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00:04
Any suggestions for a book about ODE's? I've Arnold's book but I don't know if it's appropriate for a basic course :/
@TedShifrin When you say "take an inner tubular nbhd of $\partial{U}$ in $U$, this means to me to take an open subset $Z$ such that $\partial{U} \subseteq Z \subseteq U$. Which implies $\partial{U}\subseteq U$, but $U$ here is open.
00:20
No, $Z\subset U\cup \partial U$. Get the picture in your mind and then make it work.
@SineoftheTime one dimension or systems, @Sine?
@TedShifrin let me check the programm of the course
Ah, so just take $Z\subseteq U\cup \partial{U}$ that misses $K$?
Is hospital/doctor anxiety a thing?
I was shaking having to see a doctor for a normal routine checkup. Not sure why
@ted -Motivation and examples. Basic definitions (ordinary differential equations, Cauchy problem). Basic general theory: local existence and uniqueness of the solution under Lipschitz hypothesis by the method of Picard's iterations.
- Gronwall's Lemma. Maximal solutions and extension of solutions. Continuous dependence on the initial datum.Maximal solutions and extension of solutions.
- Basic resolution methods for differential equations in normal form: separable variables equations, first order linear differential equation, 1-forms and differential equations. Bernoulli's equation. Differ
Even scarier when I get asked questions about medical history
00:23
-Qualitative study of an equation: comparison theorem, monotonicity theorem and asymptote theorem.
- Existence for the Cauchy problem in the case of continuity. Ascoli-Arzelà Ascoli theorem. Peano's theorem.
- First order linear systems: exponential of a matrix and Jordan normal form. -Linear differential equations of order n. Characteristic polynomial. Euler equation. Method of variation of constants. D'Alembert method for the reduction of the order.
00:35
Anything can become an anxiety if you think about it too much.
hades, totally a thing
@leslie any suggestion for a good book about ODE?
Test anxiety has its roots in fear of failure.
arnold's book is pretty good. what do you want? in a lot of universities, the first course with "ODE" in the title is just a mishmash of recipes to memorize with no theory. and you actually need quite a lot of prerequisites to work out meaningful theory.
I've written the program
00:45
@SineoftheTime This is a single-variable ODE course, not systems. Birkhoff & Rota is a classic text that's more theoretical than the typical books used in the US. These days, many US ODE courses are more centered on systems of ODEs (and the interplay with linear algebra).
yeah, birkhoff and rota, that's the other good one.
thank you both :)
The end of your course is systems, but it's not the bulk of the course. For that material, there's very little better than Hirsch and Smale (and new edition with Devaney, and I don't know that version).
@leslietownes Part of it could be because I'm just not good at explaining/talking but another part of it felt like I was asked by the professor to solve a problem on the board, which I can do quite well if its mathematics but I have almost 0 medical knowledge
no, anxiety around medical care is a huge and normal thing. my mom was a nurse. they take 'vital signs' like pulse and blood pressure, and write down whatever those are, but the numbers they get are often discounted mentally because people are generally somewhat worked up from being in a situation where a nurse is involved.
00:53
My BP in any doctor’s office is always too high.
they definitely know that. one of the weird things where a measurement is taken where everybody knows it might be off by 50%.
And I don’t have anxiety other than about my BP.
in our wonderful USA you can also pile economic anxiety on top of whatever else might be going on in a medical situation.
19 mins ago, by user 85795
Anything can become an anxiety if you think about it too much.
Isn't that how the brain works.
I have a rank 16 integral lattice that I know is isomorphic to $E_8\oplus E_8$ (in particular it's definite), is there an efficient way to compute an explicit isomorphism? The Gram matrix is pretty sparse, but PARI/GP is struggling quite a lot
00:59
think obsessively worry about it, actually
@TedShifrin my doctor measured my heart rate, looked at me and laughed slightly before she told me to calm down. I haven't felt more embarrassed ever since I stepped on a banana peel, and landed face first in mud
@TedShifrin I really am stuck and confused. So, just to repeat. $f=g$ on $U-K$, where $f:W\rightarrow X$ and $g:W\rightarrow X$. Define $F$ by $f(x)$ on $W-U$, $g(x)$ on $U$. We want to show $F$ is continuous.
Where $U\subseteq closure(U)\subseteq W$.
You claim that $f=g$ on a neighborhood of $\partial{U}$ in $W$ and therefore that implies continuity of $F$.
@Koro
Proof this:
-1
Q: A factorial modular equation form of the twin prime conjecture.

D Left Adjoint to U Conjecture. Let $f_q(X) = X(X - 2)(X - 3)\cdots (X - (q - 2)) = X \dfrac{ (X - 2)!}{(X - q + 1)!}$ be a function from $I = [p_n + 2, p_{n+1}^2 - 2]$ into $\Bbb{Z}$ as $q$ ranges over $2..p_n$. Then there exists a solution $X \in I$ such that: $f_q(X) = 0 \pmod q, \forall q \in\{2, .., p_n\}$ Th...

Sorry, not $g:W\rightarrow X$. I meant $g:U\rightarrow X$
@leslietownes when I was told I'm gonna need a laser surgery to correct my vision, I was visibly shaking and started feeling nauseous, the idea of a surgery is like a nightmare for me. The specialist actually had to give me a mini-medical lesson on how the surgery works in order to get me to calm down before I head home
01:10
I don't see (1). Why $f=g$ in a neighborhood of $\partial{U}$ in $W$ and (2). Why (1). implies continuity
of all of $F$.
I’ve been through 3 major surgeries and am headed for a 4th. They were all superbly done. No worry.
The two functions agree on the overlap because they agree off $K$. @monoidal
the laser eye stuff is extremely non invasive.
@TedShifrin well, not everyone can be Ted you know. And I mean that in a good way btw
@leslietownes that's precisely what the specialist explained and got me to calm down
Yes, ofcourse. But what overlap? W-U and U are disjoint @TedShifrin
I meant that to encourage you, Hades, not to be an ass.
01:13
Ofcourse they agree on $U-K$. But why does that imply $F$ is continuous
like, 30 years ago i'd think it was definitely something to be worried about. my uncle got it almost the minute after they authorized some version of it in the USA (and was fine, but, it was in a context of there not being decades of people having gotten it).
and most of the advances in it have only been to do more with less invasion.
Reread what we did. We thickened up $W-U$.
Yes. Could you just elaborate on that please? I keep rereading it but i'm confused for some reason
I think that's where i'm confused
We made $W-U$ into an open set going into $U$. Whence the overlap.
Okay. So we take an open set $Z$ such that $W-U \subseteq Z\subseteq U$. Right? okay that is clear to me
01:16
@leslietownes You know what I like about doctors and medical professionals in general? They can be really good at calming you down. Both then and today, when I was having trouble getting back to sanity, the doctor started asking about my interests, what I do, what I like to work on and so on.
Not right, of course.
Not right not right. ofcourse notright
She asked about my academic path, which I informed her of, and we got into a discussion of how I tend to approach solving problems as opposed to her as a medical professional
I got my $U$ and $W$ other way around.
In my head
Are you drawing a picture?
01:17
Yes.
So we know $f$ and $g$ agree on the overlap. The union of the two open sets is $W$.
Just a minute. 'made $W-U$ into an open set going into $U$'. in terms for $Z$ and $W-U$ and $U$, what does that mean?
I've drawn $W$, the 'largest' open set which contains $\overline{U}$ which inside contains $U$. There is some 'space' between boundary of $\overline{U}$ and boundary of $U$.
So they're nested
$W-U$ is just everything except $U$.
And we added to $W-U$ the collar around $\partial U$. Call that union $Z$.
The two open sets are $U$ and $Z$.
Okay. So by 'collar', you mean we thickened up the boundary of $U$ to an open set $C$. Then, put $Z=(W-U)\cup C$. Right?
So, the 'collar' intersects the complement of $U$, but also the interior
Aha. Now that's where i'm confused. Why is $Z$ open?
01:32
Just check.
Just curious. Are there known proofs with a similar flavor to this problem?
Let's talk Jordan canonical forms.............I don't understand what Thm 7.1 (b) is getting at. And the idea has been used in two proofs since that moment in the text.
dc3, the screenshot is missing a definition of K_lambda but as far as i can tell i'm with you, who cares
maybe it's just some boring lemma they use to establish something more important
Sorry I thought it was a general notion
if that's the case, focus on that instead of this
01:42
$K_\lambda$ are the generalized eignespaces
OK. at the moment, i honestly can't think of why you would use 7.1(b) in anything, except as some step in some other result.
so some author arranged their treatment this way and in another book you maybe never see 7.1(b). that is what i would take out of it
let me look at the proofs where it came up
@monoidaltransform This problem was cluttered with obscure red herrings. Techniques with thickening and taking tubular nbhds are all over differential topology, geometric topology, and differential geometry.
Yeah. I wonder why he chose to write that problem that way
If you want to look at their uses @leslietownes but I'll just carry on without getting bogged down by it
01:48
He seems to thrive on obscurity.
i don't know why anyone would want to set out 7.3 as something worthy of attention. 7.4 seems useful but i would regard it as a black box, who cares how you prove it
Yea that's how I'm treating 7.4 as a black box of sorts
it looks like an author made some expository choices that leaned on this lemma or whatever being used in multiple places. which, OK. but i would not give it independent importance
I disagree. 7.3 is great.
Why wouldn't it just be a corollary to 7.4 Ted?
01:51
7.4 is harder to parse. Did they not use 7.3 to prove 7.4?
that's how I interpreted it before I had read 7.4 the first time....Yes 7.3 was used for 7.4
why not just talk about direct sum decompositions, easier to parse than both of these
They summarize it up with the direct sum decomposition THor
that's nice
Okay so just to make sure I got the idea. So, We thicken up $W-U$ to $Z$ as above. For suitable collar $C$, we have $Z\cap U= C\cap U$. On $C\cap U$, $f=g$. Now $F=f$ on $W-U$ and $g$ on $U$. Define $\tilde{F}$ by $F$ on $Z$ and $g$ on $U$. Then, $\tilde{F}$ is continuous extension of $F$. So, $F$ is continuous?
it's 7:37 Am here, and I really want to get this
02:09
Yup.
Unless I missed some typographical issue.
Somewhere in there you should make sure that $C\cap K=\emptyset$.
Ah okay. Yeah. Thanks Ted. I'll make sure I got all the details down after I rest. Thanks ocne again. So much
G’morning!
Goodmorning to you too :)
All of my medical reports seem to be clear, or so the doctor claims. That's definitely reassuring
Yippee!
02:33
My post operative test was supposed to come back 0.10 or less. It came back 0.39. I talk to the doctor on Wednesday.
Does anyone know how to prove this f bijective?
I really couldn't see what else information I can get from set about the input x.
@robjohn Damn, sorry to hear that. Good thoughts.
@NotTfue Sure. If a digit isn’t $2$, what is it?
@TedShifrin 1 and if it is 1 then f(x) skips the the natural number j.
I was curious about the injection using
So, if you know $f(x)$ you know the decimal uniquely,
definition
02:44
We just argued that.
@TedShifrin how? Decimal expansion is unique of course but this is what I don't get. How did you know it is unique using function.
You know exactly what every decimal digit is if you know $f(x)$.
I can't rest. I need to figure this out. I understand the collar construction really well now. Now just the issue of continuity. Why is $F$ continuous? ($\tilde{F}$ is defined on $W$ as well, so not necessarily continuous extension).
If $j\in f(xj$, then the $j$th digit is …. And if $j\notin f(x)$, then ….
I also understand why $(W-U)\cup C$ is open. It's because the boundary points of $U$ have neighborhoods living in the collar
02:49
I will try to use pc. Can't use chat Jax.
We’re not extending, monoidal. We’re patching.
Okay, so now I have set $Z=(W-U)\cup C$. Which is clearly open. Then, $W=Z\cup U$. So the function $F$ coincides with the function defined by $f$ on $Z$ and $g$ on $U$?
If j∈f(x)
, then the jth digit is 2 And if j∉f(x), then 1.
This step is enough right?
I also made a function $g(A)=0.111...+∑_{n∈A}10^{−n}$ that extract the decimal from set produced by $f(x)$. As this step didn't look rigorous enough.
Ah right. Yes. $F$ IS the function that is defined by $f$ on $Z$ and $g$ on $U$ right @TedShifrin ?
03:10
I believe so.
@NotTfue That’s fine, too.
@TedShifrin but really how exactly does it say that it is injective though?
So if I define $\tilde{F}:W\rightarrow X$ by $\tilde{F} =f$ on $Z$ and $g$ on $U$. Then, I claim $F=\tilde{F}$: If $x\in W-U$ or $x\in U$ then clearly done. So assume $x\in C$. If $x\in C$ and $x\in U$ then we are done. If $x\in C$ and $x\notin U$ then since we are in $W$, we still are done.
And $F$ is continuous because $\tilde{F}$ is. $f$ and $g$ coincide on $U\cap Z=U\cap C$.
Is that reasoning correct @TedShifrin? or is there another way to see this?
This is the reasoning I can come up with if j∈f(x), then the jth digit is 2 And if j∉f(x), then jth digit is 1. So because every decimal expansion is unique so this implies the set f(x) is also unique. Is it the correct reasoning?
03:54
This is not quite satisfying.
04:08
@Koro I'm onto some deep NT shet:
https://math.stackexchange.com/questions/4646639/for-all-n-geq-1-there-exists-6x-inp-n2-dots-p-n12-2-st-6x-2-q
ever notice i j and k (unit vectors in the x,y,z) are aligned on the keyboard together
weird
Yes, that is weird
abstract nonsense
orthonormality?
:D
04:24
how does one construct a number using only symbols of first order set theory
and how is such a language of logic defined such that it isn't just shorthand for other languages
I believe the idea is using sets of empty sets
@D.C.theIII Would you say a countable set must exist within an uncountable set?
for infinite sets, we say one is countable if there is 1-1 correspondence with the natural numbers. For those that do not have this, can we not make further analysis on the "size" or countability of the sets?
04:41
Not sure Obliv, that is getting above my level at this moment. I'm sure one of the veterans will be able to expand on that
I guess I am not capable of doing real analysis. It is impossible for me to prove this theorem.
Real analysis require certain level of intelligence that I don't have.
What's the distinguishing feature of a computable/uncomputable algorithm for a turing machine :\
Suppose that E is a real Banach space, U is a bijection isometry. Fix a, b in E. Define D such that D(U)=$\| U(a)+U(b)-\frac{U(a)+U(b)}2\|$. Define $s_U(e)= U(a)+U(b)-e$ for every e in E. Define U’= $U^{-1} \circ s_U \circ U$. Then it is to be shown that $D(U’)=2D(U)$.
I have shown that.
Now, from here how do I conclude that D(U)=0?
@NotTfue whats the theorem
It can also be seen that $D(U)\le \|a-b\|/2$.
04:45
@DLeftAdjointtoU
Been thinking for 2 days. Can't come up with rigorous injection proof.
With this definition.
I don't know what I can do with the basic set stuff I know to build the proof.
@NotTfue hint: think about complement of f(x) in $\mathbb N$.
I don't even know what a proof is.
Are proofs subjective?
@Obliv For physics guy like us I think it doesn't matter at all lol.
04:50
I mean i've tried to form proofs but I got frustrated and gave up
I think proof writing is something you need a mentor/teacher for
Computation is what matters. I was trying out analysis because my professor said it might be used as graduate student.
so you're self studying it?
Yup. And I get really frustrated this time I just think I should give up. I wonder how math student deal with it.
Math is pretty abstract, I'm taking only math courses this semester to catch up on requisites, I miss physics lol
Actually I take it back, I think proof writing scales with your knowledge. I'm sure you could prove very simple things, but the material you're going over might be too foreign
@Koro You mean x that is in P(N) but not f(x)?
04:59
@Koro what is this called?
@Obliv I think the calculus course they(math student) take is different from ours(physics student). Also they take introduction to proof as freshman here so may be you are right.
what is that symbol before $\mathbb{N}$ in that theorem
like that function
is this a proof of countability
@Obliv power set of natural number
It's proof of cardinality.
Have you shown such a proof before @notTfue
What's your strategy
05:15
I want to prove it is bijection which means it is both subjective and invective so first I want to prove it using invective definition that I know. I know that each decimal expansion is unique but can't come up with rigorous description of what does the set say about the input x. I just know definition.
@Obliv You mean seen or shown? I have seen this kind of proof but never seen a function that maps to a set so this is the problem.
The set is the only problem.
dammit, what does invective mean?
injective
what does the hint even mean $f(x) = \{j \in \mathbb{N} : d_j = 2\}$
what does the colon mean again :
auto-correct is making me sound too smart lol
@Obliv it is stating a condition
maybe it's saying $f(x)$ is $0.2,0.22,0.222,...$
then you show $g(x) = \{j \in \mathbb{N} : d_j = 1\}$ also is bijective then you add f(x) and g(x) or something
oh well i gotta sleep good luck lol
05:25
@Obliv the element of E is not finite
finite decimal
05:38
@NotTfue ohh. I mean think of $\mathbb N-f(x)$.
@Obliv what?
nvm, I got D(U)=0.
@Koro I am not sure what this step will lead to.
This will tell you that if f(x)= f(y), then there in x and y 1 occurs in exactly the same slots after decimal.
Is uniform boundedness principal applicable in case of uncontountably many linear operators on a Banach space?
@Koro I guess this is the missing step. We already know abut 2 now we need to know about 1. But does that really prove it is injective? It makes sense. I will think about it.
05:56
yeah, I think you’re overthinking this.
@Koro yes, it is.
@Koro Yeah I think so. I spend my whole holiday looping around this question not giving fresh view. I am convinced it is injective now :)
 
2 hours later…
08:05
why is (1,1,...) an extreme point of a closed unit ball in c (the space of all convergent sequences with sup norm.)?
Suppose that $(1,1,...)= t (x_n)+(1-t) (y_n)$ for some t in (0,1) and $(x_n), (y_n)\in c$.
Then, $tx_n+(1-t)y_n=1$ for all n, so t x+(1-t)y=1, where $x_n\to x, y_n\to y$.
08:27
In this answer, this person considers the range of numbers after a given index in a sequence. He argues this range is bounded so has a supermum. I agree with that, but aren't sets and sequences fundamentally different? One allows for repetition while sets should have distinct elements: math.stackexchange.com/a/2708043/1118236
Question is multiple indices can have the same value hence the many indices can act as a supremum
 
2 hours later…
10:13
koro it makes sense to look entrywise although i'm not sure you specifically need to consider the limit. from the definition of the norm, any particular x_n and y_n will be in the closed unit ball of your field of scalars. is 1 an extreme point of that?
 
2 hours later…
11:52
I understood what I missing.
:-)
we're having "cold" weather now. 40F, may not go beyond the mid 50s today
 
2 hours later…
13:43
@leslietownes I hear that it snowed on the Hollywood sign yesterday.
And my daughter says that they got a little snow in Redlands.
0
Q: How can I show that this set is not dense in $S^1\times S^1$?

user123234 Let me define by $R_\alpha: S^1\rightarrow S^2$ to be the rotation on the circle by the angle $\alpha$. Now consider $\alpha, \beta$ and $R_\alpha, R_\beta$ and define $$M:S^1\times S^1\rightarrow S^1\times S^1;~~(x,y)\mapsto (R_\alpha x, R_\beta y)$$ Assume that $\alpha=2\beta$ I need to show t...

Can someone take a look at this question here?
14:03
suppose that H is a Hilbert space. For c in H, define H_c:= span(c). For some a,b in H, if $\|a-b\|=\inf \{\|a-x\|: x\in H_b\}=\inf\{\|x- b\|: x\in H_a\|$, then a=b.
How do I prove this?
$H_b$ is finite dimensional, hence closed. So there exists a unique element in H_b closest to a so the question makes sense.
If a lies in $H_b$, then we are done.
15:05
@AiraThunberg The $m$-th term of the $n$-th line of that "triangle" is $\sum_{k=0}^n (m+k)\choose{n}{k} = 2^{n-1}(2m+n)$
Then it's just writing $m$ and $n$ in terms of the $k$-th number of the series
I messed up the latex, here is the correct version $\sum_{k=0}^n (m+k){n \choose k} = 2^{n-1}(2m+n)$
15:22
@Astyx can you please provide it as a comment or answer there, it will really help me
16:02
Hello @TedShifrin I have asked a question a while ago ( math.stackexchange.com/questions/4646920/… ) about an exercise from "Multivariable Mathematics"; if you have the time and the inclination to take a look at it, I would appreciate some feedback about my solution, thanks.
 
1 hour later…
17:09
Are locally compact locally connected Hausdorff spaces normal?
17:56
koro: does the hypothesis imply that both a and b are orthogonal to a - b? (a-b realizing those infs is awfully close to saying that a-b is the orthogonal projection of a onto b and vice versa). this would imply that a - b is orthogonal to itself
18:35
Leslie, yes. I'd understood that. So a-b is orthogonal to a and b. So (a,a)=(a,b)=(b,b)=(b,a). (a-b,a-b)= (a,a)-(a,b)-(b,a)+(b,b)=0.
cool. i was just thinking out loud. anything to do with hilbert space you want orthogonality. or somehow it holds for some more general reason.
I asked a new question.
Maybe silly dunno
0
Q: Sign of $n$ th derivative of $f(x)$?

mickLet $f(z)$ satisfy $f(f(z)) = \operatorname{arcsinh}(z/2)$ More precisely, we construct such an $f(z)$ by using the fixpoint at $0$ and the related Koenigs function. see : https://en.wikipedia.org/wiki/Koenigs_function Now I wonder about the sign of the $n$ th derivative of $f(x)$ at point $x$ fo...

18:52
Let $B$ be a real two times two matrix with eigenvalues $\lambda\in (1,\infty)$ and $\mu\in (0,1)$ and we define $$T:S^1\rightarrow S^1;~~x\mapsto \frac{Bx}{\|Bx\|}$$
I want to show that for all $x\in S^1$, the sequence $T^nx$ converges to a fixed point of $T$.

I know that the only fix points of $T$ are $$O:=\left\{\frac{x}{\|x\|},-\frac{x}{\|x\|},\frac{y}{\|y\|},-\frac{y}{\|y\|}\right\}$$ where $x,y$ are the eigenvectors for $\lambda$ respectively $\mu$. Additionally I know that $T^nx=\frac{B^nx}{\|B^nx\|}$. But now if I take $x\in S^1$ how can I show that $T^nx\rightarrow z$ where $z\in O$.
19:32
Does there exist a proper dense subspace of a finite dimensional Banach space?
I say no. Because if it did, then being finite dimensional, it would be closed. So closure of this subspace =itself= the Banach space, contradicting properness.
Good Sunday all :)
Discrete math is such a rabbit hole. I love it. I'm trying to find a way to compute the factors $2^n$ (if they exist) for any given $x$. I generalized the "lerped log" to all bases, so this should generalize just as nicely and would provide a means for finding all factors $b^n$ in $x$.
How do we prove that kernel of an unbounded linear operator on Banach space is dense?
*functional
Let f: E-->R be an unbounded linear functional, where E is a Banach space. I prove contrapositive of the statement. Suppose that ker f is not dense in E. Then there exists e not in closure of ker f, f(e) is non zero. It follows that range f= R.e
|f(x)|=||r_x.e|| for some r for all x: ||x||=1. From here, $|f(x)|\le r_x$. Not sure, how to bound r_x from here.
0
Q: Unbounded linear functional on Banach space $X$ has dense kernel

KoroLet $f: X\to \mathbb R$ be an unbounded linear functional, where $X$ is a Banach space. Then $\ker f$ is dense in $X$. I tried to prove contrapositive of the statement. Suppose that $\ker f$ is not dense in $X$. Then there exists a $p\in X$ not in closure of $\ker f, f(p)$ is non zero. It follows...

20:05
The properties of the lerped (linearly interpolated) logarithm are quite nice. It preserves the properties from the true logarithm that proceed from the following constraint: $$\operatorname{linelog}_y(x) = \log_y(x) \leftrightarrow \{ x = y^n \land n\in\Bbb Z^+ \land \frac{\partial }{\partial x} \operatorname{linelog}_y(x) \propto \left[y^{\lfloor\log_y(x)\rfloor+1} - y^{\lfloor\log_y(x)\rfloor} \right]^{-1} \}$$. It is essentially a kind of discrete logarithm.
What does HTH, AB mean?
What we get as a result is that any integer multiple of $y^n$ has a constant fractional portion, and the integer portion tells me which $y^n$ and $y^{n+1}$ that a given number $x$ is between as well as its relative distance from either of those factors.
@Koro Who are you asking?
everyone
oh ok just making sure
8
A: Linear functional on a Banach space is discontinuous then its nullspace is dense.

martiniHint: $\overline{N_f}$ is a subspace of $X$ containing $N_f$. What can you say about $\mathrm{codim}\, N_f$? Ok. So I'll give the answer. As $N_f$ isn't closed (since $f$ is discontinuous [Do you know that?]), $\overline{N_f} \supsetneq N_f$. It follows that $\text{codim}\, \overline{N_f} < \t...

see here, for example.
I think it is 'hope this helps'.
20:11
Beyond my domain in several ways that there :)
@AMDG: it's sort form of something.
anyways, leave it who cares.
Imagine grades actually mattering and being a useful metric for measuring one's comprehension.
it's easy if you try
What is easy?
i was not substantively engaging with anything, just reacting to a statement about imagining something with follow-on lyrics from the song 'imagine'
20:20
Oh lol
so tru
kinda kills the laughs
HTH is sometimes 'happy to help,' no idea about AB
But yes, it's almost like numerically quantifying dialectics is kinda hard if you don't know dialectics.
"And here we see that the student's sentence reduces to the following subject-copula-predicate forms ... which considered collectively deviate by one essential note from the comprehension of this essence, so the idea is incomplete." I dare you to put that into the form of a bunch of discrete marks and converting to a percent. ecks dee
If you want to test one's comprehension, have him define the things you wish to know that he understands.
Insofar as the definition corresponds or deviates from the true definition, he more or less understands.
A simple number, however, isn't telling of what exactly is correct and what is not though obviously. You need to state what is correct, and what is not.
If this were done in every course, then every course would indeed have a means to measure comprehension, and then and only can the grades accurately reflect such comprehension--assuming the numbers model the right things, namely, ideas themselves, and what constitutes them.
Anyways...
20:46
What is it it that you believe "grades" are supposed to reflect?
Comprehension.
And who do you believe is the target audience of grades?
@AMDG Yeah, that isn't really the point of grades.
It is one aspect, but it is far from complete.
@XanderHenderson Ultimately the student; secondarily, parties interested in the applications of that comprehension.
@AMDG Yeah, grades are not for students. They are for administrators, and, perhaps, for interview committees.
To what end? Grades measure nothing useful, and each teacher usually has his own grading system, so the values are subjective which means they're worthless.
20:48
@AMDG I agree. Grades measure nothing useful.
But they do, to an extent, reflect a student's willingness and ability to complete work, and, to an extent, comprehend the work that they are doing.
@XanderHenderson Interesting. Then we must both be obliged to conclude that this whole system is fundamentally flawed, and that everyone out in the field is more likely to be a fraud than an expert.
An "A" sends a signal to a potential employer or administrator that a student can do a lot of work, and do it at a convincingly high level.
@AMDG I do think that grades are hugely problematic. I wouldn't go so far as to say that they are fraudulent, but I very much doubt their utility.
I really wish that I didn't have to assign grades. :/
@XanderHenderson Willingness to complete work, I can agree with. As a matter of understanding, I wholly disagree. It proves that you can do a thing in itself; it does not prove you understand a thing.
@TedShifrin As a sanity check, yesterday we didn't even need the compactness of $K$, where $f=g$ outside $K$. Right? having it closed suffices right?
@AMDG I didn't say "prove".
That's your word.
What I said was that grades "reflect a student's willingness and ability to complete work, and, to an extent, comprehend the work that they are doing."
20:52
Fair enough, but they do in fact prove in themselves that one is capable of doing something.
@AMDG Maybe?
I don't really think that grades prove much of anything.
If you can do a thing... then you do the thing, and if it's correct... then you can do the thing, so it's a definitive yes.
Eh, specifically the tests and whether or not the respective problems are answered correctly.
Anyone in principle can be taught to do something, and that's my point. A dog can be trained to do math.
In other news, I am honestly kind of shocked at how deeply effected I am by the news that Jimmy Carter is about to die. :/
Since understanding isn't required to simply do what is told (apart from the understanding of the command in itself), that one can do something does not in itself demonstrate understanding; but it is impossible to not understand a thing and completely define it except by pure chance: which concludes that holding such understanding accountable in some way is necessary to prove one's understanding.
Implicit in your assertion is the assumption that the work which earns a student a grade is the student's own work.
You also assume that the grading scheme is designed to assess "comprehension".
20:57
Yes.
I do not believe that either is a safe assumption.
@XanderHenderson well grading is more so just a reflection of what already is, in which case it belongs to implementation rather than design.
To design is to comprehend essences and specify them.
I don't think that most grading schemata are actually designed in any meaningful sense. I think that most people just do what their mentors did.
There is very little thought that has gone into grading.
A very unfortunate fact, then.
Which is not to say that there aren't folk who are really thinking about what grades are, what they should mean, and how they should be assigned.
But I think that is the exception, rather than the rule.
For what it is worth, I think that I have probably spent more time thinking about grading than most (but, like, I'm only saying that I'm in the 60th percentile, give or take---I have made some effort, I am by no means a master in this area).
21:01
Like I said though... the grades are supposed to be a reflection, and that grading is an implementation or the actualizing of the specification is manifest by the fact that usually you design tests, but the grade on it is some value of which the interpretation is dependent on the design of the test itself.
And I believe that my grades reflect a combination of (1) willingness and ability to get work done (which I think is an important skill for students to have), (2) ability to complete problems in [topic] (usually "precalc" or calculus), and (3) an actual understanding or comprehension of those topics.
My hope is that my grade represents the probability of a student being successful in the next class.
(assuming that there is a "next" class)
Well, that's contingent on the following necessary constraints: that the test is the student's own work; and that he demonstrates understanding by way of explicitly writing down what the ideas themselves are which in essence always reduces to defining essences. If you don't want to just require explicit definitions, you can require that a student justifies his answers where the sentences contain "because" which leads to implicit definitions.
@AMDG (1) has no assumption about it being the student's own work.
I wish it did, but that is beyond my ability to control.
(2) is also a mixed bag. I give a proctored final, and that is a major component of how the final grade is computed, but it is not the sole determinant. And everything aside from the final could be done by anyone. I have no ability to check if the student themself did the work.
Ditto (3), though a lot of that come out in office hours.
Does someone know what it means that a sequence of random variables $X:=(X_t)_{t\geq0}$ has right continuous trajectories?
Someone is studying random walks and cadlag functions. :P
21:08
@XanderHenderson does this mean that you can help me?
@user123234 It has been a while since I've dealt with those definitions in detail. But "right continuous" means that a function is continuous from the right, i.e. $$\lim_{t\searrow T} f(t)$$ exists for any $T$.
The wikipedia article on cadlag functions has a pretty good image:
so in my case does it mean that $\lim_{t\searrow T} X_t(\omega)$ exists for all $\omega$?
but I mean I have a sequence of functions $X_t$ with real valued index. And in the definition of right continuity I only have one function, do I then consider each $t$ separatly?
Honestly, it has been too long since I have done work here. I don't have the definitions or ideas easily to hand.
21:14
ah okey no problem, thanks anyway
But I think you want to think of $t \mapsto X_t$ as a right continuous function.
Again, $X$ should be continuous right continuous thorugh time.
(i.e. trajectories, which measure position over time, should be right continuous)
Ah so I don't consider the map $\omega \mapsto X(\omega)$ to be right continuous?
@user123234 I don't know---I don't have enough context.
What is $\omega$?
So I have a probability space $(\Omega,\mathcal{F}, \Bbb{P})$ and a sequence of random variables $X:=(X_t)_t$ and $\omega\in \Omega$
@user123234 Yeah, I am out of my depth for a Sunday afternoon. What book are you working out of?
21:26
Right continuity of $X_t \colon \varOmega \to \mathbb R$ implies that $\varOmega$ is ordered, which as far as I know is not a typical assumption
21:45
Would it make sense to say transcendental numbers don't exist?
@Obliv Would it make sense to say that natural numbers do exist?
Like in the real world, $\pi , \sqrt{2} , e, $etc are holes in the domain of the finite?
well yes, I can count natural numbers in the real world @XanderHenderson
nvm
@Obliv $\pi$ seems like a much more real, natural number than, say $100^{100^{100}}$. But the latter is "natural", while the former is not.
Nah, those transcendental numbers are merely representative of metaphysical being without limitation of imperfection. The perfect circle indeed has area proportional to pi.
@XanderHenderson Sorry for the late reply, I work with my lecture notes
21:48
You might say that everything in the rationals is quite representative of physical reality.
@user123234 Yeah, I'm no help. Maybe ask me on a Tuesday.
I feel like I have written a post on either Math SE or Math Ed SE about natural numbers and real numbers and whatnot... but I can't seem to find it. :(
@XanderHenderson Didn't want to interrupt. Well the nature of assumptions is that you necessarily must assume it is the student's own work to reckon it as the student's work; and insofar as you can reasonably guarantee that (such as by requiring a hand-written signature, comparing handwriting [and requiring it for tests], etc.), then it is safe to believe that it is. Otherwise, you're grading work which is reckoned to be "just work". :)
@XanderHenderson okey
I work at an institution where, even before the pandemic, a large proportion of the instruction was done remotely, and work was collected via email (or snail mail!).
Any work that is done outside of the classroom must be assumed to have been done by anyone, including the student, or a tutor, or Math SE, or Chegg, or whatever.
But homework is a grading component in my class. Which means that this particular component only assesses whether or not a student can get that work done.
Not that they can get it done themselves---just "can you get it done".
I am not going to pretend that I believe (or, perhaps more accurately, can guarantee) that the work is done by the student who is getting graded for that work. The only assignments that I can comfortably assume are probably completed by the student's getting the grades are the conversations we have in office hours, and the final exam (which is proctored).
21:57
@XanderHenderson My point once again being that the guarantee is lost regarding (1). The accountability mentioned is what gives moral certainty that we have a guarantee. If at some point, it is proven such a person just had someone else do it, then we have guarantee that (1) was not fulfilled.
@AMDG What guarantee?
The student got the work done.
I don't assume that the student did the work; only that they got it done.
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