Mathematics - 2024-05-04 (page 1 of 2)

12:01 AM

mm k im supposed to just look at the degrees of the exponentials i guess

makes sense

lol

nvm im still confused

$$\frac{Nk_B(\varepsilon\beta)^2 \exp(\beta\varepsilon)}{(\exp(\beta\varepsilon)-1)^2}\approx Nk_B(\beta\varepsilon)^2\exp(-\beta\varepsilon)$$ since $\beta\varepsilon\to\infty$ as $T\to 0$ idk how this follows

EE18

Do you want it the math way or the physics way?

Obliv

Either way :P

EE18

I'll let others speak to it the math way. For the physics way you simply observe that $(\exp(\beta\varepsilon)-1)^2 \to (\exp(\beta\varepsilon))^2$ in this large $\beta$ limit. Then cancel one factor of $\exp(\beta\varepsilon)$ from the top and bottom.

Obliv

ohh okay

what about the other way when $\beta$ gets to be close to 0

somehow it approximates to be $3Nk_B$

EE18

Expand $e^x \approx 1+ x$ and see what happens :)

Obliv

12:12 AM

OHH right

u can do that

bless u :P

EE18

Cheers

A "soft" question about math writing for folks here: how do you deal with nested brackets? For example, suppose I wanted to write an aside in brackets, and in that aside I refer to some equation, say equation (2.65). Would you choose square brackets in this case or is that bad form?

(This follows from equation (2.65)) or [This follows from equation (2.65)]

Thorgott

I think the former is perfectly fine, but when Xander arrives, he will probably be able to point you to the precise subsubsection in the AMS style guide where this is addressed

Obliv

or you could say (this follows from equation 2.65)

EE18

Certainly I am not writing anything so important as to deserve that :) but thanks Thorgott

That's fair Obliv, but generally the brackets are there in the label too which is why i have them

Xander Henderson

12:33 AM

@Thorgott There isn't actually a ton there, though the inference I draw is that "This follows from (2.65))" is the preferred style (note that is just "(2.65)", not "equation (2.65)").

E.g. section 6.4

> Identifying letters ((a), (b), (c)), including their parentheses, are roman in all text. The AMS will allow italic identifying letters only if consistent throughout.

Section 13.14:

> In cross-references, equation numbers are enclosed in roman parentheses to match the original label and the parentheses are always roman.

(which only says that equation numbers should always be in parentheses, not other kinds of braces).

On the other hand, the only references that the style guide has to brackets is in the context of citations and "fences" in mathematical contexts. I do not think that AMS editors would be too happy with square braces for parenthetical content.

Personally, I would seek to rewrite the phrase.

@Obliv No. This is wrong. Again, see section 13.14 of the AMS style guide.

For the reference of others: ams.org/publications/authors/AMS-StyleGuide-online.pdf

Obliv

i don't think style can be logically deduced

Xander Henderson

@Obliv Okay...

EE18

1:01 AM

@leslietownes Just quickly coming back to this conversation -- as regards the $T$-periodic space of functions which are spanned by the trigonometric functions, I guess I gave the wrong norm right? It should be induced by an inner product looking something like $<f,g> = \lim_{T \to \infty}\frac{1}{T}\int_{-T/2}^{T/2} fg$?

leslie townes

EE18 if you are looking at periodic functions the usual norm of <f,g> would be an integral of fg over one period (which is arbitrary but usually fixed, e.g. [0,T] or [-T/2,T/2]). you might be able to write out some limit over increasingly large that ends up being the same thing because of the periodicity, but i don't know why you would want that

you can also define inner products of not necessarily periodic functions in terms of integrals over all of R, usually with some kind of "decay at infinity" hypothesis floating in the background to ensure that the thing makes sense and is finite, and sometimes (not always, and not usually by definition) expressions for those sorts of norms are written as limits like that

EE18

1:18 AM

@leslietownes Ah OK I think this is what's going on, that there's some equivalence to the infinite limit version because of periodicity

I'm working with that engineering text which is keeping stuff hazy, so wanted to just clarify what was going on

user 70432

1:59 AM

@Obliv agreed, but preference should be given to consistency.

Xander Henderson

2:15 AM

@EE18 \langle and \rangle.

@leslietownes You, too. :(

user 70432

:(

Xander Henderson

@leslietownes Though in a lot of cases, those won't really be inner products, but the dual pairing (which devolves to an inner product in spaces which are self-dual. (Not that this really matters).

Obliv

2:44 AM

I personally write my inner products like this $\{\left\{\langle\left[<f,g>\right]\rangle\right\}\}$

just to be clear

more brackets means stronger seal

leslie townes

3:00 AM

xander: or maybe it is the inner product that evolves into a dual pairing. hippy music

user 70432

🎶🎵🎶🎸

Xander Henderson

3:25 AM

@leslietownes That isn't how evolution works.

Though it may be the that the inner product is the distant descendant of some common ancestor of the dual pairing.

EE18

3:40 AM

@leslietownes Actually sorry for perseverating on this Leslie, but I've been searching around and not able to find a wikipage or something like that for this. What term should I be using to learn more about this equivalence between the two possible inner products on this periodic function space?

leslie townes

maybe just prove it? i could see it as a homework problem in an analysis book. if f is periodic with period T then the average value of f on [-L, L] goes to the average value of f on [0,T] as L goes to infinity. it should make intuitive sense.

leslie townes

4:07 AM

or look at math.stackexchange.com/questions/2768853/…

there are several duplicates of that question on the site too, all with similar looking proofs. it's maybe an expository challenge to make the argument look good. it might clean things up a little to assume T = 1 or something like that

John Zimmerman

5:05 AM

I realized I have no idea how to glue together closed subsets bounded by analytic boundary S^1's when I'm not in the Top Cat. Do I operate in the Top Cat and then use some deformation or rigidification process to deform the topological surface into smooth/real analytic manifold? I think that is the right direction because there are theorems guranteeing a topological manifold and smooth manifold can be deformed to a anlaytic one. I'm lost on the details though

John Zimmerman

5:19 AM

not sure where to start exactly but maybe in Top, and generate a topological surface S^2, and then sequentially layer on smooth then analytic structures. If the S^2 here is obtained through a gluing and one can obtain an analytic metric compatible with the base (unwrapped quotient), related through the section/exact projection correspondence allowing for a direct analytic quotient metric on the S^2. I think I should go for the low hanging fruit and that would be recovering a well defined

analytic manifold with a well defined analytic quotient metric

Koro

8:09 AM

Can anyone please explain what $V\otimes_R C$ is?

leslie townes

i assume tensor product of real vector spaces, i.e. the thing in en.wikipedia.org/wiki/Tensor_product where you regard C as a vector space over R

Complexification

In mathematics, the complexification of a vector space V over the field of real numbers (a "real vector space") yields a vector space VC over the complex number field, obtained by formally extending the scaling of vectors by real numbers to include their scaling ("multiplication") by complex numbers. Any basis for V (a space over the real numbers) may also serve as a basis for VC over the complex numbers. == Formal definition == Let V {\displaystyle V} be a real vector space. The complexification of V is defined by taking the tensor product of...

Koro

thanks. So V and C are both considered over R.

I keep forgetting the definition of tensor product.

I remember it using its existence property diagram but forget it soon.

V, W - K vector spaces. V tensor W is a K- vector space along with a bilinear map f: V\times W into V tensor W such that every bilinear map out of V\times W into a K vector space U is composition of a unique linear map (from V tensor W into U) with f.

But I never really understood the motivation for tensor products.

What is their use? Why do they exist?

Mad

suppose $ R $ is a ring and $\mu: R\rightarrow [0,\infty]$ is sigma additive, how can i show that for $ A \subset \cup_i A_i: \mu(A) \leq \sum_i \mu(A_i)$ if the unity need not be in the set (ie i can not use the attribute of sigma additivity)

if it is contained, then its trivial by definition... if it is not contained?

Jakobian

9:19 AM

@Koro in this context its change of base field from R to C. That's their current use

Another use would be that they allow you to develop exterior algebra

Any time when there's an exchange from n-linear to linear you can expect tensors are involved

The usefulness of tensor products is definitely immeasurable, but I can't tell you historical reasons/from ground up motivation for their existence.

I just don't really know

John Zimmerman

9:54 AM

Why is the dirichlet divisor problem so hard?

mathworld.wolfram.com/DirichletDivisorProblem.html

see here for what the dirichlet divisor problem is all about

I don't get why it is notoriously difficult tbh

Koro

What does RHS mean?
pi_1: G---> GL(V_1), pi_2: H---> GL(V_2)
so pi_i g is a map from V_i into V_i
How can two maps be tensored?

Koro

10:20 AM

nvm, it's same as how it is done in differential geometry.

Jakobian

10:47 AM

yeah. Tensor products are, first and foremost, applied in differential geometry

"Defined as $(\pi_1\otimes \pi_2)(g, h) = \pi_1(g)\otimes \pi_2(h)$" really should be "Defined as $(\pi_1\otimes \pi_2)(g \otimes h) = \pi_1(g)\otimes \pi_2(h)$"

or in other words, the linear map coming from the bilinear map $(g, h)\mapsto \pi_1(g)\otimes \pi_2(h)$

Soumik Mukherjee

11:39 AM

Sequence -> sequential convergence. net -> netial convergence? net wise convergence?

Xander Henderson

11:55 AM

@SoumikMukherjee Net convergence.

user 70432

1:58 PM

Proofs in mathematics

►

ZaWarudo

2:40 PM

Hi, I'm studying the uniform convergence of $f_n(x)=n\sqrt{4\pi^2 n^2 +x^2}$ on $\mathbb{R}$. I proved that $f_n(x) \to \frac{x^2}{4\pi}$ pointwise on $\mathbb{R}$. My reasoning to prove that $f_n$ is not uniformly convergent to $\frac{x^2}{4\pi}$ is the following: $\sup_{x \in \mathbb{R}} |f_n(x)-f(x)| \ge |f_n(n)-f(n)|=n^2|\frac{\sin \sqrt{4\pi^2 n^2+n^2}}{n}-\frac{1}{\pi}|$.

Hence, $\lim_{n \to +\infty} \sup_{x\in\mathbb{R}} |f_n(x)-f(x)| \ge \lim_{n \to +\infty} n^2|\frac{\sin \sqrt{4\pi^2 n^2+n^2}}{n}-\frac{1}{4\pi}|=+\infty$

This shows that $f_n$ is not uniformly convergent to $x^2/4\pi$. Is this reasoning correct?

I forgot a $4$ is the denominator of $-\frac{1}{\pi}$ in my first message, sorry.

Obliv

2:56 PM

typically $\| \|$ is used for norms of vector quantities and $||$ just normal abs bars are for numbers?

Thorgott

sometimes yes, sometimes no, it doesn't really matter

Obliv

Suppose $\overline{E}$ is $\sum_s E(s)P(s) = -\mu BP_{\uparrow}+\mu BP_{\downarrow} = -\mu B(P_{\uparrow}-P_{\downarrow}) = -\mu B\tanh(\beta \mu B)$. I want to set the energy levels to $0$ and $2\mu B$ instead of the $\pm \mu B$ but I'm not sure if that's the same as $$\mu B\left(\frac{1-e^{-2\mu B\beta}}{1+e^{-2\mu B\beta}}\right)$$

which is what u get if you do that. Like is what's in the parentheses a $-\tanh(\beta\mu B)$

I think it should be. I guess there isn't much of an advantage doing it this way since $\tanh$ is more recognizable

Soumik Mukherjee

@XanderHenderson So anticlimactic

EE18

@leslietownes This is awesome, thanks so much :) \

So am working with my analysis text this morning which is on a last brief section about vector spaces before the fun with sequences starts next chapter

The book has a brief aside about polynomial interpolation, and proves that the existence and uniqueness of an $m$ degree polynomial to "match" a function at $m+1$ locations

They do it by explicit construction of the Lagrange interpolation polynomials without much motivation, fine

They then mention the following:

What could the last sentence possibly mean? I know things like "simple" or "explicit" can be informal notions but I fail to see why Gauss-Jordan wouldn't give us explicit descriptions of each $p_j$ and, putting them all together, the given polynomial?

Thorgott

3:25 PM

they would, it's just tedious

Jakobian

@Obliv depends on what you think of as a vector and what you think of as a number

Obliv

is $\sigma_X=\sqrt{\sum_s P(s)^2X(s)^2}$? Or is it just $\sqrt{\sum_s P(s)X(s)^2}$

Jakobian

@Obliv are you talking about standard deviation?

Obliv

yea, or root mean square deviation. square root of the average of the squares

so I imagine it's the latter but i'm not 100%

Jakobian

what is $P(s)$ and $X(s)$

Obliv

3:31 PM

$P(s)$ is the probability of finding $X(s)$

Jakobian

the what?

Obliv

hmm I'm gonna just post the relevant sections lol

I just wanted to write it in terms of an arbitrary variable $X$

oh wait it was average of the squares of the deviations

I didn't read a)

so a deviation can be given by $X_i - \overline{X}$

so $\sigma_X$ is the square root of the average of these squared

I'm so confused lmao

so I guess the "average of the squares of the deviations" can be $$\sum_s P(s)(X_s-\overline{X})^2$$

Jakobian

The standard formula is that $\overline{X} = \sum_k k\cdot P(X = k)$. But $P(X = k) = \sum_{s:X(s) = k} P(s)$ so the formula $\overline{X} = \sum_s X(s)P(s)$ seems correct

Obliv

yeah I guess that makes sense

Jakobian

@Obliv here it should be neither in general, unless observations are independent

Obliv

3:45 PM

yea what I wrote above was wrong

I think it could be $$\sqrt{\sum_s P(s)(X_s-\overline{X})^2}$$

Jakobian

no wait I'm spouting gibberish

I guess physics has that effect on people

Obliv

bruh

Jakobian

$\sigma_X^2 = \overline{X^2}-\overline{X}^2$ where $\overline{X} = \sum_s X(s)P(s)$ as we already observed and $\overline{X^2} = \sum_k k^2P(X = k) = \sum_s X(s)^2P(s)$

so it would be $\sigma_X = \sqrt{\sum_s X(s)^2P(s) - (\sum_s X(s)P(s))^2}$

Obliv

:( so it's not what I wrote

why'd he write it in english instead of explicit math ugh

Jakobian

If you assume $X$ has $0$ as an average, then its $\sigma_X = \sqrt{\sum_s X(s)^2P(s)}$ so its still somewhat valid

Obliv

3:54 PM

i meant like right above

Jakobian

@Obliv this one?

Obliv

yea

that one also isn't the same

"average of the squares of the five deviations"..."then compute the square root of this quantity" made it sound like what I wrote

Jakobian

well you wrote $X_s$ instead of $X(s)$ but its the same formula

as in, its also a valid formula

Obliv

is it? when u multiply out dont' you get $\sqrt{\sum_s P(s)X^2(s)+\sum_s P(s)\overline{X}^2(s)}$

Jakobian

Exercise: Convince yourself both formulas are true i.e. both expressions are equal

Obliv

3:57 PM

ok author

after I brb

Jakobian

@ZaWarudo $f_n(x) = n\sin\sqrt{4\pi^2n^2+x^2}$?

ZaWarudo

@Jakobian Yes, thanks for catching up the typo!

Jakobian

alright, then its indeed pointwise convergent to $\frac{x^2}{4\pi}$

@ZaWarudo its correct

ZaWarudo

@Jakobian thank you. Instead, if I consider the uniform convergence in $[0,a]$ with $a>0$, that reasoning above is not valid anymore because, even if it's still true that $\sup_{x \in \mathbb[0,a]} |f_n(x)-f(x)| \ge |f_n(n)-f(n)|$, I then must take the limit as $n \to +\infty$ and for fixed $a>0$ I have $n \notin [0,a]$ for $n$ big enough, right?

Jakobian

4:12 PM

@ZaWarudo Well... its not true that $\sup_{x\in [0, a]}|f_n(x)-f(x)|\geq |f_n(n)-f(n)|$

ZaWarudo

I mean, it's true for $n \in [0,a]$ but it is not true for each $n \ge N$ for some $N \in \mathbb{N}$ and this is what I need when I want to evaluate $\lim_{n \to +\infty} \sup_{x \in [0,a]} |f_n (x)-f(x)|$

Or am I missing something?

Jakobian

sure. The gist is that this argument doesn't show to disprove uniform convergence

EE18

@Thorgott weird that they would say no explicit solution though no?

ZaWarudo

Yes, I meant that. I just wanted to be sure that I understood correctly why it doesn't work. Thanks! :)

Thorgott

@EE18 they don't say that?

they just say "no simple expression"

EE18

4:46 PM

Got it, fair enough. Just seems weird, not like Lagrange polynomial is so “simple” right? Anyway, I won’t dwell

Jakobian

4:58 PM

You're saying that you understand the word "simple" might be subjective, yet completely ignore it

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