Mathematics - 2022-07-01 (page 1 of 2)

1:12 AM

So I think I finally "get" why the number of partitions of $n$ into odd parts equals the number of partitions into distinct parts

Given a partition into distinct parts, turn each term of the form $q2^n$ ($p$ odd) into $2^n$ copies of $q$

Conversely, given a partition into odd parts, if there are $n$ copies of $q$, write $n$ in binary so that $n=\sum_i2^{a_i}$, and then turn the $n$ copies of $q$ into a copy of $2^{a_i}q$ for each $i$

Koro

Given $f_n(x)=n\sin (\sqrt{4n^2\pi^2+x^2}), x\in [0,a]$. Does $f_n$ converge uniformly on [0,a]?

I know that $f_n$ converges pointwise to $\frac{x^2}{4\pi}$ but I'm stuck at showing uniform convergence.

Ted Shifrin

1:34 AM

How did you get the pointwise limit?

Koro

1:48 AM

0

Q: Uniform convergence of $f_n(x)=n\sin (\sqrt{4\pi^2n^2+x^2})$ on $[0,a],a>0$.

I want to show the uniform convergence of $f_n(x)=n\sin (\sqrt{4\pi^2n^2+x^2})$ on $[0,a],a>0$. I tried it as follows: $f_n(x)=n\sin (\sqrt{4\pi^2n^2+x^2}-2n\pi)=n\sin\left(\frac{x^2}{\sqrt{4\pi^2n^2+x^2}+2n\pi}\right)\sim n\left(\frac{x^2}{\sqrt{4\pi^2n^2+x^2}+2n\pi}\right)\to \frac{x^2}{4\pi}$....

@TedShifrin I think I figured out uniform convergence too while posting my question on 'how to show uniform convergence'.

To get the pointwise limit, I used $\sin \theta=\sin (\theta-2n\pi)$

Ted Shifrin

I would suggest using Taylor on the stuff inside sin, from the beginning.

Maybe it’s equivalent to what you did.

P.S. Do not write $\sqrt u^3$.

leslie townes

it's wine and cheese night.

Koro

Actually I was trying to avoid the series expansion. Thankfully, I didn't have to use the expansion for the pointwise convergent part.

@leslietownes Happy wine and cheese night

Ted Shifrin

Not expansion. Error term.

Whine and cheese. The SCOTUS is overthrowing majority will at full blast. Happy days.

shintuku

i could also say today is sunny

yet, tomorrow could also be sunny

Koro

1:54 AM

hi @shin!

shintuku

hello Koro

Koro

shin, I finally got the college I wished to go to for my masters in maths. :)

leslie townes

ted: at least we get a short break before the next term.

Ted Shifrin

I’m gonna make BLT to have with my yummy homemade coke slaw.

I’m pondering a 9-day trek northward. Hope my neck and car can make it.

Koro

Now, I'll try showing ufc of the above sequence on $\mathbb R$

@TedShifrin $u^\frac 32$.

shintuku

1:56 AM

@Koro many congratulations Koro

Koro

is $\sqrt u^3$ frowned upon? Why?

shintuku

from what I see you put a lot of time into math so it is deserved

Ted Shifrin

I veto strongly, koro.

I even yelled at robjohn for that, koro, and he conceded.

Mixed syntax is yucky and sometimes confusing..

leslie townes

ted: 9 days of driving? that's, what, juneau?

robjohn

@Koro It's not frowned upon in general. I know that a number of papers on relativity use $\sqrt{1-v^2/c^2}^3$ in places

Perhaps there are others that use $\left(1-v^2/c^2\right)^{3/2}$

Koro

2:10 AM

Thanks a lot @shintuku : )

robjohn

@TedShifrin "coke slaw" must be interesting... do you need a mirror to eat it?

Koro

$\sqrt {u^3}$

robjohn

Well, I know that $\sqrt{x^2+y^2}$ shows up in a lot of places.

Koro

: )

Now I am stuck at showing uniform convergence on R.

Akiva Weinberger

2:26 AM

So here's something

A function $A(x)$ is D-finite if there exist polynomials $q_0(x),\dots,q_d(x),q(x)$ such that $q_0(x)A(x)+q_1(x)A'(x)+q_2(x)A''(x)+\dotsb+q_d(x)A^{(d)}(x)=q(x)$.

robjohn

@TedShifrin EPA, Roe v Wade, Holsters for all...

Ted Shifrin

@leslietownes not 9 days of driving … although who knows how many hours a day my body will do.

Akiva Weinberger

Then, theorem: $\sum_na_nx^n$ is D-finite iff $\sum_na_nx^n/n!$ is D-finite.

Koro

I think that since $\sup|f_n(x)-f(x)|\ge |f_n(\sqrt{12}\pi n)-f(\sqrt{12}\pi n)|=6\pi^2n^2$

Akiva Weinberger

@TedShifrin $\sqrt{u^a}^b$

Koro

2:32 AM

It follows that the convergence is not uniform on R.

I mean = 3$\pi n^2$ instead of 6$\pi^2 n^2$.

Is my understanding correct in showing that ‘f_n doesn’t converge to f uniformly on R’?

f_n and f were defined earlier.

Akiva Weinberger

Also, fun fact: the number of paths from $(0,0)$ to $(n,kn)$ with steps $(1,0)$ and $(0,1)$ that do not rise above the line $y=kx$, is $\dfrac1{kn+1}\dbinom{kn+n}n$

geocalc33

why do my $5 earbuds sound significantly better than my 250 dollar microphone?

Akiva Weinberger

Is the microphone making noise, rather than just recording?

(This (my previous thing) is the so-called $k+1$-Catalan number $C_n^{k+1}$)

geocalc33

like the vocal quality after I record is better on the 5 dollar mic

Akiva Weinberger

Oh, the earbuds also record

Maybe it's about placement? Is it directional?

robjohn

2:40 AM

@Koro I think $1+\frac{x/2}{1+x}\le\sqrt{1+x}\le1+x/2$ (Bernoulli's inequality) might be useful.

geocalc33

the only thing I can think of is that it's broken in some way or I have a pre-amp and that's not very expensive so if I upgrade the pre-amp maybe the quality will improve

Balarka Sen

@AkivaWeinberger I think that every Legendrian surface in R^5 contains a contactly embedded copy of {zig-zag} x {interval}, but the length of the interval may be very small compared to the height of the zig-zag.

But I don't think this is visually clear

Akiva Weinberger

I feel like reading about enumerative combinatorics is very refreshing

No wishy-washy hypersurfaces, just "how many of these concrete things are there"

(not that I'm suddenly deciding I dislike topology)

Balarka Sen

its usually too dry for me

leslie townes

geocalc: if you're gangster rapping, it may help to put something between you and the mic to take the edge off of the harder syllables. the headphones may already be doing that.

Akiva Weinberger

2:49 AM

@BalarkaSen It's a thousand-page handbook, but (at least this section) it's surprisingly engaging

It's introducing generating functions

Balarka Sen

maybe for you

Akiva Weinberger

I may feel like it's easygoing just 'cause I actually read Wilf's Generatingfunctionology however many years ago and already know something on the topic

so maybe that's my bias here

Balarka Sen

possibly

never felt the urge to look into combinatorics too deeply. tried Stanley once very early in my undergrad, went through 5 pages in some section and stopped forever

leslie townes

ted: my daughter misbehaved at school today

Akiva Weinberger

(re: "Stanley") I'm not familiar

Balarka Sen

2:52 AM

the most interesting combinatorics i have encountered is very rudimentary ramsey theory, because of probabilistic methods

but the main theorem of erdos is more probability than combinatorics

leslie townes

enumerative combinatorics. multiple volumes.

i had a similar reaction to stanley, but it's said to be a classic

i will never know

Balarka Sen

yeah thats the book

its very rigorous algebraic combinatorics

Akiva Weinberger

Maybe I'll check it out one of these days

So apparently $\sec$ and $\tan$ are not D-finite, in the sense I defined earlier

25 mins ago, by Akiva Weinberger

A function $A(x)$ is D-finite if there exist polynomials $q_0(x),\dots,q_d(x),q(x)$ such that $q_0(x)A(x)+q_1(x)A'(x)+q_2(x)A''(x)+\dotsb+q_d(x)A^{(d)}(x)=q(x)$.

$\sin$ and $\cos$ are, though, as $(D^2+1)\sin=0$

(I mean, they solve $y''+y=0$)

Balarka Sen

pretty pictures

Ted Shifrin

A deal of algebraic geometry in Stanley’s work!

Balarka Sen

2:56 AM

i think he was the one who worked on hyperplane arrangements before Goresky-MacPherson's breakthrough

Ted Shifrin

DogAteMy — you’re visiting Seattle?

Akiva Weinberger

Indeed I am

Ted Shifrin

Gorgeous city — go forth and ‘xplore!

leslie townes

ted: my daughter was "hitting" (more like swatting at) one of her classmates, then she took his hat off and threw it in the trash. she said it was 'a joke' and that 'everybody was laughing.'

where the teacher was, she didn't say

Ted Shifrin

The boy is too nice.

Mad Spaces

3:01 AM

why do those dudes not write what the hell above the sequences should frkn stay mani ts so annoying

help me overlords

Balarka Sen

zigzag spotted!

Mad Spaces

$0 \rightarrow ker(\varphi) \rightarrow V \stackrel{\varphi}{\rightarrow} W \rightarrow coker(\varphi) \rightarrow 0$

Balarka Sen

I think this is it. Height of this zig-zag is huge, whereas you can only take so small a neighborhood of it to make a {zigzag} x {interval} chart @AkivaWeinberger

Mad Spaces

what is the function from W to coker and from ker to v??

Ted Shifrin

@MadSpaces Your babble is nonsense.

What is the definition of coker?

Mad Spaces

3:04 AM

I can only think of $w \mapsto [w]$ as prohjection and dont know what the other one is

Akiva Weinberger

@MadSpaces Define ${\rm coker}(\phi)$ for me

Mad Spaces

$W / im(\varphi)$

Akiva Weinberger

I would've (maybe) beat Ted to that if autocorrect didn't keep on trying to fight me on whether "coker" should be capitalized

Ted Shifrin

Yes, that’s right. Well, wrk backwards. To get exactness, what does the first map have to be?

Akiva Weinberger

Where does $\ker(\phi)$ live?

test: $\coker$ $\im$

aw

Ted Shifrin

3:05 AM

You spend too much time whining instead of thinking.

Nope, no coker. No image or bunches of other things.

Mad Spaces

okay, kern projection = image phi ..kern projection is the zero of W so the image of Phi needs to be zero, you what mate?

because $0 \mapsto [0]= image \varphi$

Akiva Weinberger

"Kern projection is" what now

Mad Spaces

Sorry i mean the function from $W$ to $coker$ , so $w \mapsto [w]$

Akiva Weinberger

Sure. What's its kernel

Mad Spaces

How do you define a kernel on a function going to modulo raum? whats the zero element in the modul raum? it must be $im\varphi$

Akiva Weinberger

3:09 AM

Perhaps write $w+{\rm im}(\phi)$ rather than $[w]$

Mad Spaces

Thats the same

Akiva Weinberger

@MadSpaces Yes, the kernel is ${\rm im}(\phi)$

which exactly matches what it should be, for exactness

Mad Spaces

Yes so as i said, then $0 \mapsto [0]= im \varphi$ so to have exact, we need that $im \varphi = kernel $ of projection

Akiva Weinberger

$[\phi(v)]=[0]$ for $v\in V$. Why? Because that's what we're quotienting by

Mad Spaces

Oh, you are absolutely right, so any elements of the space we use as quotient is actually a zero element

that checks out i missed that

Akiva Weinberger

3:13 AM

Now, for the map $\ker(\phi)\to V$

Mad Spaces

waaait

dont say it!!

i am trying to work it out :P

Akiva Weinberger

OK

I will hint only to write out the type of everything

Mad Spaces

Alright so we have $kern$ projection = $im(\varphi)$ that checks out. so then we move one step down the sequnce and we have $ker\varphi$ = image of the function going from $ker \varphi$ to $V$ so the function is the identity?

Akiva Weinberger

For example, $\phi$ is of type $V\to W$ (meaning $\phi$ is a function from $V$ to $W$)

and so the elements of $\ker\phi$ have type [redacted]

@MadSpaces "Identity" is only really used when the domain and codomain are the same thing - people would say "inclusion"

but yes, it does not change the inputs at all

Mad Spaces

So the identity but limited on the set Ker phi

got it.

Akiva Weinberger

3:16 AM

because $\ker\phi\subset V$

Yeah

Mad Spaces

Makes sense my dudes

spasibo bolshoi

Akiva Weinberger

Test $A\hookrightarrow B$

Mad Spaces

Its these questions that make me sometimes feel dumb

Akiva Weinberger

I personally think it's hard to keep track of types in my head. Writing out things like "$\subset V$" above $\ker\phi$ reduces the cognitive load

(One day I will properly learn type theory)

Koro

@robjohn how about the following?

52 mins ago, by Koro

I think that since $\sup|f_n(x)-f(x)|\ge |f_n(\sqrt{12}\pi n)-f(\sqrt{12}\pi n)|=6\pi^2n^2$

robjohn

3:25 AM

where $f(x)=\frac{x^2}{4\pi}$?

Koro

yes.

RHS should be corrected to $3\pi n^2$.

robjohn

it looks as if you are showing that it does not converge...

Koro

$f_n(\sqrt{12}\pi n)= \sin 4\pi n=0$

I thought that since we are analyzing behavior on R, I could take $x=$ any real number.

robjohn

I thought the question was about some interval, not all intervals

Koro

@robjohn why? I thought that showing that the sequence $\sup_{x\in \mathbb R} |f_n(x)-f(x)|$ does not converge to $0$ will do the job.

ahh, I see.

Akiva Weinberger

3:29 AM

@MadSpaces You know how traditionally matrices and sets are capital letters, u and v are vectors, a b c are scalars, etc?

"Matrix", "set", "vector", "scalar" etc are all types

and you can be more specific (eg "functions from V to W" is a type)

Those conventions with capitalization are to help keep track. In an alternate universe, we just write the types explicitly next to each variable

That's kinda what the arrows you sometimes see on u and v do

Koro

@robjohn so on R, my working for showing $f_n$ does not converge uniformly on R to f is correct, right?

automorp15m

i know this was years ago but

is water wet? (for you)

Akiva Weinberger

@BalarkaSen I'm in a "state interesting theorems without proof" section of my book

That probably helps

(They have references to other books that have proofs)

robjohn

@Koro the sequence converges uniformly on any compact subset of $\mathbb{R}$.

If you change your interval, of course you can find places out far enough that it won't converge for longer.

Mad Spaces

3:47 AM

hey akiva one more question

Koro

@robjohn I agree.

Mad Spaces

we said that the zeros of the porjection is the image of phi because you get v + image phi = image phi. (where as v is element of image phi)

but doesnt this imply that image phi is a vector space

and this works only if phi is linear

So for none linear functions, the kernel will not be equal to image phi

because it needs not to be a vector space

right?

Akiva Weinberger

Oh I was assuming throughout that we were dealing with linear functions

Mad Spaces

we are

i am just asking generally

Akiva Weinberger

Yeah if you have a general function then things start breaking

Mad Spaces

3:50 AM

Okay, so the kernel for none linear functions, does need to be the image of the function for the projection function

Akiva Weinberger

Generally you would stick with the morphisms of whatever category you're in (in the category of vector spaces, the morphisms are linear maps; in the category of abelian groups, the morphisms are homomorphisms; etc)

Mad Spaces

i didnt study any of those terms you just said catgeory what so ever. i am doing Linear algebra i am not a math student

Akiva Weinberger

@MadSpaces It feels weird to even use the word "kernel" in that context, but yeah I think so

@MadSpaces Do you know group theory?

Mad Spaces

no

Akiva Weinberger

Ah, OK. Then ignore the category stuff I mentioned

"Vector spaces and linear maps" is a special case of a more general concept, but it doesn't make sense to jump to that until you've seen more examples of that sort of thing

Mad Spaces

3:53 AM

okay, i was just wondering, thanks anyways, maybe we will talk about that again in the future when i have more knowledge

for now this is sufficient

Koro

4:33 AM

Hi @BalarkaSen

Balarka Sen

Hi @Koro

copper.hat

5:19 AM

hello

robjohn

7:11 AM

@Koro I get $\frac{x^2}{4\pi}-\frac{24x^4+x^6}{384\pi^3n^2}\le n\sin\left(\sqrt{4\pi^2n^2+x^2}\right)\le\frac{x^2}{4\pi}$ for $n\ge\frac{x^2}{2\pi^2}$

which, for bounded $x$, gives uniform convergence.

Gwyn

7:50 AM

Is it correct to say that $o((x^2/2+x^3/6+o(x^4)^4)=o(x^4)$ as $x \to 0$ because $o((x^2/2+x^3/6+o(x^4)^4)=o(x^8/8+o(x^8))$ and so $\lim_{x \to 0} \frac{o((x^2/2+x^3/6+o(x^4)^4)}{x^4}=\lim_{x \to 0} \frac{o(x^8/8+o(x^8))}{x^4}=\lim_{x \to 0} \frac{o(x^8/8+o(x^8))}{x^8/8+o(x^8)} \cdot \frac{x^8/8+o(x^8)}{x^4}=0$

I mean, is it rigorous to divide for $x^8/8+o(x^8)$? I'm asking because it contains a little o symbol, and I am dividing this as it is a known function.

robjohn

@Gwyn you have mismatched parentheses. It is uncertain what $o((x^2/2+x^3/6+o(x^4)^4)$ means

Gwyn

Thank you for noticing, I mean $o((x^2/2+x^3/6+o(x^4))^4)$.

I am noticing that it should be $x^8/16$ all along, sorry for the typos. Even if it is irrelevant for the little o.

Koro

But is my way of showing 'not uniformly convergent on R' correct? @robjohn

$\sup|f_n(x)-f(x)|\ge |f_n(\sqrt{12}\pi n)-f(\sqrt{12}\pi n)|=3\pi n^2$

and since the sequence on LHS doesn't converge to $0$, it follows that $f_n$ doesn't converge uniformly to f on R.

robjohn

@Koro It is hard to follow your argument. You post a lot of equations but it is hard to follow how you are getting from one to the next.

@Koro how do you get that inequality? never mind. As I said, it is not uniform on all of $\mathbb{R}$. It is uniform on compact subsets.

Koro

The inequality: $\sup_{x\in \mathbb R} |f_n(x)-f(x)|\ge |f_n(t)-f(t)|$ for any real $t$. In particular for $t=\sqrt {12} \pi n$.

And using this, I obtain that $\lim_{n\to \infty}\sup_{x\in \mathbb R}|f_n(x)-f(x)|\ne 0$

robjohn

8:06 AM

You are showing that something which is not asked in the question is false.

Jakobian

2

A: The diameter of a convex hull.

Let $\ell = \mathrm{diam}\;A$. Since $C(A) \supseteq A$, it is trivial to see $\mathrm{diam}\;C(A) \ge \ell$. If $\mathrm{diam}\;C(A) > \ell$, then one can find $p, q \in C(A)$ such that $d(p,q) > \ell$. By Carathéodory's theorem, we can express $p$ as a convex linear combination of $P \le n...

robjohn

The question is only asking for uniform convergence on $[0,a]$ for any given $a$

Koro

I am showing that $f_n$ is not uniformly convergent on R.

robjohn

that is not what is asked in the question

Jakobian

Why is there a need to use Caratheodory theorem here?

Koro

8:08 AM

@robjohn yes. But then another part of the question (not posted in mse but posted only in chat) is to analyze if f_n is uniformly convergent on R or not.

And I'm trying to show that $f_n$ is not uniformly convergent.

robjohn

and I have said that it is not.

and you have given a reason why not

Koro

yes, and I want to know if my reason of concluding that is correct or not.

robjohn

5 hours ago, by Koro

@robjohn so on R, my working for showing $f_n$ does not converge uniformly on R to f is correct, right?

I couldn't tell what you were asking there. it looked as if you were talking about your post.

Koro

I see. There was a confusion. :(

robjohn

6 hours ago, by Koro

I think that since $\sup|f_n(x)-f(x)|\ge |f_n(\sqrt{12}\pi n)-f(\sqrt{12}\pi n)|=6\pi^2n^2$

that shows what you asked about $\mathbb{R}$ that convergence is not uniform on all of $\mathbb{R}$

Koro

8:16 AM

yes : )

Thanks a lot @robjohn. I requested for a review of that step only.

: )

I realized just now that working in my post is completely wrong.

I have fixed that just now.

robjohn

8:45 AM

I used $x-\frac{x^3}6\le\sin(x)\le x$ for $x\ge0$.

as well as the Bernoulli inequalities I mentioned before

$1+\frac{x/2}{1+x}\le\sqrt{1+x}\le1+x/2$

Koro

10:00 AM

@robjohn: I have edited my post to show that o(1/n) has no dependency on x.

one potato two potato

10:24 AM

'The general theory of relativity asserts that the universe is a smooth Lorentzian 4-manifold'

Mad Spaces

11:40 AM

what are the base elements of the polynomial ring in multiple variables

i am thinking of $X^I$ where as $I$ is an index notation, such that for eg $I =(1,1,0,0,0,0 \cdots) $ would mean $X^I = x_1*x_2$

geocalc33

1:18 PM

Is the notation here correct? $$ T^s=e^{s\frac{d}{dx}B(x)} $$

Shift operator

In mathematics, and in particular functional analysis, the shift operator also known as translation operator is an operator that takes a function x ↦ f(x) to its translation x ↦ f(x + a). In time series analysis, the shift operator is called the lag operator. Shift operators are examples of linear operators, important for their simplicity and natural occurrence. The shift operator action on functions of a real variable plays an important role in harmonic analysis, for example, it appears in the definitions of almost periodic functions, positive definite functions, derivatives, and convolution....

Roddy MacPhee

3:09 PM

Every set of two odd prime squares is equal to twice a sum of two squares ?

Akiva Weinberger

@geocalc33 What's B? Just $T^s=e^{s\frac d{dx}}$, no?

geocalc33

@AkivaWeinberger B(x) is a function that is the input

or is that not the way it's done

I was thinking that the operator takes B(x) as an input and spits out a new functionC(x)

geocalc33

3:42 PM

cross posted math.stackexchange.com/questions/4484309/…

Akiva Weinberger

4:19 PM

@geocalc33 $T^sB(x)=e^{s\frac d{dx}}B(x)$

I was confused because it only appeared on one side of the equation

@geocalc33 If $B$ is in the exponent then it is no longer a linear operator

When I write $e^{s\frac d{dx}}B(x)$ I mean to use the fact that $e^x=1+x+\frac{x^2}2+\frac{x^3}6+\dotsb$, and so $e^{s\frac d{dx}}B(x)$ means $$B(x)+s\frac d{dx}B(x)+\frac{s^2}2\frac{d^2}{dx^2}B(x)+\frac{s^3}6\frac{d^3}{dx^3}B(x)+\dotsb$$

Note that if $B(x)$ is a polynomial, this is a finite sum, so you can try $B(x)=x^3$ for an example

$e^{s\frac d{dx}}x^3=x^3+s(3x^2)+\frac{s^2}2(6x)+\frac{s^3}6(6)$

${}=x^3+3sx^2+3s^2x+s^3=(x+s)^3$, exactly the shift of $x^3$. Pretty miraculous, no?

leslie townes

meh, i've seen more miraculous

Akiva Weinberger

Another way to make sense of $e^{s\frac d{dx}}B(x)$ is to use $e^x=\lim\limits_{n\to\infty}(1+\frac xn)^n$

and so $e^{s\frac d{dx}}B(x)=\lim\limits_{n\to\infty}(1+\frac sn\frac d{dx})^nB(x)$

and then use $B(x)+\frac snB'(x)\approx B(x+\frac sn)$ for large $n$, repeatedly

The series way (the first version I wrote) is equivalent to Taylor's theorem by the way

robjohn

@AkivaWeinberger $s$ is not a function of $x$ here?

Akiva Weinberger

@robjohn No, it is a constant.

robjohn

okay

leslie townes

4:33 PM

we're well into the point of answering questions that nobody asked at this point

geocalc dropped his question into the chat and fled

robjohn

@geocalc33 That operator can be used to motivate the Euler-Maclaurin Sum Formula.

leslie townes

akiva: is there a version of the two arcs = three arcs example that can be rendered in a single line of ascii text

Akiva Weinberger

Heh? I was answering geocalc's question. He didn't understand that the exponential was of an operator, not a function. It's an easy mistake if you haven't seen before

@leslietownes Draw a line segment through the heart of a double spiral

leslie townes

where is the double spiral in code page 437

oh, i was unclear, i meant, graphically rendered, not described.

like: /\/\/\/\x/\/\/\/x\/x\/\___\

that kinda thing

Akiva Weinberger

Oh, uh

I guess the first one I came up with was kinda like 888888 but they got smaller and smaller

The "double spiral" thing was this one

leslie townes

4:38 PM

with four or five lines of ascii text i think you could convey the idea

i'm imagining how someone would draw it on a usenet post in 1993

Akiva Weinberger

Yeah probably doable

copper.hat

i guess i had a lot of serial upvoting

robjohn

4:53 PM

@copper.hat The big correction sweep was June 27, so if it was not then, it was something else.

copper.hat

5:11 PM

The big sweep netted -10, yesterday was -40.

not a biggie. just odd.

leslie townes

that was me deleting a number of copperhat stan accounts

FrustratedBird

Location:Dispersion

sorry

copper.hat

gotta get that jump suit

FrustratedBird

if Location: dispersion, \n Mean:Standard deviation, \n Median:Quartile, then \n Mode:What?

geocalc33

5:58 PM

mode: most

the most prevalent value in the data set

schn

What is the difference between a linear metric space and a metric space?

copper.hat

never heard of a linear metric space

Koro

@schn you mean normed linear space?

schn

@Koro possibly. It is used in a certain textbook I am studying.

leslie townes

schn: koro's guess is a good one. it's not a standardized term.

it is probably at a minimum a metrizable "topological vector space" (this is a standard term), and the question might just be whether it comes with a specific kind of metric (and if so, what kind).

schn

6:09 PM

@leslietownes is "topological vector space" a topological space and a vector space?

Akiva Weinberger

It's probably a "normed vector space" (these have a topology)

because if you can define open balls, you can define open sets

geocalc33

@AkivaWeinberger So if it's a nonlinear operator it won't preserve complete information as a linear operator would do. Then $B(x)'s$ transformed version $C(x)$ cannot be manipulated "solved" in that function space and mapped back to the original space under the inverse. Am I in the right ballpark on this?

because of that i'm starting to think properties $B(x)$ may have will not necessarily translate to $C(x)$ because of the nonlinearity

Akiva Weinberger

The usual solving techniques work best with linearity, yeah

leslie townes

schn: yes, with the additional property that the vector space operations are continuous.

geocalc33

@AkivaWeinberger gotcha

Thorgott

6:28 PM

never just put two structures on a space

always assume they're compatible, whatever that means

Akiva Weinberger

So here's something

geocalc33

I wonder if it's possible to take a "nonlinear object" and use a "nonlinear operator" and still get your information preserving

thing that comes with your linear operator

Akiva Weinberger

Something random I just learned

Say we have a tree with $n+1$ vertices and $n$ edges

(all trees have one more vertex than edges)

Randomly assign it an orientation

Choose one vertex to ignore

Label the remaining vertices $1\dots n$ and edges $1\dots n$

Form an $n\times n$ matrix $M$ with the entry $M_{uv}$ being $-1$ if vertex $u$ is the source of the directed edge $v$; $1$ if it's the target; and $0$ if they don't touch

Then, fact I just learned: $\det M=\pm1$

Ted Shifrin

@copper.hat Probably Cheerios.

Akiva Weinberger

(This is the reduced incidence matrix)

If you do the same thing with any other graph with $n+1$ vertices and $n$ edges (that is not a tree) then $\det M=0$

copper.hat

6:38 PM

@TedShifrin i really think breakfasts should not be allowed to create accounts

Ted Shifrin

Only meat & potatoes?

geocalc33

I only said that because sometimes it's the case that if you have a not well behaved object it's hard to study it on a well behaved space, so if you study the complicated object on a complicated space in some way they become compatible and you can actually get traction

Ted Shifrin

DogAteMy: What’s a good way to understand this? For the latter, I’m guessing we get an obvious element of the kernel.

Akiva Weinberger

@TedShifrin If you write it in the right order, it's a lower triangular matrix with 1s and -1s on the diagonal

Ted Shifrin

Ah.

Sorta like Jordan canonical bases …

Akiva Weinberger

6:42 PM

In "transformation" terms, the image of each vertex is its "parent" edge plus "lower" edges

Ted Shifrin

Right.

Thorgott

there should only be one function vertices -> edges associating to each vertex an edge incident to it

Akiva Weinberger

Wait, did I phrase it as a $\Bbb R^V\to\Bbb R^E$ map or the reverse?

I mean, it doesn't matter

Thorgott

so you only get one summand in the Leibniz formula

geocalc33

(not relevant to the current convo)... like studying random walks on random surfaces instead of studying random walks on the plane

Akiva Weinberger

6:43 PM

The image of each edge is (plus or minus) its "daughter" vertex plus a "higher" vertex

Thorgott

function should read bijection* in my first message

Ted Shifrin

Thor: That’s not a function?

Akiva Weinberger

Parent edge <-> daughter vertex is a bijection

Thorgott

well, there could be functions that aren't bijections

Akiva Weinberger

I'm drawing the tree with the deleted vertex as its root, on the top

Thorgott

6:44 PM

or maybe my condition forces them to be bijections, but I don't wanna think about it

Akiva Weinberger

This is as much detail as the book goes into

(It's more of a survey)

Thorgott

my point is that for a root, the edge you choose is determined and the rest follows inductively

eh, this is not phrased properly because we're ignoring a vertex, but I think the idea is sound

Akiva Weinberger

Here's more discussion from the book

Thorgott

proper argument: fix a vertex $v_0$. in the Leibniz formula for the determinant, you sum over bijections $\sigma\colon\{\text{vertices except $v_0$}\}\rightarrow\{\text{edges}\}$ and i claim there is only such bijection where $v$ is incident to $\sigma(v)$ for each vertex $v\neq v_0$. any bijection where that doesn't hold only contributes a $0$ summand.
now, the point is that such a bijection is determined inductively by depth counted from $v_0$. any vertex of depth $1$ is adjacent to $v_0$ via a unique edge and has to be mapped to that edge as otherwise no vertex could be mapped to that ed…

Akiva Weinberger

Ohh, that's a really neat perspective

that you're trying to find the one good bijection

@Thorgott Now if you could also give a good proof of the Cauchy–Binet formula…

(half joking)

Thorgott

6:55 PM

wow, I've never seen that formula

it screams exterior algebra, surely that's the right perspective?

Akiva Weinberger

Oh, interesting

That does make sense

Ted Shifrin

@Thorgott yes

Thorgott

I feel like this and the Laplace expansion should be the same principle

Ted Shifrin

Inner product on $\Lambda^k$ and Hodge star not unrelated …

Thorgott

oh wow, all the machinery coming together

this seems like something worth figuring out

I'll think about it after working out a couple homotopy theory proofs I still have to do...

Akiva Weinberger

7:01 PM

When you figure it out tag me because I'm still not fully seeing the lines

@Thorgott So here's something (else): Consider the function {partitions of $n$ into distinct parts} $\to$ {partitions of $n$ into odd parts} which, given a partition into distinct parts, takes every term of the form $(2n-1)2^k$ and replaces it with $2^k$ copies of $2n-1$.

Is this a bijection? Why?

Or said another way, replaces $q2^k$ with $2^k$ copies of $q$ for odd $q$

geocalc33

7:52 PM

@MathematicalEmergency hey

Mad Spaces

8:41 PM

is there a geometric meaning for the Algebraic multiplicity of an eigenvalue

Eigenvalue is the amount of which some vector which is by a transformation on its linear span stays multiplied with, say if Evalue = 1 then the vector stays the same, but what does algebraic multiplicity relaly imply?

Other than it devides char pol and whatever

leslie townes

8:56 PM

mm, generalized eigenvectors for A corresponding to the eigenvalue k (on a space of dimension n) are honest-to-goodness eigenvectors for (A - kI)^n corresponding to the eigenvalue 0. i'm not sure if that's 'geometric' unless you can 'visualize' powers of the operator.

Mad Spaces

i understood train station.

But okay thanks for trying

usually an indication i am not mathematically mature to understand this yet.

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