Mathematics - 2019-10-28

Stan Shunpike

01:25

helllo

Ultradark

01:38

hello @StanShunpike

Stan Shunpike

01:52

I saw a problem which went as follows. Suppose $v,w \in \mathbb{R}^n$ and $c$ is a scalar. Show $span(v+cw,w)=span(v,w)$.

I don't understand what there is to prove. Both $w$ and $v + cw$ are spanned by $v,w$. So by construction, $span(v+cw,w)=span(v,w)$

Akiva Weinberger

02:03

@StanShunpike That proves that span(v+cw,w) is a subset of span(v,w)

What you need to prove it two things: ${\rm span}(v+cw,w)\subseteq{\rm span}(v,w)$ and ${\rm span}(v,w)\subseteq{\rm span}(v+cw,w)$

To prove the first one, you have to show that if something's an element of ${\rm span}(v+cw,w)$ then it's an element of ${\rm span}(v,w)$

To prove the second, you have to show the reverse

So: suppose $x\in{\rm span}(v+cw,w)$…

Stan Shunpike

02:25

@AkivaWeinberger ah ok thanks for clarifying

Silent

06:10

In this proof, I can't understand why $q-q(c_j)=(x-c_j)h$ holds.

Silent

06:27

plz help!

Rithaniel

08:23

I feel like I'm losing my mind. Is it not true that, if $X$ and $Y$ are random variables subject to $0\leq x\leq y \leq 1$, then $F_{X+Y}(k)=\int_0^1\int_x^{k-x}f_{X,Y}(x,y)dydx$?

Okay, I think my issue is that I am messing up when finding $f_{X,Y}$

This is actually $f_{X_{(3)},X_{(4)}}$ when $X_1,X_2,X_3,\ldots X_4$ are four independent random $\text{Unif}(0,1)$ random variables and $X_{(k)}$ is the $k-$th order statistic of these four.

We supposedly have that $f_{X_{(i)},X_{(j)}}(x,y)=\frac{n!f(x)f(y)F(x)^{i-1}(F(y)-F(x))^{j-i-1}(1-F(y))^{n-j}}{(i-1)!(j-i-1)!(n-j)!}$
However, plugging in $i=3$ and $j=4$ with the above random variables, we arrive at $f_{X_{(3)},X_{(4)}}(x,y)=12x^2$

Then taking $\int_0^1\int_x^{1-x}12x^2dydx$ gives me $6-4k$, which says that $P(X_{(3)}+X_{(4)}\leq 1)=-2$

Which is absurd

Shawn

10:45

Is uncountable product of compact spaces compact ?

Rithaniel

11:29

Take the cover of the space which is just each set in the product @Shawn

Alessandro Codenotti

11:40

@Shawn yes

Shawn

11:56

@Rithaniel I didn't got , can you elaborate little ?

Also in box topology or in product topology or both ?

The Terrible Puddle

Whenever I sit down to read abstract algebra some song starts to play in my head.

Loudly

Shawn

It might be possible that arbitrary cover i am referring to does not have an element having each set as product ?

@TheTerriblePuddle medicalsciences.stackexchange.com or wait for TedShifrin

@Rithaniel It's not that easy (to prove in one line), Tychonoff theorem there for help

Also Tychonoff theorem proves for product topology

Martin Sleziak

12:21

Well, the thing you have asked (product of compact spaces is compact) is precisely Tychonoff's theorem.

For box product, see: Tychonoff Theorem in the box topology or $[0,1]^{\mathbb{N}}$ with respect to the box topology is not compact.

ÍgjøgnumMeg

12:46

Afternoon all

ÍgjøgnumMeg

13:27

Sanity check: if I have a free $A$-Module $A^r$ with $A$ contained in a field $K$ then $A^r \otimes_A K \cong K^r$

just extension of scalars right?

I can send $(a_1, \dots, a_r) \otimes k \mapsto (ka_1, \dots, ka_r)$ for one direction I guess, where do I send $(k_1, \dots, k_r)$ for the other direction? lol

MatheinBoulomenos

@ÍgjøgnumMeg tensor product commutes with direct sums, so you may assume wlog $r=1$

ÍgjøgnumMeg

and then I just send $k \mapsto 1 \otimes k$?

MatheinBoulomenos

yes

ÍgjøgnumMeg

Cool :) Thanks

@Mathein also I think I got the surjectivity of that homomorphism $\operatorname{SL}_2(\Bbb Z) \to \operatorname{SL}_2(\Bbb Z/N\Bbb Z)$

MatheinBoulomenos

okay

ÍgjøgnumMeg

13:35

which is nice, since it was frustrating me lol

YOUSEFY

13:54

Hi, I have a small question: Give G=(V,E) and |V|=n, What is the maximum number of edges of path cover [vertex-disjoint path cover] problem? I believe it is n - 1. Is it possible to be n?

Here is a definition for path cover

Leaky Nun

@ÍgjøgnumMeg $\left( \frac{a_1}{b_1}, \frac{a_2}{b_2}, \cdots, \frac{a_n}{b_n} \right) \mapsto \left( a_1 b_2 \cdots b_n, b_1 a_2 \cdots b_n, \cdots, b_1 b_2 \cdots a_n \right) \otimes \frac{1}{b_1 b_2 \cdots b_n}$

or more simply, don't forget that elements of tensor product don't need to be pure tensors

YOUSEFY

Path cover

Given a directed graph G = (V, E), a path cover is a set of directed paths such that every vertex v ∈ V belongs to at least one path. Note that a path cover may include paths of length 0 (a single vertex).A path cover may also refer to a vertex-disjoint path cover, i.e., a set of paths such that every vertex v ∈ V belongs to exactly one path. == Properties == A theorem by Gallai and Milgram shows that the number of paths in a smallest path cover cannot be larger than the number of vertices in the largest independent set. In particular, for any graph G, there is a path cover P and an independent...

Leaky Nun

@ÍgjøgnumMeg $(k_1, \cdots, k_r) \mapsto (1, 0, 0, \cdots, 0) \otimes k_1 + (0, 1, 0, \cdots, 0) \otimes k_2 + \cdots + (0, 0, 0, \cdots, 1) \otimes k_r$

ÍgjøgnumMeg

Ahhh nice

@Mathein where was that room where I could get printed Skripte nochmal? :D

and am I allowed to just walk in and take a skript?

MatheinBoulomenos

first floor

yes you can just take one

Avnish Kabaj

15:20

does anyone know where do profs steal questions from

linear algebra & diff eqns

Leaky Nun

steal is a strong word

Rithaniel

Some questions tend to be standard ones that are asked across the board (they're more common in the more advanced or abstract courses) but the majority of problems are made specifically by the professor themselves

The East Wind

16:27

Can anyone help me with calculating a double integral over a general region

Semiclassical

"Just ask; don't ask to ask."

The East Wind

@Semiclassical Sorry, forgot about that.

What I am asking is calculating the double integral when the region of integration is something like this:

This is in contrast to something like this:

In the latter case, we are able to find constant bounds for the outer integral

Semiclassical

okay. so it's a region which you can't describe easily as "the area between graphs of two singe-variable functions"

The East Wind

Yup

Semiclassical

typically, you'd look for some convenient way of parametrizing the region of integration

The East Wind

16:34

@Semiclassical So basically a change of variables so that the region in the new coordinate system is easily described as the "area between graphs of two singe-variable functions"

Semiclassical

Pretty much, yes. The tricky thing is that your change of variables will affect the double integral via the Jacobian. (This is just the multivariable version of the substitution rule.)

The East Wind

Semiclassical

Okay, so it's kind enough to give you the transformation.

The East Wind

@Semiclassical Yes i read about that.

@Semiclassical I am still having trouble with the outer integral though :(

Semiclassical

Can you see what range of u and v would generate that region?

The East Wind

16:37

u obviously goes from 1 to 4

Semiclassical

Right. The v part is the problem.

The East Wind

Yes.

Any hints?

Semiclassical

I suppose I'd go like so. First, we pick some value of $u$ from 1 to 4.

We now want an expression for the range of $v$.

So we've got $u=x^2-y^2$ and $0\leq y\leq (3/5)x$.

Gross.

The East Wind

Well the answer suggests that the ratio of the upper and lower limit is 2

Semiclassical

We've got $x=\sqrt{u+y^2}$, so $0\leq y\leq 3/5 \sqrt{u+y^2}$

Actually, come to think of it:

$u$ is a difference of squares, so $u=x^2-y^2=(x-y)(x+y)=(x-y)v$

So $x+y=u$ and $x-y=u/v$. Hence $x=(u/2)(1+1/v)$ and $y=(u/2)(1-v)$

So therefore $y=(u/2)(1-v)\geq 0\implies v\leq 1$

should have been $y=(u/2)(1-1/v)$, woops

and therefore $v\geq 1$

similarly, $y\leq (3/5)x\implies (u/2)(1-1/v)\leq (3/5)(u/2)(1+v) \implies 1-1/v\leq (3/5)(1+v)\implies 1 \leq 4/v \implies v\leq 4$

so $1\leq v\leq 4$?

that seems too easy

The East Wind

16:48

but this gives the wrong ans

Semiclassical

yeah

The East Wind

this gives exactly twice the given ans

the ans is ln2*(e^4 -e)*0.5

Semiclassical

hmm, interesting

well, the jacobian in this case is 2u, isn't it

regardless, I don't trust my work

oh, yeah, I wrote it wrong

u=x^2-y^2, v=x+y thus u/v = x-y thus x=1/2(v+u/v), y=1/2(v-u/v)

then y>=0 gives v>=u/v

and y <= 3/5 x gives v+u/v <= 3/5(v-u/v) gives v+4u/v <= 0

only way that can work is if v isn't positive, in which case I think one gets $-2\sqrt{u}\leq v \leq -\sqrt{u}$

which looks an awful lot like what you suggested

still not convinced that's right tho

The East Wind

the jacobian is 1/2v

so that makes the final integral e^u/2v

u goes from 1 to 4

the problem is with v

Semiclassical

17:04

hmm. I get $\partial (u,v)/\partial (x,y) = (2x)(1)-(-2y)(1)=2(x+y)=2v$

so agreed

as I look at how (u,v) parametrize (x,y), I don't think one can write the range on $v$ as $f(u)\leq v\leq g(u)$ for $1\leq u\leq 4$

The East Wind

so is the question wrong?

Semiclassical

let me restate that. you can write the range in that form, but f(u) and g(u) will have to be piecewise functions

so different f(u), g(u) depending on what range of u you're considering

which...yuck

The East Wind

So how do i begin to write those piecewise functions

anakhro

17:30

Hey lukewarm lizards

2

Rithaniel

That is an interesting greeting

The East Wind

Does anyone else have any ideas on this^^?

Rithaniel

I am having trouble staying awake, so, unfortunately, nope

anakhro

@Rithaniel well every other time I use hot cats and no one comments. :(

Rithaniel

I've noticed when you greeted with hot cats. But those days I was in lurker mode

Maybe try tepid tapirs?

Muggy mooses?

anakhro

17:44

I like the tapir one.

Rithaniel

Magical liopleurodons?

MatheinBoulomenos

this reminds me of this: wiki.ubuntu.com/DevelopmentCodeNames

anakhro

@Rithaniel I'd ask you what math you are doing lately, but you should probably go to sleep if you are having troubles staying awake.

Rithaniel

I am in lecture. It would be very rude to go to sleep. :P

As for math: probability HW and examining groups of order $2^6$

MatheinBoulomenos

All 267 of them?

anakhro

17:50

@MatheinBoulomenos did you know that number off the top of your head?

Rithaniel

Yes, by hand

(Nah, just a couple)

MatheinBoulomenos

good luck with that

Rithaniel

$Q_8\rtimes Q_8$ and $C_2^3\rtimes D_4$ in particular

I would consider myself quite skilled if I could classify all 267 myself, though

Of course, it seems more tedious than anything

MatheinBoulomenos

iirc we don't know the number of groups of order $2^{11}$ yet

Rithaniel

Hey, open problem for me to lose my mind trying to tackle

I'm assuming it's a huge number, maybe more than $2^{17}$ groups?

Because the number of order $2^6$ is more than $2^8$, so I imagine it grows roughly exponentially

MatheinBoulomenos

18:01

the number of groups of order $2^{10}$ is $49487365422$. There are probably some heuristics for estimating the number of groups of order $2^n$, but I'm not familiar with it

Rithaniel

That's more than $2^{35}$, yeah

Also, potentially worth noting is that the highest power of 2 it is divisible by is 2.

Ultradark

19:05

When, if ever, do the isotopy subgroups of non-isomorphic abelian varieties commute?

Shine On You Crazy Diamond

20:57

@Ultradark

Anyone here good at free groups?

s.harp

@ShineOnYouCrazyDiamond i know what a free group is and i know the pingpong lemma, but thats all

Alessandro Codenotti

Just ask your question

Ted Shifrin

Hi, demonic @Alessandro

Alessandro Codenotti

Hi @Ted

Shine On You Crazy Diamond

21:13

@s.harp

:math.stackexchange.com/questions/3412931/…

s.harp

@ShineOnYouCrazyDiamond does the fat dot $ z=i \bullet i$ indicate that there is some string $x$ so that $z= ixi$?

Shine On You Crazy Diamond

Yes

that's precisely what it indicates

THere may be many versions of it inside of $s$ (ways of realizing $i \bullet i$)

s.harp

the string between them can also be empty?

Shine On You Crazy Diamond

@s.harp thx for the upvote

Yes

empty is possible as in $ababab$

$i = ab$

thus $i^4 = 1$

every element should indeed be invertible under this construction

$H$ is indeed normal.

That's a trick in free groups. Quotienting by all elements $y Ey^{-1}$ given what you really want to quotient by $E \subset F$.

s.harp

if $s= abab\, c\, ab$, is then $(ab)^4=1$ one of the relations you are interested in imposing? what about $(ab)^2c(ab)^2$?

Shine On You Crazy Diamond

21:26

On the other hand, we could just say $H = \{ x \in F : x $ is not possible $\}$ which should simplify things

Remember

that grouping on $s$ you gave

corresponds to $i^4$.

I.e. each element corresponds to many groupings of terminal variables within $s$.

*many possible

s.harp

hmm so if $i \in I(s)$ (is terminal) with $s= i\bullet i ... \bullet i$ (n times) then you want to have every $i x_1 i x_2... x_n i$ to quotient to the neutral?

Shine On You Crazy Diamond

@s.harp see edits

I changed $H$ to the simpler version

Because it is true that the set is multiplicatively closed by $xy^{-1}$ since you can't multiply two impossibilities to reach a possibility!

If $i \in I(s)$ then $s \notin F$, only elements of $I(s)$ are allowed in!

yet you wrote $s = i \bullet \cdots \bullet i$.

@s.harp is it making sense yet?

s.harp

not really, sorry

Shine On You Crazy Diamond

I'm not sure about $x \in H \implies x^{-1} \in H$ though.

s.harp

$\bullet$ is also connected to how multiplication is denoted in the Free group $F$?

Shine On You Crazy Diamond

21:38

Yes it is the free group multiplication

@s.harp how can I make the construction more rigorous / clear?

formal

s.harp

ok, one more question: for your example $I(s)$ has $11$ elements, is $F$ then a free group with $11$ generators? or do you have relations like $ab\bullet cd = abcd$ (all 3 are in $I(s)$)

Shine On You Crazy Diamond

Yes

first guess is correct

$F = \text{FreeGroup}(I(s))$.

s.harp

ok^^

Shine On You Crazy Diamond

So in essence the string nature of them is not used, the elements of $I(s)$ are just placeholders

Though, it is a string algorithm

s.harp

yes, that confused me initially, but now i understand that

Shine On You Crazy Diamond

21:42

What should I say in the post to make it more clear?

There might be a relationship between $\cdot$ and $\bullet$. Not sure

*concatenation = $\cdot$

Let $s = abab$. Then $I(s) = \{ab\}$ and $i \in I(s) \implies i^3 = 1$. Thus $G(s) = C_3$ the cyclic group of order $3$.

s.harp

yes, I think I am understanding

Shine On You Crazy Diamond

I may be wrong, it could be $C_2$

You have to determine whether each of $x y^{-1} \notin H$ for $x,y$ powers of $i$.

s.harp

the subgroup $H$ is the subgroup generated by all elements $i_1\bullet ... \bullet i_n$ ($i_1,..., i_n\in I(s)$ ) so that there is no way to inject strings into the $\bullet$ to get $s$?

Shine On You Crazy Diamond

Yes

If you label each $\bullet_i, i = 1...k$

then there is a string hom that does the replacement

Not sure how that applies though, but good guess! It's fun

What would be neat is if we could relate the group at this stage $G(s)$ with the group $G(s')$ where $s'$ is $s$ with some compressibles taken. For instance is one a subgroup of the other?

That is to say it would have to survive the construction that way

$s' = IabIabc$ and $I = ab$. Then $I(s') = \{ab, abI \}$.

I think $G(s') \leqslant G(s)$ is a subgroup!

A grammar at the end of this process would have $G(s^{(k)}) = \{1\}$ since $I(s) = \{\}$ and we call $\text{FreeGroup}(\{\}) = \{1\}$.

@s.harp would you like to collaborate with me on this question?

It's fertile research area: smallest grammar problem (SGP)

s.harp

22:02

I bookmarked your messages, I'll think about it tomorrow, right now I cannot concentrate

Ultradark

@ShineOnYouCrazyDiamond yo

how's the research going?

topologicalmagician

Hi

Shine On You Crazy Diamond

Yo @Ultradark

See my link:

2

Q: Quotient of a free group on the compressible substrings of a string.

Let $s$ be a string over an alphabet $\Sigma$, say $s = abcdabcdab$. We write $t\leqslant s$ to mean $t$ is a substring of $s$ or there exist $u, v$ strings over $\Sigma$ such that $utv = s$ where multiplication is string concatenation. Define $C(s) = \{ t \leqslant s : tvt \leqslant s,$ for s...

Ultradark

Okay

topologicalmagician

@Ultradark A function $f:A \subseteq \mathbb{R} \rightarrow \mathbb{R}$ is uniformly continuous if $\forall \epsilon>0$ $\exists \delta>0$ such that $\forall x \in A and \forall y \in A$ with $|x-y|\leq \delta$ we have $|f(x)-f(y)|\leq \epsilon$.... Is the negation then There exists and $\epsilon>0$ such that $\forall \delta>0$ there exists $x\in A$ or $y\in A$ such that $|x-y|\leq \delta$ and $|f(x)-f(y)|\geq \epsilon$??

Shine On You Crazy Diamond

22:13

@Ultradark research is going okay :)

amanuel2

Can someone ping me when yall finish answering his question? Thanks :)

Shine On You Crazy Diamond

@amanuel2 which question?

amanuel2

The one you were working on, I think Ultradark

Im gonna ask my question now, free to answer whenever since im gonna be going to class rn :)

Ultradark

@ShineOnYouCrazyDiamond that's good

amanuel2

For this MATLAB Question Questions 5C) and 5D), we are suppose to comment the answers since in (*), but I really dont seem to understand how to get the answer

So far my code:

v1 = [2; 1; 3; 1];
v2 = [1; 5; 8; -3];
v3 = [-1; 1; 1; 2];
v4 = [0; 3; 5; 5];
v5 = [3; 3; 7; 4];
A = [v1 v2 v3 v4 v5];

rref(A)

BASIS = A(1:4,1:3)
%c
%Polynomial : No Idea HELP
%d
%No idea if W = P3 HELP

Semiclassical

22:39

@amanuel2 what do you get for rref(A)?

amanuel2

23:07

1 0 0 1 2
0 1 0 0 0
0 0 1 2 1
0 0 0 0 0

nbro

Please, read this.

923

Q: Stack Overflow is doing me ongoing harm; it's time to fix it!

Over the last month, Stack Overflow has violated its own policies and precedents to cause egregious and unnecessary harm to me -- to my reputation (personal and professional), to my health, and to my safety. This harm is significant and ongoing. It is past time for the company to correct its e...

discussion moderators

And this.

28

Q: Reinstate Monica

Monica Cellio has called for Stack Overflow to address what they did to her: Stack Overflow is doing me ongoing harm; it's time to fix it! A number of users have changed their username to "Reinstate Monica" or some variation thereof and changed their avatar as follows: At CV, it appears that ...

discussion

schn

23:49

What is the intuition behind the fact that an isometry is represented by an orthogonal matrix? The proof here is clear (math.stackexchange.com/q/169923).

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$Mathematics$ Mathematics