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7:01 PM
What does $\coprod_{y\in Y} X$ mean in terms of sets
?
 
Disjoint union of $Y$-many copies of $X$
 
@AlessandroCodenotti how would an element inside that set look like?
 
an element of X indexed by an element of Y
$x_y$
 
Usually, one expresses $\coprod_{i\in I}X_i$ $=$ $\{$ $(x,i): x\in X_i$ and $i\in I$ $\}$, so $\coprod_{y\in Y}X=$ $\{$ $(x,y)$ $:$ $x\in X$, $y\in Y$ $\}$?
@Astyx
am I right?
 
i think so
 
7:07 PM
@Astyx so $X\times Y$ $=$ $\coprod_{y\in Y}X$?
 
Yup
In mathematics, the disjoint union (or discriminated union) of a family ( A i : i ∈ I ) {\displaystyle (A_{i}\colon i\in I)} of sets is a set A = ⨆ i ∈ I A i , {\displaystyle A=\bigsqcup _{i\in I}A_{i},} with an injective function of each...
 
@Astyx thanks!
 
@anakhro I got the book "Introduction to Manifolds" as you suggested. It seems way above my level however
 
@geocalc33 is that by lee?
 
@topologicalmagician It's by Loring Tu
 
7:28 PM
I'll just learn what I can from it and take it step by step
 
is there any good reason to use "$A\subset B$" rather than "$A\subseteq B$" for subsets? (taking less effort to write does not count as a good reason for the purpose of this question)
 
8:01 PM
@Thorgott no
 
@Astyx Salut mon ami :D
 
@SAJW guten Abend
 
@geocalc33 what about it seems "way above" your level?
 
@LeakyNun in Ostfriesland würde man sagen "moin" :D
 
@SAJW moin Abend?
 
8:11 PM
@LeakyNun nein, einfach nur "moin", ohne abend, zu jeder tageszeit.
 
@LeakyNun こんばんは
 
@SAJW huh, interessant
@LukasHeger べいこくでひるまです。
 
What is truly random?
 
quantum stuff apparently
 
8:33 PM
Hi @Lukas
 
Ciao @Alessandro
 
0
Q: Adjunction space quotient properties.

topologicalmagicianLet $X\cup_f Y$ be an adjunction space. Let $q:X\coprod Y \rightarrow X\cup_f Y$ be the associated quotient map. Where $\sim$ is generated by $a\sim f(a)$ for all $a$ Show that $q$ is injective. My attempt: Case 1: Let $(x_1,0),(x_2,0)$ ( I will write it as $x_1,x_2$) be in $X$. So $q(x_1)=q(...

Feedback anyone?
 
I don't think $q$ will be injective. Take any $a \in X$, then by definition $q(a)=q(f(a))$, but $a \neq f(a)$ in $X \coprod Y$
quotienting out by an equivalence relation never gives you an injective quotient map unless you quotient out by the trivial equivalence relation
 
Hi @SAJW
 
@Astyx i have read this thread: philosophy.stackexchange.com/questions/1012/…
 
8:47 PM
Very philosophical
 
i started with "what is truly random"
also on philo.se
is mathematics the philosophy of numbers?
 
Meh not sure
Maths doesn't have margin for opinion
 
I like to difer, 0 is neutral in everything but france, so math leaves room for opinion (even if that opinion is in different definitions)
 
That's just convention though, the maths behind is the same
 
Huy
any idea how I can figure out for which functions f(t) the coupled IVP

x'(t) = -y(t)^3 * f(t)
y'(t) = x(t)^3 * f(t)
x(0)=1
y(0)=0

has periodic solutions?
 
8:55 PM
What's IVP ?
 
Huy
for constants obviously but f(t) = x(t) or f(t) = t already doesn't work anymore
 
@Astyx initial value problem
 
Ok thank you
 
9:11 PM
the quadratic formula fails if a=0(because you divide by 0), but how to name this:
if (a!=0)
    use quadratic formula
else
    solve for b;
solution_to_polynomials_of_degree_less_than_3 seems a little bit long
oh, i meant solve for x!
 
Huy
f(t) = x(t)^2 + y(t)^2 works too
but f(t) = x(t) + y(t) doesn't
 
@LukasHeger you're completely right. I read the problem incorrectly. I was supposed to show $q|_X$ is injective, which is essentially the first case.
 
@anakhro I really like reading the passages and understand those but the notation is just overwhelming and I'm not used to it
and also I have to admit I'm being lazy and not really focusing on it, so I guess I should start there.
 
10:12 PM
bonsoir @Ted
 
10:28 PM
Is there a proper extension of analytic continuation that also satisfies uniqueness?
 
Doesn't analytic continuation satisfy uniqueness ?
 
Yes.
 
Oh, I can't read
 
Ha, I see what you mean. I meant "that also satisfies uniqueness (like analytic continuation does)".
Some kind of generalization of analytic continuation that retains uniqueness.
 
I what way would you want to extend this notion ?
 
10:33 PM
A possible example: Consider $\exp(-1/x^2)$ for $x > 0$. Is there a precise sense in which $f(x) = \begin{cases} 0 & x = 0 \\ \exp(-1/x^2) & \text{otherwise} \end{cases}$ is the "most natural" extension to all of $\mathbb{R}$?
 
Hmm I don't see why it would be
 
In the second question I linked to, Gammel refers to "a kind of analytic continuation which differs from the usual kind" which he attributes to Borel, but I haven't been able to determine what that is.
math.wustl.edu/files/math/imce/2001.pdf mentions "Generalized analytic continuation".
 
10:59 PM
@geocalc33 The (modern) language of manifolds will be overwhelming at the start. That's a given. It's the subject where the definitions are hard, and the theorems are easy.
You can escape a lot of the notation if you do something like Guillemin and Pollack's differential topology book, but then the notation catches up to you later.
 
11:38 PM
@user76284 No. When I'm not constructing bump functions, I'm equally fond of the function $f(x)=\begin{cases} e^{-1/x^2}, & x\ne 0 \\ 0, & x=0\end{cases}$.
 
@TedShifrin Well, the first one is more natural at least when seen on $\mathbb{C}$.
 
I am very skeptical. What do you mean, when seen on $\Bbb C$?
Oh wait. We wrote down the same thing.
My fault.
 
Ha, I didn't notice that either. I thought you meant 0 for $x < 0$.
 
$e^{-1/x^2}$ can't be extended to an analytical function can it ?
 
LOL, that's what I thought you meant. That's the one that appears most everywhere, though, because of the importance of bump functions.
 
11:41 PM
well, not on $\Bbb C$ is what I mean
 
No, @Astyx.
 
@Astyx You can go around the pole in $\mathbb{C}$.
Oops I mean the essential singularity.
 
All the derivatives at $0$ would have to be $0$, so a real analytic function would be dead $0$.
 
Yeah
 
Another example given by Shapiro in Generalized Analytic Continuation is this: Let $z_k$ be a sequence of points which is dense on the unit circumference and $c_k$ be complex numbers such that $c_k \neq 0$, $\sum_{n=0}^\infty |c_k| < \infty$. Then the function $f(z) = \sum_{n=0}^\infty \frac{c_k}{z-z_k}$ for $|z| < 1$ cannot be continued analytically across any point of the unit circle.
Because you get a dense set of poles on the unit circle.
"Nevertheless, it has nontangential limiting values almost everywhere on $|z|=1$, which coincide almost everywhere with the non-tangential limiting values of the analytic function defined by the same series for $|z| > 1$."
 
11:49 PM
If $G$ is a finite group and all its irreducible representations have a non degenerate G-invariant symmetric bilinear form, $\sum_{\rho\in\hat G}dim(\rho)$ is the number of elements of order 2 of G
 
I really, really want to know what lies on the other side of the prime zeta function.
I want to see the landscape hidden behind that natural boundary.
 
@TedShifrin I might be crazy, but say I have two orthogonal projections, $P,Q$ and $P(Q(x)) = x$. I want to show that $x\in \im P \cap \im Q$ but I am struggling to figure it out. >: ( any tips?
It's probably just some little thing I am missing, but I tried to go about looking at $\|P(Q(x))\|^2$ with the inner product, but to no avail
Like getting something like $\langle Q(P(Q(x))),x\rangle = \|x\|^2$ sort of thing.
Didn't seem helpful. I also tried to split it up into $x = v_1 + v_2 + k_1 + k_2$ where v_i are in the images of P and Q respectively and the k_i in their kernels (orthogonal complements).
 
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