Mathematics - 2024-05-31

EE18

01:09

@Thorgott Sorry Thorgott, only getting back from work now. I guess I just wanted to confirm because "identification" can preserve different properties in different contexts right?

Like as discussed, here it is not the vector space structure which is preserved

@Thorgott As relates to this comment, I completely agree they are equal as sets. But I still disagree that these are literal equalities, because they are not equal as inner product/normed spaces right?

leslie townes

EE18 a very common choice (more common than alternatives) is to define C to literally be R^2, as a set. you can get different vector and inner product spaces out of it (e.g. if you think of C as a complex vector space vs. a real vector space) but that is more about the operations on top of the set than the set itself.

[maybe i'm barging into some larger conversation i don't fully appreciate]

some books will do stuff like define the more structured objects as tuples, where you have not just "C" but (C, +, *, ., < , >)

so that you can express the distinction in notation. when you do this, "C" (without adornment) is likely an abuse of notation for some more structured thing

EE18

01:26

@leslietownes It's exactly this

The passage being discussed (just above where Thorgott responded to me) is talking abouta sort of metric space (induced from the Euclidean inner products on each) equivalence between $C^m$ and $R^{2m}$ via the natural identification

here the former is of course a $C$-vector space and the latter an $R$ vector space

leslie townes

yeah, i would see that existing more at the level of the operations you have in mind on those things than what "C^m" and "R^2m" are as sets.

there's also pedantic stuff you can get into around, like, are R^4 and R^2 x R^2 the same set, where the literal answer can maybe depend on how you encode ordered pairs and ordered 4-tuples as sets, although it is not a particularly interesting answer for any purpose

your boy enderton might even discuss this if he gets into different set theoretic encodings of the concept of ordered pair

Jakobian

01:44

@leslietownes its okay, EE18 probably needs it explained like five more times

EE18

Ya I'm not even getting into the set theory bits here, just want to pay attention to the difference between the two inner product spaces while noting their "identification" when it comes to matters of convergence

leslie townes

"identification" is a tricky concept, it is like a knife that does great things in the hand of an expert sushi chef and maybe less great things in the hands of a 5-year-old

when someone identifies two things, is is usually possible to guess at what maps are involved and what properties they supposedly have, but not always

Jakobian

I wasn't being sarcastic for the record

EE18

Here is what I wrote in the end just to summarize in my notes (as usual my writing is not parsimonious)

(1) The two equalities of sets (of the individual unit balls and of the set of neighborhoods) should be understood as meaning that the image of one under the bijection is the other (i.e. using "=" is really a shorthand).

leslie townes

jakobian i mean it's genuinely tricky, isn't it? i could cite published papers that contain errors that are buried in things being identified with other things

EE18

01:49

(2) Given that $\Bbb C$ actually is $\Bbb R^2$ when it comes to sets (recall the construction in Chapter I.11) and ignoring the operations thereon, is there even anything to be said here?

Yes: the semi-non-trivial point is that even though the two inner product spaces $\Bbb C^m$ and $\Bbb R^{2m}$ strictly speaking are different,\footnote{Having different fields of scalars, they are different as vector spaces (so obviously as inner product spaces).},

one can (easily) show that the two norms induced lead to the same balls and sets of neighborhoods in the sense of (1) above. That is the "canonical bijection" noted should be understood as a map between different vector spaces which \emph{does} identify the two (show they are "equivalent") as metric spaces in the sense of 1 above.

Jakobian you can see I used your language in the footnote :)

Jakobian

I'm trying to think of a situation where this or similar identification would cause problems

leslie townes

at the same time i would second anybody who hinted that self study time could be also be spent on almost anything other than these sorts of issues

Jakobian

It more so caused problems for me to understand a proof as an undergrad, than the proof having errors

another thing would be something like equivalence vs homeomorphism of compactifications, perhaps this would qualify as an example

or equal topologies vs topologies leading to homeomorphic spaces

leslie townes

your (or someone's) point set example of products and quotients not quite behaving the way you'd expect, i don't remember the details, is the kind of thing i had in mind

there was a guy in my field who never checked those details and had a number of false proofs because there was literal equality at the level of what the quotient object was defined to be, but not at some more relevant level

so his thinking, if you could call it that, was if the underlying set is this thing, of course the thing built on top of it will also be this thing. which maybe was even true in a lot of cases of interest, but not always

Jakobian

That one was about quotients and products not commuting, even up to homeomorphism as opposed to some canonical homeomorphism

leslie townes

01:58

i agree that 99% of learning to get nervous about these details is just so you check them and never worry about them again

Jakobian

at the end of the day, the checking process is fairly pointless

leslie townes

some of EE18's questions make me wish that they were obsessed with something like planar euclidean geometry instead of all of this other stuff, because there probably wouldn't be too many questions about what a triangle is :)

Jakobian

02:15

I don't know what triangles in (synthetic) Euclidean geometry are supposed to be, nor how does it work on axiomatic level, so I'd be trying to figure out what triangle is pretty hard

or, I suppose the first thing to do would be to actually accept some axioms

by the checking process being pointless I meant something like this

you are checking things because you aren't certain of them, so its natural
but at the same time you are wasting massive amount of time that you could just as well apply to doing some actual math

most of the time, the errors you found in a book are irrelevant

saying "oh okay they wrote it here but if I take it literally then..." that's even worse

and that's how this all pedanticism looks to me 19 out of 20 times

because there is a way to understand something for which there's no problem, but you are creating a problem

if you go to an actual math class, they do things so fast that this "proper understanding" means even less

optimize your understanding of texts, people

I don't know I might be rambling with no sense

Sine of the Time

09:20

-_-

Binky McSquigglebottom

09:32

Howcain i show that $\sum_{n\geq 0} a_n((1+p)^s-1)^n$ is analytic (i.e., can be expanded as a power series in $s$) for $a_n,s\in Z_p$? p odd, if p=2 change 1+p by 1+4??

$(1+p)^s = \sum_{k=0}^\infty \binom{s}{k} p^k.$

$\binom{s}{k} = \frac{s(s-1)\cdots(s-k+1)}{k!}.$

$For (k=0), (\binom{s}{0}=1). For (k=1), (\binom{s}{1}=s). For (k=2), (\binom{s}{2} = \frac{s(s-1)}{2}), and so on.$

$(1+p)^s - 1 = \sum_{k=1}^\infty \binom{s}{k} p^k.$

$\sum_{n\geq 0} a_n ((1+p)^s - 1)^n.
\
Substitute $1+p)^s - 1 = \sum_{k=1}^\infty \binom{s}{k} p^k$:
\[
\sum_{n\geq 0} a_n \left( \sum_{k=1}^\infty \binom{s}{k} p^k \right)^n.
\]$

Pizza

Hi

Binky McSquigglebottom

Using the multinomial theorem:
$\[
\left( \sum_{k=1}^\infty \binom{s}{k} p^k \right)^n = \sum_{m_1 + m_2 + \cdots = n} \frac{n!}{m_1! m_2! \cdots} \prod_{j} \left(\binom{s}{j} p^j\right)^{m_j}.
\].$

Therefore, the series $(\sum_{n\geq 0} a_n ((1+p)^s - 1)^n)$ can be expressed as a convergent power series in $(s)$ with coefficients in $(\mathbb{Z}_p)$. Thus, it is analytic in $(s).$

This my proof is right?

Pizza

i dont know :(

Sine of the Time

@Pizza hi

Binky McSquigglebottom

09:47

hi

can someone help me

@Pizza 🤦‍♂️

Pizza

10:07

$A=\begin{pmatrix}\frac{1}{2} & 0 & 1 \\ 1 & 2 & 0 \\ 1 & 1 & h\end{pmatrix} \Rightarrow \begin{pmatrix}\frac{1}{2} & 0 & 1 \\ 1 & 2 & 0 \\ \frac{1}{2} & \frac{1}{2} & \frac{h}{2}\end{pmatrix} \Rightarrow \begin{pmatrix}\frac{1}{2} & 0 & 1 \\ 1 & 2 & 0 \\ 0 & -\frac{1}{2} & 1-\frac{h}{2}\end{pmatrix}\Rightarrow \begin{pmatrix}\frac{1}{2} & 0 & 1 \\ \frac{1}{2} & 1 & 0 \\ 0 & -\frac{1}{2} & 1-\frac{h}{2}\end{pmatrix}$

$\Rightarrow\begin{pmatrix}\frac{1}{2} & 0 & 1 \\ 0 & -1 & 1 \\ 0 & -\frac{1}{2} & 1-\frac{h}{2}\end{pmatrix}\Rightarrow \begin{pmatrix}\frac{1}{2} & 0 & 1 \\ 0 & -\frac{1}{2} & \frac{1}{2} \\ 0 & -\frac{1}{2} & 1-\frac{h}{2}\end{pmatrix}\Rightarrow\begin{pmatrix}\frac{1}{2} & 0 & 1 \\ 0 & -\frac{1}{2} & \frac{1}{2} \\ 0 & 0 & \frac{-1+h}{2}\end{pmatrix}$

Did I reduce correctly?

if I did it right the rank of the matrix is 3 if $h \ne 1$

Binky McSquigglebottom

@Pizza No, there were some errors in your steps.

Pizza

where

Binky McSquigglebottom

So, there were some errors in the intermediate steps. The correct upper triangular shape is:
$
\begin{pmatrix}
\frac{1}{2} & 0 & 1 \\
0 & 1 & -1 \\
0 & 0 & \frac{1 - h}{4}
\end{pmatrix}
$

Pizza

$A=\begin{pmatrix}\frac{1}{2} & 0 & 1 \\ 1 & 2 & 0 \\ 1 & 1 & h\end{pmatrix} \Rightarrow \begin{pmatrix}\frac{1}{2} & 0 & 1 \\ 1 & 2 & 0 \\ \frac{1}{2} & \frac{1}{2} & \frac{h}{2}\end{pmatrix}$ here i divided by 2

@BinkyMcSquigglebottom where is my mistake

and then I did the subtraction to get a 0

Binky McSquigglebottom

The first error occurred in the step where you subtracted the first row from the third row. Quoting your passage:

From:
$\
\begin{pmatrix}
\frac{1}{2} & 0 & 1 \\
1 & 2 & 0 \\
\frac{1}{2} & \frac{1}{2} & \frac{h}{2}
\end{pmatrix}
$
to:

\begin{pmatrix}
\frac{1}{2} & 0 & 1 \\
1 & 2 & 0 \\
0 & -\frac{1}{2} & 1-\frac{h}{2}
\end{pmatrix}

Pizza

10:16

why?

:(

Binky McSquigglebottom

Here, it appears you have successfully subtracted the first row from the third row. However, if we carefully examine your result, we see that the term $\frac{1}{2}$ in the bottom left corner has not been changed correctly. It should be $0$ after subtracting the first row. So, the first error occurred precisely at this point.

Pizza

I wrote 0 in the next step...

Binky McSquigglebottom

Let's continue with the other steps to make sure everything is correct.

I apologize for misunderstanding the passage. You are right, you have correctly set the element in the lower left corner to
0
0. So, reviewing the passage, there are no errors there.

Pizza

ok then

can you see the full steps

Binky McSquigglebottom

yes

After going through all the steps, there doesn't appear to be any further errors.

Pizza

10:23

I saw that wolfram is found differently

I think the matrix depends on how I choose to reduce, right?

Binky McSquigglebottom

@Pizza can i see?

@Pizza yes

Pizza

10:42

It seems correct to me

Pizza

11:21

should be correct, right?

but I didn't consider the determinant, was I wrong?

however $r(A) = 2$, if $h = 1$, and $r(A) = 3$ if $h \ne 1$. sorry I wrote wrong in the photo.

leslie townes

12:27

pizza, without reading through those calculations, there is certainly no reason why you would have to use the determinant in a problem like that (although you could certainly use it as part of identifying values of h for which dim ker f is nonzero). case analysis based on a row echelon form is a very natural approach, provided of course that you handle the computation and the cases correctly

Pizza

@leslietownes yes in the end by rank theorem, I can say that dim Im f = rank(A) , right?

yes in the end by rank theorem, I can say that dim Im f = rank(A) , right?

leslie townes

i don't know what "the rank theorem" is; different textbooks use different definitions and names for these things. those two things are definitely the same

Pizza

Rank–nullity theorem

The rank–nullity theorem is a theorem in linear algebra, which asserts: the number of columns of a matrix M is the sum of the rank of M and the nullity of M; and the dimension of the domain of a linear transformation f is the sum of the rank of f (the dimension of the image of f) and the nullity of f (the dimension of the kernel of f). It follows that for linear transformations of vector spaces of equal finite dimension, either injectivity or surjectivity implies bijectivity. == Stating the theorem == === Linear transformations === Let T : V →...

leslie townes

"rank(A)" is very frequently defined to be dim Im(f), or when it is not, whatever it is defined to be is easily shown to be dim Im(f)

so, as many books set it up, often, the equality of rank(A) with dim Im(f) is more fundamental than the rank nullity theorem

Pizza

👍

leslie townes

13:02

shoutouts to my son for making it to 5am this morning before waking up screaming

Xander Henderson

13:12

Ha!

I made it to 3 this morning before waking up. I'm waiting for my brother to wake up and show me where the coffee is before I continue on to Tucson.

Pizza

@XanderHenderson What are your plans for the summer

Xander Henderson

Teaching.

Classes start Monday.

Pizza

👍

Good luck then !

Binky McSquigglebottom

14:02

$\int e^{-2t^2+2xt} dt$ how to solve this

leslie townes

by completing the square and a change of variable it is a multiple of e^(-u^2) du, which does not have an elementary expression, but is well tabulated and known. if you were taking a definite integral, that might have an elementary expression

Binky McSquigglebottom

thanks

Pizza

14:22

$\begin{pmatrix}\frac{1}{2} & 0 & 1 \\ 1 & 2 & 0 \\ 1 & 1 & 1\end{pmatrix} \cdot \begin{pmatrix}x \\ y \\z\end{pmatrix}= \underline{0} \Rightarrow \begin{cases}\frac{1}{2}x+z=0 \\x+2y=0 \\ x+y+z=0\end{cases}$

It is a linear system that admits $\infty$ solutions, with the choice of some parameters, how can I understand which ones?

for example

$\begin{pmatrix}2& 1 & 0 & 1 \\ 0 & 1 & 0 & 1 \\ 0 & 1 & 0 & 1 \\2 & 3 & 0 & 3\end{pmatrix} \cdot \begin{pmatrix}x \\ y \\z \\ w\end{pmatrix}= \underline{0} \Rightarrow\begin{cases}2x+y+w=0 \\y+w=0 \\ 2x+3y+3w=0\end{cases}$

It is a linear system that admits $\infty^2$ solutions, that is, infinite solutions
with the choice of 2 parameters; in this case they are the variables $z$ and $w$

How can i understand it?

leslie townes

pizza this is very difficult to talk through on chat, even simple examples involve a lot of computation. it is best to find a textbook that addresses these issues. there is sometimes an element of arbitrary choice involved, for example, in choosing which variables are 'free.' this choice is often more an artifact of the specific solution process than it is reflective of any underlying mathematical reality.

as a very simple example in the single equation in two variables x and y given by x - y = 0, you could take either variable to be 'free', with the other determined from the 'free' one by that equation. a lot of textbook methods would decide which one is the 'free' one from how the variables are ordered.

Jakobian

14:50

@XanderHenderson why so early?

Heard that waking up early is a symptom of depression

Sine of the Time

I've read that you're more productive

psie

15:44

In this answer, I just have a small comment. In the quoted part, and the first displayed equation, that is $$\int_{F[A]}f\,d\nu=\int_AF^\ast(f\,d\nu):=\int_Af\circ F\,d(F^\ast\nu).$$ What does $F^\ast(f\,d\nu)$ mean? Doesn't $F^\ast$ take a measure and outputs another measure? The way it is written looks like it takes a function.

Ah wait.

I think I understand it now.

Binky McSquigglebottom

im writing

psie

The answerer defines $F^\ast(f\,d\nu)$ to be $f\circ F\,d(F^\ast\nu)$.

Binky McSquigglebottom

In summary, $F^\ast(f \, d\nu)$ is a concise notation indicating the operation of composing $f$ with $F$ and integrating with respect to the pullback measure $F^\ast \nu$ . It does not mean that $F^\ast$ acts directly on $f$; rather, it denotes the combined effect of measure pullback and function composition.

psie

ok, thank you! :)

Jakobian

16:19

@psie $fd\nu$ would mean the measure $(fd\nu)(A) = \int_A fd\nu$

$F^*(fd\nu)(A) = \int_{F[A]}fd\nu$ is the pullback of this measure

the difference is between, if we put $A$ as an input or in the lower index of the integral, but this is essentially the same thing

I don't think that the author is right to define $\int_A F^*(fd\nu)$ to be $\int_A f\circ F d(F^*\nu)$, I think the first equality is a definition, and second is the result

I.e. the equality we are showing is a formula $$\int_A F^*(fd\nu) = \int_A f\circ F d(F^* \nu)$$

@psie @BinkyMcSquigglebottom

@peek-a-boo

see also comments about notation $d\mu$, like:

@Alexey There is a clear answer: it has no stand-alone meaning, it is just crappy notation inherited via tradition. It doesn't represent a differential form. — Chill2Macht Sep 30, 2018 at 22:32

psie

16:35

ok. I also think the quoted passage by peek-a-boo has stronger assumptions than what peek-a-boo is restating in the link, namely here. peek-a-boo assumes that $F$ has a measurable inverse, which the change of variables formula using pushforwards does not assume, i.e. we assume we have only a measurable function.

Binky McSquigglebottom

Thanks so much for the clarification. Now everything is much clearer. As you explained, the equality $\int_A F^*(fd\nu) = \int_A f\circ F d(F^* \nu)$ is a formula that proves the result of the measurement transformation under the map $F $.

Thanks again for your detailed explanation.

Jakobian

@psie the problem is that this is a pullback and not a pushforward, the measure is on the codomain

that's why it was assumed that $F$ has a measurable inverse, otherwise taking pullback might not make sense i.e. $F[A]$ does not have to be measurable

perhaps this can be weakened to "$F$ takes measurable sets to measurable sets" though

such functions definitely exist, iirc absolutely continuous functions take Lebesgue measurable sets to Lebesgue measurable sets

(though I might be wrong, I know they take null sets to null sets)

ah no, I'm right they do take (Lebesgue) measurable sets to themselves

more generally any continuous function $F:\mathbb{R}^n\to\mathbb{R}^n$ which takes null sets to null sets should map Lebesgue measurable sets to Lebesgue measurable sets

and any continuous function from $\mathbb{R}^n$ to $\mathbb{R}^m$ takes Borel measurable sets to themselves, of course, with the caveat that the sigma-algebra is not complete anymore

psie

so is the quoted passage by peek-a-boo equivalent to the statement found here? Or is it a weaker/stronger statement? I understand that we need a measurable bijection for the pullback to make sense.

Jakobian

@psie in some sense its stronger since we can just apply it to $g = f\circ F_0$

where $F_0$ is our invertible function and $F = F_0^{-1}$

i.e. we can easily prove it from that one about pushforwards

psie

ah ok

Jakobian

16:49

I mean the pushforward one is stronger than the pullback one, in this sense of being easily able to prove one from another

of course they're statements about slightly different things so you can't directly compare their strength

Sanjana

17:34

Does anyone know how to evaluate $$\int_0^\infty \frac{t^{x-1}}{t+1} dt$$ without using Contour integrals? The answer is $\frac{\pi}{\sin(\pi x)}$

This arises in a proof of Euler's reflection identity

Jakobian

$x$ is from what?

Sine of the Time

@Sanjana let $b=1$ here

Jakobian

yeah okay just beta function

Sanjana

@SineoftheTime Oh oh :( I was using this integral to prove Euler's reflection identity. Any other possibilities?

Jakobian

thought it would be something like that

Sine of the Time

17:39

didn't read your second message

Jakobian

@Sanjana what about taking derivative and solving an ODE

Sanjana

But wouldn't solving that ODE involve solving the same integral?

Jakobian

what do you mean

did you try it already and it popped up

Sine of the Time

see here, maybe you can find something

Kripke Platek

Is there a function f:[0,1]→[0,1] such that:
(1) f is everywhere continuous
(2) f is nowhere differentiable
(3) f(0)=0 and f(1)=1 (Optional)
(4) for every c∈image(f), there is only a finite number of points x[1],...,x[k] in [0,1] such that f(x[i]) = c. [1≤i≤k]

Sine of the Time

17:43

Mark Viola's answer should do the job @Sanjana

Sanjana

@SineoftheTime Yeah it does. Thanks!

Sine of the Time

np

Sanjana

@Jakobian Now, i see what you were talking about

Jakobian

@KripkePlatek no one would state condition (4) like this, to be honest. You'd just say $f^{-1}(t)$ is finite for all $t\in [0, 1]$

Sine of the Time

it's not so easy to build nowhere differentiable functions :D

Jakobian

17:54

maybe (1) and (4) implies that $f$ is increasing on some strip, hence not nowhere differentiable

Suppose for all $0\leq t_1 < t_2\leq 1$ there exists $t_1 < t < t_2$ for which $f(t_1), f(t), f(t_2)$ is not monotone.

Maybe lets start with $0 < t_1 < 1$ then $t_n < t_{n+1} < 1$

so we have a strictly increasing sequence $t_n$ such that $f(t_n), f(t_{n+1}), f(1)$ is not monotone for all $n$

and I believe now all left to do is pick some fiber and show its infinite

nah this wouldn't work (this argument)

but I'm pretty sure (1) and (4) implies increasing on some strip, intuitively

yeah like if we take $r > 0$ such that $f^{-1}(f(0))\cap (0, r) = \emptyset$

I think for every $t\in [0, 1]$ we want to take a neighbourhood $U_t$ of $t$ with $f^{-1}(f(t))\cap U_t = \{t\}$ perhaps

Obliv

I'm guessing there is no way to escape the recursive definition of relations i.e., defining a relation as a set via the "=" relation

Jakobian

and I think there has to exist distinct points $t_1, t_2$ with $t_1, t_2\in U_{t_1}\cap U_{t_2}$

and then taking $t$ with $t_1 < t < t_2$ but $f(t_1), f(t), f(t_2)$ is not monotone

I should obtain a contradiction from IVT

@Jakobian something something Baire category theorem

Binky McSquigglebottom

18:18

$\displaystyle\iiint_R (x^2+y^2+z^2)^{-2}\,dx\,dy\,dz$ where $R$ is in the region in the first octant outside the sphere $x^2+y^2+z^2 = 1$

Jakobian

but the argument relies on existence of such distinct points $t_1, t_2$

Binky McSquigglebottom

$x = \rho \sin \phi \cos \theta, \quad y = \rho \sin \phi \sin \theta, \quad z = \rho \cos \phi$
The Jacobian of the transformation from Cartesian to spherical coordinates is $\rho^2 \sin \phi.$ Thus, the volume element $dx\,dy\,dz$ becomes:
$dx\,dy\,dz = \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta$

Jakobian

I think for any open cover $\mathcal{U} = \{U_t : t\in [0, 1]\}$ of $[0, 1]$ with $t\in U_t$ there exists such distinct points

Binky McSquigglebottom

$x^2 + y^2 + z^2 = \rho^2 \quad \Rightarrow \quad (x^2 + y^2 + z^2)^{-2} = \rho^{-4}
(x^2 + y^2 + z^2)^{-2} \cdot \rho^2 \sin \phi = \rho^{-4} \cdot \rho^2 \sin \phi = \rho^{-2} \sin \phi$

Jakobian

I'd be really surprised if not

Binky McSquigglebottom

18:25

$\rho \in [1, \infty), \quad \phi \in [0, \frac{\pi}{2}], \quad \theta \in [0, \frac{\pi}{2}]$

$\iiint_R (x^2+y^2+z^2)^{-2}\,dx\,dy\,dz = \int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}} \int_1^{\infty} \rho^{-2} \sin \phi \, \rho^2 \, d\rho \, d\phi \, d\theta$

$\int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}} \int_1^{\infty} \sin \phi \, d\rho \, d\phi \, d\theta$

$\int_1^{\infty} \rho^{-2} \cdot \rho^2 \, d\rho = \int_1^{\infty} 1 \, d\rho = [\rho]_{1}^{\infty} = \infty - 1 = \infty$

Is correct?

Sine of the Time

shouldn't it be $1/\rho^2$ ?

Binky McSquigglebottom

I do not think so

Pizza

the June 18th algebra and geometry exam has been cancelled... :(

I'll have to do it in September...

Sine of the Time

@BinkyMcSquigglebottom why?

Jakobian

I'm drawing a blank about this open cover stuff above. I feel like Baire category theorem is the way to go but how do I apply it

Sine of the Time

18:32

@Pizza can't you do it in July?

Pizza

@SineoftheTime in the month of July, I cannot take exams, due to personal problems

Sine of the Time

oh

sorry to hear that

Pizza

so I will try to prepare as many exams as possible for September, before the new courses start

Binky McSquigglebottom

@SineoftheTime where?

Sine of the Time

@BinkyMcSquigglebottom here you multiplied by the jacobian and then you multiplied another time by $\rho^2$

Jakobian

18:34

Let $A = \{t\in [0, 1] : \forall_{s\in U_t} t\notin U_s\setminus\{t\}\}$ I suppose

Sine of the Time

$\frac{dxdydz}{(x^2+y^2+z^2)^2}=\frac{\rho^2\sin\phi d\rho d\phi d\theta}{\rho^4}$

Jakobian

$A = \{t\in [0, 1] : \forall_{s\in U_t\setminus\{t\}} t\notin U_s\}$

Binky McSquigglebottom

@SineoftheTime ?

Sine of the Time

what is not clear?

Binky McSquigglebottom

It seems to me it's the same thing

@SineoftheTime all clear

Sine of the Time

18:42

then you should have $1/\rho^2$ instead of $1$

Binky McSquigglebottom

then I find that it converges

so no.

This morning I was able to do all the exercises, now you've made me doubt

Sine of the Time

@BinkyMcSquigglebottom here why did you multiply by $\rho^2$ ?

Binky McSquigglebottom

When we convert the integrand $(x^2 + y^2 + z^2)^{-2}$ to spherical coordinates, we must consider that the square of the radial distance $(x^2 + y^2 + z^2 )$ becomes $\rho^2$ in spherical coordinates.

Sine of the Time

so?

Binky McSquigglebottom

so that integral comes out

Sine of the Time

18:52

$(x^2+y^2+z^2)^{-2}=\rho^{-4}$

Binky McSquigglebottom

🤦‍♂️🤦‍♂️🤦‍♂️It's TRUE, I've been doing math for 8 hours, you can understand me...

$\int_1^{\infty} \rho^{-4} \, d\rho = \left[ -\frac{1}{3} \rho^{-3} \right]_1^{\infty} = \left(-\frac{1}{3} \cdot 0\right) - \left(-\frac{1}{3} \cdot 1\right) = \frac{1}{3}$

Obliv

I was reading this and I don't fully understand the first condition: "There is an algorithm such that the set of input numbers for which the algorithm halts is exactly S". Doesn't this just mean $S$ is finite?

Binky McSquigglebottom

$\int_0^{\frac{\pi}{2}} \sin \phi \, d\phi = \left[ -\cos \phi \right]_0^{\frac{\pi}{2}} = -\cos \frac{\pi}{2} + \cos 0 = 0 + 1 = 1$

$\int_0^{\frac{\pi}{2}} d\theta = \frac{\pi}{2}$
$\frac{\pi}{2} \cdot 1 \cdot \frac{1}{3} = \frac{\pi}{6}$

now , is correct?

I've been doing this exercise for 1 and a half hours

Soumik Mukherjee

@KripkePlatek If we take the Weierstrass function W(x), take the graph (x,W(x)) in R^2 and take the projection on [0,1] in x-axis, will that satisfy 4?

Xander Henderson

@Obliv I don't see anything in that page which indicates that $S$ is a finite set.

Indeed, they author(s) seem to be working very hard to talk about "enumerable", which suggests that countably infinite sets are considered.

Moreover, I would expect that any finite set is computably enumerable, no?

Obliv

19:08

Yeah, I'm not getting this. I can just devise an algorithm that is the 0 map and halts after dumping every input to the kernel. but it says "The first condition suggests why the term semidecidable is sometimes used. More precisely, if a number is in the set, one can decide this by running the algorithm, but if the number is not in the set, the algorithm runs forever, and no information is returned. A set that is "completely decidable" is a computable set."

Jakobian

Oh like assume $V_t = (-1/n+t, 1/n+t)$ for some $t$, then let $A_n = \{t \in [0, 1] : \text{diam}(V_t) = 2/n\}$. If $t, s\in A_n$ are distinct then from assumption $|s-t|\geq 1/n$. But then $[0, 1] = \bigcup_n A_n$ is countable

Obliv

@XanderHenderson I'm confused by the "but if the number is not in the set, the algorithm runs forever, and no information is returned." seems sketchy

Jakobian

@SoumikMukherjee I'm pretty sure I just proved such function doesn't exist

Xander Henderson

@Obliv I would read the article, rather than just the lede. The lede is only there to give a high level summary.

There are formal definitions below.

Obliv

I wasn't even aware I was only reading the lede, thank you lol

Ugh, chasing my tail with these definitions that all seem vaguely similar

Binky McSquigglebottom

19:13

Is there a way to analytically get a direction vector from spherical coordinates? Currently I convert spherical coords very close to each other to cartesian sapce, and then subtract them to get this direction, but this is a fragile solution that is inaccurate at extremes.

Obliv

@BinkyMcSquigglebottom what is a direction vector?

Sanjana

Is there any book which contains lots of detailed proofs of relations involving "special functions" like Legendre, Bessel, Hermite, etc. ?

Binky McSquigglebottom

@Obliv in what dimension?

Obliv

I'm asking what a direction vector is in the context of your question, so whatever dimension you're asking about?

Jakobian

No wait I think I've made an error: that $f$ is not increasing on any strip doesn't mean that for all $t_1 < t_2$ there exists a $t_1 < t < t_2$ with $f(t_1) < f(t) < f(t_2)$ not increasing

Binky McSquigglebottom

19:18

@Obliv euclidean space

Jakobian

It means that there exist $t_1 < s_1 < s_2 < s_3 < t_2$ such that $f(s_1), f(s_2), f(s_3)$ is not increasing

Binky McSquigglebottom

@Obliv Given a fixed nonzero vector $\mathbf v$, "the direction vector of/corresponding to $\mathbf v$" would typically refer to the unit (length/magnitude) vector in the same direction: $\dfrac{\mathbf v}{\Vert\mathbf v\Vert}$.

i mean this

Obliv

Okay, so what are you asking for specifically? Spherical coordinates are just a way of writing your basis in $\mathbb{R}^3$

Binky McSquigglebottom

yes

Is there a way to analytically get a direction vector from spherical coordinates?

The current method of converting to cartrsian works, I normalize the direction vector by dividing it by its length.
My problem is that it's a brute force method that relies on adding an epsilon value to A, which is a poor idea in floating point arithmetic.
Which is why id like to avoid it altogether and find an analytical solution

Obliv

I have no idea what you're asking. The direction is specified in $\mathbf{v}$ and you yourself said you get the unit vector by $\frac{\mathbf{v}}{\|\mathbf{v}\|}$

Why are you adding/subtracting vectors

Binky McSquigglebottom

19:26

what is not clear?

Obliv

Do you not have $\mathbf{v}$?

Binky McSquigglebottom

where

it's theory

Obliv

it's not. It's definition and you haven't stated what your problem is so I don't think I can help.

Binky McSquigglebottom

wdym

I apply a rotational correction to put them back within the limit.

Pizza

@BinkyMcSquigglebottom ?

Binky McSquigglebottom

19:33

🤦‍♂️

Thorgott

21:11

@EE18 Yes, it depends on context. I'm just saying that when a text says they are doing an identification, it means precisely this: regarding things as equal with some identification (generally speaking, an isomorphism) implicit in the background

@EE18 well, each of those symbols could mean a dozen different things, so it depends on what they're supposed to denote

the topological space $\mathbb{C}$ and the topological space $\mathbb{R}^2$ are literally equal as far as I'm concerned

EE18

21:39

@Thorgott I think this is the main point for my understanding on this

And when you say "equal as far as I'm concerned" you mean in the sense of identification discussed right? in the sense of images of neighborhoods equalling neighborhoods etc

Xander Henderson

@EE18 I still feel like you are making this way harder than it should be...

Thorgott

no, I just mean that that's how I would define these things

psie

22:23

Does anyone where I can find a proof of the following lemma? It is stated without proof in my book.

I feel a bit weird about this lemma, since I'm not really sure what it is saying. The definition of an (absolutely) continuous r.v. is that it admits a density. So, I'm not really sure what this lemma is saying.

peek-a-boo

23:08

@Jakobian my bad, for the sake of presentation I should have just written $\int_{F[A]}f\,d\nu=\int_A(f\circ F)\,d(F^*\nu)$, and this is of course not a definition, but a corollary of the more general change-of-variables theorem involving pushforwards, applied to the case of pullbacks. The chain of equalities $\int_{F[A]}f\,d\nu=\int_AF^*(f\,d\nu)=\int_A(f\circ F)\,d(F^*\nu)$ is just how I ‘remember’ the theorem because it behaves so naturally with respect to pullback, just like forms.

and of course, if anything the first in this chain of equalities is essentially a definition, while the second is an application of the theorem.

@psie yup as Jakobian mentions, my quoted formula is weaker than the one in the link (hence why I said “let us consider a special case rewriting of the theorem in the link” right at the beginning).

psie

ok, good to know :)

leslie townes

23:26

psie: if the definition of a continuous rv is that it admits a function like that, seems like it's just saying that the density is uniquely determined by (1) a background assumption that the "h" satisfying that formula is a continuous function (and not some more general thing that might be integrated), and (2) the values of those integrals

which depending on your toolkit should indeed not be hard to prove and maybe more of an observation (or as they call it 'lemma') than a key result

Lucas Henrique

hey chat

psie

alright, I was leaning towards that interpretation, so more of a uniqueness statement

leslie townes

if we're all a little concerned about the "for every B subset R^n" in a probability context, we can let that slide just this once (usually there is a background assumption of measurability or even more regularity than that on the subsets you consider). if h is continuous it would be enough to have that hold on 'nice' subsets B

even just dyadic rectangles oughta do it

Lucas Henrique

so I'm trying to prove that normality is a local property - if $R_{\mathfrak m}$ is normal for every maximal ideal, $R$ is normal. I'm trying to do that by choosing an appropriate maximal ideal for which the $s$ in $r/s \in \operatorname{Frac}(R)$ belongs.

the problem is, i can always make $s$ belong to any maximal ideal by multiplying the numerator and denominator by the same element. so i'm wondering if there's any way to take a minimal pair, something like gcd = 1 elements when we're working with UFDs

does anyone know about something like that?

psie

ok, thanks leslie

Lucas Henrique

23:32

my initial idea is to consider the set of representants of the same fraction and partially order them by the following: $r/s \le r'/s'$ if $(r,s) \subseteq (r',s')$ (this last inequality might be reversed, but whatever)

something something, zorn and take a minimal element. but i'm really not sure whether this works

peek-a-boo

thanks to your comments, I have edited my answer to clarify some of these points

psie

👍

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$Mathematics$ Mathematics