7:45 AM
Looking R as a subset of R^2, we see that it has no interior but it is connected. Same with the circle in R^2, connected with no interior. these two exist in the the "intuitively speaking", higher dimension space? like an interval in R would be open relative to R, but not R^2, similar considerations for the circle aswell; are there any other examples of connected sets with non connected interiors?
@SineoftheTime Pretty good, thanks, how's the study going?
7:59 AM
@nickbros123 what do you mean by non connected interior?
interior of the set is not connected- as in- it is union of two non empty separated set
Consider the region xy>=0
The examples you wrote (circle, interval in R) have empty interior in R^2. The empty set is connected
oh shoot, I keep making these lapses
its worrysome
completely unacceptable
8:21 AM
@nickbros123 union of two closed disks touching at a point
right
really embarassing tbh the whole time i was thinking the empty set was not connected, complete lack of attention
For R such subsets don't exist. A connected set is convex i.e. interval and this has convex set i.e interval as interior
@nickbros123 people argue about that actually. But if empty set were connected then subsets of R wouldn't have this nice property for example
Under most definitions it should be connected
People also argue about it being metrizable
I wonder when they will start to argue if empty set is a set at all. Or maybe they already do
8:36 AM
0

From Linear Algebra I always knew that the standard inner product on the vector space $F^n$(, where $F=\Bbb C$ or $\Bbb R$ ) is given by $<a,b>$ and $a,b\in F^n$ such that $<a,b>=\sum_ia_i\bar{b_i}.$ This is often referred as the dot product in elementary courses. Now, say $F=\Bbb C$ and $n=1$ th...

Need some help with this :/
Confusion over X. some calculation or other. Where did I make mistake?
Same format
8:52 AM
it's not clear what is $a \cdot b$
For posts looking for feedback or verification of a proposed solution. "Is this proof correct?" is too broad or missing context. Instead, the question must identify precisely which step in the proof is in doubt, and why so.
Its not the job of people on this site to find your mistakes
@Jakobian and @SineoftheTime Yes, and that's why I edited it.
I am also sure I made no mistake
@SineoftheTime Fixed that
Do you know the definition of inner product?
@SineoftheTime yes
$\mathbb{C}$ as a inner product space over $\mathbb{R}$ and $\mathbb{C}$ is different
Over $\mathbb{R}$ we identify $\mathbb{C}$ with $\mathbb{R}^2$ by $x+yi\mapsto (x, y)$
This is not inconsistent, but I get why it might be confusing for you
9:09 AM
@Jakobian that's really confusing me badly. Is this a convention or something? Like when studying complex analysis, we must think of C as R^2 and when studying linear algebra we must think C to be a vector space over itself (,unless otherwise specified) ?
If $\Bbb C=\Bbb R^2$ then I get what you're saying. But the question is, is it?
9:25 AM
@ThomasFinley the answer is it doesn't matter
You're missing the point
@ThomasFinley if $a\cdot b=xp+yq$ is an inner product, what's the problem?
Consider a smooth (nice) function $\gamma(t)$. Can $\gamma(t)$ be a solution to two distinct initial value differential problems $(D_1, \gamma(t_0))$ and $(D_2, \gamma(t_0))$ where $D_i$ is a differential equation ?
$\mathbb{C}$ can be treated as an $\mathbb{R}$-vector space $\mathbb{R}^2$ using the natural identification defined above
With this identification comes also the dot product $(x+yi)\cdot (p+qi) = xp+yq$
Is this an R-inner product space? Or C-inner product space?
Depending on how you treat it there is a different dot product
@Jakobian this one if you treat it as R-inner product space
Or $z\cdot w = z\overline{w}$ for C-inner product space
Indeed, "dot product on C" is not precise as you see
But it should be clear once we agree on what vector space is C i.e. over C or over R
In practice this should cause no confusion. There will probably never be a situation where it matters to you
If someone says "dot product on C^n" it should be clear they are talking about C-vector spaces
Unless specified otherwise
10:17 AM
I think Thomas doesn't really read those
Probably got help from other people and left. Why post here then
You're that impatient?
@Jakobian Isn't $\sigma(X,Y)=\sigma(\sigma(X)\cup\sigma(Y))$?
@psie yes, and?
Generating set is in no way unique
@Jakobian ok, if I understood you correctly, you were saying the generating set is $\{A\cap B: A\in\sigma(X), B\in\sigma(Y)\}$.
Yes. One of many
ok, how do you see that they generate the same sigma algebra?
i.e. the collections $\{\sigma(X)\cup\sigma(Y)\}$ and $\{A\cap B: A\in\sigma(X), B\in\sigma(Y)\}$.
10:31 AM
You agreed that $\sigma(X)\cup\sigma(Y)$ generates $\sigma(X, Y)$. The family I'm proposing is larger, and clearly contained in $\sigma(X, Y)$
10:45 AM
@SineoftheTime Good enough , I have to go back to the analysis part a bit, because I've done other things :/
Is Faustino Oro the best chess player in the world???
He is only 10 years old, still a long way to go for him
Last time I saw that you wrote that you wanted to do 1 vs 1 on chess, are you there now?
Yes, but currently I am on road, I can play after 20 mins
Okay :⁠-⁠)
11:10 AM
@BinkyMcSquigglebottom Okay I can play now
1+0?
Okay
@SoumikMukherjee
What is your actual rating btw?
:⁠-⁠|
I was destroyed
@SoumikMukherjee 1300
Not really, you played well
Can we do more time?
If you still want to play
11:21 AM
Yeah
same format?
@BinkyMcSquigglebottom I did 1+2 btw, there was increment
okay, 3+2?
Okay
:⁠-⁠(
Gg
gg
@BinkyMcSquigglebottom Fide rating?
I haven't played any live tournaments
You?
11:30 AM
Same here, I mostly play online
I've played otb tournament only once at my uni

3 hours later…
2:27 PM
Hi everyone. How can I check that the analytic function $h(u) = \frac{iR + u + i \sqrt{R^2 − u^2}}{2}$, with $u = x_1 +ix_2$, defines a conformal map from the region outside a semicircle, $\mathbb{C} \setminus S_R = \left\{(x_1, x_2):|x_1|^2 + |x_2|^2 = R^2, x_2 \leq 0 \right\}$ to the region outside a disk $D$ of radius $\frac{R}{\sqrt 2}$ centred at the origin, $\mathbb{C} \setminus D$?
@BinkyMcSquigglebottom Soumik is strong
@SineoftheTime And Sine is stronger :)
Same level in my opinion
2:43 PM
:)
2:55 PM
I struggle with this concept. Let $\mathcal B_1\lor\mathcal B_2$ denote the smallest sigma algebra that contains $\mathcal B_1\cup\mathcal B_2$. Is a generating set for $\mathcal B_1\lor\mathcal B_2$ simply $$\{B_1\cap B_2:B_1\in\mathcal B_1,B_2\in\mathcal B_2\}?$$Why?
4 hours ago, by Jakobian
You agreed that $\sigma(X)\cup\sigma(Y)$ generates $\sigma(X, Y)$. The family I'm proposing is larger, and clearly contained in $\sigma(X, Y)$
Which part are you struggling to verify?
3:11 PM
ok :) I think I found a detailed explanation here. Seems pretty straightforward now that I see a proof for the first time. Anyway, thanks!
4:03 PM
I am looking for innocent looking cases of positive integers "surpringly" divisible by the square of a large prime. "Surprisingly" means that no construction is involved or hidden algebraic factors. An example is $87641^2\mid 5^{31}+24^{10}$ , but I look for more striking examples.
4:20 PM
@psie why not yourself?
@Jakobian it matters wrt this situation, you see. In the book, I am studying from, the dot product is defined by regarding $\Bbb C$ as a vector space over $\Bbb R.$ But the point is(, as you mention as well) we can do the dot product in different ways for eg: by considering $\Bbb C$ as a vector space over itself. It all depends on what context are we are talking.
@Jakobian and why do you think that? I like your explanations and on several occassions if I recall correctly, I expressed by admiration for the explanations you provide.
And I again, mention, I remain grateful to anyone who tries helping me out.
So that counts you as well. You help me often when I get confused. You try to use some of your time in helping me out, just for the reason that I have a better understanding and this really means a lot to me.
I've explained why I think that
Where?
6 hours ago, by Jakobian
Probably got help from other people and left. Why post here then
Because you left without any sort of response
4:38 PM
Oh, I understand. Actually, I was gone for a while. But as soon as I came back, I made sure to read all the notifications I have.
$388057^2\mid 20^{10}-3068$ is a bit better.

1 hour later…
5:56 PM
I hate to sound like a broken record, but reread your text, with a pencil in hand. If the authors define an inner product / dot product for you, then that is the inner product. Period. Unless the text gives you some reason to care, don't worry about whether $\mathbb{C}$ is isomorphic (in whatever sense you are imagining) to $\mathbb{R}^2$ or not. If the authors of the text you are reading don't care enough to make this point, then it likely doesn't matter for the purposes of their exposition. — Xander Henderson ♦ 5 mins ago
@XanderHenderson Thanks! I get it. That makes it clear.
How I wish that I was a student under your guidance! Not only you're a good teacher but also a kind-hearted person. Your students are so lucky to have you as their mentor.
It's always a great privilege to chat with you. Thanks, again!
Someone who can help me?
6:31 PM
6:47 PM
Hi everyone. How can I check that the analytic function $h(u) = \frac{iR + u + i \sqrt{R^2 − u^2}}{2}$, with $u = x_1 +ix_2$, defines a conformal map from the region outside a semicircle, $\mathbb{C} \setminus S_R = \left\{(x_1, x_2):|x_1|^2 + |x_2|^2 = R^2, x_2 \leq 0 \right\}$ to the region outside a disk $D$ of radius $\frac{R}{\sqrt 2}$ centred at the origin, $\mathbb{C} \setminus D$?
Can I really get a conformal map from the region outside a semicircle, $\mathbb{C} \setminus S_R = \left\{(x_1, x_2):|x_1|^2 + |x_2|^2 = R^2, x_2 \leq 0 \right\}$ to the region outside a disk $D$ of radius $\frac{R}{\sqrt 2}$ centred at the origin, $\mathbb{C} \setminus D$? I should end up with $... Looks like too much complex analysis for me. I haven't studied that much Ted could have helped Or @leslietownes 7:35 PM @SineoftheTime I don't see it... The falling factorial is defined to be the number of permutations of length$k$from an$n$element set, $$n^{(k)}=n(n-1)\cdots(n-k+1).$$ Is this simply$0$when$k>n$(just like$\binom{n}{k}$is zero when$n<k$)? I'm unsure when it comes to conventions... is$n \choose k$even defined if$k>n$? I believe it's convention to take it$0$. اهلا وسهلا 7:50 PM ok, I guess then$\binom{n}{k}$is zero when$k>n$because the falling factorial is$0$when$k>n. @SineoftheTime yes @psie yep ok, seems like a sensible convention then Buonasera Bonsoir 8:07 PM 晚安 Anyway quick sanity checK: \begin{align} x&=y\\ &=z \end{align} for some reason that's not rendering properly when trying ot write an answer on the main site and now it's workign, idfk 8:24 PM @Semiclassical try with dollars? no, that's not the issue for some reason it was ignoring the line break Is it not rendering in the preview? but when i refreshed the page it worked fine (and the issue was only on the main site, not here) 9:00 PM In the third example here what is meant byX_m$? @SoumikMukherjee looking at the page, I think it just means that$\{X_m:i\in I\}$is a family of objects indexed by$I$but that seems way too indefinite 1 hour later… 10:14 PM @Semiclassical flagged as "Spam, Exists only to promote a product or service" ;) @SineoftheTime i am shamed, shamed i tell you it is funny how many of my questions are based on other people's questions 10:43 PM about$10-15%$of my knowledge is thanks to other people's question 11:10 PM The solution to$y' = a(t)y$is$y = Ce^{\int a(t)dt}$, so if I start with a general solution and assume it's from a linear first order ODE with no forcing term, I can solve for$a(t)$to recover the ODE. I'm having trouble doing this with recurrences. The only solution I can find to$z[n + 1] = b[n]z[n]$is$z[n] = D\Pi_{i = 1}^{n - 1} b[i]$. Is there a way I can solve this for$b[n]\$?