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12:00 AM
so yeah, forgetting the twos (pull them out of the sum), we've got z^n + (-z)^n inside the sum. the summand will indeed be 0 if n is odd and 2z^n if n is even. so (reconciling the factors of 2) you'll get sum over n even of z^n, or (changing variables) sum over all n of z^(2n). that seems right to me.
 
@leslietownes. Thanks. Can you please add how we got $2$ above in $z^{2n}$?
 
a sum, over even values of n, of z^n, is a sum of the form 1 + z^2 + z^4 + z^6 + ...
you could regard this as a sum from m = 0 to infinity of z^(2m) because it yields the same result
and then you could replace m with n
 
yawn
 
12:15 AM
@leslietownes. Wow..
@leslietownes. Thank you :)
 
i feel like my life has been a lie now that i'm understanding abstract algebra
 
a lie group?
 
to be determined when i reach chapter 7
 
speaking of Lie groups
Here is a group homomorphism $\exp: \Bbb R\to \text{SO}(2) $ from the additive reals to the circle group
 
the suspense is killing
 
12:32 AM
up to isomorphism, it (SO(2)) is the unique 1 dimensional compact, connected Lie group
 
1:04 AM
so, a graph of a function f(x) looks like a curve on a square
any working definition of those functions that intersect this square on the top side?
 
$\exp: \text{SO}(2)\to H $ Can you do this or not?
 
1:19 AM
anyone here use sympy perchance>
 
@AndrewMicallef Ive used it a bit, but I just google-hacked my way through what I've wanted to do
 
haha, no worries was a stretch the folks in python room seam to have some experience :P
 
that stands to reason.
ha, just got it. killing.
 
1:56 AM
found a better question: say I have a square whose center is at origin, and this square is growing s.t. its diagonal is getting infinitely bigger. any way to know what functions would always intersect its top side?
 
Who knows what that means?
 
well, there are a couple of curves which we can say with absolute certainty that it corresponds to our intuition
x=0 always intersects it on top, y=0 never
$x^2 = y$ doesn't intersect it on top when it is small, but as it gets bigger, it does
$\sqrt{x} = y$ seems to always intersect its right side as it gets infinitely bigger, but not the top side
 
hey @TedShifrin
 
2:12 AM
found a working definition: if a function's derivative is eventually greater than 1, it can be said to intersect the top side
 
you want functions f(x) for which, for all c >= 0, the equation f(x) = c has a solution in [-c, c]? i am assuming an orientation of the square here.
 
@leslietownes that works too, thanks!
 
@shintuku: So you're just asking for curves that intersect $y=c$ for arbitrarily large values of $c$?
Why not just say that?
 
@leslietownes I think if it always intersects the top of the square, we have $f(x)\ge x$, no?
 
Oh, square. So maybe leslie has what you want. I dunno.
Hi @Thor.
 
2:24 AM
Hey, Ted! <-- no ping!
 
@robjohn I think so too
 
Hi, @robjohn. About to go eat dinner.
Thanks for the pingless hello.
 
@TedShifrin bon appetit!
 
You mean bon appétit :P
 
seems right to me.
i'm wondering why we're doing this to squares. what have squares done to you?
 
2:26 AM
@leslietownes I was called a square many times in high school
 
in my era we called them nerds. that's right. we, and them.
 
nerds, geeks, squares,
four eyes
 
i was blessed with several extremely nerdier people in my high school class. all of the focus was on them. i could just sit back and be my cool self.
 
I've come with my daily complex geometry question!
 
here we go.
 
2:30 AM
is the line bundle of holomorphic $(n,0)$-forms on an $n$-manifold always trivial?
 
2:41 AM
Leslie Townes "Bob" Hope (May 29, 1903 – July 27, 2003) was a British-American stand-up comedian, vaudevillian, actor, singer, dancer, athlete and author. With a career that spanned nearly 80 years, Hope appeared in more than 70 short and feature films, with 54 feature films with Hope as star, including a series of seven "Road" musical comedy movies with Bing Crosby as Hope's top-billed partner. In addition to hosting the Academy Awards show 19 times, more than any other host, he appeared in many stage productions and television roles and wrote 14 books. The song "Thanks for the Memory" was his...
 
outed.
the first few road movies are really good.
 
Leslie Townes Bob Hope
 
@Thor Try $n=1$?
 
most of his movies are terrible. but his comedic persona is similar to mine. hence the name
 
@leslietownes You give yourself far too much credit.
 
2:46 AM
which is exactly what a bob hope character would do! you see?
check mate.
 
Theodore Shiffrin? Who's that?
 
i'm not saying i'm as funny as he is. it's just the persona.
talk a whole lot of nonsense, run at the first sight of trouble.
 
oh duh, already fails for CP^1
 
$(\cos(u)(R+r\cos(v)),\sin(u)(R+r\cos(v)),r\sin(v))$ has curvature $\frac1{r^2+rR\sec(v)}$ The computed Mathematica form is pages long, but applying FullSimplify gives the form I quoted above.
 
so, this might seem like an very silly question. but I have two curves, $x^2 - y^2 = 0$, and $x^2 - y = 0$. how can I get the set of their intersections?
 
3:01 AM
subtract the two and what do you get? (any point in common will satisfy their difference)
 
I get $-y(y+1)$. but I'm used to working with functions, so I have no clue how to work this, what happened to the x's?
 
if $x^2=y^2=y$ then there are not many choices for $y$ and hence $x$.
 
@shintuku so that shows that any point in common has $y(y-1)=0$ (notice the $-$)
 
right, from there I know that they intersect at $y = 1$, which corresponds to the graph
 
Then you solve for $x$ (for each of the values of $y$)
 
3:05 AM
what bothers me is, I can't tell what happened to $x$
ahh
wait this was excessively obvious all along. I got scared because it wasn't a function. thanks for your guidance
 
as tuco said, "when your going to shoot, shoot don't talk”
 
@shintuku sometimes looking at it the right way is all that is needed.
 
Let $\chi$ be a nonprincipal character mod k. How can we prove that for all integers a<b, $$\left|\sum_{n=a}^{b}\chi(n)\right|\le\frac{1}{2}\varphi(k)$$
Any hints
 
3:25 AM
what's varphi? is this a standard group thing?
euler phi?
i'm assuming frac12 should be $\frac{1}{2}$
 
I fix
@leslietownes number of residue classes mod $k$ relatively prime to $k$
I assume
 
yeah, that's all i could think of. euler phi.
 
@leslietownes yes
 
let me put my representation theory hat on and see what i can do. meaning, let me look in my notes.
 
There are $\varphi(k)$ residue classes where $\chi(n)\ne0$, $\chi(n)$ sums to $0$ over all residue classes and $|\chi(n)|=1$
I don't think much more is needed
 
3:31 AM
that does seem to be enough.
i'm getting my first jab on monday so if anyone needs a 5g hotspot in southern california let me know.
 
jab?
 
coronavirus vaccine injection.
aka the first half of the bill gates 5g microchip.
 
okay, that is sort of what I thought
my son got his first this afternoon
 
that's great. my wife is three weeks past her second shot. our daughter is too young.
 
My son is 28. Is there a lower limit?
 
3:34 AM
we think she already had it and gave it to us. in october i had two weeks where i had no appetite, couldn't taste anything, and needed to lie in bed all the time. it coincided with an outbreak at her day care.
 
ah. I was very sick in Nov 2019, and was thinking maybe I had it then. I had nothing other than flu symptoms though
 
my 17yo son got his pfizer#1 at hte coliseum in oakland last week.
 
in california i think it's 16 and over in terms of eligibility now, and just a question of where you can get it.
long beach had a very well publicized event where they ran out of vaccine like 20 minutes into it.
 
the coliseum thing was surprisingly well organised.
 
because the appointment i've made is somehow personal to me, i'm assuming that won't happen.
 
3:36 AM
my son got his at CSUN, same place I got both of mine (CalState Northridge)
 
i find myself updating myself much more frequently now
 
my best friend got both of hers at dodger stadium. which was reported in the media as a shit show but apparently surprisingly well put together on the ground.
 
Last time I was in Dodger Stadium was for an Elton John concert in 1975
 
that must have been amazing.
 
i like elton john songs sung by sinead o'connor
 
3:39 AM
my dad saw bob dylan at oakland coliseum once. nice of them to do this vaccine stuff at the sites of legendary shows.
 
@leslietownes It was great. The smell of pot was thick.
 
the grateful dead had one or two good shows at the coliseum in the 80s.
 
my last show there was u2 or u3 whatever they are cslled now
say them before they were big
i think it was in the arcadia in cork
 
that's a good story.
 
didn't really get why they are so big
its very repetitive stuff
 
3:42 AM
i turned down tickets to the zoo tv tour because i didn't want to spend an hour on the road getting there.
they were free. my dad got them from somewhere.
my sister went and had a blast, but, i dunno. i don't like to stay up late.
 
my da turned down free tiks to a police concert across the street from the executive residence we were staying in because he thought we would not be interested
 
hahaha
 
we ere able to sort of hear it from the top of a structure called the wonderful barn which adjoined the exec residence.
it has wikipedia entry now i see.
hah, you can see the house we stayed in!
 
you're wikipedia famous.
indirectly
 
my da used to get up early to chase (his intent was to kill) the bantam roosters that woke him early am
my great granda has a wikipedia page
wow, i've come a long way from my privacy concerns (which is where the name copper hat came from).
see, i don't get why people are preoccupied with this sort of stuff: math.stackexchange.com/q/4106553/27978
 
4:02 AM
So, I was thinking about cutting a torus with a plane that is bitangent, and how that forms circles on the cut cross section of the torus (@TedShifrin you were talking about that before I just can't find it in transcript)
 
Yes, I mentioned it. Slice lots of bagels!
 
I guess there is clearly an analogue for a cylinder, but I'm struggling to get a visual
 
doUGHnuts
 
I'm thinking of cutting at 45 degree
 
I looked and found out I missed Elton John in LA in Sept 2019. :-(
 
4:04 AM
You get ellipses always except the slice parallel to the base
 
but I need to go 45 degrees from both x and y axis (if the length of the tube is in z)
yeah, ok, so what makes a torus special?
 
@TedShifrin and the slice through the center
 
Grrrr @ picky robjohn
 
wait are we talking about donuts or cylinders?
 
@TedShifrin did I spoil something again? Damn
 
4:06 AM
No. Just picky .
 
ok
 
Torus = hollow doughnut
 
doughnuts
 
cylinders sliced anywhere that is not parallel to the base gives an ellipse, except for parrallel to the length which gives a rectangle?
 
@AndrewMicallef two parallel lines
 
4:07 AM
Right ... or parallel lines if infinite
 
yeah okay, an infinitly long cylinder gives parallel lines, with you
 
There are fancy answers for why torus, mostly having to do with the three-dimensional sphere
 
I guess it is probably beyond me at this stage to try and work through a proof. I found a paper on it but it is behind a paywall
 
It's a beautiful exercise in trig/algebra/linear algebra to verify that what I told you (bitangent plane) works.
Some of my diff geo students have done the exercise.
 
@TedShifrin for any ratio of the circles?
 
4:12 AM
Maybe once I'm finished resovling this momentum problem I might try it out, geometry is fun, I like pictures
(also I started getting fed up with the algebra and tried using a computer algebra software to solve it, but now I'm getting frustrated with feeding the equations into the software)
Are Villarceau circles related to the property that some (most?) straight paths around a torus never return to their origin (maybe I'm mixing this up with something else)
 
@robjohn Absitively. The exercise in my diff geo text is for arbitrary $a>b$.
@Andrew: Geodesics on a torus are quite fascinating. Not related, though, so far as I can see.
 
The formula for the curvature at a point on a torus simplified to something extremely simple (which I am sure you know)
 
Yes, I know it well :)
 
The formula first returned by Mma was horrid. FullSimplify made it extremely nice
 
See my diff geo text for a general surface of revolution, if you care.
 
@TedShifrin I'll have to get a copy. Let me contact my black market contact...
 
Nah, this is free and legal, @robjohn.
Yes, @Andrew.
Linked in my MSE profile, among other places, @robjohn. That's where Andrew just found it.
 
Yeah, I just downloaded it
 
It's on my pile of books to read :P
Can I print bits?
 
Of course, @Andrew.
@Andrew: You should have stuck to physics. Hanging around here with your curiosity is a dangerous proposition.
 
4:29 AM
I'm off again, FYI, that problem I have been working on this week has gone from this:
$$
\left<p\right> = \int{\Psi^*\left[-i\hbar \frac{\partial}{\partial x}\right]\Psi}dx
$$
to this
$$
i\hbar
\left(
\int\limits_{-\infty}^{+\infty}{
\frac{-i\hbar}{2m} \dfrac{\partial^2 \Psi^*}{\partial x^2} \frac{\partial \Psi}{\partial x} + \frac{i}{\hbar}V \Psi^* \frac{\partial \Psi}{\partial x}
}\,\mathrm{d}x
+
\int\limits_{-\infty}^{+\infty}{
\Psi^* \left[
\frac{i \hbar}{2m} \frac{\partial^3 \Psi}{\partial x^3} -
@TedShifrin well actually this curiosity is a pretty dangerous thing. Don't get me started on how someone stealing my motorcycle has me trying to figure out quantum mechanics
I think you call this X- Y ing
 
Um, I don't like the looks of those integrals. Did you forget the $d/dt$ at the beginning of the problem?
 
oh yeah, sorry copied the wrong part at the start!
I meant this problem ! $$
\frac{d\left<p\right>}{dt} = \frac{d}{dt}\int{\Psi^*\left[-i\hbar \frac{\partial}{\partial x}\right]\Psi}dx.
$$
 
Oh, I see. So the third derivative shows up from the second term here + Schrödinger.
 
yeah, I could show you my full working if you'd like (though I don't want to bore you)
 
I'm not that interested, but I now do not understand how this integral is the correct thing for momentum.
 
4:34 AM
O as in short for no?
 
NO, as in typo.
 
is there any way in general to know whether a real function intersects the circle $x^2 + y^2 = 1$?
 
The graph of a real function?
 
yeah
 
it's not momentum, it's the time derivative of momentume (which if my physics is on point has something to do with energy)
 
4:34 AM
there's that
 
Oh, so you're doing the expectation of the derivative of the wave function, but I guess I don't see how the $x$-derivative of the wave function should be momentum. We need a physicist.
 
anyway it is suppose to simplify to this: $- \frac{\partial V}{\partial x}$, like I said I need to work through the paonful algebra still :P
 
but I've been trying to resolve the equation of $x^2 + y^2 - 1$ with some basic functions and it is not at all a trivial process
 
@shintuku The only plausible answer is to find some point on the graph inside the circle and some point outside (assuming the function is continuous).
What is the function you're talking about? This will most likely have to be graphical, not algebraic.
 
"I'm not that interested, but I now do not understand how this integral is the correct thing for momentum", as in short for:
"inventive miraculous natural optimistic transforming thrilling hug action thrilling innovative now terrific ecstatic remarkable effective skilled terrific elegant delight beaming upstanding truthful imagine nice okay willing dazzling open natural okay thrilling upstanding novel distinguished exquisite reliable skillful truthful angelic now dazzling heavenly one wow trusting healing imagine secure imagine nutritious truthful encouraging glamorous refined amazing livel
 
4:37 AM
smacks Leaky hard
 
well, I know there are no solutions to $\frac{1}{x}-y = x^2 + y^2 - 1$, but I was just surprised that it was not simpler to resolve
 
whoa, artwank in a maths chat?
 
That is incorrect, @shintuku. My asking about graph completely flew out of your mind.
 
new term for my vocab
 
Oh, you edited.
It's still all wrong.
 
4:39 AM
the graph has no intersection, of course
 
You're asking if $y=1/x$ intersects the circle $x^2+y^2=1$.
Your equation is garbage.
 
sorry
 
If you're at the level of proofs you need to pay careful attention to this and not write garbage.
You can rephrase this in terms of solving an equation if you eliminate $y$. Can you do that correctly?
 
off I trot, cioa all
 
Ciao, @Andrew.
 
4:41 AM
hasta la manana
 
@copper.hat "manana" for "tomorrow" doesn't use the article "la"; "la manana" means morning
 
oh, and then I would only be asking when the $x$'s are the same, right? @TedShifrin
 
and "tomorrow morning" is "manana por la manana"
 
No. You're asking if the $y$'s are the same for any values of $x$.
 
@LeakyNun it is a song.
 
4:45 AM
ah, of course: $\frac{1}{x} = \sqrt{1-x^2}$
 
oh
 
Oy.
 
thanks
 
Better. And of course there's a $\pm$, but we'll look just for $x>0$.
It might make for easier algebra/calculus if you don't do square roots in the first place. What else could you do?
 
$\frac{1}{x}y = x^2-1$
 
4:49 AM
Huh?
We eliminated $y$.
 
oh
 
you could show that the closest the graph of $x \mapsto {1 \over x}$ gets to the origin is $\sqrt{2} > 1$.
 
@TedShifrin that's Golden...
 
@copper.hat never thought of that!
@TedShifrin hm, are you suggesting there's another manipulation that's not square rooting?
 
don't let me distract.
 
4:53 AM
Isn't there a sign issue?
Yes, @copper, that's my preferred solution.
 
Yeah, I spoke too soon
 
You need to be pickier, robjohn! :D
 
5:11 AM
@robjohn $\frac{1+i\sqrt3}2$, the leprechaun golden ratio. ;)
 
ROFL
 
@PM2Ring I imagine so
 
:D But on the topic of the real golden ratio, here's some Fibonacci fun that I wrote in SVG a few years ago. A visual "proof" that 5×13 = 8×8
 
@PM2Ring I have a sliver of doubt...
 
IIRC, I first saw that "dissection" in an April Fools column by Martin Gardner.
 
5:16 AM
@PM2Ring I think that may be where I saw it first, too
I know I saw that article, in any case
The missing square puzzle is an optical illusion used in mathematics classes to help students reason about geometrical figures; or rather to teach them not to reason using figures, but to use only textual descriptions and the axioms of geometry. It depicts two arrangements made of similar shapes in slightly different configurations. Each apparently forms a 13×5 right-angled triangle, but one has a 1×1 hole in it. == Solution == The key to the puzzle is the fact that neither of the 13×5 "triangles" is truly a triangle, nor would either truly be 13x5 if it were, because what appears to be ...
Slightly different
 
Slightly, but very closely related. And connected to the fact that the golden ratio is the most irrational number, in the sense that its continued fraction has the slowest convergence. So there are plenty of rational numbers with small terms (i.e., adjacent Fibonacci numbers) that look pretty similar.
 
@robjohn what you did ?
 
 
2 hours later…
6:59 AM
Is there an example of a single fractal that transitions continuously from fractal dimension 0 to 2 (or 3) along a particular direction?
 
7:18 AM
hi, im trying to show that if I have a system of sets containing $0$ in a vector space $V$, such that the intersection of any two of these sets contains a member of this system, then the system of translates of these sets is a basis for a topology on $V$
I understand that in order to do this I need to show that if $U_1,U_2$ are members of the system of sets containing $0$, then $U_1 + x_1 \cap U_2 + x_2$ can be written as a union of translates of the system of sets containing $0$ but I don't see why this is true
 
7:51 AM
given polynomials $P_1,P_2$, is there a way to find a polynomial $X$ s.t. $P_1X= P_2$?
 
if $P_1 \mid P_2$ then sure, by long division
brave to call your polynomial $X$
 
well, there was $P_3$
 
8:19 AM
this does not seem right to me, if $f$ is a continuous linear functional on a normed linear space $V$, and $|f(x)| < \epsilon$ on some bounded set $A$, then given $||g - f||$ is sufficiently small, $|g(x)| < \epsilon$ on $A$
but apparently its true :(
here $||h|| = sup_{|| x || \leq 1 } |h(x)|$
 
 
2 hours later…
10:07 AM
@Rover $$\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\underbrace{\frac1{\frac{\mathrm{dx}}{\mathrm{d}y}}\frac{\mathrm{d}}{\mathrm{d}y}}_{\frac{\mathrm{d}}{\mathrm{d}x}}\underbrace{\left(\frac1{\frac{\mathrm{dx}}{\mathrm{d}y}}\right)}_{\frac{\mathrm{d}y}{\mathrm{d}x}}=-\frac{\frac{\mathrm{d}^2x}{\mathrm{d}y^2}}{\left(\frac{\mathrm{dx}}{\mathrm{d}y}\right)^3}$$
 
First line of content from Ted's book,
> We say a vector function $f:(a,b) \to \mathbb{R^3}$ is $\mathscr{C}^k$ ...
does $\mathscr{C}$ have a name?
@robjohn I thought you can't treat differentials like fractions?
(maybe I'll just pop off again, I'm clearly still very confused, and it is time for bed anyhow)
 
@AndrewMicallef $\displaystyle\frac{\frac{\mathrm{d}\color{#C00}{f}}{\mathrm{d}x}}{\frac{\mathrm{d}y}{\mathrm{d}x}}=\frac{\mathrm{d}\color{#C00}{f}}{\mathrm{d}y}$ The $\color{#C00}{f}$ here is $\dfrac1{\frac{\mathrm{d}x}{\mathrm{d}y}}$
 
 
1 hour later…
11:23 AM
Not so math related maybe, but is it displayed ON or IN a logarithmic scale?
 
11:40 AM
Hola! I got to prove that the convergence in $L_r$ doesn't imply the convergence in $L_s$ for s > r. I was trying different counterexamples but can't figure it out... Could someone guide me please on how to start with that?
 
12:15 PM
@porridgemathematics this is false, no? what if the system consists of a single ball.
@wklm try some fractional powers of x
 
@Thorgott I was trying sqrt, but it failed miserably :(
 
@Thorgott yes good counterexample, it is definitely false without more assumptions.. i ask because the functional analysis notes i am using for my course justifies that some system of neighbourhoods of $0$ defines a topology on a v.space (so it becomes a TVS) because the intersection of any two neighbourhoods contains a member of the system... which didn't seem like enough for me
im a bit unsure about what this means: Let $L : V \rightarrow V$ be a linear transformation where $V$ is a finite dimensional vector space, and denote by $M_L$ the associated $K[T]$-modulewith underlying abelian group $V$, and multiplication $P(T).v = P(L)v$ here $P(L) = a_nL^n + ... +a_1L + a_0 \mathbf{1}$, then if $M_l \equiv K[T]/(P_1) \oplus ... K[T]/(P_n)$ where $P_1 | ... |P_n$ are monic nonzero, thes $P_i$ are uniquely determined by $L$
is what this really means, $M_L$ as a $K[T]$-module is uniquely determined by $L$, and so its classification as a finitely generated $K[T]$-module is too?
and $M_L$ is uniquely determined by $L$ because for example, $T.v = L(v)$... ? Surely we can have different linear transformations $L,L'$ that induce isomorphic $K[T]$-modules right? What is the importance of saying '$M_L$ is uniquely determined by $L$', and if they are isomorphic $K[T]$-modules then they have the same classification...
which makes me even more confused because in algebra we normally talk about things up to isomorphism, but clearly in this case we can have different linear maps that induce isomorphic $K[T]$-modules and thus have the same associated polynomials $P_i$?
Like, lets say $L$ has matrix $A$ with respect to some fixed basis of $V$, and $W$ is another linear map that has matrix $C = BAB^{-1}$ with respect to the same fixed basis, then a presentation matrix for the $M_L$ and $M_W$ $K[T]$-modules using this basis $\{v_1,...,v_m\}$ as generators and the relations $Tv_i = \sum_{j=1}^m a_{ji}v_j$ and $Tv_i = \sum_{j=1}^m c_{ji}v_j$ are $T Id - A$ and $T Id - C$ respectively, and the latter is $B(T Id - A)B^{-1}$ which is the former
up to row/column operations
therefore both have the same smith normal form
so in general shouldn't it be two matrices are similar iff they have the same smith normal form, and we can't really talk about linear maps?
 
12:45 PM
@wklm I think what I suggested doesn't suffice, sorry. Instead, you can try playing around with functions whose graphs look like increasingly taller, but thinner triangles.
@porridgemathematics I do believe there's some way of specifying a topology on a vector space by specifying a neighborhood base of $0$, but you need more conditions.
I don't know this stuff, but it feels like this should be in standard textbooks
@porridgemathematics no clue what the "uniquely" is trying to communicate, but a K[T]-module structure on the K-vector space V is uniquely determined by a linear map L:V->V
not up to isomorphism, but up to literal equality
 
1:03 PM
@Thorgott yeah you're right, the additional conditions are a bit of a mouthful though, you need stuff like for ever member of the system $U$, there is some member $V$ so that $V+V \subset U$ and some other stuff
@Thorgott okay, literal equality, yeah I can see that much is true, just thought in algebra basically everything is up to iso so was confused
 
just read about Lagrange's four-square theorem, I have a related but different question. Given a number $n$, it can be written as a sum of four squares. e.g $52 = 7^2 +1+1+1$, but this is not the representation with the minimum number of square terms, minimum would be $2$ as $52 =6^2+ 4^2$. So given $n$, how can i find the minimum number of squares needed to write $n$ as a sum of them.
 
this is answered by the two-squares theorem and the three-squares theorem
no, I'm not joking
 
ok. i'll look that up. thanks
got it. thanks.
 
1:20 PM
@Thorgott the Euler characteristic of any connected graph with no intersections on a surface should match with the Euler characteristic of the surface, right?
I think "face" has to be elaborated on here.
 
I feel like there's a definitional issue
you can draw a triangle on any surface
 
Because if I have a torus and I draw a point on it, I can make a circle by connecting an edge in one of two ways (corresponding to the generators of the homotopy group). One encloses a disk, and the other does not. Both have an edge and a vertex (which cancel out in the graph Euler characteristic).
I think your suggested problem is the same as mine.
So I wonder if this can be easily fixed, but in a less constraining way than "triangulate the surface" or "decompose it as a cellular complex".
 
I feel like faces should be number of components of the complement
but then there's no way to make your example graph work out
 
Yeah, I think that works as long as the component is homeomorphic to R^2.
But in both these cases, it is not.
 
okay, I think I'm getting somewhere, another noob question, I got a function: f(x) = sqrt(n) * 1{[0, 1/n]}(x) for n in natural numbers. How do I evaluate the integral of that?
 
1:36 PM
the point should be Mayer-Vietoris, maybe
 
@wklm what definition of an integral are you using?
 
lebesgue measure
 
take a sufficiently small tubular neighborhood of your graph, then euler char of the surface is euler char of the complement + euler char of the graph - euler char of intersection of complement with the tubular neighborhood
where here euler char of the graph is in the algebraic topology sense
so rearranging this essentially tells us how we have to count faces
I don't like this, but it seems forced
 
1:53 PM
Seems quite forced.
@wklm how far do you get when trying to plug it into your definition for the Lebesgue integral?
 
@anakhro I'm getting lost at the multiplication by the indicator function
so right at the beginning in fact
 
 
1 hour later…
3:28 PM
the integral of c 1_E is c m(E) (c being a constant, E being a measurable set contained in your measure space, 1_E being an indicator, and m( ) being the measure)
when m is lebesgue and E is an interval it's the same as the Riemann integral, just an area of a rectangle if c > 0
 
@wklm Sorry, I had a meeting. It might help to try to draw the function.
 
n is a red herring
 
Leslie's synesthesia is a little off.
 
blue herring?
wet herring?
it's a herring, i'm sure of that
 
What have you been up to lately, Leslie?
 
3:43 PM
Can someone recommend me a textbook which properly defines integration on manifolds/boundaries? For example in this question here:
https://math.stackexchange.com/questions/619554/bounding-an-integral-on-boundary-of-lipschitz-domain?rq=1
the local surface measure of a Lipschitz boundary (which is what I am interested in) is mentioned and I was wondering about a textbook to cite for integrals along the boundary of Lipschitz domains.
 
a lot of fairly annoying work. outside of that, visiting a duck pond with my daughter. there are baby mallards this time of year, very exciting for her.
 
I think snake Lemma is still true for algebras over a cring isn’t it?
 
@leslietownes now I want to go check out the local pond for baby ducks
 
my daughter's just old enough to understand that the smaller ducks are 'children,' it's a whole lot of fun for her. there's also a goose that she likes to visit.
she's obsessed with birds generally. we have a "bird-a-day" desk calendar from something like 2012 that survived a series of cross-country moves. she makes me go through it with her a few times a week. 365 birds is a lot of birds.
but if anybody knows a better way of killing an hour with a two-and-a-half year old i'd love to hear it.
felix that is a good question. i thought spivak's series did an OK job of getting manifolds with boundary into the integration mix, but i might be wrong about that. i don't remember any lipschitz condition.
maybe his boundaries were assumed to be better behaved than that.
 
Heh, 365 birds is indeed a lot of birds.
You could teach her integration on manifolds in an hour, I think. But only compact without boundary.
 
3:53 PM
she knows dozens of birds. the most distinctive ones (flamingoes, macaw parrots, snowy owls), and then the ones she sees every day (crows, robins, doves, phoebes, house finches, house sparrows, scrub jays, juncos, and of course her ducks and geese)
we're going to build from the ground up. we start measure theory on monday.
finitely additive, because countable additivity is a crutch i don't think she will need. she can get there by taking limits if she does need it.
 
the snake lemma works in every abelian category
 
4:09 PM
@leslietownes maybe you can try this one time with her: math.brown.edu/reschwar/farm.pdf
The pages with the little corner markers I think were for "advanced kids", which you might want to skip with younger ones.
If I recall correctly...
 
ha, this is delirious. i love the art style.
my daughter's outside on the porch right now with my wife. she's reciting how everyone in the world has her first name. mom has it, dad has it, the cat (weirdly identified by name, olivia) has it. we're all her now.
 
is Leslie's daughter.
What a turn of events.
 
she brought her mother's parents into it too. they're both her.
time has become a loop
 
This is what happens when you teach your kid equivalence relations. :((((
 
4:35 PM
she understands subtraction. she always wants five crackers. if you give her fewer than five, she asks for the correct amount to make up the difference.
 
If you gave her $k$ crackers, just tell her you thought she meant $5 \pmod{5-k}$ for $1\leq k < 5$ (if you give her $0$ crackers, then you are just a bad person).
 
her number sense does seem to vanish shortly after 5.
modular arithmetic would do her head in.
 
Maybe the infinite cow is too long for her, then. :(
 
it is fun to see them learn
 
about a month ago she made the mistake of letting us know that she could turn crying on and off. she said "i'm going to cry like this" [and then a very convincing display of crying] and then she immediately shut it off.
and said "like that."
i can't trust her anymore.
 
4:47 PM
@leslietownes my sister also does that..
 
it was bath time the other day. we had warned her about it the day before, but when she found out it was bath time she lost it. then i asked if she was crying for real and she immediately stopped and said "no. let's go to the bath."
she'll be a good attorney.
 
i am a hard ass when it comes to tears.
i may have inadvertently removed a tool from my daughter's arsenal.
 
then she began crying for real when her mother wanted to take her out of the bath.
children are little psychopaths.
 
you need to keep the derivative of your reaction to a change in tears close to zero
 
yeah, it can't become a tactic
 
4:51 PM
@robjohn
 
there is no bat signal here
 
@robjohn I still don't get why you do so ..
 
sometimes if i talk nonsense about geometry it seems to summon ted. but yeah, it's not the bat signal.
 
@TedShifrin just appeared...
 
@robjohn Independent variable x is changed to y means what exactly?
 
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