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00:11
What is the relation btw the dot vector and the angle btw the vectors?
@AbhishekBhatia $a\cdot b=|a||b|\cos \phi$ where $\phi$ is the angle between the vectors $a$ and $b$.
What does a higher dot product imply-> is the angle lesser?
@Balarka: Did you eat the eggs?
@MikeMiller I found an NSA question I can answer in detail ^^
The response was quite winded though; I wasn't sure how to format a long answer
Absolutely typical of you, @Balarka.
@GBeau: Grats!
00:34
@MikeMiller I might have to edit it though; it's a little hard to judge how to write the answer when I'm assuming most of the readers aren't familiar with that particular framework
I get that
stuck in this damn airplane. my connecting flight leaves in 40 minutes, but is in a different terminal, and doors close 15 before the flight leaves
I wish they would let me out :(
@MikeMiller I'm flying tomorrow D;
In 2d space, is this always true: dot product is higher when angle between vectors is small?
I hope you don't miss your flight bud
@AbhishekBhatia: Think about the formula $|a \cdot b| = \|a \| \|b\| \cos \theta$, $\theta$ the angle between them
true till 180 and then opposite?
00:41
me too!
I am confirming since my prof. said the above statement.
When we have that $\sigma_u\times \sigma_v=J \ (\tilde{\sigma}_{\tilde{u}}\times \tilde{\sigma}_{\tilde{v}})$, where the Jacobian $J$ is nonzero, do we have that the tangent plane spanned by $\sigma_u, \sigma_v$ is equal to the tangent plane spanned by $\tilde{\sigma}_u, \tilde{\sigma}_v$ ?
@MikeMiller can you confirm please
@MikeMiller I was proving the equivalence of $\lim\limits_{x\rightarrow\infty}f(x)=a$ and $f(x)\simeq a$ for all unlimited $x$ and $a$ is standard for somebody who asked a question about the relation in NSA between "evaluating at infinity" and taking the limits at infinity (where the latter is trivially equivalent to the standard part being equal to the limit)
@MikeMiller do you have an idea abou my question above?
00:51
Made it
Hey there
hi SemiC
did you make your flight? well, your connection
01:06
It's about to take off and yup
so bye
cool. have a good flight
 
1 hour later…
02:11
I am slightly nervous about next semester.
Despite doing well, last one was awfully demanding, and this one will be even more.
Yes, but you are now better prepared for the demand.
I hope so. I've been reading my books from time to time to ease the jitters.
Idling in this chat also helps. I'm a bit weird.
02:30
Asking again(this might be naive though): In 2d space, is this always true: dot product is higher when angle between vectors is small?
It follows from the definition: $a\cdot b=\left\|a\right\|\left\|b\right\|\cos\theta$.
$\cos\theta$ is "highest" when $\theta=2\pi n$ for any integer $n$.
02:49
In the context of vector spaces, I often see the expression $\mathbb R^{n+1}$ (and similar ones). What is this exactly?
Never mind. I figured it out.
@JosuéMolina Can you explain more?
Is it decreasing function from 0 to 180 and then an increasing one form 180 to 360?
@AbhishekBhatia That is correct.
So dot product has inverse relationship with angle from 0 to 180. From 180 to 360 direct?
@AbhishekBhatia Correct. You can see that $\cos\theta=\cos(2\pi-\theta)$.
03:18
God damn
someone went on my account and postem a differential equation question
what can I do to get rid of it?
my hat is gone :(
I don't know anything about differential equations
and that question cost me all my reputation :(
no
dislikes
the guy just went on
posted with terrible formatting
and didn't explain what he wanted well
I flagged it for removal
Hah, I saw the edit for that post earlier.
And I downvoted it. Whoopsy.
03:25
@Michael Why should we believe you? :-P
It's not the end of the world. You lost a couple rep and the formatting looks fine now.
@JosuéMolina Cuz I have no idea what a differential equation is....
Like I have no clue what that person wrote
Idc about the rep, but I don't want a useless question that I don't understand to be out there
Well, to be fair, we don't know what you know. Also your bio indicates that you're studying integral calculus and soon will be doing differential equations.
I am moving to multivariable first
BUt
I am interested
First, what is a differential equation?
google? :)
03:32
It's just an equation that relates a function with derivatives of itself.
Ok so the equation I posted
A simple example is $dy/dx = y(x)$. Know any solutions to that?
$\int y(x)$
Honestly, no clue
Do you know any functions that if you take the derivative you get the same function back?
03:35
e^x
Yep, e^x is a particular solution.
oooh
ok that is cool
Is there anything else?
You'll find out :). Pretty sure there's questions on stackexchange that ask about whether $ae^x$ for some constant $a$ is the only such function.
Thank you very much sir/m'am! However, I'll stick with my very very very introductory multi-variable for now :)
I'm male. No problem.
03:57
DEs felt too algorithmic for my taste.
At least the courses I took as an undergrad.
A lot of engineers in those DE courses
I knew I was going to have a similar PDEs course when I read in the book something along the lines of: "We're going to solve a PDE $u(x,y)$ by separating variables. So let $u(x,y)=\phi(x)\psi(y)$. We give no justification for this except that it works."
My jaw dropped.
04:43
eh. when i see "we give no justification for this save that it works" i read it as "it can be justified rigorously, but for our purposes that would serve only to confuse. so we only work formally."
05:22
is every linearly ordered topological space normal?
I cant recall, and my brain is too tired to figure it out
ok I see that it is true, assuming axiom of choice
bye
 
2 hours later…
06:59
@MikeMiller suppose u were stranded on a island with no internet. What 5 books do u take?
 
1 hour later…
08:13
Hi there!
08:52
Hi everyone. Does anybody knows whether there is difference between potential and conservative field in terms of mechanics?
Huy
Huy
09:08
@mikeonly: I don't think so. A conservative field is a vector field that can be written as the gradient of a scalar field, i.e.
$$\mathbf{v} = \nabla \varphi.$$
the scalar field $\varphi$ is usually called the (scalar) potential for $\mathbf{v}$. i've often heard people say "potential field" as a synonym for conservative field.
@Huy There are equivalent in most cases, I think, though there might be a reason why there are two words.
@Huy I was taught that potential field is that has vanishing curl.
Huy
Huy
@mikeonly: some people define a conservative field as a field such that any path integral only depends on its endpoints
you can then prove that "path integral only depends on its endpoints" is in fact equivalent to "there exists a scalar field $\varphi$ such that $\mathbf{v} = \nabla \varphi$"
@mikeonly this is WRONG
@mikeonly: every potential field has vanishing curl, but not every field with vanishing curl is a potential field
@mikeonly: see this question I asked during my freshmen studies: math.stackexchange.com/questions/39231/…
@Huy Thanks, I'll read it. Although now I cannot feel why that statement is wrong.
Huy
Huy
@mikeonly: the question contains the classical counterexample
Also there is this article on MathWorld, it has the condition for a vanishing curl. I think it may explain why we have been given two terms.
Huy
Huy
09:18
@mikeonly: yes, note that if the domain is simply-connected, then vanishing curl is equivalent to the other conditions
this is also known as Poincaré's lemma
@Huy That means, that if the domain is simply-connected, what we call potential field is also conservative, doesn't it?
Huy
Huy
@mikeonly: I still don't know what *you* call a potential field and what you call a conservative field. from what I know, they are exactly the same, although *some* people define:

conservative: path integrals only depend on endpoints
potential: there exist a scalar field $\varphi$ such that $\mathbf{v} = \nabla \varphi$
these two things are equivalent.
now if the domain is simply-connected, then the condition "curl is zero" is also equivalent
@Huy I don't have any strict definition of a conservative field. In mechanics, though, it was referred to a field, where the sum of potential and kinetic energy conserves. On the other hand, potential field is clearly defined for us as that having a vanishing curl.
Huy
Huy
@mikeonly: then you are either implicitly always assuming simply-connected domains or doing something very wrong
@Huy I hope the former is the case. It was only mechanics classes where I heard about these terms, as I haven't had vector field theory yet.
@Huy Can ‘simply-connectedness’ be related to time dependency of a vector field?
Huy
Huy
09:27
@mikeonly: maybe bring it up and ask in class?
no, not really
simply-connectedness is a property of the domain only
@Huy I have recently changed university and will no longer have those classes. Teachers in my former university are unlikely able to provide a definite answer though.
Huy
Huy
@mikeonly: what are you studying? physics? some engineering?
@Huy Do we expand the domain with a time-coordinate?
Huy
Huy
@mikeonly: usually, no. just the position domain.
@Huy Starting a neuroscience program after two years of applied physics.
Huy
Huy
09:31
@mikeonly: I think it's likely they just said it's equivalent (without the simply-connected restriction) because you usually deal with simply-connected domains anyways in mechanics
@Huy I am sure they have not mentioned it, also I am satisfied if that is the difference (simple-connected restriction).
For example, we have never referred to electrostatic field as to conservative one. For us it was potential field, with a vanishing curl, as due to Maxwell's law.
09:55
@Huy Also thanks for references to your question, it clarifies a lot!
10:41
My mustache is gone!
Roh
Roh
Hi guys
-3
Q: How these fractions becomes this?

RohToday I was trying to solve an integral for a Fourier series. I looked at the solution and this was the solution: \begin{align*} C_n &= \frac{1}{2\pi} \int_0^{2\pi} x^2 e^{-inx} \,\text{d}x \\ \xrightarrow{\text{integration by parts}} C_n &= \frac{1}{2\pi} \left[ - \frac{x^2 e^{-inx}}{in}...

Roh
Roh
10:58
Is there anyone here who can help me to solve my problem?
 
2 hours later…
13:03
@robjohn lol i havent enjoyed my hats even enough !
13:21
Hi!!!
I want to make the graph of the following relations:
$6x+3y \leq 180 \\ 4x+5y \leq 200 \\ 5x+2y \geq 100 \\ 2x+4y \geq 100 \\ x,y \geq 0$
What points could I pick so that I can do it by hand? Aren't for example the points (20,20)(20,20) and (10,40) for the first line too high?
Hi @Evgeny
Are you familiar with the simplex method?
For which $\theta$ is $(\cos\theta\cos\phi,\cos\theta\sin\phi,\sin\theta)$ a parametrization of a unit sphere?
For every $\theta \in [0,2\pi]$ ?
13:46
@MikeMiller Here's a proof that closed convex subset has a vector of minimal norm implies the problem. $V$ be Hilbert, $W$ subspace and $W^\perp$ orthocomplement. $v \in V$ be a vector not in $W$. $w$ be the vector closest to $v$, i.e., $||v - w||$ is minimal (this is possible by considering the set of all vectors of the form $v - p$, $p \in W$, which is clearly closed and convex).
Then, note that $||v - w||^2 < || v - w'||^2$ where $w' \in W$ is distinct from $w$. Pick $w' = w - au$ where $u \in W$ is some other vector and $a$ is arbitrary const. $||v - w||^2 < ||v - w'||^2 = ||(v - w) + au||^2 = \langle v - w + au, v - w + au\rangle = ||v - w||^2 + 2a\langle u, v - w \rangle + a^2||u||^2$. Thus, $a(2\langle u, v - w\rangle + a||u||^2) > 0$.
I can break this by picking negative $a$ unless $\langle u, v - w\rangle = 0$, which proves $v - w$ is orthogonal to any $u \in W$, as desired.
I figured out the convexity condition from looking at a few examples. It is easy to come up with $\ell^2$ examples of closed subsets of $\Bbb R^\infty$ with no element of minimal norm, I think. Take the subset of all sequences such that $n$-th slot is $1/n$ and the rest of them are $0$.
Also, it's intuitively visualizable why convexity should make it work.
I haven't been able to come up with a proof yet.
And no, I did not eat 51 eggs.
Hi @TedShifrin
14:07
Hi @Balarka ... 51 eggs
Mike gave me another homework. 'Twas to eat 51 eggs in 2 hours.
Why is that example closed?
@TedShifrin Well, distance between any two sequences is $\sqrt{1/n^2 + 1/(n+1)^2}$, no? This is $\sqrt{n^2 + (n+1)^2}/n(n+1)$. This is greater than $\sqrt{2}/(n+1)$. So distance between two sequences grows as $n$ grows. A sequence must be eventually constant if it converges.
Huy
Huy
@TedShifrin: can you check a short argument of mine?
Limits points are all inside.
14:12
Huh? @Balarka
Why isn't $0$ a limit point?
What's that, Huy?
Huy
Huy
@TedShifrin: just a proof for a question on MSE but I'm not too confident yet whether all my manipulations are rigorous which is why I haven't posted it
On what?
Huy
Huy
a Haar measure on the semi-direct product of two metric groups with Haar measures
@TedShifrin Oops, that was nonsense. $\sqrt{2}/(n+1)$ gets smaller as $n$ increases, duh.
$(0, 0, 0, \cdots)$ is a limit point.
Yikes, @Huy .... Never thought about that.
@Balarka: I think you're making the problem too hard.
14:18
Ok, what about $0$ everywhere except the $n$-slot, where it is $1 + 1/n$?
Why bother with the 1/n's?
Ah, fair enough. $0$ everywhere except the $n$ slot where it is $1$ :)
Still not sure why you're doing this ...
But that's closed, right? I am not sure what could be easier than that.
@Huy: Does your argument give the right answer for $E(2)$?
Yeah, Balarka, but I don't see why you're thinking about this?
We're talking about a closed linear subspace?
Huy
Huy
14:24
what's $E(2)$? also, the question already gives an explicit form for the Haar measure, so it's just a confirmation that the expression is indeed left-invariant
@Agawa001 It's always that way when the hats go away.
@TedShifrin Ah, ok. Well, I want to prove that if $W$ is Hilbert, $V$ a closed subspace, then $W = V \oplus V^\perp$. I figured that to do this I need to prove that convex closed subspaces of $W$ has a vector of minimal norm.
closed subspace is linear, not a topological subspace
This problem originated from your Fourier series lectures. You said Fourier series is orthogonal projection of $C^0([-\pi, \pi])$ onto the subspace $V_k$ spanned by $1/\sqrt{2}, \sin(n \pi x), \cos(n \pi x)$ where $n = 1, \cdots, k$. But I didn't know when infinite dimensional things have orthogonal decomp.
(so why does orthoprojection even make sense? that was the question)
but that was a finite-dimensional subspace, so you write down the projection explicitly with inner products and basis elements.
14:28
@TedShifrin Yeah, but $V$ is not the issue. The set $\{w - v: v \in V\}$ is, where $w$ is an arbitrary vector on $W$ not in $V$.
Ok, so consider the inf of the lengths of those.
@TedShifrin Er, I see you have defined projection differently in your book, but I was working with the Artin definition. Artin defined it as the projection onto 1st coordinate $V = W \oplus W^\perp \to W$ and his proof of orthodecomp works only for finite dim $V$.
Hi @Akiva.
@TedShifrin Right.
Hi Homework.
Huy
Huy
14:32
@TedShifrin: if you have a minute to check: here is the argument. the upper-half is from the original post on MSE, just definitions which you probably already know and the claim.
Semidirect product is only topologically locally a product, @Huy. It's a fiber bundle.
A nontrivial one, in general, I believe.
Hi, I am trying to prove Eisenstein's criterion, I reduce the equation $P(X)=Q(X)R(X)$ to $\overline{a_n}X^n=(\overline{b_qX^q}+\cdots+(\overline{b_0})(\overline{c_r}X^r+‌​\cdots+\overline{c_0}).$
Huy
Huy
@TedShifrin: which part of the argument does that contradict?
cool, didn't know semidirect product of topological groups are fiber bundles.
I don't have time to read it now, but you started by saying it was topologically a product.
14:36
intuitively quite satisfying, because that's how i think of semidirect products.
Huy
Huy
@TedShifrin: that was the definition of the semidirect product given in the question on MSE, but nevermind then if you don't have time
The book I have said to use unicity of the decomposition in irreducible.
The representation gives a twist to the product structure, so I don't believe what's there.
ah, so $G \to Aut(H)$ is something like a transition function?
You reduce mod $p$, right, @JeSuis? Yes, you want to use unique factorization.
Well, there's no open cover, @Balarka, so I dunno.
14:40
morning chat
hm ok. thanks though. i won't think about it anymore because don't want to get distracted from what i'm doing again :P
Hi @Semiclassic
Hello
yep @TedShifrin But I don't see how... I just want to say as $a_n=b_qc_r$ mod $p$ because $p$ doesn't divides $a_n$, so all the term are $0$.
modulo p
hi @Semiclassical
14:42
oh, that reminds me
All terms except the leading term, @JeSuis
have you made sure area is const?
14:44
the widest one is $f(x) = c e^{-1/x(1-x)}$ with the constant chosen to make it have unit area
and the rest are rescalings $n f(x/n)$ confined to the interval $[0,1/n]$
yes, Semiclassic, that rescaling is the standard mollifier trick.
@TedShifrin If I take Z/4Z $(X^2+\overline{2})^2=X^4$, then (monôme in french) $X^4$ is not necessarly the product of two (monômes). I understand than I need that $A$ is an integer domain (intègre in french). But Why the unicity ?
i figured. reminded me of nascent delta functions
14:46
integral domain ^^
though usually with nascent delta functions i'm okay with the convergence being pointwise except at a single point
you need $\Bbb Z/p\Bbb Z$, p prime, @JeSuis.
then you have unique factorization.
you mentioned, though, that there's smarter ways of doing bump functions in the context of manifolds
Ok, I need to get going now.
@TedShifrin ok but why ? I don't understand why we need unicity ( Z/pZ, or A/(p), if $A$ is an UFD and $p$ irreducible)
14:49
Morning @Ted.
hi @mike. how was your flight?
@Ted @Huy It is not correct that semidirect products are nontrivial fiber bundles. Topologically they are products.
An extension of groups is a G-bundle. But when you have a semidirect product, you have a section, which trivializes the bundle.
15:05
@MikeMiller Wait a second. Let $f_n$ be the functions I and @Semiclassical worked out yesterday(graph of $f_n$ is a triangle placed at $[0, 1/n]$ of height $n$ and is $0$ a $[1/n, 1]$). $f_n$ converges pointwise to the $0$ function. But the maximums of $f_n$ are not tending to $0$, right? This contradicts your statement here.
In fact $f_n$ does not converge to $0$ in $L^2$ norm.
Ell
Ell
Hi folks
*at $[1/n, 1]$.
@Balarka: Ted called me out on that, I believe. I was thinking of Dini's theorem. You need a monotonicity assumption.
Ah, ok.
I never think about pointwise convergent stuff so I got all those facts jumbled. Sorry.
15:08
No problem. I only just figured out it contradicts that.
So pointwise convergence does not imply $L^2$ convergence.
See Dini's theorem for the correct statement.
Still not clear why the sign function example proves $C^0([-1, 1])$ with $L^2$ norm is not complete :(
Ok, thanks, looking it up.
i think you'd want it on $[-1,1]$ instead, if only because the sign function on $(0,1]$ is boring :P
Poittwise clnvergencd is not a natural convergence method for functions. It's natural to use for functionals (elements of dual spaces), but let's not talk about that
@Balarka: No more hints.
what i find myself surprised by in that triangle example is that i'd expect it to converge to a delta-function
15:10
Buddy
don't pull this delta function stuff on me.
pffft
but usually when i think about sequences of nascent delta functions i have in mind things which would converge pointwise except at one point
woops
@Semiclassical but the $f_n$'s are all $0$ at $0$.
so you're vertex is squished between $0$ and some small number which gets closer and closer to $0$ as $n$ grows.
this is why it vanishes
my prev message was false
i think it's just me being misled by the usual $\delta(0)=\infty$ handwaving one sees in non-rigorous treatments
wait no it's true
15:13
what was the statement?
Ah, yes, I can prove Dini's theorem.
Ell
Ell
Can someone guide me through getting eigenvectors from a characteristic polynomial?
I can sometimes do it out intuition, but I don't have a method that works every time
for example from this matrix:
$$\begin{bmatrix}4&1&1\\2&5&4\\-1&-1&0\end{bmatrix}$$
@SemiC: It does converge to $\delta$ in the space of distributions (which is where $\delta$ lives ya punk)
oh, i wasn't disagreeing
Ell
Ell
I got the following characteristic equations:
$$3x+y+2=0 \\
2x+4y+4z=0 \\
-x-y-z=0$$
15:16
what caught me off guard was that those functions converge pointwise to zero everywhere
pointwise convergence is very, very unnatural
i'm beginning to see that
any nice properties are by dumb luck instead of by design
unnatural in the sense that there're many pathologies?
though in some respect i'm being a bit silly. i can take any of the usual examples of nascent delta functions and modify them so as to all vanish at $x=0$
15:20
what is a nascent delta function
hi @iwriteonbananas
@Balarka: in the sense that there's no reason to care about this kind of convergence other than "well, it seems nice"
Did you eat the eggs Balarka
Hey @BalarkaSen @MikeMiller
the examples above are all of that type, actually (that's what Mike was making sure i understood)
@MikeMiller Can you give me some example of properties of say $L^2$ convergence which should be a good reason for one to care about it other than being nice? I'm curious, as I know absolutely nothing about these stuff.
No, I did not eat the eggs. I pinged you with a message saying that I didn't.
integral against an appropriate function is continuous in the $L^2$ norm. so $L^2$ convergence implies the integrals converge
15:23
for nascent delta functions: en.wikipedia.org/wiki/…
the link is to the subsection on nascent deltas, not the entire page as it was showing
in particular, the Fourier transform of $L^2$ functions (whatever that means) is $L^2$ continuous; indeed it's an isometry $L^2 \to L^2$
@MikeMiler wow, cool.
How can we check if there are umbilics on a surface?
for a cute physical interpretation: in quantum mechanics, position and momentum space representations are on equal footing
with position space being the first L^2 and momentum space the other
I agree $L^2$ norm is a good way to make the integrals converge. But I think someone told me that one can fix this in the pointwise convergence by modifying the notion of integrals (Lebsegue? I forget).
15:26
don't know what that means. All integrals are Lebesgue
what about Riemann-Stieltjes integrals? :p
those are silly and I will not allow them
ok, I don't know what it means either. I don't know what Lebesgue means anyway.
i don't actually remember properly myself
15:28
one day you'll have to know that and what Sobolev spaces are and Fredholm inequalities etc are, if you want to do Hodge theory :-)
i utterly forget what a sobolev space is. only thing i think i remember is that you can do integration by parts without worrying about the boundary term in them
which happens a lot in certain physics calculations
it's nice when you can act as though $u\,dv=-v\,du$ in your integrals :P
which is something i usually associate with quantum mechanics on the real line, come to think of it
In case there are any algebrafriends here: math.stackexchange.com/questions/1600848/…
@Krijn How's cohomology?
@BalarkaSen Being delayed
15:33
@AndrewT: Come to the dark side...
Had my first exam today (on elliptic curves) and another one on Algebraic Number Theory on thursday, but I'm gonna start (seriously for this time) on cohomology tomorrow.
@MikeMiller: We were planning on flooding the chat with topology. Now it's flooded with analysis.
@Krijn Nice. Hope you did well in the exam.
we could talk about riemann surfaces and thereby get complex analysis + algebraic topology :P
unfortunately i don't know much about riemann surfaces :(
15:52
Not sure what the algebraic topology of Riemann surfaces would be
i have in mind their (co)homology
e.g. a genus 1 riemann surface has first homology Z^2 and first cohomology R^2
not that strange, but useful for some things re: elliptic integrals
@MikeMiller Ah. Duh.
@MikeMiller maybe Hodge-Riemann bilinear relations ....
You back in LA?
16:11
Got back in lSt nigjt. My laptop no longer works. Sigh.
What did you do to it?
Who's to say? Long trip, something probably got jostled.
Screen is no longer backlit, AKA is unusable.
Will take it open at some point and see if there's something I can easily reconnect.
If we want to check how many umbilics there are on the surface, do we have to look how many different principal curvatures there are? @TedShifrin
We're having some good seminars this week; too bad you're not here to watch.
Given the title of the first talk I also might have no choice but to go to the algebra seminar this quarter.
Do people find their eyes ruined by too big computerscreens?
16:18
No
I've gone from 0.0 to -1.5 in two and a half years using a 15.6".
I would guess a change like that was not induced by your screen
(Actually not got my eyes tested, just tried different glasses with 0.25-increments.)
Someone asked a clarifying question to my question. There is hope!
Ah, just saw he was a PhD student at UCLA, was about to ask.
Thanks a bunch!
16:21
Balmer student, good friend
Jason Starr is a good Joe Harris alg geo student
and also Author of the noir crime thrillers Cold Caller, Nothing Personal, Fake I.D., Hard Feelings, Tough Luck, Twisted City, Bust, and Lights Out
@MaryStar: You have to see where $k_1=k_2$.
Probably different :)
Can you have a random variable that is uniformly distributed, say on (0, 1) that also has non-zero autocorrelation?
I went back to read some Donaldson and he started doing deformations of normal crossings singularities... I don't know that I have a choice but to learn some algebraic geometry now
Luckily I was planning on it
16:25
Btw, just when I was slightly motivated to apply for grad programs in the US I found I was already too late. Msc in Norway it is, which is probably a good thing.
@AndrewThompson: Oh, yes, way too late. I thought you were thinking about this particularly early or something.
Yeah, here one applies in April.
MYBAYBE YOU'LL JOIN US AFTER YOUR msC...
whoa caps
woo, wrote my lesson for today
now to write a talk for tomorrow and a talk for thursday...
Yes, as long as I can get out of this B-runt I've been in for the last exams. Even converted to US-grades they do not look to pretty, for someone with msc UCLA demands 3.6/4.0. A Norwegian B is a 3.66/4.0 in the US, and I suppose successful applicants are quite a lot above the minimum.
@TedShifrin I want to show that, if p, q and r are distinct positive numbers, there are exactly four umbilics on the ellipsoid $\frac{x^2}{p^2}+\frac{y^2}{q^2}+\frac{z^2}{r^2}=1$.

To show that I considered the parametrization $\sigma (u,v)=(p\cos u\sin v, q\sin u\sin v, r\cos v)$.

Then the principal curvatures are the roots of
$\begin{vmatrix}
L-\kappa E & M-\kappa F\\
M-\kappa F & N-\kappa G
\end{vmatrix}=0$.

Am I on the right way?
16:32
Couldn't say.
I just imagined Bernie Sanders being a college professor grading on a curve.
The problem these days is that there isn't massive grade inequality
Are grades as heavily inflated as people claim?
@MaryStar: Yes, I think so. When does that quadratic have a double root?
Classes started already, @MikeM? What lesson?
225b
@AndrewThompson: A lot more people should fail than do.
16:42
@TedShifrin How can we see it? I got stuck right now...
Our uni recently hired as an associate professor (in Norway := tenure) who previously worked in the US. She was far more reluctant to fail people in the US as their education is, well, less than cheap. I argued that one should actually study if one pays a crapload
I don't know. Compute E,F,G,L,M,N, of course, and simplify carefully.
I failed plenty, @AndrewT.
@TedShifrin Ok... I will do now the computations.
But I'm an old, nasty person, @AndrewT :)
Haha. In general few people fail in Norway in oral exams, mostly because oral exams are in higher division courses and students actually study for those. There have, however, been examples of students taking a course named 'Manifolds' not knowing what a manifold is.
(Not sure if its true if I keep the plural there.)
16:50
I complained loudly that my colleagues gave passing grades in calculus and linear algebra to students who knew basically none. Part of why I retired.
On written exams, what is the norm for failing students? I.e., how many percent is requried for a passing grade?
I am cold and wet. It is time to invest in an umbrella
Past time, Mike.
This is the first time I'ce ever seen a rain.
Andrew, I've rarely graded on a standard scale.
16:54
Ah, I see. I've looked at some calcexams in the US, its almost like someone is trying to make an army of integrating robots.
yup
And we don't even succeed at that

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