Hey all

hi

Hey can someone answer a real analysis problem for me?

I'm suppose to show that a set is closed and open iff the boundary is null

Is this true for open sets by themselves?

How is boundary defined in your case?

It is not true for open sets in general. Any open interval (a,b) has a boundary consisting of two elements {a,b}.

boundary is the intersection of the closures of the set and itscomplement

@anthony

ayo

Did you do anything yourself?

I was just asking if it was true for open sets by themself

Did I answer your question?

I don't think so?

@Prototank Err, what was his question?

:O

@PedroTamaroff I'm not asking for answers.

Well I am.

@Anthony Well, $\Bbb R$ is open and has empty boundary.

Just not to my homework.

$\varnothing$ is open and has empty boundary.

@PedroTamaroff Yeah, but they are also closed.

@Anthony Well, the point is $\Bbb R$ is connected.

So those are the only examples.

I don't know what connected means...

It means it cannot be written as the union of two open disjoint sets.

I see.

Two or more?

Did you show open and closed if empty boundary?

@Anthony Two.

I see.

Uhm, no, so it's iff, I thought for the forward openness was sufficient.

@PedroTamaroff Also, I figured things out with the lim sup f(x) stuff from earlier. I managed to prove some things as well.

@Prototank Cool.

For going from empty boundary, I'm a little clueless

Well, don't you have formulas for $\bar A$ and $A^{\circ}$ involving $\partial A$?

For example... $\bar A=A\cup \,?$

And $A^\circ =A\smallsetminus \,?$

Yes, I guess I do.

That's easy.

But for the forward direction, does any open set have an empty boundary?

@Anthony No.

@Prototank just have you an example.

Oh it's just things in the closure woops.

Yeah, I was confused.

But you have a formula for $\partial A$ in terms of...?

$\partial A=\bar A\cap \overline{X\smallsetminus A}$.

And $\overline{X\smallsetminus A}=X\smallsetminus A^\circ$.

So...?

I think I could deduce those.

@Prototank I'm sorry, I misunderstood, you're the (wo?)man

@Anthony what

lol

OH

thanks.

@PedroTamaroff How do those help though?

If A is closed and open then...

@Anthony If $A$ is both open and closed then $A=\bar A=A^\circ$.

So $\partial A=\bar A\cap \overline{X\smallsetminus A}=A\cap (X\smallsetminus A^\circ)=A\cap (X\smallsetminus A)=\varnothing$.

Oh.

I see.

Apparently not taking the Math GRE affects application desirability

Hello?

hello

Can someone help me understand how the Horizontal Asymptote of $\frac{x^2}{x^2-9}$ is NOT $\frac{1}{9}$ but actually just 1?

plug in a huge number for x

the numbers up top and down below will be about the same

why? because for large x, the constant 9 is small fry

but the 9 is negative

why is it at least not -1

did you read what I wrote?

yes

but why not -1

x^2-9 is more or less the same as x^2 for large x

@Brittany you might as well say "but why not green?"

i see

haha

in order to meaningfully ask "why not X" there must be a reason to expect X in the first place

i guess i have a very elementary way of seeing this

did you do what I asked? plug in a huge number for x and see what you get up top and down below

for example, plug in x=1,000,000,000

because I thought when the degrees are the same you kind of ignore the degrees and solve the fraction for whats left, in this case it woul dbe -1/9

yes i did

@Brittany sort of, but you do (leading coefficient up top) / (leading coefficient down below)

(assuming they have the same degree)

the -9 does seem not important

indeed it isn't, it could be replaced by literally any other constant for the same result

another way of seeing it is that $\frac{x^2}{x^2-9}=\cfrac{1}{1-\frac{9}{x^2}}$ obviously $9/x^2\to0$ in the limit so that leaves $1/1$.

is it safe to say then if this were that same exact equation but in the numerator instead of $x^2$ it was $2x^2$ the answer would still be one?

no, see my (lead coeff) / (lead coeff) comment

ok

$\frac{2x^2}{x^2-9}\sim\frac{2x^2}{x^2}\to2$

another way on the original problem: $\frac{x^2}{x^2-9}=\frac{x^2-9+9}{x^2-9}=1+\frac{9}{x^2-9}$ and $x^2-9\to\infty$ so this tends to $1$

the idea is that $a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0\sim a_nx^n$

any power of $x$ less than $n$ is not in "the same league" as $x^n$ is

anything in a lower league is overwhelmed to the point that it doesn't even affect the result

so $\displaystyle\frac{a_nx^n+\cdots+a_1x+a_0}{b_nx^n+\cdots+b_1x+b_0}\to \frac{a_n}{b_n}$ depends only on the leading coefficients

all of the other constants are superfluous and don't affect the result

(note in that quotient the numerator and denominator have the same degree)

ohhh okay, its because of the lead coeff wow! 9 is not in any way a lead coeff so its irrelevant

Im sorry that took me so long, I promise im not 5 years old haha

thanks for all the explaining !!

np

@seaturtles ^^^

apologies are unnecessary

just don't make soup out of me like usukidoll

@seaturtles turtle soup?

yes

=P

is it possible to solve $x^2$ for x?

that's not an equation

Ill give an example of what im getting at, Im studying rational functions

"solve for x" means find the values of x which, when plugged into an equation or claim, make the equation or claim true. the expression "$x^2$" by itself is not an equation and does not claim anything. say you plugged in x=2. you get x^2=2^2=4. is 4 true or false? that doesn't make any sense.

$\frac{x^2}{x^2-9}$ Find the X and Y int. The instructions I have to find the x int is, solve the numerator for x, if there is no v variable then there is no xint, but there is an x variable in this case

what you actually mean is "set the numerator equal to zero and then solve the resulting equation for x. check which solutions are in the domain of the rational function."

"solve for x" is not magical shorthand for "set equal to zero and then solve for x"

but anyway, where are you having trouble answering the parts of that question?

are you asking how to solve $x^2=0$ (as part of finding the $x$-intercept)?

what is the xint in this case

yes

well, what are the value(s) of x that square to zero?

0

correct

that is the only solution

woot

so 0 is the x-intercept

okay that was easy

is 0 in the domain?

oh good question

the domain is all real numbers

no

except those which will make the denominator 0

right

which are?

9

no

all real numbers except 3

closer, but incomplete

the domain is all reals except +3 and __ (fill in the blank)

all real numbers except 3 and -3

right

so 0 is in the domain

so 0 is indeed an (the in this case) x-intercept

because its squared right? and in order to solve that for x you need to get rid of that sq root

because positive numbers have both negative and positive square roots

i see

and when solving for things, you don't get to throw out perfectly good solutions

=) i hope youre a teacher somewhere

an assistant (:

good, back to what you said about solving for x and setting an eq to 0 then solving

they are not the same idea?

@Brittany what be the problem?

just wondering about this

"solve for x" implies x is already in some equation. "set equal to zero" means you have an expression with x in it, but an expression is not the same thing as an equation.

an equation equates two expressions

for example you could have x^2+1=2x-1 and want to solve for x. there is no "set equal to zero" here.

OTOH, "x^2" by itself is not an equation or a claim, so to "solve it for x" is as meaningless as for colorless green ideas to sleep furiously. if you set x^2 equal to 0, then you have an equation, namely x^2=0.

i see

@seaturtles

?

I never finished the proof that $G\simeq H\oplus K$ if those maps existed.

Remember?

The maps $\iota_1,\iota_2$ and $\pi_1,\pi_2$.

refresh

OK.

Just a sec.

@seaturtles There.

One direction is easy.

Namely $G\simeq H\oplus K\implies$ yadda yadda

the hypotheses mean $$G\to H\oplus K:g\mapsto(\pi_1(g),\pi_2(g)) \\ H\oplus K\to G:(h,k)\mapsto \iota_1(h)+\iota_2(k)$$ are inverse functions

Ah, didn't think of that.

@seaturtles you said something false earlier

oh?

i have totally slept furiously before.

@Mike WAT

Missed you Mike.

I am in the zone now.

10 bucks the answer with the word "intuition" in it will get accepted

@seaturtles Nice =)

I am not in the sone

@Mike No shit Sherlock.

@seaturtles I'm almost finishing the groups section on D&F. Phew.

did anybody here watch LOST?

@Mike is that a great movie ? :D

@Mike Did not.

no it's awful

@Mike That's what I heard.

Oh!

I earned the group-theory badge.

I'm confused about a question in Dummit & Foote. It says to show that the division subring generated by an abelian subgroup of any division ring is a field.

Isn't every subgroup of any ring abelian?

(I'm assuming they mean additive subgroup?)

addition is commutative, yes.

By definition, right?

lol

they probably mean a subgroup of the group of units, which can be nonabelian

Ah, that'd make more sense.

@PedroTamaroff Could it be

Are you still there?

Soft Question: If $x>0$, how is $\vert t ^{x+iy-1}\vert = t^(x-1)$?

I have a challenge-With a boundary being the intersection of the closure of the set and it's complement, can someone show that the boundary of the boundary of the boundary is equal to the boundary of the boundary?

@TheSubstitute I think you mean t>0

do you know how to do complex exponentials?

@seaturtles Ah that's what it is. It has been a while, I will read up on that, thanks.

oh no

@Mike I'm started to get concerned that I actually bother you

you don't

:)

and I'm easily bothered.

@Anthony why should you care?

@PedroTamaroff I mean... I dunno. It's not that I should, but I definitely do.

Also

Or am I responding to the wrong question.

$\partial A$ is closed.

Yes.

It is!

So?

∂A

damn

So it contains its bdry.

Letting A = bdry B, you get fr fr fr B is inside rr fr B

So you have to show fr fr B is inside fr fr fr B

Indeed.

Well, I leave that to you.

lol

That was the part I was stuck at

Well, me and like, 10 of the kids in my class.

I feel like it shouldn't be too bad...

@PedroTamaroff

Yes

zon

@mike off to sleep

noo

@Pedro What was the first problem of Alex?

Oh, I recall.

$x^4 + 1$ has infinite zeros over $\Bbb H[x]$

Right?

@AlexanderGruber

$x^4 + 1 = (x^2 - i)(x^2 + i)$

Unless

@BalarkaSen I just woke up now. 5 mins ago!

@BalarkaSen yes

Aha, that's where the center comes to play

the roots of $x^4+1$ over $\Bbb H$ are $a{\bf i}+b{\bf j}+c{\bf k}$ with $a^2+b^2+c^2=1$ IIRC

Why are you revealing the solution?

eh?

It was given to me as an excercise.

oh, I didn't realize

That is just foolish.

And the roots of $x^2 + 1$ are like that, not $x^4+1$.

see, your right to sojourn alone has not been damaged

okay, new exercise: explain why $f(x)=g(x)h(x)$ in ${\Bbb H}[x]$ does not imply $\forall a\in{\Bbb H}~f(a)=g(a)h(a)$

@seaturtles I don't understand that expression.

it's not an expression, I'm just using fancy words

@seaturtles Are they? Aren't those the roots of $x^2 + 1$ instead of $x^4 + 1$?

I just responded to that point

by saying "see? I haven't spoiled anything for you at all"

And I didn't understand the fancy reply.

=)

OK.

@seaturtles Let me think. It's kinda obvious, but...

@seaturtles Very nice hint.

We need commutativity to conclude that.

when one expands the product g(x)h(x) and writes it in the usual form (coefficients on the left of powers of x), we tacitly have x commuting with all scalars. but the "a" we evaluate the polynomials at need not commute with these coefficients like x does in the polynomial ring.

Exactly.

Otherwise we get things like $axb, cx^2dx$ for $x=a$, right?

Which does not result the desired in $g(a)h(a)$

Am I correct? @seaturtles?

$x^4 + 1 = (x^2 + i)(x^2 - i)$. So for x to be a root, we need $x^2$ to commute with $i$

...or not.

Let me think.

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@Sawarnik Why are you spamming?

@BalarkaSen Because I getting bored.

OK =)

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Don't do that.

You are scrolling the chat too much.

So I can't see what is in above.

BuT I aM geTting BorED. i WaNT To spAm.

BuT I aM geTting BorED. i WaNT To spAm.

BuT I aM geTting BorED. i WaNT To spAm.

roBjohn, I hoPE yoU are nT HERe.

@seaturtles I proved the factor theorem over $\Bbb H$ just now.

So a root of $x^4 + 1$ must be either a root of $x^2+i$ or $x^2-i$

yawn

yawn yawn yawn yawn

@BalarkaSen oh?

@BalarkaSen Ok, I stop. Forgiv me.

@seaturtles Yes?

How do we express $a+b$ in big pi? Anyone please?

@Sawarnik you mean big sigma?

No, big pi.

okay $\prod_{i=1}^1(a+b)$

@BalarkaSen (1+j)/sqrt(2) is a root of x^4+1 but not a root of x^2+i or x^2-i

Without plus. It would be circular isnt it?

@Sawarnik why would it be circular?

it's not claiming anything

so there is no circular reasoning

So any idea?

Nah, dunno.

yes, $\prod_{i=1}^1(a+b)$

Without plus sign.

@seaturtles Mhhmm.

$-\prod_{i=1}^1 ((a-1)(b-1)-ab-1)$

no plus sign

ask a serious question

Let's see, @seaturtles.

$g(x) = \sum a_i x^i$, $h(x) = \sum b_i x^i$

$g(x)h(x) = \sum a_ix^i \sum b_ix^i$

@BalarkaSen as you've seen, $x^4+1=a(x)b(x)$ doesn't mean $t^4+1\Rightarrow a(t)=0\vee b(t)=0$ for $t\in\Bbb H$. so factoring $x^4+1$ is not helpful/

Wait, let me think.

$g(x)h(x) = \sum a_i x^i h(x)$

We don't really need the expansion of $h$....

Let $g(x)h(x) = f(x)$ be zero at $t$

I don't see why you're doing that

you might as well just write $x=a+bi+cj+dk$, plug that into $x^4+1=0$ and then simplify the resulting system involving $a,b,c,d$

We can just commute the $h$ with the $x$

@seaturtles Too tedious.

I am trying to simplify it.

$f(x) = \sum a_i h(x) x^i$

Yes, that's it.

$x$ commutes with $h(x)$, but $a$ may not commute with $h(a)$ for quaternions $a$

$f(t) = \sum a_i h(t) x^t$

@seaturtles I know that.

I am commuting x with h(x)

what are you trying to accomplish? the problem is to solve $x^4+1=0$ no?

Then simply plugging in t there in f(x)

but you can't do that

@seaturtles I am trying to avoid tedious multiplication.

@seaturtles Why?

$xh(x)=h(x)x$ does not mean $th(t)=h(t)t$ for $t\in\Bbb H$

I never contradicted it.

I am simply computing $f(t)$

you switched $x$ and $h$ around and then evaluated it

=)

@seaturtles I am not computing $g(t)h(t)$.

I already told you that you can't switch the order of things and then evaluate and expect to get the same thing you would have originally gotten evaluating in the first place

@seaturtles You are misunderstanding my goal.

you have not stated it

Let me think first and then I will state.

yawn yawn yawn

Okay, isn't $(aba^{-1})^n = ab^na^{-1}$ in $\Bbb H$?

yes

Done then.

so what are the solutions?

I am working on that still. I have only proved that if $f(x)$ has a root $t$ then either $h(x)$ has root $t$ or $g(x)$ has root $h(t)th(t)^{-1}$

Correct, @seaturtles?

yawn yawn yawn

Now I have to solve a system of complicated equations.

i.e., I need to find roots of $x^2 - i$ and $x^2 + i$

And then the roots of $x^4 + 1$ is either the former or a conjugate of the latter.

Someone find me the roots of those two.

Anyone? @seaturtles?

@AlexanderGruber @Pedro Let $f(x) = g(x)h(x)$ and $g(x) = \sum a_i x^i$ and $f(t)=0$. Then $f(x) = g(x)h(x) = \sum a_i x^i h(x) = \sum a_i h(x) x^i$ and $0 = f(t) = \sum a_i h(t) t^i = \sum a_i h(t) t^i h(t)^{-1} h(t) = \sum a_i \left ( h(t) t h(t)^{-i} \right )^i h(t) = g(h(t)th(t)^{-1})h(t)$ Thus, either $h$ has the same root as $f$ or $g$ has it's conjugate thingy (or whatevs the name, I forget things). $x^4 + 1 = (x^2 - i)(x^2 + i)$ and enough for now. Am I walking in a plausible direction?

I suppose that's plausible

I would use exponentials

if $x^4+1=0$ then $x=ye^{(a{\bf i}+b{\bf j}+c{\bf k})/4}$ where $a,b,c\in\Bbb R,n\in\Bbb Z$ with $a^2+b^2+c^2=2\pi n$ and $y$ is a given solution to $x^4+1=0$

fun stuff

I must off to sleep

@seaturtles Exponential are not just my stuff.

I keep things as algebraic as possible.

I'm wondering if I do this correctly

no, it should be $\frac z3$ out front

Greetings

Heya

I have a series $\sum_{n=-\infty}^{n=\infty} f_n(z)$, and I want to show that this converges uniformly on $K$. I know that $\sum_{n=R+1}^\infty f_n(z)$ converges uniformly. And also that $\sum_{n=-\infty}^{-R-1} f_n(z)$ converges uniformly. Now the middle part, is constant. How can I now justify that the whole series must converge uniformly ?

@Kasper That is uniform convergence of the Cauchy Principal Value.

...

When you take a sum over all of $\mathbb{Z}$, $\lim\limits_{n\to\infty}\sum\limits_{k=-n}^na_k$ is the Cauchy Principal Value

...

as with integrals

@Sawarnik Now you're just being a fool.

...

:14118825 No, it's just that $$\#\{\text{Sawarnik is online}\}\cap\#\{\text{Balarka is online}\} \geq\!\!\geq \log(n)$$

What does that mean?

Read a bit about asymptotic analysis then come back.

Search [big-O notation]

≥≥?