12:00 AM
because the algebraic values for $f$ likely form an empty set
Oh I have a question: Can a function $f$ no matter what the input is, only be transcedental?
so for any real $x$ $f$ always returns a transcendental number

No (assuming continuous, nonconstant). Nearly the same reason - its image contains an interval, every interval contains an algebraic number

okay that makes sense
there exist transcendental functions that only produce transcendental numbers for a transcendental input

Is $f(x)=x$ not one of them?

yeah I guess that would be an example

I think $e^x$ maps algebraic->transcendental (except for 0)
so $(x,e^x)$ avoids points of the form (algebraic,algebraic) (except for (0,1))

12:14 AM
yeah

Can you have a function avoid points of the form irrational,irrational)?

12:28 AM
@Ultradark Re: my question, no. You can't have countably many rationals subjectively cover uncountably many irrationals in an interval
(so the same answer happens for avoiding (transcendental,transcendental))
*surjectively
And any set with countable complement

Hello friends!

$f(x)=\exp(1/\log(x))+\exp(2/\log(x))-\exp(3/\log(x))=0$ is equivalent to golden ratios $f(x)=\phi+\phi^2-\phi^3=0$

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?

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

1 hour later…
1:52 AM
You've been hit by
You've been hit by
A C¹ criminal

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?

2 hours later…
3:55 AM
@MadSpaceMemer If you have a nonsingular matrix $A$, where does multiplication by $A$ send the standard basis vectors?
heya @Eric

@ted
I got it nvm :d

OK, ignore that.

@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

4:16 AM
one of my eigenvektors is the zero vector . what does that mean>
is that even ok to happen ?

@TedShifrin Hey Ted!

4:52 AM
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.

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

2 hours later…
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/…

3 hours later…
11:05 AM
0

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...

2 hours later…
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?

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

1:59 PM
$\mathit{ŋηɳηŋηɳη}$

2:16 PM
check out this graph I made

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

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?

check out this graph I made (2)

Cool, what is it

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

What is the blue?

in CSIR-TIFR-ISI-NBHM, 4 mins ago, by N. Maneesh
0

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...

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

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...

Here's more context.

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

it's the prime counting function

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

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

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

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 $2$

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

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

@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

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

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 $f(x)=k$

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 $n$?

yeah, pretty much

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

yeah but $n$ is fixed

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!

3:31 PM
I have to go now, please make this more precise :P

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)

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

(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

@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

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

3:33 PM
You should have some $\sigma$-algebras fixed such that $\phi$ is measurable though

@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

How is it defined?

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

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?

3:57 PM
can someone give me a clue how to mathjax this ?

Are the numbers subscripts of the C's?

yes, this is 'combination'

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

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.

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

4:08 PM
that's it. thanks! I'll save that text and work on problem.

@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

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

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

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

local compact + hausdorff

4:12 PM
Shouldn't that be $C_0(X)\oplus\Bbb C$?

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 $1$

What exactly is a unitisation of $C_0(X)$?
@s.harp Aha, that makes sense

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)

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

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

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

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

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

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"

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"?

(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

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)

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

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

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)

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

@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)

4:40 PM
@schn No and no.

Sure, I didn't want to distract you!
I might ask a question on MSE later to find a reference or understand properly how this construction works

i like the topic of compactifications <-> unitisations a lot
(so no worries about distracting :P)

Any one on this knows about SVM ?
(Support Vector Machines)

5:06 PM
Why is $x\mapsto (x,x)$ called the diagonal function?

Hi Ted
Hi The Terrible

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

Just a function say $\delta :M \rightarrow M \times M$
woops, wrong chat

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

see the link to latex in the chatroom description

5:11 PM
You have to load mathjax into your browser to make it render. See--what leaky said

I need to install it ? From which website ?

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

Oh Thanks , its working
I used to think how these people understand such complicated expression
:(
$E$
$E$
Oh cool it's working

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

5:20 PM
(SO)They could not code this page properly to avoid all these bookmarking etc ?

I mean I can read mathjax source code

Mathjax isn't enabled by default, no. Not sure why but that's always been the case.
9

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

@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)?

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 $s>2t^3$?

6:28 PM
Yes, of course the $2$ should be there.

The "solution" says $s\geq t^3$ ...
But why the greater than or equal sign? Is this also incorrect?

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

7:00 PM
Hi, demonic @Alessandro. It is wiser than you realize.

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.

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

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

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

8:17 PM
@AlessandroCodenotti it's just a section of the orientation bundle
@AlessandroCodenotti also, drinking game: take a shot every time a mathematician says "it's just"

3 hours later…
11:02 PM
@Leaky that is a DANGEROUS game you propose