Ok. Now apply to distributive law again to the left side

(With the understanding that P=P(1) )

= (p + bP) + (p + dP)

No.

That is not a valid use of the distributive law.

@philmcole Wir haben in Ana 2 das Poincaré Lemma bewiesen, daraus folgt als Spezialfall, dass quasi wenn du eine Menge in $\Bbb R^3$ hast, die eine bestimmte Bedingung erfüllt (konvex reicht zum Beispiel, aber es geht allgemeiner), dann folgt für ein Vektorfeld $F$ mit $\operatorname{div} F = 0$, dass ein Skalarfeld $\Phi$ existiert mit $\operatorname{grad}(\Phi)=F$. Also ich glaube Physiker würden sagen, dann existiert zu einem quellfreien (z.B. Kraft)feld ein Potential

ab+ac = a(b+c) and ac+bc=(a+b)c.

if you're not using one of those, you're not using the distributive law.

How does this relate to adding 1 to the left side?

When did I tell you to do that?

@philmcole ah nein, lol ich meine $\operatorname{rot}F=0$ also "wirbelfrei", sorry

die Vektoranalysis verwirrt mich

in the example, the formula looks like: P + bP − dP = (1 + b − d)P.

Ok.

if you do the distributive law on the first two terms on the left, what do you get?

you've got P+bP. how else can you express that?

@MatheinBoulomenos In response to your suggestion, this seems to be a bit too advanced for the course in question.

@MatheinBoulomenos Ja genau, rotationsfrei ist in $R^3$ die Kondition dafür, dass es ein Pot. gibt. Allgemeiner müssen glaube ich die part. Ableitungen $\partial_k\partial_j f$ kommutieren oder so ähnlich. Teilweise witzig, aber in unserem Ana Skript gibt es tatsächlich ein paar physikalische Beispiele. Z.B. wurde das Snelliussche Brechungsgesetz gezeigt oder im Zsmh. mit Integralen untersucht welche Rotationskörper am schnellsten eine schiefe Ebene runterrollen.

Der Witz ist, dass die Beispiele besser sind als die im Physik Skript.

@Antonios-AlexandrosRobotis it was a joke

I figured :P

p(bP)

No.

@philmcole dass partielle Ableitungen kommutieren hat tatsächlich was damit zu tun, aber das gilt immer für offene Teilmengen von $\Bbb R^n$

es hat mehr was damit zu tun, dass die Menge auf der deine Felder definiert sind keine Löcher hat

Also z.B. $\Bbb R^3$ ohne eine Gerade ist schlecht

What are the a,b,c here?

da gibt es so rotationsfreie Vektorfelder, sodass wenn du einmal im Kreis um die Achse integriest, nicht null rauskommt

(sowas wird in Funktionentheorie super wichtig)

I'm sorry @Semiclassical

Also dass du einmal im Kreis um eine Singularität integrierst, meine ich

@Antonios-AlexandrosRobotis but do you work over arbitary fields? I think it's super cool if you do that

Hab noch mal nachgelesen: In unserem Thm steht Menge muss offen und sternförmig sein. Dann ist $f$ konservativ genau dann wenn die ersten partiellen Ableitungen gleich sind, also $\partial_kf= \partial_jf$ (sorry, nicht die zweiten).

@Semiclassical I need to go now, but thank you for giving me a clue (distributive law) for me to investigate further

kann man so formulieren, ja

Ohne Löcher ist noch allgemeiner denk ich

aber es gibt auch nicht-sternförmige Mengen für die das gilt

ja genau

man braucht da eigentlich Homologie um das genau zu formulieren

Noch ein Teilbereich den ich noch nicht kenne :P

Aber ja, die Aussage, dass die Rotation verschwindet ist dasselbe, wie dass die partiellen Ableitungen gleich sind

Ich geh mal in Richtung Bett. Tschau @MatheinBoulomenos!

Gute Nacht! @philmcole

can someone help me with an integral

the integral is the integral of x/sqrt(4-(x-2)^2)

i think the antiderivative family with involve arcsin x

to get rid of the 4, i factored it out in the sense:

x/sqrt(4(1-(x-2/2)^2

then i took the sqrt(4) out of the integral leaving me with:

1/sqrt(4) * integral of x/(1-(x-2)/2)^2 dx

i am having trouble getting rid of the x on top im not sure what u-sub to do

anyone on here

@MatheinBoulomenos sorry my phone gets notifications like 15 minutes late from this chat, idk why

and no, just over $\mathbf{R}$ and $\mathbf{C}$ later in the course.

ah :-/

the students have only had one semester of calculus as a prerequisite

in all likelihood they don't really even understand that

but do you start with abstract vector spaces?

lol nope

that's like week 2

okay, week 2 is still early

im thinking about letting u = (x-2/2)

well the course is 6 weeks and is 8 hours of lecture per week

at least it's not mindless matrix computations

nah I won't let it be mindless gaussian elimination

so i can get an easier form of the integral u+1/sqrt(1-u^2) and then i will split the integral into 2

is that a good idea guys

of course the integral has a factor of 4/sqrt(4)

@Antonios do structure theory for modules

lol

someone please

@MacroGuy I'm not gonna do the computations but if you do $u=x-2$ you end up with $\int \dfrac{u}{\sqrt{4-u^2}} + \dfrac{2}{\sqrt{4-u^2}} du$ which should be easy

@Antonios @Daminark definitiion: a ring is a one-object abelian category. If $R$ is a ring, a left-module over $R$ is an additive functor $R \to \mathbf{Ab}$

LOL

ok thank you im already trying the u=(x-2)/2 i think it's working but maybe your way is shorter

is asking hw questions in this manner alowed in chat

Yes

if that won't make them understand vector spaces, I don't know what will

ok thanks

@Mathein lmaoooo

Just don't spam the room with them :P

I need to learn about abelian categories tbh

They're pretty cool

I didn't really understand the point of abelian categories until doing a bit of AG

and of course, the next definition after that: If $R$ is a ring and $F,G: R \to \mathbf{Ab}$ are left-modules, then a $R$-linear map from $F$ to $G$ is a natural transformation $F \Rightarrow G$

abelian categories were love at first sight for me

Right now all I've got in mind is that abelian categories are supposed to be the context where exactness makes sense, basically generalizing R-Mod

that's not too far off

you can talk about exactness in more generality

but yeah, abelian categories formalize what's so nice about R-Mod

hey how useful is mathematics in neuroscience would anyone know

at a graduate school level

last night something strange happened to me. I thought I had proved some crazy stuff in my dream, but then the moment I woke up I realized that somehow in my dream I failed to differentiate between hereditary rings and semi-hereditary rings, so it was all wrong

im just a sophmore undergrad so i wouldn't know

@MatheinBoulomenos nice

@MatheinBoulomenos you might want to check out my latest question

I kinda want to know what a psychoanalyst would make of that if I ask for a deam interpretation

@LeakyNun I asked myself the question before

@MatheinBoulomenos what did you answer

it's true

Do prefer the answer on main or here in chat?

I'll just answer on main, gotta get that rep

oh

is it because of this

hmm

but go get your rep

I don't think so

ok

maybe it's possible but I don't see how right now

@LeakyNun can I assume that the functor is faithful?

that's kind of obvious

can you prove it?

The induced map on Hom-Sets doesn't change the morphisms as maps at all

does it hurt to include that sentence to your answer

sure