Ooooh

We actually never got to stoke's theorem in my calc 3 class. We just did Green's and the prof said that Stoke's was a generalization.

@leslietownes 'pde-infected' XD ..

Surface integrals and the rest of the fundamental theorems of calculus are quite important in a lot of mathematics, physics, engineering ...

@RandomVariable YO!

If you watch my later videos, you will learn this stuff in the "appropriate" setting (i.e., manifolds and differential forms), although I went back and talked about the classical formulations and their physical interpretations.

Alright, cool. I just realized it would be of some benefit for me to cover that on my own.😅

Depends what you're eventually going to study and use ... but yes.

don't fall into the cesspit of stokes's theorem.

It seems to be following me.

smacks leslie and muzzles him

millions of murderous zombies running around and infecting people with it doesn't make it right

@leslietownes loll

we just call everything integration by parts .. Green's/Stokes all of it ..

Why?

No, that's the reverse of the correct answer.

Integration by parts is the fundamental theorem of calculus PLUS the product rule. The FTC gets replaced by Stokes's Theorem in higher dimensions.

How do I show that there is no real valued function such that $f'(c;u)>0$ for fixed $c\in \mathbb R^n$ and every non zero $u\in \mathbb R^n$?

don't let him bite you and infect you with the virus

That's a silly question, @Koro. Stop and think.

koro ^^^

$u=v$, @Koro, of course.

@r9m Hello. I haven't seen you around here in a long time.

It wouldn't hurt to think about what directional derivative actually means in a physical example. But you don't need this.

@TedShifrin but we do prove all that using chain rule and FTC/by parts at the end of the day right?

@RandomVariable ya .. just dropping by :) Happy New Year btw! :))

@r9m Chain rule? What are you talking about?

Clearly the question statement is false when n=1.

Why is it false when $n=1$, @Koro?

If I take $f(x)=x$ then f'(x)=1

my wife's school has unexpected low enrollments and withdrawals in a number of courses taught by adjuncts. is this happening more generally? a covid thing?

@leslie I am way too far removed to have any idea.

Are these adjuncts the same people who had good enrollments and retention a few years ago?

i'm wondering if the pool of transfer students is drying up, or if people are just avoiding mathy stuff, or what.

yes, they are.

Oh, you're talking about math courses at her school? I was assuming sociology

these are intro to stat courses.

Is there a control for this experiment? Are there any standard faculty teaching who have good enrollments/retention?

usually not taught by anyone cognizant of math or stat, but, as mathy as it gets as this liberal arts school.

i'll ask.

@TedShifrin differentiating $f(x + t\nu)$ for a direction $\nu$ .. etc

Someone was posting on main the other day about a probability problem his instructor couldn't solve (just said what the answer book told him to say). The instructor was a business person. Ha.

@r9m What the **** does this have to do with integration by parts and/or Stokes's Theorem?

higher level classes apparently have normal enrollments but there is some kind of disruption in the pipeline. the lower level classes are under enrolled.

i worked with a business prof who was very savvy about his domain of savvy-ness but would never defer to a textbook.

Ah, I think covid has made it hard on students who struggle, and intro stat is often populated by such.

Probability is subtle and tricky, even for those of us trained in math. I worked very hard teaching that probability course!

if you're choosing between college and your car payment, maybe you choose college if you're a senior but not if you're a freshman. i get it.

Well, plus the fact that learning is just REALLY hard with remote blah.

So math/stat takes a bigger hit because it's harder than intro civics.

prob and stat are like the hardest and subtlest things in the universe.

i internally groan whenever anyone says, let's have this replace calculus in high school.

I would expect intro physics has issues, too.

you have no idea how easy calculus is, in comparison to that stuff.

Well, the AP stat course is apparently quite routinely a joke.

For solid math students.

I admit I know essentially no stat, although I co-advised a masters thesis on the linear algebra/geometry in stat that the statisticians forget to know.

i believe it. my wife is nobody's fool but could not teach stat 1 except at the level of the very formulaic. which might be stat 1 for all i know.

and she has a masters degree in stat.

I kicked myself after teaching probability the one and only one time (and then retiring a semester later). I wish I had taught it several times in my career.

one of my friends with a similar educational background as my wife responded to a question about rolling two dice with a formulaic answer with pi in it.

I'm taking a code-based stat course this upcoming semester.

But the standard text I used, Ross, had its failings. The use of the indicator function was delayed way too long. I would do things differently.

What does code-based mean?

@robjohn I've noticed that MathJax-only titles display fine on the Activity page. It's titles that are mostly MathJax with a little bit of text that are all messed up.

From what I've gathered, we're gonna be using R to run simulations.

you get an A if your code compiles, even if it just prints "Hello World" and exits, perhaps with memory allocation errors.

or that.

Not sure what kind of simulations exactly.

Oh, well, using R in a stat course seems highly appropriate.

Actually doing real-life problems requires some computer assistance.

R is the best platform for that stuff. my wife knows stata, but most 'real' folks in her field don't use it, and she doesn't know enough about general programming to make the transition.

It'll be my first stat course, but maybe this will make things more intuitive.

I don't believe this answer.

f is not given to be differentiable.

You do not need differentiability, @Koro. Just existence of directional derivatives. Do I need to give a hint?

@leslietownes This was essentially my intermediate programming course using C++.

This is very clearly done in my YouTube lectures, BTW.

But the statement is false for n=1

The professor kept saying the implementation doesn't matter, it just needs to run and do what he told us to make it do.

There exists $f(x)=x$

such that $f'>1$

Write down the definition of the directional derivative. You're making an important mistake.

^

winking ^

^_^

;)

gets ready to start smacking ...

;_;

I'm gonna get smacking now, I think. Instead of writing directional derivative, I made emoji.

$f(c;u):=\lim_{h\to 0}\frac{f(c+hu)-f(c)}{h}$. Here the problem is that even if I take $-u$ instead of $u$, I don't get any contradiction as h takes care of any sign changes in $u$.

Be very careful. You should know better than to be glib like this when Leslie and I have told you to be careful.

In particular, what exactly is $f'(c;\lambda u)$? This is quite important in understanding what directional derivative means. Once again, see my lecture(s).

or his tiktok dances

I meant $f'(c;u)$ in my earlier message. Sorry for the typo.

Maybe Munchkin can work on those

I got it now.

@RandomVariable However MathJax only titles are discouraged because you cannot use them as a link

so if that is true, it is of little use. We need the core problem fixed.

@Koro how does $h$ take care of sign changes in $u$?

I meant that if that limit exists then it doesn’t matter what sign u is of, if we keep u fixed throughout the limiting process of h.

That says nothing.

Ted, here’s is what I was missing:

Earlier I said (wrongly) that f(x)=1 was a counterexample.

@Koro are you saying that we get the same result if $u$ is replaced by $-u$?

It is not because. The total derivative in this case is a 1 by 1 matrix which is [1]. Now what’s the directional derivative along u? It is matrix multiplication [1]u= u

And therefore the directional derivative is not positive for all non zero u

@Koro Ok. I was having trouble trying to figure out what you were claiming.

(I can take u = -5 to get contradiction).

What about $f(x)=|x|$ at $x=0$?

Rob: my earlier comment was in reference to my earlier comment.

recursive overload

for avoidance of doubt, i should point out that my later comments (after this) are in reference to my later comments (also after this).

@leslietownes Thanks, that will clear a lot up later.

I concur on the basis that whatever will be said will be justifiably false.

If f(x)=|x| then f is not diff. at x=0. But we can speak of directional derivatives: along positive x axis and negative x axis.

So, what is $f'(0;u)$?

@robjohn oops, sorry. I have switched to phone now. :(

Ted, so far am I going in right direction?

We'll see.

Ok. So say $u>0. f’(0,u) =\lim_{h\to 0}\frac{f(0+hu)-f(0)}{h}$ which doesn’t exist. As h can approach 0 from negative and positive both.

@robjohn It seems that you can now use them as a link. But yes, the core problem needs to be fixed.

Right, even though $f$ increases in both directions! :)

Yay!

So my last to last to last comment was wrong.

We can’t speak of directional derivatives of |x| at 0.

@robjohn I was wrong as the example f(x)=x shows. I realized it now.

Now the question boils down to: can we have a vector that gives a positive dot product with every non zero vector?

So you should observe that $f'(c;\lambda u) = \lambda f'(c;u)$. When we think of rate of change in a direction, this is why every standard calculus book makes one take $\|u\|=1$ to calculate a directional derivative. But in analysis/geometry we want linearity, so we remove that restriction.

No, the question does not boil down to that unless the function is differentiable.

And of course you know the answer to that question, anyhow.

(Question says prove it, so must be true somehow :D)

Homogeneity in second slot does not seem trivial to me.

Huh? Question says prove what!

Ohh, never mind, I understood how that lambda comes out.

@RandomVariable Oh, if you can use the MathJax as a link now, that is a good thing.

Work out from the definition (nothing else) the "you should observe that" I gave you up there.

OK.

Now I can take lambda negative.

:-)

Of course.

He now has lambda chops...

So what is a correct physical interpretation of $f'(c;u)$?

smacks @robjohn

oh, oh, oh, Mr Kotter!!

Go put some on the grill for me. Plenty of olive oil, garlic, parsley. I'll be by shortly.

Let v=-5u, so $0<f’(c;v)= f’(c;-5u)=-5f’(c;u)$

it's like magic

which is absurd as f’(c;u)>0

OK, we're done with this question. Answer mine.

I think in terms of tangent for physical interpretation.

@robjohn You are still the Christmas grinch, even after New Year's.

@TedShifrin I figured that if the Bash Hats are still on, I'd leave my hat on.

@Koro: I don't quite know what that means. Be explicit.

@robjohn Oh, of course. But mine has disappeared.

I still see the finger on your avatar

let me refresh

gone from mine.

ah, they are gone

Well, I just put mine back on, but does it show?

Given graph of f, I cut it along direction v. Then, along that curve I find derivative. But I think of this only when I have f(x,y)=z

wait, I still have my Fu Manchu

Ha, it’s not visible now.

OK, @Koro, but I still do not know what "along that curve I find derivative" means.

I see Koro's, Ted's, and mine.

I added mine back, and I see it in my profile, but not here.

Ted, I also don’t see it here. I think that’s due to iOS.

I did a deep refresh and it all looks the same to me. I see them next to the avatars next to the text. They never showed on the attendance bar.

By along that curve, I mean that I find a tangent along v.

I'm on my iMac. ... You're still not helping me, @Koro. What does "find a tangent along v" mean?

I'm using Firefox

Let me put it this way: I have f(x,y) for example. And I want to talk about f(c;v). This means that v is parallel to some straight line (knife) which slices the graph and makes a crack. On that crack, I choose (c,f(c)) and draw a tangent to that crack.

A tangent line to the curve. Now what?

@Koro can you write $f'(c;v)$ in any other way?

Let's finish this before I go to lunch.

Lunch??

It’s late night here.

@TedShifrin I almost missed that it is lunch time.

thanks

12:23 here

@robjohn sadly directional derivative is not linear in the second slot so I’m not sure about other way than the definition.

@Koro: So let's finish so you understand my point. What am I doing with this tangent line?

wipes mouth good luck finishing this before i go to lunch.

smacks leslie for being rude and not offering us lunch

@leslietownes I'm with you there.

Then I find slope of that tangent line.

And is that slope the answer?

Yes.

🥲

So what's the difference between taking $v$ and taking $2v$?

I get the same curve and the same slope.

@Koro I am confused. The directional derivative is not linear in the second slot?

Ahh, so what I said is valid for unit vector v.

@robjohn professor Rob, I meant that $f’(c;v+u)$ does not have to be equal to $f’(c;v)+f’(c;u)$.

I am not going to go down this path. Answer Ted's question.

I already did.

?

The difference between v and 2v is that: if I take 2v I get twice the slope that I got during geometrical interpretation discussed above.

@Ted is your question answered? If so, I missed it.

So it's not slope that's the right interpretation, @Koro. We need a better physical interpretation.

Here's a hint. Suppose $f$ gives the temperature in space (and it does not vary as a function of time).

yes.

@TedShifrin do you know if there is a well-known name for the algebra $so(n;A) = \{X \in M(n) \mid XA + AX^\top = 0\}$, where A is an invertible nxn matrix? It's basically so(n) depending on this A.

I don't know one, @anakhro.

Ted, are you going to say temperature gradient now? ^_^

While studying the definition, I did see a disclaimer that some authors define this for ||v||=1 but it’s not assumed here. But I didn’t think much on that.

Ala Kazam

abracadabra

robby bobby johnny jack

the main man

No, I want you to understand the difference between $v$ and $2v$, @Koro. Think about walking on the line with velocity $v$ and with velocity $2v$.

Look at the actual definition. What is it calculating?

I just applied for 5 tutoring positions. Do I expect to get hired: not a chance. -_-

Back after lunch.

@PenAndPaperMathematics be more like $e^x$. Be more positive.✨

You mean convex?

Was going to say, I don't see what positive has to do with $e^x$.

e^x > 0, guys

Sure

A number >0 is often said to be positive.

Tough crowd today

So true

It's just an odd thing to say about $e^x$.

That it is a positive function?

You mean that the range of the function for real values of $x$ is positive.

we are dealing with a mr. smarty pants here

I see how it is.

lol I'm just being precise.

I imagine we’re just going around in circles!

bah dum tiss

Pedantic to the point of redundancy

Ted, what was for lunch?

If I've learned anything being in this chat for over a year, I'm skeptical of mixing adjectives.

That was a fast lunch, was it just a breadstick

Anyways, back to working on modulus

I liked $\text{AMDG} \pmod{\text{pedantry}}$ better

I was not being pedantic in the least!

@TedShifrin don't worry, that wasn't directed at you.

But I still want to know your lunch.

Basically pastrami sandwich w/ Russian dressing, cheese, tomato.

Sounds pretty good.

Allegedly the reason it is called Russian dressing, despite being american, is because of the traditional use of caviar.

I always forget the difference between Russian and 1000 Island.

Ah, my finger hat is back.

@TedShifrin finally got refreshed.

I assume my Fu Manchu is visible?

Yes, indeed.

It looks like @Koro gave up on understanding what directional derivative actually means.

Maybe they needed a break.

Sometimes you get burnt out when you get stuck on something for awhile.

No, not stuck.

not flowing freely, either.

LOL, it didn't help that we were both pestering :P

I defer to you for the next round :)

Sorry to have interrupted. You think that something will help and it only served to sidetrack.

No, no, but it does tend to make people feel more under attack :P

Especially since you and I both can get rather insistent sometimes :P

@TedShifrin often Russian has a hint of horseradish and Thousand Island has pickle relish

Russian spicier, Thousand Island sweeter.

Yeah, I never put in pickles, although I like dill pickles. I didn't put horseradish this time, but I usually do for pastrami. And I definitely put in Worcestershire.

I use chili sauce, never ketchup.

I have a very old bottle of chili sauce in the fridge from when I tried making flygande jakob.

I find ketchup/catsup too boring. flygande jakob?!!

You've never heard of it, or have you heard of it and that is why you are surprised?

It's bizarre, so both are plausible.

No, I've never heard of it. I figured you were being humorous.

Like fliegende Jakob?

Flying Jacob in Swedish.

Ah, yes, actual German.

"a Swedish casserole composed of chicken, cream, chilli sauce, bananas, roasted peanuts and bacon"

I wonder if the chili sauce there is the same one we have. I suspect not.

There are all sorts of asian chili sauces that are completely different. Who knows.

AFAIK, they often use the Heinz chili sauce.

Not sure if that's the one you use.

It's like a slightly spiced ketchup.

Yes, whatever brand.

Ketchup is just too sugary for me.

You'd probably hate banana ketchup then

Bananas instead of tomatoes?

Yes.

I'm thinking yuck.

It's actually good, but extremely sweet.

Yuck.

There was a filipino place I got some banana ketchup and spaghetti sauce from. The spaghetti sauce that they use is basically filled with the banana ketchup. It's really weird and has this characteristic flavour that I am unsure on how to place exactly.

Have you tried filipino spaghetti?

If so, it's likely you already opted for it. :P

Not that I know of, no.

I don't get any banana flavour from it, anyway.

It disappears somewhere in the spices.

"Jufran" is the popular brand for the ketchup.

I've had adobo and kare-kare.

Maybe pancit, too. It's been a long time.

When covid is done (if?) ... I should go to some Filipino restaurants here.

Jollibee is a popular filipino place with restaurants in north america that sells filipino style spaghetti. Have you heard of it?

Nope.

It's comfort food, from my understanding.

Oh man I love food.

Off to errands. Bubye for now.

c u l8r

does anyone know how to contact the moderators of math.stackexchange? Thanks.

@pisoir moderators are the users with a diamond next to their name.

you can also flag a post of yours for moderator attention and ask something.

How do we prove that I, Ted, or robjohn are not just AI's with a chat interface?

When Ted says "Off to errands". They're really retraining an AI on the past 24 hours

@PenAndPaperMathematics are you?

That never happens to me on MSE, but on MO and SO it does happen to a lot of people all the time, and it's wrong.

I ask one question. One or two upvotes => suspended account for not having "high enough quality". That's BS. Check out my post history.

@robjohn The Sun's far too small to fuse iron. In the later stages of being a red giant it will fuse some helium to carbon, but not much. From en.wikipedia.org/wiki/Carbon-burning_process Stars of below 8–9 solar masses never reach high enough core temperature to burn carbon, instead ending their lives as carbon-oxygen white dwarfs after shell helium flashes gently expel the outer envelope in a planetary nebula.

I put a nebula pic in the bg of my site

@PM2Ring I was responding to his comment about stars and iron. I was not being star specific, but I guess if you follow the chain back far enough, they were talking about the Sun, so yeah.

The comments are not linked, but the trail is short.

Hey, @Ted!

@robjohn Fair enough. But the conversation was definitely about the Sun. ;) We have some good questions (and great answers) on why fusion stops around iron / nickel, eg astronomy.stackexchange.com/q/21284/16685 & the links on that page. Of course, very few stars get to the stage of burning silicon, and most stars are red dwarfs, which won't even burn helium, although they will burn a higher percentage of their hydrogen than any other type of star.

@PenAndPaperMathematics That sounds really disturbing. I know Claude is an elderly person with harsh health conditions and since he devotes a lot of his time in the MSE, the suspension can be really devastating for him. It also doesn't help the community. If the suspension is done only by some bot, without human intervention/moderation, it sounds horrifying to me.

@pisoir we are not supposed to discuss the details of suspensions. Claude is not under the same restrictions. Please do not pester him, but if he wishes to discuss this with you, he can.

I saw that. I’ve always been an irrational ribot!

@pisoir: for your own information, answering low quality questions can also lead to a suspension. I am not saying anything about your posting history as I have not looked at it.

This is part of the EoQS.

Or robot.

@robjohn Sorry, I didn't want to defend anyone or investigate the reasons. I have no interest in knowing the reason. I believe it was done in a good will, but still the reason sounds like an error (user with top reach-out has "low-quality" answers)? I noticed it only by accident and I am in no contact with him (except for MSE). I just felt really surprised.

or robbit $\unicode{x1F407}\times\unicode{x1F916}$

@robjohn Can you clarify that point about automatic suspensions? I thought that all actual suspensions (where the rep is temporarily set to 1) were manual. I know there are automatic question & answer bans. And of course chat suspensions are a different category.

I love Claude's work, and almost always upvote his answers. IIRC, he's now blind.

(I do not intend to discuss Claude's suspension).

@PM2Ring There are automatic bans from certain activities, but suspensions that reduce your rep to 1 and keep you from posting, etc, are manual. Manual suspensions have a duration. Automatic bans don't time out, but require action to be lifted.

@PM2Ring I do too, and I am sorry to see that he has been suspended.

Thanks

That doesn't mean I agree or disagree with the action. I understand it.

@robjohn Thanks, @robjohn. I just wanted to be sure that there was no unjustified wrongdoings on Claude (like a malicious bot who flagged him, or an error in automatic ban system).

@pisoir Thank you for your concern

@PM2Ring Isn't it mainly that fusing iron takes more energy than it produces?

I've heard some people say that as soon as a star starts fusing iron, it dies almost instantaneously. I find that hard to believe given that it can take tens or hundreds of thousands of years for a photon produced in the core to escape the star.

It would seem that there would be some decent delay.

@robjohn Not quite. Fusing Iron, nickel, & even zinc (with helium) are exothermic reactions. But the helium used in those reactions comes from photodisintegration, and that reaction (which is essentially fission) is highly endothermic.

The entire silicon-burning "ladder" is quite short-lived. IIRC, in a 20 solar mass star it lasts about 5 days, but the processes leading to core collapse start within a day or so after silicon burning starts.

Does core collapse cause a collapse of the star faster than the photons travel to the surface?

and therfore the (super)nova

So it's not exeactly the production of iron (etc) causing the collapse. It's just that for the core to be hot enough to burn iron it's really hot throughout the whole core. A high proportion of the blackbody photons are energetic enough to blast nuclei to pieces.

@robjohn Hell, yeah. :) When the collapse occurs, you suddenly get a bunch of electrons & protons combining into neutrons & neutrinos, and those neutrinos can carry away about 20% of the energy released at almost lightspeed.

@PM2Ring they don't encounter all the matter that was blocking the photons. Or they do, with extreme prejudice?

The neutrinos would escape of course

Under normal circumstances, the star is fairly transparent to neutrinos. But their cross-section isn't zero, and gets bigger at higher energy. So as well as the neutrinos produced by protons & electrons merging you get a big flux of thermal neutrinos & antineutrinos. So the collapsing core does slow down the neutrinos a little, but they still manage to escape at highly relativistic speeds

Even the fairly sedate neutrinos produced in the Sun by the p-p chain are highly relativistic. Our best neutrino detectors can only detect neutrinos with $\gamma>10^6$ or so, i.e., their kinetic energy is a million times their rest mass. That's nothing compared to the neutrinos from a supernova.