Mathematics - 2017-05-06 (page 1 of 4)

Alberto Andrenucci

00:01

Good evening boys.

May I ask for a little help?

If someone is here.

TheGreatDuck

ok

Alberto Andrenucci

May I write in LaTeX here?

TheGreatDuck

sure

Alberto Andrenucci

$$\lim_{x\to\infty} \int_{x}^{x^2} e^{-t^2} dt$$

I'm trying to study this limit, but I'm having difficulties.

Semiclassical

to enable it, use the latex in chat link in the room desc

Ted Shifrin

00:04

Great question, @Alberto.

What do you think the answer is?

Alberto Andrenucci

Trying the link.

$$\lim_{x\to\infty} \int_{x}^{x^2} e^{-t^2} dt$$

Damn, it doesn't work.

TheGreatDuck

lol

Ted Shifrin

It works fine for us. We see what you have typed nicely.

Alberto Andrenucci

Oh, nice. I don't see it.

Semiclassical

You'll want to add the link listed there as a bookmark, then click said bookmark while in this chat.

TheGreatDuck

00:06

i dont see latex rendered

Ted Shifrin

For some people, the MathJax bookmark needs to be updated, as in the 6th starred item to the right.

Semiclassical

Huh, odd.

Ted Shifrin

I haven't had to do that yet, which is weird.

Semiclassical

Nor I.

Ted Shifrin

Did you load the bookmark, @Alberto?

TheGreatDuck

00:06

i don't use the link though so I don't really count.

just wanted to make it clear that not everyone sees it rendered

:p

Ted Shifrin

Ducks render a lot of fat if you cook them right.

2

Alberto Andrenucci

I'm really having trouble doing it, sorry.

Ted Shifrin

Well, don't worry, @Alberto. What do you think your limit is?

TheGreatDuck

@TedShifrin im a human though

Alberto Andrenucci

@Ted

TheGreatDuck

00:07

cooking humans to render fat is a crime

Ted Shifrin

I think of you as a duck, Duck.

Alberto Andrenucci

I don't know, really.

TheGreatDuck

i don't think of myself as a duck

Alberto Andrenucci

I previously studied the integral function.

Ted Shifrin

Have you drawn a picture of the function?

TheGreatDuck

00:08

but i do think of myself as a quack

Semiclassical

And yet you call yourself one :P

Alberto Andrenucci

I've only drawn a picture of my integral function.

Ted Shifrin

Draw $e^{-x^2}$ and what is an upper estimate on that integral?

Alberto Andrenucci

Before we continue, can you please tell me how to see rendered LaTeX? :')

Ted Shifrin

LOL, you have to use that link to get a bookmark, and then click on the bookmark when you're in here. (Are you on mobile or on a desktop?)

TheGreatDuck

00:10

mobile

Alberto Andrenucci

Desktop. I'll try it. Gimme a second.

Ted Shifrin

I got it to work on iPad and iPhone, Duck, but it was sneaky. I think robjohn put up the instructions that I passed on to him.

Alberto Andrenucci

Oh, finally I got it.

Ted Shifrin

Yippee :)

If only math were that easy :)

Hi @TimThe

Semiclassical

lol.

Tim The Enchanter

00:12

@TedShifrin Hello

Alberto Andrenucci

$lim_{x\to\infty}e^{-x^2}=\sqrt(pi)

If I put a dollar at the end it'd be better.

Ted Shifrin

No, no, @Alberto. But the math is wrong.

Semiclassical

You can edit what you wrote.

Alberto Andrenucci

Oh damn, I'm stupid.

I wanted to say a different thing.

It's late, and I'm tired.

Ted Shifrin

Relax.

Alberto Andrenucci

00:13

Obviously the limit is $0^+$

Ted Shifrin

Yeah, you shouldn't be doing math at 2 AM.

Semiclassical

^

Alberto Andrenucci

I wanted to say that the integral converges at $\sqrt{\pi}$

Ted Shifrin

Right. But you have a decreasing function. So what is always an upper bound on the integral from $a$ to $b$?

The integral does that from $-\infty$ to $\infty$. That's not relevant here.

Alberto Andrenucci

$1$ is always an upper bound.

Ted Shifrin

00:14

On the function, yes, but what about on the interval $[x,x^2]$?

Semiclassical

Wouldn't it also work just to write $\int_0^{x^2}-\int_0^x$ and compute the two (equal) limits separately?

Simple

Anzone here knows the upwind scheme

Anyone

Ted Shifrin

@Semiclassic — yeah, but that's overkill and doesn't lead to the best learning.

Alberto Andrenucci

$f(x)>f(x^2)$

(Talking about $x>0$)

Ted Shifrin

Right. In fact, $f(x)>f(y)$ for all $y>x$.

Alberto Andrenucci

00:16

True.

Ted Shifrin

So the area we're looking at is contained in a rectangle with base $[x,x^2]$ and what height?

Alberto Andrenucci

$f(x)$

Ted Shifrin

So what is an upper bound on your integral?

Alberto Andrenucci

$[x,x^2]*f(x)$

Ted Shifrin

Um ...

Write down the explicit number (function of $x$)

Alberto Andrenucci

00:19

$(x(x-1))e^x$?

Ted Shifrin

What's $f(x)$ here?

Gnight, @MikeM.

Alberto Andrenucci

$e^x$?

Ted Shifrin

No, look at the problem again, @Alberto.

Alberto Andrenucci

Oh, I want an upper bound for my entire function. So I'd say.. $e^{-x^2}*(x^2-x)e^x$

Ted Shifrin

Why do you have $e^x$ in there?

Don't use * in math.

This isn't a computer program. :)

Alberto Andrenucci

00:22

I think I have $e^x$ because I'm taking the upper bound of the area and then multypling it for the other piece.

Ted Shifrin

Huh? The maximum height is $f(x) = e^{-x^2}$ and the base of the rectangle is $x^2-x$, as you said.

Alberto Andrenucci

Oh, sure. You're completely right.

So.. I'd say the upper bound for my general function is $e^{-2x^2}(x^2-x)$

Ted Shifrin

Why the $2$?

Tim The Enchanter

So in this answer math.stackexchange.com/a/2267129/385075 he uses a functional equation for $f(x-1)$ in terms of $f(x)$ to show that each class of extrema of a function has a different magnitude. As I see it, this would require the x values of the extrema to each differ by 1, or am I missing something?

Alberto Andrenucci

Because without the 2 it's only an upper bound for my integral. Then I'm multiplying it for $e^{-x^2}$

Uhh, now I understand.

I wrote in a wrong way the limit.

Tim The Enchanter

00:27

Never mind, I'm just being stupid again.

Ted Shifrin

So do we agree it should just be $e^{-x^2}(x^2-x)$?

I can't process that fast, @TimThe.

Alberto Andrenucci

$$lim_{x\to\infty} e^{-x^2} int_{x}^{x^2} e^{-x^2} dx$$

Ted Shifrin

Oh, wait, @Alberto. Are you changing the question now?

Tim The Enchanter

@TedShifrin No need Ted, I figured out what I was missing. Thanks anyway.

Ted Shifrin

I bet that is not the question, @Alberto. Let's be sure now.

Alberto Andrenucci

00:29

I don't know why my LaTeX is not computing well. And sorry for my mistake, I knew I was missing something.

I'll write it again.

Well.

$\lim_{x\to\infty} e^{x^2}\int_{x}^{x^2} e^{-t^2} dt$

Now it's right.

Semiclassical

That's not what you wrote earlier, though.

Ted Shifrin

OK, this is more subtle.

Alberto Andrenucci

I know. I'm not good at writing in LaTeX and I always do mistakes.

Ted Shifrin

Now we have to use L'Hôpital's rule.

Can you see how to set this up with L'Hôpital's rule?

Alberto Andrenucci

The limit I asked for before it's easy, that's why I was having doubts with your answers.

Ted Shifrin

00:33

Yeah, our earlier limit was obviously 0. Now we have $\infty\cdot 0$, so we have to work harder.

Alberto Andrenucci

I want to put it in a $\frac{\infty}{\infty}$ or $\frac{0}{0}$ form to use Hopital.

Ted Shifrin

Well, we already know that part of it goes to $0$, so let's go for $0/0$. :)

Alberto Andrenucci

Uhh. If I multiply and divide for $e^{-x^2}$ I should obtain $\frac{\int_{x}^{x^2} e^{-t^2} dt}{e^{-x^2}}$

Ted Shifrin

Proceed.

Alberto Andrenucci

Now It's suitable. Thanks a lot.

Ted Shifrin

00:38

And, what's more, now you know how to read ChatJax in here.

Alberto Andrenucci

I own you a favour! :') Thanks again, you're gentle.

Ted Shifrin

LOL, you're welcome.

Alberto Andrenucci

So the limit should be $-\infty$, right?

Ted Shifrin

No, that can't be possible.

Are you doing the Fundamental Theorem of Calculus correctly?

Alberto Andrenucci

I'll try again before I ask you for a further hint.

Ted Shifrin

00:41

Good :)

But positive/positive can't go to $-\infty$ ... :)

Alberto Andrenucci

Uh, I had a minus I shouldn't have had. I think the limit is 0.

Ted Shifrin

That is the correct answer.

Alberto Andrenucci

Uh. Finally.

Thanks again.

Semiclassical

A good follow up question is what you'd need to have as a prefactor in order to get a finite limit.

Ted Shifrin

It's quite reasonable, based on our earlier discussion, because the exponential decreases so fast that that rectangle was a terrible overestimate.

Good question, @Semiclassic.

Alberto Andrenucci

00:45

What do you mean exactly?

Ted Shifrin

Me, or Semiclassic?

Semiclassical

Suppose you were to have also put $x$ out in front. Would the limit still be zero?

And if so, what about higher powers etc.

(I happen to know what power is needed because I cheated with Mathematica first.)

Ted Shifrin

I was thinking about changing the $e^{-x^2}$ in the denominator, instead.

Semiclassical

I think that if you make it $e^{-ax^2}$ with $a>1$ then it diverges.

Alberto Andrenucci

I'll think about it. Thanks for the question! It's always nice when I log in on Stack. I always learn things in a better way.

Ted Shifrin

00:50

Sleep well, @Alberto. Buono notte.

Semiclassical

night.

Ted Shifrin

So what do I do to get a nonzero finite limit, @Semiclassic?

Alberto Andrenucci

I'll not sleep right now. I'm a night student, you know. AHAH Thanks again! And.. Ted, it's "Buona Notte", with the final A. The O means you are talking about a masculine thing, night, in italian, is feminine.

Semiclassical

$\displaystyle \lim_{x\to \infty} xe^{x^2}\int_x^{x^2}e^{-t^2}\,dt=\frac12$.

Ted Shifrin

Oops, sorry, @Alberto. I'm good with French and decent with German, but know no Italian :(

Alberto Andrenucci

00:52

Just telling you so you can learn, you tought me how to solve that limit, I'll teach you how to speak in Italian. :lol:

Ted Shifrin

LOL, keep teaching. Now that I think about it, "night" is feminine in every language I can think of (except for English, which is too ridiculous to have genders for nouns).

Semiclassical

I did use Mathematica to facilitate finding that, though.

Ted Shifrin

@Semiclassic: I don't cheat the way you do. I did it by thinking. You still haven't answered my question :P

Semiclassical

Which? To get a nonzero finite limit, one has to multiply by $xe^{x^2}$ rather than $e^{x^2}$.

Ted Shifrin

No, I want to change the $e^{-x^2}$ in the denominator, once we've rewritten.

(And don't tell me $e^{-x^2}/x$.)

Semiclassical

00:56

Okay, I'll just tell you $e^{-x^2-\ln x}$ instead :P

Ted Shifrin

puts Semiclassic on ignore

Semiclassical

pfft

I don't really see what you're looking for, though.

Ted Shifrin

Well, I guess, having worked the problem (unlike you), I knew what I wanted the derivative to be to get a finite limit.

Semiclassical

So how would you modify it?

Ted Shifrin

You know me better than that — I don't give away answers.

Semiclassical

00:59

roll

I don't see what's wrong with $e^{-x^2}/x$ as an answer, to be honest.

Ted Shifrin

I was thinking more simply of something whose derivative would be just $-e^{-x^2}$.

Semiclassical

So you prefer the transcendental error function?

That...ehhhh

To me, -that- feels like cheating.

Ted Shifrin

It's sort of interesting that it and $e^{-x^2}/x$ are "on a par."

In fact, that seems totally wrong.

Semiclassical

The asymptotic behavior of erf is kinda weird.

Ted Shifrin

No, I guess it doesn't seem wrong, after all.

Interesting in terms of average value.

Semiclassical

01:03

I actually have run into that question of erf's large-x behavior before, in fact.

Ted Shifrin

I haven't.

Semiclassical

Lemme find the slide I know.

John Jiang

@TedShifrin Thanks a lot for the Griffith reference. I should just work through it.

Ted Shifrin

There's some beautiful stuff in there, @JohnJ.

I gave a bunch of geometry seminar lectures on it years ago.

Semiclassical

@TedShifrin I was hoping to find the slides, but couldn't.

Ted Shifrin

01:11

I forgive you, @Semiclassic.

Semiclassical

But I did find the bit from the author's book on Google books: books.google.com/…

John Jiang

@TedShifrin Oh wow. I am eager now!

Ted Shifrin

We need to get together next time I drive up to the Bay Area, @JohnJ. Probably this fall sometime.

John Jiang

Yes you are always welcome to my suburb! Maybe right after my Greece trip in early September.

Semiclassical

Basically, they construct the sectionally analytic function $$\psi(z)= \left\{\begin{array} {e^{z^2}}(2-\text{erfc}(z)),&\text{Re }z>0\\-e^{z^2}\text{erfc}(z),&\text{Re }z<0 \end{array}\right.$$

there we go.

Ted Shifrin

01:14

@JohnJ: I'm going to Europe in June. We'll see if the Orange Cheeto lets me back in.

John Jiang

@Ted

Semiclassical

sigh. gone

John Jiang

we'll see if he quits before then

Ted Shifrin

Oh, too much ego to quit.

@Semiclassic: What's erfc?

Semiclassical

bah. Should be $e^{z^2}$ in front of the first function.

complementary error function i.e. $\int_z^\infty e^{-t^2}\,dt$.

Ted Shifrin

01:16

Oh, OK

John Jiang

erfc = 1 - int_0^x gaussian or something

int_^{-\infty}^x

Ted Shifrin

Well, if there's a $\sqrt{2\pi}$ normalizing or something.

OK, I don't want to think about this any more.

John Jiang

lol

Ted Shifrin

Hell, I'm retired. This is too much work.

Semiclassical

Basically, they then go on to argue that $\psi(z)$ is a solution to a Riemann-Hilbert problem

Ted Shifrin

01:18

@JohnJ: The fun news is that I may well teach a class for AoPS starting in the fall. They're building a brick-and-mortar school about 20 minutes from me.

3

Semiclassical

With the punchline being at the end of the second page, namely that $\psi(z)$ can be found globally from a single fast Fourier transform

(the slide actually talked about that in a little more detail, whereas the book refers to a later section. so not as helpful in that regard.)

John Jiang

@TedShifrin That would be such a great thing! Just don't get burned out by those kids.

Ted Shifrin

Well, all but one of my attempts to volunteer to work with kids who really need help have been ignored. I'm bummed.

John Jiang

@TedShifrin I wish I have the your energy level today.

Semiclassical

See this page for the details of that, though I don't consider that terribly clear.

Ted Shifrin

01:21

@Semiclassic: Let it go. :)

Semiclassical

pff

Ted Shifrin

@JohnJ: Trying to raise kids takes too much energy :)

Semiclassical

It is neat, though, that it converges uniformly and superalgebraically.

John Jiang

@TedShifrin there is some truth to it. Though my kids are certainly not the worst, since they probably inherited my low energy levels :p

Ted Shifrin

LOL ... tell your wife to energize you a bit :)

OK, I'm outta here. Bye, all.

John Jiang

01:24

I am hoping to get revitalized by Archimedes in the fall.

ok hope to see you soon! good night

Ted Shifrin

Archimedes, huh? Cool. Talk soon, John.

John Jiang

yep, or pythagoras

though I have less respect for the altter

latter

Simply Beautiful Art

01:40

Frivolous theorem of arithmetic

12

Michael Albanese

01:58

Has anyone here been able to access nLab recently?

Simple

02:36

In the context of a Bernoulli process, does $P(S_3 = 2\,|\,S_5 = 3)=0$

John Jiang

depending on whether you have idle probability

Simple

idle probability?

John Jiang

@Simple if S_{i+1} is always S_i +/- 1, then this probability is clearly 0

is that what you meant by bernoulli process?

Simple

ok

John Jiang

I usually call that simple random walk

Simple

02:45

there is possibility that $S_4=3$

John Jiang

then you are looking at a lazy simple random walk

mean with nonzero idle probability (nonstandard terminology)

so then your original probaiblity is not 0

Simple

Thanks

Suppose $T_i: T_i=\min n\,|\,S_n=i$, $S_n$ is a bernoulli process. How to compute $E(T_i\,|\,T_{i-1})$

John Jiang

I don't think you can, not enough info

the min could have been achieved very recently or a long time ago

in other words, T_i is not Markov

oh actually I misread

Simple

$T_i$ is the ﬁrst time the Bernoulli process $S_n$ has $i$ successes.

John Jiang

right, then its' doable

Simple

02:57

$E(T_i\,|\,T_{i-1})=E(S_n\,|\,S_{n-1})?$

John Jiang

that should just be E(T_1) right?

Simple

I am not sure that

John Jiang

I mean E(T_i | T_{i-1}) = E(T_1) + T_{i-1} I think

Basically because T_i - T_{i-1} has the same distribution as T_1

Simple

$T_i-T_{i-1}=S_n-S_{n-1}=1$?

John Jiang

nono, n is a dummy variable

in the definition of T_i

try to mentally or on paper simulate what T_i is supposed to mean

Simple

03:01

$T_i$ is the first time to reach $n$

John Jiang

no it's the first time to reach i

n is time here

that's why n is dummy

Simple

yes, I misread it on the problem

John Jiang

ok cool. glad you didn't miscopy it:)

Simple

does $T_i=T_{i-1}+T_1$

John Jiang

good question

it depends on what you mean by the right hand side. We need to know the joint distribution of what you meant by T_{i-1} and T_1

they need to be independent, then yes

Simple

03:09

We know $T_i=\min n\,|\,S_n=i$, $T_1$ and $T_{i-1}$ are independent

John Jiang

if T_1 and T_{i-1} refer to the random variable associated with the same sample paths, then they are dependent

because knowing T_1 will restrict the value that T_{i-1} can take

Simple

Oh, right. $S_n$ and $S_{n-1}$ are dependent, so, $T_{i}, T_{i-1}$ are dependnet as well

John Jiang

yes, so saying T_i - T_{i-1} equals T_1 in distribution is the most natural thiing

since you do want T_i and T_{i-1} to be dependent here

of course you can also say T_{i-1} + T_1 equals T_i in distribution for random walks

but this is not true for markov chains in general

independent T_{i-1} and T_1 that is

Simple

03:30

I still confuse your answer $E(T_i\,|\,T_{i-1}) = E(T_1) + E(T_{i-1})$

John Jiang

you understand conditional expectation?

Simple

$E(T_i\,|\,T_{i-1}=\sum iP(T_i,T_{i-1})/P(T_{i-1})$

John Jiang

not really

conditional expectation is a bit different from conditional probability, although the latter is a special case of the former

when the random variables are indicators

Here is an intuitive explanation: imagine T_{i-1} is already known to you, what do you expect T_i to be

Simple

$i$

John Jiang

that's E(T_i | T_{i-1}). But this is itself a random variable, since you want to view this from the perspective of you at T_0

Hatshepsut

03:38

I just learned about the bolzano weierstrass theorem and i came up with an interpretation. Is this right? In frogger, No matter how many lives you start with, if you dont move you'll eventually lose.

John Jiang

no! remember i is not the same as n

that expectation should be T_{i-1} + E(T_1)

T_{i-1} is the part of T_i you already know

the remaining part of T_i is equi-distributed with T_1, so when you take expectation, you just get E(T_1)

Simple

03:55

$P(T_i\,|\,T_{i-1})=1/i$ @JohnJiang

John Jiang

the LHS makes no sense. E.g what do you mean by P(T_i)?

Hatshepsut

and no matter how many lanes

John Jiang

T_i is not an event, it's a random variable

so you can't take its probability

unless you meant E(T_i | T_{i-1}), but the RHS is incorrect. It should just be E(T_1)

Simple

I still don't see the reason behind it. Could you show me all the steps?

BAYMAX

is this the graph of $log(x) / x$

How do I plot it for natural numbers only

John Jiang

04:04

S_n = sum_{i=1}^n X_i, where X_i = +/-1 with probability 0.5, and X_i's are independent

is this your definition of S_n?

Simple

$X_i=1$ with probability $p$, $X_i=0$ with probability $1-p$

John Jiang

ok

Then do you agree that T_i - T_{i-1} has the same distribution as T_1?

P(T_1 = 0) = 0, P(T_1 = 1) = P(X_1 = 1) = p, P(T_1 = 2) = P(X_1 = 0 & X_2 = 1) = (1-p) * p, etc

Simple

We start from $T_0$, then up to $T_1$, so $T_1-T_0$ is the same as $T_1$

yes

John Jiang

P(T_i - T_{i-1} = 0) = 0, P(T_i - T_{i-1} = 1) = \sum_{j=1}^\infty P(T_i - j = 1|T_{i-1} = j) * P(T_{i-1} = j) = P(X=1) * \sum_{j-1}^\infty P(T_{i-1} = j) = P(X=1) = p

for i > 1

so to be very pedantic, you need to write down all the possibilities of T_{i-1} and apply Bayes rule to each. There are infinitely many such

note that P(T_i - j = 1 | T_{i-1} = j) = P(X_{j+1} = 1). So I wrote X to stand for any of X_{j+1}, since they are all identically distributed

BAYMAX

Maximum value of Log(n) / n ?

for which n?

n is a natural number!

and what happens when I extend these natural numbers case to real number case?

like maximum value of Log(x)/x where x is a real numbere (positive)

John Jiang

04:22

@Simple yea probability is often easy to reason but laborious to write down completely rigorously

Simple

$E(T_i\,|\,T_{i-1})=E(T_i-T_{i-1})=E(T_1)$?

BAYMAX

nice one there @SimplyBeautifulArt

John Jiang

nope, that was my typo above. It should be T_{i-1} + E(T_1)

as stated earlier

Simple

You mean $E(T_{i-1})+E(T_1)$

John Jiang

nope that's not waht I meant

I really meant T_{i-1} + E(T_1)

the most tricky concept in probability theory is "conditional expectation"

there is a measure theoretic way to understand it, and a more intuitive way.

E(.. | T_{i-1}) means what is expected value of .. if you already know the value of T_{i-1} (and that's all you know)

Simple

04:28

$E(T_i\,|\,T_{i-1})=E(T_i-T_{i-1}+T_{i-1}\,|\,T_{i-1})=E(T_1)+E(T_{i-1}\,|\,T_{i‌-1})=E(T_1)+i+1$

John Jiang

all correct except the last step

E(T_{i-1}|T_{i-1}) = T_{i-1}

read my sentence above

the expected value of something you already know is that something itself!

Simple

$E(T_{i-1})=np$

John Jiang

you seem confused about n and i

Simple

ip

John Jiang

n is always a dummy variable here.

Simple

04:30

(i-1)p

John Jiang

yea except (i-1) p I think

yep!

Simple

the final result is $p+(i-1)p$

John Jiang

but when you write E(T_{i-1}) you are from the perspective of the beginning of the process

not the middle of the process when you already know the value of T_{i-1}

anyway I hope I conveyed enough of the subtleties in conditional probability here that you have some faith in the theory

Simple

Although I still have a little confusion, I will study more about it. Thanks your help

John Jiang

no problem. Just note that E_(T_i) is the same as E(T_i | T_0)

Anonymous

05:40

@BalarkaSen If $f(x)f(y)=f(x+y)$ [$f$ is a real valued differentiable function]. Then can $f(x)$ be anything other than $f(x)=0$ and $f(x)=e^{kx}$ ? [I solved using the differential eqn method but bit confused about the $f(0)=0$ case]

Balarka Sen

@blue $f(0) = 0$ means $f(x) = f(x+0) = f(x)f(0) = 0$ for all $x$.

Anonymous

@BalarkaSen Oh, that's great. So no other possible solutions. yay

Balarka Sen

Well, no, that's just the $f(x) = 0$ solution. There's no other possible solutions.

Ah, you edited.

Anonymous

(edited^)

Anonymous

:P

Anonymous

05:49

These functional equations can be confusing

Anonymous

More or less I see that the differential eqn method works for most

Balarka Sen

Yeah you have to play with the symmetry of the equation.

Anonymous

Yup, okay thanks :)

Balarka Sen

@blue Only if the function is differentiable. I guess if you're just writing an answer then you don't care as long as the answer is correct.

Anonymous

@BalarkaSen Oh yes, and that. We have only functional equations for smooth differentiable and real valued functions as of now(in syllabus)

Balarka Sen

05:52

OK.

Anonymous

BTW if a function is infinitely differentiable at a point and all its derivatives vanish then must it be a constant function ?

Balarka Sen

Nope :)

Anonymous

But from the Taylor polynomial it seems so

Anonymous

(Assuming it's smooth and derivable)

Balarka Sen

The crucial point about Taylor polynomial is that it always has a remainder term. It does not express the function as a convergent power series - that is a far stronger notion of analytic functions.

Bump function

In mathematics, a bump function is a function f : Rn → R on a Euclidean space Rn which is both smooth (in the sense of having continuous derivatives of all orders) and compactly supported. These are also called test functions [φ]. The space of all bump functions on Rn is denoted C 0 ∞ ( R n ) {\displaystyle C_{0}^{\infty }(\mathbf {R} ^{n})} or ...

Anonymous

05:58

@BalarkaSen Umm, so the function must have a convergent Taylor series for that claim to be valid?

Balarka Sen

Yes, if a function has a convergent power series expansion on a neighborhood, then vanishing derivatives implies it's constant.

This is the thing.

Anonymous

@BalarkaSen "on a neighborhood", does that imply in other places it might not have a convergent taylor series but yet it will be constant if in a certain neighbourhood its all derivatives vanish?

Balarka Sen

No, it will vanish on that neighborhood.

Not necessarily anywhere else.

Anonymous

I see. Interesting

Balarka Sen

But you're analytic everywhere then you're constant everywhere.

Anonymous

06:03

Is there any method to check where a function is analytic ?

Balarka Sen

You can recognize most analytic functions to be analytic by understanding if they have convergent Taylor series

Eg $e^x$ is clearly analytic.

Anonymous

@BalarkaSen I see. Right. I can check if Taylor series is converging by putting some values in the initial terms anyway. Cool!

Jo Be

06:35

Hey, is the nLab down for you guys too? ncatlab.org

Anonymous

"ncatlab.org’s server DNS address could not be found"

Anonymous

"This site can’t be reached"

Jo Be

damn

thanks

Anonymous

np

satyatech

07:50

I don't know how the above image got flipped ::but try the above summation.It seems to be arithmo -geometric progression and I have no idea how to talk to this guy!!

Maybe you guys can talk so help me to interact with this guy!

Balarka Sen

Let $S$ be that sum. $S - 1/2S = 1/4 + 1/8 + 1/16 + \cdots$, which is a geometric series.

satyatech

thanks @BalarkaSen

gansub

08:55

@TedShifrin - am I right in saying that the difference between an ordinary vector field and a "material" vector field is defined by the following relation V(a,t) = Jij(t) V(a). That is a material vector field is defined in terms of it's neighboring points in space and time but a ordinary vector field is a mapping from a particular origin to a point in space ?

The above definition I gave was in terms of a Lagrangian reference

but you could take an Eulerian definition similar to the above

this point is not emphasized in fluid dynamics texts books

the difference between a material vector field and a ordinary vector field

people just state the definition of the material derivative and leave it at that

the above definition helps you in relating the material derivative to the Lie derivative

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