12:00 AM
You can factor any 2nd degree polynomial over $\mathbb{C}$

@MikeMiller comedy gold

what do you mean

I meant that even something like $x^2 + 1$ factors as $(x + i)(x - i)$

nlab is a very serious website for those who have taken the red pill

No, although you can factor any 2nd deg polynomial, I am talking over the reals

12:03 AM
You can factor it over the reals when the discriminant is not less than $0$

@xxxx036 what identity do you mean

like x^2+6x+6, x^2+13x+30
you can change the plus's into minus's and can be factored over the real

@Thorgott pretty bummed out right now. there's no nlab article on "based categories"

well, a based category should be a pair $(\mathcal{C},C)$ with $\mathcal{C}$ a category and $C$ an object in $\mathcal{C}$, no?
brb writing my first nlab article

pointed topological spaces and pointed simplicial sets are important in homotopy theory (where they are often called based)
@Thorgott important contribution
I really hope you actually start a stub
or just make a small edit here ncatlab.org/nlab/show/…

12:07 AM
@MikeMiller is this too much? math.stackexchange.com/questions/4001252/…

I need an academic source to cite
brb getting a PhD, writing a paper on based categories, then gonna cite myself on the nlab

@Thorgott based

So you mean it is related to topology

commutative dogeram

12:11 AM
@xxxx036 That's only because the polynomial $x^2+ax + b$ can be factored over the real iff $a^2-4b \ge0$, hence the sign of $a$ does not matter

are anagrams automatically commutative?

@Astyx I know that

Then I don't understand your question

@user2103480 I got a laugh when I pointed out that this got weirder to say after Lil B
One laugh
But it was a hearty one

wholesome
got a patrician in your class, it seems

12:13 AM
like x^2+5x+6, if you change the plus signs into minus signs, although you get four polynomials, you can factor them all

so what's the question?

$x^2+x+2$ cannot be factored but $x^2+x-2$ can?

argh physics is killing me, how do their definitions work

they don't lol

12:15 AM
the problem is to find a generator, which can generate all of these

Do physicists have definitions?

youtube.com/watch?v=W67kVhDwzdg anybody care to look at the diagram at 15:40

lol what is 'time'?

@Astyx yes that's what I mean

this is... just integration... but how does this plot come out? If I variate the angle and integrate along the ray it's constant in that diagram

12:17 AM
well it plots I(q) against q[nm^{-1}], seems clear

since the outward slices of the intensity are radially constant

@xxxx036 I don't understand what your question is
@user2103480 Which plot?

Do you know what the identity is at first?

15:40 there's this 2-dimensional function which is invariant under rotation

what identity ? Be specific

12:19 AM
then it says we integrate along the azimuth angle
ahhh I think I get it now

like x^2+10x+24 ,x^2-10x+24 ,x^2+10x-24 ,x^2-10x-24, can be factorised over reals

we integrate spherically for each radius q
so, in this case, over circles of radius q

@KarlKroningfeld doesn't exist

I think you do a mean of that integral
(ie you divide by the surface of each sphere)

if you can have two moments that are identical to every one of your senses then the moments are the same

12:21 AM
Otherwise it would (most likely) go to 0 at 0

I think the black disk around 0 is excluded
that's an artifact of the experiment

@MikeMiller + memory eraser

it sends a highly focused beam through a medium and to not damage the detector, the center of the beam is stopped by a physical barrier before it hits the detector
so only the scattering is measured

@Astyx the problem is to find a generator equation, which can find all of the polynomials having this identity

I still think you'd want to do an average
And considering the next slide, it seems they are doing an average

12:25 AM
I don't understand what you mean

I'm not talking to you
@xxxx036 I've given you the equation above
Almost

@xxxx036 A quadratic is factorable over the reals iff b^2 - 4ac >= 0 as pp said. You want to know when this remains true even if you negate a,b, or c. All that can do is change the inequality to b^2 + 4ac >= 0. Moving to the other side you want to know that b^2 is >= both 4ac and -4ac, aka, b^2 >= 4|ac|

okay i get what you mean. I was meaning it can be factored over the integers

That's a much harder problem

OK, but think quadratic equation again and see what each case forces on the coefficients.

12:29 AM
for x^2+ax+b, (a,b) are like (5,6),(10,24),(20,96),(40,384),.... and (13,30)
I got this so far
But there's a problem, as the fractions isn't really capable of every positive p and t

1:19 AM
how am I even supposed to get intuition for this physics course lmao the instructor told me I only need probability theory and all this is only applied probability

and then he defines this without batting an eye https://en.wikipedia.org/wiki/Dynamic_structure_factor
I mean, mathematically no problem, but how am I supposed to know what information this gives me when the experiments that give us the functions are called like this
"Experimentally, it can be accessed most directly by inelastic neutron scattering or X-ray Raman scattering."
"the intermediate scattering function and can be measured by neutron spin echo spectroscopy" ah yes thank you
From my perspective this is theology

Hello guys. This looks rather basic but anyway...
Suppose $f$ is a $C_0^\infty(\mathbb{R}^n\times [0,\infty))$ function with compactly supported in $\mathbb{R}^n$ (i.e. for each fixed $t$, $f_t:\mathbb{R}^n\to \mathbb{R}$ has compact support.

May I write
$$\lim_{t\to \infty}\int_{\mathbb{R}^n}f(x,t)\,dx = \int_{\mathbb{R}^n}\lim_{t\to \infty} f(x,t)\,dx?$$

If the limit exists of course

consider $f(x,t)=1_{[t,t+1]}(x)$
ok that isn't smooth, but that's a non-issue

1:35 AM
@Thor: You have to modify for $n>1$, but yup.

yeah, this is just n=1
what goes wrong for n=1 goes wrong for n>1 too

Sure.

except for being disconnected after removing a point :P

Highly germane.

1 hour later…
2:38 AM
Hi @TedShifrin I wanted to know if you could verify my thinking. It involved two questions from Sec 2.1: 2b and 3:

2b) Assuming I got the original parametrization correct (it feels like I did) the second part asks to produce infinitely many solutions. So to do that I replaced $1$ with $z$: $(x,y) = (\frac{z-t^{2}}{z+t^{2}}, \frac{t}{z+t^{2}})$ where $t \in \mathbb{Z}$ and $z \in \mathbb{R}$.

3) For this one I used the idea of arc length and adding vectors. So based on the diagram in you text it appears that the arc length of the circle up to the point $a$ is equivalent to the straigh

Hi All.

1 hour later…
3:49 AM
@dc3rd: If you replace $1$ with $z$, you're no longer on the circle. The issue is to think about putting in rational values of $t$ !! For #3, of course you want to add vectors. The $a(\cos\theta,\sin\theta)$ is clear, yes. But the point is that the vector from the edge of the reel to $P$ is orthogonal to the first vector (that's what pulling taut does). So you have length, but the direction you used was totally random.

4:14 AM
@TedShifrin for 3) I just scribbled something down form your suggestion of orthogonality. So using that idea what came to mind was taking the dot product of the radius with the direction vector going from $a$ to $P$. In other words:

Letting $P = (x,y)$

$(a \cos(\theta), a \sin(\theta)) \cdot (x - a \cos(\theta), y - a \sin(\theta)) = 0$. But of course solving for $x$ and $y$ becomes hard because I only have one expression.

You already told me you know how to rotate $\pi/2$ in the plane. Just have to go the right way.

4:30 AM
Hmm...interesting. I see it, now it is a matter of "mathefying" it. So in english, I would be rotating the segment $\overrightarrow{aP}$ that attaches to the radius $a$ from the origin. Just going to go work on the correct expression.

5:12 AM
@TedShifrin

So here is how I thought it out:

$(x,y) = (a \cos(\theta) + a \theta, a \sin(\theta) + a \theta)$ would give me the full vector from the origin to $\overrightarrow{OP}$ in a straight line, but I need to rotate the fragment $\overrightarrow{aP}$. This fragment is: $\overrightarrow{aP} = ( a \cos(\theta) + a \theta - a , a \sin(\theta) + a \theta - a)$. But in this form it is pointing in the same direction as the full vector $\overrightarrow{OP}$, so to rotate it in the correct direction the coordinates would be $(y, -(-x))$, since the original vector $OP$ points on the negative

This is crazy wrong.
Forget your wrong stuff and start over. The first vector we know. The second vector is orthogonal to it and has length $a\theta$. What then is the second vector?

Good day!

5:44 AM
Dr. Shifrin! For part (c) of the limits exercise, where you wrote that b) and c) are related, this is how I solved it earlier: I let $P$ be the statement that the limit for $f$ exists, $R$ that the limit of the sum $f + g$ exists, and $Q$ be the statement that the limit for $g$ exists. Then from b) we know that $P \land R \implies Q$. We can write this as $\neg P \lor \neg R \lor Q$.
Then notice that $P \land \neg Q \implies \neg R$ can be written as $\neg P \lor Q \lor \neg R$, and so by commutativity of the logical or, we know both statements are true.

@TedShifrin

First vector: $(a \cos(\theta), a \sin(\theta))$

Second vector: $(a \theta\ \sin(\theta), a \theta\ \cos(\theta))$

Then the parametric equation would be the sum of those two vectors

6:00 AM
@TedShifrin: good evening. It's nice to see you here so late.
Maybe it's not so late, but I often get here in the evening too late to see you.

2 hours later…
7:33 AM
hello

1 hour later…
8:37 AM
-5

The paper's link is https://arxiv.org/abs/1704.05747. In this paper, we give a proof to Riemann Hypothesis, please feel free to indicate any omissions. It's my pleasure to communicate with anyboby who has any questions.

9:08 AM
Hello!! Does somone of you have an idea about my question : math.stackexchange.com/questions/4001644/… ?
It is about homogeneous linear system of first order equations

9:32 AM
@MaryStar Is likely to be closed for lack of context.

9:50 AM
@TedShifrin 5am late night epiphany while laying in bed:

Second vector is actually: $(a \theta\ \sin(\theta), - a \theta\ \cos(\theta))$

user496045
10:12 AM
Hello

user496045
:)

11:19 AM
@DragonairGirl hi

3 hours later…
1:56 PM

2:11 PM
my prof has got to be kidding me. If I understand this correctly he first uses $\vec R_k$ as time-dependent processes governed by differential equations given by the poisson-bracket... and then proceeds to use brackets to signify that some function has inputs $(\vec R_1, \vec P_1,..., \vec R_n, \vec P_n)$? Literally one line afterwards?

2:33 PM
Also, $R$ and $P$ are treated as functions and as variables in which we differentiate? And $\vec R_i(t)$ is differentiated with respect to $\vec R_k, k = 1,...,N$? I just don't get this... Is this only a complicated phrasing of $\frac{d}{dt}\vec R_i(t) = (\partial_{r^k_1}H,\partial_{r^k_2}H,\partial_{r^k_3}H)^T$ where $(r^k_1,...,r^k_3) = \vec R_k$ is the $k$-th variable of the Hamiltonian
Ah yes that seems to be it
goddammit
Oh, must be $\frac{d}{dt}\vec R_i(t) = (\partial_{p^k_1}H,\partial_{p^k_2}H,\partial_{p^k_3}H)^T$
Anyways this is all horrible notation

Argh physics.
Your professor has very nice handwriting though.

lmao someone says that literally every time I post an excerpt
en.wikipedia.org/wiki/Liouville%27s_theorem_(Hamiltonian) what is $\frac{d \rho}{d t}=\frac{\partial \rho}{\partial t}+\sum_{i=1}^{n}\left(\frac{\partial \rho}{\partial q_{i}} \dot{q}_{i}+\frac{\partial \rho}{\partial p_{i}} \dot{p}_{i}\right)=0$ even supposed to mean
why is a regular derivative with respect to $t$ different than a partial derivative

2:54 PM
why are they using the dot notation and $\frac{\partial}{\partial t}$??

because one is representative of the actual evolution wrt time, the other is only wrt to each coordinate

@Thorgott good question as well

Because the dot notation is the $\mathrm{d}$ derivative

@Astyx thankfully they dont define what an actual time evolution is
@Astyx whats that
total derivative?

The idea is that $\rho$ is a function of $\rho(\{q_i(t)\}, \{p_i(t)\}, t)$
@user2103480 yea, that's the word

2:56 PM
@Astyx that makes more sense

So when you derive by the time, you get the above equation

I'll never take a physics-related course again, the theorem above is trivial and understanding the notation takes an hour due to lack of explanation

You can see $\rho$ as a function of 2n+1 variables, which you compose with every $q_i$, which is a function of time
then this is just the chain rule

yeah no, this is horrible

tfw finance people have more intuitive and accurate notation than physicists

2:59 PM
Do they?

They do. If you want good stochastic calculus explanations with semi-formal but enlightening calculations, go to finance SE
Do not go to physics SE, they will hit you with white-noise calculations
Function not even measurable? who cares

"measur-what?" :)

physics.stackexchange.com/questions/28630/… "The best way to define this free of mathematical annoyances"... [goes on to define some lattice limit and then integrates over non-measurable, extremely incontinuous function]

"This account is temporarily suspended network-wide. The suspension period ends on Mar 18 '92 at 16:28."
das a long suspension

To quote a set of lecture notes "Mathematically, the process $\xi_t$ suggested in this paragraph is a mess. One could, at least in principle, construct such a process using a technique known as Kolmogorov's extension theorem.
However, it turns out that there is no way to do this in such a way that the sample paths $t \mapsto \xi_t$ are even remotely wellbehaved: such paths can never be measurable [Kal80, Example 1.2.5]. In particular, this implies that there is, even in principle, no way in which we could possibly make mathematical sense of the time average of this process.
loool accidentally stumbled across a ron maimon post
Tbf he's smart. And does nothing which isn't completely physics standard in that post, I think
@Thorgott yeah physics SE is weirdly combative
@Thorgott also, physics theorem: all measurable functions are differentiable
A is only supposed to be measurable (observables are complex-valued elements of $L^2$)

3:26 PM
that's a weak theorem, we proved the much stronger "every function is more or less analytic" recently

@user2103480 when I did my set theory seminar on the consistent generalization of Fubini the professor (Luecke) commented that in some set theoretic universes the physicists are right

@Thorgott kek
@AlessandroCodenotti also kek

@Thorgott do you know which '92 it is

@Thorgott But I think we can generalize this
all linear operators are defined on the whole space
and of course continuous
"oh yeah that was proven in 1932 by Dirac"

2092, I assume lol
@user2103480 the inverse laplace transform always exists

3:32 PM
@Thorgott ah well, you're out by a good 200 years
2292 = 99,999 day suspension

@NiharKarve looool

lmfao

nuking things isn't always the solution, physicists

Ok so say that we have $(C_f)_M := (\overline{k}[x,y]/(f(x,y)))_M$ where $f$ is irreducible, and $M$ is generated by $x$ and $y$. Assuming the maximal ideal of $(C_f)_M$ is generated by one element, $z$, we can eventually write $rx - sy = 0$ in $(C_f)_M$ because we can write $x = zs, y =zr$.

Since we can assume wlog $s$ a unit in $(C_f)_M$ we can pick polynomials $\tilde{s}, \tilde{r}, g \in \overline{k}[x,y]$ such that:

$f g = \tilde{r} x - \tilde{s} y$

Ultimately I want to show that $f_y (0,0) \neq 0$, but I get to a step where $f_y (0,0) = -\tilde{s}(0,0)/g(0,0)$ and I am not so s

my favorite moment in my physics lecture this semester was when the professor, after months of never worrying about whether we actually divided by 0 at any point, stopped during a calculation to worry momentarily about whether we are dividing by 0.. and it was a calculation in which we divided by an exponential

3:35 PM
Also should mention it is clear we can pick $\tilde{s}$ such that it has nontrivial constant term
Lol, and here I am, worrying about if I divided by 0

@Thorgott looool

@Thorgott yes but the exponential gets very small so it's zero for all practical purposes

We can pick on physicists, but let's bring up some of the awful notation and names for things mathematicians use.

it's so random. At one point they're starting to get all mathematically and then at the next moment they drop all that
@anakhro we're normal, regular people

lol
@user2103480 you forgot "good"

3:44 PM

But how do they not flinch at this kind of thing: "$\partial_t A = i \mathcal{L} A$ so formally, $A_t = e^{i\mathcal{L}t}A$
I think a good amount of semigroup theory is swept under the rug here right

looks correct tho?

it's a differential operator given by the poisson bracket
Unbounded

4:03 PM
this is possible using the spectral theorem since the operator is self-adjoint
I thought nonnegativity or compactness was necessary but this post mentions that its not necessary math.stackexchange.com/questions/1697318/…
@BigSocks scholze got luried mathoverflow.net/questions/382270/…

bruh moment

I don't get it.

NO ONE IS SAFE

4:44 PM
I'm appalled

@anakhro no one knows what it means, but its provocative

@Thorgott can you write the content of these two books briefly?

He would probably find the contents by the same method you would (googling).
He's making fun of the fact that abelian moves lines between volumes

5:01 PM
Lol
I have taken it seriously

@Thorgott they should name one "Infinite abelian groups" and the other "Abelian infinite groups"

@Astyx i first thought this was the case

@Thorgott Apparently the author gives nu Fuchs

lol

5:35 PM
This way they'd be miles apart in the library

0

This is a follow-up on this post, which is based upon this paper. First, let me set up some definitions, etc. A Semigroup $S$ is said to be an inverse semigroup provided that for every $x \in X$, there exists a unique element $x^{-1}$ such that $x^{-1}xx^{-1} = x^{-1}$ and $xx^{-1}x=x$. The se...

5:48 PM
@anakhro Agh!

6:24 PM
Hi @TedShifrin. I believe I got the answer to the question we were discussing yesterday, but I wanted to go through my thought process to make sure it is the right way to approach these things. I also drew a crude diagram to illustrate the main part of my idea.

So to start off, the second vector is: $(a \theta\ \sin(\theta), - a \theta\ \cos(\theta))$

How I got this was the following:

Knowing that the arc length of the second vector is $a \theta$, I could think of this as another vector emanating from the origin with coordinates given by $(a \theta\ \sin(\theta), a \theta\ \cos(\theta)) No. First, don't call length of a vector arclength. It's just length. :) You corrected yourself at 5 AM, so don't fall back in the same mistake. What are the two unit vectors orthogonal to$(\cos\theta,\sin\theta)$? You're making it way too hard. Just think basics. Don't worry about lengths until the end. There are NO reflections anywhere here. Howdy, @Rithaniel. those two vectors would be:$(-\sin(\theta), \cos(\theta))$and$(\sin(\theta), -\cos(\theta))$Heya Ted. How goes it on your side? Righto. Now we know we want to go "to the right" rather than "to the left" — so it is the second one. (Rotating$-\pi/2$rather than$\pi/2$.) Anyhow, you do that, then you multiply by the correct length,$a\theta$, and then you add the two vectors. Done. Bumbling along, @Rithaniel, and you? I see I have a weakness when it comes to understanding orthogonality and using the right triangles.... 6:33 PM Also bumbling. Teaching again this semester Fewer assignments, though, so more me time Yeah, @dc3rd, my emphasis on geometric thinking may not be good for you :) That said, statistics is really full of projections and geometry, even though statisticians don't teach it that way. What are you teaching this time, @Rithaniel? Well, technically it's the same course. They split calculus for business majors over two semesters. First semester we cover derivatives and second we cover antiderivatives Oh, but it's all new prep for you. No chance to learn from what you did right/wrong last term. Well, maybe next year. The only issue is that I have some people who haven't taken the first half for several semesters, so I can't really rely on their calculus knowledge being fresh in their minds. (Also, the lack of prep, as you say) @TedShifrin more the reason for me to integrate this form of thinking. Give myself the opportunity to broaden my thought processes. That's one of my underlying motivations in pursuing math......I also have noticed all of the geometrical objects that appear in statistics hence why I actually do want to understand a geometric way of thinking. Because sure we could only "picture" a distribution in 3 dimensions, but having the understanding of how these objects behave is invaluable. Also Halmos's essay on math as art comes to mind in my efforts 6:38 PM Yeah. We found at UGA that people that waited more than one (or at worst, two) semesters to go on to the next course were almost always in trouble. I think we even got a flag on the class roll so that teachers could alert the students, but we couldn't force them to retake the course. Well, @dc3rd, in fact so many concepts in stat are about projections and dot products. Even regression is solving the normal equations (you'll get to that in chapter 5) and finding the projection onto the column space of an appropriate matrix. I co-directed a masters thesis on the geometric linear algebra in standard statistics stuff. I can send it to you if you are interested (eventually?). Yes. I'm also doing a linear regression course at the moment and have seen the basic use of matrix for simple linear regressions. So I know the use of linear algebra is only going to get larger as I look at multivariable systems. Yes I would appreciate looking at the thesis. I always welcome sources of knowledge @TedShifrin :) Well I'm off to go do some more studying. I'll probably be back in the room in a few hrs asking questions or just absorbing what you masters of the art form discuss I want the least natural number for which$(n!)^2$is greater than$n^n$how can I solve it? The qn I got was if this number is more or less than 1000. What have you tried? I tried to rewrite n! in reverse order and multiply it over to compare with individual terms I think you're going to need to do something a bit more numerical than that. Do you know Stirling's formula? 6:49 PM the stirling approximation is the only part I know. Did you try to use that?' actually, no I hadn't thought of it. It might give you some hints. yes let me check. Actually, now that I do this, I think your first approach might have been just fine. 6:52 PM how so? I didn't find it of much use? Did you try starting with actual$n$values? I did try it for that Intuitively I do get that numbers beyond 1000 would have the inequality proven but I can't really show that properly. . . does the following necessarily hold if$(x_n)_{n \in \mathbb{N}}$is a real valued markov process?$\mathbb{E}[\mathbb{E}[x_{n+1} | x_{n-1}] | x_n ] = \mathbb{E}[x_{n+1} | x_{n-1}]$. It feels like the answer should be no, but I can't find an easy counterexample Try some small$n$, @napstablook. Even Stirling will tell you to start small, which is sorta contradictory, but, nevertheless. I have only encountered stirling only once in a beginner statistical thermodynamics book, and I am presently in high school so I'm afraid I don't get what you meant by that 6:57 PM @porridgemathematics that's just the law of total probability, no? OK, don't worry about Stirling. Just try some actual small numbers. @Thorgott is it? this equality is one of random variables whats the law of total probability you're referring to? huh it's weird lower numbers hold this quite true as inequalities Yup. 7:00 PM Actually I will get back to you after trying some more. thanks @TedShifrin oh okay, the tower property, so how would that apply here? @napstablook: If you know a bit of mathematical induction, you can actually show that once it starts holding, it will hold forever more. since its a markov process, we do have$\mathbb{E}[x_{n+1} | x_{n-1}] = \mathbb{E} [\mathbb{E}[x_{n+1} | \sigma(x_0,...,x_n)] | x_{n-1} ] = \mathbb{E}[ \mathbb{E}[x_{n+1} | x_n ] | x_{n-1} ] $, but that doesn't give us what we want we want to show$\mathbb{E}[x_{n+1} | x_{n-1}] = \mathbb{E} [ \mathbb{E}[x_{n+1} | x_{n-1} ] | x_n]$which is the equality on the first line, with$x_n$and$x_{n-1}$swapped (or disprove this) 7:29 PM I think its false in general but even with a simple random walk you need to go to the third step to see a counterexample, I think 7:50 PM @robjohn Not to leave you empty handed, this is what I had in mind, e.g., as an image of a mean heart: Or This seems about right. Brilliant! @amWhy pushed me into it! 8:07 PM @TedShifrin Hilarious! For St. Patties Day, @robjohn, some inspiration: (you're on the left, I'm on the right): I probably out-grinch robjohn. And for a pink angry easter bunny, some inspiration (not a lot of pink though): I could adopt him. @TedShifrin Go for it!! ;D @TedShifrin Aww, you're a nice guy! But then again, while robjohn flaunts his "mean square", deep down he's probably a teddy bear! You don't have substantive evidence on either one of us! I'm getting meaner and meaner to the people who post homework/exam questions with zero effort. 3 8:22 PM @TedShifrin Me too! And here's our teddy bear robjohn: Now wait a minute! Don't be usurping my Ted name. @TedShifrin You are so correct! So, you are now the owner of the Ted(dy) Bear, that captures the justified meanness you sometimes feel! @TedShifrin Is your given name Theodore? 8:44 PM The Cruel Heart Crappy Valentine's Day! @robjohn Love it!!!! Keep it handy for to wear near Valentine's Day! or... I'll have to see if one looks better as an avatar actually, in between looks best: 9:10 PM @robjohn I voted on which of the three! 9:52 PM I concur. Yes, @amWhy, I'm officially Theodore. Given a variety$X$, is an algebraic subset then a subvariety? And would a subvariety be something such that it's a variety on its own with the inclusion map being a morphism? My notes skip over many definitions :c Actually I'm not good enough at algebraic geometry to get into this Ah no, I didn't see what you wrote Hello! I was hoping I could just find a def online that would fit my notes, but there's so many sort-of-equivalent formulations, so I'm a bit unsure 10:20 PM If we have$\int_a^b f(x) ~ dx$, why is it that we can solve$\int_a^b x' f(x) ~ dx$to get the same answer? what is$x'$?$x'\$

true, but not interesting

Never mind, this solution abused notation. That's why

10:24 PM

10-4

11:09 PM
@amWhy: getting a bit'a'head in my avatars
I reused the heart some

@robjohn Perfect! See, I always know it you can pull it off! And to date, you are the most seasonally relevant user on math.se!!

@amWhy I added a border to the heart and changed it to orange.

@robjohn Gotta stay true to yourself! ;D

@robjohn I'm just green with envy!

11:29 PM
@TedShifrin WTL (what the luck)

11:42 PM
a St Patrick's anagram