Mathematics - 2017-04-25 (page 1 of 4)

Meow Mix

00:00

@Daminark I think I've noun'd Ted. It now means "a professor you consult when you need a hint or a smack."

Ted Shifrin

LOL ... I am more liberal with both hints and smacks than most professors.

Daminark

Idea: Let's start now by cracking open a book on algebraic geometry and whenever we see a word we don't know, we just go down the wikipedia chain until it makes sense

Proposition: This will probably not go any faster than the normal track

Topologicalife

The problem is the notes I have are in spanish language.

Ted Shifrin

I do not recommend learning by Wiki googling.

Topologicalife

If they would be in english I already would had linked them

Ted Shifrin

00:02

So, in general, a solution is of the form $c_1e^{\lambda_1 t}v_1 + c_2e^{\lambda_2 t}v_2$, where $v_j$ is the eigenvector corresponding to eigenvalue $\lambda_j$.

Daminark

Lol it was kind of a joke, wikipedia articles can be a bit tricky to glean information

Mike Miller

@MeowMix Please stop.

Meow Mix

This band-aid is tight.

Topologicalife

Yeah, I know. I'm just trying to help a friend. He does it diagonalizing.

Ted Shifrin

When you convert from complex to real form, as in this case, you still think through the same matrix stuff.

Meow Mix

00:03

Oops, I meant "BAND-AID® Brand Adhesive Bandages"

Mike Miller

I don't know whether I should be telling you inspirational stories about success to convince you not to worry or whether I should just say that self-deprecation isn't good for anybody in this room but in either case don't do it

Ted Shifrin

I support @MikeM in this order.

Meow Mix

Alright, Mike.

And @Ted too, sure.

Daminark

@Mike In total fairness, feelings of utter incompetency are very hard to shake, even when unjustified, especially in the right company. But I do agree that it's probably best not to worry too much about it right now

Mike Miller

I know that. Every single graduate student I know has impostor syndrome, including myself.

Ted Shifrin

00:06

@Topologicalife: This example is sort of confusing for the usual matrix set-up. When you convert the complex diagonal matrix to its corresponding real form with respect to that real basis, you just get your original matrix back: $\begin{bmatrix} 0 & 1\\-1& 0\end{bmatrix}$, so we go in a big circle. No pun intended.

Meow Mix

Okay, the band-aid made everything better.

I just think of it as a deep paper cut.

Topologicalife

But where is the minus sign?

Ted Shifrin

@MikeM: I just had dinner last night with an old student/friend of mine (now at UCSD) and 6 of his graduate classmates. Quite fun.

Topologicalife

I don't see where I failed.

Mike Miller

This is bad! It's a widespread mental health issue that plenty of people aren't getting therapy/medication for. And one should either try to combat it themselves, with the help of friend, and/or seek professional help. I just don't like self-deprecation in this room in general because it seems to spread as rapidly as a yawn.

Martin Sleziak

00:07

Since it seems that there are some differential geometers around, I'll copy this here (I am not sure how much following that room has):

in Geometry & Topology, 3 mins ago, by Zophikel

@MikeMiller do you any books that I could refer too for dealing with DG on an advanced undergradute level

Mike Miller

@TedShifrin Do I know him?

Ted Shifrin

Nope, @MikeM.

Mike Miller

Do you know I don't know him?

:D

Ted Shifrin

He does interesting applied stuff, despite having been taught by me in 4 courses :P

Mike Miller

Ah. Then yes, you know I don't know.

Ted Shifrin

00:08

@MartinS: We've actually addressed that question with that person in this chatroom.

Martin Sleziak

Oh. I did not know that.

Ted Shifrin

@Topologicalife: I don't really want to type everything out here, but I think you have something confused somewhere. Note that your $C$ matrix is wrong, as I explained.

Topologicalife

I think one can understand what the notes say without to know spanish: dropbox.com/s/5ozv4pl1omvp29p/EDO.jpg

Ted Shifrin

Oh, no, it's OK.

Daminark

That is true, I guess it's just that the feeling of incompetency coupled with the feeling that to get anywhere in life you need to be a god leads to quite some anxiety/panic, and sometimes some slack ought be awarded

Topologicalife

00:10

So where is the problem? :(

Meow Mix

autovalor = eigenvalue?

Topologicalife

Yeah.

Meow Mix

This isn't that hard to understand

Topologicalife

Math are universal :-)

Daminark

And in contexts like this room, it really catches up to you. In part from here, and in part from seeing a few people in classes, I begin to really think that I'm kind of slow for not having seen algebraic topology 3 years ago.

Ted Shifrin

00:12

@Topologicalife: I don't understand what the issue is.

Daminark

Or solved an open problem if I look at professors... Either way, while self-depracation ought not be encouraged, it is somewhat understandable, ish

Topologicalife

@TedShifrin If I compute $Ce^{\mbox{Re}(\lambda)} \begin{bmatrix}{\cos (wt)}&{-\sin(wt)}\\{\sin(wt)}&{\cos(wt)}\end{bmatrix} \begin{bmatrix}{c_1}\\{c_2}\end{bmatrix}$, my system doesn't hold this solution

and it must hold it, so somepart is wrong

if I compute $\begin{bmatrix} 0 & 1\\1& 0\end{bmatrix}e^{\mbox{Re}(\lambda)} \begin{bmatrix}{\cos (wt)}&{-\sin(wt)}\\{\sin(wt)}&{\cos(wt)}\end{bmatrix} \begin{bmatrix}{c_1}\\{c_2}\end{bmatrix}$ I get as solution $$\left[\begin{array}{cc}{x(t)}\\{y(t)} \end{array}\right] = C e^{tB}=C e^0 \begin{bmatrix}{cos t}&{-sen t}\\{sen t}&{cos t}\end{bmatrix} \left[\begin{array}{cc}{c_1}\\{c_2} \end{array}\right] = \left[\begin{array}{cc}{c_1 sen t + c_2 cost}\\{c_1 cos t - c_2 sen t} \end{array}\right]$$

which doesn't hold the initial system of ODE's.

Err, sorry

I meant $\begin{bmatrix} 0 & 1\\1& 0\end{bmatrix}$

Ted Shifrin

Sure it works, @Topologicalife. You have $x'=y$ and $y'=-x$.

Topologicalife

I have $\begin{cases} x'=-y \\y'=x \end{cases}$

Ted Shifrin

Guess again.

Topologicalife

00:17

What should I guess?

Ted Shifrin

Oh, you have something wrong in what you typed.

Topologicalife

Mm where?

Ted Shifrin

Oh, no, it's right. You did multiply by $C$. And if you look at your final thing $x'=y$. What is your issue?

Topologicalife

but my system of ODE's is $x' = -y$ and $y' = x$

note the minus sign in the first equation

it isn't $x' = y$, $y' = -x$

Ted Shifrin

Right, so you have to use $-1$ for the imaginary part, not $+1$.

Go back through the derivation in the first place.

Topologicalife

00:20

Yeah, but I don't see why.

What I did is this:

Ted Shifrin

You have to go through the derivation to see it.

Topologicalife

Yeah, here it is:

I solved $(A-\lambda I)w=0$ where $w=(x_1, x_2)$, then I get $\begin{bmatrix}{-i}&{-1}\\{1}&{-i}\end{bmatrix} \left[\begin{array}{cc}{x_1}\\{x_2} \end{array}\right] = \left[\begin{array}{cc}{0}\\{0} \end{array}\right]$

i.e: $\begin{cases} -i x_1 -x_2 =0 \\ x_1 - i x_2=0 \end{cases}$

and solving, $x_1=ix_2$ with $x_2 = s$, where $s$ is arbitrary

if $s=1$ then $w=(i,1)=(0,1)+i(1,0)=u+iv$

so $C = \begin{bmatrix} 0 & 1\\1& 0\end{bmatrix}$

Ted Shifrin

But whether you rotate $+\pi/2$ or $-\pi/2$, the eigenvalues are $\pm i$ either way. Here's the exercise I gave my students: Prove that if $A$ is a $2\times 2$ real matrix with eigenvalues $\alpha\pm i\beta\in\Bbb C$, then there is a basis $\mathcal B$ for $\Bbb R^2$ so that
$$[A]_{\mathcal B} = \begin{bmatrix}\alpha&-\beta \\ \beta&\alpha\end{bmatrix}.$$
(Hint: There are $x,y\in\Bbb R^2$ so that $A(x+iy) = (\alpha-i\beta)(x+iy)$.

Topologicalife

Yeah, I know.

Ted Shifrin

Note that you have to pick the right sign on the imaginary part for it to come out right.

Topologicalife

00:26

and that is the $B$ matrix in my notes.

Uhm, wait.

Yeah, I did it correctly.

Ted Shifrin

But I'm telling you you should be looking at the $-i$ eigenvector.

That will change a sign.

Topologicalife

I did $tA = C [A]_{\mathcal B} C^{-1} = C \begin{bmatrix}\alpha&-\beta \\ \beta&\alpha\end{bmatrix}C^{-1}$

Ah, uhm...

I don't get why.

Ted Shifrin

@Topologicalife: At this point, I'm lost. The text you're using may have messed up the sign as I'm indicating.

Topologicalife

Probably. I'm seeing the notes i am using doesn't even talk about what you are talking about

Ted Shifrin

With the differential equations, the usual way is to solve them using the complex stuff and then take real and imaginary parts at the end.

Topologicalife

00:39

Yeah, that is what I usually do.

But a friend asked me this problem and I couldn't solve it. So here I am.

He sent me that image.

I solved it in the usual way but I can't find where is the error in what he is doing (that's what I talked here)

I think I will try to find another textbook which talks about this

Ted Shifrin

@Topologicalife: If you follow my exercise, the eigenvector for $-i$ will be $(-i,1)$, and then it'll all work.

In fact, you'll get that $C$ is the original matrix :P

Topologicalife

Yeah, I know. But I want to know why we take the eigenvector $-i$ instead of $i$.

Ted Shifrin

Because of that exercise ... Seriously. Tell you friend to work it if you don't want to.

Otherwise you end up with the minus sign in the wrong place in the matrix when you do the change of basis.

Topologicalife

I'm trying :/

Ted Shifrin

You just have to write $T(x+iy) = (\alpha-i\beta)(x+iy)$ in terms of real and imaginary parts.

Topologicalife

00:48

Yeah, I proved it.

But I mean I don't see the point of that exercise, sorry.

Ted Shifrin

It's how you convert complex diagonal matrices to the real rotation form.

So you follow through how you use the diagonalization to solve the ODE with this change of basis instead.

Topologicalife

But $C$ is the "step matrix", no the "diagonal matrix"

Ted Shifrin

Right. It's the change of basis matrix which would ordinarily have the eigenvectors as columns.

That's how you get $Ce^{\Delta t}x_0$ for your solution. You get columns of $C$ multiplied by $e^{\lambda_i t}$, and this is the form I gave above.

Topologicalife

Yeah.

Yeah, indeed. But that form is the same as $Ce^{\mbox{Re}(\lambda)} \begin{bmatrix}{\cos (wt)}&{-\sin(wt)}\\{\sin(wt)}&{\cos(wt)}\end{bmatrix} \begin{bmatrix}{c_1}\\{c_2}\end{bmatrix}$

Ted Shifrin

Right. I'm just saying you have to make up $C$ correctly. And that's it.

Topologicalife

00:52

I.e: I use what you said to end up with that expression.

Ted Shifrin

Or else signs get messed up.

Hence my exercise with the surprising minus sign.

Anyhow, I've lost interest in this.

Topologicalife

Yeah, I understand it.

Thanks for your patience and help.

Excalibur42

Doing ODEs in my analysis course atm and feeling like I've forgotten everything from first year calculus...why is it that if $x'(t_0) = f(t_0,x_0)$, then for small $h >0$, $x(t_0+h) \approx x_0 + hf(t_0,x_0)$?

where $x(t_0) = x_0$

Ted Shifrin

That's just linear approximation (think first degree Taylor polynomial).

Topologicalife

Taylor series.

Excalibur42

00:57

oh ffs thanks so much. I need another coffee

Ted Shifrin

LOL, sure, Excalibur.

BAYMAX

01:28

Image in $w$ plane of $|z-1| \leq 1$ under the mapping $f(z) = z^2$ ?any help on this I tried but it goes too long and I am lost in between!

Julius

Hi. When two random variables are discrete, their almost sure equality becomes equivalent to equality for every $\omega\in\Omega$, right?

BAYMAX

I got this but

1

Q: Find the image of $|z-1|=1$ under the mapping $f(z)=z^2$

I'm trying to find the image of $|z-1|=1$ under the mapping $f(z)=z^2.$ I know that this is a circle of radius $1$ centered at $(1,0),$ given by $r=2\cos\theta.$ So I have $$z=2\cos\theta e^{i \theta}$$ and so $$f(z)=4\cos^2\theta e^{i 2 \theta.}$$ Is this correct and how do I interpret this geo...

complex-analysis

In the answer how he got that $\rho$

Actually what is $\rho$ there

onelessproblem

01:43

Continuity of $f(x)= x^{2}$ when $x$ is rational and $2-x^{2}$ when $x$ is irational? $f:(0.1)\rightarrow \mathbb{R}$

Hello, This function is divided for rational and irrational values of x within (0,1)

So is it continuous anywhere ?

I have an exam, i really need help.

arctic tern

what do you think?

onelessproblem

I think no

Because at no point is x^2 and 2-x^2 equal

But the distinction based on rational and irrational

is confusing

Will Njundong

how do you find the average value of a function across an interval?

arctic tern

is there an analogue of summing values for a function on a continuous interval?

Will Njundong

summing?

arctic tern

01:51

@onelessproblem hint: for every x, there is a sequence of rationals converging to x, and there is a sequence of irrationals converging to x

@WillNjundong yes. if you average the values of a function on a finite set, you must sum the values up. (addition). what is the analogue of that for functions on an interval?

Will Njundong

sigma?

Mike Miller

there are uncountably many points on an interval, so just adding all those up doesn't make sense

onelessproblem

@arctictern So the value of the function will keep oscillating between the rational and irrational values, So there will be breaks in the function, therefore, not continuous anywhere on (0,1)?

arctic tern

@onelessproblem it's not important that the outputs are rational and irrational

onelessproblem

@arctictern So what's the answer? Umm

But there will be areaks in the function definitely

breaks*

Will Njundong

01:55

@MikeMiller how about using the "area" of the function then

onelessproblem

@WillNjundong Maybe using the uniform distribution formula and integrating over the interval will help?

arctic tern

@WillNjundong yes, integrate. then, instead of dividing by the number of values, what do you think you have to divide by instead?

Will Njundong

@arctictern difference between lower and upper bound!

arctic tern

yes

onelessproblem

@arctictern is the function continuous or not?

arctic tern

02:12

I gave you a pretty serious hint. Think about it.

user123733

03:40

-1

Q: Sphere inscribed in a tetrahedron

How there can be two centres? the centre would be the centroind of tetrahedron. and volume between two planes should be infinite as first octant is infinte

geometry 3d

Anyone help me in this

@TedShifrin

Cows

start ChatJax

Iced Palmer

03:58

If $B_t = (B_t ^1,...,B_t ^n)$ is an $n-$dimensional Brownian motion, what is meant by $|B_t|$?

Tim The Enchanter

04:22

@onelessproblem It is untrue that $x^2 \ne 2-x^2 \forall x \in \mathbb R$

Anonymous

04:46

If $$f(x)=\int_{0}^{x} t \sin(\frac{1}{t})dt$$ in $(0,\pi)$ then is it okay to directly differentiate f and check if $f'(x)$ is continuous? Or it would be wrong to differentiate using leibniz rule without knowing if f is differentiable?

Ted Shifrin

05:45

@blue: You don't need the Leibniz rule — just the usual Fundamental Theorem of Calculus that tells you you'll get the value of the integrand at $x$ at any point where the integrand is continuous. (What is happening at $t=0$?)

Topologicalife

Hi @TedShifrin.

Anonymous

@TedShifrin Umm, t=0 is only one point. So I think we can neglect it during integration? The function in t becomes undefined at t=0...

Ted Shifrin

Oh, I just noticed you only are considering $x\in (0,\pi)$, not even asking at $x=0$?

Topologicalife

I solved the problem. There was so much mistakes (in the notes, i mean)

Ted Shifrin

But it's not undefined, truly, @blue. What can you say about $\lim\limits_{t\to 0} t\sin(1/t)$?

Good, @Topologicalife. It's always annoying to find mistakes in textbooks/notes. I have always tried very hard to be careful in my books.

But I still say that what I was trying to tell you should have helped. :P

Topologicalife

05:48

Indeed, it did.

Ted Shifrin

Good. :) I'm glad.

Did you explain it all to your friend?

Topologicalife

Now I understand what you were saying. At first I was like... confused, because the mistakes in my notes.

Yes.

Ted Shifrin

Good :)

Topologicalife

I wrote a pdf and send it to him.

Anonymous

@TedShifrin The limit exists. I agree. But at exact t=0 it is undefined. A removable discontinuity.

Ted Shifrin

05:49

Right, @blue — it's removable. The usual thing is to define $g(t)=\begin{cases} t\sin(1/t), & t\ne 0\\ 0, & t=0\end{cases}$ and discuss integrating $g(t)$.

Topologicalife

Well, it is a simple point.

Ted Shifrin

Anyhow, $g$ is everywhere continuous, so ... ?

Anonymous

@TedShifrin That's what I was thinking. If a function is continuous in a certain domain is it a must that it is integrable in that domain?

Ted Shifrin

@Topologicalife: I'm sure writing it all out carefully was helpful for your understanding, too. I always understood things much better when I had to teach them to someone.

Topologicalife

Yeah, me too :)

Ted Shifrin

05:51

@blue: Of course. But the Fundamental Theorem of Calculus (as I said above) tells you that $f$ is differentiable at $x$ whenever $g$ is continuous at $x$.

Topologicalife

That's one of the reasons (not the main one) of why I love to help other people with their problems.

Anyway, I was thinking on your exercise...

We could try to prove the matrix $A_{\mathcal{B}}$ has the eigenvalues $\alpha \pm i\beta$

Ted Shifrin

Well, sure, but that's not good enough.

Topologicalife

I think it is a bit hard.

Ted Shifrin

No, not at all. You just need to remember how to find the matrix for a linear transformation with respect to a given basis $\mathcal B$.

Do you remember how that's defined?

Topologicalife

No, I don't.

Anonymous

05:54

@TedShifrin Okay. So g is continuous in $(0,\pi)$. So its integral must exist and hence f must be differentiable. Is that reasoning fine?

Ted Shifrin

$g$ is continuous on all of $\Bbb R$, in fact, the way I defined it, @blue. Yes, $f$ is differentiable. Now, is $f'$ continuous?

@Topologicalife: The columns of the matrix are what $T(v_j)$ are expressed in terms of the basis $v_1,v_2$.

Anonymous

@TedShifrin If f is differentiable in $(0,\pi)$ then f' must be continous in that region too, isn't it?

Ted Shifrin

So we have $T(x+iy) = (\alpha-i\beta)(x+iy)$. What are $T(x)$ and $T(y)$?

@blue: Not necessarily. $f$ must be continuous, not $f'$. You have to actually look at $f'$ and decide.

Anonymous

@TedShifrin Uh oh, I get it now!

Anonymous

Yes, you are right

Ted Shifrin

05:58

In fact, something very close to this is the famous example of a function that is everywhere differentiable but the derivative is not continuous.

Topologicalife

$T(x) = \alpha x + \beta y$ and $T(y) = (\alpha y- \beta x)$ if I'm not wrong

err

Ted Shifrin

Not quite right, @Topologicalife. Fix it. No $i$'s in the answer.

Aha.

Now it's right.

So the first column of the matrix is $(\alpha,\beta)$ and the second column is $(-\beta,\alpha)$. Done.

Topologicalife

:-P Nice.

Ted Shifrin

You get it?

Topologicalife

And that is the way I suppose to prove your exercise too.

Yeah.

Ted Shifrin

06:00

We just were doing my exercise :D

Topologicalife

Yeah, I see.

This is the first time I study diagonalization. It seems fun.

Ted Shifrin

Oh ... Well, I assumed you'd done this kind of stuff before. If you want at some point you can look at my YouTube lectures on change of basis and diagonalization, eigenvalues, etc. There's even some of this differential equations stuff in one or two of the lectures.

Topologicalife

!?!?

Do you have youtube lectures? :D

Ted Shifrin

Yup. 112 or so of 'em. Linear algebra + multivariable calculus/analysis.

Topologicalife

Before of this exercise I considered myself an expert in differential equations. :-P

Nice, which* is the link?

Ted Shifrin

06:06

You can get the link in my profile.

Topologicalife

It seems interesting, thanks, I will give it a look :)

Ted Shifrin

Just thought you might find it useful for some things.

Topologicalife

Wow, you are so friendly teacher.

Kindly

Ted Shifrin

Don't let it fool you :D

Topologicalife

Meh.

I've never had a teacher as you.

That's the main reason why I avoid to go to class.

Ted Shifrin

06:12

That's a shame :(

I know my style is not the typical "European" style. I have always liked students to talk and give ideas.

Topologicalife

Well, I actually learn a lot more of textbooks than of lectures.

Ted Shifrin

I always liked learning more from lectures.

Topologicalife

Well, I think I have attention deficit

:D

Ted Shifrin

Meh. :)

Topologicalife

Mm I wonder if your book is on e-book format :P

Ted Shifrin

06:19

Not yet.

Daminark

Hey guys!

Ted Shifrin

Heya Demonark.

Topologicalife

It will?

HI @Daminark.

Daminark

Lol, the lecture versus textbook debate... I honestly can't say for sure yet what I prefer

I mean probably lectures

I'm really an audial person mostly

Ted Shifrin

I haven't asked the publishers that. Right now getting some corrections made in the book is a major battle because some of the LaTeX files that produced the book are gone forever.

Yeah, me too, Demonark. But I really tried hard to get students to be an integral part of the class experience. Some never really were, but a lot were.

Daminark

06:22

I mean really I'm best off when a few of us are battling it out on a chalkboard

Ted Shifrin

Good.

Daminark

For some reason talking about it has always made it more clear in my mind than listening or reading

I mean "for some reason", it's because I'm more actively thinking about it

Ted Shifrin

That's not unusual. Good reason to teach people things you understand, because then you'll understand way better.

Daminark

Definitely

But yeah, also lectures where it's more interactive (which is part of why I'm also not terribly fond of how Europe does things)

Ted Shifrin

I'm excited 'cuz I might get to start teaching smart kids in Art of Problem Solving — they're building an actual brick-and-mortar school not too far from me.

Daminark

06:29

Oh, that'll be very fun

Art of Problem Solving is more puzzle-based math, right?

Ted Shifrin

Some, but they actually have an enriched regular curriculum (through calculus and group theory) for smart kids.

I actually was talking to a grad student last night who took most of that curriculum before he went to college.

Daminark

Ah, what'd he think of it?

Ted Shifrin

He loved it. Said it was challenging and engaging.

Daminark

Nice

Ted Shifrin

(Most people do this stuff on-line.)

I actually hadn't known about it until recently.

But it's been around quite a while.

Daminark

06:34

I have known about it for a bit over a year now, I think

But yeah, that's on the whole a really nice package

Topologicalife

Teaching is the best way to learn :D

Daminark

I tend to emphasize the puzzle nature a good bit, Laci's influence is starting to take hold on me

Eric

hi chat

Daminark

(Honestly though the best arguments that I've ever came up with have been with Hungarian mathematicians)

Hey @Eric!

Topologicalife

Actually, writting your problem with a pencil and paper it helps you to solve it.

Ted Shifrin

06:38

I'm not so much a puzzle person, however. But I did spend 15-20 years co-writing the UGA high school math tournament, which had some pretty challenging stuff on it.

Topologicalife

Hi @Eric.

Ted Shifrin

heya @Eric

Eric

I see AoPS is being discussed

their books are hard :)

Ted Shifrin

My fault.

I'm hoping to start teaching it in the fall.

Balarka Sen

Hi @Ted.

Isn't it past your bedtime? :)

Daminark

06:39

Hey @Balarka!

Ted Shifrin

Heya @Balarka. Not quite yet, but thanks for caring about me.

Eric

I remember I picked one up at the beginning of high school and I struggled through it with no contest/any real math experience. it was pretty rewarding in retrospect

Balarka Sen

Heya.

2

Topologicalife

@TedShifrin Here is my answer to my friend: rinconmatematico.com/foros/…

We use a spanish math forum.

:D

user123733

-1

Q: Functions with 3d geometry

I assumed the functions to be 5x,2x and 3x respectively. Solving through this, I got option (C) as the answer. However, the answer is given as (B,C,D). Please help

functions 3d

Can anybody help me in this

Balarka Sen

06:41

@TedShifrin Gotcha.

I just woke up. (It's 12 PM now)

Ted Shifrin

Sleeping through school?

Balarka Sen

Ah, school's off.

Topologicalife

Here its 8 am, almost 9.

Daminark

Here it's 1:45

AM

Probably gonna want to sleep soon because manifolds

(Why do morning classes exist? :( )

Ted Shifrin

Because some of us like to learn/teach in the mornings.

Eric

06:45

@Daminark I will never take a 9am again unless I'm forced to

my brain just doesn't function properly that early

Topologicalife

Haha

Ted Shifrin

And there aren't enough classrooms to do everything later in the day, and faculty have lives so they don't want to teach at night. (Plus seminars usually start at 2-3 PM, so faculty don't want to be teaching that late.)

Topologicalife

My brain only works at nights.

Eric

for me 10:30 or 12 are ideal times for classes

Ted Shifrin

I decided I didn't like 12 or 1 ... people are either starving or just ate and fall asleep.

Eric

06:47

I always take my lunch at 11:30 since the math building is right next to a food court here at chicago :)

very convenient

Daminark

^^^^

Anonymous

$f(x)$ and $g(x)$ is defined for reals. If f is not differentiable at say x=a and x=b and g is not differentiable at x=c and x=d. Then we define $h(x)=f(x)g(x)$. Can we say that $h(x)$ is not differentiable at $x=a,b,c,d$ given that $f(x)\neq0$ at x=c,d and $g(x) \neq 0$ at x=a,b. Is that a sufficient condition? Ideas anyone?

Daminark

I usually have food right after analysis

Especially because when I have a class I almost never wake up early enough to go to breakfast beforehand

Ted Shifrin

Do you know that $a,b$ and $c,d$ are different, @blue?

Anonymous

@TedShifrin Yes they are different

Daminark

06:49

Unless I sleep in the Barn

Ted Shifrin

@blue: You can answer your own question by contradiction, essentially. Assume $h$ were differentiable and show that $g$ would have to be at $c,d$.

Eric

@Daminark I can't even do math before I eat breakfast

it just doesn't work out well

Ted Shifrin

Yeah, I can't function at all without breakfast.

Balarka Sen

Is it really true? $f(x) = x^{1/3}$ and $g(x) = x^{2/3}$. $f, g$ are both not differentiable at $x = 0$; $f(x)g(x)$ is. Can you not modify this so that $f, g \neq 0$ at $x = 0$, etc etc?

Anonymous

@TedShifrin Can I use product rule for that or do I apply the limit definition of derivative for that?

Ted Shifrin

06:52

@Balarka: That's why I asked specifically if $c,d$ had to be different from $a,b$?

@blue: Product/quotient rule are fine.

Balarka Sen

Ah, I see now.

Anonymous

a,b,c,d are all different @BalarkaSen

Balarka Sen

$c \neq a$ and $d \neq b$ are necessary.

Anonymous

I'm trying Ted's method

Topologicalife

If $f: S_1\to S_2$ is a smooth map where $S_1$ and $S_2$ are regular surfaces, I want to prove $df(p): T_p S_1\to T_p S_2$ is a generalization of the concept of differential.

Ted Shifrin

06:53

More precisely, @Balarka, $c\ne a,b$, $d\ne a,b$.

Balarka Sen

Yeah :P 4 is too many

Thanks.

Topologicalife

Should be enough to define the differentiable curve $c:(-\epsilon,\epsilon)\rightarrow{}\mathbb{R}^2,\quad c(0)=p$ which represents the vector $x = c'(0)$? so that $df(x)$ is equivalent to $f\circ c$?

Daminark

Lol I think I function well enough without breakfast

Though dysfunctionality is very hard to recognize

Ted Shifrin

You're close, @Topologicalife. :)

We recognize it in you better than you do, Demonark.

Balarka Sen

@Daminark But do you function continuously?

Topologicalife

06:56

Ah, $(f\circ c)'(0)=Df(c(0))\cdot c'(0)$.

Balarka Sen

(Bonus: Do you function smoothly?)

Ted Shifrin

On that note, good night, all.

Topologicalife

And I'm done.

Balarka Sen

Night, @Ted

Topologicalife

Night @TedShifrin.

Balarka Sen

06:56

Yep, @Topologicalife

Ted Shifrin

Night, @Topologicalife, Demonark, @Eric, and @Balarka.

Daminark

It often seems to come up more by ridiculous thought patterns that are both entirely wrong and also impossible to understand than by just "how do I measure theory?"

That's true @Ted, and good night!

Anonymous

Yeah, so I get $h'(x)=f(x)g'(x)+g(x)f'(x)$. $h'(a)=f(a)g'(a)+g(a)f'(a)$. If $f'(a)$ doesn't exist then $h'(a)$ doesn't either unless $g(a)=0$. Similarly for the other ones. So the claim I made was true ? (Let me know if I am missing something) @TedShifrin @BalarkaSen

Eric

night @Ted

Daminark

@Balarka It's $C^{37}$

Jk it's the Weierstrass function at best

Eric

06:57

wow so specific

Balarka Sen

lol at C^37

Eric

I wonder what the highest $k$ is for which an interesting theorem requires something to be $C^{k}$

Daminark

Lol the one day I slept until 3PM and missed analysis, Soug decided to work in $\mathbb{R}^{16}$ for the heck of it

Balarka Sen

@blue Your claim seems to be right to me. I don't buy your proof; you can't just write $h' = fg' + f'g$ if $f', g'$ doesn't exist at $a$.

Anonymous

@BalarkaSen Even I was thinking so....But since we are just proving existence or non existence of derivative I think that works. @TedShifrin Any ideas on the ambiguity pointed out by @BalarkaSen ?

Eric

06:59

@Daminark I seem to recall there was a point last year where Marianna used 19 as a variable.

Daminark

Beautiful, and I actually do wonder this, it seems like the main numbers we encounter in analysis are 1, 2, and $\infty$

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$Mathematics$ Mathematics