 0$ right?

Ted Shifrin

Yes. So that is one orbit. Now what?

Govind75

So $O(0)$ is a singular point

Then I get one more orbit for the other values of $x$

Can $x$ ever be $0$ though?

Ted Shifrin

No, you can't get just one more orbit. Think about it. Orbits have to be connected, so one can't jump over $0$.

Govind75

Lemme get the definition of an orbit straight, for any general solution $\lambda(t,t_0,x_0)$, we define the flow to be $\phi(t,x) = \lambda(t,0,x)$ - which is basically the general solution but around the initial condition $(0,x)$. Then the orbit is $O(x_0) = \{\phi(t,x_0) | t \in J_{\textrm{max}}\}$

where $J_{max}$ is our maximal time interval about $t=0$

Ted Shifrin

18:57

OK, fine :) Note that $e^t>0$ for all $t$, so the sign of $Ae^t$ is completely dependent on the sign of $A$. Thus ...

Govind75

So for $A > 0 $, our orbit is $(0,\infty)$, for $A = 0$, our orbit is $0$, for $A < 0$ our orbit is $(0,-\infty)$?

Ted Shifrin

Bingo. I would write $(-\infty,0)$, but yes.

So three orbits. Now what happens with my more complicated example?

Govind75

I get $log|x-1| - log|x| = t+C$, but I feel like I need to be careful from here

Ted Shifrin

OK ...

Govind75

I wanna just combine the logs and then exponentiate giving $\frac{x-1}{x} = Ae^t$

Ted Shifrin

19:06

Provided $x\ne 0,1$ you're OK.

Govind75

Yeah and then $1 - \frac{1}{x} = Ae^t$

Ted Shifrin

So you work on it and tell me how many orbits.

Govind75

But my general solution was wrong

Ted Shifrin

You have the same issue with your first one. There you get $\log |x|=t+c$ and you have a problem with $x=0$.

You need to observe that if $x_0$ is a singular point (i.e., $\dot x=f(x)$ and $f(x_0)=0$), then $x_0$ is automatically an orbit.

Govind75

That makes sense

Ohhhh

I guess up to this point I've only ever been looking at the solution

So $x = 0 $ or $ 1$ gives a singular point

Ted Shifrin

19:15

Yup.

Govind75

And then the rest will be governed by $\frac{1}{1+Ae^t}$

Ted Shifrin

So, in summary, how many orbits?

Govind75

4?

Ted Shifrin

Yup.

Go back and talk to your mate now :)

Govind75

Cheers

So it is impossible in $1D$ to have infinitely many different orbits

Ted Shifrin

20:31

Can you write a continuous function that has infinitely many zeroes?

Xander Henderson

Oh! Oh!

I can!

Pick me!

Govind75

21:05

$sin(x)$

Im confused, so are there infinite orbits or not, dies $\dot{x} = sin(x)$ product infinite singletons?

Govind75

22:10

I've spent 3 years on this question. If I have two subpopulations $A_1,...$ and $B_1,..$, where $A \sim N(\mu_1,\sigma_1^2)$ and $B \sim N(\mu_2, \sigma_2^2)$ and the

copper.hat

What do you mean by producing singletons?

Govind75

Now if I take a random sample $X_1,...,X_n$ where there is $\alpha$ probability $X_i$ is from $A$ and $1-\alpha$ probability $X_i$ is from $B$, what would the joint distribution of $X_1,...,X_n$

Points $x$ that give $\dot{x} = 0$?

leslie townes

infinite robots?

oh, orbits. nvm

Govind75

Any idea for the stats one, my first though was $f_{X_i}(x_i) = \alpha f_1(x_i) + (1-\alpha)f_2(x_i)$ where $f_1, f_2$ are the pdfs of the respective normal distributions above

copper.hat

somehow infinite singletons seems contradictory

Govind75

22:16

Is it because the maximal time interval would be finite?

copper.hat

A singleton means one of something.

Govind75

makes sense

leslie townes

there's a lot of this around 'finite'/'infinite' as adjectives.

every open cover has a finite subcover. here 'open' refers to elements of the subcover, and 'finite' refers to the subcover (and not its elements).

elements of the cover, i should have said in the first part.

it's sometimes helpful to insert words between finite/infinite and what they modify. e.g. instead of 'infinite singletons,' 'infinite collection of singletons' clarifying that 'infinite' refers to the family itself, not to a property of individual members of the family.

Govind75

Or is the distribution of $X_i = \alpha A + (1-\alpha)B$, then it's a lot simpler :)

But idk which one to choose agh

Ted Shifrin

22:55

@leslietownes You missed the earlier discussion of how many orbits $\dot x = f(x)$ has. The word singleton is inappropriate. We're talking about singular points, equilibrium points, or zeroes of $f$.

And then @Koro was thinking about group actions and orbits. Orbits must be in the air.

@leslie Here is one for you to answer quickly.

leslie townes

23:15

looks like others are on it. my answer would have been that there's no reason for an operator to have eigenvalues no matter what the field is although i hesitated to answer because i was afraid the response would be "well what if they do??? where are they?" getting back to his fundamental discomfort with the field-dependence of the definition.

once you stop distinguishing between vector spaces, or allow yourself to embed the thing you care about in a larger space where the operator might have some natural extension, you can create all kinds of things, including eigenvalues where there were none in the smaller space. but probably not what this guy was asking.

Derivative

my complex analysis book wants me to prove that if the coefficients of two polynomials differ by less than $\epsilon$ and the first doesn't have any roots on the unit circle, then they have the same number of roots in the unit disk

I know how to prove that, but not using any complex analysis.

leslie townes

seems like some quantifiers are missing. e.g. what does it mean for coefficients (lists of numbers) to 'differ by epsilon' and maybe even what epsilon is. 1 and z aren't that far apart as lists of coefficients but have different numbers of roots in the unit disk.

look in the index for rouche's theorem.

Derivative

ooooh right Rouché's theorem

thanks. I hope I manage to learn this type of problem solving skill

Ted Shifrin

23:46

@Derivative Oh really?

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