Mathematics - 2018-04-30

Joe Stavitsky

00:08

Could someone please explain how they get F_x,F_y, and F_z ?
https://i.imgur.com/CbHu6CC.png

Akiva Weinberger

(re-moved)

So that's ∂F/∂x, ∂F/∂y, and ∂F/∂z

Joe Stavitsky

@AkivaWeinberger yea I think I see the rule

Akiva Weinberger

The derivative of tan is sec^2

The derivative of tan(y+z) with respect to x is 0, because there's no x in it

Joe Stavitsky

@AkivaWeinberger yea I got it this crap is embarassingly simple

@AkivaWeinberger I really wonder why they don't cover this in calc I

Akiva Weinberger

@JoeStavitsky It happens to all of us

Partial derivatives? I dunno

Joe Stavitsky

00:12

@AkivaWeinberger They teach implicit, but for whatever reason partials are too scary

shurg

Samuel Yusim

more letters is more spooky

ÍgjøgnumMeg

@Akiva I most certainly did not

Shaun

Hey :)

Akiva Weinberger

0 out of 3 then?

ÍgjøgnumMeg

^

I furthered my own knowledge, understanding, and wisdom though

so

?/9

Shaun

00:24

I must be suffering from the Imposter Syndrome because, several months into my PhD in Mathematics, I don't feel good enough :/

Ah shit, I've got to go. Never mind!

Akiva Weinberger

@Shaun Those are the symptoms yes

Ted Shifrin

Normally partials aren't discussed until multivariable calculus, @JoeStavitsky, although the formula for implicit differentiation, of course, is simple with partial derivatives.

Joe Stavitsky

@TedShifrin The more classes I take the more I think we waste a lot of time by putting things into categories. For instance I think more physicists should visit chem labs. And vice versa.

Daminark

Hello!

Ted Shifrin

It's not a matter of categories, Joe. But there's just too much math to cover, and putting some of Calc III into Calc I means you're kicking out other stuff to make time ... and you can't do it justice.

Hi, Demonark.

Akiva Weinberger

00:33

@JoeStavitsky People should be free to explore, at least

How that's implemented, I dunno

Daminark

How's everything going?

Akiva Weinberger

Obviously I can't shove a bunch of randos in a chem lab and say "Play", that's a horrible idea

Ted Shifrin

I just taught a class today where there just wasn't enough time for stuff to soak in, given the level at which the students had mastered preceding stuff (I substituted for the usual teacher) ... so we didn't come close to doing all the things the powers that be had planned to be covered.

Joe Stavitsky

@AkivaWeinberger at [insert my undergrad chem program here], the physics guys could help so much with the hardware but the professor doesn't want them coming around because they will try to run shit. Sadly, he's probably right

Akiva Weinberger

Why would a physicist be better at the hardware?

Sha Vuklia

00:37

guys, are 1-dimensional manifolds always orientable? I don't really need a proof, but intuitively I would think yes?

Akiva Weinberger

Yes

Ted Shifrin

You be right, Sha.

Sha Vuklia

cool

Akiva Weinberger

Unless you have non-Hausdorffian madness, but usually those aren't considered manifolds

Ted Shifrin

slaps DogAteMy silly

Sha Vuklia

00:37

yea I'm still working in Euclidean space

Joe Stavitsky

@AkivaWeinberger many of the chemists know next to nothing of pressure and temperature (orgo).

Akiva Weinberger

I think I learned about pressure and temperature only in my chem class, and not at all in my physics class

(high school)

Ted Shifrin

At college level, DogAteMy, both chem and phys teach thermodynamics.

Akiva Weinberger

Physics was drawing force lines on the thing and seeing how fast the cylinder will roll down the mountain

Ted Shifrin

I took the p chem version, but my friend in Athens has been teaching the physics course the last years.

There's also E&M and quantum, DogAteMy.

Joe Stavitsky

00:39

@AkivaWeinberger, I'm not actually "at" this school, I just "work" here, so I don't know what goes on in the classes. But these people seem scary ignorant of gas laws for people who use vacuum pumps on the regular

Akiva Weinberger

@TedShifrin Oh, yeah, I didn't take those, but a bunch of people did

Well, E&M, not quantum

So I still think electricity is magic

Ted Shifrin

high school E&M is a waste of time. You need multivariable calculus to do it justice.

That said, a lot of college courses do it garbage, too.

Akiva Weinberger

Why would you need multivariable calculus to figure out what goes on in a loop of wire

Ted Shifrin

To do it "right" you need line integrals, surface integrals, div, curl, etc.

Akiva Weinberger

Holy hell

Ted Shifrin

00:42

Don't swear at me.

Joe Stavitsky

@TedShifrin, strongly agree

Akiva Weinberger

I won't swear again, I swear to thee

(/s)

Ted Shifrin

Thank you, DogAteMy.

Joe Stavitsky

@TedShifrin, I had to take that course before calc 3, now I will have to do a lot of work by myself.

Akiva Weinberger

Fuckin' sweet

Joe Stavitsky

00:44

OK guys, gtg, c u tomorrow

Akiva Weinberger

See ya

Ted Shifrin

Should I leave?

Akiva Weinberger

Unless you have what to talk about

Ted Shifrin

I ain't doin' no talkin'.

Akiva Weinberger

No, Mr. Ted, I expect you to sci

is what I would say if we were in the h bar

Joe Stavitsky

00:48

@AkivaWeinberger excellent

Nicholas Roberts

Hi all. Question, any hints or steps in the right direction would be appreciated. We have the polynomial $x^3 -3x + 1$. The splitting field for this polynomial is $\mathbb{Q}(\theta)$ where $\theta$ is a root of the polynomial. This spitting field is a degree 3 extension over $\mathbb{Q}$. Now it says that the other two roots of the polynomial can be expressed as $a + b\theta + c\theta^2$ for some $a, b, c \in \mathbb{Q}$. Determine the other roots explicitly in terms of $\theta$.

Was thinking of using vietas formulas to try and get a system of linear equations but it got really ugly. Want to see if anyone knows of an easier/more elegant approach.

Ted Shifrin

@NicholasRoberts: So do you know about the resolvent?

Nicholas Roberts

LaGrange resolvent? Barely.

Ted Shifrin

So what have you learned?

Nicholas Roberts

We used a lagrange resolvent in a few proofs in class but thats pretty much it. Should I look more into it?

Ted Shifrin

00:58

Well, when is the splitting field of degree 3 rather than of degree 6?

Nicholas Roberts

If the galois group is the cyclic group of 3 elements

?

Ted Shifrin

And what's the condition for that in terms of the coefficients of $x^3+px+q$?

Nicholas Roberts

The discrimnant need be a perfect square in Q

In this case, it is. The discriminant is 81

Ted Shifrin

Aha.

So can't you get the other roots if you know that discriminant?

Nicholas Roberts

Hm, not sure how to go about that. The disriminant is an expression that is squares of the differences of the roots.

Ted Shifrin

01:02

But you can relate the square root of it to the other roots if you know one of 'em.

Nicholas Roberts

Ah, ok :)

Well, we abstractly know a root is theta.

Is that what you mean when you say we know one of them?

Ted Shifrin

That is, if you know $\alpha_1$ and $\delta$, you can write down $\alpha_2$ and $\alpha_3$ in terms of them and $p$ and $q$.

Right.

Nicholas Roberts

Ok, will give this a shot! Thanks

Ted Shifrin

Most welcome.

Samuel Yusim

02:22

how do I read jacob lurie's ridiculous books without going nuts

in the foreword of his higher algebra book he says "prerequisites should be minimal apart from my other book" but his other book is 1000 pages

and the higher algebra book is 1500

so like, huh?

Mike Miller

02:52

@SamuelYusim i think you don't, you pick up ideas from other sources (or try to get a feel for the yoga of doing things sort-of-simplicially) and use it as a reference

but i am not a category

Samuel Yusim

arg math is so hard

Mike Miller

yeah

Samuel Yusim

every time I look at this stuff I feel like I've made a wrong turn

why are higher category people such heirophants

Mike Miller

i would be glad to talk more about that elsewhere, email in profile if you want to; i think such convos often turn relatively quickly into things that can be hurtful to the people who like that kind of math

best to avoid hurting feelings

i think that once you get your teeth sufficiently into (some relatively concrete and precise application of this) it's easier to get a feel for what's going on

when someone tries to prove something in homological algebra the proof ends up looking like diagram chasing but people think about things more subtly; they have mental models for the things they're working with and the operations they use on them

true everywhere, i think

Samuel Yusim

03:09

yeah, you're right

Mike Miller

(but i sympathize with you)

i have an MO question asking why anybody bothers with this stuff

that might help

if your experience is like mine when i posted it, most of most of the answers will be opaque and unhelpful but there might be some kernels that you like. as i go back later i agree with more of it (and still disagree with some

Samuel Yusim

I've just landed on it from a pretty unexpected direction so it's confusing to see so many people saying the thing I care about is higher-categorical even though nobody's saying how

Mike Miller

yeah that's fair

Samuel Yusim

like there's a page on the nlab that gives a bunch of related articles that just 100% don't seem related

Mike Miller

do email me, i'm glad to bullshit arbitrarily much about this

Secret

03:15

AI learns to play Google Chrome Dinosaur Game || Can you beat it??

►

Samuel Yusim

03:37

@MikeMiller I sent you an email

Mike Miller

My email has this irritating send-receive delay so it'll be a bit before I see it, but duly noted

Samuel Yusim

sure thing

0celo7

04:02

@BalarkaSen from Kirby's new paper:

Semiclassical

@0celo7 holy mother of contraction batman

0celo7

@Semiclassical it's the most impressive use of indices i've seen in a while

ah, this is not Rob Kirby, but Mike Kirby

I actually know him

Guy is doing interesting stuff with curvature of high-dim data sets

Semiclassical

it sorta seems unnecessary tbh: the first 15 terms are the ways to partition 6 objects into 2+4, and the second is the way to split 6 into 2+2+2

BAYMAX

Hi chatos

.

At least I know something about differentiability in $\Bbb{R}$
but how to extend the definition of differentiability to $\Bbb{R}^2$
?
or in general to higher dimensions like $\Bbb{R}^m$
here what will be the criteria of differentiability in $\Bbb{R}^m$ ?

0celo7

04:18

@BAYMAX en.wikipedia.org/wiki/Fr%C3%A9chet_derivative

@Semiclassical sounds right, I dunno why they're doing that

I clicked on the paper because I recognized a name

Semiclassical

hmm

BAYMAX

@0celo7 Nice! thanks :)

Semiclassical

I mean, I don't mind it being explicit. just seems strange

skull

04:42

(removed)

Sharath Zotis

06:02

Hello

Is this claim true? If T(T(V)) is a zero transformation then T(V) is a zero transformation as well where V is a finite vector space and T is a linear operator.

Tobias Kildetoft

@SharathZotis $T(V)$ is a vector space, not a transformation. But if you mean whether $T^2 = 0$ implies $T = 0$ then the answer is no (think of strictly upper triangular $2\times 2$ matrices).

Isa

06:22

Hello, Is this true? Symmetric BC for a pair of functions X and Y have the property of $XX'|_a^b\le 0$ ?

feynhat

07:43

hello

Tobias Kildetoft

hi

feynhat

Let $F$ be a set of bounded, continuous, real-valued functions on a compact space X. If $F_x = \{f(x): f \in F\}$ is bounded (wrt. the usual metric in $\mathbb{R}$) for each $x \in X$, then can we say that $F$ is bounded wrt. sup-metric?

I think its true because, if $M_x = \sup F_x$, and $M = \sup_{x\in X} M_x$, then $\sup |f| \leq M \ \ \forall f \in F$.

Any thoughts?

Alessandro Codenotti

08:11

The crucial step is to justify why $M$ is finite, which you haven't done

feynhat

08:22

@AlessandroCodenotti Each of $M_x$ is finite.

hmm...

But that doesn't necessarily imply $M$ will be finite.

How do I do this?

Astyx

You use the fact that your space is compact

feynhat

08:37

Oh. $X$ is compact, then so is $f(X)$ for each $f \in F$. But $f(X)$ is in $\mathbb{R}$ so it is bounded...

I still don't understand how this will imply $M$ is finite.

@Astyx

Astyx

What does it mean that $M$ isn't finite ?

Wait actually I'm starting to doubt

$M$ doesn't have to be finite

mb

Exercise : prove it

feynhat

Wait are you saying that $F$ may not be bounded?

Astyx

Yeah

feynhat

Oh no.

Mary Star

09:00

Hello!! Could you give me a hint how we could check if the set in $\mathbb{R}^2$
$S=\{x\in \mathbb{R}^2: 1\leq x_1^2+x_2^2\leq 4\}$ is connected?

Tobias Kildetoft

@MaryStar First, draw it to arrive at a guess

feynhat

@MaryStar Or you could define a map $f: [1, 2]\times[0, 2\pi] \to S$ as $f(r, \theta) = (r\cos\theta, r\sin\theta)$. $f$ is continuous and $[1, 2]\times[0, 2\pi]$ is connected.

Mary Star

@TobiasKildetoft We have the following:

This set is connected if the only open subsets is the whole set, isn't it? @TobiasKildetoft

@feynhat How do we know that $[1, 2]\times[0, 2\pi]$ is connected?

Tobias Kildetoft

no

feynhat

@MaryStar Product of connected spaces is connected.

Mary Star

09:14

Ah.. Isn't the definition the following?
Let U and V be open sets with $U \cup V= A$ and $U\cap V=\emptyset$, then A is connected iff A=U or A=V? @TobiasKildetoft

Tobias Kildetoft

@MaryStar Yes, and that is not what you wrote

Mary Star

@feynhat Ah, and each interval is connected because a close interval is connected?

feynhat

@MaryStar Yes. A subset of $\mathbb{R}$ is connected, if and only if it is an interval.

It may be closed, open, semi-open, unbounded, ... whatever.

But it should be an interval.

Tobias Kildetoft

@feynhat You meant connected, not closed

feynhat

@TobiasKildetoft yeah

Mary Star

09:24

@TobiasKildetoft Can we check if the given set is connected with this definition? How?

@feynhat So, do we have to wite always the given set as a map and check if the domain is connected?

feynhat

@MaryStar That's one way, yes.

Otherwise, you could show that every continuous function from $S$ to the discrete 2-point space $\{0, 1\}$ is constant. But, I am not sure how this method will be helpful in your problem.

Mary Star

Ok! For example, if consider the set $M=\{x\in \mathbb{R}^2 : x_2\cos x_1=\sin x_1\}$, what function can we consider here? @feynhat

feynhat

@MaryStar wait. Isn't this graph of $y = \tan(x)$?

Graphs of continuous functions are connected in $\mathbb{R}^2$

oh wait, $\tan(x)$ is not continuous, on $\mathbb{R}$.

Then, M may not be connected.

3

Q: Is the graph of a continuous function from a connected space also connected?

Let X be connected and $f:X \rightarrow Y$ continuous. Is the graph $G$ of this function connected? My thought is yes, which seems pretty intuitively clear. For contradiction, assume $G$ is not connected and $G= U \cup V$. I pick $(x,y)$ in $G$, so WLOG assume $(x,y)$ is in $U$. But since $U=A...

general-topology

@MaryStar ^

@MaryStar $M$ is not connected. But if you restrict it over $(-\pi/2, \pi/2)\times\mathbb{R}$, then it is connected.

Mary Star

09:52

Ah ok! I see!! So, M is not connected because the function tan(x) is not continuous on the whole R, right? @feynhat

feynhat

Yes.

Mary Star

At the set $A=\{x\in \mathbb{R}^2 : x_1^2-x_2^2=1, x_1, x_2>0\}$, do we consider the map $f(\theta)=(\cosh(\theta), \sinh(\theta))$ ? We have that $f:(0,\infty) \to A$, or not? The domain is an open interval, which is a subset of $\mathbb{R}$ and so it is connected. This the function is continuous, it follows that A is connected.

Is this correct? Have I understood correctly the way? @feynhat

feynhat

@MaryStar Yeah. It looks fine.

Mary Star

10:07

Great!!

One last set, at which I don't know if we have to find a function.. $B=\{x\in \mathbb{R}^2 : x_1 \text{ or } x_2 \in \mathbb{Q}\}$.
What do we have to so in this case? @feynhat

feynhat

@MaryStar $B = \mathbb{Q}\times\mathbb{R} \cup \mathbb{R}\times\mathbb{Q}$

Alex

@TobiasKildetoft Are you familiar with crystal bases and/or B(\infty)?

feynhat

@MaryStar $\mathbb{Q}$ is not connected. Write each $\mathbb{Q}$ in the above expression as a disconnection, say, $\mathbb{Q} = ((-\infty, \sqrt{2}) \cap \mathbb{Q})\cup((\sqrt{2}, \infty)\cap\mathbb{Q})$. Then you'll have a disconnection for $B$.

Willtech

10:33

Afternoon all, would this question be any good over here?

Mary Star

10:44

@feynhat Can we write $\mathbb{Q} = ((-\infty, \sqrt{2}) \cap \mathbb{Q})\cup((\sqrt{2}, \infty)\cap\mathbb{Q})$ because $\sqrt{2}$ is not a rational number?

Astyx

11:14

Yeah but you can't do that with $\Bbb R$

feynhat

11:50

@AlessandroCodenotti Will I have to use the fact that a bounded function on compact set does attain its extremities?

Shobhit

@TedShifrin need your help here : chat.stackexchange.com/rooms/76739/…

feynhat

Or do I have to use Zorn's lemma...

pilko

12:15

Hello, what is the english translation for "polynôme scindé sur K" i.e. the polynomial has all its roots in K. I cannot seem to find an equivalent for "scindé", can I say "split" ?

Balarka Sen

@pilko The polynomial splits over K, yes

user193319

Problem: Suppose that $X$ is a Banach space with norm. Let $X_0$ be a dense subspace of $X$. Assume that $X_0$, when normed by the norm it inherits from $X$, is also a Banach space. Prove that $X = X_0$. Proof: Let $x \in X$. Since $X_0$ is dense, there exists a sequence $\{x_n\} \subseteq X_0$ converging to $x$. This means that $\{x_n\}$ is Cauchy. But it must also be Cauchy in $X_0$, since $x_n - x_m \in X_0$ for all $n,m$, because $X_0$ is a subspace.

Hence, $\{x_n\}$ converges to a point in $X_0$. But, as $X$ is Hausdorff, these two limits must be the same, which shows that $x \in X_0$.

How does that sound?

pilko

@BalarkaSen thanks.

Mary Star

12:40

@Astyx We can't do that with R, because then we would loose a number, or not?

Silent

@feynhat, why is $1+1+1+1+4$ not class equation for group of order $8$?

sorry, got it.

Prototank

13:11

The ring $R=\mathbb{R}[X,Y]/(X^2+Y^2-1)$ is generated by the elements $X$ and $Y$, right?

because $X\cdot X+Y\cdot Y\in (X,Y)$

Akiva Weinberger

@Prototank Yes

Wait, actually, I'm not sure

$\Bbb Q[X,Y]/(X^2+Y^2-1)$ is generated by $X$ and $Y$

but I'm not sure if that's still true if we change $\Bbb Q$ to $\Bbb R$

As an $\Bbb R$-module rather than just a ring, it should be true

(or, as an $\Bbb R$-vector space)

@Prototank

Prototank

I think that the set $\lbrace rX+sY: r,s\in R\rbrace = R$

pick your favorite $a\in R$, then choosing $r=aX$ and $s=aY$ gives that $a$ is in the left hand side.

Akiva Weinberger

13:29

$aX^2+aY^2=a(X^2+Y^2)\equiv a(1)=a$

Yeah, that works, cool

Abra001

13:56

the youtube channel 3blue1brown is the best i'v known in dedication for offeriing visualised understanding of mathematical facts.

vanaghka

how do I show that $l^1 \subset l^2$?

Szabolcs

I'm looking for advice on this:

0

Q: Terminology: customary name of graph "smoothing"

The Wikipedia page on graph homeomorphism describes an opposite operation of edge subdivision that it calls smoothing. What is the customary name of this operation in mathematics? I have seen the term subdivision used in many places, but I have not seen the term smoothing anywhere else than on ...

graph-theory terminology

As well as on the following: homeomorphism is defined as: two graphs are homeomorphic is they have isomorphic subdivisions. I believe it is also true than that they have isomorphic topological minors. Is that not the case? I cannot see why not, but than I don't see why the definition isn't ever stated like this.

Akiva Weinberger

14:20

@Abra001 So much this

agaitaarino

In case someone over here cares about these things, there's a bounty on this tricky question:

10

Q: What is the minimum integer value to make quantum factorization to be worthwhile?

Let us assume that we have quantum and classical computers such that, experimentally, each elementary logical operation of mathematical factorization is equally time-costing in classical and in quantum factorization: Which is the lowest integer value for which the quantum proceeding is faster tha...

quantum-computer quantum-algorithms quantum-speedup

Sha Vuklia

@Akiva mind if I ask you something about manifolds?

Akiva Weinberger

Sure

Sha Vuklia

could you explain why it doesn't go wrong in this definition if our coordinate patches overlap?

I personally thought we would need to restrict our coordinate patches, such that they are disjoint

but apparently that's not necessary

I know that later on they will show that the integral doesn't depend on the choice of coordinate patches, so in a sense that's an answer

but I was looking for a more direct explanation (is it because of the partition of unity?), even if it's only on the intuitive level

GFauxPas

What's an intuition for all the nice properties of symmetric maps in a finite dimensional space? I've seen proofs, but I'm looking for an intution. Symmetric real is diagonalizable, orthogonal eigenbases iff wholly real eigenvalues, commuting iff simutaneously diagonalizable.

Akiva Weinberger

14:33

What do they mean by "dominated"? @ShaVuklia

Sha Vuklia

That each support of $\phi$ is contained in a coordinate patch

oh

that's the answer then I guess

so the overlapping parts are sort of 'extracted' separately

loch

The point is they sum to 1

Sha Vuklia

do you think that's the main point of the partition of unity here? Enabling us to work with overlapping coordinate patches?

or is there some other thing I'm looking over

loch

Yeah - and also the fact that they are supported in a coordinate patch so working with each $\phi_if$ is the same as working on $\R^n$ - where you know how to do calculus

Akiva Weinberger

I'm back

At any point $p$, $\sum\phi_t(p )=1$

so $\sum\phi_t(p )f(p )=f(p )$

Prototank

14:46

partitions of unity are magic

Akiva Weinberger

so $\sum\phi_tf=f$

Prototank

studying for my differential topology comp

loch

I only realised how important they are when I'm doing alg geom / complex geom where you no longer have partition of unity..

Jannik Pitt

15:50

If $E$ is a vector space over $K$ and $L(K,E)$ is the set of all bounded operators from $K$ to $E$, why is the map $T: A \mapsto A1$ injective? This is a really dumb question but I really don't see why.

Jake Rose

16:06

Express $sin(\frac(pi)_(12))^2$ in terms of $sin(\frac(pi)_(6)$

Having troubles with this one

Thanks

Prototank

you need $\sin(\frac{\pi}{12})^2$ in terms of $\sin(\frac{\pi}{6})$

Jake Rose

Yeah thanks

Im always struggling with latex

Prototank

so fraction works like this: \frac{numerator}{denominator}

Jake Rose

Ohhhhh

Prototank

special symbols like sin and cos are simply just \sin and \cos

Jake Rose

16:09

I was putting {}_{}

Prototank

you were doing ()_()

Jake Rose

Makes more sense

I changed it to that when it originally didnt work

Prototank

even more egregious

haha

to answer your question, I think it might help you to search for double angle identities

Jake Rose

I used the sin double angle formula to begin with

But im ending up with a $4\cos(\frac{\pi}{12})^2$

On the denominator

Prototank

I think you just want to apply

Jake Rose

16:14

I think ive got it

Just saw a different step I could have taken near the end

But gimme a sec lemme check

Prototank

nice!

Jake Rose

Know anyway I can check this?

Oh wait

A calculator

I forget we have them for maths

They unfortunately did not come out as equal

Ah I see my mistake

I cant see how this will work anymore

Any help is appreciated

vanaghka

bit of an odd question, but I'm wondering does anyone recognise the book that this link is taken from? ma.huji.ac.il/~razk/iWeb/My_Site/Teaching_files/Banach.pdf

Alessandro Codenotti

16:30

here is chapter one and here is chapter three, they were on the author's website

vanaghka

@AlessandroCodenotti thank you

GFauxPas

16:57

@Prototank write $2 \frac {\pi}{12} = \frac \pi 6$

Prototank

@JakeRose

GFauxPas

oh oops i misread who asked the question sorry

Prototank

I was playing with it for a bit, but I don't think there's a nice way to get what he was looking for

GFauxPas

hm, lets see

Prototank

I think the question should have one of them being sin the other cos

as @JakeRose discovered

but yeah absolutely you split like that and apply double angle

but that was the end of my attention as far as that goes

GFauxPas

17:02

yeah its fine

I hthink

you can write $4 \sin^2 \frac \pi 6 \cos^2 \frac \pi 6$ in terms of whatever

if I have a bijection from a set $X$ with $+$ and scalar multiplication into a vector space $V$ and I prove the morphism property, is that enough to prove $X$ is a vector space?

$f(x+_X y) = f(x) +_V f(y)$?

Twink

17:23

to solve the equation $(x^2-3x+3)^{(x^2-7x+12)}=1$ we just need to solve the equations $x^2-3x+3=1$ and $x^2-7x+12=0$?

Balarka Sen

"or" not "and".

Captain Bohemian

@robjohn excuse me, how to solve the problem that MathJax can't be installed when I use Microsoft Edge?

robjohn

@CaptainBohemian I don't know much about MS Edge, but can you install javascript links in the toolbar?

Prototank

@GFauxPas You need to show that $X$ is an $F-$module for some field $F$, right?

Twink

why or?

if I need to find all the solutions

Prototank

17:29

@GFauxPas then I think you are done if $X$ itself is an abelian group. I don't think $X$ itself needs scalar multiplication. You can *define* scalar multiplication of $X$ to be simply this:

$\alpha\cdot x:=\alpha f(x)$ for any $\alpha$ in the field attached to $V$

Silent

@0celo7, The class equation of a group $G$ is $1+4+5+5+5$. I have to show that centralizer with $4$ element conjugacy class, which has order 5, is normal subgroup. The proof I have goes on proving that there is only one subgroup of $G$ that has order $5$, so it has to be normal. I can't understand this last claim.

GFauxPas

Yes proto, specifically a unitary module

Ah nice, good point about not needing explicit scalar multiplication

@Twink you should consider cases where the base is negative one and the power is an even integer, though i dont know if that's possible in your case

Prototank

prove a group is a semidirect product: normal brain
find all isomorphism classes of [some product of primes]: big brain
find all characters $G\to \mathbb{C}^\times$: G A L A X Y B R A I N

Twink

17:45

@GFauxPas and that's all I have to consider?

I mean for excercises like this

could be another equation

skull

Welcome back @robjohn

GFauxPas

Well 1 only has two roots for any power if you're only looking at real roots

And if you're working with real numbers only then (-1)^x is only well defined for integer powers

So for real solutions, you have three cases

$1^x$ for any $x$

$(-1)^x$ for $x$ an integer

Twink

then all the soltuons of $(x^2-9x+19)^{(x^3-3x^2+2x)}=1$ are 1,2,3,4,5 and 6?

GFauxPas

And $x^0$ for any $x$

Twink

I don't wanna miss any solution

GFauxPas

17:52

How'd you get those

Captain Bohemian

@robjohn I just read the Wikipedia for javascript, but it doesn't give the link of javascript. so what is the link of javascript?

GFauxPas

I didn't work out the problem I just looked at it

Twink

factoring

I considered $x^2-9x+19=1$, $x^2-9x+19=-1$ and $x^3-3x^2+2x=0$

GFauxPas

Does the second one have solutions?

Twink

when $x^2-9x+19=-1$ the exponent is even

yes

it's the equation $x^2-9x+20=0$

$(x-4)(x-5)=0$

GFauxPas

17:56

And 5 and 4 give an even integer in the power, good :)

That's all

Twink

yes

so the solutions are from 1 to 6

GFauxPas

You missed one , the power has 3 roots

Set the power = 0

Anonymous

18:16

Hi. Does anyone here happen to know what exactly "open edges" and "closed edges" mean in the context of graph theory? Also, "open cluster" and "closed cluster"

Jannik Pitt

18:32

Asked this a while back, maybe someone can help me:
If E
is a vector space over K and L(K,E) is the set of all bounded operators from K to E, why is the map T:A↦A1 injective? This is a really dumb question but I really don't see why.

Alessandro Codenotti

18:50

What are A and A1?

Jannik Pitt

T is a map from L(K,E) to E and A is an element of L(K,E)

Tanuj

Can someone help me with a bit of chemistry , the chemistry room seems dead currently

Waiting

19:08

What is the MSE news today?

(well, maybe there is no news)

Leaky Nun

19:30

@AkivaWeinberger if you still remember the degree problem (degree of map of unit quarernion q to q^n), I chose the wrong point. once you pick a different point I think you can show that the degree is n

it was a month ago (search for “quaternion”)

0celo7

@Silent sorry I have no idea what a class equation is

Waiting

Time to go.

(out)

Sha Vuklia

19:52

@Balarka would you mind checking an argument about volume elements with me?

DarkRunner

21:17

Hi, I have a question:

Does the inradius of a triangle always lie on it's altitude, since they're both perpendicular to the opposite side?

/

Corellian

21:35

Are $\mathbb{Z}_n$ and $\mathbb{Z}/n\mathbb{Z}$ essentially the same?

@DarkRunner No. Generally, they can be parallel.

Leaky Nun

21:56

@Corellian they're just different notations

but some use the former for a different object, so that's a caveat

@Silent because the conjugate of any subgroup is also a subgroup. if there is only one subgroup, then any conjugate gives you the same subgroup

@AlessandroCodenotti buongiorno

Dattier

Salut Gérard n'est pas là ?

Mais je vois qu'il y a Anna, comment vas-tu ?

alors tu penses les avoir les 2 médailles ?

Salut Edward

toi tu les as eu les 2 médailles..lol

Abra001

@DarkRunner only if the triangle is isocele

for them all to lie on the altitudes, the triangle should be equilateral.

Dattier

Bon aller quand vous aurez fini vos gamineries on pourra parler comme des grands

tchuss

Lil

22:25

Anyone available for a combinatorics question?

Abra001

dont ask to ask, just ask

Xander Henderson

23:03

I mean, I'm available, and there are one or two combinatorics questions that I could probably answer, but the combination of the two, well, that's just too much.

user193319

23:53

Problem: Show that the collection of sets of natural numbers is uncountable and conclude that $\ell^\infty$ is not separable...The hint is oddly phrased. Sure, the powerset $\mathcal{P}(\Bbb{N})$ is uncountable, but how am I to represent sets in $\mathcal{P}(\Bbb{N})$ as points in $\ell^\infty$?

I mean, I can't just look at the set $\{(x_n) \mid x_n \in \Bbb{N}\}$, since this contains unbounded sequences...so I don't really know what the hint means...

Anyone?

Transcript for

$Mathematics$ Mathematics