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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
Can you have a function avoid points of the form irrational,irrational)?
(Not sure - will think about this)
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))
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
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
@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.
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
Q: How to derive the formula for coefficient (slope) of a simple linear regression line?

Always ConfusedThere 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
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
Q: Several variables, differentiablity, continuity and primitive function

Mabud Ali SarkarLet $F_1 , F_2 : \mathbb{R^2} \rightarrow \mathbb{R}$ 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
Q: Growth of the number of values of $k$ that solve this equation for each $n$ $f(x)=\pi(x)\pi(n-x)=k.$

UltradarkFor 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
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!
That was very helpful, thanks
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 ?
See the link.
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
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.
A: Any chance of MathJax in chat?

REINSTATE MONICA -Jeremy BanksFrom 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
@MadSpaceMemer yeah
@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

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