Mathematics - 2015-05-18 (page 1 of 3)

It probably is, I just hope mine is not wrong

Wait i cant edit my msg anymore

hm

1:23 AM

time limit.

damn

didnt put absolute values properly, thought that was the right way

and forgot $+C$ :p

I just use the |key

ah

What about this one:

$\int{\frac{1}{cosx-1}dx}$

I immediately see that you should multiply by $\dfrac{\cos x +1}{\cos x +1}$

1:25 AM

Okay.

Thomas Andrews

You can always use the Weirstrass substitution. en.wikipedia.org/wiki/Tangent_half-angle_substitution

I did what Hushus suggested.

Then I split the integral into two parts

And expanded the denominator to see if I could do anything with it.

Weirstrass also works too

The only idea I have right now is to make something sin

$\int \dfrac{1}{\cos x -1} dx = \int \dfrac{\cos x +1}{\cos^2 x-1} dx = \int \dfrac{\cos x +1}{-\sin^2 x} dx = \int \dfrac{-\cos x}{\sin^2 x} dx +\int -\csc^2 x dx $

1:34 AM

Wait, what?

I might have done something totally absurd, im dead tired :P

is there anything that is weird?

$(cosx-1)(cosx-1)$ isn't $cos^{2}x-1$

$(\cos x-1) (\cos x +1) = \cos^2 x-1$

It's $cos^{2}x - 2cosx + 1$, I'm sure.

of course $2\cos\equiv 0$

1:36 AM

in the first step i multiplied the numerator and denominator by $\cos x +1 $

Oh.

Yes, I had the wrong sign.

I have a question about the later steps

ya

What happened to the numerator, to make $\int{\frac{cos x + 1}{-sin^{2}x}}$ become $\int{ \frac{-cosx}{sin^{2}x}}dx$

$\dfrac{a+b}{c} = \dfrac{a}{c} + \dfrac{b}{c}$

Oh, you split it there.

1:44 AM

yea

the negative comes from the denominator

just moved it up

I thought you had used some obscure trick.

no haha

2:31 AM

Ugh, I forgot about partial fractions

Ben Lim

@TedShifrin Hello.

3:19 AM

Hi@TedShifrin

Hi@AlexClark

user147690

@Paul No it seems I might not do my central extensions talk, but in that case there is no chance I don't do my next talk[on inverse limits of rings probably]

user147690

@Rememberme Hey Sayan

@AlexClark Thats sad

user147690

@Rememberme It is a little, but it wasn't as ready as I would have liked it anyway

@TedShifrin, you here and active?

Anyone here know the technique for solving $\sin x = \sin 2x$? I know the technique for cosines, but sine is a bit different.

3:34 AM

@Jeff try bringing the sin2x to this side and use the value for sin2x

@Rememberme There is a technique (that TedShifrin taught me a few days ago) for solving $\cos x = \cos 2x$. By letting $z=2x$, and solving $\cos x = \cos z$ as if $x$ were fixed. You get that $z=\pm x + 2k\pi$. Then, you "unsubstitute" for $z$ and you have your answer.

I want to use that method for $\sin x = \sin 2x$. But when $z=2x$, the solution isn't a clean plus-or-minus $x$.

Use the identity $\sin(2x)=2\sin x\cos x$.

@KarlKronenfeld, I get $\sin x = 2 \sin x \cos x \Rightarrow \sin x (1-2 \cos x) = 0$ and eventually that $x=k\pi, \pm \frac{\pi}{3}+2k\pi$. I'm checking the answer now. TY

@KarlKronenfeld, it checks out. I always forget about the trig identities (probably because I'm so terrible at memorization).

3:52 AM

Sorry @Jeff I was having breakfast. I was talking about the identity karl told you

Using that it becomes very easy

@Rememberme, no problem. Got what I need (and you got breakfast). Though I do still wonder if the other way will come to the same answer. I'll do that next.

Ted's approach is likely best for solving any equation $f(x)=f(g(x))$, but you can also solve $\cos x=\cos(2x)$ in similar way to what I suggested for $\sin$. Indeed, $\cos (2x)=2(\cos x)^2-1$ so $0=(2\cos x+1)(\cos x-1)$ and you're solving $\cos x=1,-1/2$

Yes thats fine@KarlKronenfeld

But ted's way is very witty

i rather like ted's way, too. it's enlightening. But @KarlKronenfeld's way is easier.

@anon I came across this question which you have answered, and I am wondering about the related question, if $I$ is an ideal for which there exists $J\ne R$ and $IJ=I$, then is $I$ idempotent? I found a counterexample when $R$ is not an integral domain, but I am unsure how it works in integral domains. Do you have anything?

abel

3:57 AM

@jeff, ted must have meant this. $\sin 2x = sin x$ implies $2x = x+2k\pi $ or $2x = \pi -x +2k\pi$ now solve these twi linear equations.

@anon Counterexample: $\mathbf F_2[x,y]/(x^2,x+y-1)$ and $I=(x)$, $J=(y)$.

@abel yeah, i'm doing that now to see if it compares with using the trig identity (it better!)

@abel ted's idea was easier for the original problem i asked him about: solve cos(x)=cos(2x).

abel

if you insist on using $\cos,$ then you can write $\sin x = \sin 2x$ as $\cos x- \pi/2 = \cos 2x - \pi/2$ and go back

@abel yeah. But i got the same answer doing it like you suggested in your message before that

abel

@jeff, good. this method breaks down when the coeffs are not the same. like $\cos x = 2\cos 2x$

4:05 AM

@abel right, i see that now.

Well any method suffices until you can give the logic why you did it and why is it true@Jeff

4:22 AM

Is it true that for open sets $G_1,G_2,......G_n$ $$\bigcup_{k=1}^n{G_k}$$

That ^ is also open

You're in a metric space @Rememberme?

Well if i am talking about some set being open then it should be in some context@Karl

There is a good reason why I ask. But anyway, you should recall/prove that if $S$ is a set with the property that for any $x\in S$ there exists an open set $U$ such that $x\in U$ and $U\subseteq S$, then $S$ is open.

So in some metric space $\mathcal{X}$
.. I feel$@KarlKronenfeld

anon

@KarlKronenfeld not sure

any union of open sets is open

4:28 AM

spoiler: metric spaces are topological spaces :D

well i have not been yet introduced to topological spaces

what anon said happens to be one of the defining features of a topological space (generalization of metric space), which is why this question is so awkward

Ahh

Because it is a definition?@Karl

user147690

Are you doing all the exercises @Rem?

I havent yet come across the exercises they are at the last end of the chapter@AlexC

4:30 AM

@Rememberme yeah, when you don't say something like, "in a metric space blah blah" one would assume it's about topological spaces

user147690

@Rememberme For previous chapters I meant

Yes only one chapter that was fairly easy

Because there is only one chapter

before topology@AlexC

@anon Presumably a counterexample would not be finitely generated.

@ThomasAndrews Weierstrass substitution is really nice and easy for the question

Lucas

hello

@DavidCardozo you know the unite circle for cos and sin?

4:47 AM

Hi one question I have the following equation in a group: g^{-1}xg=a{-1}xa, can I conclude that g = a?

no @DavidCardozo.

@k

See if you can find an $x$ for which we cannot draw that conclusion.

@Karl any simple counterexamples that comes to your mind?

Hint: yes

4:50 AM

yeah x be the identity. Sorry awful question

Lucas

@karl can you help me with a draw to show that $cos(\frac{\pi}{2}+x)=-sin(x)$ ?

I don't know to draw

I want to understand without using formula cos(a+b)

because is easiest

recall that f(x+a) is f(x) but shifted horizontally a units to the left

knowing this, try drawing both graphs

user4904589

Would be great if anyone can guide. math.stackexchange.com/questions/1287402/…

i.e., the graphs of cos(x+ pi/2) and -sin(x)

Is there a simplification for $\int_a^b f(x) dx + \int_b^c (-f(x)) dx$ so I don't have to do the same integral twice?

Note that $a \le b \le c$.

anon

5:04 AM

no

I didn't think so, (even though it feels like it should).

If both functions had been the same, then we could do one big integral from $a$ to $c$, but the negative sign throws a wrench into the mix.

If the values of $f$ are in a field of characteristic $2$, then you can just integrate from $a$ to $c$. :P

@KarlKronenfeld I have no idea what that means. I'm preparing a lesson plan for a calc 2 class for finding the area between curves. I don't think that applies.

Totally kidding around.

@Karl, oh. i missed the emoticon afterwards. :D

5:09 AM

In some places $2=0$, so if $F$ is an antiderivative you get $-F(c)+F(b)+F(b)-F(a)=F(c)-F(a)$ because $-1=1$ as well.

Would that be an integral in a binary ring or similar?

I have actually never heard of the term binary ring

but yeah, a type of ring

i was thinking of the objects from abstract algebra, mod 2 rings or something like that.

a ring mod 2

i was also making a joke, since i thought you still were.

in other words, if you want a joke to sound completely serious, make it about math and state it on the internet.

exactly!

i have a document of math jokes, most of which are over my head, but i suspect you would like it.

5:20 AM

Well @KarlKronenfeld why did a man name his dog cauchy?

something about urinating on poles

Because he always use to leave some residue near the pole :D

user147690

5:37 AM

Balarka's average day on chat starts in about 10 min

user147690

@SamuelYusim Your new gravatar actually hurts my eyes more than the last one

huh?

it hasn't changed

also it's basically pure white so I don't know how that could be the case

user147690

@SamuelYusim It was gray a few days ago

definitely hasn't changed

user147690

@SamuelYusim It's yellowy cream now

5:43 AM

I can guarantee it hasn't

user147690

@SamuelYusim Yes it has, it's not even the same picture let alone color

want to show me a screenshot of what it shows up as for you?

user147690

@SamuelYusim Sure

user147690

I am on the uni computers, so it could be that

user147690

imgur.com/hxcZb4I

5:45 AM

it's possible that there are different contrast settings or something, yeah

yep that's definitely what it's always been

user147690

Hey @Balarka chat.stackexchange.com/transcript/message/21693236#21693236

@AlexClark lol

user147690

@SamuelYusim Ahhh sorry then, I'll look when I go home and you are probably right

Hello @KarlKronenfeld

user147690

@BalarkaSen I was about 2 min off, so not too bad

user147690

5:46 AM

@Balarka Why does the subring test work?

what's an example of an injective continuous function from $\mathbb{R}$ to $\mathbb{R}$ that's neither open nor closed

user147690

Or is it easy enough for me to solve if I think about it enough

What's the subring test?

user147690

@BalarkaSen Checking a subset of a ring is a subring only requires checking closure of multiplication and subtraction

oh, ok. didn't know it even had a name. think about it.

user147690

5:48 AM

ok

@BalarkaSen Oh hi

$R$ be a subring of $R'$. Then $R$ is closed under addition, multiplication and taking additive inverses wrt operation of $R'$.

Then the subring test agrees with me.

But can you show the converse?

depends on if you require that a subring contain 1

I don't care about rings without multiplicative identity :D

user147690

rngs 'rungs'

Mike Miller

6:17 AM

@SamuelYusim pick a homeomorphism $f: \Bbb R \to (-1,1)$. then do $x \mapsto f(x)^2$, as a map $\Bbb R \to \Bbb R$

user147690

6:55 AM

@Balarka Okay yes, we are expected to do one talk over the two assignments, so it looks like I will be doing inverse limit of rings as my only talk. So I've gotta kick arse :P

user147690

(since my original talk was an extension of other other students talk)

nice. so prepare this talk thoroughly.

fix your decision that you'll do inverse limit of rings, so you don't change the topic at the last moment. and then study hard.

you can't provide good talks unless you know why you should care about the topic you want to talk about. i, for one, don't care about central extensions :P

user147690

Ahaha

user147690

Nor do I unfortunately

so make sure you do care about inverse limits, as you should :)

7:16 AM

@AlexClark Have you thought out a skeleton for the talk, i.e., what are you going to talk about, etc?

user147690

Not yet, I am trying to understand rings first, so I think it will be Wednesday the main planning starts

oh, I see.

user147690

Some things I need to learn are ideals, maximal ideals, prime ideals and principle integral domains

definitely.

user147690

This is what you were looking at for the second example? individual.utoronto.ca/jordanbell/notes/profinite.pdf

user147690

7:21 AM

Solenoids at the end, was it you who was talking about them afterwards

what was the second example, again?

user147690

You were giving a topological example, but I remember much less of that one than the p-adic one

oh, yeah, solenoids. those are good examples of inverse limit of rings.

i think you should give 3 or so examples in your talk : p-adic rings, formal power series rings and solenoids.

that should constitute a good beginning of the talk.

user147690

I think it is still a 5 min talk ahaha

user147690

That sounds like a 20 min talk

7:23 AM

nah. you just talk about what they are.

just the definitions, i.e.

user147690

maybe then, but I think most of the talk is meant to be defining what an inverse limit of rings is

definition of inverse limit should take about 1.5 mins, examples about 2 mins, and the rest of 2.5. mins you can spend on proving that inverse limits are unique upto isomorphism

user147690

That's 6 min haha

damn it :P

user147690

I could probably sneak an extra minute in anyway

user147690

7:25 AM

Is that proof at the end really so easy that it is 2.5 min?

Mike Miller

the inverse limit is defined by a universal property, and things defined in this way are essentially automatically unique up to isomorphism

the only interesting part is that whatever construction you do satisfies this

i would bet that you're both severely underestimating the time things take in a talk; again, you really should practice this while you're writing it

it's like 5 minutes so not much of a timewaster

i guess. 5 minute talks are really bad.

user147690

@MikeMiller I think Balarka is underestimating it. I practiced my other talk and it was about 13min long and it wasn't very extensive

user147690

So now I know I can say practically nothing...

Mike Miller

yes

7:29 AM

meh.

user147690

@BalarkaSen Haha it's alright, I am going to write the long version for my own sake, and then compact it into something crappier and shorter

as you should. it'd have been nicer to let the students talk a bit more, they'll essentially learn nothing just by writing the definition.

user147690

@BalarkaSen Indeed. I think 5 min was probably alright for central extensions to be a semi-good talk, but 5 min for this one seems pretty bad since just covering the p-adic numbers is a little time consuming(we have never covered these in any class)

might be : i have never tried talking about anything with a time constraint (most of my talks have been huge, like 1 or 2 hours or so). maybe change the topic depending on what fits the time?

user147690

Nah locked in now

user147690

7:36 AM

It looks interesting

user147690

@BalarkaSen You've given 1hr and 2hr talks?

yeah.

user147690

Nooice

hello, @Kaj

@AlexClark those have been un-nice talks, though, asked from me to make sure i really understand what i think about (one was about proof of PNT, the other about proof of fundamental theorem of galois theory). i screwed up my PNT talk a bit, but i passed anyway. :P

i didn't go "oh noes, i got O(xloglogloglogx/logloglogx) instead of O(xlogloglogloglogx/logloglogx)" when my error terms got screwed though :P

8:10 AM

Hey @BalarkaSen

thinking about anything interested, then, @Kaj?

@BalarkaSen, I'm giving the primitive element theorem some thought. A specific case says that if $K$ is a finite extension of $\mathbb{Q}$, $K = \mathbb{Q}[\alpha]$ for some $\alpha$ in $K$. I'm starting with easy extensions first, like only adjoining two elements, and trying to determine which elements will suffice as our $\alpha$, then I'm looking to generalize.

oh, nice.

So for example, I can show that $\mathbb{Q}[\sqrt{p}, \sqrt{q}] \cong \mathbb{Q}[\sqrt{p} + \sqrt{q}]$ for distinct primes $p$ and $q$.

right. in general, $\Bbb Q(\alpha, \beta) \cong \Bbb Q(\alpha + c\beta)$ for some $c$, where $\alpha, \beta$ are algebraic.

8:25 AM

$c \in \ $ ?

$\Bbb Q$

Oh very cool. I'd strongly suspected it was true that $\cong (\alpha + \beta)$ only because I'd seen so many problems and noticed the pattern.

It's not related to anything though. This problem from a few minutes ago just got me thinking about it: math.stackexchange.com/questions/1287482/degree-of-extension/…

well, that construction is the basis of the proof of primitive element theorem

it works for any characteristic 0 field, for one.

And of course things get more complicated if char \neq 0

yeah, that's a mess.

8:34 AM

3

Q: Uncountable set of reals

Let $E\subseteq \mathbb{R}^1$ be an uncountable set. Can we obtain some subset $F\subseteq E$ which is closed and uncountable? Basically, I want to construct some set containing only irrational numbers which is also uncountable and closed, in a sense, I want to know a general process to c...

real-analysis

This question intrigues me

@Kaj Have you read Morandi's Galois theory? I skimmed through the chapters, and it seemed really good.

I haven't heard of it

here

it does a bit of galois cohomology and infinite galois theory too, which is nice.

@KajHansen Well, trivially, $F = E = \Bbb R$ works :P

I'm taking algebra in the fall. I'm fairly excited to get back into it.

oh, nice.

8:44 AM

@BalarkaSen, I'm interested really in the text underneath the yellow box: Does there exist an uncountable set of only irrational numbers that is closed?