Mathematics - 2019-10-24 (page 3 of 3)

user193319

3:01 PM

Here's a theorem in a book I'm working through: If $B$ is a unital $C^*$-algebra and $A$ is a unital $C^*$-subalgebra of $B$ (same unit), then $\sigma_{A} (a) = \sigma_{B}(a)$ for all $a \in A$.

Alessandro Codenotti

@user193319 exactly what it sounds like intuitively

user193319

Here's part of the proof I don't quite understand: "But, if $a$ is invertible in $B$, so is $aa^*$. Then $aa^*$ is invertible in $A$, since $\sigma_{B}(aa^*)$ is real and thus has no holes. Therefore $a$ has a right inverse in $A$. But this right inverse must equal the inverse of $a$ in $B$ and so $a$ is invertible in $A$."

Alessandro Codenotti

So simply connected should work as a formalization

user193319

So, $\Bbb{C} \setminus \sigma (x)$ is simply connected>

Alessandro Codenotti

I think it's more like $\sigma(x)$ is a disjoint union of simply connected sets

user193319

3:09 PM

In the finite case that's true, because it's a union of singletons.

Leaky Nun

@Rithaniel we have $G \to S_4$ by permuting the 4 Sylow 3-subgroups; letting $v$ be the order of the image we have $v \mid \gcd(36,24) = 12$; the image is a transitive subgroup of $S_4$, so $v \ge 4$; so $v=4,6,12$; if $v=4$ then the kernel has order $9$ and kernel is normal so $n_3=1$, so this cannot be the case; by analyzing the transitive subgroups of $S_4$ (they are $C_4$, $V_4$, $D_8$, $A_4$, $S_4$) we see that the image must be $A_4$

so our group has $A_4$ as a quotient, and the kernel is a group of order $3$, so $G/C_3 = A_4$

for $A_4$ we have $n_3=4$ so this is also sufficient, and $C_3$ is the intersection of all the Sylow 3-subgroups

Rithaniel

Ah, that's pretty slick. No matter what, you end up with a normal subgroup of order 3.

Leaky Nun

so I must have made an error somewhere since you say that $(C_2 \times C_2) \rtimes C_9$ exists

actually $\operatorname{Aut}(C_2 \times C_2) = S_3$

so they should still intersect non-trivially?

Rithaniel

Alright, but where do you go from there. I imagine $C_3\times A_4$ falls automatically, but it might be possible to say that there is a group with a normal subgroup isomorphic to $C_3$ such that the quotient by $C_3$ is $A_4$, right?

Leaky Nun

what's wrong with $C_3 \times A_4$?

it's necessary and sufficient that $G/C_3 = A_4$

Rithaniel

3:18 PM

Well, I mean it "falls out" as in "is a valid option"

Leaky Nun

why? it does have 4 Sylow 3-subgroups

oh sorry I misread it as "fails"

Rithaniel

Ah, you're good. I could have been more clear. "Falls out" is still ambiguous

Ultimately the goal is to show that any group of order 36 is the semidirect product of it's Sylow subgroups

Leaky Nun

ok $V_4$ is a normal subgroup of $A_4$ so you get $C_3$ as a quotient

$G/(C_3 V_4) = C_3$

we have an order $3$ element so $1 \to C_3V_4 \to G \to C_3 \to 1$ has a section, so it is a semidirect product

by $C_3 V_4$ I mean $H$ such that $H/C_3 = V_4$

$H$ is a normal subgroup of order $12$

and $G = H \rtimes C_3$

so we have $1 < C_3 < H < G$ where $C_3$ and $H$ are both normal in $G$ and $C_3$ is normal in $H$

have you classified groups of order $12$?

Rithaniel

Ah, also the preimage of a normal subgroup under any homomorphism is a normal subgroup, so $V_4\times C_3$ should work, right?

Leaky Nun

the preimage of $V_4$ in $H$ is just $H$

$H/C_3 = V_4$

Rithaniel

3:26 PM

True, I'm thinking about in the original $G$. So, $H=V_4\times C_3$

Leaky Nun

how do you know $H = V_4 \times C_3$?

if $n_2=1$ for $H$ then indeed $H = V_4 \times C_3$

if $n_2 = 3$ then...

Rithaniel

Well, we know that $G/C_3\cong A_4$, and the preimage of $V_4\leqslant A_4$ under the natural projection should be $V_4\times C_3$

Leaky Nun

I'm not so sure about that

it might not be a projection at all

there are non-trivial maps $V_4 \to \operatorname{Aut}(C_3) = C_2$ fwiw

which gives you $C_3 \rtimes V_4$ that are not isomorphic to the direct product

Rithaniel

But are there non-trivial maps from $A_4\rightarrow\text{Aut}(C_3)$?

Leaky Nun

it doesn't need to be a semidirect product

Rithaniel

3:30 PM

I'm not sure, I've been working on this problem all morning.

Leaky Nun

if $n_2 = 3$ then $H \to S_3$ has image $C_3$ or $S_3$; for the former case the kernel is of order $4$ so $n_2 = 1$; so the image is $S_3$ and $H/C_2 = S_3$; $C_2$ and $C_3$ are normal subgroups of $H$ so $C_2 C_3 = K$ is a subgroup having $C_2$ as a normal subgroup, so $K = C_6$, so $H/C_6 = C_2$ having a section, so $H = C_6 \rtimes C_2$

so $G = (C_3 \times V_4) \rtimes C_3$ or $G = (C_6 \rtimes C_2) \rtimes C_3$

$C_6 \rtimes C_2$ is either $C_6 \times C_2$ or $D_{12}$

so $G = (C_3 \times V_4) \rtimes C_3$ or $G = (C_6 \times C_2) \rtimes C_3$ or $G = D_{12} \rtimes C_3$

hey $C_6 \times C_2 = C_3 \times V_4$

so $G = (C_3 \times C_2^2) \rtimes C_3$ or $G = D_{12} \rtimes C_3$

one can see that $\operatorname{Aut}(C_3 \times C_2^2) = \operatorname{Aut}(C_3) \times \operatorname{Aut}(C_2^2) = C_2 \times S_3$

a direct product would have $n_3 = 1$ so we look at the one sending $g \in C_3$ to $g \in C_3 < S_3$

well $C_3$ is not permuted so it's really $C_3 \times (C_2^2 \rtimes C_3) = C_3 \times A_4$

and for the latter case, i.e. $G = D_{12} \rtimes C_3$

Rithaniel

I'm copying and pasting this stuff to read later. I'm feeling that I'm at the limit of what my brain can absorb in one go. However, I did find this: math.berkeley.edu/~wodzicki/257/G36.pdf

It's talking about adjoint actions inducing a homomorphism from $A_4$ to $\text{Aut}(C_3)$ and I don't even know what an adjoint action is (Some googling implies it has to do with Lie algebras, though)

Leaky Nun

if $G/N=H$ then $H$ acts on $N$ via $(gN) \cdot n := gng^{-1}$; this is well-defined when $N$ is abelian

adjoint just means conjugation

Rithaniel

Ah, how do you show well-defined-ness? (I can probably figure that one out)

Leaky Nun

you can figure that out

let $\varphi$ be an automorphism of $D_{12}$ of order $3$; then $\varphi(r) = r$ and $\varphi(s) = r^2s$ or $r^4s$ (which are really the same)

Rithaniel

3:48 PM

Yeah, so if $gN=hN$ then there exists $c\in N$ such that $g=hc$ and so $gxg^{-1}=hcxc^{-1}h^{-1}=hxh^{-1}$ for $x\in N$ (I almost forgot to use the fact that $N$ is abelian)

Leaky Nun

how about you use $N$ instead of $C_3$

but ok

$D_{12} = D_6 \times C_2$ and $(1,g)$ corresponds to $r^3$

Rithaniel

Fair, generality is better than specificity

Leaky Nun

so $\varphi$ is constant on $C_2$, so $D_{12} \rtimes C_3 = C_2 \times (D_6 \rtimes C_3)$

this feels fishy

because $C_2$ is not a normal subgroup of $A_4$

but I don't want to think about this anymore

Rithaniel

(ditto)

It's a headache-inducing thing

Leaky Nun

well I don't really have a headache but I have other stuff to do

Rithaniel

4:00 PM

Yeah, I imagine for you that it might be more about time consumption than difficulty

ÍgjøgnumMeg

Evening

Rithaniel

Heya Meg

ÍgjøgnumMeg

How's it going @Rithaniel ?

Leaky Nun

hi @ÍgjøgnumMeg

ÍgjøgnumMeg

Hey @Leaky

Rithaniel

4:09 PM

Headache, taking a break from math. Pretty good because I got a lot of feedback on a problem I was having trouble with, though.

ÍgjøgnumMeg

Nice :)

But bad about the headache

lol

I'm typing to solutions to the first two problem sets for the semester, and we'll get the second ones tomorrow T_T

Rithaniel

What sort of problems? Harder or easier than you're used to?

ÍgjøgnumMeg

lol waaaay harder than I'm used to, but not really hard

it's just that my previous university had questions like "integrate $x$ between 0 and 1"

Rithaniel

Wow, that's pretty easy

ÍgjøgnumMeg

rofl well maybe not that easy, but there was no thought involved in most of the problems, usually just a lot of computation

Rithaniel

4:14 PM

Well, yeah, that's what you hope to see.

ÍgjøgnumMeg

The problems here are fine though, I assumed that because it was the first exercise sheet everyone would find it super super easy and so I thought I was gonna immediately fail because I didn't get the answers instantly lol

but maybe everyone thinks that

Nat

5:54 PM

"Ask HN: Are There Books for Mathematics Like Feynman's Lectures on Physics?", Hacker News.

schn

6:11 PM

Why does dropping the $y^2$-term work here? Isn't the function $x^2+y^2$ taking on different values than the function $x^2+x^2$ in the domain? Given $x\geq 0, y, z\leq 1$, doesn't that imply it is in an infinite rectangular box?

Rudi_Birnbaum

6:55 PM

Hi all! Servus @ÍgjøgnumMeg!

Alessandro Codenotti

Is it obvious that a Borel subset of $\Bbb R$ of positive Lebesgue measure contains a compact subset of positive measure?

Rudi_Birnbaum

special case: if its just one open set, then it should be obvious right?

Alessandro Codenotti

Sure, the only interesting case is when you have a set of positive measure containing no intervals

Rudi_Birnbaum

I feel I miss many theorems in that field ...

ÍgjøgnumMeg

@Rudi habe deeehri

Hey @Alessandro

Rudi_Birnbaum

7:09 PM

@ÍgjøgnumMeg Hawedere aa!

ÍgjøgnumMeg

@Rudi alls roger? :)

Rudi_Birnbaum

Logo, und bei Dir??

ÍgjøgnumMeg

jo goht schooo

muss etz aufgaben ufschrieba

Rudi_Birnbaum

Alemanisches understatment

wos fir aufgaben?

Just thought again about 1D-quasicrystals

Alessandro Codenotti

Hi @ÍgjøgnumMeg

ÍgjøgnumMeg

7:12 PM

ja fia d'vorlesung zu Modulformen, wobei ma bis etz eig nur über möbiustransformationa greadet hond

Rudi_Birnbaum

@ÍgjøgnumMeg head si cool oo

Ted Shifrin

@schn The region is symmetric about $x=y$, so the integral of $x^2$ and the integral of $y^2$ over the region are equal. Infinite box? Huh? It's a 1x1x1 cube.

Rudi_Birnbaum

Take a 2D lattice (e.g. pairs of integers).

Ted Shifrin

Hi @Rudi @ÍgjøgnumMeg

ÍgjøgnumMeg

Heya @Ted :)

Rudi_Birnbaum

7:13 PM

Hi @Ted!

schn

@TedShifrin But where are the lower and upper bounds for $x$ and $z$, respectively? And what about $y$?

Rudi_Birnbaum

and "intersect" it with a line of non-rational slope. Where intersect means to determine to orthogonal projections of the nearest points in orthogonal direction.

Ted Shifrin

What are you talking about? Everybody goes from 0 to 1.

This is the easiest sort of problem.

schn

Aren't they saying $x\geq 0, y, z\leq 1$?

Rudi_Birnbaum

The resulting points are a quasi crystal.

Ted Shifrin

7:16 PM

What's typed there is $0\le x,y,z\le 1$. That means they all go from 0 to 1.

schn

Confusing.

Ted Shifrin

You would need to write $x\ge 0$ and $0\le y,z\le 1$ for yours.

No, not confusing.

ÍgjøgnumMeg

@Mathein the second Übungsblatt for Modulformen is up, first question is showing that congruence subgroups are finite index in $\operatorname{SL}_2(\Bbb Z)$

Rudi_Birnbaum

In the sense that their Fourier Transform (however that might be defined) is discrete.

Ted Shifrin

Otherwise, with your interpretation, you have no upper limit on $x$ and no lower limit on $z$ and I don't know what you do with $y$.

No logic to yours.

Besides, the problem should be stated clearly in the book. You don't learn the problem by reading the answers.

schn

7:17 PM

@TedShifrin Got it. It is one big interval.

apt45

Hey everybody.. I have a very stupid question.. How can I define the Imaginary part of $f(z)=Log(-z)$ evaluated on z+i\epsilon? So I want to compute $Im[f(z+i\epsilon)]$... I should revise complex analysis....

Rudi_Birnbaum

What I now thought of is if there is a way to choose a "grid" such that we get a bijection between the quasicrystals and the reals.

Ted Shifrin

Where is $z$ and what branch of Log are you using?

Does Log mean the principal branch? What does that mean?

Rudi_Birnbaum

Maybe not today.

Alessandro Codenotti

Hi @Ted

Ted Shifrin

7:18 PM

Hi, demonic @Alessandro.

apt45

@TedShifrin $z$ is on the real axis and the branch of log is on the positive real axis

schn

@TedShifrin Although when writing $x\geq 0, y, z\leq 1$ (which is also written in the problem statement!), how does it imply that $ y,z \geq0$

Ted Shifrin

You mean the branch cut is on the positive real axis? Really?

What I read was $0\le x,y,z\le 1$, @schn. One of us is dreaming

schn

I am!

Ted Shifrin

nods

schn

7:20 PM

Thanks.

Ted Shifrin

LOL, you're welcome.

apt45

@TedShifrin is that not correct?...

Ted Shifrin

It's customary to make the branch cut on the negative real axis, @apt45, and let the imaginary part go from $-\pi$ to $\pi$ (not inclusive). Double-check.

But maybe your book is different.

Most people want the log of positive real numbers to be consistent with what we all know.

I suspect I'm right because your problem is about $f(z)=\text{Log}(-z)$.

apt45

@TedShifrin ok.. I understand this. However, i get confused when I have to compute an integral like $\int_{z_1}^{z_2} dz Im((log(z+i\epsilon)+log(-(z+i\epsilon$)))$ where $z_2 > z_1$ and $z_1>0$.

accordingly to where I choose the branch cut, I might get different answers.. no?

Ted Shifrin

Yes. But if you wrote Log, that specifically means the principal branch. You're going to have some arctans in there, but roughly the first one is a bit more than $0$ and the second one is a bit less than $-\pi$.

You really need to look at the definitions carefully. Whether $-\pi<\theta<\pi$ or $0<\theta<2\pi$ is crucial.

apt45

7:28 PM

@TedShifrin why the second one is a bit less than $-\pi$?

Ted Shifrin

Sorry, I meant a bit greater than $-\pi$.

apt45

yes ok.. but why $-\pi$?

MatheinBoulomenos

@ÍgjøgnumMeg I mean that's just obvious from the definition

Ted Shifrin

Because I'm using the branch with $-\pi<\theta<\pi$, as I said earlier.

ÍgjøgnumMeg

@Mathein thumbs up ...

Ted Shifrin

7:29 PM

You're just below the negative real axis there.

heya @Mathein

MatheinBoulomenos

Hey @Ted

set theorists have the best terminology:

Mouse (set theory)

In set theory, a mouse is a small model of (a fragment of) Zermelo–Fraenkel set theory with desirable properties. The exact definition depends on the context. In most cases, there is a technical definition of "premouse" and an added condition of iterability (referring to the existence of wellfounded iterated ultrapowers): a mouse is then an iterable premouse. The notion of mouse generalizes the concept of a level of Gödel's constructible hierarchy while being able to incorporate large cardinals. Mice are important ingredients of the construction of core models. The concept was isolated by Ronald...

apt45

@TedShifrin thank you!

Alessandro Codenotti

@MatheinBoulomenos Also the best results

Ted Shifrin

Sure :)

MatheinBoulomenos

@Alessandro that's debatable

Alessandro Codenotti

7:33 PM

There's also the concept of a Weasel which should also be related to extender models somehow (which is where mice come up as well)

@MatheinBoulomenos Sure, we can debate that, but I'm right

MatheinBoulomenos

the best results are obviously in non-associative algebra

tigre

hey there. just a quick question. if i have $A = [(12)]+[(23)]$ in $\Bbb{C}[S_3]$, and i consider $\alpha:\Bbb{C}[S_3]\to \Bbb{C}[S_3]$ such that $\alpha(v)=Av$, so its just left action. someone wrote down on my page that $tr(\alpha) = tr(A)=tr([(12)])+tr([(23)])$. is this true? i feel like it makes no sense, and you would first have to write the left-action operator $A$ in a basis for $\Bbb{C}[S_3]$, and its not this simple, in particular $\rho: \Bbb{C}[S_3]\to \text{GL}(\Bbb{C}[S_3])$ isn't -

-additive, so $\rho(A)=\rho([(12)])+\rho([(23)])$ makes no sense, so i can't see why the trace would sum over these

Tobias Kildetoft

@tigre Where did $\rho$ come from?

tigre

i'm considering $\Bbb{C}[S_3]$ as a $\Bbb{C}[S_3]$-module over itself, and just using $\rho$ to denote the action

Tobias Kildetoft

don't

tigre

7:39 PM

i guess $\rho(A)=\alpha$ in my dumb notation

Tobias Kildetoft

because in that case, $\rho$ is indeed additive and has a different image than you wrote

tigre

okay ignore $\rho$. i'm just wondering about computing the trace of left-action operators, in the case that the left-action operators come from sums in the group algebra

Tobias Kildetoft

trace is additive, so what is written is correct, except you obviously still need to compute those individual traces

tigre

wait, so if $L_A:v\mapsto Av$ denotes the left-action operator by $A$, then $tr(L_{A+B})=tr(L_A)+tr(L_B)$?

Tobias Kildetoft

yes

tigre

7:43 PM

why?

Tobias Kildetoft

just write out the definition

tigre

oh right, i think i realised where im messing up all along

$\rho:\Bbb{C}[S_3]\to End(\Bbb{C}[S_3])$ rather than the image being $Aut(\Bbb{C}[S_3])$ i guess, giving me back the additive structure

MatheinBoulomenos

you actually have map $\Bbb C[S_3] \to \mathrm{End}_{\Bbb C}(\Bbb C[S_3])$ (call it $\rho$ if you want) given by $a \mapsto L_a$ where $L_a:v \mapsto av$. This is additive, multiplicative and $\Bbb C$-linear

yeah

tigre

yeah, i thought for some reason they had to be automorphisms, which made it seem like nonsense, my bad lol

ÍgjøgnumMeg

7:55 PM

lines going through zero in $\overline{\Bbb C}$ are of the form $\beta z + \bar{\beta}\bar{z} = 0$ right? ($\beta \in \Bbb C$)

So if I wanna find the inverse image of a line going through $0$ under a Möbius transformation I can just apply $\varphi_M^{-1}$ to both sides of that equation; where's the error

rofl

MatheinBoulomenos

@ÍgjøgnumMeg I don't see an error

ÍgjøgnumMeg

@Mathein okay so.. I'm getting $(\beta z + \bar\beta \bar z)z_1 = (\beta z + \bar\beta \bar z)z_2$ out where $z_1, z_2$ have respective images $0$ and $\infty$ under the relevant möbius transformation

which seems wrong to me but idk

MatheinBoulomenos

the inverse of a Möbius transformation is another Möbius transformation (you can easily invert the matrix, too), so let's just work with applying a Möbius transformation

ÍgjøgnumMeg

yeah I inverted the matrix, got the inverse of the transformation and just applied that to both sides of the equation

and in theory that should give me a generalised circle through $z_1$ and $z_2$

although now that I think about it a line through 0 should go to a line through 0 right

MatheinBoulomenos

8:11 PM

@ÍgjøgnumMeg you have to make a case distinction if your Möbius transform is of the form $\frac{az+b}{z}$ or not

ÍgjøgnumMeg

8:28 PM

@Mathein well my Möbius transform is $c\frac{z - z_1}{z - z_2}$

c a non-zero constant

MatheinBoulomenos

and can $z_2$ be zero?

ÍgjøgnumMeg

hmm

Well the image under that transformation of $z_2$ should be $0$ so I guess it can be zero

erm

$\infty$ I mean

makes sense

MatheinBoulomenos

so if $z_2=0$, you have to be careful

ÍgjøgnumMeg

Cool :) but if $z_2$ is non-zero then I still seem to get some nonsense out

Ultradark

9:13 PM

0

Q: Looking to solve the dynamical system via differential equation or lack thereof

Given the boundary of the following phase portrait (which the interior can easily be filled in), find the differential equation solution or prove that the solution is independent of a differential equation. I've gone through different scenarios of phase portraits, such as the hyperbolic phase po...

Take a look!

JohnnyApplesauce

What size of infinity are the complex numbers?

Rithaniel

I believe they are the same as the reals. Though someone else will probably be able to give more insight

ÍgjøgnumMeg

I'd say the same naively lol

Ultradark

same as the reals, but if you take the complex field and compare it to the reals then you see that the reals are a subfield

JohnnyApplesauce

What makes it bigger?

Ultradark

9:21 PM

it's four dimensional

JohnnyApplesauce

I thought reals were two dimensional and complexes third?

Rithaniel

Yeah, it's two dimensional as a vector space, at the very least

1 and i serve as a basis

Well, as a vector space over the reals

Ultradark

how could the complex numbers be three dimensional?

there is no algebraically coherent triplex

Rithaniel

complex numbers are 1 dimensional as a vector space over the complex numbers

JohnnyApplesauce

I’ve commonly seen graphs explaining complexes and attempting to concretize the concept by showing it as a sphere

Ted Shifrin

9:27 PM

The "extended" complex numbers (including $\infty$) can be thought of as a sphere, yes.

JohnnyApplesauce

Back to infinity

krauser126

Having a hard time with this question

0

Q: What is the probability given another event is occurring?

I have Box A with $3$ red balls and $1$ blue ball. I have Box B with $1$ red ball and $4$ blue balls. I randomly take a ball from Box A and put it into Box B. I then randomly draw a ball from Box B and it happens to be a red ball. What is the probability that the ball taken from Box A was red? ...

probability conditional-probability uniform-distribution

JohnnyApplesauce

I know that aleph one is bigger than aleph null, and aleph two in turn larger than both

But what is bigger than reals/complexes?

I know, even, that there are infinite sizes of infinity, but what has those sizes?

Ultradark

quaternions, octonions,...

JohnnyApplesauce

Ah, the set of all n-ions has size n-2?

Rithaniel

9:33 PM

Quaternions have the same cardinality as the complex numbers (unless we're talking about the group)

Ultradark

these number systems are bigger than the complex numbers but they don't have the same favorable properties like the complex numbers

Rithaniel

(If we're talking about the group, there are only 8 elements)

JohnnyApplesauce

What’s a group?

Rithaniel

A set with a binary operation satisfying closure, associativity, invertibility, and identity

JohnnyApplesauce

I know what associativity and identity are. Is invertability similar to being able to be negated?

Rithaniel

9:35 PM

Abstract algebra stuff

JohnnyApplesauce

The generalization?

Ted Shifrin

@Rithaniel: Be careful. A vector space is also a group. You mean the group of unit quaternions, of course, under multiplication.

Rithaniel

Yeah, basically identity means that you have an element e such that ea=ae=a. Invertibility means that for all a, there exists a b such that ab=ba=e.

Also, very good point, I hadn't thought about that

JohnnyApplesauce

I’ve taken two discrete math courses and one calculus course. Is there any way that I have been exposed to invertability before? It seems strange.

Rithaniel

Yeah. Negative numbers.

They are the "additive inverses" of positive numbers

Ultradark

9:40 PM

you've probably found the inverse of a function

JohnnyApplesauce

Ah, that makes sense.

Rithaniel

In the integers under addition, 0 is the identity and, for any integer $a$, then $-a$ is it's inverse.

JohnnyApplesauce

Is the closure that you mentioned earlier

Like closure of a set?

Rithaniel

Well, not exactly. Basically it means that if $a,b\in S$ (where $S$ is the group) then $ab\in S$

So if you stick two things together, you get something else from the set

JohnnyApplesauce

Is that the generalization of multiplication?

Rithaniel

9:44 PM

Like, you don't add 1+2 and get a matrix. (It's like a generalization of many operations)

So, addition can be a group operation, multiplication can be a group operation, composition of functions can be a group operation, etcetera

JohnnyApplesauce

Ok, so what are some closure that I haven’t heard of but are in my immediate grasp?

Rithaniel

So, if you have the integers. Add two integers. What can you say about what that sum?

It's another integer, right?

That's an example of a binary operation on a set obeying closure

It might be more insightful to see what something disobeying the closure axiom would look like

Like, if you somehow added two numbers together and got a cheeseburger and fries, that would definitely not be a binary operation obeying closure

If you're looking at a set of real numbers and your binary operation somehow takes two reals and spits out a complex number, then that binary operation is breaking closure

Kind of an idea of "what happens in the set, stays in the set."

krauser126

9:59 PM

Anyone?

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Q: What is the probability given another event is occurring?

I have Box A with $3$ red balls and $1$ blue ball. I have Box B with $1$ red ball and $4$ blue balls. I randomly take a ball from Box A and put it into Box B. I then randomly draw a ball from Box B and it happens to be a red ball. What is the probability that the ball taken from Box A was red? ...

probability conditional-probability uniform-distribution

Semiclassical

10:17 PM

@krauser126 I like to do such problems by considering a hypothetical sample set with those frequencies

krauser126

I'm not sure how that helps me

I already got started with it if you read the post. I just don't know if Im proceeding correctly and/or what Im missing @Semiclassical

Semiclassical

So, suppose I perform that drawing 600 times. In 150 cases, I get a blue ball and so the composition of box B is 1 red : 5 blue. Hence I’d get red from box blue in 25 drawings and blue in the other 125

In the other 450 cases, the ball drawn from box A is red so the composition of box B is 2:4. So the frequency of red draws to blue is 150 to 450

So: there are 25+150=175 drawings in which the red ball is drawn from box B. In 150 of those, the red ball was drawn from box A

So the probability that a red ball was drawn from box A, given that a red ball was drawn from box B, is 150/175 = 6/7

In retrospect, I could just as well have taken 12 samples altogether

krauser126

10:32 PM

"the frequency of red draws to blue is 150 to 450"

Isn't it hte other way around? Frequency of blue to red?

Semiclassical

Yes, you’re right. This is the peril of doing math on my phone @krauser126

That said, one nice bit of this calculation is that it’s not that hard to spot issues like that if you’re looking for them

As you just demonstrated :P

user10478

11:41 PM

The sum of two different differential operators applied to the same operand can be combined in the obvious way, right? So $(D^2 + 2iD - 8D - 8i)[f(x)] + (D^2 - 2iD - 8D + 8i)[f(x)]$ simplifies to $(2D^2 - 16D)[f(x)]$?

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