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20:00
because if i've calculated them, I can show whether it holds that $f_{X,Y}=f_Xf_Y$, and if so, then I can simply use the convolution formula?
Does any one know how to write an infinite geometric series in sigma notation in wolfram alpha ? i don't know how the syntax goes to write it
No, it holds true in general
I said "useful", not "true" :P
wait
oh wait, I think I get it
no, actually not
you don't need independence to get marginal densities
I know, but how am I going to use the marginal densities in case of dependence?
20:02
I don't understand why you ask this
say $X$ and $Y$ are dependent, what use is there to calculate the marginal density functions?
there is no use, right? only in the case of independence
because then we can factorize our joint density function into the marginal functions
I'm not really following... Regardless of independence, computing the marginal densities can tell you the distribution of X and Y, which is certainly useful
$C_\beta(\alpha)_0=\{\gamma,\Omega_{\beta+1}|\gamma\le\Omega_\beta\},\Omega_0=\omega$
$C_\beta(\alpha)_{n+1}=C_\beta(\alpha)_n\cup\{\gamma+\delta,\gamma\delta,\gamma^\delta,\psi_\Gamma(\eta)|\gamma,\delta,\Gamma,\eta\in C_\beta(\alpha)_n,\eta<\alpha\}$
$C_\beta(\alpha)=\bigcup\limits_{n<\omega}C_\beta(\alpha)_n$
$\psi_\beta(\alpha)=\min\{\gamma|\gamma\notin C_\beta(\alpha)\}$
Anybody have any ideas on what $\psi_0(\Omega_2)$ would be?
that's a lot of greek letters
Yup
Oh yeah, $\Omega_\alpha=\omega_\alpha$ for most people.
20:11
@SimplyBeautifulArt you understand my problem ?
@Vrouvrou Which problem? And probably not, I'm not a real analysis type of guy.
Lol, any ideas on my problem...?
@LeGrandDODOM ohh, I see where the confusion comes from :(
I didn't write down the question correctly
the question stated: find the density function of $X+Y$ (and not "$X$ and $Y$")
@SimplyBeautifulArt what does $C_\beta(\alpha)_0=\{\gamma,\Omega_{\beta+1}|\gamma\le\Omega_\beta\},\Omega_0= \omega$ do, does it increment $\beta$ by one?
@Secret It says to consider the set of all ordinals less than or equal to $\Omega_\beta$ and also add in some $\Omega_{\beta+1}$.
20:21
Then you consider repeatedly applying addition, multiplication, exponentiation, and lots of $\psi$ functions on those sets
@ShaVuklia one possible trick is to get the density of $(X+Y,X-Y)$ and then compute the marginal density to get that of $X+Y$
Since $(x,y)\mapsto (x+y,x-y)$ should be a $C^1$ diffeomorphism
@SimplyBeautifulArt ok please just why any equation $x^3-y=0$ has a real solution ?
thank you
@LeGrandDODOM Is it also correct to compute the following?:
$$
\frac{d}{dz}\int_0^z\int_0^{z-x}\frac{1}{2}(x+y)e^{-x-y}dydx
$$
@Vrouvrou I'd usually say the solution is $x=\sqrt[3]{y}$ and say that's real. Not sure if I can deliver the argument you want.
I'm not really familiar with your proposal, and my test is tomorrow, so I'm guessing it's better to stick to what I know:d
am I supposed to do partial integration with my approach?
20:30
$C_0(\alpha)_0=\{0,1,2,\dots,\omega,\omega+1\}$

$C_0(\alpha)_1=C_0(\alpha)_0 \cup \{0\dots,\omega+1+\omega+1,(\omega+1)^2,(\omega+1)^{\omega+1},\psi_{\omega+1}(\eta),\eta<\alpha\}$

$C_0(\alpha)=\textrm{sup}_{n}(C_0(\alpha)_1\cup C_0(\alpha)_2 \cup C_0(\alpha)_3 \cup \cdots \cup C_0(\alpha)_n)$

$\psi_0(\alpha)=C_0(\alpha) \cup \{\textrm{sup}(C_0(\alpha))+1\}$

Therefore...
No...
$C_0(\alpha)\ne\sup_n(\dots)$
And $C_0(\alpha)_0=\{0,1,2,\dots,\omega,\omega_1\}$
@ShaVuklia yes, this is sound. As for the computation, I would just bruteforce it
Let's take an example:
$C_0(0)_1 = \{0,1 ,2,\dots ,\omega,\omega +1,\omega+ 2,\dots,\omega2, \omega2+1, \dots, \omega^2,\dots, \omega^\omega, \dots , \beta:\beta\ge\omega_1, \beta = \text{ some combination of all the previous stuff...}\}$
You keep doing this, and then $C_0(0)$ is the set of all ordinals less than $\varepsilon_0$ with some uncountable ordinals thrown into the mix. Thus, $\psi_0(0)=\varepsilon_0$.
@ShaVuklia the double integral yields $1-\frac{1}{2} e^{-z} (z (z+2)+2)$ and the derivative is easy to get
what is $C_0(0)_0$?
20:36
did you do that by partial integration? @LeGrandDODOM
$C_0(0)_0=\{0,1,2,3,\dots,\omega,\omega_1\}$
@ShaVuklia I computed the inner integral and then the outer one
So what is the $\alpha$ controlling in $C_{\beta}(\alpha)_0$, I don't see any alphas in the definition of it?
$\eta<\alpha$
@ShaVuklia if they're independent the joint distribution is just the product of the marginal ones
20:39
But $\eta\in C_\beta(\alpha)_n$ as well, so you can't just take arbitrary $\eta<\alpha$.
yea but they're not i think
@ShaVuklia If $X=(X_1,\cdots ,X_n)$ is a vector of random variables and $A\in GL_n(\Bbb R)$ do you know how to compute the density of $Y=AX$?
@AlessandroCodenotti i don't think so..
i'm just confused
i don't understand how they can put $\frac{\partial}{\partial z}$ within the integrals?
because the outer part also depends on $z$
$C_0(0)_0=\{0,1,2,3,\dots,\omega,\omega_1\}$

$C_0(0)_1=\{0,1,2,3,\dots,\omega,\omega_1\} \cup \{\dots,\omega_1+1,\omega+1,\dots,2\omega_1,\dots,\omega_1^{2},\dots,\omega_1^{\omega_1},\dots,\psi_{\omega_1}(0)\}$

?
@Secret You also need to include $\omega+1,\omega+2,\dots,\omega2,\omega2+1,\dots$
All the way up to $\omega^\omega$.
So it looks good so far
20:47
@LeGrandDODOM to compute the inner integral, we need to do partial integration right? that seems like way too much work
Ack typo:
$C_0(0)_1=\{0,1,2,3,\dots,\omega,\omega_1\} \cup \{\omega+1,\dots,\omega_1+1,\dots,2\omega_1,\dots,\omega_1^{2},\dots,\omega_1^{ \omega_1},\dots,\psi_{\omega_1}(0)\}$
It is also taken upon trivial assumption that $\psi_\beta(\alpha)\ge\psi_\beta(0)\ge\omega_\beta$
Yup that's the veblen hierarchy
Uh, no
It goes way beyond the Veblen hierarchy
No I mean that inequality you just wrote, not $C$
20:50
$\psi_0(\omega_1^{\omega_1^{\omega_1}})$ is already the Large Veblen ordinal.
Oh, okay
Focusing on the largest element of each hierachy helps to understand the growth rate of these functions IMO, but I am not really an expert in growth, so...
hence all those dot dot dot
Because in the end, how far it can go is determined by the largest element
@ShaVuklia yes, you have to integrate some things by parts, but it's really not that heavy and there are some nice simplifications at the end.
oh okay, I will try it then
So...
$C_0(1)_1=\{0,1,2,3,\dots,\omega,\omega_1\} \cup \{\omega+1,\dots,\omega_1+1,\dots,2\omega_1,\dots,\omega_1^{2},\dots,\omega_1^{ \omega_1},\dots,\psi_{\omega_1}(0),\dots,\psi_{\omega_1}(1)\}$

hence

$C_0(\alpha)_1=\{0,1,\dots,\omega,\dots,\omega+1,\dots,\omega_1+1,\dots,2\omega_1,\dots,\omega_1^{2},\dots,\omega_1^{ \omega_1},\dots,\psi_{\omega_1}(0),\dots,\psi_{\omega_1}(\eta <\alpha)\}$
20:55
@ShaVuklia hm, nevermind then. You might have seen this problem before, but it's an interesting one, of $X$ is uniform in $[-1,1]$ and $Y$ is uniform in $\{-1,1\}$ what's the distribution of $X+Y$?
@Secret Notice that you must also include $\psi_0(0)$ if $\alpha>0$
Ah right...
If I may,
$\psi(\alpha)=\varepsilon_\alpha\forall\alpha\le\zeta_0$
@AlessandroCodenotti i'll come back to your question in a sec, I have to finish this integral first:(
$C_0(\alpha)_1=\{0,1,\dots,\omega,\dots,\omega+1,\dots,\omega_1+1,\dots,2 \omega_1,\dots,\omega_1^{2},\dots,\omega_1^{ \omega_1},\dots,\psi_{0}(0),\dots,\psi_{\omega_1}(0),\dots,\psi_{\omega_1}(\eta <\alpha)\}$
?
20:58
@Secret But beware that you can only take $\psi_0(\eta\in C_0(\alpha)_0,\eta<\alpha)$. This will probably trip you up later.
Sure, there's no hurry
@Secret To speed things up, I recommend writing $[\omega_1]$ to indicate all ordinals beyond that point are larger than $\omega_1$, since they are unimportant for now.
Ok let me write them more neatly...:
So I have:
$$
\begin{aligned}
f_Z(z)&=\frac{\partial}{\partial z}\frac{1}{2}\int_0^z\int_0^{z-x}(x+y)e^{-x-y}dydx=\\
&=\frac{\partial}{\partial z}\frac{1}{2}\int_0^z-ze^{-z}-e^{-z}+e^{-x}dx\\
&=-\frac{1}{2}ze^{-z}.
\end{aligned}
$$
but that's now the correct answer
oh wait
I'm close, because the answer should be $\frac{1}{2}z^2e^{-z}$
@ShaVuklia Maybe you made a mistake somewhere :-(
21:11
oh yea, I see a mistake
$C_0(0)_0=\{0,1,2,3,\dots,\omega,\omega_1\}$

$C_0(0)_1=\{0,1,[\omega],[\omega_1],[2\omega_1],[\omega_1^2],[\omega_1^{ \omega_1}],\psi_{\omega_1}(0)\}$

$C_0(0)_2=\{0,1,[\omega],[\omega_1],[2\omega_1],[\omega_1^2],[\omega_1^{ \omega_1}],[\psi_{\omega_1}(0)],\psi_{2\omega_1}(0)\}$

Is this correct so far (I am too lazy to write all the ...)?
@Secret Well it should certainly include things like $\omega k,\omega^k,\omega^\omega$.
Well you can stop after $[\omega_1]$, since the rest is certainly larger than it
The important parts are the stuff that come before $\omega_1$
Question: how does ${\frak so}(8,\Bbb C)$ act on $(V_1\otimes V_2)\oplus(V_3\otimes V_4)$ here? A priori it only seems like a rep of ${\frak sl}(2,\Bbb C)^{\oplus 4}$ to me...
I would think that $\int_0^{z-x}(x+y)e^{-x-y}dy=-ze^{-z}+xe^{-x}-e^{-z}+e^{-x}$, but that would mean that
$$
\begin{aligned}
f_Z(z)&=\frac{\partial}{\partial z}\frac{1}{2}\int_0^z\int_0^{z-x}(x+y)e^{-x-y}dydx=\\
&=\frac{\partial}{\partial z}\frac{1}{2}\int_0^z-ze^{-z}+xe^{-x}-e^{-z}+e^{-x}dx\\
&=0.
\end{aligned}
$$
21:16
@SimplyBeautifulArt I think I don't fully get this $[x]$ notation. Mind show me an example?
@Secret $\{\omega_1,\omega_1+1,\omega_2,\omega_\omega,\dots\}=\{[\omega_1]\}$
Some set of ordinals greater where the elements are greater than or equal to $[\cdot]$
ok so you are interested in the behaviour of this function before $\omega_1$ since everything before that is countable?
@Secret Yes, until we hit $\psi_0(\omega_1)$ and $\psi(\omega_2)$.
Then we need to consider the $[\omega_1]$ stuff
@Secret So we should have:
$C_0(0)_1=\{0,\dots,\omega,\dots,\omega+k,\dots,\omega k,\dots,\omega^k,\dots,\omega^\omega,[\omega_1]\}$
Hi @Ted
Are you still sick?!
0
Q: Find joint density function for $f(x,y)=\frac{1}{2}(x+y)e^{-x-y}$

Sha Vuklia If $X$ and $Y$ have joint density function $$ f(x,y)=\begin{cases} \frac{1}{2}(x+y)e^{-x-y}&\text{if }x,y>0,\\ 0&\text{otherwise,} \end{cases} $$ find the density function of $X$+$Y$. I did the following: $$ \begin{aligned} f_Z(z)&=\frac{\partial}{\partial z}\frac{1}{2}\int_0^z\int_0^{z-...

21:26
Yup, Zach. Thrilling news, I know.
Blegh.
And I have standardized testing for the rest of the week
$C_0(0)_0=\{0,1,2,3,\dots,\omega,\omega_1\}$

$C_0(0)_1=\{0,1,...,\omega,...,\omega+k,...,\omega k,...,\omega^k,...,\omega^{\omega},...[\omega_1]\}$

$C_0(0)_2=\{0,1,...,\omega,...,\omega+k,...,\omega k,...,\omega^k,\omega^{\omega},...\omega^{\omega}+k,...,\omega^{\omega}k,...,\omega^{\omega^{\omega}},...,[\omega_1]\}$ ?
Well, that's always lots of fun. Just think — eventually, you can get back to fun math :D
@Secret Almost, but it should be $\omega^\omega k$ and $\omega^{\omega^\omega}$
We can't go beyond $\omega$ raised to itself $n+1$ times for any $C_0(0)_n$
@Secret Looks good
21:32
What is $Z$ in your question on MSE? @sha
$C_0(0)_3=\{0,1,...,\omega,...,\omega+k,...,\omega k,...,\omega^k,\omega^{\omega},...,\omega^{\omega}+k,...,\omega^{\omega}k,..., \omega^{\omega^{\omega}},...,\omega \uparrow ^2 3+k,...,(\omega \uparrow ^2 3) k,..., (\omega \uparrow ^2 3)^k,...,\omega \uparrow ^2 4,...,[\omega_1]\}$
So skipping ahead...
@Secret so messy lol
Yeah because blah + k, blah k and blah $^k$ keeps repeating for each increment in $n$
$C_0(0)_n=\{0,1,...,\omega,...,\omega+k,...,\omega k,...,\omega^k,\omega^{\omega},...,\omega^{\omega}+k,...,\omega^{\omega}k,..., \omega^{\omega^{\omega}},...,\omega \uparrow ^2 3+k,...,(\omega \uparrow ^2 3) k,..., (\omega \uparrow ^2 3)^k,.......,\omega \uparrow ^2 (n-1),...,[\omega_1]\}$
@Secret So what's the smallest ordinal that will never be in $C_0(0)_n$? It should be $\varepsilon_0=\sup\{1,\omega, \omega^\omega,\omega^{\omega^\omega}, \dots\}$
indeed yes (meanwhile Knutt arrow still survives, $\uparrow ^3$, but not for long once we start ramping up the other index of the $C$ function)
21:42
hello i need some help, i have (27x - (1/81)*y)^9 and i am trying to find the coefficient for x^6y^9 but i can't seem to find the answer. I set up my term to be:

9Ck * (-1)^k * 27^(k-9) * 81^-k * x^(k-9) * y^k

I can get x^6 if k is 15 but then that gives me y^15 so thats obviously not correct..what am i doing wrong here?
@Secret Likewise, the difference in $C_0(1)_n$ is that we may summon $\psi_0(0)$, so then $\psi_0(1)=\varepsilon_1$
@Secret This keeps going with $\psi_0(\alpha)=\varepsilon_\alpha\forall\alpha\le\zeta_0$
And so $\psi_0(\zeta_0)=\zeta_0$
But then $\psi_0(\zeta_0+1)=\zeta_0$
The reason is we can't summon $\psi_0(\zeta_0)$ into our $C$ until we reach $\zeta_0\in C$, which will never happen...
And so $\psi_0(\alpha)=\zeta_0\forall\zeta_0\le\alpha\le\omega_1$
Would you agree on that @Secret ?
I am going to skip any instance of blah + k blah k and blah ^k from this point on, it is tiring to write them again and again
Start ramping up the $\alpha$ then
What do you mean?
21:47
fair enough i'll have to ask else where
computing $C_0(1)_n$, $C_0(2)_n$ ... $C_0(\alpha)_n$
@Secret Uh... doing that right now?
Well I am computing them now, please wait
@Secret You sure you don't want me to just give you the answers?
I want to see the function grow to better understand it, I 'll check ans with you
21:50
Okay
@Secret But you haven't even seen the function truly grow yet
@Sha is it possible that $X+Y$ has density $f_Z(z)=ze^{-z}$? (I calculated that without pen and paper and a lot of wolfram alpha so it's likely to be wrong)
@Secret When you finish, check in on why $\psi_0(\Omega)=\zeta_0$.
Hello, everyone. Is it true that a union of overlapping path connected sets is path connected? If so, I have a problem that relates to it.
$C_0(1)_n=\{0,......,\omega,......,\omega \uparrow ^2 (n-1),...\epsilon_0=\psi_0(0),...,\epsilon_k=\psi_0(k),...,\psi_0(\omega \uparrow ^2 (n-1))\}$ ?
o wait, I jumped too far...
@Secret How did you take $\psi_0(k)$? I only allow $k<1$ here
21:59
ah sorry...
$C_\beta(\alpha)_0=\{\gamma,\omega_{\beta+1}|\gamma\le\omega_\beta\}$
$C_\beta(\alpha)_{n+1}=C_\beta(\alpha)_n\cup\{\gamma+\delta,\gamma\delta,\gamma^\delta,\omega_\gamma,\psi_\Gamma(\eta)|\gamma,\delta,\Gamma,\eta\in C_\beta(\alpha)_n,\eta<\alpha\}$
$C_\beta(\alpha)=\bigcup\limits_{n<\omega}C_\beta(\alpha)_n$
$\psi_\beta(\alpha)=\min\{\gamma|\gamma\notin C_\beta(\alpha)\}$
@Secret Don't mind any changes I may have added, any changes do not affect your current workings
And brought this here so you could easily see it
@AlessandroCodenotti yea almost. i've kind of given up tho. got too much to do (test in 18 hours) thanks for your help anyhow
$C_0(1)_n=\{0,......,\omega,......,\omega \uparrow ^2 (n-1),...\epsilon_0=\psi_0(0),...,\psi_{\omega \uparrow ^2 (n-1)}(0)\}$
I have a test on probability in 2 days as well I'm supposed to know how to do that...
@Secret Trivially by definition, $\psi_\alpha(0)\ge\omega_\alpha$
So kinda toss them all away with $[\omega_1]$
22:05
Nobody knows the answer to my question?
@user193319 lol, sorry, it got lost xD
Haha That's okay. Here it is, again: Hello, everyone. Is it true that a union of overlapping path connected sets is path connected? If so, I have a problem that relates to it.
$C_0(1)_n=\{0,......,\omega,......,\omega \uparrow ^2 (n-1),...,\epsilon_0=\psi_0(0),\epsilon +1,...,[\omega_1]\}$
@Secret I shall be back in a bit
@Secret Don't forget to apply operation onto $\varepsilon_0$
typo"
$C_0(1)_n=\{0,......,\omega,......,\omega \uparrow ^2 (n-1),...,\epsilon_0=\psi_0(0),\epsilon_0 +1,...,[\omega_1]\}$
22:09
@Secret So the smallest ordinal you can't reach will be $\varepsilon_1$, an infinite tower of $\varepsilon_0$'s
But $\epsilon_1=\psi_0(1)$ thus $\epsilon_1 \not\in C_0(1)_n$?
@Secret Yes...
You can't construct $\varepsilon_1$ using arbitrary finite iterations of addition, multiplication, and exponentiation on $\varepsilon_0$
yeah, we need the veblen to get beyond that point
22:12
Well no
You can construct every ordinal less than $\varepsilon_1$
so $\psi_0(1)=\varepsilon_1$ without any Veblen function
Likewise, when we go back, we can take $\psi_0(1)\in C_0(2)_n$
$C_0(2)_n= \{ 0,......,\omega,......,\omega \uparrow ^2 (n-1),...,\epsilon_0=\psi_0(0),...,\epsilon_1=\psi_0(1),\epsilon_1+1,[\omega_1] \}$ ?
@Secret pretty much
(guess I'm not leaving for a bit)
$C_0(\alpha)_n= \{ 0,......,\omega,......,\omega \uparrow ^2 (n-1),...,\epsilon_0=\psi_0(0),...,\epsilon_{\eta}=\psi_0(\eta < \alpha),\epsilon_{\eta}+1,[\omega_1] \}$
22:17
@MikeMiller Do you know if Rolf Berndt's introduction to symplectic geometry is good? It doesn't appear to have J-holomorphic curves, but maybe that isn't really necessary to have in an introduction. Anything you would recommend?
@Secret $\eta\in C_0(\alpha)_{n-1}$
^^^ Crucial condition
I shall be back in about 20-ish minutes
For real this time
I think I will go to sleep, cuase it's 8:18 am now
@AlessandroCodenotti I GOT IT NOW!!!!!
i got it, i got it, i got it
finally. what a torture.
and hi @KasmirKhaan @Ted
Hi ...
@ShaVuklia Hi :)
22:25
@TedShifrin I was wondering do you recommend a rigorous linear algebra text for me ? There is some gaps that I am missing.
@TedShifrin advanced linear algebra text.
What gaps?
Jordan form
Hi everyone. Does anyone understand what the negative degree hermite polynomial mentioned in this answer is? Answer: math.stackexchange.com/questions/2237761/…
maybe a second course into linear algebra I guess
or third
Use module theory — see Artin or Dummit and Foote.
Or Jacobson.
22:27
okay awesome thanks
@Secret Okay good night
could someone explain to me why the last equation is equivalent to the other two above it?
I've seen it on several places, but nobody explains why this is equivalent
I see that the second equation is simply the definition of conditional probability
oh I think I see it
it's sort of "overlapping". That's like asking; what is the chance that $y>2$ and $y>1$; you might as well only ask what the chance is that $y>2$
22:46
Hello everyone!
Hi Dami
Hii @Daminark
How's it going?
good enough:p and you?
It's ok. Just schematic-ing a computer
22:47
Pretty well, I have officially ceased believing the Riemann integral
@Meow I know that anything can be a verb if you verb it, but what exactly does this entail?
@Daminark congrats, I guess :P
You should see something called the Lebesgue integral, we did it in class today and it is so much better
@Sha You may do this in analysis pretty soon
ahh yea, I keep hearing about it (like the Lebesque measure and stuff). I hope we get it soon :P
It's not too hard to construct
So with measure, you basically try to define a system of sets that preserves complements and countable unions, and should contain the empty set
This is called a $\sigma$-algebra
Then a measure is a function mapping a $\sigma$-algebra on a set to $[0,\infty)$
@Daminark I bought a Zilog Z80 microprocessor
22:52
yea, I've had that in probability
Now, there's this process called Caratheodory extension, so if you have some reasonably well behaved system of sets, along with a nice function mapping them to $[0,\infty)$, you can extend it to a measure
now I'm going to hook it up to some programmed EEPROM and RAM to make a computer. But first I have to make a schematic of it.
ahh alright, I guess :P
Lebesgue measure is when your sets are finite unions of intervals (or boxes in $\mathbb{R}^n$), and the measure is exactly what you expect, the length
We did not verify this in class because it's boring and tedious, but this satisfies the well-behavedness conditions, so you can extend it to get the measure
@Meow OK, schematic here being just design or is it a specific technical term?
A... schematic.
22:55
aha :P (you've kind of lost me, but that doesn't matter)
But that's Lebesgue measure, the Lebesgue integral can be defined on any measure space
@Meow So just in the everyday sense, then. Cool!
@Sha The idea behind the Lebesgue integral is that you take a simple function as being one which takes on finitely many values
Equivalently, this is saying that $f = \sum c_i\Bbb 1_{X_i}$ where the $X_i$ are measurable sets
Then the integral of $f$ is just $\sum c_i\mu(X_i)$
For general non-negative functions, you take the supremum of integrals of simple functions approximating it from below
And for general (integrable) functions, you separate positive and negative and work from there
Does anyone know the answer to my question (about the Hermite polynomials)?
@Daminark aha :P well I guess it sounds cool XD
just out of curiosity; do you watch Pyrocynical? @Daminark
I have not seen that
I want to know if I can make certain references of not :P
ah too bad
23:00
(I haven't seen most media lmao, I intend to get into Rick and Morty at some point in life)
Darn
but you said you were into memes:'(
Well, memes is different, I meant like, TV stuff
Pyro is a youtuber
Oh, somehow I assumed that was a TV show
haha no:P YouTube $\gg$ TV :)
23:01
This is truly the case
I'll check it out!
yeaa, cool!
:)
Well, what have you been up to? It seems like now you're doing probability
yea, that's true. tomorrow I have a test on probability, and the day after on analysis (series, convergence, taylor)
notices series What fun! /s
Lol jk it's important but MERP
In analysis this year our professor skipped all that stuff
Which was nice because we would've gotten 20 problems from Rudin chapter 3 which were just painful
Instead he's just like "Yeah so you know sequences, series, and complex numbers already, so let's just roll from there"
hahaha :P I like series now, because I finally understand it :P
I can't wait to show convergence of series tomorrow :P
it will be one of the last things I understand, because from then on I will have to do a lot of new stuff, because I'm lagging behind
I'm off to bed now, btw! it's 01.12 here :P see ya later!
23:12
Oh lol, alright, see you!
23:27
@Secret Don't click the following link until you want to see what happens around $\zeta_0$ and $\omega_1$ (which is $\omega_1=\Omega$ in the link)
http://chat.stackexchange.com/transcript/message/36744016#36744016
Zach in the dark would be Dazachrk
Nice @Zach
Hello
Hey @Hardeep!
Also hey @Adeek!
23:41
hi @Daminark there is one part of a proof in hatcher which is bothering me
lol I was shocked when I heard the sound, I did not know that tagging works like this in chat
Hahaha, rip
@Hardeep @Hardeep @Hardeep
Hello!
You are late :P I have already put my laptop on mute
I didn't think it'd triple ping if you triple tag, would it?
23:43
@Hardeep :-(
@Daminark Probably not lol
@Adeek Which is it?
Demonark: Yes, I believe it triple-pings.
A quick question, are lower and upper riemann sums same as lower and upper darboux sums right?
Yes, @Hardeep.
23:45
so the part I don't understand why are the fibres of the form g_1(U)
Oh no...
Well, @Hardeep, not quite. A Riemann sum must be formed by using the value of the function at some point on the interval. For functions with discontinuity, you may not have the sup or inf.
And I'll check it out @Adeek
@TedShifrin If we select partitions in such a way that function is continuous in those partitions then they are same?
Low quality, but the question is interesting:
-4
Q: trying to study the convergence of a serie $\sum_{n=0}^{\infty}\frac{\sin(\cos(n))}{n}$

Hptunjy PrjkeizgI was doing some exercises and this one just stunned me had to study the convergence of this serie: $$\sum_{n=0}^{\infty}\frac{\sin(\cos(n))}{n}$$ i tryed alot of diffrent things, but i got no where can anyone please help me with an idea or a clue just to start ? thanks in advance !

23:50
@Daminark okay
@Hardeep: Yes, but if you have a function that is not everywhere continuous, this won't be possible.
So wait are you asking why $g_2g_1^{-1}$ takes $g_1(U)$ to $g_2(U)$?
@TedShifrin thanks
(I'm just trying to identify here what part of the proof you're not happy about @Adeek)
@Daminark I mean for something to be a normal cover
$p : \bar{X} \rightarrow X$ is normal cover iff for each $\bar{x}_1,\bar{x}_2 \in p^{-1}(\{x\})$ we have a deck transformation taking $\bar{x}_1$ to $\bar{x}_2$
but what I don't understand why the fibres of Y/G are of the form $g_i(U)$?
23:58
I don't know if it's necessarily saying that, given how it just mentioned that $g_i(U)$ mapped to an open set in $Y/G$
Maybe it's that a fibre would need to be contained within $g_i(U)$?

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