Mathematics - 2024-04-27

Jakobian

12:13 AM

I still don't even know why I'm learning Clifford algebras. What if this is all pointless

Lukas Heger

@Jakobian Clifford algebras are related to K-theory

you can prove Bott periodicity with them

not sure if K-theory is enough to solve an existential crisis, though...

Thorgott

12:34 AM

sounds like an awkward approach to Bott periodicity

I think Balarka tried telling me something along these lines once upon a time

Jakobian

12:45 AM

@LukasHeger nah I'm trying to help my friend so its going to be something applied

I've boiled down how to derive all Clifford algebras, at least the ones over $\mathbb{R}$

$\mathbb{R}^{p, q}$ is the quadratic space (i.e. vector space with quadratic form) I've been talking about above

and $\mathbb{R}_{p, q}$ is the corresponding Clifford algebra

Then one just has to know $\mathbb{R}_{1, 0}$ and $\mathbb{R}_{0, q}$ for $0\leq q \leq 7$ as well as three formulas

1) $\mathbb{R}_{p, q}\cong \mathbb{R}_{q+1, p-1}$ for $p > 0$

2) $\mathbb{R}_{p+1, q+1}\cong \mathbb{R}_{p, q}\otimes \mathbb{R}(2)$

3) $\mathbb{R}_{p, q+8}\cong \mathbb{R}_{p, q}\otimes \mathbb{R}(2^4)$

if you know those, you can write $\mathbb{R}_{p, q}$ as an algebra of matrices over one of the $9$ algebras I've mentioned above

here $\mathbb{R}_{1, 0} = \mathbb{R}^2$, $\mathbb{R}_{0, q}$ is $\mathbb{R}, \mathbb{C}, \mathbb{H}, \mathbb{H}^2, \mathbb{H}(2), \mathbb{C}(4), \mathbb{R}(8)$ and $\mathbb{R}^2(8)$

oh I guess one actually only has to know $\mathbb{R}_{1, 0}$ and $\mathbb{R}_{0, q}$ for $0\leq q\leq 3$ if one apples formula 1) above

but yeah everything is an algebra of matrices over either $\mathbb{R}, \mathbb{R}^2, \mathbb{C}, \mathbb{H}$ or $\mathbb{H}^2$

@Thorgott Bott periodicity seems to be related to 3) ?

Thorgott

1:02 AM

yeah, that should be it

real K-theory is 8-periodic

Jakobian

1:16 AM

@Thorgott is complex K-theory 2-periodic?

this would be what I'd guess from how complex Clifford algebras only change from matrices over $\mathbb{C}$ to matrices over $\mathbb{C}^2$ and conversely

leslie townes

yes, it is

Jakobian

oh, are you knowledgeable in the topic leslie?

leslie townes

not really, i took a class that covered it 20 years ago

Jakobian

More knowledgeable than me at least

Magnus Alexander

1:43 AM

Hello everyone! Some easy PT task that I can't solve, maybe someone might give me a hint...
Given $\{X_i\}$ -- uniformly distributed independent random variables. Let $X_{n} = max{X_1, X_2, ..., X_n}$. Find the limit by distribution of $n\cdot (1-X_{(n)})$

Magnus Alexander

2:27 AM

Solved it, but thanks for anyone who considered it !

Lucky Chouhan

3:21 AM

@leslietownes so you're in your 40s?

leslie townes

i'm in at least my twenties :)

Bumblebee

Is it possible to evaluate this integral without a substitution? $$\int \frac{\sin(x) \cos(x)}{\sin^{4}(x) + \cos^{4}(x)} \, dx$$

Jakobian

3:36 AM

@leslietownes 20 years old genius progidy that has been listening to (and understanding!) math lectures before he was born

leslie townes

the change of variables is so fundamental that i'm not sure what would count as on the table if that tool is taken off. if there were bounds, at least some definite integrals involving that can be evaluated without computation (it is an odd function)

Magnus Alexander

@Bumblebee What do you mean by integrating without a substitution? I don't see how it's possible, unless the function comes from the table of integration

It's certainly not the easiest one even considering the substitution. It's obvious, but seems quite technical.

Bumblebee

@MagnusAlexander I was just curios to see if it were possible.

Magnus Alexander

Well, theoretically, you can integrate it via substitution, find the answer, and then prove that the difference between this answer and the Riemann sums is epsilon, but that's dumb xd

X4J

5:33 AM

Suppose $f$ is a real function differentiable at $[1, \infty)$ s.t $\lim_{x \to \infty} f'(x) = 0$. Does it imply that $\lim_{x \to \infty} f(x)$ exists (namely, that limit is either finite or infinite)?

Intuitively I dont see why it won't be true and struggle to pick a counter-example too but by logic I cant prove it

Bumblebee

@MagnusAlexander would a substitution of $\sin^{2}(x)$ work?

Soumik Mukherjee

6:10 AM

I will evaporate soon

Because of this scorching summer heat

X4J

By trying to find a counter-example, I expect to look after functions $f$ such that $f'$ is 'as slow as' $\frac{1}{x}$ but gets negative values too. Can someone please guide me on this

Lucky Chouhan

6:59 AM

@SoumikMukherjee haha,

@SoumikMukherjee yeah, it is very hot outside.

Soumik Mukherjee

7:14 AM

Yeah

Soumik Mukherjee

7:31 AM

@X4J Hint, take a function g such that lim x tends to infty g(x)=infty and limit x tends to infinity g'(x)=0

Take another function h such that lim x tends to infty h(x) does not exist

Now make a new function f out of these two such that limit x tends to infty f'(x)=0 but limit x tends to infty f(x) does not exist

X4J

8:34 AM

@SoumikMukherjee Yeah thanks, so we can do $sin(\sqrt{x})$ for example

Oh then in some sense a discrete analog is a series like 1 - 1 + 1/2 + 1/2 - 1/2 - 1/2 + 1/3 + 1/3 + 1/3 - 1/3 -1/3 - 1/3 + ...

Soumik Mukherjee

@X4J yes

@X4J I don't get it though, how are you gonna define differentiation of a series of real numbers?

X4J

9:47 AM

Maybe its just me then

Pizza

hi

Let $f,g : [a,b] \to \Bbb R$ be two continuous functions in $[a,b]$ and differentiable in $(a,b).$ If $g'(x) \neq 0 \ \ \ \forall x \in(a,b)$, then there exists $c \in (a,b)$ such that
$$\frac{f(b) - f(a)}{g(b) - g(a)} = \frac{f'(c)}{g'(c)}$$

How can I represent this graphically?

Soumik Mukherjee

12:02 PM

@Pizza See this

Jakobian

I never had the need to use Cauchy's mean value theorem in any way

psie

12:51 PM

I'm reading about the Stieltjes integral. There is a passage where the author goes:

> We observe that $$\int \chi _{\left[a,b\right]}\, dF(x)=\lim _{n\to \infty }\int \chi _{\left(a-\frac{1}{n},b\right]}\,dF(x)=F(b)-F(a-)\tag1.$$In particular, taking $b=a$, the function $F$ is continuous if and only if $\mu$ is diffuse.

First, does the first equality above follow from the monotone convergence theorem? Second, why is $F$ continuous if $\mu$ is diffuse? Recall, $\mu$ is diffuse if it has no atoms, i.e. $\mu(\{x\})=0$.

Jakobian

@psie yes

Note that a corollary of 1) is that $\mu(\{a\}) = F(a)-F(a_-)$ by letting $b\to a^+$

so we see that $F$ is continuous at $a$ iff $\mu(\{a\}) = 0$

psie

ok, but is $\chi _{\left(a-\frac{1}{n},b\right]}$ increasing towards $ \chi _{\left[a,b\right]}$? Seems to me that we get a smaller and smaller interval $\left(a-\frac{1}{n},b\right]$ as $n\to\infty$.

Jakobian

(also, recall that a random variable $X$ is continuous iff $P(X = x) = 0$ for all $x$)

@psie okay sorry my bad

not monotone convergence theorem but dominated convergence theorem

psie

ok 👍

Jakobian

I mean you can use a version of monotone convergence theorem that says something something too

that would work for decreasing sequences

$f_n$ are non-negative, decreasing and $f_1$ is integrable

psie

1:01 PM

ah cool

John Zimmerman

1:35 PM

Can $\Bbb {\hat C}$ be a bounding sphere analogous to how $S^2=\partial \Bbb B^3$?

Jakobian

whats the hat

John Zimmerman

Riemann sphere

Jakobian

Riemann sphere is $S^2$

as a differentiable manifold

John Zimmerman

$\Bbb C \cup \lbrace \infty \rbrace$

@Jakobian Okay so the answer is yes, but - what does $\Bbb {\hat C}$ bound then?

the unit complex ball?

Jakobian

@JohnZimmerman your question makes no sense

John Zimmerman

1:41 PM

@Jakobian It does make some sense actually. I correctly denoted the riemann sphere

Jakobian

@JohnZimmerman no

We came to a conclusion that $\hat{\mathbb{C}}\cong S^2$ at which point we end the discussion with, it bounds $\mathbb{B}^3$ in the sense that $\partial \mathbb{B}^3 = S^2\cong \hat{\mathbb{C}}$

the discussion about what it bounds, makes no sense now

we already gave answer to this loosely defined question

John Zimmerman

It bounds a 3d ball

that's the answer

so it is analogous to how $S^2$ bounds a 3 ball

Jakobian

2:14 PM

If we define what it means to bound such a ball

Then its tautological

Pizza

Let $f,g : (a,b) \to \mathbb{R}$ be differentiable, with $g'(x) \neq 0$ for all $x\in(a,b)$.
If these three conditions are met:

$\lim_{x\to a^+} f(x)=0, \ \lim_{x\to a^+} g(x) = 0$
there is a finite or infinite $\lim_{x\to a^+}\frac{f'(x)}{g'(x)}$
Then, $\lim_{x\to a^+}\frac{f(x)}{g(x)} = \frac{f'(x)}{g'(x)}$
$Proof:$
We extend by continuity the definition of $f$ and $g$ to the entire interval $[a,b)$ by setting $f(a)=0$ and $g(a)=0$.
In this way, the resulting functions $f:[a,b)\to\mathbb{R}$ and $g:[a,b)\to\mathbb{R}$ are continuous in $[a,b)$, and differentiable in $(a,b)$ with the same…

why do the two functions extend continuously?

Xander Henderson

@Pizza Well, start with the definition. What does it mean for a function to be continuous?

Pizza

ok

A function $f(x)$ is continuous at a point $x_0$ if $\lim_{x\to x_0} f(x)=f(x_0)$.

Xander Henderson

@Pizza Great. That'll do. What do you know about the function $f$ in your theorem?

Pizza

it's differentiable, so it's continuous???

Xander Henderson

2:27 PM

@Pizza No. You just said that a function $f$ is differentiable at $x_0$ if $\lim_{x\to x_0} f(x) = f(x_0)$. You asked about why you could extend your functions to be continuous at $a$. So what do you know about the functions at $a$?

Pizza

wait

a function is continuous in [a,b] if it is continuous at every point $x_0\in[a,b]$ , if $x_0=a$ then only the right limit is considered.

Xander Henderson

I have not asked about continuity on the interval---you already know that it is continuous on $(a,b)$ because it is differentiable. YOU asked why we know that it can be extended to a continuous function on $[a,b)$. The only thing that can possibly go wrong is if it fails to be continuous at $a$. So how do you know that $f$ can be extended to a continuous function at $a$?

Jakobian

@XanderHenderson how do you find those doi numbers for citation purposes again? I forgot

Xander Henderson

@Jakobian No idea in general.

Google will generally list them in their "scholar" search results.

Jakobian

oh I found one. Thanks!

Pizza

2:34 PM

@XanderHenderson i dont know

Jakobian

wait how does doi work. Does it just link you to a site

Xander Henderson

@Pizza You asked why the function $f$ can be extended to a continuous function at $a$.

What do the hypotheses in the theorem tell you about $f$?

@Jakobian It is just a number.

That number uniquely identifies a particular document in a database. The database is maintained by doi.org .

So, given a DOI, you can look up the document in the database.

Jakobian

@XanderHenderson because I have a link of the form doi.org/A/B where A and B are numbers, and it links me to a site taylorfrancis.com and not sure if thats correct

Xander Henderson

@Jakobian Yes. doi.org will automatically redirect you to the original document.

It is a little like a url shortener.

E.g. most bit.ly links automatically redirect you to the original document.

But if you have just the DOI, which will be a string like 10.xxxx/xxxx, you can search for it on the doi.org homepage, and be redirected to the original document.

Jakobian

Does that mean the same book can have multiple doi's leading to different sites

Pizza

2:40 PM

$f(x)$ and $g(x)$ are both differentiable on an open interval $(a,b)$, except possibly at $a$.
$g'(x) \neq 0$ for all $x \in (a,b)$.
$\lim_{x \to a^+} f(x) = \lim_{x \to a^+} g(x) = 0$, or one of the limits is infinite, or both.

Xander Henderson

@Pizza I am not asking you to simply parrot back the hypotheses. THINK about them.

Which of those are relevant to extending $f$ to a continuous function at $a$?

Hint: anything involving $g$ is completely irrelevant.

Pizza

$f$ is differentiable on $(a,b)$ and $lim(f(x)) = 0$ when $x->a+$

lim(f(x)) = 0 at a
so i define f(a) = 0
by definition, lim(f(x)) = f(a) when x-> a

Jakobian

> DOIs are standard, so there would not be more than one for any article or book.

according to guides.library.oregonstate.edu/c.php?g=285973&p=2442706

Xander Henderson

@Pizza So do you see now how the function is extended to a continuous function?

Pizza

2:55 PM

@XanderHenderson we can do it because lim x->a+ f(x)=f(a)=0????

(defining the function in a)

so it is continuous at point a

for definition

Jakobian

3:14 PM

@SineoftheTime youtu.be/GaXuMC_GHEE?si=qstyS79j1_9iooyl

why don't you get to know them

Lucky Chouhan

@Pizza is pineapple

Greetings @XanderHenderson How is life?

Pizza

3:50 PM

I got to the end of the demonstration, but I can't understand the last sentence, pls help

Jakobian

@Pizza what do you need help with?

Pizza

Since $c\in(a,x)\subseteq(a,a+\delta)$, we have
$\frac{f'(c)}{g'(c)}\in V$
and therefore
$\frac{f(x)}{g(x)}\in V$
as desired. $\Box$

@Jakobian

Jakobian

@Pizza why would $f'(c)/g'(c)\in V$ imply $f(x)/g(x)\in V$

and what is $V$

giving me an excerpt from the proof you are reading doesn't really classify as answering my question

Pizza

Let $f,g : (a,b) \to \mathbb{R}$ be differentiable, with $g'(x) \neq 0$ for all $x\in(a,b)$.
If these three conditions are met:

$\lim_{x\to a^+} f(x)=0, \ \lim_{x\to a^+} g(x) = 0$
there is a finite or infinite $\lim_{x\to a^+}\frac{f'(x)}{g'(x)}$
Then, $\lim_{x\to a^+}\frac{f(x)}{g(x)} = \frac{f'(x)}{g'(x)}$
$Proof:$
We extend by continuity the definition of $f$ and $g$ to the entire interval $[a,b)$ by setting $f(a)=0$ and $g(a)=0$.
In this way, the resulting functions $f:[a,b)\to\mathbb{R}$ and $g:[a,b)\to\mathbb{R}$ are continuous in $[a,b)$, and differentiable in $(a,b)$ with the same…

Jakobian

you have to use words

Pizza

4:02 PM

this proof

Jakobian

I'm sorry but you can't communicate without words

@Pizza well if you can't understand the last sentence then you need to explain what do you not understand about it

using words

otherwise I won't help you

Pizza

$V=(L-\varepsilon,L+\epsilon)$

Jakobian

not because I'm trying to be mean but it's just required if you really want someone to help you

you need to explain yourself for that

the other person needs to understand you, and needs to be given enough information to understand you

Pizza

I don't understand why $\frac{f'(c)}{g'(c)}\in V$ implies
$\frac{f(x)}{g(x)}\in V$

Jakobian

yes, but why do you not understand that?

I'm not able to read other people's minds and so I won't be able to help you if you don't explain why this step causes problems for you

Someone asking a question has just as important responsibility to explain themselves as the one answering has the responsibility to answer in a clear and if possible easy to understand way

Pizza

4:12 PM

okok

wait

now i write

Xander Henderson

@Pizza Since it is taking you a LONG time to respond, my guess is that you are typing out a very long argument, again.

I would suggest that you respond one sentence at a time.

Point out the specific things that you are having trouble understanding. One sentence at a time.

Pizza

ok

"Since $\frac{f(x)}{g(x)}=\frac{f'(c)}{g'(c)}$, where $c\in(a,x)\subseteq(a,a+\delta)$ and $\frac{f'(x)}{g'(x)}\in V$."

what I'm thinking is that for every x of V the two functions are equal to their derivatives at point c

so both functions and derivatives belong to V

Xander Henderson

Either you are using $x$ to mean two different things, or you are misunderstanding something. $x$ is not a element of $v$...

$f'(x)/g'(x)$ is in $V$, not $x$.

But I don't know what $V$ is, so I'm a bit lost there, too.

Pizza

Let $V\subseteq\mathbb{R}$ be a neighborhood of $L$.$V=(L-\varepsilon,L+\epsilon)$

Xander Henderson

Okay, so $V$ is just some interval around $L$.

Pizza

4:25 PM

yes

Xander Henderson

Again, $x$ doesn't live there. $V$ is in the codomain of $f'/g'$.

Pizza

if you go above there is the full proof

Xander Henderson

@Pizza If you want help, don't place the burden of reading an entire proof on other people. Explain what YOU understand, and where YOU are stuck.

Pizza

@XanderHenderson Ah ok, no because I think these things are connected to the previous ones in the proof, so I didn't know if it was good to send separate things

Xander Henderson

@Pizza In any event, perhaps the problem is that you are not seeing the dependence of $c$ on $x$? For each $x$ in $(a,a+\delta)$, you can (via the the Mean Value Theorem) find some $c_x \in (a, x)$ such that $$ \frac{f(x)}{g(x)} = \frac{f'(c_x)}{g'(c_x)}. $$

X4J

4:50 PM

Let $f$ be differentiable at $[0, \infty)$ s.t $lim_{x \to \infty} f'(x) = 0$. I'm trying to prove that $\lim_{x \to \infty} \frac{f(x)}{x} = 0$. I know it follows directly from lHopital but the specific variation of lHopital wasn't introduced to us and hence I need to try to reason it in a different way. I assume that there exists such a way, since lHopital proof applied here seems to overkill it.

I've proved the result for $\frac{f(n)}{n}$

using Cesaro-mean and MVT

Jakobian

@X4J I believe that any argument working for this specific example will boil down to a re-proof of l'Hospital

Pizza

5:13 PM

@Jakobian @XanderHenderson Since $\frac{f(x)}{g(x)}=\frac{f'(c)}{g'(c)}$ and $c\in(a,a+\delta)$,$ x\in(a,a+\delta)$ then it is true that I can always find some x of that interval to which to apply Cauchy's theorem in $[a,x]$ for which the equality is verified. So it is true that they belong to the same $V$

Is this correct reasoning or not?

Soumik Mukherjee

Massey has 2 books on AT, can one go directly for the 'basic course' book without reading the 'introduction' book?

Jakobian

@Pizza Since $x\in (a, a+\delta)$ you can apply Cauchy's mean value theorem to $f, g$ to find some $c = c_x\in (a, x)$ for which $\frac{f(x)}{g(x)} = \frac{f'(c)}{g'(c)}$ holds

So those quotients are exactly the same, because of how $c$ was chosen

Pizza

yes

Xander Henderson

@Pizza you seem to have it backwards. Reread my last comment.

Pizza

ok, all clear

user 70432

5:26 PM

💯✅

Let's all celebrate with pizza 🎉🍕🎉

🥂

Jakobian

@user70432 I'm not sure what I'm celebrating, but a free pizza is a free pizza...

user 70432

🍕🍕🍕

Help yourself

🍕🍕🍕

Xander Henderson

5:44 PM

@Jakobian I'm planning on making pizza tomorrow. The sour dough starter is doing its thing right now. I'll be able to split it in a couple of hours and get the dough proofing.

X4J

6:41 PM

@Jakobian I dont think so because the question was given although we havent seen the proof or the variation at all

Lucky Chouhan

@user70432 be independent. :)

ZaWarudo

For a function $f$ defined on $[a,b]$, set $f_n(x)=\lfloor nf(x) \rfloor/n$ for $x \in [a,b]$ and $n\in\mathbb{N}\setminus\{0\}$. Prove that $f_n \to f$ uniformly on $[a,b]$. My work: from the properties of the floor function, we have $nf(x) \le \lfloor nf(x) \rfloor <nf(x)+1$ and so $|f_n(x)-f(x)|<1/n$. Since $1/n \to 0$ uniformly on $[a,b]$, this shows that $f_n \to f$ uniformly on $[a,b]$. Is this correct?

Jakobian

@X4J doesn't sound like this invalidates what I said

@ZaWarudo no because the inequality $nf(x)\leq \lfloor nf(x)\rfloor < nf(x)+1$ is not true

but the inequality $|f_n(x)-f(x)| < 1/n$ is true, and the argument is valid

ZaWarudo

6:57 PM

Oh yes, it was a typo: I meant $\lfloor nf(x) \rfloor \le nf(x)<\lfloor nf(x) \rfloor+1$

And from this I can infer again that $|f_n(x)-f(x)|<1/n$

user 70432

31

Q: University is killing my passion [for mathematics]

In high school, before having begun with a degree in mathematics, I derived pleasure from studying [mathematics]. At that time, I was naturally looking forward to university, since it meant no more subjects that I found uninteresting, such as languages, and instead complete focus on what interest...

Jakobian

7:13 PM

@user70432 so they're not complaining about mathematics, but the actual process of studying that every student must go through

EE18

Happy weekend all

user 70432

Thanks ee18

EE18

Given that I am self-studying, it's sometimes hard for me to know which exercises are "worthwhile". Hoping for some advice to that end:

Is this exercise sequence useful elsewhere, or just to get me using the definitions? I'm vaguely familiar with the topological notion of closure but it's not clear to me how this would be related

user 70432

@Jakobian there are some created chatrooms in the comments' section

Jakobian

> Maybe the legitimately higher workload, expectations, and stress of university are dampening your feelings of enjoyment in general (rather than specifically for mathematics).

sounds like it

leslie townes

7:17 PM

EE18: well, it's primarily a definition chase, and not something that you would, like, 'use' anywhere. but it's a set theory book (that's enderton, right)? where the definition chase is the point. and there is obvious general value in being able to show that differently defined objects turn out to be the same.

EE18

For sure, and I've worked out the exercises up till now in order to solidify that, but don't want to work every single Enderton (of course you are correct :) ) exercise lest I remain in intro set theory purgatory forever

leslie townes

EE18: as a rule of thumb i might generally skip, as a self-studier, any exercise of the form "Formulate an analogue to [one exercise] for [another setting]."

Jakobian

@EE18 its related in the sense that $A\mapsto C$ is an example of a closure operator

EE18

So funny, I was just reviewing a convo we had a month ago here about a similar exercise (on functions being compatible with relations so that we could define unique functions on the quotient set)

Jakobian

but this is in a sense more general than topological one

Closure operator

In mathematics, a closure operator on a set S is a function cl : P ( S ) → P ( S ) {\displaystyle \operatorname {cl} :{\mathcal {P}}(S)\rightarrow {\mathcal {P}}(S)} from the power set of S to itself that satisfies the following conditions for all sets X , Y ⊆ S {\displaystyle X,Y\subseteq S} Closure operators are...

leslie townes

7:21 PM

EE18: i guess one observation i would make about this specifically is that (1) you will find the abstract concept of a closure operator in other settings, (2) i would not infer from exercises 10 and 11 that these sets are things one can easily compute for any function 'that has a formula you can write down.' if you switched the function a little, you could get very difficult problems

EE18

OK, interesting, thanks Jakobian and Leslie. I will file this one away in (hopefully) nonvolatile memory and come back to it whenver I encounter this notion of closure later

leslie townes

EE18: so he's putting specific examples in there maybe as a guide to seeing what the definitions do in the clean setting where you can "write it all down" or "compute it," but this is maybe a relatively rare setting

Jakobian

@EE18 usually there are two ways of viewing closure operators, abstractly as intersections of some sets, or more concretely with a description. For example, closure in the sense of taking a subgroup generated by a subset of a group has two descriptions. The exercise is about defining a closure operator and proving that this more concrete description of it is correct.

Soumik Mukherjee

Leslie: do you know about the Massey book thing that I asked above?

EE18

Is the recursive definition meant to be the concrete one here?

Jakobian

7:24 PM

yeah

leslie townes

soumik: i'm not sure i even realized he had two different books

EE18

I guess "closure of a set under a function" is meant to intuitively capture "closure under infinitely many iterations of that function"?

In that sense, neither definition is that surprising I guess

Jakobian

@EE18 it'd just mean $A$ is a set such that $f(x)\in A$ for $x\in A$

EE18

for a subset $A$ to be closed, I agree. But for non-closed we then extend to this hazier notion of smallest containing set which is itself closed under $f$ ($C^*$)?

Soumik Mukherjee

I ordered the basic book online. Hopefully the introduction book is a subbook of this. At least looking at the contents it feels that way.

Jakobian

7:28 PM

@EE18 It'd just mean smallest set which contains $A$ and is closed under $f$ in the above sense

imbAF

Can anyone help me with something. In my laboratory course I got some date (x and y) and I am trying to have a cauchy fit for those. The thing is that, while I am able to do a fit which gives me values for the coef. a b c and d, I don't have errors for these coeficients i.e a=23.45 and I need something of the sort a=23.45 +/- 0.2. How can I get errors/standard deviations of the coef. ?

leslie townes

EE18: one thing i might have done, had i been enderton, is not used the very overloaded word "closed" at the beginning of the exercise with only an implicit definition

EE18

7:41 PM

I think he did define closed on p70 to be fair (on a second glance)

leslie townes

okay. well, out of context, scare quotes around a term that you're not defining in an exercise sucks.

the other way you run into stuff like this, outside of self study, is you're teaching and assign some exercises assuming (after skimming them) that they are self contained, and they aren't, for some reason like this that you wouldn't maybe notice the first time around.

which is also laziness on the instructor, but one of the reasons to teach out of a textbook is to outsource at least some of the organization/presentation of the material

i'll write this up in some notes and post them on enderton's grave

EE18

8:36 PM

I glanced back and you are in fact correct, Leslie. "Closed" was defined, but not "closure"

well taken on everything else

@leslietownes Along with the comment "your set theory text actually had some interesting comments"?

Jakobian

@imbAF isn't this better for a statistics/physics chat?

(physicists do a lot of statistics)

John Zimmerman

8:51 PM

I am trying to define a function that counts the number of functions less than a given function

only plotted 30% of it

Jakobian

discrete?

John Zimmerman

yeah.. but

it depends on how many red curves you choose to randomly generate. My conjecture is that as the number of curves tends to infinity the step function will oscillate around the identity

I am not sure how to formalize this yet...

Jakobian

good luck

John Zimmerman

it's a function of the rationals (at least for any finite number of curves)

EE18

9:20 PM

I am surprised, I would have expected that we need some statement like $0 \in A$ here

What am I missing?

Jakobian

@EE18 every number less than $0$ is in $A$

EE18

ah, vacuously. and so $0 \in A$

ok, thank you jakobian. the wording in english was confusing me with all of the quantifiers and implications

I was getting confused cause I remembered this from my other book

And here it seems that the supposition that $N \neq \emptyset$ was necessary

Jakobian

unconfuse yourself

EE18

9:35 PM

Would it be possible to elaborate? The proof clearly uses the supposition that $N \neq \emptyset$, no?

Jakobian

Doesn't elaborate. Leaves

@leslietownes what else can you tell me about periodicity of K-theories?

that you remember

Jakobian

9:56 PM

Bott Periodicity Models Consciousness? Preliminary Exploration

►

I found some weird crankery

leslie townes

10:28 PM

@Jakobian i certainly didn't know that it models consciousness

Jakobian

neither did I

psie

There is a well-known result in measure theory that relates an increasing, bounded and right-continuous function to a unique finite measure on $\mathbb R$, and vice versa. I've seen a proof of this theorem and I understand it. However, I'm reading that this theorem can be extended "easily" to measures that are only finite on compact sets. Does anyone have any idea how one would go about proving this extension? I imagine using what one has already proven, but I'm not sure how.

The theorem I'm talking about is the following:

I know that, in some way, it involves $$F(x)=\begin{cases} \mu((0,x]) & x\geq0 \\ -\mu((x,0])& x<0.\end{cases}$$

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$Mathematics$ Mathematics