Mathematics - 2017-10-06 (page 2 of 5)

Eric Silva

04:01

for a (pseudo)riemannian manifold you have the one that in local coordinates looks like: $\sqrt{|\det(g)|}dx^{1} \wedge \cdots \wedge dx^{n}$ which is like, a natural one to prefer, and it comes from the metric $g$

to use notation and not just words: for a general smooth manifold, if $\omega$ is a volume form, you really have no reason to prefer $\omega$ vs $f \cdot \omega$ where $f \in C^{\infty}(M), f > 0$.

John P

I don't understand why that is true - why would I not prefer $f=1$ everywhere?

Semiclassical

$\omega_1=2\omega_2$, $\omega_2=\frac 12 \omega_1$

which one of them is preferable?

Eric Silva

^exactly what Semi is pointing out

Semiclassical

one is a rescaling of the other, and vice versa

Eric Silva

an important point i glossed over: things need to be orientable for any of this to make sense, otherwise there wont be any volume forms lol

but like the point is that, if you have a volume form, then you have literally infinitely many volume forms, so if you're just working with a smooth manifold with no extra structure, then they're all equally good

Semiclassical

04:13

if the manifold has finite volume then you presumably could scale so that $\int \omega = 1$

Eric Silva

that strikes me as kind of weird but i guess it's fine

Semiclassical

i imagine there's cases where something like that is sensible. but the fact is that it's a choice

John P

Sorry guys, I still don't get it :(. But I'm starting to think this is a consequence of manifolds not really being coordinate based?

Eric Silva

it strikes me as weird cuz im from Riemann land where the volume is a thing that actually matters i guess @Semi

@John i'm not really sure what you mean by "coordinate based"

John P

04:29

I thought maybe it had something to do with being able to choose a different atlas/map, but nope I see that doesn't really change the value of an integral on a manifold.

Eric Silva

@John do you at least see why there are many volume forms?

John P

Not really, I don't know what it is :(.

Yeah nevermind, I'm too new to this stuff and barely know what a manifold is, I'll get it later on.

Eric Silva

yeah maybe familiarize yourself more with the basics first

John P

I'll remember "you can scale it" and "there's infinitely many volume forms" and try to get it later. Thanks for helping, and sorry that it was a bit of waste xD.

Semiclassical

04:51

@EricSilva I think this is more-or-less the kind of example I had in mind: math.stackexchange.com/questions/1637831/…

funny thing is, the place where I've seen integral differential forms be of interest was in cases involving a Kahler manifold

so riemannian to begin with

Eric Silva

the terminology is so unfortunate

Semiclassical

(something something geometric quantization.)

my attempts to understand geometric quantization can basically be summed up as

I understand about half the words involved in these definitions, and I'm pretty sure this doesn't even get you to the end of it.

Eric Silva

im too tired to read this lol

Semiclassical

hah

Daminark

05:27

Hey

user52932

Hey guys, I have a quick question as an intermediary step to a problem I am trying to solve. I am not quite sure what E(E(Y|X)|X) evaluates to. Is this object just equal to E(Y|X)?

Shea

Can someone help me with this question?

Suppose that 30% of all students drive to school, 50% take the bus, and 20% walk. Of those who drive, 20% are usually late for their first class of the day. Of those who take the bus, 10% are usually late for their first class of the day. Of those who walk, 15% are usually late for their first class of the day. What is probability that a randomly selected student is regularly late for their first class?

angussidney

05:51

@Shea have you tried asking a question on the main site? If you do, I can answer it for you

Shea

I figured it out it's all good

user84215

06:02

THIS USER WANTS THE AUTHORITIES TO UNFREEZE AND UNDELETE THIS ROOM AND THIS ROOM, RESPECTIVELY, AS SOON AS POSSIBLE.

user84215

06:14

@mixedmath ^

Leaky Nun

and of course "this user" is yourself

Daminark

06:32

Hey everyone!

Fun puzzle I found recently if anyone wants to try it out with me

I've got about half of it down

It's proving that if $p$ is a prime and $p^t \mid \dbinom{n}{k}$, then $p^t \le n$

Leaky Nun

$$\begin{array}{rcl}
\displaystyle \int_{B(i)} \frac{\cos^2(z)}{1+z^2}\ \mathrm dz
&=& \displaystyle 2i\pi \lim_{z \to i} (z-i) \frac{\cos^2(z)}{1+z^2} \\
&=& \displaystyle 2i\pi \lim_{z \to i} \frac{\cos^2(z)}{z+i} \\
&=& \displaystyle 2i\pi \frac{\cos^2(i)}{2i} \\
&=& \displaystyle \pi \cosh^2(1) \\
\end{array}$$

am I doing it wrong?

@Daminark equivalent to there's no prime power greater than $n$ that divides $\dbinom n k$

Daminark

Yup

I'm guessing you induct on $t$, since it's clear that if a prime divides it, then it has to divide some number between $n-k+1$ and $n$

What'd be funny is if there's a group theory way of doing it

Leaky Nun

@Daminark nah, it's more combinatorical

Daminark

Something like, a p-group cannot have a faithful action on the set of subsets of a smaller set

Leaky Nun

count the multiplicity of the prime in $P^n_k$ and then in $k!$

@Daminark oooh that might be good

Secret

06:39

...Well.... that surprising amount of traffic in mathworks due to the set theory construction does somehow turned that room into more like the vision of mathematics workshop, which is not what I originally intended

but anyway, this is not the most important thing

The most important thing right now is that we might need to fall back on logic before dealing with set theory, but these will be dealt with later. I am still trying to comphend the latest batch of messages

It seems trying to explore infinity is more difficult than I thought

Leaky Nun

the group action of $S_n$ on $\dbinom n k$ is transitive, and the stabilizer of each element is of order $k! (n-k)!$

[I'm obviously treating $\dbinom n k$ as a set here]

Daminark

Presumably the set $\{1,\ldots,\dbinom{n}{k}\}$, yeah?

Leaky Nun

no, the set of subsets with k elements of n letters

Daminark

Okay having toyed around with things a bit more I've come to the conclusion that there's no good way to ignore the denominator

Unless we can play the $\dbinom{n}{k} = \dbinom{n-1}{k} + \dbinom{n-1}{k-1}$

If $p^t$ fails to divide one of them, it will fail to divide either, and... maybe we can conclude that $p\mid n$?

Leaky Nun

06:56

$S_4$ has a $2$-subgroup of order $8$...

what is it

Daminark

Yup that'd force it since $p\mid \dbinom{n}{k} = n\dbinom{n-1}{k}$

So if $p^t \nmid \dbinom{n-1}{k-1}$, then $p\mid n$

Leaky Nun

$\dbinom n k = \dfrac n {n-k} \dbinom {n-1} k$

Daminark

Ah fuck

Okay I mean $p\mid \frac{n}{n-k}$ so in particular $p\mid n$

So we're still okay

This might actually be beneficial

Leaky Nun

$|S_4| = 24 = 2^3 \times 3$

It has either 1 or 3 Sylow 2-subgroups

Daminark

This is true

Leaky Nun

07:02

$\langle (1,2,3,4),(1,3) \rangle$ is the only 2-subgroup of $S_4$ according to groupprops

but the problem is I can't guarantee $p$-subgroups of $S_n$ in general to be normal

Daminark

If a subgroup is unique of a given order, then it is normal

(And if it's Sylow, those two conditions are equivalent)

Leaky Nun

yes, but it only so happens for $S_4$ that there is just one $2$-subgroup

Daminark

Well what exactly are you trying to do?

Leaky Nun

I don't really know

if it's normal then I can consider the group structure of the quotient

Daminark

Anyway I'm gonna sleep on my problem

Leaky Nun

07:10

@Daminark what a strange place to sleep on

FuzzyPixelz

Hello

Leaky Nun

07:33

$$\begin{array}{rcl}
\displaystyle \int_{-\infty}^{\infty} \frac{\cos^2(x)}{1+x^2} \ \mathrm dx
&=& \displaystyle \frac12 \int_{-\infty}^{\infty} \frac{1+\cos(2x)}{1+x^2} \ \mathrm dx \\
&=& \displaystyle \frac12 \int_{-\infty}^{\infty} \frac{1}{1+x^2} \ \mathrm dx + \frac12 \int_{-\infty}^{\infty} \frac{\cos(2x)}{1+x^2} \ \mathrm dx \\
&=& \displaystyle \frac12 (\arctan(x))_{-\infty}^{\infty} + \frac12 \Re \left[\int_{-\infty}^{\infty} \frac{\exp(2ix)}{1+x^2} \ \mathrm dx\right] \\
&=& \displaystyle \frac12\pi + \frac12 \Re \left[\int_{-\infty}^{\infty} \frac{\exp(2iz)}{1+z^2} \ \mathrm dz…

How can I argue that the contribution from the last integral vanishes?

Daniel Guldberg Aaes

Good morning ;)
Do you guys know if f(x):=1/x is consideres a one to one, where x is a set of all real numbers.

Leaky Nun

@DanielGuldbergAaes it isn't even defined

Daniel Guldberg Aaes

@LeakyNun sry but you lost mere there..... what do you mean by it isn't defined?

Leaky Nun

08:06

what is f(0)?

Daniel Guldberg Aaes

0

Gabriel Romon

@user52932 yes, E(E(Y|X)|X) = E(Y|X). One reason would be that E(Y|X) is already measurable with respect to the sigma algebra generated by X

Daniel Guldberg Aaes

@le

@LeakyNun
\[
f(x)
\left \{
\begin{tabular}{ccc}
1/x if x \neq 0 \\
0 if x = 0 \\
3 & 3 & -8
\end{tabular}
\right \}
\]

Gabriel Romon

@user52932 another proof would follow from the tower property

FuzzyPixelz

Mysterious urge to share this; “Freedom is the freedom to say that two plus two make four. If that is granted, all else follows.”
― George Orwell, 1984

Secret

08:26

2+2=5

universe implodes

FuzzyPixelz

Haha

Mathmore

That's quick maffs

FuzzyPixelz

You just broke all the laws of physics and the universe ceased to exist, therefore reality isn't real and it was just figments of our minds before you declared that 2+2=5.

Secret

therefore, the universe still exists because the fact that the universe ceased to exist is a figment of our minds as described by the reasoning above

user8469759

08:48

hi guys

what can I use to view formulas in the chat?

Secret

read upper right corner

user8469759

ah ok

latex in chat xD

haven't seen it

test $a_k \to 0$

ok

Leaky Nun

@DanielGuldbergAaes yes it is injective

Secret

09:13

[Chemistry] More complicated molecule sent. Now to wait for a bit, thus have some time to do some maths

If this run has no problem, then the first batch can be sent without worries

Mathmore

@LeakyNun Consider $|\int_{0}^{\pi} \frac {\exp(2iRe^{i \theta})}{1+ Re^{i \theta}} iRe^{i \theta} d\theta| $

$\le \int_{0}^{\pi} \frac {|\exp(2iRe^{i \theta})|}{|1+ Re^{i \theta}|} |iRe^{i \theta} d\theta|$

Since $\sin \theta \ge \frac {2 \theta}{\pi}$ for $ 0 \le \theta \le \frac {\pi}2$,

$|\int_{0}^{\pi} \frac {\exp(2iRe^{i \theta})}{1+ Re^{i \theta}} iRe^{i \theta} d\theta| \le$

$\frac {2R}{R^2 -1} \int_{0}^{\frac {\pi}2} e^{-2R \sin \theta} d\theta \le$

$\frac {2R}{R^2 -1} \int_{0}^{\frac {\pi}2} e^{-2R \frac {2\theta}{\pi}}=$

$\frac {2R}{R^2 -1} \frac {\pi}{2R} (1-e^{-R})$. Now apply limit $R \to \infty$.

lol

user84215

@mixedmath chat.stackexchange.com/transcript/message/40386700#40386700

Mathmore

09:29

@LeakyNun I just forgot to add square in the denominator for the term $(Re^{i \theta})$ in first 2-3 steps. In the last 3 steps, it is there.

Mohamed Ziad

10:50

when trying to solve an ordinary linear differential equation, why is it when deriving the integrating factor we assume the equation is a non-exact one and apply the formula to make it exact? help, please.

Secret

11:03

If the ODE is exact, then the integrating factor will be just 1

Mohamed Ziad

@Secret yeah, I understand what you said, but I don't understand what does it have to do with treating Linear ODEs as if it were nonexact in order to derive the formula for the integrating factor?

Secret

11:28

The formula given to you is expected to make any linear ODE exact. Therefore, we assume all linear ODEs are inexact, use that formula. If the ODE is exact, it gives 1 as the integrating factor, if it is inexact, then the formula will find the integrating factor for you

Mohamed Ziad

11:42

@Secret Right! So this integrating factor that I get (which happens to be the same as the one that I can find by inspection) would make the equation exact. Then this means that I could also solve it as if it were an exact equation?

nbro

12:14

Hi!

The following question is for people with a good experience writing proofs by contradiction, maybe teachers that also have to correct them.

Do you usually explicitly state at the end of proofs by contradiction something as follows "(But/and) this is a contradiction"?

Secret

It is common to have the word "contradiction" to appear somewhere in a proof by contradiction, followed by the conclusion

Balarka Sen

@EricSilva I always understand the Riemannian volume form using Gram's formula

Secret

12:32

[Random]

I wonder what will an inverse relation of the thomae function be like...

should probably be more than just multiplying each element by their reciprocals

user193319

Would someone mind taking a look at this question? math.stackexchange.com/questions/2459538/…

I already received one response but I am not entirely pleased with it.

Secret

@MohamedZiad yes, once you multiplied by the integrating factor, the resulting ODE becomes exact and thus methods for exact ODE can be used

Balarka Sen

12:53

Hi @Daminark

Mohamed Ziad

13:25

@Secret Thanks for clarifying things! However; I have one last question, (I think this is where my ignorance kicks in but ..) could the ODE then have 2 solutions?

Secret

The maximum number of independent solution an ODE can have depends on its degree. Therefore an ODE with 2nd derivative can have at most two independent solutions

Mohamed Ziad

@Secret Oh, I only mean ones with first derivatives only. If I apply the method of solving exact eqns I should get the solution of a constant function right? which is different from the solution from solving it as a linear ode?

Secret

Nope, the solution of both the exact and linear ODE will be a function or relation of x. It does not seemed that apparent for exact ODE because the function is expressed in the form f(x,y)

Secret

13:41

Confluence (abstract rewriting)

In computer science, confluence is a property of rewriting systems, describing which terms in such a system can be rewritten in more than one way, to yield the same result. This article describes the properties in the most abstract setting of an abstract rewriting system. == Motivating examples == The usual rules of elementary arithmetic form an abstract rewriting system. For example, the expression (11 + 9) × (2 + 4) can be evaluated starting either at the left or at the right parentheses; however, in both cases the same result is obtained eventually. This suggests that the arithmetic rewriting...

I frequently use confluence diagrams in algebra proofs

because it is more visual

Leaky Nun

@Mathmore thanks

user193319

Would someone mind taking a look at this question?
https://math.stackexchange.com/questions/2459538/linearly-independent-sets-in-abelian-groups

I already received one response but I am not entirely pleased with it.

Silent

14:00

Hi

Mathmore

@LeakyNun Welcome.

Secret

Hmm... type theory is more flexible than set theory in terms of constructing new things, but not as much when it comes to simplifying terms and defining function

I think I need to dig some introductory material on it and see how to use it in constructing infinite objects

Silent

I am stuck with this:

>Let $F$ be an ordered field.
Suppose $S$ is a subset of $F$ and $y$ an element of $F$, and let $T = \{s+y | s∈S\}$. If $S$
has a least upper bound, sup $S$, then $T$ also has a least upper bound, namely $(\sup S)+ y$. <P> Deduce from this that if $x$ is a nonzero element of $F$ and we let $S = \{nx : n\text{ is an integer}\}$, then S
has no least upper bound.

Please somebody help!

Leaky Nun

@Silent which part?

you just need to use definitions

Silent

@LeakyNun, I don't know how to deduce that $S = \{nx : n\text{ is an integer}\}$ has no supermum from sup({s+y : s∈S}) = (sup S)+ y

Leaky Nun

14:15

if it has, then so would S+x

but S+x = S

so sup(S) + x = sup(S)

x=0

contradiction

Silent

@LeakyNun Aha! But S+x = S

@LeakyNun, thank you very much :)

anakhronizein

@Secret HoTT supposedly will be a great basis for algebraic topology

Since it implies we won't need to work in the category of CGWF spaces.

Mohamed Ziad

14:32

@Secret So far so good! now correct me If am wrong, please. When solving it as an exact ode the solution is a function in terms of x and y (which is dependent on x), hence resulting in a constant function. But solving as a linear ode the solution gives the function in terms of just x?

Secret

14:54

yup, they are still functions of x so they give the same type of solutions. For the exact ODE, the solution look like a constant function simply because it is f(x,y(x)) which for some of these f, you can rearrange such that only y is on the LHS

Mohamed Ziad

@Secret

thank you so much!

Leaky Nun

15:09

can there be a set of sets S such that every pair of sets in S is disjoint but $\bigcap S \ne \varnothing$?

anon

15:21

yes, {0}

Balarka Sen

did you mean {{0}}

Hi @anon, haven't seen you around in a while.

anon

ah, yes, {{0}}

Secret

Ordinal question: Is it possible to construct $\omega_1$ and prove that $\alpha < \omega_1$ where $\alpha$ countable ordinal in any set theory?

Related question: Is hartog's number predicative?

Kasmir Khaan

@LeakyNun Hello

@anon Hi

Guy Fsone

can any explain why is this a duplicate? math.stackexchange.com/questions/2460291/…

anon

15:27

hello

Kasmir Khaan

anon :D

i got some abstract algebra question that i need help with

anon

ok

Kasmir Khaan

how is G/N naturally isomorphic to the image of G under f in H

Alessandro Codenotti

What's your new avatar? @Balarka

Kasmir Khaan

provided we have a homomorphism from f from G to H

anon

15:30

and N is the kernel of f?

that's the first isomorphism theorem no?

Kasmir Khaan

yes N is kernel

well the thing is the book states that 2 sections before the iso theorems

anon

the map gN->f(g) is a well-defined isomorphism

Kasmir Khaan

hmm

Balarka Sen

@AlessandroCodenotti It's a scene from Yuri Norshteyn's "Tale of Tales" :)

Kasmir Khaan

so if we have N is normal subgroup of G , we can find a group H and a homomorphism pi : G---> H such that ker (pi ) = N

anon

15:32

yes, H=G/N and pi(g)=gN

Alessandro Codenotti

I see. I was expecting Tarkovsky

Kasmir Khaan

@anon can you put it in better words? like i think i understand what it sais but not sure I got it right

anon

put what in better words?

Balarka Sen

@Alessandro Norshteyn's animations are the closest to Tarkovskian style in the world of animation. "Tale of Tales" is based off of the style in "Mirrors", I believe.

Kasmir Khaan

the relation between a homomorphism from G--> H and the quotient groups from G to other groups @anon

Leaky Nun

15:36

@KasmirKhaan maybe you can think about it better

instead of relying on our spoonfeeding

Kasmir Khaan

@LeakyNun I dont ask before trying

anon

If $f:G\to H$ is a homomorphism with kernel $N$, then $N$ is normal and the map $gN\to f(g)$ is a well-defined group isomorphism $G/N\to f(G)$, where $f(G)$ is a subgroup of $H$. Also, if $N$ is any normal subgroup of $G$, then there is an onto group homomorphism $G\to G/N$ with kernel $N$ given by $g\to gN$.

Kasmir Khaan

this is the first time i take a course that i have absolute no previous knowledge of

@anon that makes sense thanks anon :)

I got one last question about the preimage

anon

I am skeptical your book expects you to know the first isomorphism theorem before it states it. I would guess it either is saying that for a specific example or is conversationally giving a sneak peak of a fact it will explain pages later.

ok

Kasmir Khaan

phi : G--> H is a hom , let E be a subgroup of H , i need to prove that phi inverse (E) is a subgroup of G

Let me show u what i tried=p

@anon the book is by dummit and foot , page 83, its avaible online , stated that before the iso theorems

since E is a subgroup of H , E is not empty

let x and y belong to E

the element of E are of the form phi (- )

x = phi ( a) and y = phi ( b) where a,b are in G

phi^-1 ( xy ) = ab

because phi is a hom so is phi inverse

correct me if am wrong, what this exercice trying to say is that, if we take a subgroup of the image of a homomorphism

then its preimage is also a subgroup of G

anon

15:44

(i) why does E being a subgroup of H imply E is not empty? (checking you know why.)

(ii) phi needn't be one-to-one, so you can't treat phi^-1 like a function

in particular, phi^-1(xy)=ab makes no sense if phi has no inverse

Kasmir Khaan

ah makes sense =p

well for i)

anon

(iii) you also haven't explained why a and b are any elements of the preimage. you started with two elements of E; you should be starting with two elements of the preimage.

Kasmir Khaan

well for i ) the identity of G maps to identity of H

since this is a hom

João Pedro

Hi, is there any book on the historical development of Laplace transform (or Fourier transform), it is widely used in Electrical Engineering and I want to get a intuition on how/why it Works and what was the motivation for its development. Appreciate any help.

Kasmir Khaan

E being a subgroup of H , means that E has at least the idenity of H

anon

15:47

yes

Kasmir Khaan

if we would draw this , just to make sure i understand the notion of a preimage

G----> H

E subgroup of H

then phi ^-1

would take elements of E and send them back to G

the preimage of E under phi

will be a subset of G

anon

phi^-1(E) is notation that means the set of all elements in G that are mapped into E by phi. that doesn't mean there exists a map phi^-1 from E to the preimage though.

Kasmir Khaan

Yes i see why my argument was wrong

hmm need to argue in a different way

well i havent used that the map is a homorphism , and i knew something were wrong from the start

anon

call the preimage D

pick two elements a,b of D

show ab^-1 is in D

(one-step subgroup test)

a,b being in D means phi(a)=x and phi(b)=y for some x,y in E

since E is a group...

Kasmir Khaan

yes that part i know , but what comfuses me is that without using that phi^-1 makes sense

i cant use that f^-1 (f(a) f(b) ) = f^-1 ( f(ab) ) = ab

anon

15:52

you're trying to verify ab^-1 is in D right? what does it mean to be in D? it means gets mapped to an element of E by phi. you don't have to use an inverse map, you use phi itself.

Kasmir Khaan

i change phi to f here

anon

to verify ab^-1 gets mapped to an element of E, consider what phi(ab^-1) must be in light of the fact phi is a homomorphism

Kasmir Khaan

so if i got this right

D is a subset of G

we pick two elements of D , a and b

phi (a) and phi (b) are in E by definition

i need to show that phi ( ab^-1) is also an element of D

i feel like we have to send G-->H --> G , but the way you did it was G-->G

anon

you need to show ab^-1 is also an element of D, which means showing phi(ab^-1) is an element of E

Kasmir Khaan

ahhh

got it =p

now when i think about it , it makes more sense that way, easiar than working backwards from E -->D

@anon but what i can be sure of is that phi (a) phi (b) ^-1 are in H

anon

15:57

you mean how can you be sure?

phi(a) is in E, phi(b) is in E, and E is a subgroup...

Kasmir Khaan

i mean phi (a) phi (b) ^-1 are in H , does not garantee that they are in E

or what is it am missing

anon

32 secs ago, by anon

phi(a) is in E, phi(b) is in E, and E is a subgroup...

Kasmir Khaan

okay =p i was thinking of something else

we defined those to be in E to begin with

so makes sense

anon

we assumed a,b were in D, which by definition means phi(a),phi(b) are in E

Kasmir Khaan

the way i thought of it at first, we take D subset of G , then we take their image under phi

Yes :) thanks Anon :D

too many new concepts with abstract algebra , am going crazy

like why do we need the notion of a preimage

anon

16:01

haha

preimage is a purely set-theoretic concept

the fact the preimage is a subgroup is the abstract algebra fact

even without any group structure, you can talk about equivalence relations, surjections, and set partitions all being the same thing. also, have to leave now so bye.

Kasmir Khaan

hmm I think with more examples ill get the hang of it :D

@anon okay thanks again ! bye :)

CausingUnderflowsEverywhere

16:23

Gilbert has a music service subscription where he pays (dollar sign) 10 monthly to listen to anything but has to pay an additional $1 per song he wishes to download and keep.
The function is $C(s) = s + 10$ where $C$ is the monthly cost, $s$ is any additional song bought. Describe how the average rate of change in his monthly cost changes as the number of songs he buys increases.

Is this a trick question? The rate of change clearly does not change. Unless they mixed up the wrong function with the wrong question.?

oh lol fail

Gilbert has a music service subscription where he pays \$10 monthly to listen to anything but has to pay an additional \$1 per song he wishes to download and keep.The function is $C(s) = s + 10$ where $C$ is the monthly cost, $s$ is any additional song bought. Describe how the average rate of change in his monthly cost changes as the number of songs he buys increases

Leaky Nun

16:38

@KasmirKhaan is doing everything too quick

which is not your fault

and I don’t see any way to fix the situation

Kasmir Khaan

@LeakyNun I dont really have the time to think about everything , thats y i ask , i dont want to ask ofc, i want to solve problems on my own, but the rate we doing things on this course is very fast that one just cant spend hours on unsolved problem because then would miss alot

we starting with ideals and rings next week then we have exam

like we did group theory , sylow , group actions, and ring field nd ideals and many something else , all topics are new and too many new concepts

this is the first algebra course itake too, i understand that you feel that I ask because am lazy , but thats is not true , and i cant prove to you otherwise so =p

I appritiated and still appriciate your help ofc @LeakyNun

Leaky Nun

@KasmirKhaan prove that any group of order 255 is abelian

255 = 3 x 5 x 17

and hence cyclic

Kasmir Khaan

well this can be solved using sylow but I am not 100 % get how to use it atm

still doing chapter 3

Leaky Nun

just make any progress

Kasmir Khaan

isomorphism theorems

Leaky Nun

16:44

the destination is not the point

the journey is the point

if you look at the end goal too much you won’t achieve anything

tell me everything you know about G a group of order 255

Kasmir Khaan

well we proved some stuff about groups of order 30 and 60 that they are simple

but i really did not have the time to do more than being in the lecture

I planned to go though that between today and monday

next lecture is on tuesday so i think i can catch up

I ll think about it and come back to you with something =p

Leaky Nun

ok

Balarka Sen

18:48

Hi @Ted, @Eric

Eric Silva

sup

Leaky Nun

hi @BalarkaSen

Balarka Sen

hey

Leaky Nun

Let S be a hamel basis

what is the outer measure of S?

CausingUnderflowsEverywhere

hi balarka

Eric Silva

19:16

@Balarka what is Gram's formula

Balarka Sen

It says if you have a $k$-dimensional parallelpiped in $\Bbb R^n$, spanned by vectors $v_1, \cdots, v_k$, then the volume of that thing is determinant of the $k \times k$ matrix with entries $\langle v_i, v_j \rangle$.

Eric Silva

oh i didnt know it had a name lol

Balarka Sen

I learnt it from Ted

Eric Silva

i always just called it the volume formula

Balarka Sen

aha

Well, I missed a square root

Eric Silva

19:18

ya ya

welp i gotta go to the geometric analysis seminar

Balarka Sen

njoy

Transcript for

$Mathematics$ Mathematics