@TedShifrin My condensed notes usually abandon full proofs in favour for providing an overview to the ideas and intuition.

I understand, but char. classes require serious examples and sophistication.

@TobiasKildetoft actually, what does a void lecture is like, is it just silence or you hear voices?

Part of the fun is trying to come up with a way to convey the serious/sophisticated examples in an easy-to-read way.

Definitely will be harder with characteristic classes, though.

does "desuetude" count?

Forms and integration was pretty easy since they are both algebraically and geometrically not difficult.

Yup.

OK, I'm taking a lunch break. Bye all.

Enjoy!

Danish has several expressions of the form "x and y" which just mean "x" (or sometimes y), and where the other part is not actually a thing in itself (at least not a word in use any longer)

There is also a variation on that in the word "strømpesokker" which literally translated means "sockssocks" and which is what you are wearing if you are only wearing socks (usually with emphasis on the fact that you are not wearing shoes).

strømpesokker, makes me think of:

because of the first part "strøm"

but there is no "strøm" there (strøm means electricity)

right

Evening all

$\frac{d^2}{dt^2} x = k e^{-iwt}$

The solution to this is of the form $x=x_0 e^{-iwt}$

but it’s a second order differential so where is the second constant?

oh there’d be a plus c term

which goes away by bc

Adding a constant is like moving a hill

all the gradients remain the same

*translating

In $M \otimes_R N$ you have $m \otimes 0 = 0 \otimes n = 0$ for all $m \in M$ and $n \in N$ right... just by the linearity in each variable

yep

hey guys. is there an elementary way to determine all group homomorphisms from the additive rationals to the unit circle? (elementary, as in, not resorting to representation theory)

You need $\varphi(a/b + c/d) = \varphi(a/b)\varphi(c/d)$

?

Let $n \in \Bbb{Z} \subset \Bbb{Q}^+$ then $\varphi(n) = \varphi(1)^n$

yea, I had that too

but I didn't really know how to proceed

$a/b \mapsto \varphi(1/b)^a$

neccessarily

yes

for positive $a$ at least

$1/b = b^{-1}$ so $1/b \mapsto \varphi(b)^{-1}$

Nvm

that's not true

yea, it's additive~

In the rationals

We have math.oregonstate.edu/home/programs/undergrad/…

yep

power series?:0

So you take $1/b = 1/(x - 1)$ and $x = b + 1$

So then

Nvm

that doesn't work either

Unless we set $x = 1/b$

can we claim that $\phi(1)$ determines the entire homomorphism?

where $\phi$ would be our homomorphism

I don't think so

You have to prove it

since $\phi(n)=\phi(1)^n$ and $\phi(1/n)=\phi(1)^{1/n}$

How do you 1/n in something in an abelian group?

Your saying $1/n \times 1 = 1/n$

I'm confuse

I'm saying $\phi(1)=\phi(n\cdot 1/n)=\phi(1/n)^n$, hence $\phi(1/n)=\phi(1)^{1/n}$

however, the $n$-th root of unity is not unique

That's a tough one :)

it is:d

Hey @ShaVuklia!

hey there~ @BalarkaSen

let V be a region containing 0 and bounded by S. Then there cannot be any function/distribution G such that ∇²G=δ(r) in V and ∂G/∂n=0 on S, right? @Semiclassical @ÉricoMeloSilva @BalarkaSen

@LeakyNun obviously not

how obvious was it? :P

very

divergence thm

yeah same thought here

it's fine if u make the boundary condition a constant tho

but by divergence theorem, ∫ ∂G/∂n dS = 1 right

so ∂G/∂n, if it is a constant, must be the reciprocal of the total length

yes you put in the right constant lol

I hope this doesn't come up in tomorrow's exam

I probably don't have the guts to write contradiction though

wut is ur class on

multivariate calculus

and fourier series / transform

and wave equation / heat equation

oh cool

In my textbook, when determining the convergence of a definite integral over a quite complicated function (with limits 0 to 1), the author uses the phrase $x\approx 0$ to intuitively better estimate the convergence. Why is it justified to approximate x in that way?

I can see why, when studying infinite series, one only cares about the tail of the series and so one can simplify the function or sequence by approximating "for large k". Yet I can't see the analogy here.

@schn maybe something like this: you can split the integration interval into [0,x] and [x,1]. At that point it may be sufficient to examine the behavior on the first interval