Mathematics - 2015-03-26 (page 4 of 4)

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8:05 PM

@Kaj I just went through your lecture on the proof of Ramsey's finiteness theorem. Pretty interesting stuff.

Indeed it is

I haven't really thought about it for a long time

I still am really curious about how the piles of number theory jumps in the play though.

It seems fascinating that such combinatorial stuff can even relate to the arithmetic structure of primes.

8:32 PM

Kaj, going to watch those. Love watching lectures online. But why are the blackboards swaying like that? I'm getting some vertigo. :)

I have to calculate the limits of some functions, if they exist.

One of the functions is the following:
$$\lim_{(x, y) \rightarrow (0, 0)} (x^2+y^2+5)$$

To calculate this limit what do we have to say??

That since it is continuous we can sustitute $x$ and $y$ with $0$??

Or is there a better way to formulate it??

@DanielFischer Do you have an idea??

For any $\epsilon > 0$ try choosing $\delta = \sqrt{\epsilon}$

Then $\|(x,y) - (0,0)\| < \delta$ means $\sqrt{(x-0)^2 + (y - 0)^2} = \sqrt{x^2 + y^2} < \delta$

for every $\epsilon > 0$ there exists a $δ > 0$ such that for all $(x,y)$ with $0 < ||(x,y) − (p,q)|| < δ$, then $|f(x,y) − L| < \epsilon$

so $x^2 + y^2 < \delta^2 = \epsilon$

$|f(x,y) - 5| = ?$

8:50 PM

$|f(x,y)-5|=|x^2+y^2|<\delta$ Is this correct??

How could we calculate the limit besides using the definition?? @DavidWheeler @Theorem

@MaryStar $|f(x,y) - 5| = |x^2 + y^2 + 5 - 5| = |x^2 + y^2| = x^2 + y^2 < \delta^2$, not $\delta$

Since $\delta = \sqrt{\epsilon}$ we have shown that if $\|(x,y) - (0,0)\| < \delta$, then $|f(x,y) - 5| < \epsilon$, which means that the limit is 5.

And yes, since $f$ is continuous, $\lim_{(x,y) \to (0,0)} f(x,y) = f(0,0) = 5$

9:08 PM

I see...

So an other way besides using the definition is to say that "since $f$ is continuous, $\lim_{(x,y) \to (0,0)} f(x,y) = f(0,0) = 5$" ?? @DavidWheeler

Of course, KNOWING $f$ is continuous is dependent on knowing that the functions: $g(x,y) = x^2, h(x,y) = y^2, k(x,y) = x+y$ are continuous, and that the composition of continuous functions is continuous.

since $f = [k\circ (g \times h)]$

these "itty-bitty" steps in proving "complicated expressions" are continuous functions are usually proven as theorems in a decent course, because they are important.

9:28 PM

Ok... Or is it better to say that since $(x, y) \rightarrow (0, 0)$ we have that $x \rightarrow 0$ and $y \rightarrow 0$. So, we have that $x^2+y^2+5 \rightarrow 0^2+0^2+5=5$ ?? @DavidWheeler

9:49 PM

Hey @ThomasAndrews @AlecTeal
Are you familiar with matlab?

10:04 PM

If we have some homomorphism $\lambda:G\rightarrow G'$ on two Lie groups, does it induce some sort of isomorphism $\lambda_*:\mathfrak{g}\rightarrow\mathfrak{g}'$ on the Lie algebras?

Hi @MikeF Are you familiar with matlab?

Morning @Ted

I think they probably meant a different mike. Mike Miller? I'm really never in chat.

hahaha

@evinda

10:06 PM

hi @MikeF

good night, @Mike ... Have you processed my earlier answer?

No, I meant you :D @MikeF

@evinda: It's time you stopped asking every single person for help on every single thing.

@TedShifrin Ok, I am sorry...

Yes, @Ted. It was helpful.

10:07 PM

OK, @Mike. Then my job is done. Now I can retire with a clear conscience.

You should write MikeM, since MikeF is getting pinged.

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@0celo7 Hmmm I believe so

oy ... it's already too much work. I just won't talk to anyone any more.

hi @Committing

Hi @MikeMiller. Yes it looks like I get your pings

You people needed more rarified names :P

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10:09 PM

@TedShifrin Hey Ted

I'll bug people to use my initials, @MikeF

I've tried with Alex, as we have so many of them.

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I always ping when I have enough characters. Also Ted I am also Alex :P

Is there a way of calling in English $1/\log(x)$? Some idiom or sorta?

oy, @Committing ... I'm actually a Ted.

nope, just $1/log(x)$.

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10:10 PM

@TedShifrin I thought you were Shifrin Ted wow mind blown

@Committingtoachallenge How does one construct this isomorphism?

smacks @Committing

@TedShifrin I need to find a name for integrals with $1/\log(x)$ in my book ...

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@0celo7 Jordan-chevelley decomposition, I will check it out(not an expert here)

Would you have a special name for integrals with $x^{-3}$? @Chris'sssis?

10:11 PM

@TedShifrin Not really, but these ones are special for some reasons.

maybe say they involve Li.

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@0celo7 Actually I am busy right now with Functional, but I believe J-C decomposition is the right way to go

@TedShifrin you mean li, right?

in the US it's usually capitalized

@Committingtoachallenge Will check it out.

10:13 PM

OK

isn't it on Wiki?

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@0celo7 Tell me what you find if you have time

mathworld.wolfram.com/LogarithmicIntegral.html

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@Chris'ssis Yeah right at the bottom of that link

Hello @TedShifrin !!

I have to calculate the limits of some functions, if they exist.

One of the functions is the following:
$$\lim_{(x, y) \rightarrow (0, 0)} (x^2+y^2+5)$$

Which way is the best one to solve such exercises??

To say that since it is continuous we can sustitute $x$ and $y$ with $0$??

Or is it better to say that since $(x, y) \rightarrow (0, 0)$ we have that $x \rightarrow 0$ and $y \rightarrow 0$. So, we have that $x^2+y^2+5 \rightarrow 0^2+0^2+5=5$ ??

Or is an other way better??

10:14 PM

@Chris'ssis: Oh, interesting.

Committing to a challenge

user image

@TedShifrin Yeah, I was referring to integral logarithm.

@MaryStar: You're still trying to get us to do all your homework. It's out of context for us. If it's a rigorous analysis course, the answer is different from when it's a basic calculus course. I have no idea what you are expected to do. Ask your professor.

@Committingtoachallenge Indeed. I always used li for avoiding the confusions with polilogarithm.

I know, @Chris'ssis. Apparently, Wiki uses Li for a special case. I hadn't realized that.

So go with li.

10:16 PM

@Committingtoachallenge My knowledge of abstract algebra is rather sparse...I'm doing a lot of definition chasing right now.

@TedShifrin I think li is better for avoiding a possible confusion with the polylogarithm.

You would know far better than I, @Chris'ssis.

@TedShifrin Well, I'm always open to learn new things, it's risky for me to think there is nothing new to me to learn. Hope to never reach this conclusion.

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@0celo7 Fair enough, best of luck!

I think I am starting to 'get' epsilon delta stuff, I don't know why it has taken me so long...

It takes practice, @Committing.

10:19 PM

@Committingtoachallenge I need this specifically for spinor analysis in general relativity, but the physics chat is barren for some reason.

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@0celo7 Yes, that is strange, it is a peak time for H-bar normally

well, I'm deserting ... dinnertime. You all teach everyone everything without me.

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@TedShifrin Is it actually 'harder' to learn for students? I mean have you taught beginning algebra and/or set theory?

I find it much harder than algebra/set theory beginning stuff that I just 'got' instantly. Well have fun @Ted

I've taught abstract algebra plenty of times. And our introductory proof course, too. The order of quantifiers in analysis makes it one finger harder.

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Yeah not knowing where quantifiers sit was a major problem, very true

Or when you pick an $x$ before your epsilon as well

10:23 PM

I used to ask students, when I taught our calc with theory class ... If you want to build a rectangle (let's say a square here) with area 4 with maximum possible error $\epsilon$ (so $4\pm\epsilon$), how accurately must I measure the side length?

No, you don't pick $x$. You make a preliminary stipulation to restrict $x$. :)

Kind of a silly question, anyone got a moment for it? So if $U$ is an open subset of $R^n$ and $f:U \to \mathbb{R}^n$ is a $C^1$ map with an invertible Jacobian at every point in $U$, then $f$ is an open map. Okay fine.

So, if $f$ is also 1-1, then $f$ is a diffeomorphism onto its image $f(U)$ (which is also open)?

Yup @MikeF

Thanks. Just slightly thrown off by the wording of the 1st theorem here... they don't say that $f$ is a diffeomorphism between two open sets in $\mathbb{R}^n$, but this is what is happening, right?

oops...

link

@MikeF: Notice that they're keeping the jargon to a minimum, so they're not using the word diffeomorphism (or immersion or submersion or ...).

argh @ diffeomorphism being used to mean $C^1$ diffeo

10:28 PM

it's isomorphism in the appropriate category, @MikeM; stop your fussing.

@Committingtoachallenge Yeah, and there's only one guy who frequents the H-bar and knows anything about group theory. (Might be a slight exaggeration.)

@Ted: Makes sense that the writer would want to do that. Although they do say "nonzero Jacobian" where I think they mean "nonzero determinant of Jacobian".

i was confused as to how $C^1$ was promoted to $\infty$ for a second, @Ted

but I will go back to my cave.

@MikeF: Many authors refer to the Jacobian determinant as Jacobian. I tend to do that. I say derivative matrix when I mean the matrix.

I'm leaving, anyhow, @MikeM. You're in charge.

yikes

10:30 PM

@TedShifrin: bye Ted. If Mike's in charge, does that mean diffeos are C^infinty now?! ;)

I usually mean $C^\infty$ unless otherwise noted. :P

You two arm-wrestle over it.

I missed one complex analysis question on my exam :(

missed = forgot to do? did incorrectly?

Didn't know how to do it at all so left it blank

It's probably a question I'd post on MSE with absolutely no work showed because I don't know how to approach it

You don't even imagine how many victims this very nice problem produced (I added to my book)

I love it terribly much!

10:51 PM

@TedShifrin do you find working with indices cumbersome? Or is it something you get used to. I'm trying to write out how we come to the definition of the covariant derivative, the one that includes a correction term to the partial derivative of a tensor....and the indices are just such a pain! I feel like there must be a way to simplify the notations.

Good evening all

@StanShunpike You can't do any concrete calculations in index-free notation. The reason you'd write $ V^j{}_{;i}= V^j{}_{,i}+\Gamma^j{}_{ik}V^k$ is if you want to do an actual calculation. Otherwise $\nabla V$ works just fine.

@StanShunpike Note also that the connection coefficients are not coordinate invariant and thus it does not really make sense to have them in an index free notation.

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11:14 PM

Can someone explain the end of this: math.stackexchange.com/a/1202440/142198

Here is a question I consider it interesting. Find $f_n(x)$ such that the integral converges $$\int_0^1 \frac{1}{\log(x)}+ \frac{1}{\log^2(x)}+\cdots + \frac{1}{\log^n(x)}- f_n(x) \ dx$$

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We get $\frac{2|f(x_n) - f(x)|}{|f(x)^2|}\lt \frac{2\epsilon}{|f(a)|^2}$

Can we just take that $\epsilon$ to be $\delta$ for notation reasons and take $\epsilon = \frac{2\delta}{|f(a)|^2}$?

Since we can make $\epsilon$ as small as we want, when we take $n\geq N$ to be as big as we want?

Hi @Committingtoachallenge

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@ᴇʏᴇs Hi

@Committingtoachallenge You are already better at analysis than I am

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11:20 PM

Doubt it

I looked at some of your posted problems and I can't prove them myself

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My recent one?

Yes

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They took me a sad amount of time

How long is that

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11:22 PM

Hours...

I suck

Oh, that's not long

I would probably spend a week or two on that one

Can we make $\frac{xy}{x^2+y^2}$ continuous, defining it appropriately ay $(0, 0)$ ??

What does it mean to make a function continuous??

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@Mary What is the limit of that? What have you tried?

@MaryStar Usually you will want to attempt something for a long time before asking for help, otherwise you will learn nothing

I would set y=mx and see what happens ...

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Try $x=y$

Try $x=2y$

11:28 PM

Try $x = x$

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Try $x=\int_0^\infty \frac{y}{\ln y} dy$

Try $x= \sum_{n=1}^{\infty} \frac{\psi\left(\frac{n+1}{2}\right)}{ \binom{2n}{n}}$

Plug in a solution the Navier-Stokes equation, should work.

:-)

@Committingtoachallenge @Chris'ssis @ᴇʏᴇs

$\frac{xy}{x^2+y^2}$

For $y=mx$ we have the following:

$\frac{mx^2}{x^2+m^2x^2}=\frac{m}{1+m^2}$

Since it is dependent on $m$, the limit doesn't exist, right??

@MaryStar Exactly. It shouldn't depend on the direction you consider. But in this case it depends on since it's dependent on $m$. Thus, the limit doesn't exist.

It's also important to try to visualize what happens there.

11:40 PM

Since the limit doesn't exist, does this mean that we cannot make the function continuous?? @Chris'ssis

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What do you think Mary?

user image

@MaryStar ^^^

So that a function is continuous it should stand that $$\lim_{x \rightarrow a} f(x)=f(a)$$ so if the limit doesn't exist it cannot be continuous. Are my thoughts correct?? @Committingtoachallenge

@MaryStar Yes, if the limit exists then the discontinuity is a removable one.

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$\forall \epsilon \gt 0, \exists \delta \gt 0, |x-a|\lt \delta \implies |f(x)-f(a)\lt \epsilon$

11:44 PM

@MaryStar Think about the "hole in the graph" from freshman calculus.

user image

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Obviously I dropped my right | by mistake, but you get the point

@Committingtoachallenge I think you're missing a quantifier

@Committingtoachallenge This is the definition of the limit, right??

Analysis is pure, concentrated pedantry.

Me no likey.

11:50 PM

What's pedantry mean

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I hate analysis so far

Algebra is my (strawberry) jam

@ᴇʏᴇs Here ya go.

Link won't load

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Don't worry the link is unhelpful

@ᴇʏᴇs google.com/…

11:53 PM

@0celo7 thanks

infinitesimal simplicio

It means overly formal

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Hi @MikeM

@infinitesimalsimplicio Where the heck is everyone in the H-bar?

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Dede

o no

infinitesimal simplicio

11:55 PM

Teachers who try too hard are guilty of it. @ᴇʏᴇs

Dunno @0celo7

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