Hi @Ted (even though I guess you're not really here)

No, I'm not here.

Thought so

Oh.

@nbro: I haven't read more than the first few sentences of your question. If you're working in $\Bbb R^3$, you need to say so explicitly.

@TedShifrin Opps. Yeah, I'm going to explicitly say that.

Hi @Ted

@YOU: You must work in radians, not in degrees. But $\pi/2$ is wrong, regardless. It should be $\pi/6$.

Hi @Balarka

@TedShifrin Edited.

how 30 become -pie/2

?

Wait, @YOU. You're confusing two different problems. $-\pi/2$ is the answer to $\sin^{-1}(-1)$.

Yes

Question is Sin(inverse) (-1)

And answer in key is: -pie/2

but my answer is 30

?

No, you're working the wrong problem.

What do you mean?

$\theta=-\pi/2$ is the UNIQUE angle $\theta$ between $-\pi/2$ and $\pi/2$ with $\sin\theta = -1$.

Where are you getting $\cos\theta = \sqrt3/2$?

OPPS!!!!!!

I think i am posting two different questions

in same solution

I am sorry.

Well, and the answer book's correct answer is for the first problem, not yours.

Hey

\('-')/

The trick here is degrees vs radians

I answered that the first time, @YOU. I already told you you need to use radians, not degrees.

gets tired of this

Yes how can i can i convert it?

30 to pie/6

If you have degrees, you multiply by $\frac{π}{180}$

frac?

pi/180

π/180

what is pi in degrees?

Sniped @Balarka

pi radians =180 degree

@You Ok. So what is pi/6 in degrees?

Ummm

180 ÷ 6 = 30

so 180 = pie

So pi/6 in radians = 30 degrees.

Also, please write pi and not pie. The latter is a different object :P

@nbro: You're working way too hard for your question and I don't have the patience to read it. Forget about $\cos\theta$. The point is that $\|n\times v\|$ is always the length of the projection of $v$ onto the plane with normal $n$. You can see this by using $\|n\times v\| = \|n\|\|v\|\sin\theta = \|\v\|\sin\theta$, or you can write $v=v_1+v_2$, where $v_1$ is parallel to $n$ and $v_2$ is orthogonal to $n$, and note that $n\times v = n\times (v_1+v_2)=n\times v_2$

and so the magnitude is just $\|v_2\|$.

Great!

I am stuck on this from last 8 hours

o/ @Dami

now i know how to convert them

@Astyx how goes it?

I'm good and you ?

But what if answer is 45?

I'm good, thanks!

180 ÷ 45 = 4

but answer in keybook is 1/√2

?

@Astyx, Demonark: Any thoughts on this question? Maybe I missed something.

how?

any thoughts on this question?

?

Hi chat

answer my question alessandro

did you work out that squares mod p thing? @Dami

see above

Hi Alessandro

hello @AlessandroCodenotti

Hi Alessandro

@TedShifrin Lol, I knew I was trying to prove it in a hard way. Btw, I had already noticed that $\| n \times v \| = \| n \| \| v \| \sin \theta = \| v \| \sin \theta$, the thing is that I wasn't sure that $\| v \| \sin \theta$ was the scalar projection of $v$ onto the plane... Btw, you're not forced to read it or help me, clearly.

Draw the picture, @nbro. When you write $v=v_1+v_2$ as I said, you'll see that $\|v_2\| = \|v\|\sin\theta$ immediately.

@TedShifrin I actually drew the picture and I was still not convinced...

How so?

$v$ is the hypotenuse, and $v_1$ and $v_2$ are the legs, $v_1$ along the normal, $v_2$ in the orthogonal plane.

I don't know

@Ted Your answer seems good to me. I guess one can't say much when $V$ is not Banach, maybe one could find a counterexample (I can't) ?

@AlessandroCodenotti not quite

@TedShifrin Yea, but, for some reason, in that moment I may have had a blackout ...

I didn't know this property of linear operators concerning their image being closed, that's nice

LOL, @nbro. I leave you to resuscitate.

Clearly the author intended finite-dimensional in this question.

Anyway, mine, if correct, is an alternative proof. Let's look at the bright side of it..

And yeah @Ted it was definitely meant to be finite, too many considerations of Hilbertness and Axler doesn't often deal with them to my understanding

If at all

@Daminark What's this about squares you're talking about ?

Right, Demonark. I've never much liked the book, anyhow.

I mean yeah, it's bizarre

I'm not terribly fond of the whole, every field is R/C

Also I've heard it misses some important content

@Astyx finding which prime numbers are also prime in C

All linear algebra should be Hoffman and Kunze

:P

In $\Bbb C$ ?

Yup

@TedShifrin Thanks!

You mean $\{a+ib, a,b \in \Bbb Z\}$ ?

Yeah

Right

hi chat

Hai

hi @semi

hi @Semiclassical

So @Dami you know that squares mod $p$ satisfy $a^{\frac{p-1}{2}}=1$. Now show that if an element of $\Bbb F_p$ satisfies that then it must be a square

Suppose that $A$ and $B$ are normal in some group $G$. Is it true that $\langle A \cup B \rangle$ is normal in $G$? If so, I could use a hint on how to prove it. If not, is in possible to find a counterexample in the symmetric group?

So here's what I was going for, if two numbers a and b between 1 and (p-1)/2 have the same square mod p, then p divides a+b or a-b

But a+b is positive and less than p

Hi Semi

So then a-b=0

Never mind. I found this post: math.stackexchange.com/questions/1449104/…

Wait actually I'm not sure what reservations I had now that I think about it

How do I make my text be one the right after an equation on Latex ?

Not sure I'm understanding

Do you mean like: $$a^2+b^2=c^2 \tag{Pythagorean Theorem}$$

Yes

\tag

Thanks

@Astyx been thinking about it. If we hold that $V$ is hilbert space, then the statement is true for the infinite dimension case too. Managed to prove it :)

Oh cool

What's Hilbert again ?

inner product space?

...That is complete with respect to the metric induced by the inner product.

Ah right

And if you're Kolmogorov-Fomin, you define them to be separable

@Damin that really narrows them

You can have orthonormal base even if they're not separable after all

Oh yeah it totally does, everything becomes l2 if you ask for separability

Yeah I read that somewhere, haven't reached the proof yet..

Probably later in the course

Yeah

Wait hold on

To have a Schauder basis in a Banach space you need separability

Salut @Vrouvrou

So actually if you have an orthonormal basis in a Hilbert space, this is equivalent to separability, which is equivalent to being l2

salut @Astyx

, s'il te plait si j'ai la distance $d(f,g)=\int_0^1 |f(x)-g(x)|dx$ comment trouver la boule ouverte B ?

Oh yeah, that's true.

My mistake

Laquelle ? @Vrouvrou

We can have a non-complete separable space though, right?

With countable orthonormal system

@Astyx n'importe quel ouvert $B_{d}(f_0,r)$

Oh I meant Hilbert

C'est l'ensemble des fonctions $g$ telles que $d(f_0, g)\lt r$, il n'y a pas de caractérisation plus simple a priori

@Ted Who's Adler?

@Astyx $B_{d}(f_0,r)=\{g\in E, \int_{0}^1 |f_0(t)-g(t)|dt <r\}$ je ne peux pas simplifier ?

Oops. Axler. Typo @PVAL

It seems I'm sick again.

Of course you are, @Balarka. I'm finally healthy.

lol oops

Did you give it to me?

Jokes apart, I'm glad you're better.

:D

After so long

Let's hope it doesn't immediately reflare

Demonark: Easy Jumble again. "The college tennis star planned to join the army and was ..." (5/2/5) PERVSOTREDOU

(I suspect the 40C temperature is to blame in this case though)

@Ted Good to know your vertebrae are all in alignment (since that is of course the root of all sickness)

Je ne crois pas @Vrouvrou

NO, PVAL, they most definitely are not.

My hips will hurt all through my trip.

@TedShifrin Glad to hear it !

I'm getting over my cold

Good to know. I got sore throat this morning and now I have fever.

I got an email today saying that at the upcoming graduation ceremony, "umbrellas are not allowed for security reasons".

presses F to pay respects

@MikeM: Allergies are crazy this year, apparently. I finally gave in and bought zyrtec.

Umbrellas could be guns, PVAL.

Lol @PVAL

Especially if, like at Notre Dame, Pence is the speaker.

@Ted AFIK You are allowed to carry guns.

At a coliseum filled for graduation?

Outdoors in front of the tower in the middle of campus

It would seem to contradict state laws.

Well, I guess it won't rain.

If I can't ban them in my classes, I don't see how they can be banned in a graduation ceremony.

thinks Astyx and Demonark are awfully slow on this Jumble

@Astyx merci

proud to serve?

Wait it's proud to serve

Balarka isn't too sick ... :)

Dammit Balarka, sniped

:P

OK, back to math.

Hi chat

Heya Eric

@Ted I just got my final Riemannian problem set :(

initiates process of turning this room into one of number theory and commutative algebra

Why so sad, Eric?

Never, Demonark.

@Daminark nah

Cause it means the course is finishing

:P

It was a fun class

"nothing gold can stay..."

I still need to do the homework for Ciprian's class this quarter

Other than my reading course with him I think it's been the most fun course I've taken in math

Well, that's great, Eric. But you still have zillions of my problems to do, so you shouldn't wallow in too much sorrow :P

This is true

The beginning of Summer will be fun :D

Does anyone here know how to prove that $\pi/4\pm\log 2$ is irrational?

Is that even known?

Well, I can think of trying $e^{it} = \sqrt2(1+i)$.

Hermite-Lindemann will tell us then that since $e^{it}$ is algebraic, $it$ must be transcendental.

I suspect that proves it's irrational.

But I doubt that's legal for this question.

If it's rational then e^(Pi/4) * 2 = e^r for some rational r. So e^(Pi/4 - r) is rational, no? That cannot happen, by a well known theorem, I am sure

t is what?

$t=\pi/4+\log 2$ in my case, PVAL.

You're bouncing around the same thing I did, Balarka.

Then e^it isn't what you say it is,

It's argument is $\pi/4 +\log 2$

whatever the hell that is

it's probably Gelfond-Schneider theorem

Well its implied by the same theorem @Ted quoted

So @Balarka 's way seems to work.

Oh yeah fair enough. I forgot all the relevant theorems.

What, @PVAL? $e^{\log 2} = 2$, multiply by $\frac{1+i}{\sqrt2}$.

Oh, rats.

I see.

Crazy question for an elementary class, irregardless.

@Ted: Did I tell you I explicitly worked out the details of Stokes' theorem after we talked?

I'm starring your use of irregardless

for your shame @ted

I did it on porpoise, of course. My math shame is far worse.

So if you throw in a i into @Balarka's argument

No, @Balarka, but I'm not too surprised. Of course you have to get from cubes to manifolds with boundary ... or corners.

@TedShifrin True. The point is to, by a partition of unity, boil it down to a differential form compactly supported on an open subset of the cube. If it's an interior chart, it's just 0. If it's the boundary chart it's what you get by integrating on the boundary face modulo signs

omg

watch out

Actually I still don't see how to do it.

@TedShifrin I was eating :(

Hey there

what does x (down arrow) a mean?

Does it mean $x\to a^+$ or $x\to a^-$?

context

limit from right/left, I'd guess, but context is needed.

its is in context of limits

As an example, consider the sign function i.e. equals +1 if x>0 and -1 if x<0.

If the arrow points down

it means $x$ is decreasing

so the right limit

If I approach zero from the right, my values of x are all positive and I'll get a limit of 1 as $x\to 0^{+}$

If I approach from the left, all my values of $x$ are negative and I'll instead get -1 as $x\to 0^{-}$.

Since they disagree, the limit as $x\to 0$ doesn't exist.

heres a link to my problem: imgur.com/a/G42sa

it seems to be doing the opposite to what you guys are saying

unless im missing something

$$\lim_{x \substack{\to\\>} a} = \lim_{x \downarrow a} = \lim_{x \searrow a} = \lim_{x \to a^+}$$

@PVAL-inactive The word 'irregardless' appears in merriam-webster.com, but of course, it is marked as nonstandard and people are advised to use 'regardless' instead.

@mrnovice It seems to be doing the opposite indeed.

The way they're doing it seems very counter-intuitive, so I'm just going to specify $x\to 2^-$, thanks for the help

I was reading a 3 yr old answered question for bits I could use, and realized the first step is invalid because a restriction is missing (and a circular-logic one at that). Should I bother pointing it out?

I'd say yes

Okay, thx

It's probable no one will react though

Hullo people

(@Astyx how's undertale going?)

Hi @Daminark

Hi Dami, it's being postponed after work is done unfortunately

Sad reax only

Hopefully "after work is done" is not in too long a time

Haha, hopefully

Hi chat

I have a program I wrote some months ago, I need to make it readable for some people and comment why it's good etc

ie tedious work I don't enjoy

Darn

Readability is overrated

(proceeds to write some textbooks)

But yeah @Alessandro so yesterday we had gotten the upper bound, I think I've found the way to establish this lower bound

Wait till you want to do research and your promotor gives you the "scripts the previous phd students left behind"

You will curse them for not making their code readable :p

@Daminark how?

So if you take $a,b \in \{1,\ldots,\frac{p-1}{2}\}$, and assume $a^2 \equiv_p b^2$. Then $p \mid a^2 - b^2 = (a+b)(a-b)$

However, $0 < a+b < p$ so that can't work

Meaning, $a = b$

So you've found that their squares are all different mod p, hence the lower bound

@Astyx i find just adding line breaks at key points increases readability dramatically and is easy to do.

hm, looks fine

My biggest problem is I don't remember wether my code is finished/optimized or not, I need to rewrite it to make sure of that @heather

So this actually helps a lot, since if $p \equiv_4 1$, then the number of squares mod p is even.

Thankfully I did comment a few things so that'll make things easier for me

so now you know that there are $(p-1)/2$ squares in $\Bbb F_p$ and that an element of that field is a square iff $a^{(p-1)/2}=1$

that's enough to decide when $-1$ is a square mod $p$

I generally break up my code with line breaks and add a simple comment for each section explaining what the code is doing, which is pretty quick. i also find that variable names and function names, which should be created while coding, are important for readability. I also (@Astyx this is probably most relevant to your problem) add a header with notes like what needs to be fixed or improved or whatever.

I did that

I now need to fix/improve/whatever those things :p

the last thing i comment is little tricks i didn't know about but for google that i added in to improve efficiency or whatever - i'll add a link in a comment for where i got it, or a quick explanation for how it works.

@Astyx improve what things?

So then $-1^{\frac{p-1}{2}} = 1$ and we're good

Thanks!

I thought you were done with the code...

I am, but some things are sloppy and not quite how I'd like them to be

Anyway, I need to work

yesterday you were hoping that if $a$ is a square then $-a$ is as well and I told you that's false in general. When is it true? What happens in the other case?

This is true when $-1$ is a square, and in the other case, I think that's when you'll be a Gaussian prime

By the way I never learned the right terminology, but usually "squares" mod $p$ are called quadratic residues and $\left(\frac{a}{p}\right)=a^{(p-1)/2} \text{ mod } p$ is called the Legendre symbol and has value $1$ for residues, $-1$ for nonresidues and $0$ for $0$

@Daminark yeah, what can you say about $a$ and $-a$ in the other case?

So I've heard

And what can we say about them... regarding what?

I mean, if $-1$ is a square than $a$ is a square iff $-a$ is a square

Oh that type of statement

Then it's mutually exclusive

In a paper I'm reading, the algorithm says "set $S(\rho) = \frac{1}{|R|}\sum\limits^{|R|}_{i=1}F_i(\rho)$. Here, $R$ is a set, $\rho$ is a generic term referencing the elements of $R$, and $|\cdot|$ has been used in the paper previously to reference the length of the list inside. What exactly does it mean to have something on top of the $\sum$ symbol?

confusingly enough, the paper doesn't define $F_i$ (at least I can't seem to find it)

The something on top is usually the limit - I would read that as i = 1 to size of R

okay, thank you.

now i just need to figure out what the heck it's summing =P

I am always suspicious when they use $F_i(\rho) vs. F(\rho_i)$

Right now it looks like they have one function per element

So that if you just apply all the functions for each element to a given one and divide by the order of the set, this is what's giving you $S(\rho)$

@Daminark "all the functions for each element" - what do you mean?

Like, you have $F_1,\ldots,F_{|R|}$, so you're taking an element and just summing $F_1(\rho) + \ldots + F_{|R|}(\rho)$

@user16839 i always have so much trouble knowing what subscripts mean in various contexts.