Mathematics - 2021-10-15 (page 1 of 3)

12:00 AM

yeah. maybe there is something 'characterized' by the property that it achieves that minimum, or involves that minimum, which leads it to be called something like that.

Avra

YeS. It's indeed written as follows charchterization equation for $N_{i,j}$

So we need to find minimum attained to $N_{i,j}$?

Ted Shifrin

@BalarkaSen you are dead to me!

leslie townes

@BalarkaSen me too

Avra

:0

Hope all is good

copper.hat

12:36 AM

wow

Ted Shifrin

Hi @copper

copper.hat

Hi @TedShifrin! I'm collecting my son for the weekend tomorrow. Looking forward to 2hrs and 5 mins with him. 2hours stuck on the freeway and 5 mins after he gets home :-)

Ted Shifrin

LOL … The drive on 17 on a Friday late afternoon? Crazy.

leslie townes

2 hours if you're lucky. :)

i drew another cat monster with my daughter. that seems to be what we draw these days. it's only a matter of time before day care sends a letter home.

Ted Shifrin

I did that drive in my Saab 96 with a badly leaking brake master cylinder. That was one of the most nerve-wracking drives of my life.

Designing Halloween costume?

leslie townes

12:43 AM

yes, i think so. these draft pictures are pretty crazy.

Ted Shifrin

Who’s the tailor/seamstress?

copper.hat

the saab 96 was a wonderful car.

Ted Shifrin

I loved it, but it caused me anxiety to the end.

copper.hat

the roads on the hill are cambered really badly, it is nerve wracking to drive even in my wrx.

Ted Shifrin

And too many repairs for a grad student trying to pay Berkeley rents.

copper.hat

12:45 AM

ahh yes. for me car+love=money

that said, i did love my geo metro which cost $6.2k new.

Ted Shifrin

But it made it (towing after dead fuel pump on several occasions) to a few years in Athens. Finally bought a new Corolla in 83.

leslie townes

the geo metro might be the only gas car that is as fuel efficient as my prius c.

copper.hat

it got 50mpg on average. that was why i bought it. of course i ruined it all with my wrx.

leslie townes

i can get 60 in good conditions, usually 55ish. i don't know of a non-hybrid that does remotely as good.

Ted Shifrin

Electric :)

leslie townes

12:48 AM

of course the metro would collapse like a concertina in an accident

copper.hat

saves space

Ted Shifrin

And living driver.

copper.hat

solid geo engineering

leslie townes

the prius c basically is a geo metro hatchback. same shape.

copper.hat

i didn't get aircon so in summer i would drive 80/17 with the door ajar

Ted Shifrin

12:49 AM

You are a veritable idiot.

I had no AC. I used windows. Yours obviously didn’t work.

leslie townes

i noticed that people don't use windows in southern california

i do, unless it is really hot

copper.hat

it had the seatbelts that attached to the door so it wouldn't swing open. when traveling at 15mph tere

leslie townes

even if it means i have to hear some csulb student blasting sublime for 30 seconds at an intersection

copper.hat

there's not much risk

someone stole my radio in saratoga and the window repair was not up to scratch.

leslie townes

someone took a shot at me with a pellet gun in my current car, but the pellet gun hole is not visible while i'm driving (it's right behind the rear view mirror) so i haven't fixed it. doesn't rain often enough.

Ted Shifrin

12:54 AM

Leslie: for you

copper.hat

wow. i got shot at with paint guns. pellet guns can hurt.

i expect to find c*s in an asian restaurant.

one where the meals are swimming around in tanks in the restaurant

leslie townes

this is the kind of thing that i will halfway understand and then martin argerami will post an answer to. just as i've figured it out.

i do wonder if it was a pellet, from the hole it could have been a .22 but it did ricochet and not go all the way through so could not complete my ballistic analysis.

copper.hat

that's a bit scary. we had an air rifle that could kill.

leslie townes

i didn't like it at all

when i was growing up our neighbors would fire guns into a woodpile next to the fence between our house and theirs. i didn't like that either.

copper.hat

a few times farmers shot at us. they might have been firing blanks, but we didn't care to investigate.

leslie townes

1:03 AM

too busy fitting the timer and shipping something off to london

copper.hat

1:57 AM

i suspect those particular farmers were more likely to be on that side of the fence...

leslie townes

3:11 AM

this giants-dodgers game is about as boring as soccer.

Koro

3:34 AM

Given that V is finite dimensional and that $\phi_i$'s ($1\le i\le n$) is a basis of $V'$, how can it be shown that $\phi_i$'s are a dual basis of some basis of $V$?

I understand that dim $(\cap _{i=1}^n null \phi_i )=0$

leslie townes

ted's suggestion to look in terms of matrices, transposes, etc. might be really good about now.

Joe Shmo

what is $V'$? the dual space?

leslie townes

nobody tell him i said that.

Koro

yes, V' is the dual space.

Leslie, I'll think in terms of matrices because as of now, I'm not getting any intuition whatsoever. Every exercise problem in 3F chapter seems like a completely new thing to me.

Joe Shmo

well, if the $\{\phi_i\}$ were projections onto the $i^{\text{th}}$ coordinate, can you see how they are a dual basis?

Koro

3:40 AM

no :'(

Joe Shmo

let's see what wiki has to say...

Dual space

In mathematics, any vector space V {\displaystyle V} has a corresponding dual vector space (or just dual space for short) consisting of all linear forms on V {\displaystyle V} , together with the vector space structure of pointwise addition and scalar multiplication by constants. The dual space as defined above is defined for all vector spaces, and to avoid ambiguity may also be called the algebraic dual space. When defined for a topological vector space, there is a subspace of the dual space, corresponding to...

let me know if something's unclear

Koro

In Axler's LADR, I don't think projection has been introduced by chapter (3F) :'(

Joe Shmo

I called them a fancy name

Ted Shifrin

4:02 AM

@leslietownes mutters inverse, cough cough.

leslie townes

cough, cough, hack, wheeze, louder cough, death rattle

Ted Shifrin

Sounds worse than poor Screech.

leslie townes

how's she holding up? we didn't have to deal with livvy post-op. does she get treats?

Ted Shifrin

It is a sLow recovery. Mostly sleeping, less eating. Apparently it can take up to 10-14 days to return to normal, and the cone is no help.

Avra

4:36 AM

@TedShifrin. Hope she/he recovers. No idea anyway

leslie townes

the munchkin refused most of her lunch at day care today and ate a handful of grapes. second day in a row she's done this. we are not fans of it.

Avra

@leslietownes. what is munchkin pls

leslie townes

it's a slang term for my daughter.

Avra

Oh!

I thought cat

Sorry to hear that :(

leslie townes

i have a cat too. she eats everything.

Avra

4:40 AM

children though have strong immune system

So hopefully she will recover quickly 🙏

leslie townes

this is true. she brings everything home only long enough to give it to us.

Avra

:(

love_sodam

The fact that two manifolds $M,N$ are orientable if and only if $M\times N$ is orientable is well known.

I actually proved this in my differential geometry class

But only using the definition of orientable manifold using homology, I think it's hard (I doubt if it's even possible) to prove.

nice properties all assume the considered manifold is closed

leslie townes

4:55 AM

what's the homology definition? something about the top degree homology class?

love_sodam

@leslietownes Definition using local orientation and local consistency.

leslie townes

the "only if" has to be easy. what goes wrong with the if?

love_sodam

I only tried 'if' so I don't know 'only if' is easy but the problem is that it's not guaranteed $H_i(M;R) =0$ for $i>n$ if $M$ is $n$ dimensional manifold.

leslie townes

isn't it possible to prove that using not much more than homology, if M is connected, closed, whatever else would be normally expected?

all of my memories of this stuff are blended up, i could be way off base

love_sodam

Ahh wait why only if is easy? I actually tried to prove only if statement

I think to prove $M\times N$ is orientable, I need to choose some open nbd $U\times V\subset M\times N$ that satisfies some local consistency

And choosing $U$ and $V$ the open balls that satisfy local consistency for $M$ and $N$ respectively would be natural.

During the proof I need Kunneth formula and in that, I need $H_i(M;R) = 0$ for $i>n$.

Anyway I'll first prove this assuming $M,N$ are both closed orientable manifolds and next prove the general case using orientable cover.

(I actually tried to avoid using orientable cover)

shintuku

5:23 AM

@PM2Ring hey i noticed you use sagemath, so i have a question for you if you have time. How do you get nice axes through origin for 3d plots?

Koro

5:37 AM

@JoeShmo I got that now. Thanks.

Ted Shifrin

@love_sodam Why do you think this isn’t true?

Think simplicially, for example.

Rajorshi Koyal

6:01 AM

Please assist me with this problem set..

I am unable to write the values in the two flowcharts on my own..

I am looing for a step by step guidance

SAJW

6:23 AM

What is a good book about the surface area of smooth things?

BannedUser

Today I was saved by a physicist

leslie townes

SAJW a lot of intro to differential geometry books discuss the geometry of surfaces, which includes surface areas. is there some particular application you have in mind?

copper.hat

was the physicist a life guard or something?

leslie townes

last night a dj saved my life

SAJW

6:38 AM

@leslietownes Ah ok. I don't have something in mind, just find the surface area of a ball/sphere interesting. I'll check it out.

leslie townes

there's a well known formula in terms of double integrals for the surface area of a surface that happens to be a portion of the graph of a two-variable function, z = f(x,y). the subject is somewhat subtle because unlike in the 1D case, where rectifiable curves have lengths that are limits of approximating line segments, limits of approximating triangular planar chunks need not converge to the area of a surface.

fractal freaks like xander may know more about this than i do.

SAJW

can a surface be fractal and smooth at the same time?

leslie townes

probably depends on how you define those things, most likely not. but computing areas can be subtle if you veer too far away from the calculus textbook setting

Koro

Operators on a finite dimensional, non zero, complex vector space have an eigen value.

leslie townes

no objection here :)

Koro

6:56 AM

This statement is not valid for complex vector spaces of characteristic 2. Right? I'm referring to theorem 5.21 of LADR. But I recall that by "complex vector spaces", LADR means that a vector space $V$ over field $C$= the set of all complex numbers. Is my understanding correct here? I say that the theorem is not valid for complex vector spaces of characteristic as the proof goes something like this: Since V (with dim n) is non zero, there is a non zero v in it, then consider

leslie townes

i don't give a s # ! + about characteristic 2

probably safe to assume that most of linear algebra breaks down for that, except, like, elementary row operations. but who cares

if a government (any government!) wants to pay me to do cryptography, i will care about characteristic 2

but i will not care about characteristic 2 for free

Koro

$v,Tv, T^2v, ..., T^{n}v$. These must be LD. And hence there exist $a_i\in F$ (not all zero) such that $a_0 v+...a_nT^nv=0$, now if all $a_i=0$ for $i\ge 1$ then $a_0v=0$ does not give $a_0=0$.

copper.hat

do you like characteristic two?

Koro

@leslietownes ah, so one should not be worried about char 2?

leslie townes

this is entirely my opinion and not reality, but the urge to generalize everything to every possible case should be avoided

linear algebra over finite fields, maybe. linear algebra in characteristic 2 is notoriously bad

there are enough hard problems over C

why be bothered

Koro

7:00 AM

@copper: i think that there should be a footnote after each theorem wherever the theorem breaks in case of char 2.

copper.hat

The assumptions of the theorem should preclude such guesswork.

gn folks!

Koro

it's only 12:30 PM here @copper. :P

leslie townes

you can generalize almost anything to some greater context, that's not a bad impulse, but it helps to have a reason for doing it that isn't just "wait, what can i generalize this to"

because that is a neverending abyss

unless you do category theory, in which case it is how you get grant money until your advisor's best friends become emeritus professors and don't serve on those committees anymore

Koro

I got your point Leslie but this gives another confusion: nobody knows before hand what use some generalization shall have. Often, the use/application part my be discovered later. Right?

leslie townes

sure, but i don't have to be the person who does it

the interesting stuff tends to come from people who are generalizing for a reason, and not just generalizing to generalize

if i don't have a reason, i can stay away from it

this is highly subjective, others may disagree

this came up a little bit earlier. you can do some of calculus with definitions of limits, derivatives, etc. for fields like Q that are not complete

some of the theorems still hold, others break down, but generalize

should we all do calculus on Q now? i don't know, i would certainly be interested if someone had some reason to care about it

but without that reason, i'm happy to ignore it

PM 2Ring

7:21 AM

@shintuku There's an axes keyword, which you can pass directly to show, or you can supply it as a keyword in any plot & it will get passed to show. It creates a very basic RGB axes object.

If you want something fancier, you need to build it yourself from line3d. There's also a ruler object, but it's ugly. Here's an axes demo:

sagecell.sagemath.org/…

Koro

is sagemath like matlab @PM ?

PM 2Ring

Sort of.

From https://www.sagemath.org/ SageMath is a free open-source mathematics software system licensed under the GPL. It builds on top of many existing open-source packages: NumPy, SciPy, matplotlib, Sympy, Maxima, GAP, FLINT, R and many more. Access their combined power through a common, Python-based language or directly via interfaces or wrappers.

Mission: Creating a viable free open source alternative to Magma, Maple, Mathematica and Matlab.

I mainly use it as a way to write Python & create plots on my phone.

leslie townes

GAP is really good.

PM 2Ring

So if you already know how to use any of those systems, you can access them via Sage, utilising your existing skills. But if you don't know them, you can still harness a lot of their power through the Sage wrapper.

Eg, if you know how to do matrix stuff with Numpy, your Sage script can make Numpy matrix calls. Or you could just use Sage's matrix functions & under the hood Sage will use Numpy to do the number crunching.

With Numpy, your code tends to focus on manipulating the actual arrays. With Sage, you're operating at a slightly more abstract level, focusing on the mathematical entities.

Also, Numpy only knows about crunching numbers. But a Sage matrix can contain symbolic values.

SAJW

7:52 AM

has a wave with an arbitrary small frequency a finite length? that would be in a sense fractal and smooth at the same time

the waves height is also arbitrary

i think that's called amplitude

love_sodam

8:50 AM

@TedShifrin I believe it's true. Don't know how to prove.

robjohn

11:09 AM

Belief without proof... where have I heard that before?

Astyx

Covid mandates

ducks for cover

love_sodam

believer

I found the proof that uses cohomology with compact support which I don't know

love_sodam

11:35 AM

0

Q: Fundamental class of $M\times N$

(Hatcher 3.B.4) Show that the cross product of fundamental classes for closed $R$-orientable manifolds $M$ and $N$ is a fundamental class for $M\times N$. Assume $\dim M = m,\dim N = n$. Let $[M]$ and $[N]$ be fundamental classes of $M$ and $N$ respectively. Since $M,N$ are closed orientable ma...

love_sodam

11:46 AM

Essentially asking local orientation of $M\times N$

Koro

12:26 PM

@love_sodam: any progress on your GPA question?

I thought about it. To get the desired GPA, I think you’re supposed to finish that course first. But having infinitely many classes for the course makes it impossible I think.

Also, I don’t know about GPA system. I know SPI (semester performing index) or CPI( cumulative performing index). Each of these are awarded after every semester and not after every class so the question should have “infinitely many semesters” I think.

love_sodam

@Koro Not a proof but I believe that any rationals between $0$ and $4$ can be obtained. $\pi$ can be obtained as it has a decimal expansion.

Koro

We do have a sequence of rationals converging to $\pi$. But that’s not the problem. The problem (or rather my confusion) was the correctness of the question: to get GPA/CPI, one needs to finish a course, right? But since there are infinitely many classes, can the course ever be finished?

love_sodam

@Koro Then I would say infinitely many courses

Koro

That of course makes the question better but again can anyone finish infinitely many courses?

athing

12:46 PM

hi, For a linear vector space $X$, we have a seminorm $N$ defined over it. Let's define an equivalence relation over $X$ such that $x \sim y$ if $N(x - y) = 0$.

Now I'm trying to show that the function $\| [x] \| := \inf_{y \in [x]} N(y)$ is a norm for the quotient space $X/\sim$. So I have 3 properties to satisfy to be a norm and I'm stuck at "norm is 0 iff the vector is null" part. What is a null vector in a such quotient space? Aren't the elements of the set equivalence classes? How is addition defined for them so that I can find a null vector?

user193319

1:05 PM

I am asked to prove that the function $$\rho (f) = \left(|f(1/3)|^2 + \int_{0}^{1} |f'|^2 \right)^{1/2}$$ defines a norm on the Sobolev space $W^{1,2}[0,1]$. But I don't think this actually defines a norm. Let $f(x) = 1_{\{1/3\}^c}(x)$, where $\{1/3\}^c$ denotes the complement of $\{1/3\}$ in $[0,1]$. I'm pretty certain $f'=0$ almost everywhere (i.e., it has a weak derivative and it is equal to $0$ a.e.) and so I guess $f'' = 0$ almost everywhere too, so I think this means $f \in W^{1,2}[0,1]$.

Yet, $\rho(f) = 0$. Does this seem right, or am I mistaken?

Oh, wait, I don't care about the second derivative. I still think this example shows that $\rho$ isn't a norm.

@athing Given a seminorm, I thought the induced norm on the quotient space was simply $||[x]|| = N(x)$, no? Why do you need the infimum?

athing

I'm given that; trying to show it is indeed a norm :p

But I'm confused what even is a null vector in a space where vectors are supposed to be equivalence classes. How do you even add two equivalence classes?

user193319

No, you seem to be trying to show some other function is a norm. But the usual way (or so I thought) was to define $||[x]|| = N(x)$ rather than $||[x]|| = \inf_{y \in [x]} N(y)$. If you define it the former way, I think it's easier.

$\alpha [x] + [y] := [\alpha x + y]$ where $x,y \in X$ and $\alpha$ is a scalar. That's the vector space structure on the quotient.

athing

Oh thanks I didn't know that.. Then $[0]$ is the null vector... shame on me

user193319

Yup. That's okay. No need to shame yourself. All this stuff just takes getting used to.

Funnily enough, the function I defined above $\rho$, I think, is a seminorm.

athing

:d

user193319

1:19 PM

So, that's weird that we both asked related questions.

Thorgott

@user193319 those are the same thing and one effectively has to prove that anyway in order to obtain the result

user193319

Ah, okay. Good to know.

I still can't see why $\rho$ is a norm...hmm...

Thorgott

your example seems correct to me, but take that with a grain of salt

user193319

Okay. Thanks. I'm carefully working through all of the definitions to determine whether this is in fact an example.

love_sodam

1:42 PM

@Thorgott How can I show the cross product $H_n(\Bbb R^n,\Bbb R^n\setminus\{0\})\otimes H_m(\Bbb R^m,\Bbb R^m\setminus\{0\})\to H_{m+n}(\Bbb R^{n+m},\Bbb R^{n+m}\setminus\{0\})$ gives an isomorphism?

because Tor term vanish?

No then I'm using general Kunneth formula

Thorgott

2:16 PM

Fix the standard generator $e_1\in H_1(\mathbb{R}^1,\mathbb{R}^1\setminus\{0\})$. Then you only need to know that $e_n=(e_1)^n$ is the standard generator of $H_n(\mathbb{R}^n,\mathbb{R}^n\setminus\{0\})$, since then associativity implies this map maps the standard generator $e_n\otimes e_m=(e_1)^n\otimes(e_1)^m$ to the standard generator $(e_1)^n\times(e_1)^m=(e_1)^{n+m}=e_{n+m}$.
To prove this, you can use the isomorphism $H_n(\mathbb{R}^n,\mathbb{R}^n\setminus\{0\})\cong H_{n-1}(\mathbb{R}^n\setminus\{0\})\cong H_{n-1}(\partial I^n)$, where $I=[-1,1]$, that simply maps $(e_1)^n\mapsto\par…

Alternatively, there is a funny little argument that taking cross product with $e_1$ in fact gives an iso $H_n(X,A)\rightarrow H_n(X\times I,X\times\partial I\cup A\times I)$ for any pair $(X,A)$ of spaces (and this implies the above via induction). You can find that in the appendix of Milnor-Stasheff

user193319

2:51 PM

@Thorgott Maybe it doesn't work, unless I am screwing something up. Here's the definition of weak derivative I'm working with: Let $f,F \in L^1_{loc}(\Omega)$, where $\Omega \subseteq \Bbb{R}^n$. If for all $g \in C_0^{\infty}(\Omega)$ it holds that $$\int_{\Omega} F(x)g(x)dx = (-1)^{|\alpha|} \int_{\Omega} f(x) D^{\alpha} g(x)dx$$, then $F(x)$ is called the weak derivative of order $\alpha = (\alpha_1,...,\alpha_n)$ of $f(x)$.

Note, $|\alpha| = \alpha_1 + ... + \alpha_n$. If $F$ denotes any function which is $0$ a.e., then $\int_{[0,1]} F(x)g(x) =0$ for all $g \in C_0^{\infty}([0,1])$. But I don't think $\int_{[0,1]}f(x) g'(x)dx=0$ for all $g$...unless I am making some stupid mistake...

Still, I don't see how $\rho$ is postive-definite. If $\rho(f) =0$, then this implies $f(1/3)=0$ and $f'=0$ almost a.e...But why does that imply $f=0$?

I believe $f'=0$ a.e. implies $f$ is constant a.e.

Thorgott

the right definition when working over a closed interval is that you want $g$ to vanish at the boundary

user193319

Hmm...so, what does that say about my example?

Would ∫[0,1]f(x)g′(x)dx=0 in that case? It doesn't seem like it.

Thorgott

of course it would, that's the fundamental theorem of algebra

robjohn

3:17 PM

$g\in C_0^\infty([0,1])$ implies that $g$ and $g'$ vanish at the boundaries.

oh, that is $C_0$ not $C_C$, but I think that $g$ and $g'$ vanish at the boundaries.

user193319

Oh, shoot. You're right because $f$ is $1$ a.e., so $f(x) g'(x) = g'(x)$ almost everywhere...and you guys are saying $g'(0)=g'(1)=0$, right?

cOnnectOrTR 12

Can we show mathematically that taking limit does make the value of the average velocity (for example ) into instantaneous because we know there are infinite entries between any two numbers. We can forever decrease deltat and still get a value close to what we usually assign

I know geometrically slope of secant converges to slope of tangent!

leslie townes

think of the limit as the definition of instantaneous velocity

what else is it going to be

cOnnectOrTR 12

It’s not the question of what else. It’s is it accurate because we can forever go on decreasing deltat

Do we really get the limit at certain point in the process of decreaseing the deltat or is it always a assumption

leslie townes

3:36 PM

it depends on what you are studying. in 'real life' matter is quantized at small scales and a lot of things governing movement at really small scales do not reflect what we perceive as velocity

the limits defining velocity probably do not exist for individual molecules

it's a model, not reality, hence, no harm in taking it as the definition of instantaneous velocity

if only to ask, does this model line up with whatever it is i'm studying. at least it gives you something to check with and compare against

cOnnectOrTR 12

@leslietownes So it’s not exact but we still use it in Newtonian mechanics because it’s the best we have got!

Right?

leslie townes

kind of. i don't know that i'd say it isn't 'exact,' if only because that suggests there's some actual thing out there that it's failing to match up with, and maybe could match up if we fixed it. it's a model. it might or might not capture what you are interested in and if it doesn't you might switch the model

same for all of newtonian mechanics, really :)

user178758

"As far as the laws of mathematics refer to reality, they are not certain; and as far as they are certain, they do not refer to reality.” ― Albert Einstein.

cOnnectOrTR 12

Can we show that in some large number of steps the quantity does converge to actual limit ?

leslie townes

time to go shut physics.se down by asking what is waving in the transmission of electromagnetic waves

cOnnectOrTR 12

3:48 PM

@user178758 if science and maths are not accurate what is?

@leslietownes can we?

&

user193319

Hmm...so, how do we prove that $g$ vanishes at $0$ and $1$ if $g \in C_{0}^{\infty}([0,1])$?

leslie townes

in what setting? there will be examples where they don't. turbulent fluid flow is not modeled by newtonian analysis of individual particles (as far as i know). the question is, is there anything inherently wrong with a model where these limits do exist. not really

what function space is that?

user193319

The space of all infinitely differentiable functions on $[0,1]$ with compact support.

But $[0,1]$ is compact, so that second part is redundant.

leslie townes

the constant function g(x) = 1 is in there and does not satisfy the boundary conditions. usually boundary conditions like that are separately imposed and do not fall out of differentiability conditions

unless you're doing something very weird to define differentiability at the endpoints

user193319

Hmm...I'm not sure exactly how differentiability is defined in this context.

leslie townes

3:55 PM

hence my choice of example, i'd hope that a constant function has that property wherever it is defined :) but who knows

user193319

Hmm...I see...This problem is becoming quite vexing...I'm suppposed to show that $\rho$ (defined above) is a norm, but I don't think it has the property of being positive-definite.

$$\rho (f) = \left(|f(1/3)|^2 + \int_{0}^{1} |f'|^2 \right)^{1/2}$$

for $f \in W^{1,2}[0,1]$.

cOnnectOrTR 12

Limit is a weird concept

leslie townes

i guess if rho is zero, f' is zero a.e. and hence (via some regularity assumptions) there is a constant that it is a.e. equal to, and then the condition on f(1/3) (where f has point evaluation i guess because of some differentiability assumptions in the "W^1,2") implies that the constant is zero?

i never studied sobolev spaces and don't remember what the 1,2 in W^1,2 means

cOnnectOrTR 12

It’s not as obvious as 1+1=2

leslie townes

i agree. limits are subtle things

Thorgott

4:02 PM

it means it has weak derivatives up to first order and those are in $L^2$

user193319

@leslietownes Yeah, that's what I was thinking. But the derivative doesn't exist everywhere; it exists almost everywhere or something weird like that, so I wasn't sure if $f(1/3)$ being $0$ meant it was constant $0$a everywhere.

Thorgott

but I actually am confused overall

leslie townes

i mean, okay, f is represented by something that is differentiable a.e. but you can change the value of the representative at 1/3 without changing its equivalence class in your family of L-norms

Thorgott

because Sobolev space is usually defined as subspace of $L^2$

so $f(1/3)$ does not make sense

leslie townes

so "f(1/3)" must be interpreted in some subtle sense

Thorgott

4:03 PM

unless you wanna work with the space of representatives instead, but then it's obviously not a norm

leslie townes

it's so weird when thorgott and i share our thoughts

we could not be less alike

user193319

Ah, you're right. It could be infinite at $1/3$.

Thorgott

don't worry, the algebra isn't contagious

user193319

I didn't think about that.

leslie townes

and i take a daily pill to prevent functional analysis breakouts

Thorgott

4:05 PM

I used to pretend to be an analyst like two years ago, some thought patterns seem to have stuck with me

user193319

When I type up my solutions, I think I'll just say that the definition of $\rho$ doesn't make sense because $f(1/3)=\infty$ is possible and call it a day because it's Friday and I don't care anymore.

cOnnectOrTR 12

@leslietownes why don’t we all agree with limit as some vodo dodo and just prevent a mental breakdown from happening?

leslie townes

works for me :)

cOnnectOrTR 12

Anybody else :)

leslie townes

i think the resolution passes by majority vote

i woke up this morning to my daughter asking me why a "cat monster" was on the coffee table

cOnnectOrTR 12

4:10 PM

Sadly I don’t own a wife

user193319

Lol I hope no one owns a wife.

cOnnectOrTR 12

🍫☕️

leslie townes

yeah, i don't own one either

or maybe legally we both own half of each other, it's a community property state

BannedUser

4:23 PM

life world be so much better if physics were taught by mathematicians

they are more careful at choosing what words to use

user178758

But, that would leave no time for the lab.

Slate

That's fine, we can replace physics labs with mathematics labs. Students will be able to experience first hand what it's like to experimentally observe a maths in action.

leslie townes

that's what makes mathematicians so annoying

Xander Henderson

@Slate But... mathematics isn't empirical. :(

Lukas Heger

welcome to our Galois theory experiments

leslie townes

4:31 PM

xander, labs mean lab funding. don't ruin this.

we couldn't construct the math lab for less than, i dunno, $900 million.

Xander Henderson

@leslietownes Ah. Okay, then.

leslie townes

we need to supercollide the maths together.

Slate

If we collide the maths together at high enough velocity they'll split apart into axioms, and then we get a Field medal.

leslie townes

we're gonna need another $500 million to construct the barriers that prevent the math from getting out

user, i saw the swipe at lawyers. you're on the list.

user178758

🙊🙉🙈

btw, $500 million is peanuts compared to the $5.5 billion they spent on the LA stadium

During the height of covid

Holee Cannoli

4:40 PM

what's good my dudes

leslie townes

we're also going to need a gondola cable car thing to transport people to and from the math lab

user: this is why math is a good deal.

Holee Cannoli

gondola cable cars are the future ngl

I'm almost 100k rep O.o

leslie townes

i have a 10000k rep in my own house. i have all the privileges.

Holee Cannoli

same

you only need 25k for all the privileges tho

user178758

"Just label yourself maximally difficult, leslie."

who said that^

leslie townes

4:49 PM

who do you think? he's not here. we can say his name.

hyper-neutrino

i think there is something that theoretically maxes out at 35k

flags maybe? though you can get that sooner with a good flag record

user178758

🙊

Holee Cannoli

yeah there's the pedantic narrative for 35k

but if you're maxing out your votes you're alredy doing it wrong

Ted Shifrin

@leslie Well, the consensus of opinion is that I don't know the rules of "formal logic." Meh.

Holee Cannoli

haha sit

Wolgwang

4:56 PM

> Let $p(x)$ be a polynomial of degree $3$ and $p(n) =\frac{1}{n}$ for $n=1,2,3,4$. Find the value of $p(5)$

How to solve this by finite difference? (I have solved this by polynomial transformation)

leslie townes

@user178758 ^^^ that guy.

ted: this must be awful to learn so late in life. i hope you can somehow get through this.

Xander Henderson

@HoleeCannoli I'm not your dude, guy.

Ted Shifrin

On top of my cat's being misgendered, this may be more than I can handle.

Holee Cannoli

I wasn't talking to you obviously, you shill wiley

leslie townes

you did, in your unconscious but still perhaps infinite wisdom, choose a relatively ungendered name. dustin diamond's character notwithstanding.

Ted Shifrin

4:58 PM

Yes, Dustin was not in my thoughts.

Holee Cannoli

who is dusting what?

user178758

Hoffman?

Transcript for

$Mathematics$ Mathematics