the no-normie zone - 2017-08-30 (page 1 of 2)

Balarka Sen

17:38

@Blue Are you up for LA today?

Anonymous

@BalarkaSen Yes...just one min...having dinner :p

Balarka Sen

Sure, sure

Anonymous

@BalarkaSen I'm back...btw I had some analysis questions also which I was having trouble with. Can we do both Analysis and LA today? How long will you be free/awake? I think can stay awake till 3-4 if required since I had slept today afternoon :d

Anonymous

Do let me know if you have some other work..we can do Analysis later then

Balarka Sen

For sure. Actually I'm finish dinner on this end now so you should ask the analysis question

(s)

I don't have much work

Anonymous

17:48

Here's the first question: I need to find $\delta(\epsilon)$ for showing continuity of $\frac{1}{x^2+1}$ at $x=0$

Anonymous

I need to get done with epsilon-delta for once and for all :P

Anonymous

Here's my attempt:

Anonymous

$|f(x)-L|=|\frac{1}{x^2+1}-1| <\epsilon$

Anonymous

I need to somehow bring this to the form: $|x-0|<\delta$

Balarka Sen

Right

Anonymous

17:51

$|\frac{x^2}{x^2+1}|<\epsilon$

Balarka Sen

Well, you want a find a $\delta$ such that $|x| < \delta$ implies $|x^2/(x^2+1)| < \epsilon$

Anonymous

Yep!

Anonymous

I

Anonymous

I'm not sure what to do next

Anonymous

Should I pick a range around x=0?

Anonymous

17:53

Like [-1,1] ? But that would not be a general solution

Balarka Sen

$$\frac{x^2}{x^2+1} = \frac{1}{1+\frac1{x^2}} \leq \frac{1}{1+\frac1{\delta^2}} = \frac{\delta^2}{\delta^2+1}$$ right

so it suffices to pick a $\delta$ such that $|\delta^2/(\delta^2+1)| < \epsilon$ I guess

Just say $\delta = \sqrt{\epsilon}$?

Anonymous

@BalarkaSen Wait...that's what I don't get. How do books all of a sudden get $\delta=\sqrt{\epsilon}$ from the previous relation? I suppose the subset of the $\delta$ which satisfy the first equation will also satisfy the second

Balarka Sen

So you agree that given an $\epsilon$ it suffices to find a $\delta$ such that $|\delta^2/(\delta^2+1)| < \epsilon$, right?

$\delta = \sqrt{\epsilon}$ is such a $\delta$.

You have to find just one such $\delta$.

Anonymous

@BalarkaSen Aha...that's what I was confused about!

Anonymous

One such

Balarka Sen

18:00

The thing you have to keep in mind is "for every $\epsilon$, there is a $\delta$"

I once found a very nice post on MO or MSE about the intuition for epsilon delta. Let's see if I can find it.

By a very good mathematician

Anonymous

Right...right! Lemme try out one or two more examples to drill this into my mind

Balarka Sen

First paragraph of the first answer in this question

Anonymous

Say $x+\sin(x)$ and $\sqrt{x}$

Balarka Sen

The question is by one of the best topologists of the century, William Thurston. RIP.

Anonymous

@BalarkaSen Checking!

Balarka Sen

18:03

Sorry I meant first paragraph in the top answer

ah and the answer is by... Terry Tao :P

Anonymous

One specific mental image that I can communicate easily with collaborators, but not always to more general audiences, is to think of quantifiers in game theoretic terms. Do we need to show that for every epsilon there exists a delta?

Anonymous

Then imagine that you have a bag of deltas in your hand, but you can wait until your opponent (or some malicious force of nature) produces an epsilon to bother you, at which point you can reach into your bag and find the right delta to deal with the problem.

Anonymous

One specific mental image that I can communicate easily with collaborators, but not always to more general audiences, is to think of quantifiers in game theoretic terms. Do we need to show that for every epsilon there exists a delta?

Anonymous

Ah, that's nice way to think

Balarka Sen

Right, so $\forall \epsilon \exists \delta$ has that game theoretic interpretation.

Anonymous

18:08

I am holding a bag of deltas...I need to find one which will satisfy a given epsilon!

Balarka Sen

It's like you're playing a game against an opponent bothering you with epsilons

Anonymous

Cool interpretation :D

Balarka Sen

Yup

And notice you just need to find one delta from your bag to counter an epsilon

Anonymous

@BalarkaSen Ah, Terence Tao :P Saw him in many Youtube videos :)

Balarka Sen

@Blue Right. The thing I like about these sort of questions in MO is it helps you understand how professional mathematicians think.

And those intuitions are usually top class intuitions

Anonymous

18:10

Gotcha...lemme try out the two examples I stated above then

Balarka Sen

OK

Anonymous

Anonymous

One sec...

Anonymous

$\delta=\epsilon$ won't work

Anonymous

Something smaller than that is needed

Anonymous

18:15

Aaaahhhh

Anonymous

Got it

Anonymous

$|x+\sin(x)|<\delta+1$

Anonymous

So $\epsilon-1=\delta$

Anonymous

Should be fine

Balarka Sen

There's an issue with that.

Anonymous

18:17

Oh..yes

Anonymous

The mod sign

Balarka Sen

Well, no, not quite. I mean if you pick $\delta = \epsilon - 1$, you risk $\delta$ to be negative.

So that works only if $\epsilon > 1$

Anonymous

Let's see: $|x+\sin(x)|<|\delta|+|1|$

Balarka Sen

You have proved "for all $\epsilon > 1$ there is a $\delta > 0$"

What about the case $0 < \epsilon \leq 1$?

I like what you did though.

Anonymous

Umm...$\delta=|\epsilon-1|$ ?

Anonymous

18:21

$||\epsilon-1|+\sin(\epsilon-1)| < |\epsilon-1|+1$

Balarka Sen

Anonymous

That doesn't seem right

Anonymous

I should divide it into cases perhaps

Balarka Sen

Me too.

That's the way I'll go about it

You dealt with case 1: $\epsilon > 1$

Need to deal with case 2: $\epsilon \leq 1$

Anonymous

Let's take $0<\epsilon \leq 1$

Anonymous

18:24

$|x|<\delta$

Balarka Sen

Uh, that's not the assumption.

The assumption is $|x + \sin(x)| < \epsilon$.

You have to find $\delta$ such that $|x| < \delta$ implies the statement of the assumption holds.

Anonymous

Right..sorry

Anonymous

Let's see:

Anonymous

$|x+\sin(x)|<\epsilon$....then if I take $\delta=1-\epsilon$...then I get |1-\epsilon+ some sine term which may be positive or negative|

Anonymous

But that thing can add up to greater than 1

Anonymous

18:30

So such a delta is useless

Balarka Sen

You could directly use $|\sin(x)| < |x|$, can't you? $|x + \sin(x)| < |x| + |\sin(x)| < 2|x|$

Anonymous

$x>\sin(x)$ for small positive x

Anonymous

Wait...we are thinking in the same direction I guess

Balarka Sen

Not just for small $x$. For all positive $x$.

Yeah.

So if I pick $\delta = \epsilon/2$, that means $|x| < \delta \implies |x + \sin(x)| < \epsilon$ I think.

Anonymous

Of course....$|\epsilon/2+\sin(\epsilon/2)|<\epsilon$

Anonymous

18:36

@BalarkaSen Exactly...from the graph :)

Balarka Sen

Right, so that's the thing.

Anonymous

We can prove it mathematically also

Anonymous

$f(x)=x-\sin(x)$

Balarka Sen

Sure.

Anonymous

$f'(x)=1-\cos(x)$ and $f(0)=0$

Anonymous

18:37

So always increasing :)

Anonymous

Yay!

Anonymous

So we need to apply these algebraic and trigo properties whenever needed

Balarka Sen

Alternatively $\sin(x)$ is the perpendicular of the right triangle with angle radian $x$ and hypotenuse of length $1$

Anonymous

I see

Balarka Sen

And $x$ is the length of the arc, which is clearly longer than the hypotenuse.

Anonymous

18:38

Ah! That's a cool interpretation

Anonymous

:D

Anonymous

alright...next problem then!

Anonymous

$\sqrt{x}$

Anonymous

$|\sqrt{x}-0|=|\frac{x}{\sqrt{x}}|$

Anonymous

hmm

Anonymous

18:40

I don't think I need this

Balarka Sen

This one is not hard.

Anonymous

$\delta =\epsilon^2$

Balarka Sen

mhm

Anonymous

$|\sqrt{x}|<|\sqrt{\delta}|$....so $\epsilon=\sqrt{\delta}$

Anonymous

Alright...done

Anonymous

18:42

Phew

Anonymous

Should I try for $\sin(1/x)$ ?

Balarka Sen

That's a good example to work out.

Anonymous

$f(0)=0$

Anonymous

$f(x)=\sin(1/x)$ when $x\neq 0$

Anonymous

Okay...thinking

Balarka Sen

18:46

You want to prove $f$ is discontinuous at $x = 0$.

Anonymous

Yes...the limit won't exist

Anonymous

Then I shouldn't be able to find a $\delta$

Balarka Sen

It's good to write down the negation of "for every $\epsilon$ there is a $\delta$"

Anonymous

$\exists\epsilon>0$ such that $\forall\delta>0$, $0<|x|<\delta$ and $f(x)>\epsilon$

Balarka Sen

Yes, good

Delete the "if" (there is no "then")

Anonymous

18:53

$|\sin(1/x)-L|<\epsilon$ (should be) whenever $|x|<\delta$

Anonymous

$L$ is the limit when $x\rightarrow 0$

Anonymous

I think we can take $|L|<1$ as it is a sine's limit

Anonymous

Not sure about this step though

Balarka Sen

@Blue Oh by the way I didn't catch something here. You mean to add "there exists some $x$" before $0 < |x| < \delta$.

Anonymous

Right

Anonymous

18:59

Sorry :P

Anonymous

$|\sin(1/x)-L|<|\sin(1/x)|+|L|<|\sin(1/x)|+1<\epsilon$

Balarka Sen

I am not sure what you're doing with the $L$ here.

$L = f(0) = 0$

Anonymous

$L$ is the limit

Balarka Sen

Why would you care about the limit?

You want to prove continuity of $f$. It suffices to show $L$ is not $0$.

It doesn't matter in general if sin(1/x) has a limit as x --> 0 or something (it doesn't as a matter of fact).

Anonymous

@BalarkaSen I thought we were proving discontinuity

Anonymous

19:03

at $x=0$

Balarka Sen

Yes, so it suffices to show $\lim_{x \to 0} \sin(1/x) \neq 0$

You don't have to deal with $L$

That makes it complicated

Anonymous

Oh, that's a good idea

Anonymous

Let's see:

Anonymous

$|\sin(1/x)-0|<\epsilon$

Balarka Sen

Why don't you write out what it means to say $\lim_{x \to 0} \sin(1/x) \neq 0$ in full using epsilon-delta instead of writing bits and pieces like what you wrote?

Like, you want to choose a $\varepsilon$ such that for all $\delta$, there is some $x$ with $|x| < \delta$ and $|\sin(1/x)| > \varepsilon$

Take $\varepsilon = 1/2$ say.

Is it true then?

Anonymous

19:10

There exists an $\epsilon>0$ such that for all positive $\delta$, there exists some $x$ such that $0<|x|<\delta$ and $|\sin(1/x)-0|>\epsilon$

Balarka Sen

Right.

So since we say "there exists an $\epsilon$", to demonstrate this we better choose an $\epsilon$ to begin with.

I'm choosing $\epsilon = 1/2$. I can't choose something bigger than $1$ being sin(stuff) is always less than 1.

"|sin(1/x)| > epsilon" won't be satisfied then

Anonymous

Huh...wow...got it!

Anonymous

Phew...the negation of the statement is helpful

Balarka Sen

it is

Anonymous

I was thinking of a more complicated method using $L$ :P

Anonymous

19:13

Anyhow

Balarka Sen

So, you found out why it's true that "For all $\delta$, there is an $x$ such that $|x| < \delta$ and $|\sin(1/x)| > 1/2$"?

That is, why we can choose arbitrarily small $x$ so that $\sin(1/x)$ is bigger than $1/2$?

Anonymous

Yes...of course

Anonymous

Got that

Anonymous

For some choices of $x$ that'll hold

Anonymous

So the negation of the epsilon-delta statement becomes true

Balarka Sen

19:15

Well you haven't produced such an $x$ yet

Anonymous

There would be millions probably...it oscillates so rapidly near 0

Anonymous

Let's see

Anonymous

I think I can pick one

Balarka Sen

Ok, that's good intuition

We can move on if you want, it's not too hard

Anonymous

Anonymous

19:18

So many :P

Anonymous

Maybe we can get a general soln

Balarka Sen

Yeah you can do like $x = 1/(2n\pi + \pi/2)$

For big $n$'s

Then $\sin(1/x) = 1$

and you can choose $n$ so that for any $\delta$, $|x| < \delta$

Anonymous

Right...okay...let's move on

Anonymous

I've got a few more questions...

Balarka Sen

Ok?

Anonymous

19:23

For stuff like $\lim_{x\rightarrow 1} \frac{x^2-1}{x-1}$ we have to find the limit by the algebraic method first I guess...we can't use epsilon-delta to find limits...only prove limit is correct...right?

Anonymous

After getting $L$ algebraically, then we prove it's true

Balarka Sen

Mhm. Once you have a candidate for a limit, epsilon/delta is used to check if it works or not.

Anonymous

alright...that's cleared then...lemme check if I have any more problems

Balarka Sen

Alrighty

Anonymous

Huh...found two good problems :

Anonymous

19:32

$$\lim_{x \rightarrow \infty} \frac{[1^2x]+[2^2x]+[3^2x]+...+[n^2x]}{n^3}$$

Anonymous

[x]<x...perhaps I should use that

Anonymous

Ah no...this is easy only

Anonymous

lol

Anonymous

Prove that $\lim_{n\rightarrow\infty}(\lim_{m\rightarrow \infty} \cos^{2m}(n!\pi x))$ is nowhere continuous

Anonymous

19:35

hmm

Balarka Sen

Oh yeah I think that is a well-known function, that limit

Hm

Anonymous

If I take cosine to the power infinite

Anonymous

Then it should be 0

Anonymous

I must be going wrong

Balarka Sen

cos(N*pi) = 1 for any integer N right? (lol)

Ah, no plus or minus 1

I see, got it

@Blue Take $x$ to be rational.

Then $\cos(n! \pi x)$ is plus or minus 1, agree?

Anonymous

19:38

@BalarkaSen Sure...rational will be like p/q

Anonymous

And there's that n! thingy

Anonymous

It will cancel out the denominator

Balarka Sen

Right well uh I guess you need to exchange the limits before that

Anonymous

And the rest of the terms in n! will remain

Balarka Sen

To get n as large as possible so that q appears in n!

Anonymous

19:39

@BalarkaSen Uhhh..is that allowed?

Balarka Sen

Yeah you can exchange limits

Anonymous

I'm not conceptually clear about repeated limits it seems ;_;

Anonymous

@BalarkaSen Why?

Balarka Sen

It's non-trivial. I think it's called the dominated convergence theorem or something.

There are conditions for when you can exchange.

But let me blackbox that for a bit, ok?

Anonymous

@BalarkaSen Alright...I'd like to know the conditions though

Anonymous

19:42

Okay

Anonymous

go ahead

Balarka Sen

Suppose you can exchange limits; then the cos term becomes +/- 1

There is a power of 2m above

So after taking all the limits it should be 1

For irrational x it should be 0 I think

Anonymous

@BalarkaSen Yeah...even I think so

Anonymous

Coz it won't be of the form p/q

Anonymous

I was stuck with that exchanging of limits part

Anonymous

19:44

Is there any online article regarding that?

Balarka Sen

I am making a handwaving argument here, but I think it should be possible.

I am not actually sure if exchanging limits are needed.

@Blue Ok, here's something. Consider $a_m = \cos^{2m}(n! \pi x)$

Anonymous

@BalarkaSen Okay

Anonymous

Then?

Anonymous

(My net connection sucks, today :d)

Balarka Sen

If $|\cos(n!\pi x)| < 1$, then $a_m$ converges to $0$. This is because for any $t < 1$, $t^n$ converges to $0$.

Anonymous

19:48

Yes

Balarka Sen

The only other possibility is $|\cos(n! \pi x)| = 1$, in which case $a_m$ converges to $1$.

Anonymous

Agreed

Balarka Sen

Now notice $b_n = \lim_{m \to \infty} a_m$ is another sequence.

Which, by our analysis, can take values $0$ or $1$ depending on whether $|\cos(n!\pi x)| < 1$ or $|\cos(n! \pi x)| = 1$.

Anonymous

Oh...so the limit should not exist

Balarka Sen

If $x$ is a rational, then there is an $N$ such that for all $k > N$, $b_k = 1$.

Because; take $N$ to be the denominator of the rational form of $x$. We have $b_k = \lim_{m \to \infty} \cos^{2m}(k! \pi x)$. Given $k > N$, the denominator of $x$ cancels with $k!$, so we remain with an integral multiple of $\pi$ in the argument of $\cos$, so $b_k = \lim_{m \to \infty} ((\pm 1)^2)^m = 1$.

Makes sense?

So if $x$ is rational, the sequence $\{b_n\}$ is eventually $1$. That means $\lim_{n \to \infty} b_n = 1$.

On the other hand, if $x$ is irrational, notice that $|\cos(k!\pi x)| < 1$ for any $k$ whatsoever (because solutions to $\cos(a) = \pm 1$ are precisely the integer multiples of $a$, but $k!x$ is not integer for any $k$ by irrationality of $x$).

Hence $\lim_{m \to \infty} a_m = 0$.

Anonymous

19:57

Awesome

Balarka Sen

That is to say, if $x$ is irrational $b_k = 0$ for all $k$, hence $\lim_{n \to \infty} b_n = 0$.

Anonymous

That was a good treatment of sequences

Anonymous

I get it now

Balarka Sen

Yeah it's fiddling with sequences.

Anonymous

It seemed a bit hazy at first

Anonymous

19:58

I think I need to learn limit as sequences formally someday soon

Balarka Sen

That means the function you wrote down is exactly the function which is 1 on rationals, 0 on irrationals

That's nowhere continuous brah

Anonymous

@BalarkaSen Yep! :D

Anonymous

So we're done with this

Balarka Sen

@Blue You need to get Rudin's little book

and just work through it at some point

the first 3 chapters or something covers everything you need

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♦ $Mathematics$ the no-normie zone

Transcript for

♦ the no-normie zone

♦ $Mathematics$ the no-normie zone