Mathematics

12:00 AM

@MatheinBoulomenos Why question is that "Is is true that $\int_a^b \lim\limits_{n \to \infty} f_n(x)=\lim\limits_{n \to \infty} \int_a^b f_n(x)$ even if $f_n$ does not converge uniformly?" We know that uniform convergence implies the integral limit exchange. But is it true the other way around?

MatheinBoulomenos

oh

now I understood your question

I thought you wanted an example where you can't interchange the integral

the example $x^n$ on $[0,1]$ from ultradark works. This converges non-uniformly to the function which is $0$ on $[0,1)$ and $1$ at $1$ (convergence has to be non-uniform since the limit is not continuous), but the integral limit exchange works still

if you go further in real analysis, you will get to stronger theorems which allow exchanging limits and integrals, such as dominated convergence and monotone convergence

user330477

@MatheinBoulomenos What about the example $f_n(x)=\frac{2x}{n}+\frac{1}{n^2}$? Does it work?

user193319

$Mathematics$ Group Theory

Let's discuss group theory!

user330477

@MatheinBoulomenos Thank you so much.

MatheinBoulomenos

@user330477 this converges uniformly to $0$ on each closed interval $[a,b]$

Ultradark

12:21 AM

Is it kosher to raise a number to a probability

Joe Shmo

what's a probability

Ultradark

a number between $0$ and $1$

Joe Shmo

more generally, for $a, b \in \mathbb{R}$, is $a^b$ well defined?

hint: yes

Ultradark

okay good

so if $p(x)$ is a function that gives a probability, how does one interpret $n^{p(x)}$?

is there any interpretation?

Joe Shmo

12:45 AM

well, what is $n$, a natural number?

Ultradark

yeah

Joe Shmo

you actually have the burden of defining $n^{p(x)}$

and we haven't answered what $a^b$ is

venture a guess

Ultradark

1:03 AM

I am going to guess that raising a natural number to a probability function, acts as an isometric map on $p(x),$ such that a bijecition can be formed between $p(x)$ and a target region. This target region encodes information about $p(x).$ The benefit of such a map is that there is a way to go back and forth between the probability function and the encoded probability in the target region

Joe Shmo

yeah.. can you define it explicitly?

forget the probably function for a second. What's $a^b$?

Ultradark

its a number raised to another number

Joe Shmo

i know what $2^3$ is. but on the face of it, i don't really know what $\pi ^ e$ is

does it even make sense to raise an irrational to the power of another irrational?

Ultradark

oh

Rithaniel

Or $2^{1-2i}$, for that matter.

Joe Shmo

1:12 AM

yeah, that's another story in and of itself. complex exponentiation is a bit different

Ultradark

okay let's say $p(x)$ is some function $1/\ln(x)$

Rithaniel

(Yeah, my mind is on generalizations. Like, how do you know what you're doing can even make sense in the first place? How do you define $2^s$ where $s$ is an element of the dihedral group on the n-gon?)

Joe Shmo

yuh

Ultradark

taking $n^{p(x)}$ yields transcendental numbers

Joe Shmo

@Ultradark you haven't circumvented the issue

Ultradark

1:17 AM

what's the issue

Joe Shmo

you dont know if $\pi ^ e$ is even well defined

Ultradark

$n$ is a natural number

Joe Shmo

and $\frac{1}{\ln x}$ is certainly going to give you nasty numbers

unfortunately the base isn't the issue

Rithaniel

Like, how do you know if it yields transcendental numbers? How do you know it yields a number at all? What if it yields a chicken?

Joe Shmo

so, ok, what does $n^e$ mean?

Newbie

1:19 AM

@TedShifrin maybe whenever you get a chance you can look at my question.

https://math.stackexchange.com/questions/3122924/comparing-sigma-algebras-with-topologies

Ultradark

oh wait, even $1/\ln(x)$ gives transcendental numbers

that might be a problem

it is unknown whether $n^{1/\ln(x)}$ is transcendental

solution: round the number to $4$ decimal places

now it is rational

Joe Shmo

that's the solution of the physicist

Ultradark

so now we have transformed $p(x)$

Joe Shmo

im not sure im following

hows this. whats the verbatim problem statement you are trying to solve

Dair

sounds like you would do: step 1: define exponentiation for nats, step 2: try defining for positive rationals, step 3: try defining for positive reals. and then it would take some more work for showing there is some natural way to raise $n^{P(x)}$ where P(x) is some distribution...

doesn't sound easy lol.

Ultradark

1:33 AM

well this allows us to set up a bijection between the target region and the region outside, which I'll denote, $\zeta$ and $L$ respectively

$\zeta \in (0,1) \times (0,1)$

and $L \in (1,\infty) \times (1,\infty)$

these two regions have the same size

Joe Shmo

@Dair yeah..

@Ultradark let's drop all the fancy language

let's follow @Dair 's suggestion instead

Ultradark

but the bijection

Joe Shmo

forget the bijection

you haven't even defined the object at question properly

and you were almost there

Ultradark

really?

Joe Shmo

you can define it for 4 decimal places

(can you?)

(...how?)

Lucas Henrique

5:24 AM

@Ted: I'm having a hard time on proving that basis of different finite cardinalities don't exist; i.e., that theorem you told not to use matrices.

I guess I'll have to use some equivalent statement of the rank-nullity theorem

Ultradark

I think I just solved the Tao conjecture again

Ted Shifrin

5:43 AM

@Lucas: You can write out the proof of Theorem 4.1 just with summations and without using coordinates (in those matrices) for the vectors $v_j$ and $w_i$.

Lucas Henrique

I don't think I know how.

I said that "has basis of cardinality $n$" iff bijective linear map $\phi: \Bbb F^n \to V$

Ted Shifrin

Just write $w_j = \sum a_{ij} v_i$ and consider $\sum c_jw_j$. Rewrite that as a linear combination of the $v_i$.

Lucas Henrique

analogy to linear independence, but only injective linear map

Ted Shifrin

I'm proving specifically the theorem I stated, not a different approach. From that you conclude that dimension is well-defined (as in Theorem 4.2).

Lucas Henrique

I'm sorry beforehand since it's 2:45am here and my classes start... tomorrow.

Ted Shifrin

5:45 AM

Well, I'm not staying around long, anyhow.

Happy classes.

Pretty much every treatment of linear algebra I know shows that if you have linearly independent vectors and spanning vectors, then there are fewer (or equal) of the first than of the second.

Lucas Henrique

@Ted: I'm not currently with the book. We have more $w_j$ than $v_i$, is that right?

Ted Shifrin

Right.

Lucas Henrique

Hmm, ok. So $\operatorname{Span}(w_1, \dots, w_n) \subseteq \operatorname{Span}(v_1,\dots,v_m)$

That's basically what we get when we rewrite a linear combination of $w$ in terms of $v$

Or being explicit (no pun intended), $\sum c_jw_j = \sum\sum a_{ij}c_jv_i$

Ted Shifrin

Right ... And because the matrix $A = (a_{ij})$ has more columns than rows, there's a nonzero vector $c$ ....

(So we're not avoiding matrices entirely. But this proof works fine in abstract vector spaces ...)

Lucas Henrique

6:08 AM

thanks god lmao

just the way I was thinking. thanks, @Ted. :)

Ted Shifrin

You're welcome, Lucas.

Ultradark

is there a bijection between these two parts?

Mosab Shaheen

Hi guys, I have a simple question https://math.stackexchange.com/questions/3124499/are-the-points-that-make-the-function-fx-non-differentiable-the-same-as-the-po

I am sure many of you know the answer as it is from very basics of mathematics. Could you answer it please?

Akiva Weinberger

6:47 AM

Quick what's 40.44% of 25 (do it in your head)

Group Theory

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$Mathematics$ Group Theory

$Mathematics$ Mathematics