2:53 AM
Do all the us school record their math classes?
I have seen so many videos about them and it is unusual in my school at least to math
Many of them even have prepared printed note for distribution.

3:50 AM
@oscarmetalbreak Absolutely not.
My 112 YouTube lectures are there only because a few students wanted to do the work to do it.

hmmm, maybe it was because there are too many US schools

There’s also a ton of crap on the internet.

I only watch the recording of lecture of other schools, I am not very interested into youtuber though

There are lots of university lectures on YouTube.

2 hours later…
5:56 AM
most or at least a majority of the lectures that are part of the standard syllabus get recorded here, but they're just for internal use and not uploaded to YT

6:25 AM
I imagine Covid and remote classes changed lots of things, and I’m an out-of-date fossil.🤷‍♂️

covid, remote classes, electricity, the printing press, indoor plumbing

1 hour later…
7:57 AM
Uniformly integrable condition seems very good in martingale theory.

why???
I think this is plain wrong.
I see that Jakobian is also not here.

1 hour later…
9:23 AM
0

In $R^2,$ let $L$ be the line $y = mx,$ where $m\neq 0.$ Find an expression for $T(x, y),$ where $T$ is the projection on $L$ along the line perpendicular to $L.$ (See the definition of projection in the exercises of Section $2.1.$) This was a problem given in the book, "Linear Algebra" by Steph...

Hey guys! I need some help with this, please.

9:49 AM
thomas: in your question, you write: "Complying with this definition, I considered the subspace of R^2 consisting of all the points in line L and the subspace consisting of all the points lying on the line perpendicular to L" (emphasis added), which makes sense, as the book only defines the projection on W_1 along W_2 for complementary subspaces W_1, W_2 of a vector space.
you then begin talking about the set of all lines perpendicular to L. this is where you get distracted. there are indeed infinitely many lines perpendicular to L, but only one of them is a subspace of R^2 - the one going through the origin.
the exercise is asking you to consider that line only, and not any of the other ones.

Given any set in R^2, can it always be written as a product of two sets in R?

koro: are you asking if a subset of R^2 must be a cartesian product of subsets of R^1 (obviously not)? or some other thing?

yes

when A has more than one element, the "diagonal" {(a,a): a in A} is a good example of a subset of A x A that is not a cartesian product of subsets of A

@onepotatotwopotato Vitali's convergence theorem?
@Koro Fubini

9:58 AM
@leslietownes thanks.

@Koro no

@Jakobian still the image is wrong?
2 hours ago, by Koro
Shouldn't it be the set {(w,x): 0<= x< X(w)} here instead?

ah yes. Thats a typo
good eye
Also technically I should have said Tonelli

Why I asked it was because prior to that image, this was an exercise.
and this was used there in image before that somehow.

what's an improper random variable?
@Koro yeah... measurability issues

10:03 AM
So it's a measurable function $X:\Omega\to \bar R$ such that $P(X^{-1}\{\pm \infty\})=0$

I see.

But even without this exercise, I can prove it, I think.

When using Tonelli you ought to check the functions involved are measurable
thats why they ask you to prove this, because otherwise integrals wouldn't make sense
technically for the above it should be $0 \leq X(\omega) \leq x$ I guess but either way

$E(X)= \int X(w) P(dw)=\int \int_0^\infty 1_{0\le x\lt X(w)} dx P(dw)=\int_{\{(w,x): 0\le x\lt X(w)\}} P(dw)\otimes dx$.

The last inequality is strict

10:08 AM
The set $\{(w,x): 0\le x\lt X(w)\}=\cup_{r\in Q^+} (X^{-1}([r,\infty))\times [0, r))$ is measurable in the product space.
so I can apply Tonelli's.
2 hours ago, by Koro
which gives me the last equality here.

@Koro yep
perfect
Of course its also true that $E(X) = \int_0^\infty P(X\geq x)\mathrm{d}x$ when $X\geq 0$

indeed, X is non negative here. (I forgot to mention that)

yeah. I didn't question that you didn't mention it, I just assumed it because I know the problem

thanks a lot.
my food is getting cold 😅

There is a more general version of this theorem which goes something like $E(f(X)) = \int_0^\infty P(X\geq x) f'(x)\mathrm{d}x$
Try that one too. The proof is the same way
well, you need some assumptions on $f$ but you will know what they are once you try to prove it. It should work for $f(x) = x^p$

10:15 AM
ok.
I know that $E(f(X))= \int_0^\infty f(x) d(P\circ X^{-1})(x)$

@leslietownes For reasons similar to this, in the product topology $X\times Y$, not every open set is of the form $U\times V$ for open $U, V$

10:56 AM
If $G$ is a Lie group and $G\to G$ is a group homomorphism, then it's automatically a diffeomorphism?
@Jakobian I don't know if they're related.

@onepotatotwopotato There exist homomorphisms $f:\mathbb{R}\to \mathbb{R}$ which are non-measurable
However note that this question implies that continuous homomorphisms are smooth
Of course they don't have to be diffeomorphisms... idk why you're asking for that

oh group isomorphism not just homomorphism
I wonder $Aut(G)$ is also a Lie group where $Aut$ is a group of group automorphisms

The homomorphisms $f:\mathbb{R}\to\mathbb{R}$ by construction are isomorphisms

does that map maps $0$ to $0$?

yes? Its a group homomorphism after all
$f(x+y) = f(x)+f(y)$ automatically implies $f(0) = 0$

11:12 AM
I mean if $f(1) = a$ then $f(x) = ax$ isn't it?

No
Why would it be
$f(x) = ax$ only for rational $x$

hmm
what if I add continuity condition? continuous group homomoprhsm
still not diffeo?

well, I already said that this implies smoothness

Continuity of the inverse should follow from invariance of domain
I'm not sure if this holds for every topological group, let me see

11:16 AM
so if $G$ is a discrete Lie group then $Aut(G)$ is a Lie group

@onepotatotwopotato I think you mean continuous group isomorphism here?

@Jakobian thats fair, Im totally ignorant to those subfields

@onepotatotwopotato what is meant by $\text{Aut}(G)$?

underlying group automorphism group

automorphisms in what sense
in what category

11:20 AM
group

Hmm.. I see. Well, if you have some construction that allows you to equip the set of continuous automorphisms of a Lie group with a structure of a Lie group, I see no issue

i recently came across a neat result, if a function $f : U \subset \mathbb{C} \rightarrow \mathbb{C}$ is an open map ($U$ is open), and has partial derivatives a.e. (in both $x$ and $y$ directions), then its also differentiable a.e.
the same result holds in any dimension

@porridgemathematics There is no people that know everything. I'm ignorant at many things too, so I often try to not formulate an opinion about subjects and things I know nothing about. Most people don't even notice when they're doing this.

@Jakobian what exactly are you asking whether it holds for all topological groups here?

@AlessandroCodenotti That continuous isomorphism implies homeomorphism, but I googled it so I know its not true and the example is pretty trivial
so invariance of domain was pretty important

11:26 AM
@Jakobian yeah, I knew I was ignorant to the minutiae of what professional general topologists do, Id like to think I had some understanding of what it might look like though

I mean. Its fair to assume that all we do is some bizarre set theory since a lot of it is that

@Jakobian Ah yes there's many counterexamples

@porridgemathematics differentiable in what sense? Complex differentiable?

I am struggling to see why the method of lagrange multipliers works, I am looking at this figure en.wikipedia.org/wiki/Lagrange_multiplier#Single_constraint.

@Jakobian in the R^2 sense

11:31 AM
Oh I see. I'm pretty sure this is weaker than complex differentiable?

oh yeah, a lot weaker
complex differentiable is R^2 differentiable plus the derivative matrix is a scaling + a rotation
so its a much smaller class of functions

I agree that as you move along the constraint $g=c$ either you cross a contour of $f$ or you are tangent to a contour of $f$. However I dont see why $f$ is always at a maximum on the constraint when they are tangent. e.g. this example on desmos (the blue curve is the constraint) desmos.com/3d/80883fd44d

something's wrong

12:15 PM
0

I am looking for trig identities of the form $\sin(a+b)^{2k} = \sum a_i \sin^{2 n_i}(b_i a) \cos^{2 m_i}(c_i a) \sin^{2 l_i}(d_i b) \cos^{2 j_i}(e_i b)$ where $k,n_i,m_i,l_i,j_i$ are integers $> -1$ and $a_i,b_i,c_i,d_i,e_i$ are rational numbers$>0$. Im not sure they exist for all $k$. For insta...

Perhaps a hard question ?

@mick maybe Euler formula can help

12:37 PM
Is there any name for $\Bbb{R}_k\times \Bbb{R}_k$?

@SoumikMukherjee whats $\mathbb{R}_k$?
@mick You need to make clear where the exponent are, and that this isn't composition
to make sure if we're exponentiating the function, the argument, or what

@Jakobian R with the k topology

whats k topology
@SineoftheTime I've noticed that you were browsing questions about sequentially closed sets. Maybe I should add that "sequence has too few points" is only one possible reason for why sequences are not enough. If you only consider nets indexed by ordinals, this might still not be enough to describe the convergence in a topological space.
Even though, you could say, such nets have as many points as you would want

Why do you expect it has a name

Well as R_l×R_l has a name so..

1:09 PM
@Jakobian I was looking at the proof that $A^{\perp}$ is closed using a sequence in $A^{\perp}$ and proving that its limit point is in $A^{\perp}$. then I've seen that sequentially closed does not imply closed

1:25 PM
I struggle to see why this is true, I tried to prove inductively but I am pretty much stuck at the step
Does someone can see a convenient way to approach this?

@SineoftheTime Is this is a normed space?
I've seen proofs on this site about normed spaces for weak/weak* topology that indeed shouldn't use sequences. Seems to be a common mistake

Yes, $A$ is in a Hilbert space

1:44 PM
If a function $f:\Bbb R\to\Bbb R$ is smooth everywhere except a point, then its weak derivative equals the classical derivative except that point?

You can check by hand that a function defined to agree with the derivative where the latter is defined (and to be whatever on the point of nonsmoothness) satisfies the definition of weak derivative

@onepotatotwopotato If $c\in [a, b]$ is the point where it lacks smoothness, then you can consider some small neighbourhood around $c$, outisde of it everything will work fine, inside of it, the integral can be made arbitrarily small

finitely many points doesn't affect the integral, so I can just perform integration by parts if the derivative already exists so the weak and the classic are the same?

basically yes, you can move the $\partial$ operator to $f$ as long as you are integrating over a region where $\partial f$ makes sense and you can apply Stoke's theorem on

@onepotatotwopotato No?

1:56 PM
so you can do integration by parts on $[-1,-\epsilon]$ and $[\epsilon ,1]$
(here assuming $0$ is the singular point)

Well yes. The integration by parts where $f$ is actually smooth, but you still need to isolate the bad points

then let $\epsilon \rightarrow 0$ and you get what you want

yes two people are saying the same thing

2:11 PM
@X4J Try row reduction

@X4J hint: consider two cases 1) A_1 is full tank. 2) A_1 is not full rank.

2 hours later…
4:43 PM
12

As you wrote, you have already proved $\lim_{n \rightarrow \infty} \mu^*(A_n) \leq \mu^*(A)$. It remains to show that $\mu^*(A) \leq \lim_{n \rightarrow \infty} \mu^*(A_n)$. Your idea to complete the proof is essentially correct, all it needs is a small adjustment. Given any $\epsilon > 0$. We...

so tricky

1 hour later…
6:03 PM
how is this inequality proven $|a+b|^n\le 2^n(|a|^n+|b|^n)$, n>0 is not necessarily an integer?
I forgot how it is proven.
23

For any real numbers $a$ and $b$ and $1 \leq p < \infty$, prove that $$|a+b|^p \leq 2^p \{ |a|^p +|b|^p \}$$ This inequality is given in the the book Real Analysis by Royden, Chapter $7$, page $136$. I don't understand how the author comes to this inequality. Can anyone provide some hints?

Maximize $(a+b)^p$ subject to constraint $a^p+b^p=1$?

it's an application of triangle inequality.

ted do you have $5$ minutes to check whether my solution is correct or not? It's an exercize on Laurent series

What are you doing with Laurent series?

@TedShifrin Feeding it to my pet tarantula?

6:11 PM
The tarantula may get an upset stomach.

Possibly.

$f(z)=1/\sin(z)$, I have to find the Laurent decomposition of $f$ on the discs $\{n\pi<|z|<(n+1)\pi \}$ for $n\ge 0$. By Laurent decomposition I mean $f=h+g$ where $h$ is analytic for $|z|>n\pi$ and $g$ is analytic for $|z|<(n+1)\pi$
I'll write what I've done
$1/\sin z=\frac1{z-z^3/3!+z^/5!+\dots}=\frac1{z(1-z^2/3!+z^4/5!+\dots)}=\frac1z (1+(z^2/3!-z^4/5!+\dots)+(z^2/3!-z^4/5!+\dots)^2+\dots)$

So you're doing the case $n=0$? I'm fine with just doing that, anyhow. The general case is immediate from that.

I choose $h=1/z$ and $g=\frac1z ((z^2/3!+z^4/5!+\dots)-(z^2/3!+z^4/5!+\dots)^2+\dots)$
@TedShifrin that is my doubt, I don't know if it is valid only for $n=0$ r for all $n$

Your $g$ has a removable singularity at $0$ and is analytic everywhere, no?

6:18 PM
yes

So how do we see that there's a pole at $\pi$?

because $\pi$ is a simple zero of $\sin z$, so it is a pole of $1/\sin z$

I know that, silly. How do we see it from your series?

it's analytic, so there isn't but $f$ has a pole. So this representation is valid only for $0<|z|<\pi$

6:24 PM
So I guess to find the general case I have to expans around $n\pi$

If we think of this as $\dfrac1{z(1-h(z))}$, like $\dfrac1{z(z-1)}$, how would we look at it?
No, the expansion is still around $0$. The annulus is centered at $0$.
So how do we do the Laurent series of $\dfrac1{z(z-1)}$ on the annulus $0<|z|<1$?

using the geometric series

Partial fractions first.

@Koro Apply Jensen's inequality to $(a+b)^p\leq 2^{p-1}(a^p+b^p)$, $a, b \geq 0$

6:27 PM
So does that generalize to $\dfrac1{z(1-h(z))}$?

@Jakobian p>1 is needed here, I think. But in my case, p>0 works as well.

Partial fraction and then geometric series

using triangular inequality

For $0 < p < 1$ we have $(a+b)^p\leq a^p+b^p$
So the more optimal constant is $c_p = 1$ if $0 < p < 1$ and $c_p = 2^{p-1}$ if $p\geq 1$
You never said you want the inequality for $p > 0$, that's not my fault but yours

No, I guess my idea doesn't work.

6:29 PM
ohh I did say that.
26 mins ago, by Koro
how is this inequality proven $|a+b|^n\le 2^n(|a|^n+|b|^n)$, n>0 is not necessarily an integer?

Yes, I will defend Koro's honor in this one case.

the point is that I don't know how to modify it to generalize it since $h$ and $g$ I found work for $0<|z|<\pi$

You were using $n$ instead of $p$, and the answer you cited wanted proof for $p\geq 1$
either way

that's not the point

Assume $b\neq 0$ and divide to obtain $(a/b+1)^p\leq (a/b)^p+1$. Prove that $f(x) = (x+1)^p-x^p-1$ is increasing and $f(0) = 0$

6:32 PM
you replied to my question (1st comment)

@Sine I was thinking we just use $\sin(z+2\pi) = \sin z$, but then we end up with an annulus centered at $2\pi$, not at $0$. Interesting problem.
We might have to think about the integral representation of the Laurent coefficients.

with the integral formula?

Yeah. Integrate on a circle of radius $(n+1/2)\pi$.
Fascinating question.

@Jakobian Correction, $f(x) = x^p+1-(x+1)^p$ is increasing. For this you simply state that $f$ is continuous and $f'(x) = p(x^{p-1}-(x+1)^{p-1})\geq 0$ for $x > 0$

ted maybe this can help

6:38 PM
Oh yeah, and of course I was stoooopid to say that $g$ converges everywhere. We of course need $|h(z)|<1$.
Yes, that answer is the way to do it.
It amazes me that I've never seen this problem in any of the books I have taught out of ....

so in general $h=\frac1z-\frac1{z-n\pi}-\frac1{1+n\pi}$ and $g=f-h$

No, to use their method, you have to subtract out all the principal parts within the disk.

Make sense? (I wouldn't want to write this mess out for general $n$.)

to be honest, it's still a bit uncler. Let me reread the answer

6:47 PM
You want to remove all the poles within the disk of radius $(n+1)\pi$ to get something analytic in $|z|<(n+1)\pi$.

ok, I get it now. Thank you

Thanks for teaching me something :)
2

Suppsoe that $X_n\to X$ a.s. and suppose that $X_n'\to X$, then $P(\{w: X_n(w)=X_n'(w)\})=1$ for all $n$. How to prove this?
Let's write $\{w: X_n(w)\ne X_n'(w)\}=:E_n= \{X_n\ne X_n'\}$ for brevity. I want to show that the probability measure of E_n, i.e., $P(E_n)=0$.
but no idea how to do it

Why would this be true
Its not true

6:54 PM
Because if $P(E_n)=0$, then $P(E_n^c)=1$.

For every real number there is multiple different sequences converging to it
so why would this be true?
missing context

I see.
I probably misunderstood that statement.
it is to be proven that $P(\{w: X_n(w)= X_n'(w) \text{ for all$n$}\})=1$ and I wrote it differently above.

its not true

@Jakobian yes, but here we have a.s. convergence in $X_n\to X$.
that is, P({$w: \lim X_n(w)= X(w)$})=1

See my earlier comment

7:00 PM
and then we have pointwise convergence in $X_n'\to X$.

Doesn't matter

ohh

Consider what if $X_n, X_n', X$ are constant and the convergence is in real numbers

yes, you're right, thanks.
But what I can say is there exist such $X_n'$ such that $X_n'\to X$ (. wise) and that $P(\{X_n=X_n'\})=1$.

That you can do

7:05 PM
Let $E:=\{w: \lim X_n(w)= X(w)\}$. On $E$, I define $X_n'(w):= X_n(w)$.
Since P(E)=1, it doesn't matter how I define $X_n$' on E^c.

It does matter since you want $X_n'\to X$ pointwise

ohh right.
So I define $X_n'(w):= X(w)$.
for all $n$ and for all w in E^c.

@Koro Do we assume that $X$ is measurable?
Okay we are most likely assuming this

7:35 PM
@Jakobian yes, X, X_n's are all random variables.

1 hour later…
8:35 PM
2

Prove that if the derivative $f'(x)$ of a function exists on the measurable set $E$, then $f'(x)$ is measurable on $E$. We are told to only consider 1 dimensional spaces,that f is a measurable function in one variable.that is, f is a measurable function in one variable. Following is my s...

I think this can be solved in the following way
1) Show that for arbitrary $f:[a, b]\to\mathbb{R}$ the upper $\overline{D}f$ and lower $\underline{D}f$ derivatives are measurable.

@Jakobian yeah
or using \liminf or \limsups

2) Conlcude that $\{x : f'(x) \text{ exists}\} = \{x : \overline{D}f(x) =\underline{D}f(x)\in\mathbb{R}\}$ is always measurable.
3) Conclude that $f'\restriction_E$ where $E\subseteq \{x : f'(x)\text{ exists}\}$ is measurable for measurable $E$
Since $f'\restriction_E = \overline{D}f\restriction_E$, a restriction of a measurable function to a measurable set
@Koro Unless I'm missing something, lower and upper derivatives are defined using those
Its almost unbelievable to me that the lower and upper derivatives are always measurable
could someone confirm this
I will confirm it soon if no one does
I'm not sure if assuming that $E$ is measurable is important here
Pretty sure there is nothing preventing $E$ to be non-measurable

9:04 PM
@Jakobian yes. I wasn't sure of the names so I said \liminf \limsup.
@Jakobian yes, \liminf of measurable functions is measurable.
\limsup too

@Koro I never said $f$ is measurable

with the assumption that f is measurable.
Ohh, but f is differentiable on E so continuous.
hence measurable?

That shows $f$ is measurable on $E$, sure

yeah, so \lim(inf/sup) are measurable.

But we want $f'$ to be measurable on $E$
No?

9:10 PM
Do you have Folland's right now?

No
Just tell me what it says

it says that if $f_j$ are $\bar R$ valued measurable functions, then $\sup_j f_j$, $\limsup_j f_j$ are also measurable.

How is that relevant
To define $f'$ you need values outside of $E$
so that $f$ is measurable on $E$ doesn't help
neither does this theorem

because if f is differentiable at x, then f'(x)= $\lim g_n(x)$, where $g_n(x)= \frac{f(x+1/n)-f(x-1/n)}{(2/n)}$

9:15 PM
but f may not be defined at x+1/n?
there is some fix to it, I don't remember it right now.

There are far greater issues
You're trying to assume $f$ is measurable - your approach doesn't work

f is measurable!!

$f$ is not measurable

Because f is differentiable hence continuous.

$f$ is not differentiable

9:17 PM
f'(x) exists on E, no?

ONLY ON $E$

hmm
and E may not be an interval

$f$ is arbitrary function
$f\restriction_E$ is continuous, so measurable. But $f$ doesn't have to be
so your limit doesn't have to be a limit of measurable functions

@Jakobian yes, right.

1 hour later…
10:35 PM
hi all
0

Let $x,y,z$ be real. Consider a scalar field $$z = f(x,y)$$ More specific; $f(x,y)$ is a (given) real polynomial in $x,y$ of degree at most $5$ such that For all $x,y$ $$f(x,y)> 0$$ For a given pair of points $(a,b),(c,d)$ ($a,c$ are the x-part and $b,d$ are the y-part) consider the line integral...

10:55 PM
0

If $f:[a, b]\to\mathbb{R}$ arbitrary function, then the $$\text{upper derivative }\overline{D}f(x) := \limsup_{y\to x} \frac{f(x)-f(y)}{x-y}$$ and $$\text{lower derivative }\underline{D}f(x) := \liminf_{y\to x} \frac{f(x)-f(y)}{x-y}$$ are always measurable. Thus the set $E = \{x\in [a, b] : \over... here's my answer that the set of points where a function is differentiable is measurable, and the derivative on that set is measurable too It uses Vitali's covering theorem The tricky part is$\bigcup_n \bigcap_k E_n^k\subseteq E$. For this write$\frac{F(d)-F(c)}{d-c}$as a convex combination of$\frac{F(d)-F(x)}{d-x}$and$\frac{F(c)-F(x)}{c-x}\$

11:17 PM

Is this question clear ?
Is it hard or easy ?
true or false ?

it sounds like you are looking for a path with minimal potential between two fixed points which indeed is calculus of variation as ted indicated. Maybe you can check gelfand's calculus of variations.
Why you have the limit to the maximum degree of the polynomial being five?

11:34 PM
@mick unclear
First paragraph is unclear

what is unclear ?
its like octions or so

11:57 PM
Would the path be minimal potential in topology equivalent space
A rectangle is ismorphisc to a torus
My guess is not the minimal potential seems to depend highly on the path
Since I just studied algebraic topology, just wonder if there anything I can play with it