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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 $i$
 
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.
 
1:34 AM
How did you get the pointwise limit?
 
1:48 AM
0
Q: Uniform convergence of $f_n(x)=n\sin (\sqrt{4\pi^2n^2+x^2})$ on $[0,a],a>0$.

KoroI 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 $\sin \theta=\sin (\theta-2n\pi)$
 
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 $\sqrt u^3$.
 
it's wine and cheese night.
 
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
 
Not expansion. Error term.
Whine and cheese. The SCOTUS is overthrowing majority will at full blast. Happy days.
 
i could also say today is sunny
yet, tomorrow could also be sunny
 
1:54 AM
hi @shin!
 
hello Koro
 
shin, I finally got the college I wished to go to for my masters in maths. :)
 
ted: at least we get a short break before the next term.
 
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.
 
Now, I'll try showing ufc of the above sequence on $\mathbb R$
@TedShifrin $u^\frac 32$.
 
1:56 AM
@Koro many congratulations Koro
 
is $\sqrt u^3$ frowned upon? Why?
 
from what I see you put a lot of time into math so it is deserved
 
I veto strongly, koro.
I even yelled at robjohn for that, koro, and he conceded.
Mixed syntax is yucky and sometimes confusing..
 
ted: 9 days of driving? that's, what, juneau?
 
@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 $\left(1-v^2/c^2\right)^{3/2}$
 
2:10 AM
Thanks a lot @shintuku : )
 
@TedShifrin "coke slaw" must be interesting... do you need a mirror to eat it?
 
$\sqrt {u^3}$
 
Well, I know that $\sqrt{x^2+y^2}$ shows up in a lot of places.
 
: )
Now I am stuck at showing uniform convergence on R.
 
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 $q_0(x)A(x)+q_1(x)A'(x)+q_2(x)A''(x)+\dotsb+q_d(x)A^{(d)}(x)=q(x)$.
 
@TedShifrin EPA, Roe v Wade, Holsters for all...
 
@leslietownes not 9 days of driving … although who knows how many hours a day my body will do.
 
Then, theorem: $\sum_na_nx^n$ is D-finite iff $\sum_na_nx^n/n!$ is D-finite.
 
I think that since $\sup|f_n(x)-f(x)|\ge |f_n(\sqrt{12}\pi n)-f(\sqrt{12}\pi n)|=6\pi^2n^2$
 
@TedShifrin $\sqrt{u^a}^b$
 
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.
 
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 $\dfrac1{kn+1}\dbinom{kn+n}n$
 
why do my $5 earbuds sound significantly better than my 250 dollar microphone?
 
Is the microphone making noise, rather than just recording?
(This (my previous thing) is the so-called $k+1$-Catalan number $C_n^{k+1}$)
 
like the vocal quality after I record is better on the 5 dollar mic
 
Oh, the earbuds also record
Maybe it's about placement? Is it directional?
 
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.
 
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
 
@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
 
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)
 
its usually too dry for me
 
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.
 
2:49 AM
@BalarkaSen It's a thousand-page handbook, but (at least this section) it's surprisingly engaging
It's introducing generating functions
 
maybe for you
 
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
 
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
 
ted: my daughter misbehaved at school today
 
(re: "Stanley") I'm not familiar
 
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
 
enumerative combinatorics. multiple volumes.
i had a similar reaction to stanley, but it's said to be a classic
i will never know
 
yeah thats the book
its very rigorous algebraic combinatorics
 
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 $y''+y=0$)
 
pretty pictures
 
A deal of algebraic geometry in Stanley’s work!
 
2:56 AM
i think he was the one who worked on hyperplane arrangements before Goresky-MacPherson's breakthrough
 
DogAteMy — you’re visiting Seattle?
 
Indeed I am
 
Gorgeous city — go forth and ‘xplore!
 
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
 
The boy is too nice.
 
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
 
zigzag spotted!
 
$0 \rightarrow ker(\varphi) \rightarrow V \stackrel{\varphi}{\rightarrow} W \rightarrow coker(\varphi) \rightarrow 0$
 
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
 
what is the function from W to coker and from ker to v??
 
@MadSpaces Your babble is nonsense.
What is the definition of coker?
 
3:04 AM
I can only think of $w \mapsto [w]$ as prohjection and dont know what the other one is
 
@MadSpaces Define ${\rm coker}(\phi)$ for me
 
$W / im(\varphi)$
 
I would've (maybe) beat Ted to that if autocorrect didn't keep on trying to fight me on whether "coker" should be capitalized
 
Yes, that’s right. Well, wrk backwards. To get exactness, what does the first map have to be?
 
Where does $\ker(\phi)$ live?
test: $\coker$ $\im$
aw
 
3:05 AM
You spend too much time whining instead of thinking.
Nope, no coker. No image or bunches of other things.
 
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 $0 \mapsto [0]= image \varphi$
 
"Kern projection is" what now
 
Sorry i mean the function from $W$ to $coker$ , so $w \mapsto [w]$
 
Sure. What's its kernel
 
How do you define a kernel on a function going to modulo raum? whats the zero element in the modul raum? it must be $im\varphi$
 
3:09 AM
Perhaps write $w+{\rm im}(\phi)$ rather than $[w]$
 
Thats the same
 
@MadSpaces Yes, the kernel is ${\rm im}(\phi)$
which exactly matches what it should be, for exactness
 
Yes so as i said, then $0 \mapsto [0]= im \varphi$ so to have exact, we need that $im \varphi = kernel $ of projection
 
$[\phi(v)]=[0]$ for $v\in V$. Why? Because that's what we're quotienting by
 
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
 
3:13 AM
Now, for the map $\ker(\phi)\to V$
 
waaait
dont say it!!
i am trying to work it out :P
 
OK
I will hint only to write out the type of everything
 
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?
 
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
 
So the identity but limited on the set Ker phi
got it.
 
3:16 AM
because $\ker\phi\subset V$
Yeah
 
Makes sense my dudes
spasibo bolshoi
 
Test $A\hookrightarrow B$
 
Its these questions that make me sometimes feel dumb
 
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)
 
@robjohn how about the following?
52 mins ago, by Koro
I think that since $\sup|f_n(x)-f(x)|\ge |f_n(\sqrt{12}\pi n)-f(\sqrt{12}\pi n)|=6\pi^2n^2$
 
3:25 AM
where $f(x)=\frac{x^2}{4\pi}$?
 
yes.
RHS should be corrected to $3\pi n^2$.
 
it looks as if you are showing that it does not converge...
 
$f_n(\sqrt{12}\pi n)= \sin 4\pi n=0$
I thought that since we are analyzing behavior on R, I could take $x=$ any real number.
 
I thought the question was about some interval, not all intervals
 
@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.
 
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
 
@robjohn so on R, my working for showing $f_n$ does not converge uniformly on R to f is correct, right?
 
i know this was years ago but
is water wet? (for you)
 
@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)
 
@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.
 
3:47 AM
hey akiva one more question
 
@robjohn I agree.
 
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?
 
Oh I was assuming throughout that we were dealing with linear functions
 
we are
i am just asking generally
 
Yeah if you have a general function then things start breaking
 
3:50 AM
Okay, so the kernel for none linear functions, does need to be the image of the function for the projection function
 
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)
 
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
 
@MadSpaces It feels weird to even use the word "kernel" in that context, but yeah I think so
@MadSpaces Do you know group theory?
 
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
 
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
 
4:33 AM
Hi @BalarkaSen
 
Hi @Koro
 
5:19 AM
hello
 
 
2 hours later…
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.
 
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.
 
@Gwyn you have mismatched parentheses. It is uncertain what $o((x^2/2+x^3/6+o(x^4)^4)$ means
 
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.
 
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.
 
@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.
 
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 $\lim_{n\to \infty}\sup_{x\in \mathbb R}|f_n(x)-f(x)|\ne 0$
 
8:06 AM
You are showing that something which is not asked in the question is false.
 
2
A: The diameter of a convex hull.

achille huiLet $\ell = \mathrm{diam}\;A$. 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...

 
The question is only asking for uniform convergence on $[0,a]$ for any given $a$
 
I am showing that $f_n$ is not uniformly convergent on R.
 
that is not what is asked in the question
 
Why is there a need to use Caratheodory theorem here?
 
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.
 
and I have said that it is not.
and you have given a reason why not
 
yes, and I want to know if my reason of concluding that is correct or not.
 
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.
 
I see. There was a confusion. :(
 
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 $\mathbb{R}$
 
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.
 
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
$1+\frac{x/2}{1+x}\le\sqrt{1+x}\le1+x/2$
 
 
1 hour later…
10:00 AM
@robjohn: I have edited my post to show that o(1/n) has no dependency on x.
 
10:24 AM
'The general theory of relativity asserts that the universe is a smooth Lorentzian 4-manifold'
 
 
1 hour later…
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 $X^I = x_1*x_2$
 
 
2 hours later…
1:18 PM
Is the notation here correct? $$ T^s=e^{s\frac{d}{dx}B(x)} $$
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....
 
 
2 hours later…
3:09 PM
Every set of two odd prime squares is equal to twice a sum of two squares ?
 
@geocalc33 What's B? Just $T^s=e^{s\frac d{dx}}$, no?
 
@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)
 
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?
 
meh, i've seen more miraculous
 
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
 
@AkivaWeinberger $s$ is not a function of $x$ here?
 
@robjohn No, it is a constant.
 
okay
 
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
 
@geocalc33 That operator can be used to motivate the Euler-Maclaurin Sum Formula.
 
akiva: is there a version of the two arcs = three arcs example that can be rendered in a single line of ascii text
 
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
 
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
 
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
 
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
 
Yeah probably doable
 
i guess i had a lot of serial upvoting
 
4:53 PM
@copper.hat The big correction sweep was June 27, so if it was not then, it was something else.
 
5:11 PM
The big sweep netted -10, yesterday was -40.
not a biggie. just odd.
 
that was me deleting a number of copperhat stan accounts
 
Location:Dispersion
sorry
 
gotta get that jump suit
 
if Location: dispersion, \n Mean:Standard deviation, \n Median:Quartile, then \n Mode:What?
 
5:58 PM
mode: most
the most prevalent value in the data set
 
What is the difference between a linear metric space and a metric space?
 
never heard of a linear metric space
 
@schn you mean normed linear space?
 
@Koro possibly. It is used in a certain textbook I am studying.
 
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).
 
6:09 PM
@leslietownes is "topological vector space" a topological space and a vector space?
 
It's probably a "normed vector space" (these have a topology)
because if you can define open balls, you can define open sets
 
@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
 
The usual solving techniques work best with linearity, yeah
 
schn: yes, with the additional property that the vector space operations are continuous.
 
@AkivaWeinberger gotcha
 
6:28 PM
never just put two structures on a space
always assume they're compatible, whatever that means
 
So here's something
 
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
 
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: $\det M=\pm1$
 
@copper.hat Probably Cheerios.
 
(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 $\det M=0$
 
6:38 PM
@TedShifrin i really think breakfasts should not be allowed to create accounts
 
Only meat & potatoes?
 
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
 
DogAteMy: What’s a good way to understand this? For the latter, I’m guessing we get an obvious element of the kernel.
 
@TedShifrin If you write it in the right order, it's a lower triangular matrix with 1s and -1s on the diagonal
 
Ah.
Sorta like Jordan canonical bases …
 
6:42 PM
In "transformation" terms, the image of each vertex is its "parent" edge plus "lower" edges
 
Right.
 
there should only be one function vertices -> edges associating to each vertex an edge incident to it
 
Wait, did I phrase it as a $\Bbb R^V\to\Bbb R^E$ map or the reverse?
I mean, it doesn't matter
 
so you only get one summand in the Leibniz formula
 
(not relevant to the current convo)... like studying random walks on random surfaces instead of studying random walks on the plane
 
6:43 PM
The image of each edge is (plus or minus) its "daughter" vertex plus a "higher" vertex
 
function should read bijection* in my first message
 
Thor: That’s not a function?
 
Parent edge <-> daughter vertex is a bijection
 
well, there could be functions that aren't bijections
 
I'm drawing the tree with the deleted vertex as its root, on the top
 
6:44 PM
or maybe my condition forces them to be bijections, but I don't wanna think about it
 
This is as much detail as the book goes into
(It's more of a survey)
 
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
 
Here's more discussion from the book
 
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
 
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)
 
6:55 PM
wow, I've never seen that formula
it screams exterior algebra, surely that's the right perspective?
 
Oh, interesting
That does make sense
 
@Thorgott yes
 
I feel like this and the Laplace expansion should be the same principle
 
Inner product on $\Lambda^k$ and Hodge star not unrelated …
 
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...
 
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 $q$
 
7:52 PM
@MathematicalEmergency hey
 
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
 
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.
 
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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