Mathematics - 2023-12-11 (page 1 of 2)

Soumik Mukherjee

04:49

@Koro yes

one potato two potato

05:12

why do people say conditional "expectation" even though it's a random variable not a value

Soumik Mukherjee

06:28

@Jakobian Once you wrote the argument, I was able to understand the rest:)

My biggest problem is that I get too nervous during exams, and start doubting obvious things. Like there was this question yesterday. There exists a metric space with exactly $2024$ open sets(True/false). I checked the question at least $5$ times to get sure it was written $2024$ and not $2048$.

copper.hat

07:20

@onepotatotwopotato the undergraduate version is a number, i presume know that

one potato two potato

yeah so people just abuse the term I guess

conditional random variable is more suited

copper.hat

It is a pity the same term is used, it is very confusing for those making the transition

Thomas Finley

15 hours ago, by Sine of the Time

@ThomasFinley It's still valid and I answered it assuming $e_3=(0,1,1)$

@SineoftheTime Thanks a lot!

4

A: The Newton-Raphson Method for finding a correct monthly interest rate

For intuition, you might be interested in reading Why does Newton's method work? and Math Insight. As for this specific example, we cannot find a nice closed-form solution, so we are stuck using numerical methods instead and will choose Newton's Method, but many other root finding methods will wo...

I think tge iterations are impossible to do manually unless some technology is implemented.

copper.hat

there is this thing called a compoopoo

one potato two potato

09:39

@Jakobian Let $\Lambda\subset\Bbb S^2$ be a closed subset such that $\Bbb S^2\setminus\Lambda = \Omega_1\coprod\Omega_2$ where $\Omega_1$ and $\Omega_2$ are connected, simply connected open subsets of $\Bbb S^2$. Then $\Lambda\cong \Bbb S^1$?

Daniel Donnelly

0

Q: Linear independence of functions $[d \mid x \pm 1]$ on an interval $[1, b]$ such that $b \geq$ any involved divisor $d$?

For $d \in \Bbb{N}, d \geq 1$, define for $x \in \Bbb{R}$: $$ [d|x] := \begin{cases} 1, \ \text{ if } x \in d\Bbb{Z}; \ \\ 0, \ \text{ otherwise} \end{cases} $$ defines a family of linearly independent functions $B = \{[d \mid \cdot ] : [1, b] \to \{0,1\}; d \in I\}$ over any field $K$, whenever ...

one potato two potato

I guess topologist sine curve works

Thorgott

10:08

the topologist's sine curve has connected complement, but the "Warsaw circle" (topologist's sine curve plus an arc joining both ends) should work

Jakobian

10:48

@onepotatotwopotato I mean you can take $\Lambda$ to be a thicker strip is all. Unless you meant homotopy equivalent?

@Thorgott people from Warsaw clearly have never seen a circle

X4J

11:09

Can someone help me to understand why is it trivially claimed that $KerJ^k=ImJ^k$? math.stackexchange.com/questions/2269721/…

I struggle to see it

Or is there an equivalent way to describe this so it looks more intuitive?

Thorgott

11:20

@Jakobian or you've never been to Warsaw

ZaWarudo

Let $X$ be a topological space. If $X$ is T1, then for each $x \in X$ the set $\{x\}$ is closed. My attempt to the proof: let $x \in X$ and $y \in X \setminus \{x\}$ be arbitrary. Since $X$ is T1, there exists a open neighborhood $V_y$ of $y$ such that $x \notin V_y$. This means that $\{x\} \subseteq X \setminus V_y$, and so taking the complement $V_y \subseteq X\setminus \{x\}$.

This shows that each $y \in X\setminus \{x\}$ has a open neighborhood $V_y$ contained in $X \setminus \{x\}$, that is $X \setminus \{x\}$ is open and so $\{x\}$ is closed. My lecturer proves this differently, by noticing that $X\setminus \{x\}=\bigcup_{y \in X\setminus\{x\}} V_y$. Is my proof correct as well?

giannisl9

12:23

Hello! I am an electrical engineering student and I want someone to explain some concept’s related to vector spaces, measure and change of variables. No need for formal proofs, I will ask my own questions. I want the intuition for now but I don’t have the time to study topology and functional analysis. I can pay for it. Thanks

Soumik Mukherjee

13:15

"A topological space has the fixed-point property if and only if its identity map is universal."

What is the definition of universal map?

Jakobian

12

Q: Universal maps between topological spaces

Let $X,Y$ be topological spaces. We call a continuous map $u:X\to Y$ universal if for every continous map $f:X\to Y$ there is $x\in X$ such that $f(x) = u(x)$. If $u:X\to Y$ and $v:Y\to Z$ are universal, is $v\circ u: X\to Z$ universal? (I am thinking about this question because of an inspirati...

gn.general-topology

@ZaWarudo Sure its correct

Soumik Mukherjee

Okay so the statement holds as for any map $f$ we have a point $x$ such that $f(x)=id(x)=x$

Jakobian

Yep. Its literally what it means for $X$ to have fixed point property

Soumik Mukherjee

13:34

And fixed point property is topologically invariant so this is another way of checking whether two spaces are homeomorphic or not

Alessandro Codenotti

Sure but checking the FPP for a space is often nontrivial

one potato two potato

arxiv.org/abs/2312.05226

Just one crazy paper that relates PDE and AG

giannisl9

15:26

So no help I take it

Not that I deserve it in some way but its sad that the world wide math space with so many people over the world gives no answer

Koro

:-)

Jakobian

@giannisl9 its not the whole world and people prefer to help people for free

without any sort of obligations coming from being paid

giannisl9

I said I could pay

Not that I will

@Jakobian I just don't have the knowledge and I would like someone to help me because I don't want to go in depth, I want to get a feeling from someone who understands it for now, because I don't have the time

@Jakobian and because its not one question and certainly not a question to ask in the main questions

@Jakobian I just want to chat with someone one-on-one

@Jakobian that takes time and so if someone wishes one can get some compensation for his time

that's all

Koro

15:42

You can ask your questions here.

as you said -you'll ask your own questions.

Jakobian

I'd love to chat with someone one-to-one that would explain my doubts about everything, sadly that's not possible

Koro

it's possible in UG courses.

but not in PG courses like algebraic topology etc.

UG-undergraduate, PG- postgraduate

Jakobian

well yeah, the accessibility of people of knowledge starts to dim after some point

Koro

I say it's all demand and supply.

giannisl9

I will try

Koro

15:44

There are so many books on real analysis.

giannisl9

And just remember I want the intuition

Jakobian

I mean most people here want to learn on their own and discuss something potentially interesting

giannisl9

I am an electrical engineer student

Koro

ask a question on mse on 'how to compute this limit?' and see the upvotes and answers.

giannisl9

late in my studies

and don't have the time to study topology

rn

and so on

Koro

15:45

@giannisl9 no problem. Just write your questions and ask them here.

giannisl9

I know somewhat
General idea of vector space
Functions as members of vector space
Domain
Set, Subset
Topology, topological space
L^2 Norm, Inner product

I want to know better
Change of variables in integrals
Measure
change of measure -> change of space?
Field vs Space
How to go from topological space to vector space for functions
Notation

Koro

someone will most likely answer.

giannisl9

Example

∫_1^4▒〖dx=3〗

Here the function is 1 the set is (1, 4) and the measure is x?
Is area 3 a characteristic of the function regardless of space/transform? Is the area related to how we measure it? ie x?

Measuring the same set with x^2 so in terms of x^2 is (1, 2)

u(x)=x^2

∫_1^2▒du=1

∫_1^2▒〖2dx=2〗

Now I am lost..

I think u is the measure in another space where the same object (1,4) is something else?

Koro

1 min ago, by Koro

ask a question on mse on 'how to compute this limit?' and see the upvotes and answers.

Jakobian

are you on PC? See the channel description on how to use MathJax in chat

Koro

15:46

add to it 'do the limit without L'Hopital', that's more response.

giannisl9

Can you relate some terms I know to the change of variables case?

Jakobian

I mean for measure spaces?

Koro

But ask a question on Algebraic topology etc. and wait for it to be marked as duplicate or be closed.

or deleted

Jakobian

there's no such thing as measure is "x" too

giannisl9

I want an explanation

Not just there is no such thing

and so on

Koro

15:47

there are only a few books on AT unlike real analysis. This is purely due to low demand.

giannisl9

what is x?

Jakobian

the standard measure on $\mathbb{R}^n$ is called Lebesgue measure, the one for which $\lambda([a_1, b_1]\times ...\times [a_n, b_n]) = (b_1-a_1)\cdot ...\cdot (b_n-a_n)$ and which we call the $n$-dimensional area

Koro

profit from 1 book is say 1$ and if no. of students studying AT is only say 1000 in the world, then that's only 1000$ with taxes.

Jakobian

If you want to talk about integral $\int_a^b f(x)\mathrm{d}x$ then one way to interpret it is as integration over the Lebesgue measure $\int_a^b f(x)\mathrm{d}\lambda(x)$

but you can also interpret it without measure theory

Koro

but in case of real analysis the profit is much more due to huge demand.

that's why plethora of real analysis books.

giannisl9

15:52

Can you relate that to the example in more simple language?

Let me ask specific questions

If we change variables does that mean that we change space?

Jakobian

see chat description on how to see latex in chat

its better if you write things in latex

Soumik Mukherjee

@Koro Also one of the best books in AT is free so why would someone bother to write new books in AT

Jakobian

@giannisl9 we change the area of integration

do you know integration on $\mathbb{R}$?

Koro

@SoumikMukherjee that's propaganda. It sure is not one of the 'best'.

Jakobian

Hatcher fans will surely argue: But its free, and this is for a certain audience!

giannisl9

15:55

@Jakobian Rieman integral of a function?

Jakobian

for example, yeah

giannisl9

@Jakobian most of it

you also said on R

you also said on $\mathbb{R}$

Jakobian

I mean details of how it works are irrelevant, but the point is that if you change variables, you are integrating on a different domain, the transformation maps some set to some other set you integrate on

giannisl9

what is meant by on $\mathbb{R}$?

Jakobian

it works in multi-dimensional case how it works in $\mathbb{R}$ is my point

giannisl9

15:58

@Jakobian the domain is part of the definition of the function. is the domain the same as the set for which the topological space is defined?

@Jakobian does different domain mean different space?

Jakobian

Why are topological spaces relevant here?

giannisl9

I am trying to think this way

are they not relevant?

Jakobian

Maybe its better if you just told me what problem you're thinking of

topological spaces are relevant to integration and measure theory if we want to mix them in together

but the two things can be discussed independently

that usually comes up with when we're trying to talk about Borel and Radon measures

giannisl9

Let me walk you through how those questions came up then

Maybe you get more understanding

Jakobian

alright

giannisl9

16:01

I studied the Fourier series in one of my courses

And I liked the idea of basis functions and expansion and two different representations of the same thing

so

I thought where are the complex exponential defined

is it the circle

is it time

Is it a different space angle and time

\int_{X}fdμ(x)

$\int_{X}fdμ(x)$

Then I read about L^2

Jakobian

Complex exponential is defined on complex numbers

giannisl9

@Jakobian how you mean

by on you mean the domain?

Jakobian

yes

$\exp(z) = \sum_{n=0}^\infty \frac{z^n}{n!}$ is the definition of the exponential

by theorems from real analysis, this defines a continuous function on complex numbers

giannisl9

is time some sort of parametrisation?

Jakobian

what do you mean?

giannisl9

16:07

I mean one sees $e^{j2\pi t}$

Jakobian

thats just a certain volue of the exponential

but $\exp(z) = e^z$ is defined for all complex numbers $z$

giannisl9

okay my bad

Jakobian

for example, we can talk about it for $z = j2\pi t$

giannisl9

lets see it as a function

Jakobian

arguably its the most important function in complex analysis

giannisl9

16:08

$f(t)=e^{j2\pi t}$

so one question that comes up is

Jakobian

for $t\in [0, 1]$?

giannisl9

@Jakobian ok

Now $[0,1]$ is what?

we say L^2 on [0,1]

Jakobian

$[0, 1] = \{x\in\mathbb{R} : 0\leq x \leq 1\}$

its an interval

giannisl9

lol 😂 I like you

okay

I rephrase

$L^2$ on $[0,1]$ means square integrable l functions that have domain $[0,1]$

Jakobian

So are we discussing Fourier transform or something? You've studied Fourier series and discovered Fourier transform?

giannisl9

16:11

?

now just the Fourier series

is my statement okay

Jakobian

$L^2[0, 1]$ is the set of equivalence classes of functions $f\equiv g$ iff $f = g$ everywhere except for a set of measure zero, such that $|f|^2$ is Lebesgue integrable

Thorgott

@Koro there's a lot

giannisl9

@Jakobian that doesn't help me at all

at least intuitively

is it okay to say that?

all the square integrable functions on that interval?

Jakobian

Alright so first define $\mathcal{L}^2([0, 1]) = \{f:[0, 1]\to\mathbb{C} : f^2\text{ is Lebesgue integrable}\}$

giannisl9

@Jakobian okay

Jakobian

16:14

Where the latter means that $f$ is measurable, and $\int |f|^2 < \infty$

where the integral is a Lebesgue integral

giannisl9

@Jakobian ok

now my question is what is the measure? you didn't include it in the integral

Jakobian

now we want to define a "norm" $\|f\|_2 = \sqrt{\int |f|^2}$

giannisl9

I am ok with norm

Jakobian

but the issue is that $\|\cdot \|_2$ is not a norm on $\mathcal{L}^2$

$\|f\|_2 = 0$ if and only if $f = 0$ everywhere except for a set of measure zero

thats why we need to identify functions which are equal everywhere except for a set thats irrelevant

giannisl9

wait let me think of that for a moment

Jakobian

16:17

Thus we define $L^2 = \mathcal{L}^2/\equiv$, the set of equivalence classes, where $f\equiv g$ iff $f = g$ everywhere except for a set of measure zero

this makes $L^2$ into a normed vector space

For example, if you modify the zero function, $f(x) = 0$ for $x\in [0, 1]$ on a finite set, the integral doesn't change

its still zero

giannisl9

I don't think I can understand this because for me the norm is some kind of energy and apart from the strict definition of the norm which is not satisfied as you say I think of defining the norm in the sense that the functions have finite energy on that interval

Do you think this is important for my intuition ?

my basic understanding minimal

of what is happening here?

Jakobian

You can still think of elements of $L^2$ as functions, but keep in mind they're not precisely functions and when you talk about this space as well as $L^p$ in general, you identify functions up to set of measure zero

giannisl9

@Jakobian okay

now in the integral $\int_{X}fdμ(x)$

which will become

Jakobian

When we want to talk about functions and not equivalence classes, even mathematicians usually forget this, but the usual notational distinction is $\mathcal{L}^p$ for functions and $L^p$ for equivalence classes of functions

giannisl9

now in the integral $\int_{[0,1]}fdt$

Jakobian

16:25

If $X = [0, 1]$ and $\mu$ is the Lebesgue measure on $X$, sure

also $f$ has to be Lebesgue integrable

giannisl9

yes what is this t

?

is it a way to measure the interval?

Jakobian

or $f$ can be measurable and non-negative (if we allow for infinite values of integrals)

@giannisl9 integration?

oh $t$

giannisl9

yes $t$

what about if $t$ was $t^2$

Jakobian

Personally I'd never write $\int_{[0, 1]} fdt$, rather $\int_{[0, 1]} f(t)\mathrm{d}t$

giannisl9

that's how I thought about change of variables as measure

change

Jakobian

16:27

$dt$ just means we are using Lebesgue measure

so its the standard integral

giannisl9

when we define as space

Jakobian

if you were to use $d(t^2)$, then I'd assume you're talking about Lebesgue-Stieltjes integrals

which can be defined as integral over the Lebesgue-Stieltjes measure

giannisl9

$(L^2 functions, [0,1], t)$?

this is where I get confused

Jakobian

One that would assing value $b^2-a^2$ to a subinterval $[a, b]\subseteq [0, 1]$

giannisl9

yes I get that

I mean we have a function $f$

defined on that interval

and $t$ is used to scan that interval no?

Jakobian

16:30

@giannisl9 what do you mean

giannisl9

is that what we intuitively mean my measure?

I mean that in Rieman integral lets say

I think of the interval [0,1]

as some set

and then t

$f(t)$

wait

Jakobian

okay well, whats the goal here actually

what do you want to understand

concretely

giannisl9

what is $t$

Jakobian

a letter

giannisl9

thanks

Jakobian

16:32

no problem

giannisl9

come on you can do better pls

I really appreciate it

in the integral

what is the relationship between what we call measure

Jakobian

you know, mathematics isn't about having this mush of things you throw at someone

giannisl9

and $t$

Jakobian

so you need to be precise about what you want

giannisl9

does the measure "measure" the interval?

I think I luck intuition about distance and measure

and the interval as a mathematical entity

Jakobian

16:34

there are different measures

a measure is just a name for a certain concept

what you want is generalization of length

thats called Lebesgue measure

psie

sorry to butt in here, but my knowledge in topology is limited, so I have a few beginner's questions about the following proof. 1) Why is the codomain of $F$ the set $\{0,1\}$? 2) why do we need to show that $F$ is constant? 3) Why is $f$ continuous?

Jakobian

you can measure intervals and you can measure various other subsets, not all of them, but a lot of them

giannisl9

@Jakobian so $t$ is vaguely here some kind of time length?

Jakobian

the Lebesgue measure of interval $[a, b]$ is $b-a$

there is no concept of time here

giannisl9

@Jakobian okay

Jakobian

16:36

time is for physicists I don't care about time

giannisl9

@Jakobian okay

so what is $t$

Jakobian

most subsets of real line, the ones we call measurable, for we can measure them

or Lebesgue measurable

have assigned some value we call their Lebesgue measure

giannisl9

okay

Jakobian

For example, rationals numbers in $[0, 1]$ have Lebesgue measure zero

but irrationals in $[0, 1]$ have Lebesgue measure $1$

giannisl9

the thing for me here is that $[a,b]$ has a measure of $b-a$

Jakobian

16:38

The $t$ in $\int f(t)\mathrm{d}t$ is basically a leftover from other definitions of integrals

one interpretation is the $t$ comes from how we measure the length of an interval

giannisl9

@Jakobian yeah that's what I want

please explain this a bit

that's what I thought

how we measure it

Jakobian

if it were $dt^2$ we could interpret it as Lebesgue-Stieltjes integral, and the measure of $[a, b]$, its "length" with respect to the corresponding measure would be $b^2-a^2$

giannisl9

as a function of $t$

yea that was my thought

Jakobian

and if we had some increasing function $g$ then $dg(t)$ would mean we integrate with respect to a measure that assigns length $g(b)-g(a)$ to $[a, b]$

For example if you look at Riemann integrals, we have Riemann sums $\sum f(t_i)(x_{i+1}-x_i)$

giannisl9

so here the $t$ plays the role of how we measure things ?

Jakobian

16:41

but for Riemann-Stieltjes ones it would be $\sum f(t_i)(g(x_{i+1})-g(x_i))$

giannisl9

things = intervals

Jakobian

so the $dt$ here also indicates how we take the difference to be $x_{i+1}-x_i$ instead of the other, more exotic differences

of course this is for Riemann integrals which are simpler

Thorgott

@psie they are using the characterization that a space is connected if and only if all continuous functions from it to the discrete two-point space are constant

Jakobian

without measures

@giannisl9 you can interpret it this way, sure

$dt$ just means its integration with respect to Lebesgue measure

psie

ah, I'll have to look that up

Jakobian

16:43

but thats if this is a Lebesgue integral, there can be other integrals

which don't necessarily come from integration over some measure

psie

so we could replace $0$ and $1$ with any other numbers, as long as it is a discrete two-point space I take it

Jakobian

@psie yeah

or any disconnected space

Thorgott

yeah, it doesn't matter what you call the two points

people just gravitate towards $0$ and $1$

giannisl9

So can we relate this to a Rieman sum where we change variables ?

lets say

$f:[0,1]->R$

and

we take the $\int_{0}^{1}f(t)dt$

and we make a change of variables

$u(t)=t^2$

Jakobian

change of variables is more or less glorified chain-rule

giannisl9

16:51

I want to think of it analogously

please bear with me

now what happens is we change how we measure the interval $[0,1]$

so in essence there is a change in space

Jakobian

That depends on your point of view

giannisl9

I meant $t(u)=u^2$

so $f(u)$

is another function

Jakobian

I mean there'll always be going to change of interval but you can (usually) keep on integrating with respect to the Lebesgue measure

giannisl9

at another space

because the measure is different

Jakobian

You can write substitution as $\int f(x)\mathrm{d}x = \int (f\circ \Phi)(x) \mathrm{d}\Phi(t)$

but you can also just write it as $\int f(x)\mathrm{d}x = \int (f\circ \Phi)(x)\cdot \Phi'(x) \mathrm{d}x$

giannisl9

16:56

My point of interest is if I am correct that the space changes

we have a different function

at a different space

that is related to the old function

Jakobian

what do you mean by space more precisely

giannisl9

at the old space

space in the sense of $L^2[0,1]$

analogous

Jakobian

analogous to what exactly

giannisl9

let me state it again

Jakobian

you mean you integrate over different measure?

giannisl9

16:58

yes

because we have $t^2$

and also the interval changes somehow

Jakobian

well, like I said you can usually (and this is how we write it) keep on integrating with respect to the Lebesgue measure

giannisl9

this is what I don't understand

Jakobian

and the difference is accounted by the differential $\Phi'$

giannisl9

but we changed how we measure the interval

its not $t$ anymore

its $t^2$

okay we can write it in terms of $t$

because they are related

Jakobian

I feel lack of precision here

precise example

giannisl9

17:02

$f(t) = t, t belongs [0,1]$

Jakobian

yeah sure but whats the question

is there a question?

giannisl9

take the integral $\int_{0}^{1}f(t)dt=\int_{0}^{1}tdt$

suppose a change of variable

$t(u)=u^2$

look at the integral

$\int_{0}^{1}f(u)dt(u)$

Jakobian

well it won't be the same integral

oh okay

giannisl9

okay now

let me tell you what I think

Jakobian

well like I said, you can look at it as $d(t(u))$ but you can also look at it as $\int_0^1 f(u)t'(u)\mathrm{d}u$ and thats usually how we do it

with the differential $t'(u) = 2u$ accounting for the change of variables

giannisl9

17:07

wait

that's not my point of interest

wait please

sorry

:)

about the first integral: we have a function on [0,1] and we measure this by t

about the second integral: we have another function on [0,1] and we measure this by $t(u)$

and actually if the interval was $[0,4]$

it would become $[0,2]$

it means different function different interval different measure?

or different function same interval different measure

its like mapping a function from one space measured by t

to another function from another space measured by $t(u)$

I say different space because the same things is measured differently

Anyway there is my confusion as you can see

I want some clarification along those lines

Not definitions or completely irrelevant ideas

se can have some functions on $[0,1]$

integrate them like this and get a number

do a change of variables

get some new set of functions on some other interval (or the same measured differently)

and integrate some new function over this to get (the same?) number

when I say integrate them I mean integrate one of them

Jakobian

17:25

I'd have to repeat myself again

giannisl9

no please

no reason

we can view it the other way

but what if we view it this way

just look at it this way

giannisl9

17:45

@Jakobian Anyway, I thank you with all the thanks in the world more infinite than you can write down if that is possible

you helped a lot

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