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00:06
Can someone explain me why $f^*u=u\otimes 1$
00:27
@SineoftheTime $f^* u(\phi) = (u\otimes 1)(\Phi)$, different argument
I can't understand why
 
8 hours later…
08:17
There's a theorem that says if $E|X|<\infty\implies EX=g_X'(1)$, where $g_X$ is the probability generating function. Now, consider the r.v. $S_N$, which is the sum of a random number $N$ of terms of i.i.d. r.v.s. $X_1,X_2,\ldots$ (everything's nonnegative integer-valued). Then we know $g_{S_N}(t)=g_N(g_X(t))$. I am now reading a theorem that states;
> If $EN<\infty$ and $E|X|<\infty$, then $ES_N=EN\cdot EX$.
What I don't understand is why we require $EN<\infty$ and $E|X|<\infty$. For $ES_N$ to exist via generating functions, we require $E|S_N|<\infty$. Is this clear to someone?
08:43
Would it be correct to write $$E|S_N|=E(|X_1+\ldots +X_N|)\leq E(|X_1|+\ldots +|X_N|)=E (N|X_1|)=EN E|X_1|?$$
So we see that $E|S_N|<\infty$ if $EN<\infty$ and $E|X|<\infty$, as demanded by the theorem.
09:23
Grateful if anyone could confirm if my reasoning is correct? I am doubting $$ E(|X_1|+\ldots +|X_N|)=E (N|X_1|).$$
 
1 hour later…
10:27
Can someone please explain to me how they do encoding in 4.4. of this paper: eprint.iacr.org/2018/153.pdf. I am absolutely about to tear my head off.
The usual case is just the inverse of the canonical embedding after some "nice" rounding operations and projection to the image of the canonical embedding but in this case the polynomials in $X^{N/n}$ I cannot wrap my head around. What exactly do they do to encode a vector in $\mathbb{C}^{n/2}$ to a polynomial in $\mathbb{Z}[X^{N/n}]/(X^N+1)$?
 
3 hours later…
13:35
This is the part if anyone wants to help but don't want to check the paper.
 
1 hour later…
14:42
@Jakobian I've not needed or proven that fact before, but after considering it now, I think the proof along this line is straightforward.
15:08
@Thorgott yeah. Oddly enough this seems to remove the problem of struggling with the open sets in $X$ because I can extend a function on $X$ which is continuous by gluing lemma. Something that ends up being confusing with open sets alone but becomes somewhat more straightforward with functions
yeah, it's interesting
It's not clear to me if the inclusion of a closed subspace into a $T_4$ space is automatically "relatively $T_4$" in the sense that if you have disjoint closed subsets and have separated them with neighborhoods in the given closed subspace, you can extend that separation by neighborhoods to the total space
15:37
Consider the r.v. $S_N$, which is the sum of a random number $N$ of terms of i.i.d. r.v.s. $X_1,X_2,\ldots$ (everything's nonnegative integer-valued, and $N$ is independent of $X_1,X_2,\ldots$). What random variables is $S_N$ a function of? Certainly $N$, but is it correct to say it is also a function of $X_1,\ldots, X_N$? That would seem kind of odd to say I think, since we don't know how many $X_i$'s we got.
16:07
@Thorgott I'm pretty sure this holds in $T_5$ spaces since you can exploit that if $A_i\subseteq A$ where $A_i, A$ are closed and $A_i\subseteq U_i\subseteq A$ are relatively open disjoint sets, then $\overline{U_1}\cap U_2 = U_1\cap \overline{U_2} = \emptyset$ so there are disjoint open $V_1, V_2$ with $U_i\subseteq V_i$. In $T_4$ but not $T_5$ spaces, I suspect the answer might be no, although this is not certain since closed subspaces of normal spaces preserve normality and its the open ones
c.t. that lead to failure of hereditary normality
16:24
@Thorgott its easy to see that this is about extension of disjoint open sets $U, V$ from a closed subspace $A$ to disjoint open $U', V'$ with $U\subseteq U'$ and $V\subseteq V'$. This is equivalent to hereditary normality
This property implies $T_5$, since for two $A_1, A_2$ with $\overline{A_1}\cap A_2 = A_1\cap \overline{A_2}$ we can take $C = \overline{A_1\cup A_2}$ and $U_i = C\setminus\overline{A_{3-i}}$ and even though $A_i\subseteq U_i$ are not necessarily closed, it doesn't matter since $U_i$ separate some closed sets
this shows that one needs to pick those open sets very carefully in the proof that $Y\cup_f X$ is $T_4$ (if they are proving it by the open sets route)
16:41
@Jakobian ah, very nice, so the open set route is indeed subtle
 
2 hours later…
19:01
0
Q: How to find this pgf in a game of roulette?

psieHere's an exercise in An Intermediate Course in Probability by Gut. The probability generating function (pgf) of a nonnegative integer-valued random variable $Y$ is $g_Y(t)=Et^Y$, and the pgf of a sum $S_N$ of a random number $N$ of i.i.d. random variables $X_1,X_2,\ldots$ is $g_{S_N}(t)=g_N(g_X(...

Russian roulette
@SineoftheTime I hope not :)
 
1 hour later…
20:45
pretty sassy for a statement being made in paragraph 1464 of something
ikr
the rest of the paragraph is pretty contemptuous towards Ivory's writings on fluid mechanics
(it's from Todhunter's A History of the Mathematical Theories of Attraction and the Figure of the Earth)
you have to be forgiving with fluid mechanics though, haha, if there's any subject that ought to keep one humble
imVho
todhunter just salty about spending 1400 paragraphs trying to derive the equation for why people aren't attracted to him
lol
here's the google books link for further context: google.com/books/edition/…
i kind of like a numbering scheme where you number literally every paragraph, makes citation easy
21:01
yeah, tho here it's strictly speaking more like a subsection
@leslietownes I also like this type of numbering
not only because of citations but because it makes it easier to quantify information in a book for learning
21:17
as in "i'm going to read 20 more paragraphs of todhunter and then walk into the sea" ?
like I'm a Dagon worshiper?
@leslietownes You'd like Wittgenstein. :)
some local courts here require require line numbers on every page, which seems like it would help in citation, except for anything other than a direct quote you end up having to cite such wide swaths of line numbers that you would have been better off with paragraph numbering, and with things that are direct quotes it's pretty easy to tell whether they're there or not using only a paragraph number
its not like you have to search all that carefully for a direct quote anyway
@XanderHenderson from the way you say that i'm suspecting that i wouldn't
@leslietownes He numbers every sentence.
21:40
If there exists a way to determine primality of numbers by checking numbers no higher than 200 million or so for their own primality, what could that entail?
absolutely nothing
much, if not everything, would depend on exactly how it does that
@ZacU. I don't understand the question.
That's what I think too Leslie
@XanderHenderson so say you could guarantee a number as prime or non prime by checking a smaller number for being prime. There would be a process / algorithm required obviously
my intuition is, you can't abstract away almost everything that might matter about an algorithm, and be able to say very much about what it would allow you to do. you could hypothetically imagine an algorithm that broke one number into 10^20 zillion numbers of your small size, and somehow got a primality analysis out of that. that would probably not be a good algorithm even though it satisfies the description.
or if that would be a good algorithm, or if that hypothetical is somehow ruled out, maybe more needs to be said about why that is. if you assume infinite space and time you basically don't need algorithms anymore for anything
21:51
@XanderHenderson How I understand it, is Zac U. is trying to invite us to a discussion, but without any indication of a direction in which they want to take it
 
1 hour later…
23:11
Jakobian I'm not so sure. Your first response was that 'absolutely nothing' would entail
23:23
I'm looking to prove that this function:

$$\zeta_{L}(x) = \lim_{k \to \infty} \frac{\int_0^x \sum_{n=1}^k e^{\frac{\ln(n)}{\ln(t)}} \, dt}{\int_0^1 \sum_{n=1}^k e^{\frac{\ln(n)}{\ln(t)}} \, dt}$$

converges on the interval $x=[0,1].$ The difficulty is showing that as $k\to \infty$ the function remains well-defined. Any insights would be much appreciated :)
For more context, the end goal is to ascertain which region(s) this function can be maximally continued to on $\Bbb C$.

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