Mathematics - 2024-12-16 (page 1 of 2)

Joe

00:00

Hmmm, I didn't think it looked so bad :(

Ben Steffan

it's huge

Joe

Thanks for the help @Jakobian

Ben Steffan

I'd put that in words, or introduce some auxiliary notation

copper.hat

00:25

i think obsession with replacing clear text by complicatd set builder notation is a fundamental part of every mathematicians growth

Thorgott

01:10

$\mathrm{dim}(X)=\sup\{n\in\mathbb{N}\vert\exists Z_0\subsetneq\dotsc\subsetneq Z_n\subseteq X\text{ of irreducible closed subspaces}\}$

@Ben I've written a particularly illuminating remark

Jakobian

02:25

@XanderHenderson opinions on flavoured coffee beans?

or maybe @leslietownes though I don't know if you even like coffee

Xander Henderson

@Jakobian gross. Just get whole beans. Lightly roasted.

Grind then when you plan to use them.

Jakobian

@XanderHenderson what do you mean whole beans

I was talking about whole beans, just flavoured

Xander Henderson

02:52

@Jakobian JUST whole beans. Lightly roasted.

If you want some flavors, throw in a stick of cinnamon or a couple of cardamom pods when you grind the beans.

But flavored beans are for Karens.

Jakobian

oh, I see

I don't have cardamom pods I only have cardamom powder

that's good too?

and if I wanted to make coffee with that do I just put it all into my moka pot together

Xander Henderson

@Jakobian probably a bad idea, depending on your brewing method.

Powdered spices will generally clog up your filter and cause problems.

EM4

or you can drink tea.

Xander Henderson

03:16

@EM4 gross. Who would want to drink dirty leaf water?

:P

EM4

03:28

hehehe is the best dirty water ever HAHAH.

Xander Henderson

dirty bean water > dirty leaf water

EM4

no my friend.

Xander Henderson

I'm not your friend, pal.

Jakobian

if coffee beans are fruit then we're basically tasting a compote?

Xander Henderson

@Jakobian They aren't the fruit. They are the seed.

leslie townes

05:45

@Jakobian i am not a huge enthusiast, i have a cappuccino most weekend mornings but basically do not drink coffee outside of that. i do not like to add flavors to coffee. i do drink a lot of weird herbal "teas" though, which are sometimes a mix of herbs and things

Jakobian

@leslietownes mhm. You never tried coffee with cardamom?

I'd say its actually pretty good, that or other root spice

not sure but I think even ginger would be good to add

basically what I was talking about is flavoured coffee beans, which is what they do is they either coat coffee beans using oils, or they infuse them with flavors (say, cinammon)

leslie townes

07:13

i don't think i've tried coffee with cardamom, no. i do like ginger tea, i drink it every day

one potato two potato

08:39

If $\phi:G\twoheadrightarrow H$ is a surjective group homomorphism, then for $K<H$ with $[H:K]<\infty$, $[G:\phi^{-1}(K)] = [H:K]$?

if $K$ is normal in $H$, then it follows from the isomorphism theorem but I'm not sure for general finite index subgroup

Jakobian

08:54

The map $g\phi^{-1}(K)\mapsto\phi(g)K$ is easily seen to be well-defined and a bijection (assumption $[H:K] < \infty$ not necessary)

Ben Steffan

@Thorgott "category object $X_\bullet$ in $\mathbf{Spc}$"

shouldn't your Yoneda embedding go from $N(\mathcal{P}(\Delta^{\times n}))$ to $\mathrm{Fun}(N(\mathcal{P}(\Delta^{\times n}))^\mathrm{op}, \mathbf{Spc})$?

mo-_-

09:14

$e^z = 1 + i \quad z \in \Bbb C$, what is the fastest way to find $z$?

Ben Steffan

polar coordinates?

Jakobian

If $z = x+yi$ then $e^x = |1+i|$, then solve for $y$

leslie townes

mo it is odd to talk about 'fastest' with something like this, it might depend on the person and the differences between 'different' methods are likely not going to be substantial. conceptually a simple way of solving this and similar problems is to write 1 + i in "polar form" which easily allows for a representation of it in the form e^w for some w

and the multi valued nature of the solution is then relatable to the many but perhaps better known complex solutions t to the equation e^t = 1

Jakobian

sure, if you write $1+i = e^w$ then you can use that $e^w = e^z$ iff $w-z=2\pi i k$ for some $k\in\mathbb{Z}$

mo-_-

@Jakobian $z = x + iy$ yes

Jakobian

09:19

this should be straightforward since $\frac{1+i}{\sqrt{2}} = e^{i\pi/4}$

mo-_-

How did you do it?

I was thinking of writing

Jakobian

do what

mo-_-

$e^x(\cos(y) + i\sin(y))=1+i$

@Jakobian Sorry, I meant how would you solve the problem?

55 secs ago, by mo-_-

$e^x(\cos(y) + i\sin(y))=1+i$

Jakobian

I wouldn't try to solve it

mo-_-

So $\begin{cases} e^x \cos(y) = 1 \\ e^x \sin(y) = 1\end{cases}$

So I move from the complex field to the real

Jakobian

09:24

but if I had to then I'd do whatever comes to my mind... I don't really know what to say to you

mo-_-

@Jakobian Because there are different methods to find z , maybe you know something interesting, I don't know

Jakobian

I'd probably just normalize it like I suggested i.e. $|e^z| = e^x$ and translate the problem onto solving something in a circle i.e. $e^{iy} = \frac{1+i}{\sqrt{2}}$

Ben Steffan

I'll gently tap this sign

leslie townes

there aren't that many "methods" of expressing a point in the plane in polar coordinates

Ben Steffan

11 mins ago, by Ben Steffan

polar coordinates?

yeah

mo-_-

09:28

Using the method I was writing I found $z = \frac{1}{2} \ln 2 + i(π/4 + 2kπ) , k \in \Bbb Z$

Ben Steffan

yes

leslie townes

within the world of all ways of finding a polar form of a complex number, the best one, the fastest one, the simplest one, whatever, is the one that you can remember and implement in less time than it took you to have this conversation :)

i mean that in all seriousness

Ben Steffan

the best way to find the polar form is the friends you made along the way

leslie townes

i don't think there's some trick that will materially simplify the problem, such that finding a complex logarithm somehow becomes "simpler" than finding a polar form. at a very minimal level of abstraction, the two problems are almost the same

although nobody asked for one

and with things like hand computation, most of the time the angles involved in the polar forms of numbers with 'nice' coordinates are not going to be 'nice' unless you allow for arctangents and stuff in what you allow as 'nice.' i don't know a way around this either

mo-_-

$\log(z) = \log\left[r e^{j{(re + 2kπ)}}\right] = \ln(r) + j(\theta + 2kπ)$

$Arg(z) = \theta$

$\log(1+j) = 1/2 \ln 2 + j π/4$

$z_0 = 1/2 \ln 2 + jπ/4$

$z_k = 1/2 \ln 2 +j(π/4 + 2kπ)$

@leslietownes Did you mean that?

psie

10:05

Let $(f_n)$ be a sequence of nonnegative measurable functions on $E$. Then $$\int\left(\sum_{n\in\mathbb N}f_n\right)\,\mathrm{d}\mu=\sum_{n\in\mathbb N}\int f_n\,\mathrm{d}\mu.\tag1$$Is it possible to derive from (1) the identity $$\sum_{k\in\mathbb N}\left(\sum_{n\in\mathbb N}a_{n,k}\right)=\sum_{n\in\mathbb N}\left(\sum_{k\in\mathbb N}a_{n,k}\right)?\tag2$$

My book says this follows from putting $\mu$ equal to counting measure on $E=\mathbb N$, but I can't quite work out the details. If we put $f_n(k)=f(n)\mathbf1_{\{n\}}(k)$, then (1) becomes $\int f\,\mathrm{d}\mu=\sum_{n\in\mathbb N} f(n)$. How do I obtain (2)?

psie

10:31

I think I've worked out the details :)

Thorgott

11:27

@BenSteffan a simplicial object satisfying the Segal condition that $X_n\rightarrow X_1\times_{X_0}\dotsc\times_{X_0}X_1,\,n\ge2$ is an equivalence

@BenSteffan this is Lurie's notation, $\mathcal{P}(\mathcal{C})$ is the category of presheaves on $\mathcal{C}$

nickbros123