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3:43 AM
Can a flow be defined over the numbers?
such that their order is preserved but they all move as an ensemble
maybe I'm thinking of translating sets
2 hours later…
5:29 AM
@Ted @Mathein ich bin nach USA angekommen
bleh, why isn't my logic working
I can't convince myself whether the following two statements are equivalent. Let $A,B$ be linear subspaces of equal dimension in $V=\mathbb{R}^n$.
(1) For every $b\in B$, there is some $a\in A$ which is not orthogonal to $b$.
(2) For every $a\in A$, there is some $b\in B$ which is not orthogonal to $a$.
6:03 AM
Does this work? idk how to do latex in here so ill use A* to denote the orthogonal complement
so (A* \cap B)* = A** \cap B* = A \cap B^*
@Semiclassical they're equivalent because they're both false
er obviously exclude a or b = 0 right
oh jk my thing is false lmao
4 hours later…
10:33 AM
Really really really long line:
Take the union of the following:
$\alpha \times [0,1)$ where $\alpha \in \mathbf{On}$
This line is SO LONG that there's a point which all increasing sequences of any ordinal length will not converge to it
Probably a nice impression on how large Cantor's Absolute Infinite is
I need some help regarding analysis
I am trying to construct a perfect set of real numbers that does not contain a rational number.
What are you trying? @justanotheruser
I am taking the interval $[e, \pi]$. $\{x_n\}$ is the enumeration of rationals in it. Let $I_n = (x_n-\frac{r_n}{2^{100n}}, x_n+\frac{r_n}{2^{100n}})$
Then $A= [e, \pi] \setminus \bigcup I_n$
I need some verification on the part that any $c \in A$ is a limit point. I need someone to check it.
What's $r_n$?
10:48 AM
$r_n = \min\{|x_n-e|, |x_n- \pi|\}$. Being always an irrational, the endpoints remain irrational
For $c \in A$, we try to show that $(c-\delta, c)$ and $(c, c+\delta)$ cannot be simultaneously "removed" by our approach for any $\delta>0$ (which would ensure that $N'_\epsilon(c) \cap A \neq \phi$ ). Can it work at all?
11:05 AM
Attempt: Suppose, that for some $c \in A$, it happens. At least for two intervals $I_s$ and $I_t$, it happens. For this to occur, the intervals must lie on two "opposite sides" of $c$ since $c \in A$ and $c \notin \bigcup I_n$.

Now, any of the following pairs hold: $\{r_s= x_s-e, r_t=x_t-e\};\{r_s= x_s-e, r_t=\pi-x_t\};\{r_s=\pi- x_s, r_t=\pi -x_t\};\{r_s= \pi-x_s, r_t=x_t-e\} $.

We must have $\displaystyle x_s-\frac{r_s}{2^{100s}}=x_t+\frac{r_t}{2^{100t}}$. But it is not possible in any case since $x_s-x_t$ is rational and $\displaystyle\frac{r_s}{2^{100s}}+\frac{r_t}{2^{100t}}$ is irra
@AlessandroCodenotti could you kindly check it out
The set $A$ is closed as well. So it must be perfect.
11:58 AM
Every linear map from the zero space to a Banach space is bounded, right? (vacuously)
1 hour later…
1:20 PM
Yes, you should write down the vacuous proof
3 hours later…
3:57 PM
@Semiclassical It seems you can rewrite (1) as $A\not\subset B^\perp$. Likewise (2) as $B\not\subset A^\perp$. Certainly not equivalent.
4:18 PM
Hrm. I was trying to write down $A\cap B^\top=\{0\}$ and $A^\top \cap B=\{0\}$ respectively. But it was late so it’s entirely possible I was being silly
4:32 PM
"Form matrix equation $\textbf{L}\vec{f}=\vec{u}, \quad \vec{f},\vec{u} \in \mathbb{R}^n, \textbf{L}\in \mathbb{R}^{n \times n}$ with basis $(e_1,\dots,e_n)$ from equation $ \langle Lf_d, v_d \rangle = \langle u, v_d \rangle \quad \forall v_d \in V_d $
any linear transformation can be expressed with matrix multiplication right?
I need to somehow transform equation $\langle Lf_d, v_d \rangle = \langle u, v_d \rangle \quad \forall v_d \in V_d$ to matrix equation
Like how do I get from equation with inner product to one with matrix
@Tuki: What do you know about a vector that is orthogonal to every vector in $V$?
that inner product is 0?
@Semiclassic: No, I don't think that's right. You mean perps, of course. For $A\cap B^\perp = \{0\}$, you have to say: If $a\cdot b = 0$ for all $b$, then $a=0$.
@Tuki: Read my question carefully. every vector.
it has to be in $V^\perp$ then
@Semiclassic: So, if $a\ne 0$, there is some $b$ with $a\cdot b\ne 0$.
4:45 PM
Oh. Yeah, that makes more sense.
Ah, quantifiers different and nonzero.
@Tuki: Right. But $V$ isn't the whole thing?
Quantifies are annoying. Important, but annoying
They do annoy neophytes, yes, @Semiclassic :D
whole thing?
4:48 PM
@Tuki: The whole vector space, not just some subspace. I can't tell with your notation.
Did you read/understand the comment I posted to you days ago about the right way to understand projection?
I did read it but understanding is different
If $V$ is a subspace of $\Bbb R^n$ (say), $\vec v = \text{proj}_V \vec x$ is the unique vector $\vec v\in V$ with the property that $\vec x - \vec v$ is orthogonal to $V$. (Draw a right triangle picture.)
You can also watch some of my video lectures — they might help.
yes I'm familiar with this proprety
But doesn't the basis we do the projection with need to be orthogonal if we have this property?
You don't need any basis at all to understand this.
But yes lectures might help. I've only been reading the course material for this one. I haven't been to a single lecture yet.
4:55 PM
If you want to write a formula for the projection in terms of basis vectors, then it's easiest to have an orthogonal basis for the subspace $W$, yes.
Hello guys!
The general definition for projection was something like $P(v)^2=P(v)$
@TedShifrin actually, what was wrong with the quantifies? I see that the statements should start as “For any nonzero $b\in B$”
I need to check if for all $a,b\in\Bbb{Z}$: $$2\mid ab\implies 2\mid a\vee2\mid b$$
@Tuki: That is only one of the conditions. $P^2 = P$ (writing $P(v)^2$ is wrong).
4:58 PM
I think it is true, but I could not find any proof for that
But that doesn't say it's an orthogonal projection, @Tuki.
yes this was the thing I was trying to imply
I think it is equivalent to say: $$ab=2k_1\implies a=2k_2\vee b=2k_3,\;\forall k_1,k_2,k_3\in\Bbb{Z}$$
You need $P^2 = P$ and $P=P^\top$ for that.
Oh, we're doing different statements, but I guess you're right, @Semiclassic.
@manooooh: It's hard to know what you know.
5:00 PM
@TedShifrin well, the basics I guess
Do you know about gcd's and writing them as a combination of the two numbers?
I would start from: $ab=2k_1$ implies $2(ab)=2(2k_1)$
No, this is a bad approach.
@TedShifrin if $\gcd(a,b)=1$ then $1=ma+nb$
OK, so good.
Assume $2\nmid a$. What does that say about $\gcd(2,a)$?
And why?
Your quantifiers are all messed up in the stuff you put up there.
5:03 PM
@TedShifrin it says $\gcd(2,a)=1$?
Yes, and this is because ... ?
Because... $2$ and $a$ does not have any common factor
Would it be true if we used $6$ instead of $2$?
@TedShifrin hm, $6=2\cdot3$... probably yes
So, let's see. $6$ does not divide $9$, but is $\gcd(6,9) = 1$?
5:07 PM
@TedShifrin no, it is $\gcd(6,9)=3$
OK, so it doesn't work for $6$. What is special about $2$?
@TedShifrin it is a prime number
There you go.
OK, now use what you told me about the gcd.
5:08 PM
Ok, so we suppose $2\not\mid a$. Then $\gcd(2,a)=1$ because $2$ is a prime number. Then $\gcd(2,a)=1=2x+ay$, for $x,y\in\Bbb{Z}$
OK, now figure out how to use that equation to show that $2|b$.
Remember that you haven't used the hypothesis yet.
Ok, thanks
@Semiclassic: I think I must have messed up translating your original sentence(s). Mea culpa. i dunno.
Our hypothesis says that $2\mid ab$ i.e. $ab=2k$ for $k\in\Bbb{Z}$. We can assume $x=k$ so our equation becomes $1=ab+ay=a(b+y)$. Right @TedShifrin?
5:12 PM
Forget about that equation.
You have $1=2x+ay$. What can you do with this equation?
@TedShifrin substract $2x$ on both sides?
Who's missing from the story here?
We need to show that $2\mid b$ i.e. $b=2m$
5:13 PM
Mostly I just wanted a statement of $A\cap B^\perp=\{0\}$ that was in terms of A and B directly
So, $b$ is missing. And we want to use the hypothesis that $2|ab$.
So $b$ is missing from the story
So how do you put him in the story?
We can put him by multiplying $b$ on both sides i.e. $b=b(2x+ay)=2bx+aby$
OK. Excellent. Now why does this tell you that $2|b$?
5:16 PM
You killed me XD
$2\mid b$ means $b=2m$, so if $b=2bx+aby$... I am not able to conclude
@Semiclassic: Here is how I got what I got. For every (nonzero) b, there is $a\in A$ so that $a\cdot b \ne 0$. This means that $A\not\subset b^\perp$. This holds for all nonzero $b$, so $A\not\subset\bigcap b^\perp = B^\perp$. Something must be wrong ... or not?
Oh, oops. That last step is wrong.
Oh we didn't use the hypothesis yet
Bad Ted.
So $ab=2k$ so $b=2bx+2ky=2(bx+ky)$, so $b=2m$ where $m=bx+ky\in\Bbb{Z}$
Implies that $2\mid b$. And now we have a contradiction? I don't see
OK, @manooooh.
No contradiction.
How do you prove $A\implies B\vee C$?
5:19 PM
I don't see why we started from $2\not\mid a$ to conclude $2\mid b$ (shouldn't it be $2\mid a$?)
@TedShifrin corrected
If $2|a$ we're done. If $2\nmid a$, then we must show $2|b$.
More generally, as I just asked, how do you prove $A\implies B\vee C$?
@TedShifrin I don't see why. Are we using contradiction proof, right?
No, no. Focus on the question I've asked you twice now.
5:20 PM
The funny bit of this is that this is for a question where, if I assume that all the inverses I’m writing down are well-defined, then I can do the formal proof
@TedShifrin I asked this on my mind minutes before I asked here (with no results)... Erm, to prove $A\implies B\vee C$ we need to show that both $B$ and $C$ are TRUE
But I’d like to actually understand why those inverses are well-defined, and when I convert from matrices to subspaces I get questions like this
@Semiclassic: I guess I don't know to which inverses thou doth refer.
NO, @manooooh.
Oh, certainly. I am being vague af
@TedShifrin AT LEAST ONE of both $B$ and $C$ are true maybe?
5:23 PM
Assuming $A$ is true, you mean?
yes of course
Write down the truth table to understand this carefully. It's logically equivalent to proving $A\wedge\lnot B \implies C$.
The question is this one, for context: math.stackexchange.com/questions/3331782/…
To understand this: If $B$ is true, then we're done. If $B$ is false, then we must show that $A\implies C$ (assuming $B$ is false, if you need to).
@Semiclassic: I hate the notation. What does $\beta_\perp$ have to do with $\beta$?
@TedShifrin the truth table of $A\implies B\vee C$ says that the implication is false if and only if $A$ is true but $B$ and $C$ are false
5:25 PM
yeah, it’s kinda annoying but it is appropriate to the literature
Good, @manooooh. Now do the other one I gave you.
LOL, @Semiclassic. Annoying or not, are they related at all?
(Including the use of prime for transposition, ugh)
I am OK with that.
I obviously never do that because I mix linear algebra with calculus all the time.
Yeah. Start with $\alpha$ as some $N\times R$ matrix with full column rank
Yes, and $\alpha_\perp$ ?
5:27 PM
@TedShifrin yes it is the same because $A\to B\vee C\equiv \neg A\vee B\vee C\equiv \neg(A\wedge\neg B)\vee C\equiv A\wedge\neg B\to C$
OK, so this is the logic of the proof we just did, @manooooh.
@TedShifrin I just did* Thank you!
You should remember this, because it shows up a lot in mathematics.
You have helped me a lot about understanding the idea of proofs
I try :)
5:29 PM
Do we have a proof theory or something like that?
Then select some (non-unique) $N\times (N-R)$ matrix $\alpha_\perp$ of full column rank such that $\alpha’\alpha_\perp=0$
@manooooh: There are lots of books that do basic proof techniques. I'm fond of recommending one by Houston called How to Think Like a Mathematician.
@Semiclassic: OK, so the notation is horrible, because we can choose uncountably many such matrices. We're just saying the columns of $\alpha_\perp$ span the orthogonal complement of the span of the columns of $\alpha$.
Right. And both are full rank
Yes, right. By the way, in your comment, you put multiple $\top$s instead of $\perp$s.
I blame the OP for using \bot instead of \perp :P
(But yeah, silly of me)
5:34 PM
OK, so what specifically am I trying to prove (cuz I'm not going to read all that ...)?
Well, the OP wanted to prove a certain decomposition of $I$ under those conditions
But their premises aren’t enough to guarantee that the inverses invoked actually exist
LOL, so you're trying to find sufficient conditions to make it correct?
@TedShifrin something related to this: I was thinking about adding a new condition to say that "continuity implies derivability" on $\Bbb{R}$, but couldn't find any
Right. And looking at the literature on Granger’s theorem (see the deliberately rolled-back edit on my answer for details) I think that it’s sufficient to further assume $\alpha’_\perp \beta_\perp$ is full rank
i.e. we know that $D\implies C$, but $C\not\Longrightarrow D$. How can we find $\text{?}$ such that $(C\wedge\text{?})\implies D$?
5:40 PM
@manooooh: A new condition? I don't understand.
Oh, you changed it.
You keep doing that.
@TedShifrin Sorry corrected
No, there is no condition to put there other than differentiability. Even Lipschitz won't work.
@TedShifrin ok, good to know. Can we prove this?
Is this stuff i.e. "We cannot put any condition to guarantee a result" part of Proof Theory? Or am I just being a little silly?
No, I don't know how to prove such a thing. There are hard theorems that you may eventually learn. Like Lipschitz implies differentiable almost everywhere, but if you want everywhere ...
I mean you can try to camouflage the definition of differentiability and put that. But logically it's no different from assuming differentiability.
In particular, I think $\alpha’_\perp \beta_\perp$ is full rank iff $\alpha’\beta$ is full rank
5:43 PM
@TedShifrin ok. I forgot to say that of course we are excluding the tesis itself: i.e. I am not looking for a trivial $C\wedge D\implies D$
And this should be independent of any of your $_\perp$ choices.
@manooooh: Understood.
So you're right, @Semiclassic, that this should just be about the spaces and not about the matrices.
I understand the motivation for your questions now.
right. Ideally, one should be able to deduce the desired identity from the spaces themselves
Rather than the algebraic legerdemain I used
@TedShifrin thank you!
As a side note, even tho we can use the Mean Value Theorem to guarantee that the only function $f$ that meets $f'=0$ is $f=\text{constant}$!
5:53 PM
@Semiclassic: So what we were fiddling with should boil down to $C(A)^\perp = N(A^\top)$.
@manooooh: Assuming continuity on the closed interval, yes.
I am amazed by this result. We can't find any function except a constant such that the derivative is $0$. Amazed
It's totally intuitive.
@TedShifrin I can't see it
You can also deduce it from the Fundamental Theorem of Calculus.
i.e. we know that $f(x)=x^3-\pi x^2$ is $0$ but at $f^{(IV)}$
5:55 PM
Instead of using the Mean Value Theorem, you can make some sort of least upper bound argument, but I don't think that's the right argument.
@TedShifrin ok, don't worry about that
@TedShifrin right
@TedShifrin I mean: we can find functions where higher levels of its derivative is $0$. For example: $f(x)=x^2$. It is $f'''(x)=0$
Well, sure.
But we can't find any function (except a constant) such that $f'(x)=0$
This is an awesome result, I guess
5:59 PM
I mean, what I was doing algebraically wasn’t so absurd: $(\alpha,\beta_\perp)$ being full rank should be equivalent to $V=A\oplus B^\perp$. (I am probably not being consistent about whether \perp is a sub- or superscript)
MVT is a powerful theorem (particularly in multivariable). But it should be intuitive that in order for $f$ to change values, it must have nonzero derivative somewhere.
@manooooh you can generalize your result to higher order polynomials
Hi, demonic @Alessandro.
yeah, I keep doing that
Here is some dark magic: let $f\in C^\infty(0,1)$ such that for all $x\in[0,1]$ there is an $n\in \Bbb N$ with $f^{(n)}(x)=0$. Then $f$ is a polynomial
6:01 PM
@AlessandroCodenotti hello! I don't understand your generalization. Could you write down your new result, please?
@AlessandroCodenotti oh
Right, and that's equivalent (by dimensions) to $A\cap B^\perp = \{0\}$.
@manooooh if the second derivative vanishes you have a first degree polynomial, etc.
@Semiclassic: I agree that that is totally equivalent to the matrix product having maximal rank.
6:04 PM
@AlessandroCodenotti wow, amazing!! The translation of that theorem would be that if the function is infinitely continuous in a closed and the derivative is null, then necessarily the function is a polynomial?
@Semiclassic: And the "symmetric" condition comes just from transposing the matrix product and using the fact that the transpose also has maximal rank.
@manooooh: Huh?
@manooooh If some derivative is null, yes
Look at $f^{(n)}(x)=0$, integrate both sides $n$ times
Knowing that two antiderivatives (ON A CONNECTED INTERVAL) differ by a constant is still the MVT result, @manooooh.
6:06 PM
As I warned my calculus students repeatedly, you can have lots more antiderivatives of a function like $\sec^2 x$. They are not all of the form $\tan x + C$. Why?
Anything good lately?
Nah, just bad stuff.
Oh well. Hope you're doing alright.
LOL, other than degenerating disks/vertebrae, I'm doing great :) I hope you're doing well and are excited for the next chapter.
6:12 PM
I am having trouble understanding this two statements: $$\forall a,b,c\in\Bbb{Z}(a\mid bc\implies a\mid b\vee a\mid c)$$ (false: pick $a=6$, $b=4$, $c=9$), and $$\forall b,c\in\Bbb{Z}(2\mid ab\implies 2\mid a\vee2\mid b)$$
I mean the two statements differ by a number: $a=2$
@manooooh: You already told me what was special about $2$.
@TedShifrin yes, but the first statement is false. Meanwhile the first one is true
You keep typing wrong stuff. :)
But the first is more general than the second one. If something is false in general, isn't false for any particularization?
If I say every shirt in your closet is purple, how do you prove I'm wrong?
6:15 PM
@TedShifrin you have to give an counterexample: go to the closet and pick a shirt that it is not purple
All you need is one that's not purple. You could still have some purple ones, right?
@TedShifrin right
So for a general statement to be false all it takes is one instance where it is false, not every instance.
Ya, understood
Regarding this question:
Q: Prove that the empty relation, defined on a non-empty set, is asymmetric

manooooh Prove that the empty relation, defined on a non-empty set $A$, is asymmetric. My work: We need to show that if $\mathcal{R}=\varnothing$ is a relation on $A\times A$, with $A\neq\varnothing$, then for all $x,y\in A$, if $(x,y)\in\mathcal{R}$ then $(y,x)\notin\mathcal{R}$. Proof Since $A\ne...

6:24 PM
I am not able to understand how to prove it. I supposed $x,y\in A$ but it is wrong
Because $x,y\in A$ does not imply $(x,y)\in R$ but the converse is true
First, how do you know you can find $x,y$? Maybe $A$ has just one element.
@TedShifrin well in that case $x=y$
You can prove anything you want about the empty relation, can't you? It's symmetric and its asymmetric :P
Hi @TedShifrin long time no see.
heya @Jasper!
6:26 PM
Hi @JaspervanLooij! Remember that you have helped a lot, too. Thank you!
You'll be pleased to know I had dinner a few months ago with your idol, @Jasper.
Although I forgot to mention you :P
@TedShifrin yes of course... But I can't just say "Because the empty set is funny I say the statement is true" :P
@TedShifrin well, it's not reflexive
@TedShifrin It must be John Lee. Yeah, I told him I was crazy. =)
@manooooh I remember your nice duck pic! =)
No, I guess it's not reflexive. I hate s*** like this.
hi @Mathein
6:27 PM
Hi @Ted
If things are going to hold vacuously, let's make them interesting :P
@JaspervanLooij ^_^. A user that motivated me to change my profile pic to a duck had a discussion and deleted his account. Now I am sad, but happy to see you
@manooooh: For me it should be a Peking duck. Yum. :P
Oh no!
Oh yes.
6:32 PM
@TedShifrin so how would you start the proof of $R=\varnothing\implies\forall a,b\in
A((a,b)\in R\implies(b,a)\notin R)$?
$(a,b)\in R$ is false, so you can deduce anything from it.
@TedShifrin the main question that I am asking here is: how do you pick $a$ and $b$?
The statement says they are taken from $A$, so...
You don't. You let $a,b\in A$ be arbitrary (using that $A$ is nonempty).
@TedShifrin ok, so we let $a,b\in A\neq\varnothing$. Right. Then we can't imply $(a,b)\in R$
No, but think about the logic of how you prove symmetry (or asymmetry). It's an implication.
6:36 PM
@TedShifrin yes we start from $aRb$ and prove $b\not Ra$, but $a$ and $b$ have to be taken from a set
I can't say I let $a,b\in B$ because $B$ is even not defined. I must say $a,b\in A$ somewhere, right?
The relation is always a subset of $A\times A$.
@TedShifrin yes but an element of $R$ could not be in $A\times A$
@TedShifrin I mean that $R\subseteq A\times A$ means that everyting is in $R$, is in $A\times A$. But not the converse
Of course.
6:40 PM
Go to the exercise #3 of this page: fernandorevilla.es/blog/2014/02/03/…
It is written by a spanish mathematician
The point is that the statement of symmetry (or asymmetry) is an implication. Whenever $(a,b)\in R$ it follows that $(b,a)\in R$ [or $(b,a)\notin R$].
You can read e.g. the symmetric property: "FOR ALL $a,b\in\Bbb{R}$...". He says "For all" first, then he proves the property
@TedShifrin we need to show the implication yes
But since $(a,b)\in R$ is false, any implication will be valid.
@TedShifrin yes I know that. I understand it. But I do not understand why we are not letting $a$ and $b$ be elements of $A$
You suppose $a,b\in A$ and prove that if $(a,b)\in R$, then blah.
6:46 PM
@TedShifrin hm. I am seeing two different parts. One of them is "We suppose $a,b\in A$". The other is "We know $(a,b)\in R$". As said, we can't deduce if $a,b\in A$ then $(a,b)\in R$. So how do you connect these two parts?
Did you read what I wrote?
What you just typed seems to have nothing to do with it.
Where did "we know $(a,b)\in R$" come from?
@TedShifrin yes. I thought it was like I wrote
@Leaky!! Welcome to the land of the crazies.
@TedShifrin well I should say "We suppose $(a,b)\in R$"
Give my love to my alma mater. And, if you stop by 2-177, that was my office for two years :P
6:48 PM
@TedShifrin in HK, if you want a bus to stop at the next station, you press the stop button
NO, still wrong, @manooooh.
@TedShifrin @.@ are you saying that I am a crazy person? hehe
@TedShifrin whose office is it now?
No, the US is the land of the crazies.
@TedShifrin so how would you say
6:48 PM
I have no earthly idea, @Leaky. Since I left in 1981, I imagine dozens of people have been there.
@TedShifrin ohh
I think I learnt the purpose of the stop button in the bus in USA the hard way
@TedShifrin where the temperatures are wrong, the spellings are wrong, and the numbering of the floors are wrong
@Leaky: Here usually there's a cord to pull that rings a bell, when you want the next stop.
it was so embarrassing lmao
Well, the buildings I lived in before now had the first floor up from the entrance level — European style, but that's unusual in the US.
6:55 PM
As helpful as Celsius is for scientific purposes, I do think 0F much better captures “it’s f***ing cold” than 0C
0C is fine for me if I’m dressed right
I agree, 0ºC isn't bad at all.
I think you should just try 0ºK, @Semiclassic.
Send me a postcard when you do :)
Similarly 100C is so hot as to be irrelevant for how hot it is outside, whereas 100F lands pretty nicely there
I don't think "send" is possible :P
7:00 PM
So Celsius was invented just for science?
I actually don't know the history.
My joke is that 0/100 C is the freezing/boiling point of water, but 0/100 F is my brain’s freezing/boiling point
As far as I recall, Celcius originally had the scale reversed
iirc it was named after Celsius because he was the first one to have a consistent temperature scale
like the freezing point of water was essentially constant (at his time)
as well as the boiling point of water
nowadays they define degree Celsius via the absolute zero and the triple point of water
Rankine is a pretty silly temperature scale imo
7:06 PM
tries to remember who started this discussion
that would be me :P
Rankine : Fahrenheit :: Kelvin : Celsius
oh no
Time for Semiclassic to take his SATs.
An obscure one is electron-volts, or more properly eV/kB (Boltzmann’s constant)
Not useful for everyday computations but really useful for stuff like low-temp physics
7:12 PM
how do you latex it?
To justify it, note that entropy is measured in multiples of k_B
@TedShifrin can't wait to go to greenland without visa on thanksgiving
And change in entropy = heat transfer / temperature
7:20 PM
@Semiclassical I thought temperature was the derivative of entropy or something like that
You’re thinking of the thermodynamic definition: $T = (\partial U/\partial S)_{V}$
So temperature has units of energy / entropy
It’s not that different than people giving masses as eV when they mean eV/c^2
Ted would say that's not derivative lol
Sure it is. The internal energy U is a function of the entropy S and the volume V
So T is the partial derivative of U w/r/t S. The V is to indicate that we’re thinking of U(S,V).
oh nvm misread it as $\delta$
btw there's a rocking chair in wood in my accommodation
@Semiclassical hot take: $\partial A / \partial B$ is bad notation because it supposes values are somehow dependent on other values unlike in maths where you have functions with a fixed number of inputs and take partial derivative in the first / second / third etc variable
7:32 PM
$\dfrac{\mathrm dz}{\mathrm dt} = \dfrac{\partial z}{\partial x} \dfrac{\mathrm dx}{\mathrm dt} + \dfrac{\partial z}{\partial y} \dfrac{\mathrm dy}{\mathrm dt}$ shows this exact problem
what exactly is $z$ depending on?
and how is this mathematical
$D(f \circ g) = D(f) \circ D(g)$ is so much nicer in aesthetic and rigour
8:03 PM
@LeakyNun Well, don't make your reservations just yet.
@Leaky: That's why physical chemists use the notation they do. They indicate specifically which variables are fixed. It's actually very useful notation.
8:18 PM
Yeah, since there’s an operational difference between things like $T(\partial S/\partial T)_V$ and $T(\partial S/\partial T)_V$
Both are heat capacity, but at constant volume vs constant pressure.
Plus you get stuff like Maxwell relations
8:39 PM
Except for typo.
9:27 PM
what use do hilbert $C^*$ modules have? I've been seeing introductions to them everywhere but they seem totally esoteric to me
I guess in particular, whats an example of a useful hilbert $C^*$ module?
9:53 PM
@s.harp have addition and scalar multiplication
@s.harp probably $\Bbb C$
2 hours later…
11:32 PM
in The h Bar, 3 hours ago, by bolbteppa
7 Easy Hacks to Improve Your Math Skills http://karagila.org/2019/quick-hacks/

> 1. Get a graduate-level degree in mathematics!

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