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01:59
2 hours wow.
This place is dead all the time now.
 
2 hours later…
03:58
@Dodsy yeah, I know what you mean.
How do you make a pullback diagram (with pullback morphisms) on MSE?
04:24
@FruitfulApproach IIRC the only diagram environment supported by MathJax is AMScd. Personally using presheaf and posting here both image and link to presheaf seems like a reasonable solution to me. Presheaf website was mentioned in one of the answers here: How to draw a commutative diagram?
I'll repost this here, in case somebody is interested:
in Linear & Abstract algebra, 9 mins ago, by toddler
My question is: How one can easily understand a dual space in simplest way? I want to see this in $R^3$.
why would you start with $R^3$ lol.
What is going on with this community? he chat is dead compared to the one 3 years ago?
@Dair If you have time (and you have insights) probably you can talk directly to the user who asked that.
@VividD I think quarter system has finals now?
@VividD Room info says 14.7k messages last week and 8.8k this week. I do not visit chat too often, but based on those numbers, it does not seem to be entirely dead.
From the room info you can also see typical activity of the room throughout the day - and you can see from that graph that slow hour is comming.
04:31
ok thanks @MartinSleziak
@Dair I think $R^3$ is at least easy to visualize. I am sitting in a room.
@toddler Sorry, I probably came off as a little rude, but it is kind of hard to draw $R^3$. $R^2$ is a little easier to draw (when it comes to giving examples).
no problem...@Dair..
So do you understand what a linear functional is?
A functional is a function of function. like integral operator...Am I right?
and it must hold linearity property.
04:37
so in the case of the definite integral you take in a function and return a scalar... but...
to quote wikipedia: a linear functional or linear form (also called a one-form or covector) is a linear map from a vector space to its field of scalars
like a norm?
yes a norm is an example of a linear functional
04:39
Can someone explain how the winding number formula is derived? Source Eq. (1.23) from arxiv.org/pdf/1510.07698.pdf
@Dair...is one-one chat possible here or anywhere?
@toddler I believe so? but there are people here that are probably better equiped to talk about linear algebra than I am...
I am new here, so dnt actually know about the rules.
ok..no problem..so norm is an example of linear functional..I understand it.
04:43
@MikeMiller? could you please help with the winding number formula. See post above.
well there are a bunch of linear functionals. So you have some vector space like $R^2$ (say we equip it with the field $R$). And you have a bunch of linear functionals: $f, g : R^2 \to R$
ok..
@toddler Sorry, as I said I'll have to go but I simply did not resist. Norm is not a linear functional. For example, you don't necessarily have $\|x+y\|=\|x\|+\|y\|$.
yes ...sorry..I was wrong @Martin Sleziak.
a definite integral will work.
sorry my bad...
04:47
No problem. I just wanted to point out the mistake, so that you do not get confused.
Good luck and see you later!
right, so forgot about that $|x + y| \leq |x| + |y|$ by the triangle inequality...
You need strict. but $\int_a^b (f + g)(t) dt = \int_a^b f(t) dt + \int_a^b g(t) dt$
@zed111 I find that sort of notation very hard for me to read. Sorry.
No problem. I somehow feel the integral is evaluating the solid angle covered by the map n:S^3 -> S^3. But unable to verify.
S is 3D?
04:55
Yeah
I know I've seen that kind of formula before, but I can't remember at all how it's proven.
SBM
SBM
Hello
Morning people
@zed111 Are you sure? The cross product is best behaved in 3D, which is only where Sn would live if it was 2-dimensional...
$\epsilon^{\mu\nu}$ is presumably the Levi-Civita symbol on two indices.
I think one of the grad students I worked with (who has since graduated) actually was at the summer school where Ed Witten gave these lectures.
Oh actually n is a map from S^2 to S^2. Sorry for the misinfo @MikeMiller
04:59
@zed111 In general, if you have a map $f: M \to N$ of closed oriented manifolds of the same dimension, and you pick a volume form on $N$, then $\text{deg}(f) = \int_M F^* \omega$. Now the standard volume form on $S^2$ is given by taking the (oriented) length of $v \times w$, and so now you need to take an explicit formula for the pullback of $\omega$ under the Gauss map.
Specifically, $S_{\vec{n}}$ is a small sphere around the point $\vec{n}$.
Nice to hear @Semiclassical. These were at Princeton Summer School.
In general the Gauss map sends a vector $v$ to $\partial_{v} n$ (which is a vector in R^3!), and so the formula for the pullback is the oriented length of the cross product $\partial_v n \times \partial_w n$.
I should also say that to call this a winding number is a bit strange. As Witten notes previously to that, the fact that this formula always gives an integer reflects the fact that $\pi_2(S^2)\cong \mathbb{Z}$.
05:01
Seems like that's probably exactly what the formula is saying
So second homotopy group, not the fundamental group.
@Semiclassical I would call it how much S^n winds around itself
Hmm. I guess I'm just used to winding being used to describe curves.
@Semiclassical I've heard people use winding number for such maps.
Terminology, though.
The other reason I find 'winding number' to be weird is that homotopy groups don't have to be abelian.
And it feels weird to call anything that doesn't need to commute a 'number.'
But again, terminology.
05:04
higher homotopy groups do have to be abelian
it's the fundamental group that doesn't
I always forget that.
It always seemed odd that only the fundamental group could be non-abelian.
I imagine there's a good reason for that, but if I knew it at one point I don't know it now.
SBM
SBM
Abelian groups?
if you draw a picture of a disc and then some smaller discs inside it you can swap the two discs by moving them around
can't do that with intervals in a line
That'd do it.
Sounds rather like the fact that every knot in 4D is the unknot.
05:28
Thanks @MikeMiller
 
3 hours later…
08:23
Are all approximations of the sign function actually cheats so that they involve the sign function in its primary definition in some form?
08:48
I have tried the power series for ArcTan(x) but there is always the reciprocal function 1/x somewhere, and that does not have good enough radius of convergence.
The same goes for Tanh(x)
One can of course divide x in the series for 1/x with a constant, but then again when the function fluctuates too much, this not applicable as well.
One could instead of a constant divide by the function itself plus some small number, but that is tautological and does not solve the problem.
@MatsGranvik You may want to mention what you understand under an approximation
@SteamyRoot f(t) = Sign(Im(Zeta(1/2+It)))
wait what
And the approximation must be integrable
That's not an answer to the question, and where do you suddenly get RZ from?
08:54
From here: https://oeis.org/A058303
And here: https://oeis.org/A286707
"We can compute 105 digits of this zeta zero as the numerical integral..."
Okaaaaay... and what does this have to do with approximating the sign function?
And again, what do you mean with "approximation". Do you want a family of functions that converge to the sign function pointwise? Or uniform? And on what set do they have to converge, etc...
@SteamyRoot One form of the numerical integral that gives the first 126 zeta zeros involves the sign function.
When integrating this:
((Floor[RiemannSiegelTheta[t]/Pi + 1]) + (Sign[Im[Zeta[1/2 + I*t]]] -
1)/2)
appropriately, the exact formula is more like....
I don't know what pointwise means, but the way I view the integral is as a tree structure where the leaves are the parts to be integrated by generalized integration by parts.
09:28
I think it is obvious I still know nothing about logic:
Laws of Form (hereinafter LoF) is a book by G. Spencer-Brown, published in 1969, that straddles the boundary between mathematics and philosophy. LoF describes three distinct logical systems: The "primary arithmetic" (described in Chapter 4 of LoF), whose models include Boolean arithmetic; The "primary algebra" (Chapter 6 of LoF), whose models include the two-element Boolean algebra (hereinafter abbreviated 2), Boolean logic, and the classical propositional calculus; "Equations of the second degree" (Chapter 11), whose interpretations include finite automata and Alonzo Church's Restricted Recursive...
09:42
Hi guys
Let $0.5 < \rho \leq 1$, $r = 2^l$ , $l >= 1$ integer, and $y = \sum_{j=0}^{k-1} y_j 2^j$
assuming there's a quantity $\omega$ such that $ | \omega | \leq \rho r^{k+1}y$
How can I bound the number of bits to represent omega?
the bound I derived is given by taking the ceil of $\log_2 ( \rho r^{k+1}y )$
which it happens to be $(k+1)*(l+1) - 1$
is it correct?
10:19
$\int_{0}^{1} sin(x) dx^2 = ?$
actually what does this integral actually mean?
it means you're integrating to $x^2$
Or, to rephrase:
$$\int_0^1 \sin(\sqrt{u}) du$$
with $u = x^2$
ohk
which is $2(\sin(1)-\cos(1))$ I believe
Integral of sin is -cos ?
SBM
SBM
10:40
Yes why wouldn't it?
@BAYMAX
unless you have another function
yeah
SBM
SBM
Hello back
hello back ?
SBM
SBM
oh
11:00
$$\int \sin (f(u))du$$
Let $u=g(v)$ such that $f(u)=v$. Then $f'(u)du=dv$ and thus:

$$\int \sin (f(u))du = \int \frac{\sin (v)}{f'(g(v))}dv$$
Now suppose $\int \frac{\sin (v)}{f'(g(v))}dv = h(\cos v)$. This demands:
$\frac{\sin (v)}{f'(g(v))} = -h'(\cos v)\sin v$

$f'(g(v)) = -\frac{1}{h'(\cos v)}$
SBM
SBM
11:16
hmm
$\int_{0}^{x}(\int_{0}^{x} y(x) dx)dx = \int_{0}^{x} y(t) dt^{2}$ is this true ?
11:33
You probably meant:
$\int_{0}^{x}(\int_{0}^{x} y(t) dt)dt = \int_{0}^{x} y(t) dt^{2}$
Could someone help me with this: X is the random variable with the normal distribution with parameters m, s. Knowing that P(X<5.5)=0.72 and P(X>=3.2)=0.98 calculate P(4.1<X<5.2)
oh ya
11:56
Suppose $f(x)=\sqrt{x}$. Then

$f'(g(v)) = -\frac{1}{h'(\cos v)}$

$\frac{g'(v)}{2\sqrt{g(v)}} = -\frac{1}{h'(\cos v)}$

$\frac{g'(v)}{\sqrt{g(v)}} = -\frac{2}{h'(\cos v)}$

$\int \frac{g'(v)}{\sqrt{g(v)}} dv=2\sqrt{g(v)}= -\int\frac{2}{h'(\cos v)}dv$

$\sqrt{g(v)}= -\int\frac{1}{h'(\cos v)}dv$

Therefore
$$g(v)= \left(\int\frac{1}{h'(\cos v)}dv\right)^2$$
o.O
If you want to integrate $\sin(\sqrt{x})$, just substitute $u = \sqrt{x}$ to get $\int u \sin(u) du$ and then integrate by parts.
I am trying to understand why that substitution work. I know that it works and how it works
12:17
@MatsGranvik
@BAYMAX hi
Hi
it was with he double integral!
:)
1 hour ago, by BAYMAX
$\int_{0}^{x}(\int_{0}^{x} y(x) dx)dx = \int_{0}^{x} y(t) dt^{2}$ is this true ?
It's really confusing if you keep using $x$ on the left-hand side.
Either way, this is definitely not true.
Let $y(x) = x^2$. It's easy to see the left-hand side will be $x^4/12$ and the right-hand side will be $x^2/2$
12:36
@SteamyRoot There's a modified version of this statement that's true, I just can't figure out how to write it or what the conditions on the integrand are. For instance, this is the classic way to solve $\int_{-\infty}^{\infty} e^{-x^2}\;dx$.
yeah I too remember that similar thing
It may look similar, but it's very different.
Agreed.
I just want to know what the modified, true version of the statement actually is.
The idea is that you square the integral, and use that both integrals are independent
$\left(\int e^{-x^2} dx \right)^2 = \left(\int e^{-x^2} dx \right)\left(\int e^{-y^2} dy \right) = \int \int e^{-(x^2 + y^2)} dx dy$
and then you use polar coordinates
nice @SteamyRoot
12:39
Ah, I have it. $$\int_{a}^{b}\left(\int_{c}^{d}f(x)g(y) dy\right)dx = \left(\int_{a}^{b}f(x) dx\right)\left(\int_{c}^{d}g(y) dy\right)$$
Are there any additional conditions on $f$, $g$ besides integrability?
And are there any conditions on the bounds?
Well, integrability and the bounds are intertwined, of course.
Agreed.
If you assume that $f$ is integrable on $[a,b]$ and $g$ on $[c,d]$, then that statement is well-defined. And it's always true.
I meant stuff like "the interval has to be symmetric about 0" or something. You never know until you know, y'know?
Okay, great.
I don't see why that would even matter o.O
12:43
It doesn't to BAYMAX's question.
I just sincerely wanted to know.
Ohhhhhhhh I c what you were talking about
shuts up
I wonder if there are any functions for which BAYMAX' question holds.
$f(x) = \sqrt{x}$ comes "close", in the sense that the LHS and RHS only differ up to a constant...
:) dozing off ,will discuss later with you is that ok ?
I probably won't be online much today, but if I am, sure
SBM
SBM
13:49
Oh, lots of interesting things happened
For a partition $\omega$ of $n$, say $\omega = (n_1, \ldots, n_k)$ you could say the length of $\omega$ is $k$
So this author writes $l(\omega) = k$
Then a moment later he writes $|\omega| - l(\omega)$
What could he mean by $|\omega|$
Where is this from?
Hello everybody
SBM
SBM
Hello s.harp
@AlessandroCodenotti Nakajima' s Lecutres on Hilbert schemes
14:01
no idea
SBM
SBM
Hilbert schemes ?
I don't think explaining Hilbert schemes is relevant to the question tho
why is this true? gcd(a,b) = xa + by for some integer x and y
SBM
SBM
Are you sure it's true?
@SBM yep..
it is called Bezout's identity I just found out
Bézout's identity (also called Bézout's lemma) is a theorem in elementary number theory: let a and b be nonzero integers and let d be their greatest common divisor. Then there exist integers x and y such that a x + b y = d . {\displaystyle ax+by=d.} In addition, the greatest common divisor d is the smallest positive integer that can be written as ax + by every integer of the form ax + by is a multiple of the greatest common divisor d. The integers x and y are called Bézout coefficients for (a, b)...
anyone interested in tricky counting problems and the OEIS just let me know :)
14:29
@Lembik You can prove this by noticing that every subgroup of $\Bbb Z$ is of the form $k\Bbb Z$ for some $k$, so that $a\Bbb Z + b\Bbb Z$ must be generated by a single element. Once you prove that that element is $gcd(a,b)$, you're done.
@Fargle thanks
15:27
@Daminark @Astyx @Akiva Here you go (last one for a month). "The locksmith had great customer service, which he considered to be a ..." (3/2/7) OCUSEECTYSSK
Just curious, is there a way to uniquely (up to isomorphism) label the vertices of an undirected multigraph?
Anonymous
Anonymous
What does the second part mean?
Anonymous
How come the distance is $2/(1+|z|^2)^{1/2}$ when $z'=\infty$?
SBM
SBM
15:31
@blue Um pardon?
What do you mean with "how come"?
that's how it's defined ?
Anonymous
@SteamyRoot But why? I can't make any sense out of it....
Anonymous
Shouldn't the distance be infinite?
SBM
SBM
A function's only as good as you can define it
@TedShifrin Sorry if I'm intruding on a private conversation, but what is this about? It looks super cryptic
Anonymous
15:33
@SBM I'm pretty sure there's some logical reasoning behind the definition
@s.harp I'd imagine that's largely the point.
if the distance is infinite, you no longer have a metric.
Anonymous
@SteamyRoot What does a metric mean?
SBM
SBM
Metric spaces
Anonymous
I just started with complex analysis btw
15:35
A metric is the same as a distance
SBM
SBM
In mathematics, a metric or distance function is a function that defines a distance between each pair of elements of a set. A set with a metric is called a metric space. A metric induces a topology on a set, but not all topologies can be generated by a metric. A topological space whose topology can be described by a metric is called metrizable. An important source of metrics in differential geometry are metric tensors, bilinear forms that may be defined from the tangent vectors of a differentiable manifold onto a scalar. A metric tensor allows distances along curves to be determined through...
A distance between two points is not allowed to be infinite.
@TedShifrin I came here just for the Jumble. KEY TO SUCCESS
I'm off.
Either way, you can probably get the metric you have by using stereographic projections from the plane to the Riemann sphere...
Anonymous
@SteamyRoot I see. So how did that particular definition of distance come up when $z'=\infty$ ?
15:36
@TedShifrin Oooo that's a good one!
Way to not include me though
bad form
Anonymous
I mean there should be some logic behind it..
@BalarkaSen You bastard.
@blue That is likely the Euclidean distance of the north pole of the sphere to any other point.
SBM
SBM
@Dodsy Rude.
Anonymous
@PVAL-inactive I see. Interesting. So what is the use of defining it that way?
Anonymous
15:38
In case $z'\neq \infty$ it's taken as the distance between $z$ and $z'$
Also, I imagine $d(z,z')$ between $z,z' \neq \infty$ is not the standard metric on $\mathbb{C}$, or things will go wrong...
Anonymous
@SteamyRoot Wrong?How?
It won't satisfy the axioms of a metric...
Anonymous
I assume you mean "distance" when you say "metric"...?
Anonymous
@SteamyRoot What are the axioms of metric?
15:40
google stereographic projection.
SBM
SBM
check this
In mathematics, a metric or distance function is a function that defines a distance between each pair of elements of a set. A set with a metric is called a metric space. A metric induces a topology on a set, but not all topologies can be generated by a metric. A topological space whose topology can be described by a metric is called metrizable. An important source of metrics in differential geometry are metric tensors, bilinear forms that may be defined from the tangent vectors of a differentiable manifold onto a scalar. A metric tensor allows distances along curves to be determined through...
I aint doing ur thinking for you.
If you don't know the axioms, you probably shouldn't be trying to solve a question involving metrics.
Anonymous
@SteamyRoot I just started complex analysis an hour back :P I'm a noob
anybody can help with markov chains?
Anonymous
15:41
@SBM Thanks, will read that now
0
Q: Prove that limiting distribution is uniform

llamaDenote $S_n$ as set of all strings of length $n$ with elements from 4 symbol alphabet [A,G,T,C]. I have Markov Chain that picks one element in previous string $s_i$ uniformly and changes it to uniformly picked letter from alphabet. How can i prove that limiting distribution of this chain will b...

@Krijn I have found it and it was stupid
Don't start complex analysis before you have basic knowledge of real analysis (including metrics).
Anonymous
@SteamyRoot I am doing both side by side. Is that a bad idea?
I'd say so.
15:42
Change of variables 1D
Given known integrand of the form: $a(f(u))$ Find:

$$\int a (f(u))du$$

Let substitution be $u=g(v)$ such that $f(g(v))=h(v)$. Then $f'(u)du=h'(v)dv\Rightarrow du=\frac{h'(v)}{f'(u)}dv$ (If $\exists u,f'(u)=0$, then $\exists v, h(v)=C$)

Therefore: $\int a (f(u))du = \int \frac{a(f(g(v)))h'(v)}{f'(g(v))}dv= \int \frac{a(h(v))h'(v)}{f'(g(v))}dv$. Let $A(x)=\int a(x) dx$. Then using integration by parts

$$\int \frac{a(h(v))h'(v)}{f'(g(v))}dv=\frac{A(h(v))}{f'(g(v))}+\int \frac{A(h(v))}{(f'(g(v)))^2}f''(g(v))g'(v)dv=\frac{A(h(v))}{f'(g(v))}+\int \frac{A(f(u))}{(f'(u))^2}f''(u)du$$
Anonymous
I see...okay...will try to complete real analysis first
SBM
SBM
@Secret Oh
In short, change of variables produce the derivative of the inner function where inverse chain rule can act on, and this result in the outer function to be integrated, and the second integral made simpler because it goes down by one "unit" due to differentiation
"I started learning complex analysis an hour back"
Anonymous
@Dodsy ?
15:46
Just never heard that sentence before. :)
SBM
SBM
:(
What's wrong SBM?
Anonymous
@Dodsy 99.9% sentences you hear/see on this chat are ones you've never heard/seen before :P What's so novel about that?
Sorry, I didn't mean any harm.
The above formula also give us an idea when change of variables don't work. Note how the second integral is independent of what $g$ is, thus we can use that as a sanity check to see whether it can be integrated via a change of variables
Anonymous
15:49
@Dodsy I know :P
(Next step: Start milling down most of the integrals with integrands of the form $f(g(x))$)
SBM
SBM
@Dodsy I know that too I'm sorry
With the whole "real analysis vs complex analysis" first thing- I'd recommend following a college syllabus to see how they learn it in post-secondary institutions. This may help you if you're a high school or middle school student trying to figure out what to learn first and when. imgur.com/a/bcrLp
For instance in that link, you can see that you can learn third year real analysis with third year complex analysis
but you wouldn't take a complex analysis course with your first real analysis course.
the course numbering for that school is important, a course starting with a 1 is a first year course, a course starting with a 2 is a second, etc.
However, I was asking these questions just yesterday.
hey @Daminark
Anonymous
@Dodsy Ah, I get your point. I should at-least do two semesters worth of real analysis before I start with complex analysis. I'll keep that in mind. In real analysis I just started with posets and all.
SBM
SBM
oh I'm also learning analysis
Anonymous
15:56
@SBM You are in 11th or 12 th?
SBM
SBM
12th.
Anonymous
Nice :)
Anonymous
BTW I don't think you need it for school purpose :P
SBM
SBM
I don't need it for a school purpose but I find Mathematics interesting
Anonymous
That's great. I suppose you are through with differential and integral calculus?
SBM
SBM
15:59
Somewhat? Haven't done a lot of multi-variable integration, but as far as the school syllabi is concerned; almost.
Anonymous
Cool!
Is it always true that the value of a metric is a finite number; i.e., $d(x,y) < \infty$ for every pair of points $x,y$ in the metric space?
A metric is defined to be finite.
@SteamyRoot Thanks!
16:18
I hope people who ask questions and delete them as soon as you answer all die in painful and slow ways
D:
hi chat
@JorgeFernándezHidalgo My preferred epithet for such people: youtube.com/watch?v=ZfYFx6MOTYU
Hey
On an entirely different note, I find questions like this quite irritating: math.stackexchange.com/q/2304384/137524
Whether or not it's intended, it reads to me as "Please do my HW. Thanks, bye!"
Lmfao
I posted a homework question on here before.
But I fully disclosed that it was a homework question and posted all of my work
16:28
Sometimes homework is interesting
eh, it wasn't very interesting.
It was a systems of equations question.
It's the latter point that's important, I think.
I hate any question with a photo on it
It depends on the photo for me.
unless its a tikz or pgfplot made by the same user
16:29
If it's a photo of a diagram, I'm more or less okay with it.
If it's of equations, I hate it.
well, only if it is accompanied by a mathjaxed question
I mean if you want to understand how something works then ask a question, it doesn't matter if its homework or not
its sort of important that the spirit of the question is "I want to understand this, how can I understand this"?
To me it's a matter of due diligence.
rather than "I need to solve this, please solve this"
Exactly.
16:30
@s.harp yeah, but you should distil the question properly and try to make it interesting
Stackexchange is notorious for answerers to tear the askers a new a-hole if they don't comply with the social norms of the board, which is alright.
stackoverflow is way way harsher though
Oh I believe it.
If you've got something you don't understand, that's fine. Just make clear what you think you understand
I think its funny, imho mathoverflow is more lenient than stackoverflow but the people at mathoverflow are like uber geniuses and on stackoverflow most of the downvoters don't even know how to help you
16:31
I guess I should join mathoverflow then?
;)
Well, MathOverflow is just more likely to say "this isn't for here"
Physics.SE is apparently very harsh on HW questions.
well yeah, but most of the questions that are too hard for here, even if not so much "research related" seem to do well on MO
It depends on the type of problem, but yes.
**Example**

$$\int e^{x^2}dx$$

Using Formula No. 1 with $a=e^{\cdot},f=\cdot^2$

$$\int e^{x^2}dx=\frac{e^{x^2}}{2x}+\int\frac{e^{x^2}}{2x^2}dx$$

$$2\int e^{x^2}dx=\frac{e^{x^2}}{x}+\int\frac{e^{x^2}}{x^2}dx$$
I find most questions on MSE hard for me to understand.
16:33
My criterion for MO questions is whether or not I need to provide references to understand what's going on.
If I can just link Wikipedia or Mathworld, then probably it should stay on MSE.
If it requires citations of actual articles, though, then that's getting into MO territory.
(Textbooks are a boundary case.)
like this question: https://math.stackexchange.com/questions/2304400/why-doesnt-derivative-show-the-complex-maxima-minima

At first I'm nodding like "right, right" then I get to the end and I'm like wait what is this even asking.
Well, there's hard to understand because of content and hard to understand because of how it's been written.
A lot of the time on MSE it's the latter.
That may be the case.
Which again goes to due diligence: Have you made the effort to clearly articulate what you understand/don't understand?
I think you save yourself a lot of annoyance if you stick to reading a few tags
16:37
That too.
I almost never open questions from the front page
If the tags are (obviously) wrong, that's not an encouraging sign.
Oh that's a good suggestion.
And it's also another due-diligence thing, whether you've spent the moment to figure out three relevant tags.
As an example of one which doesn't pass my "due diligence" threshold, see this one: math.stackexchange.com/q/2304397/137524
lmfao yeah that's pretty crazy.
16:40
By contrast, this one does pass my 'due diligence' threshold even though I think they could say more: math.stackexchange.com/q/2304371/137524
I don't know how to answer either of them, but at least with the second it leaves me wanting them to get an answer.
Hi, sorry can i ask a question here ? maybe it's be so elementary but i don't know
Yes.
As per the room description: "Just ask, don't ask to ask."
I always wondered what that part "rarely if ever expressible as a ratio of integers" means.
Rarely rational.
i.e. we're a bunch of weirdos :P
16:47
Oh that's a good one!
Def1) A map $X : \Omega \to \Bbb R$ is $A$ – $\mathscr {B}(\Bbb R)$-measurable if and only if
$X^{-1}((−\infty, a]) \in A$ for any $a \in \Bbb R$.

Def2) A map $X : \Omega \to \Omega'$ is called $\mathscr{A}$ –$ \mathscr{A'}$-measurable (or, briefly, measurable) if
$X^{-1}(\mathscr{A'}) := \{X^{-1}(A') : A' \in \mathscr{A'}\} \subset \mathscr A$; that is, if
$X^{−1}(A') \in \mathscr{A}$ for any $A' \in \mathscr{A'}$.

I think that def2 is general case but we now $\mathscr {B}(\Bbb R)$ is $\sigma((−\infty, a] : a\in \Bbb R)$ so $(−\infty, a]$ form intervls are a part of $\mathscr {B}(\Bbb R)
Formula No. 1
Let $u=g(v)$ and $A'(x)=a(x)$ Then
$$\int a(f(u))du=\int \frac{a(f(g(v)))(f(g(v)))'}{f'(g(v))}dv=\int \frac{a(f(u))f'(u)}{f'(u)}dv$$
$$=\frac{A(f(u))}{f'(u)}+\int \frac{A(f(u))}{(f'(u))^2}f''(u)du$$
Ah, measure theory stuff?
Then I know for sure that I can't help :/
SBM
SBM
befuddled
@Semiclassical yes measure theory :(
16:49
Now... Let $a=e^{\cdot}, f=\cdot^2$. Repeatly applying Formula No. 1 gives:
@W.R.P.S I like your archimedes quote on your profile.
hey @MikeMiller
@Dodsy Thanks :)
@Semiclassical Got a minute?
Go ahead.
0
Q: Generalized divergence theorem over vector field

Lozansky Show that $$\iint_S (\mathbf{A} \cdot \hat{n}) \mathbf{B} \:dS$$ can be transformed into a volume integral over the volume enclosed by the the surface $S$. $\hat{n}$ is the outwards pointing normal for $S$. What requirements must the vector fields $\mathbf{A}(\mathbf{r}),\mathbf{B}(\mathbf{r})...

I'm just wondering if that last step is okay
16:54
@W.R.P.S you can prove that the set $J$ consisting of all sets $s$ such that $X^{-1}(s)\in A$ is a sigma algebra.
More specifically, I'm unsure if I can go from $\mathbf{A} \cdot \nabla B_i$ to $(\mathbf{A} \cdot \nabla) \mathbf{B}$
so if $J$ contains a set $W$ then $J$ contains the sigma algebra generated by $W$.
this is what is happening in your case
Well, you should just need to multiply by $\hat{e}_i$ and resum
in this case $W$ is $\{[a,\infty) | a\in \mathbb R \}$ and the sigma algebra generated by $W$ is $\mathscr B(\mathbb R)$
SBM
SBM
sigma algebra?
16:57
isnt that how it is said in english?
@Lozansky To put it a little differently, you could have written that first step as $$\int_S(\mathbf{A}\cdot \hat{n})\mathbf{B}\,dS=\left(\int_S (\mathbf{A}\cdot \hat{n})B_i \,dS\right)\hat{e}_i$$
SBM
SBM
@JorgeFernándezHidalgo I'm sorry I do not know about that yet.
bye chat

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