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12:08 AM
Hi folks. Wonder if anyone can offer a clue about what technique to use to integrate
$$\int \frac{5 \sqrt x}{\ln 6x} dx$$.
you can use the technique called "wolfram alpha"
It gives an answer I don't recognize and offers no help with how to integrate it.
@Jeff This has no elementary antiderivative.
12:23 AM
@Tec Oh. I was starting to suspect that.
I've never heard of the 'exponential integral' that Wolfram refers to. I'm looking it up now.
it's a particular definite integral that doesn't have an elementary closed form
the integral you wrote is equivalent to it
(by appropriate change of variables)
$\int e^x/x\,dx$
@TedShifrin TY. Got it and "did" it.
What use is the exponential integral?
12:57 AM
There are various integrals that show up, like erf (most well-known from probability), $\sin x/x$, etc., but have no elementary expressions. So they get names and have been tabulated numerically.
3 hours later…
3:28 AM
Howdy mathematics. I'm Chris. I'm I need some help in this question regarding discrete quanitties based on time [4041120] this is the code of the question
perhaps if someone has seen this topic before or a similar could help me?. I don't know exactly what to do because the only answer which has appeared there has left me more confused that how i was feeling at the beginning
Thus if someone could enlighten me with a wordy description or maybe a guide on which went wrong I will appreciate the help.
I understand that it might be a simple question but it has got me to think too much and round in circles. Thus could anyone please take a look into question [4041120]
Thanks in advance
1 hour later…
4:52 AM
N i c e
I wonder who is controlling math
I am reading a paper where the word Curvature(a real number) frequently comes; the paper deals with only Riemannian surface, i.e., dimension 2. Do I assume the author means to say sectional curvature?
I know this is a weird question, but I searched on the word curvature in whole .pdf but got no word like sectional.
5:11 AM
You need to learn basics. For surfaces there’s only one curvature.
Yeah, I know this: for a surface sectional curvature of is equal to its Gauss curvature.
Oh @TedShifrin I forgot to mention/ask with regards to my reasoning yesterday about the circle being split in three parts. So I came to that conclusion because all three circles had the same length of radius which implies that the three circles are congruent, as a result I could "combine" how the different radii split up the three circles. In other words the way RobJohn drew the picture, but have all three of those angle measures "merged" into one circle...
and from that I could find the "angle" measure for one portion.
A lot more verbose than I wanted to describe it....but yea...
3 hours later…
8:23 AM
@TedShifrin do you mean one intrinsic curvature? Otherwise, there's mean curvature, which is different, right?
1 hour later…
9:43 AM
Q: Algebratization of the twin prime conjecture. Twin primes true if and only if constructed group $G$ has infinite order.

StudySmarterNotHarderDefine $g : \Bbb{Q}^{\times} \twoheadrightarrow \Bbb{Z}$ by $\pm\dfrac{p_1\cdots p_n}{q_1\cdots q_m} \mapsto p_1 + \dots + p_n - q_1 - \dots - q_m$. The kernel $K = \ker g$ kind of measures the failure of summation of primes to be unique. Except it infinitely counts one example say $\dfrac{2\cd...

10:20 AM
The hierarchy is clear
Math is only a subfield of forklift operating
forklift theory must be rich
A fair coin is tossed three times. X is the number of heads in the first two tosses and $ Y$ the number of H in the last two tosses.

Find Cov (X,Y)
I find the distribution for X and Y E(X) AND E(Y) but how to find E(XY)
10:52 AM
@user2103480 lmfao
11:04 AM
if I engineer a coin so that it always lands on the opposite side of the previous flip, is it a fair coin?
what is fair?
is it fair that the big companies get to control the price of the stock market blatantly?
Hi! I'm bard, sorry to interrupt: I've posted some days ago the following question, but altough I've recevied some upvotes nobody has answered me. Since I'm pretty confident there's at least an algebraic-geometer here, if you want to gain some rep please take a look, it's a pretty simple problem:


Sorry again, and have a nice day!
@mathsstudent Just write out all the possibilities:
$$\sum X=8\quad\sum Y=8\quad\sum XY=10$$
11:50 AM
$$\nabla^2f=u\frac{\partial}{\partial u}\left(u\frac{\partial f}{\partial u}\right) + v\frac{\partial}{\partial v}\left(v\frac{\partial f}{\partial v}\right)=0$$
I don't know if anyone likes differential equations, but if anyone can help me solve it that'd be cool
12:02 PM
is that supposed to be the laplacian?
12:12 PM
in a different coordinate system
It's proving to be harder to solve than I thought
... does that matter?
In mathematics, mathematical physics and the theory of stochastic processes, a harmonic function is a twice continuously differentiable function f : U → R, where U is an open subset of Rn, that satisfies Laplace's equation, that is, ∂ 2 f ∂ x 1 2...
this should be what you're looking for
yeah it matters, the solutions will be different in different coordinate systems
Is $\log _3 \lfloor {3^{\frac{1}{n}}} \rfloor = \frac {1}{n}$?
12:28 PM
no, plug in 2
sqrt(3) = 1.7 something so the floor of that is 1 and log_3(1) = 0
Im trying to disprove the following statement:

Given $f,g>1:\mathbb N \rightarrow \mathbb N$ prove\disprove the following:

$f=O(g)\Rightarrow \log_fg=\Omega(1)$

Take $f=2,g= \lfloor3^{1/n} \rfloor$. Let $c=5$, then for sufficiantly large $n$ we have

$f=2 \leq 5\cdot \lfloor3^{1/n} \rfloor=c\cdot g \Rightarrow f=O(g)$


$\log_fg=\frac{\log_3g}{\log_3f}=\frac{\log_3\lfloor3^{1/n}\rfloor}{\log_32}=\frac {0}{\log_32}=0\neq\Omega(1)$
1:09 PM
I have proved that $$\sum_{m=0}^{\infty}p(5m+4)x^{5m+4}=cx^{4}\frac{\phi(x^{25})^5}{\phi(x^5)^6}$$
How can I prove that $c=5$?
(Ramanujan's identity)
The solution would be easy
In a parallel universe
Or is it easy?
Btw it is an exercise from a book
1:27 PM
evaluate f(x)/x^4 at 0 ?
Is f(x) the RHS?
I don't if that works because I don't know anything about $\phi$
Euler totient function
The RHS becomes c and the LHS becomes a
Divergent sum
Ramanujan loves divergent sums
1 hour later…
2:48 PM
hi, I know this result to be true, but I can't see it using elementary notion of conditional probability, namely, let $X,Y,Z$ be random variables , with $\sigma(Y) \subset \sigma(X)$, then $\mathbb{P}(Z = z | X = x, Y = y) = \mathbb{P}(Z = z | X = x)$ , $(X,Y)_{\ast} \mathbb{P}$ almost surely. This makes total sense if I use the factorization lemma to see it, but suppose I have $\mathbb{P}(X = x, Y=y) > 0$, so these now become elementary conditional probabilities, how can I see it?
uh sorry, I think I also need to say that we are working on a countable state space for these to be elementary conditional probabilities
im not sure if its true if I don't restrict to a countable state space
3:01 PM
hm, I guess one way to see it is that $Y$ is $X$-measurable, so it is of the form $\phi \circ X$ for some $\phi : \mathcal{X} \rightarrow \mathcal{X}$ where $\mathcal{X}$ is the state space, so by asking that $X = x$ and $Y=y$ we are really saying we need $X=x$ and $X \in \phi^{-1}(\{y \})$, and if this happens with nonzero probability then it is equivalent to asking for $X = x$
so conditioning wise, we can forget about asking for the additional $Y=y$
its almost surely redundant extra information
this is still really the factorization lemma way to see it fleshed out, i'm wondering if there is an even simpler way?
So I dont really get youer question. $\Bbb P(A \mid X, Y)(\omega) = \Bbb P(A \mid X)(\omega) = \Bbb E[1_A \mid X](\omega) = g(X(\omega))$, and in the case that $X(\omega) = x$, we say $g(x) = \Bbb P(A\mid X=x)$
the value of Y is determined by the value of x so either X=x and Y=y does not happen or always coincide
The factorization lemma is not really avoidable up to rephrasing
(to my knowledge)
yeah, the determined by the value of x thing essentially comes from the factorization lemma
okay, thats fine
@LeonhardEuler what are $p$ and $\phi$?
In particular, what is $p(4)\phi(0)$?
@porridgemathematics Yes and no
@user2103480 let me rephrase, assuming the state space were working on is $\mathbb{N}$ with the discrete sigma algebra, is there a way to go from $\sigma(Y) \subset \sigma(X)$ to $\frac{\mathbb{P}(Z = z, X=x, Y=y)}{\mathbb{P}(X=x,Y=y)} = \frac{\mathbb{P}(Z = z, X = x)}{\mathbb{P}(X = x)}$, assuming these quotients are well-defined
3:07 PM
the factorization lemma is not the reason, but its a good intuition
without factorization lemma
i.e. just manipulating elementary probabilities via sums
probably, but you see I didn't use the factoriation lemma up there?
The sigma-algebra generated by (X,Y) should be just the one generated by X
yeah, but you used that $\sigma(X,Y) = \sigma(X)$
yeah but that's just the condition
I don't think you need the factorization lemma for that
probably some good sets principle or something
and another way of defining things would be, $\mathbb{E}[X | Y = y]$ is the $Y_{\ast} \mathbb{P}$ almost surely defined $\phi : \mathcal{X} \rightarrow \mathcal{X}$ such that $\mathbb{E}[X | Y] = \phi \circ Y$ almost surely
if you use this definition, you can't immediately go from $\sigma(X,Y) = \sigma(X)$ to what we want
because $\mathbb{E}[X | Y = y, K = k]$ is now a map $\phi : \mathcal{X} \times \mathcal{X} \rightarrow \mathcal{X}$
so you inadvertently need to show this is almost surely equal to some map $\psi : \mathcal{X} \rightarrow \mathcal{X}$ (the second component is redundant, essentially)
3:11 PM
@porridgemathematics What do you want/prove?
Except what you stated above there
basically I want to know how to do it in the countable state space case, without explicit use of formal definitions of conditional expectation that apply in the uncountable state space case
if $\sigma(Y) \subset \sigma(X)$ on a countable state space then $Y^{-1}(\{a\}) = X^{-1}(B)$ for some countable set B and then you can probably distribute over that
because in te countable case, you can show conditional expectation exists without any use of the radon-nikodym theorem
you can just define it via elementary conditional probability
ah okay, yeah that does work
yeah and the $B_a$ should be disjoint yada yada
yeah I can see how we can turn this into an argument involving what we've been talking about essentially (in the most general case) distilled to the countbale case
that does answer my question :)
3:15 PM
I dont recall the proof of the factorization lemma but that should also be the reasoning behind it
modulo some fine details like measurability, which I know I messed up when I tried proving it in an exercise
@porridgemathematics ok nice
well $\sigma(Y) \sigma(X)$ immediately gives you $Y^{-1}\{a \} = X^{-1}(B)$ for some countable $B$ and then we can just say $Y = a, X=x$ is the same as $X = x$ and $X \in B$ and if this happens with nonzero probability then we can effectively ignore the $X \in B$ term in our summation
so we don't even need to use the factorization lemma, just the definition of $\sigma(X)$ for some rv $X$
*$\sigma(Y) \subset \sigma(X)$
You don't ever need the factorization lemma if you reprove it everytime, but its quicker to just use it :P
we dont even need to reprove it in the countable case, really
Fair, fair
the proof of it that I know of is just the usual start from indicator functions nonsense
3:21 PM
@porridgemathematics the indicator function basically only consists of the above step
oh yeah haha, thats true :)
its noticing what $\sigma(X) \subset ... $ entails
and then induction of course
I agree that it's hard to see all that at first
I found conditional expectations to be very unintuitive
I still can't say I find them intuitive per se, but I'm used to them now and know some heuristics
How do representants of generators of top cohomologies look like?
For manifolds. Hm. Ah. I know my fault in thinking
we only defined orientability for rings
So I guess a cochain represents the generator of top cohomology of an R-orientable manifold if it maps a chain that represents the fundamental class to the unit in the ring?
Or am I mightily wrong with that
4:23 PM
Cohomology of disjoint unions is just the product of the cohomologies?
This should just work via $C_\ast(\coprod_i X_i) \cong \bigoplus_i C_\ast(X_i)$ and then dualizing
@user2103480 yes, H^n is dual to H_n since H_{n-1} is torsion-free for closed orientable manifolds
@user2103480 yes
ok nice
you mentioned that example of computing the cross product of the generators of first cohomology of spheres to get a generator of second cohomology of the torus
I mean I get how it should work
$(\mathrm{pr}_1^\ast[S^1]^\ast \cup \mathrm{pr}_2^\ast[S^1]^\ast)(T^2) = \langle S^1, T^2|_{(0,1)} \rangle \cdot \langle S^1, T^2|_{(1,2)} \rangle = \langle S^1,S^1 \rangle \cdot \langle S^1,S^1 \rangle = 1_R$
In very suggestive notation
But to actually obtain this I need to work with an explicit chain that represents the fundamental class of the torus, right?
So that restricting that chain directly yields two circles
not gonna do that
although it should be possible with two singular simplices, or probably one if one is smart
gotta be careful about orientation though
Okay, no, it's just the obvious thing
4:48 PM
what's the easiest example of an open manifold that's not an interior?
is that directed at me
I dont even know what that means
general question
@Thorgott Take the Loch Ness Monster
as in surface with infinite genus?
5:15 PM
Is it obvious that if $P$ is a prime ideal and $p\in P\setminus P^2$, then $p^n\in P^n\setminus P^{n+1}$?
Are your prime ideals maximal by any chance
Yes, I'm in a Dedekind domain
You may be using Nakayama's lemma
Yeah so that's it
If $p^n$ is zero in $P^n/P^{n+1}$ then $p$ is zero in $P/P^2$
I'm not seeing it
Or some variation. Let me say it correctly.
5:17 PM
I have no intuition for Nakayama
this is some end-theoretic obstruction, I wager
@Astyx 4Z is not prime though
Lol I should read
french primes
There's a long tradition of French primes starting with Grothendieck's 57
5:27 PM
@Alessandro $P/P^2$ is 1-dimensional as a $A/P$-vector space, so actually $P = (p) + P^2$. Now multiply by $P$ again to get $P^2 = (p)P + P^3 = (p^2) + P^3$, etc etc
I am using full Dedekindness
Ugh, this is not telling you much, sorry. It's something along these lines, one second
I am not an algebraist
The localization at P is a DVR, and you're stating that the valuation of p^n is n times the valuation of p
Ah ok, the point is if $p^n \in P^{n+1}$ then $P^n = P^{n+1}$
Now you use Nakayama to boil it down to $P = 0$
Ok, there you go
@BalarkaSen wait where did you get this from?
Because by calculation above $P^n = (p^n) + P^{n+1}$
Oh ok I see
5:40 PM
All these quotients are 1-dimensional, because Dedekind domains are 1-dimensionals, basically
At this point one can also use the $\bicap P^n=\varnothing$ to conclude instead of Nakayama then I guess
s'pose so
does suspension do anything complex on cohomology rings
for connected spaces
If not then I'll try proving that
It destroys the ring structure usually
oh sheesh
so trivial everything
okay I'll try to show that
5:44 PM
That's actually a Hatcher exercise IIRC good to show it
I see, thanks
still trying to find a slick proof that for connected $X, Y$, where $(X,x_0), (Y, y_0)$ are good pairs, the sum of the maps induced by the collapsing maps gives an isomorphism $H^k(X;R) \oplus H^k(Y;R) \rightarrow H^k(X \vee Y;R)$
you can give the inverse explicitly
it's trivial to calculate it's an inverse on one side, but it's necessarily a two-sided inverse then, because M-V will tell you that the map is an iso
*path connected

LES goes in the wrong direction, didn't find one via universal coefficient SES, meh. Maybe there's some algebra I'm overlooking? Both collapsing maps are retractions and muh
No need, just consider the pair $(X \sqcup Y, x_0 \sqcup y_0)$
This will also tell you the ring structure agrees
5:50 PM
@Thorgott yeah that'd be the LES map I assume
@BalarkaSen that's the purpose
(it's a three step exercise)
but isnt there some diagram I can draw to avoid explicit calculation
@BalarkaSen or does this not refer to the LES
not sure which LES you're looking at, tell me what the inverse map is explicitly
@user2103480 by any chance do you know what the Scott rank of a countable model is?
$H^k(X \vee Y;R) \rightarrow H^k(X \sqcup Y;R) \rightarrow H^k(\text{pt};R)$
@AlessandroCodenotti was that the choice business?
which choice business? I don't think so
Defining cardinalities unambiguously when choice fails
Then no
5:55 PM
that's Scott's trick
ah nah that was scotts trick. close enough lol, it's about ranks and scott
So no, don't know it
ok, but that map has an easier description
it's the tuple of inclusions
I am still too lazy to calculate, but I'll do it now then
yes, that's what I wanted to hear
@robjohn Yes, of course, but no embedding of $M$ as a surface in $\Bbb R^3$ was provided. So, no.
Isn't @Thor supposed to be studying physics today?
6:13 PM
I am, actually
OK, just checking up on you.
@Thorgott You mean interior of a manifold with boundary?
@Thorgott This case is much easier. But in general, yes.
Howdy, a @Balarka.
@TedShifrin yeah
There are famous four papers of Frank Quinn which solved the problem "Which noncompact manifolds are interiors?" completely.
Ends of Maps I-IV
6:25 PM
I know this was also studied by Siebenmann in his thesis
I forgot that actually.
@TedShifrin . o O ( Thor's mom )
Topology of ends is an extremely deep branch of mathematics. For example, Stallings is famous for his theorem that a contractible PL $n$-manifold is PL-homeomorphic to $\Bbb R^n$ iff it is simply connected at infinity.
PL can be ignored due to later developments in mathematics.
I'd rather have ends of loaves of baguette. :)
6:30 PM
Ah, also, $n \geq 5$.
$n = 4$ is Freedman, I suppose.
yeah, Siebenmann actually gives a complete classification in dimensions >5
this preceded the work of Quinn you mention
Smart guy
I tried to read Quinn's paper in some occasions, but he's an extremely messy mathematician
Everything's all over the place. It's a very technical theory
Odd to hear a @Balarka complaining about lack of neatness :)
@BalarkaSen is this related to Stalling's ends of groups theorem or are they just both by Stallings and both on ends
Haha, I suppose he thought about ends a lot
We can try to read Stalling's paper. If I recall correctly it's short and neat
6:34 PM
Siebenmann's thesis looks pretty readable, but it's far beyond my background
@BalarkaSen what's the name
@TedShifrin I'm not a mathematician, so I don't have to be neat :)
You sound like you want to be qualified to be a doctor because your handwriting is so messy.
If I was actually communicating mathematics as a job I'd try to be neat
But I am not so I have no obligation to be
Oh, engulfing shows up.
man, I don't know any PL topology
6:36 PM
Me neither.
It's a good place to start
Ah, I think I now have the same problem I had last time. The pairs are $(X,x_0)$ and $(Y,y_0)$. I send $([\varphi], [\psi])$ to $(i_X^\ast p_X^\ast([\varphi]) + i_X^\ast p_Y^\ast([\psi])$ and want to show the last thing is zero. $i_X^\ast p_Y^\ast = (p_Y i_X)^\ast = (X \rightarrow \{y_0\})^\ast$.

I think I understand something wrong. I try to calculate using a representant $\psi$, on cochain level. Then for an arbitrary sum of simplices $\sum_i a_i \sigma_i $, I get $ i_X^\ast p_Y^\ast \psi(\sum_i a_i \sigma_i) = \sum_i a_i\psi(\Delta^k \rightarrow y_0)$ and duh is there an immediate reaso
@TedShifrin Neat ^
fucc formatting
6:40 PM
is chat broken again
ok no
constant map induces 0 on homology in degrees >0
cause it factors through H(pt)
you can do this calculation explicitly, but it will just be the same one as when you calculated the homology of a point
I tried to do the factoring but somehow went wrong
thanks I'll try again
@EdwardEvans What's your point?
The formatting was broken lol
okay yeah I tried not hard enough with the factoring. Reminder to myself not to write $X \subset Y$ instead of $X \hookrightarrow Y$
7:40 PM
This is annoying. I wanted to prove the tower law for relative different but I'm having trouble with one of the inclusions.
for relative different?
$E/L/K$ be a tower, $\delta_{E/K} \supset \delta_{E/L} \delta_{L/K}$ is clear. Easier to do with inverse, let $\alpha \in \delta_{E/L}^{-1}$, $\beta \in \delta_{L/K}^{-1}$ then $\text{tr}_{E/K}(\alpha \beta \mathcal{O}_E) = \text{tr}_{L/K} \text{tr}_{E/L} (\alpha \beta \mathcal{O}_E) = \text{tr}_{L/K} (\beta \text{tr}_{E/L}(\alpha \mathcal{O}_E)) \subseteq \text{tr}_{L/K}(\beta \mathcal{O}_L) \subseteq \mathcal{O}_K$
This shows $\delta_{E/L}^{-1} \delta_{L/K}^{-1} \subseteq \delta_{E/K}^{-1}$
What about the other direction lol
This stuff is so confusing
So I want to prove $\delta_{E/K}^{-1} \subseteq \delta_{E/L}^{-1} \delta^{-1}_{L/K}$, or alternatively $\delta_{E/L} \delta_{E/K}^{-1} \subseteq \delta^{-1}_{L/K}$, which I imagine will be easier
is there any subject except logic that you dont study
Any $\alpha \in \delta_{E/L}$ multiplies with (anything which pairs with anything in $\mathcal{O}_E$ and spits out something in $\mathcal{O}_L$) and spits out something in $\mathcal{O}_E$
This is logic bro
Look at those recursive statements
lOgIC oF sPacEs
@BalarkaSen basically programming
7:48 PM
But anything in $\delta^{-1}_{E/K}$ pairs with anything in $\mathcal{O}_E$ and spits out something in $\mathcal{O}_K$, which is actually a subring of $\mathcal{O}_L$, so I guess $\alpha$ multiplies with such a thing and spits out something in $\mathcal{O}_E$
This is useless
Maybe $\delta_{L/K} \delta_{E/K}^{-1}$
Any $\alpha \in \delta_{L/K}$ multiplies with (anything which pairs with anything in $\mathcal{O}_L$ and spits out something in $\mathcal{O}_K$) and spits out something in $\mathcal{O}_L$
But elements of $\delta^{-1}_{E/K}$ are such a thing because $\mathcal{O}_L$ is a subring of $\mathcal{O}_E$, so $\delta_{L/K} \delta^{-1}_{E/K} \subseteq \mathcal{O}_L$.
Isn't that nonsense
What is happening so confusing
@EdwardEvans I have decided that I would quit too if I was a Masters student in ANT
Help lol
i never actually learned about the different lmao
How do you show $\delta_{E/K} \subseteq \delta_{E/L} \delta_{L/K}$
So fucked up man
I'd usually just default to Lukas
7:56 PM
that's just transitivity
but he disappeared again
@Thorgott Ok how
there's like E/K on the left and E/L/K on the right
as usual, thanks for being of no help
ontological proof by symbolics
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