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00:10
@usukidoll One thing which you can do to make the integral less annoying is the substitution $u=y/\theta$
that eliminates most of the $\theta$'s from the problem, and all you really end up needing to compute is the definite integral $\int_0^1 u(1-u)^{n-1}\,du$
If $X=Y$ almost everywhere, can $X$ and $Y$ be independent random variables?
Another way to phrase that: If $H=0$ almost everywhere, can $X+H$ and $X$ be independent?
(may or may not be useful, but that seems like it might be productive)
I was trying to construct an independent process $(A_i)_i$ (for $i$ in an interval) with each $A_i=1$ a.e.
00:20
can a r.v. be ind. with itself?
independence means $P(X\leq x,Y\leq y)=P(X\leq x)P(Y\leq y)$
@LeakyNun I can take an independent copy of a RV.
if $Y=X$, that becomes $P(X\leq x,X\leq y)=P(X\leq x)P(X\leq y)$
does that answer your question
No, because that independent copy will only be equal in distribution
Suppose without loss of generality that $x\leq y$. Then $P(X\leq x,X\leq y)=P(X\leq y)$
In which case the condition for independence becomes $P(X\leq x)=1$
Which would mean that, for every $x$, the probability that you're less than it is 1
and thaaaat's not happening
So no, $X$ and $X$ are not independent
00:24
if X=Y a.s. then in terms of P(X<=x) that's just the same as P(Y<=x) etc right
I think so? I'm not so good with that
thanks, that clarifies things
If $P(H=0)=1$, can $P(X=x,X+H=x+h)=P(X=x)P(H=h)$?
That should be the question in the form I'm posing it.
But $P(H=0)=1$ means that $P(H=h)=0$ if $h\neq 0$
So when $h=0$ that's $P(X=x,H=0)=P(X=x)P(H=0) = P(X=x)$
and when $h\neq 0$, $P(X=x,H=h)=0$
That seems pretty darn strong.
But it doesn't seem to exclude it...
@Semiclassical have you heard of the hypergame?
00:36
so this is game between two players, Alice and Bob
Alice first names a game that ends in finitely many steps
then they play that game and it is Bob's move
that's it
yeah, googled it
the question is: does Hypergame finish in finitely many steps?
the link I saw spoiled the appearance of Russell's paradox, alas
i.e. if it's a finite game, then choosing to play hypergame is a legal first move. but then you can play infinitely by just offering to play hypergame again
if it's not infinite, then choosing to play isn't a legal first move...in which case it's finite.
so... what went wrong?
well, since it's like russell's paradox
it presumably doesn't make sense to talk about "the set of all finite games"
00:42
so how to salvage game theory lol
shrug
I don't think you run into the problem if, for instance, you limit yourself to games which finish in less than 100 million turns (picking an arbitrary cutoff)
since then you could only choose the hypergame finitely many times
there's probably some more principled way of doing it
but frankly idc
I would be curious if there's a game-theoretic version of the Yablo paradox
that one is neat
 
1 hour later…
01:53
@AkivaWeinberger hmm, regarding your digamma question, this answer looks like it might be handy: math.stackexchange.com/a/2075586/137524
See in particular the equation after "Let $t\to 1^{-}$..."
In fact, looking at it, I want to say that your result is identical to theirs if the following is true:
$$\sum_{\substack{\large\rho^b=1\\\large\rho\ne1}} \ln(1-\bar\rho) = \ln b$$
I guess you can get rid of $\bar\rho$ in there, since $\overline{\rho}$ is itself a root of unity
in any case, the sum of logs can be written as the log of a product: $\displaystyle \ln\sum_{\substack{\large\rho^b=1\\\large\rho\ne1}}(1-\rho)$
But that's the same as $\ln f(1)$ where $f(z)=\dfrac{z^b-1}{z-1}=1+z+\cdots +z^{b-1}\implies f(1)=b$
So I think that confirms the agreement between your answer and theirs
(I'll try to post that as an answer at some point)
02:24
what happens if I need to find $E[Y^{2}]$ ? could I do the same u sub but then there will be an extra y in it?
$u = \frac{y}{\theta}$
followed by
$uy(1-u)^{n-1}$
0
Q: What's that on Euler's head? Is there a name for the head covering shown in Emanuel Handmann 1753 painting?

uhohThis may be borderline off-topic but this is the only place to ask. I've always seen images of Leonhard Euler with a "hat" or head covering that is unfamiliar to me. Is there a name for the head covering shown in Emanuel Handmann 1753 painting of Euler? Does it cary any specific significance o...

 
1 hour later…
03:31
@usukidoll yeah, and that'll change the overall factors of $\theta$ accordingly. but you'll still get a definite integral, just slightly different
hey forum
"This site is all about getting answers. It's not a discussion forum. There's no chit-chat." :P
where are those guildelines?
Almost 100% certain there is a nearly identical page for every stackexchange site.
i dont htink those apply to the discussion forums
03:43
you mean chat? :P
haha
but hey, how's it goin?
anyhway syppose $x = c_1 + \ldots + c_nx_n$. Does $\langle x, y \rangle =0 \implies \langle x_i, y \ranle = 0$?
i think not..
Do you mean $x = c_{1}x_{1} + \ldots + c_{n}x_{n}$?
mhm
boy oh boy
i messed the whole thing up
03:47
and $\rangle$
$x = c_1x_n + \ldots + c_nx_n$. Does $\langle x, y \rangle =0 \implies \langle x_i, y \rangle = 0$
Can $c_1 \ldots c_n$ be $0$?
no
can $c_i =0$? answer: no
its only true if the basis is orthogonal to start with
in which case its obvious..
oy. perhaps i dont need that conclusion.. or perhaps i could start out with an orthogonal basis..
It's been a while since I've done this, but consider $x = c_1x_1 + c_2x_2$ and I think you can construct a counterexample from that...
yup. i did
thats why i (think) its false
03:56
If you have a counterexample to a claim it is false.
right. remains to show the counterexample is valid :)
are you a mathematician @Dair? what do you do?
@JoeShmo I was once a math major in college. I do CS now.
i see..
What issue are you having with illustrating the counterexample is valid?
no issues. just trying to get someone else here to think for me. im juiced out..
04:29
I'm slammed with math homework rip
but at least I'm almost done with stats... just feel burned out from all the previous days
Did you manage to get the integral done?
04:54
I just got back :/
05:05
and my dad is fixing a telephone jack so I'm connected to my relative's router
05:45
MathJax is not working on my pc. Can anyone help?
 
4 hours later…
09:53
the integral of a ball centered at (0,0,0) with radius d, if the integrand is $1/|z|^3-3z_3^2/|z|^5$, why the integral is finite?
i tried to use the polar transformation but it seems divergent
but the mathematica gives a finite result...
 
1 hour later…
10:57
What is it with the main page today?
Or rather the questions and the answers.
 
1 hour later…
12:22
Not my photo, but - this is homeomorphic to a solid triple torus, right?
Rather than a solid double torus
Should be a double torus with a handle
I wouldn't want to drink anything out of it
imagine cleaning the bottom of the cup
4
Q: Continuous Deformation Of Punctured Torus

AndrewThis is problem 11 (b) from the first chapter of "Basic Topology" by M.A. Armstrong. The author hasn't had time to develop many theorems or mathematical machinery, so this problem should be able to be solved by just picturing a series of intermediate steps. It goes Imagine all the spaces sh...

alternatively, it is homeomorphic to a puctured double torus
12:44
It's a closed epsilon-neighborhood of a punctured torus wedged with $S^1$
which should be homeomorphic to a closed epsilon-neighborhood of three circles wedged together
which is a solid triple torus
ok I give up, too complicated to do this intuitively
But what I got is basically 3 rings glued together in horizontal, vertical, horizontal fashion, with the middle ring have an extra loop glued onto it
More precisely
I don't know what's the middle thing is homeomorphic to
13:10
@Secret …What about the mug's handle
Well the mug and its handle forms a double torus right, and the opening of the mug thus act as the puncture?
@Secret You have the image on the bottom-left of this picture plus a handle
(and also take an epsilon-neighborhood of everything because the mug isn't infinitely thin)
so... three rings glued together?
Yeah
= solid triple torus
I see
It's terribly hard to reason about these things pictorically, I guess that's why we need the algebraic formulation
Still, I am curious what that 3 fold ring thing is homeomorphic to
ah wait, a triple punctured sphere
which is also homeomorphic to a twice punctured $\Bbb{R}^2$
 
2 hours later…
15:18
hi chat
Hi
hi semi! did you see that 90's layout of main site? :D
If I perform a change of variable in an integral whose bounds are $-\infty$ and $\infty$, do these bounds change?
For example, if the integral variable is $t$ and I perform the change of variable $t' = t - \delta$
15:34
what happened to math.SE?
You can go to the future
By clicking on the button on the top menu (on the right, the first button from the left)
i'm not seeing anything like that on my end
??
it seems a new CEO...
Today is the day before 1st of April
There must be some connection
oh, allright
15:39
okay, now i'm seeing it
well
i've seen it show up on one page so far
so there's something strange going on
okay, it only seems to show up when i'm on an actual question page
Seems like they modernized.
I see "go to the future" at the left side of the page, click that, then all will recover...
15:56
could anyone verify my proof here math.stackexchange.com/questions/3169453/… ?
@RithikKapoor Good question, but sorry wrong room. Moving your request to the main chat.
2 messages moved from CRUDE
um guys?
never even expected that integral has a summation form
so is it correct?
$$\int x^{x^{x^{.^{.......}}}} dx= \int\frac {-W(-\ln(x))}{\ln(x)} dx$$
I do not even have idea where this line came from
how can a infinite tetration tower be representable in terms of Lambert W
16:01
okay well go to wolfram alpha and search y=x^y
over there you would see it in alternate forms
did you get it?
it is under solutions over here^
@Secret you there?
anyone else here who could help?
I don't think you have made the correct substitution, why is the stuff in ln finite and not a tower of x?
what is your y?
y=x^x^x….
did you check the above wolframalpha link?
ok now step 1 makes sense
lol
I really need to know if this is correct
so don't leave me before you complete checkin xD
you there secret?
16:15
3 mins ago, by Rithik Kapoor
so don't leave me before you complete checkin xD
You knew I am checking when I am not responding yet
any more pestering and I will just leave this question be, because I dislike help vampires
so please be patient
If it's right, does the series form help in some way?
$$- \int \frac {W(t)}{te^t}dt= -\int \frac{1}{te^t}\sum_{n=1}^{\infty}\frac {(-n)^{n-1}}{n!}t^n = -\sum_{n=1}^{\infty} \int\frac {(-n)^{n-1}}{(n!)te^t}t^n dt$$
hmm... I don't know how one plot that series to check if it diverges, but I am sceptical of this operator interchange here
well I wanted to show my professor that there was a relation between the intefral of x^x^x… and the incomplete gamma function.
Ah, yeah it's cool. Probably has been done/thought of before if correct.
Assuming that interchanging is valid, I do not see anything wrong in the proof
16:23
@secret
I don't know how to check if that sum diverges and hence not sure if the interchange is valid
I think this integral could shed light on the very nature of space and time
other than that, everything else look fine
@Secret I am sorry if I offended you in any way. Iwas just a bit anxious
@RithikKapoor I am sorry for my rude attitude too, it's just there are too many help vampires around and thus I got annoyed
but yeah, my only skepticism is whether that interchanging of operators is legal
16:25
thank you so much for your help
@KarlKronenfeld Well, gamma functions are definitely easier to handle than tetrations
it is just that i have been trying to solve this integral for a long time but have run into some or the other problem always
plus I haven't been seen this relation anywhere I have researched
When you have a dark theme extension installed so you don't even know they're trying to pull an april fools joke.
The terms in $\sum_{n=1}^{\infty}\frac {(-n)^{n-1}}{n!}t^n$ tends to zero, but they do not do so monotonically, meaning that alternative series test cannot be used to check its convergence
it diverges, meaning that the interchange cannot be valid
someone I know took calculus and didn't know what prime numbers were
16:41
@Secret Perhaps :P
@Ultradark Tbf, I don't remember utilizing prime numbers in any calculus course I took.
do you mean to say that the series I used for the lambert W function was wrong?
no I mean you cannot interchange $\int$ and $\sum$ in the step $$- \int \frac {W(t)}{te^t}dt= -\int \frac{1}{te^t}\sum_{n=1}^{\infty}\frac {(-n)^{n-1}}{n!}t^n dt = -\sum_{n=1}^{\infty} \int\frac {(-n)^{n-1}}{(n!)te^t}t^n dt$$
@Secret This comment breaks ChatJax for me.
@RithikKapoor because the series in the integrand diverges thus violating Fubini/Tonelli theorem, meaning you cannot get the gamma function that way
16:44
what is information divided by time?
@Dair I am not sure why in your case, cause there isn't anything overly long in there
@Secret Ok, I jsut refreshed the page everything works now for some reason.
what is information divided by space?
entropy density
Why all these random googlable questions
16:58
@Secret well it looks like some questions are never meant to be solved :(
@RithikKapoor Well, by just looking at that integrand, the chance for it to have a closed form is highly unlikely. In fact, that tetration only converges for some values
0
Q: Infinite tetration convergence

NameI came across infinite tetrations on wikipedia (https://en.wikipedia.org/wiki/Tetration) it says that the infinite tetration converges if and only if $\ e^{-e} \leq x \leq e^{1/e}$. I was wondering if there is a proof of this.

Hi all, my professor mentioned in class that one cannot use L'Hôpital's rule to solve the well known lim x-> 0 (sin x)/x = 1. I was wondering what condition of L'Hôpital's rule it failed.
Nope I think your professor probably mean not to solve this limit using L'Hôpital, this limit is perfectly fine to solve under L'Hôpital
which make sense as this is one of the first limit you will learn in class thus solving it using first principles is an important skill
Ok, @Secret
THanks
12 mins ago, by Secret
Why all these random googlable questions
17:12
@Secret I'm interested in knowing if there's an equation expressing a relationship between space time and information
I must not be a good googler because I had trouble finding "entropy density"
ok make sense
Have referred that question to the h bar, i don't think we have quantum gravity researchers in this chat room
oh cool
17:36
Question: What is the origin of the word "magma" when talking about sets with binary operations?
9
Q: Etymology of "magma" in abstract algebra

KazarkMagma is one of those beautiful words of Greek origin (μάγμα) that arouses the child and the wild in me, making me think of volcanoes. I just found out, though, that it is also used in mathematics to mean a type of algebraic structure (a set paired with a binary operation on it)! I am very curiou...

Turns out it basically means "precursor to a group"
Now that makes me curious, what is the abstract algebraic version of a volcano
An infinite magma. :P
good ol' Peter Shor
Taking Bourbaki's joke even further:
Definition: An algebraic structure $(V,\cdot, E(\cdot))$ is called a volcano if its elements $M \in V$ are magmas such that there exists a unary operator $E$ such that $E(M) \in V$ are groups
(where $E$ follows the joke as it represents eruption, and thus cooled down magmas E(M) are groups)
I like it.
17:51
The group algebra of a topological group G is the collection of all continuous functions $f : G \to \Bbb{C}$ with compact support, denoted by $C_c(G)$? And when $G$ is discrete, compact support implies finite support, so we identify $f : G \to \Bbb{C}$ with the sum $\sum_{g \in G} f(g) g$, which shows that the group ring and group algebra are isomorphic for discrete groups. Does this sound right?
18:03
I can verify everything except the identification step
hey I am back again
i plotted a graph of the integral any it is pretty close to the graph of y=x^y
and since the graph of y=x^y shoots up pretty quick i think its integral would also too but then again there is no sure way to verify it.
18:31
If $G$ is a topological group, what does $C_0(G)$ denote?
 
1 hour later…
If a person wanted to learn some "basics" in non-associative algebra, what would be a good place to start? Anyone know that field well enough to offer some advice?
 
1 hour later…
20:39
hello someone here ?
Not always. Though, it's usually best to just drop a question in if you have one and see if anyone responds.
21:28
hi chat
Greetings Dair
how goes it?
Grading linear algebra homeworks at the moment. Also, looking over the results of a search algorithm I put together yesterday. So, pretty good. How about you?
Kind of tired, just reading up on the history of abstract algebra because it didn't look too taxing mentally, and sometimes it's nice to get a historical perspective on these things.
Indeed. I don't actually know much of that topic. Maybe I should read some of it, myself
21:42
Is $\int \left( x^{x^{x^{.^{.......}}}} \right) dx = \sum_{n=1}^{\infty} \frac {(-n)^{n-1}}{(n!)}\Gamma(n, -\ln x) $ ?
as by looking at the desmos graph desmos.com/calculator/8qufo3tuco , you can see that both the graphs match in the range of $e^{-e} \lt x \le e^{1/e}$. This is the same range at which the integral of $x^{x^{x^{.^{.......}}}}$ is defined.
any help guys?
@Rithaniel you there?
is anyone on?
Yeah, but a lot of people are off doing other things and checking back here every now and again.
oh okay
btw chatjax doesn't seem to be working for me
is it working for you?
In your plot, the two lines do appear to very closely approximate each other. What is the goal in this exercise, though?
Also, yeah, you have to restart chatjax each time you get on the chat, if that might be the issue.
I wanted to check if I had derieved this relation correctly as seen in this question math.stackexchange.com/questions/3169453/…
could you check it for me?
Well, I'll take a look. Can't promise I'll give you a definitive yes or no, though.
21:56
okay np
how do we restart chatjax?
Your estimate of $e^{-e}$ for the lower bound of the approximation seems off. It's closer to 0.68. (If you're on mobile, I don't know, but on desktop you should have the link in your bookmarks and you just select it from there.)
Though, your manipulation of the functions seem solid enough.
great
thank you
now I could show this to my high school teacher
There is a bit of an issue with interchanging integral and summation, but I always forget the rule on it.
22:11
But doesn't the graph prove it to be correct?
Graphs are useful, but they aren't proof. Don't take them for granted.
Like, without a proof, for all we know, as $a$ increases, one function could steadily closer approximate a second function, until $a\geq 10^{10^{100}}$, at which point it begins to decrease in how good the approximation is.
But, I do believe you've provided a proof, so long as I've not missed anything.
okay
I was actually skeptical because my HS math teacher is a PHD and she said the the integral of x^x^x… needed its own function any couldn't be defined by any other function for example in this case the incomplete gamma function.
*and
Well, she might have some evidence that I missed. Do show her your work towards it, though.
okay
thanks for the help though :)
 
1 hour later…
23:39
Hey there. Given the Laplace transform of a function, can I find the function's "total squared error" (the quantity $\int_0^\infty f(x)^2\ dx$)?

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