Mathematics - 2019-11-12

I have given a rotating matrix. and I am asked to give "the representation of the new basis vectors in regard of the old ones. I only have the rotation matrix and no more information. How can I proceed?

Ultradark

using $x=\exp\bigg(\frac{1}{\log(\phi)}\bigg)$

Akiva Weinberger

1:52 AM

You've been hit by
You've been hit by
A C¹ criminal

Rithaniel

Better than a $C^\infty$ criminal

Q: Can there exist a ring such that multiplication of two positive elements is given by the minimum or maximum of the two elements?

Ted Shifrin

3:55 AM

@MadSpaceMemer If you have a nonsingular matrix $A$, where does multiplication by $A$ send the standard basis vectors?

heya @Eric

MadSpaceMemer

@ted

I got it nvm :d

Ted Shifrin

OK, ignore that.

Akiva Weinberger

@Rithaniel So does $a+a$ have to equal either $2$ or $a$?

Similarly, $(a+a)(a+a)$ would be either $a+a$, but FOIL says it's $a+a+a+a$. So $a+a=0$

Hm… if you do $\Bbb Z_2$ with $0<1$, then the ordering doesn't respect addition but you do get multiplication to work the way you want it to

In fact I think I'm seeing that $\Bbb Z_2$ is the only one that works

MadSpaceMemer

4:16 AM

one of my eigenvektors is the zero vector . what does that mean>

is that even ok to happen ?

Stan Shunpike

@TedShifrin Hey Ted!

Akiva Weinberger

4:52 AM

@MadSpaceMemer That shouldn't happen

To find an eigenvalue, you find where $\lambda I-A$ has determinant zero. This means it has a nontrivial kernel (it sends some nonzero vectors to zero). So solving $(\lambda I-A)v=0$ should give you a nonzero $v$ for such a $\lambda$

Also: the definition of an eigenvector is a nonzero vector $v$ such that there exists a scalar $\lambda$ such that $Av=\lambda v$. Why is "nonzero" included in the definition?

Because otherwise the zero vector $v=0$ would always be an eigenvector! And in fact it would would with any eigenvalue $\lambda$, since $A0=\lambda0=0$.

It doesn't tell us anything about $A$, and it has all eigenvalues, so we exclude it from our definition for convenience.

F.White

5:56 AM

hello my friends

If $M/L\cong A$ and $N/K\cong B$, is $M\oplus N/L\oplus K \cong A\oplus B$ for $R$-modules, $R$ commutative

I certainly have a map $A\oplus B$ to $\operatorname{coker}(L\oplus K\to M\oplus N)$

Really I want to know if I can direct sum two composition series, term wise, to see that the function that tells me the number of times a specific simple module appears in a composition series is additive with respect to the direct sum

More Anonymous

8:27 AM

@AlessandroCodenotti I;m not a pro with measure theory. But I think I have an interesting idea: math.stackexchange.com/questions/3419921/…

Always Confused

11:05 AM

0

Q: How to derive the formula for coefficient (slope) of a simple linear regression line?

There is a formula for calculating slope (Regression coefficient), b1, for the following regression line: y= b0 + b1 xi + ei (alternatively y'(predicted)=b0 + b1 * x); which is b1=(∑(xi-Ẋ) * (yi-Ῡ)) / (∑ ((xi- Ẋ) ^ 2)) ---- (formula-A) source: https://stattrek.com/statistics/measurement-s...

Alessandro Codenotti

12:57 PM

@s.harp I have some doubts about the relationship between compactifications of $X$ and C*-algebras in $C_b(X)$, are you familiar with this?

tigre

Can we say anything about the cycle type of an element $g\in S_n$, when $g=hr$ and we know the cycle types of $h,r\in S_n$?

As an application, say we identify $\Bbb Z_2\times \Bbb Z_2\cong \{(1),(12),(34),(12)(34)\}\subset S_4$, and we want to find out which elements are sent where when we take $S_4/(\Bbb Z_2\times \Bbb Z_2)\cong S_3$

I imagine this would be easy if I knew how the cycle types behaved under multiplication

Although, I imagine the answer is 'they don't behave well at all, since we have things like $(1234)(12)=(134)(2)$ and $(1234)(13)=(14)(23)$'

Actually, given that transpositions generate the group, such a thing is hopeless

Nvm, my confusion with $S_4/V_4\cong S_3$ stems from the fact that my choice of $V_4\subset S_4$ isn't normal.

$\{(1),(12)(34),(13)(24),(14)(23)\}$ is the correct subgroup

Akiva Weinberger

1:59 PM

$\mathit{ŋηɳηŋηɳη}$

Ultradark

2:16 PM

check out this graph I made

ÍgjøgnumMeg

2:31 PM

@Alessandro since you're in the room and I know you know a thing or two about measures; if I have euclidean vector spaces $V, W$, an isomorphism $\varphi : V \to W$ and a measure $\mu_V$ on $V$, can I "transport" the measure on $V$ to a measure on $W$? I assume the isomorphism "changes" the measure by a factor of the determinant of $\varphi$ (notice I'm handwaving like crazy here)

actually I think I'm talking out of my ass

tigre

3:00 PM

I could find a class function for a finite group $G$, such that it is orthogonal to all but one of the characters, and it is unit length, and still satisfies $\chi(id)>0$, and yet is still not the character of a representation right?

Ultradark

check out this graph I made (2)

tigre

Cool, what is it

Ultradark

the light green is the prime counting function, and the blue is something else

tigre

What is the blue?

N. Maneesh

in CSIR-TIFR-ISI-NBHM, 4 mins ago, by N. Maneesh

0

Q: Several variables, differentiablity, continuity and primitive function

Let $F_1 , F_2 : \mathbb{R^2} \rightarrow \mathbb{R}$ be functions defined by $F_1 (x_1,x_2 )=-x_2/(x_1^2+x_2^2)$ and $F_2 (x_1,x_2 )=x_1/(x_1^2+x_2^2)$ Then (i) $∂F_1/∂x_2=∂F_2/∂x_1.$ (ii) there exist a function $f: \mathbb{R^2}-{(0,0)}\rightarrow\mathbb{R}$ such that $∂f/∂x_1=F_1$ an...

derivatives

Ultradark

3:13 PM

It's calculated using $\pi(x)\pi(n-x)=k$ and plotting all solutions for each $n$

blue oscillates up and down while light green only increases

and light green seems to be the upper bound of blue

1

Q: Growth of the number of values of $k$ that solve this equation for each $n$ $f(x)=\pi(x)\pi(n-x)=k.$

For a successive natural numbers $n$ starting with $n=2$ how many values of $k\in\Bbb N^+$ solve the prime counting function equation? I made a sequence of the number of values of $k$ that solve the equation for each $n,$ from $n=2,$ to $n=16.$ $f(x)=\pi(x)\pi(n-x)=k.$ Here's what I tried: when...

elementary-number-theory functions prime-numbers visualization

Here's more context.

tigre

What's $\pi(x)$ sorry?

Ultradark

it's the prime counting function

tigre

You're comparing $\pi(x)$ vs $\pi(x)\pi(n-x)$?

Ultradark

it's a function that keeps count of the number of primes less than an integer x

tigre

Isn't $\pi(-1)=0$, so that it's the zero function for $x>n$?

Ultradark

3:20 PM

not quite, I'm comparing $\pi(x)$ to the number of $k$ that satisfy $\pi(x)\pi(n-x)=k$ for each $n$ where $n$ is positive natural numbers starting at $2$

tigre

You're plotting $f^{-1}(n)$ for $f(x)=\pi(x)\pi(n-x)$ for each $n$?

You're plotting the point $(x,y)= (x,|f^{-1}_n(x)|)$ for each $n$, where $f_n(x)=\pi(x)\pi(n-x)$? @Ultradark

schn

Can the matrix associated with change to polar coordinates be seen as a rotation about the origin?

Ultradark

@tigre I don't know where your getting an inverse, I'm taking the concave function $f(x)$ and seeing how many horizontal lines $y=k$ solve it

so when $n=50$ exactly $14$ horizontal lines solve the equation

tigre

I don't get how $n$ is factoring in

Ultradark

n starts at $n=2$

and counts up one at a time, 2,3,4,...

and at ever $n$ you record how many horizontal lines solve the equation $f(x)=k$

tigre

3:27 PM

You are computing the cardinality of the set
$$S_{n,k}=\{x\mid f_n(x)=k\},\qquad f_n(x)=\pi(x)\pi(n-x)$$
for each $k$ and $n$?

Ultradark

yeah, pretty much

tigre

My confusion is "The number of $k$ that satisfy $\pi(x)\pi(n-x)=k$" which is dependent on two things, $n$ and $x$

Ultradark

yeah but $n$ is fixed

tigre

And I don't even know what that could mean "The number of k" where k is a value, how can there be a number of k...

So you are counting the number of $x$?

Sounds like you want $|f_n^{-1}(k)|$ after all!

Ultradark

tigre

3:31 PM

I have to go now, please make this more precise :P

Akiva Weinberger

Let $G$ be a graph, and let $s_{G,n}$ be the number of spanning forests with $n$ components (so $s_{G,1}$ is the number of spanning trees for example and $s_{G,|G|}$ is just $1$ because it's a discrete graph)

Ultradark

so when $n=10$ $|k|=4$

Akiva Weinberger

(where $|G|$ is the number of vertices in $G$)

Conjecture: $|\sum_n s_{G,n}(-2)^n|$ is a power of $2$

I have a proof that works for planar graphs and I'm not sure if it works for nonplanar graphs

Alessandro Codenotti

@ÍgjøgnumMeg You can look at the pushforward measure $\phi_\ast\mu(U)=\mu(\phi^{-1}(U))$ and I think you do get a determinant factor

Akiva Weinberger

and unfortunately I don't actually have a quick way of finding $s_{G.n}$ where $G$ is something nonplanar like $K_5$ or $K_{3,3}$

but I'm 80% sure it should be true for those too

Alessandro Codenotti

3:33 PM

You should have some $\sigma$-algebras fixed such that $\phi$ is measurable though

ÍgjøgnumMeg

@Alessandro fair :P I think I was being silly; I have the measure "vol" on an $n$-dim. euclidean vector space and the claim was that it doesn't depend on the choice of basis

Alessandro Codenotti

How is it defined?

ÍgjøgnumMeg

It's handwaved, it just says "the Lebesgue measure on $\Bbb R^n$ gives us a measure $\operatorname{vol}$ on $V$"

well not hand waved

just not explicitly defined

anyway the "reason" stated was that orthogonal matrices have determinant $\pm 1$ so I just wondered if a different linear transformation would dump a factor of the determinant out front

lol

schn

Are there rotation matrices with determinant greater than 1? Can the matrix associated with change to polar coordinates be seen as a rotation about the origin despite the determinant not equaling 1?

JTP - Apologise to Monica

3:57 PM

can someone give me a clue how to mathjax this ?

Alessandro Codenotti

Are the numbers subscripts of the C's?

JTP - Apologise to Monica

yes, this is 'combination'

Alessandro Codenotti

$\frac{_{20}C_5 {_{40}C_1}}{_{60}C_7}$ this works but it's kinda ugly

JTP - Apologise to Monica

yes, the 5 and 4 are touching.... hmmm. Thx for trying. I have a question I hoped to post, but wanted to offer what I've tried. And can waste far more time just trying to show my own work with math jax. I know jpg images are frowned upon.

Alessandro Codenotti

You can add spacing with \, or \quad or similar commands depending on how much space you need

JTP - Apologise to Monica

4:08 PM

that's it. thanks! I'll save that text and work on problem.

s.harp

@AlessandroCodenotti Every compactification of $X$ corresponds to a unitisation of $C_0(X)$, and im pretty sure every unitisation of $C_0(X)$ is contained a sub-algebra of $C_b(X)$

$C_b(X)$ is the multiplier algebra of $C_0(X)$ (a special kind of unitsation) and $C_b(X)$ is also equal to $C(\beta X)$ as an example

Alessandro Codenotti

Right, $\beta X$ is homeomorphic to the spectrum of $C_b(X)$

s.harp

whereas the algebra generated by $C_0(X)\cup\{1\}$ in $C_b(X)$ (this is called adjoining a unit) is the same as the algebra of $C(X^*)$ wehre $X^*$ is the one-point compactification

Alessandro Codenotti

(There's some niceness assumptions on $X$ which I'm ignoring)

s.harp

local compact + hausdorff

Alessandro Codenotti

4:12 PM

Shouldn't that be $C_0(X)\oplus\Bbb C$?

s.harp

it can be

thats the procedure of adjoining a unit to a $C^*$ algebra, which for $C_0(X)$ is the same as the subalgebra of $C_b(X)$ genereated by $C_0(X)$ and the constant function $1$

Alessandro Codenotti

What exactly is a unitisation of $C_0(X)$?

@s.harp Aha, that makes sense

s.harp

If $A$ is a $C^*$-algebra then a $C^*$-algebra $\tilde A$ is a unitsation of $A$ if $\tilde A$ is unital and $A$ is big in $\tilde A$ (I think that should be given by $A$ being a maximal ideal, but im not entirely sure)

Alessandro Codenotti

Ok I wasn't sure about the relationship between $A$ and $\bar{A}$

I think there should some correspondence between subalgebras of $C_b(X)$ containing the constant functions and separating points (or maybe some other adjectives...) and compactifications

s.harp

that makes sense, if it seperates points I expect Stone-Weierstraß to tell me it contains all of $C_0(X)$ and if it contains the constant functions its unital

Alessandro Codenotti

4:19 PM

So given a unitisation $A$ of $C_0(X)$ the way to get a compactification of $X$ out of it so to look at its spectrum with the weak* topology I suppose?

Hi @Ted by the way

s.harp

yes, the thing that prevents the spectrum of $C_0(X)$ from being closed is that $0$ may be a limit point of characters. But the algebra is unital then $\omega(1)=1$ for all characters and $0$ cannot be weak* limit point, hence the characters are closed in weak* + compact since they are bounded

Alessandro Codenotti

This makes sense

I just don't see how $X$ embeds densely into this space now

s.harp

thats a question of extending characters, the extension of a charcter to the constant functions is clear, and I'm not sure how it extends to the other "extra functions"

Eric Scöerg

Hi :) on my question math.stackexchange.com/questions/3425649/… would it be ok if I opened yet another question on what I called "little note"?

s.harp

(cont) I think at that point it needs to be clear what exactly a unitisation is. ie what $A$ being big in $\tilde A$ means

here is what i found about the correct condition: The essentialness is captured by stipulating that every nonzero ideal in the unitization intersects 𝐴 nontrivially. This is equivalent to the condition that 𝑏𝐴={0} implies 𝑏=0

from that i can read off that extension of characters is unique if it exists

Alessandro Codenotti

4:35 PM

Hmm but wait, so I guess that $x\in X$ should correspond to the character of $C_0(X)$ which evaluates at $x$, right? And then we want to extend that to the unitisation?

(Do you know a good reference to read about those things? It's turning out to be surprisingly hard to find a book containing this constructions)

s.harp

yes, if everything is in $C_b(X)$ its clear that it works

I don't know any book unfortunately, I think I read parts of this in Murphy

Alessandro Codenotti

It's clear to me that gives an injection from $X$ into the spectrum if we're working with a subalgebra of $C_b(X)$, the fact that the image of this injection is dense is still not clear to me

Thanks I'll check it out

s.harp

there are some super advanced $C^*$algebra books and notes around but I never made much progress reading them (Garth Warner's notes is the one I made the most in, they are nice)

Alessandro Codenotti

Maybe my mistake has been to look at topology books dealing with compactifications instead of books on operator algebras

s.harp

@AlessandroCodenotti i cant come up with a reason for denseness atm, ill think about it (but right now i need to get to correcting exercise sheets)

Semiclassical

4:40 PM

@schn No and no.

Alessandro Codenotti

Sure, I didn't want to distract you!

That was very helpful, thanks

I might ask a question on MSE later to find a reference or understand properly how this construction works

s.harp

i like the topic of compactifications <-> unitisations a lot

(so no worries about distracting :P)

Brandon

Any one on this knows about SVM ?

(Support Vector Machines)

The Terrible Puddle

5:06 PM

Why is $x\mapsto (x,x)$ called the diagonal function?

Brandon

Hi Ted

Hi The Terrible

Leaky Nun

@TheTerriblePuddle think about $\Bbb R \to \Bbb R^2$

The Terrible Puddle

Just a function say $\delta :M \rightarrow M \times M$

woops, wrong chat

Brandon

@LeakyNun Why Mathematics equations does not appear properly ? Like what you have written is not clear (not formatted)

Leaky Nun

see the link to latex in the chatroom description

Semiclassical

5:11 PM

You have to load mathjax into your browser to make it render. See--what leaky said

Leaky Nun

:P

Brandon

I need to install it ? From which website ?

Semiclassical

See the link.

It's not an install: more like a bit of code which your browser will execute from the address bar

Brandon

Oh Thanks , its working

I used to think how these people understand such complicated expression

:(

$E$

Oh cool it's working

Semiclassical

yeah. note that you will have to refresh it every time you return here, hence why it's a good idea to bookmark the bit of code

Brandon

5:20 PM

(SO)They could not code this page properly to avoid all these bookmarking etc ?

Leaky Nun

I mean I can read mathjax source code

Semiclassical

Mathjax isn't enabled by default, no. Not sure why but that's always been the case.

9

A: Any chance of MathJax in chat?

From Jeff's related answer: This is implemented on http://math.stackexchange.com -- you can check it out there. It will never be on Stack Overflow, though, as it is an extremely heavy dependency. MathJax is a client-side solution, but it uses a relatively large amount of bandwidth/time to l...

Another reason to avoid having it by default is that you can do some goofy stuff in mathjax, like having $\huge{\text{huge text}}$

so you can imagine the shenanigans that can produce, and why not being able to disable would be an issue

schn

@Semiclassical If the matrix associated with the change of coordinates to polar doesn't represent a rotation about the origin, what does the author mean in the last sentence to the solution 1a (see attached picture)?

schn

6:08 PM

Consider the change of variables $s=2x^3+3y^2, t=x$. Why is the codomain $s\geq t^3$ for $y>0$? Shouldn't it simply be $s>2t^3$?

Ted Shifrin

6:28 PM

Yes, of course the $2$ should be there.

schn

The "solution" says $s\geq t^3$ ...

But why the greater than or equal sign? Is this also incorrect?

Alessandro Codenotti

Hi @Ted

My phone really didn't want to greet you, I had to give it a few attempts

Ted Shifrin

7:00 PM

Hi, demonic @Alessandro. It is wiser than you realize.

MadSpaceMemer

7:36 PM

hello peoplz. how does the following look like ? the polynom $1-X^2$ I know that the set of all polynoms $ \mathbb{R}[X]$ is basically all $k_0 . x^0 + k_1 . x^1...………$

I am not sure I understand what $1-X^2$ represents.

Leaky Nun

$k_0 = 1$, $k_1 = 0$, $k_2 = -1$, $k_3 = 0$, $k_4 = 0$, ...

MadSpaceMemer

So basially it is only ONE element which is namely $1-x^2$? I got confused by the big $X$

Alessandro Codenotti

Do you have time (and patience) to help me sort out some confusion surrounding the definition of orientation for a manifold? @Ted

Leaky Nun

8:17 PM

@AlessandroCodenotti it's just a section of the orientation bundle

@MadSpaceMemer yeah

@AlessandroCodenotti also, drinking game: take a shot every time a mathematician says "it's just"