well, it's not continuous with respect to $C^0$-norms, but it is continuous wrt to the $C^n$ and $C^{n-k}$-norm. taking derivatives is a thing we like, so it's probably smart to pick norms where that's a continuous map, because we like continuous maps.

@Thorgott Might have step back a bit, why does taking the kth deriv yield a C^n -> C^{n-k} map?

each time you take a derivative, you take one piece of information away about what you assumed about the original function. if you only asssume a function has 10 continuous derivatives, when you take 3, you only have 7 continuous derivatives left.

Subtraction is evil.

Division is satanic.

@leslietownes oh ok. All the "examples" i keep thinking of are smooth functions. That's why it didn't make sense to me.

yeah, depending on the application, the members of interest in your function space might be smooth for some other reason, unrelated to the definition of the space.

I have this question I am struggling to paste it all here 'cos it's long but here is a link to it pastebin.com/raw/CB6aRZu2 I migt be able to paste it here

Suppose you have a scenario of people eating burgers.. You have a ratio telling you ow many people can eat how many burgers. And you have info on ow many people you have and how many burgers you have. And you want to see if you run out of burgers or if you have an excess of burgers. So suppose the ratio of burgers to people is 4:1 meaning that 4 burgers can be eaten by one person. And in terms of the amount of burgers and amoutn of people you have, you have 6 burgers and 3 people.

So you want to calculate, do all the burgers get eaten , or is there an excess of burgers. So I take the ratio of 4:1 (4 burgers to one person), And from there I work out that 6 burgers would be eaten by 1.5 people. So that tells me that if 1.5 people can eat 6 burgers, 3 people can certainly. So the burgers are all consumed, therere's noe enough burgers, we have an excess of people relative to burgers. That's my way of working it out and I think it's right.

But there's a guy I know that has a weird way of working it out and I don't see the logic of his method, and I wonder if is method is flawed and wouldn't always work. His method which he could show the method to me but couldn't explain why it worked. He said take the raio of burgers to people , in this case 4:1 Then take the number of burgers and divide it by the ratio number for burgers i.e. 4. And take the number of people and divide it by the ratio number for people which is 1.

Then write the results So Burgers- 6/4=1.5 And the result People- 3/1=3 Then he said the smaller number is the limited one.

^ there that's the question!

so in that example he gets the right result.. i'm wondering why his method works and if any examples he would get the wrong result.

He’s doing a formula that he can’t explain. You did the same reasoning conceptually. You win.

Neither of you said that you run out of burgers. Suppose every 2 people eat 5 hamburgers. Or 3 hamburgers. What happens in those cases?

Anyhow, his method is to compare the fractions $6/3$ and $4/1$, which is the same as comparing $6/4$ and $3/1$. Why?

Show that $a/b < c/d$ if and only if $a/c < b/d$.

A= (6/4)/(6/3) = (3/1)/(4/1)= 0.75

comparing two numbers X and Y, is the same as comparing AX and AY

But that’s not quite what he’s doing.

Oh, I see.

what did you have in mind?

What I wrote above ….

well, doesn't what I wrote show why his method works? as in , my numbers work and his method happens to be some multiple of my numbers.

taking my numbers as the one on the left.. i guess.. which maybe isnt right..

I thought I answerwed your question though of why comparing the first set of fractions, gives the same result of comparing the second set of fractions.

granted it's not what he is doing, but it does answer your question doesn't it?

If you’re fine with your way, that’s fine. It’s correct. What I wrote is a bit more direct if you can understand arithmetic with fractions.

I am fine with my way but I want to understand why his way works. I see another pattern maybe..

A=6 B=3 C=4 D=1. A/B and C/1 compared with A/C and B/D So maybe there's an algebraic rearrangement you are getting at?

You’re effectively looking at $\dfrac{a/b}{c/d} = \dfrac{ad}{bc} = \dfrac{a/c}{b/d}$.

Your way is the left. His way is the right.

ok thanks

It’s a good question, btw.

thanks!

And your explanation is a valid one, just wasn’t as clear to me as mine.

false alert for the convex beacon.

Have you renounced convexity and adopted concavity?

never

#unboundedabove

I want to know if the set S={(x, x,..., x) : x is real} with subspace topology of $R^\omega$ (with product topology) is connected or not.

To this effect, define f that takes x in R to (x, x,...) in S.

f is continuous and hence maps connected R to connected space.

Hence S is connected.

did you mean the elements of $S$ to be $(x,x,...)$ instead? Or did you mean 'finite' sequences?

Yes, not finite.

Probably, one subtle point here is: I have assumed here that S with subspace topology from product topology is a product topology. This enabled me to claim continuity of f via the result that f is continuous iff its coordinate functions are continuous.

Is $f$ a homeo to its image?

What you just typed is nonsense.

This result holds in product topology but not necessarily in box topology. That's why I am confused.

“Is a product topology”?

I mean: is it same as to have come from $A\times A\times... $ for some set A?

Why are you even thinking that

I want to claim continuity of f via the result that f is continuous iff its coordinate functions are continuous. This valid if the codomain is product topology but not necessarily in box topology)

$f$ is continuous in the product topology but not in the box topology. Take $U = \prod_n (-{1 \over n}, {1 \over n})$, then $0 \in U$, but $f^{-1}(u) = \{0\}$.

yes, that's what I said.

the question is: is S connected?

it seems no as per the argument I gave above. But I think the argument is wrong.

sure, since $\mathbb{R}$ is connected, then so is $f(\mathbb{R}) = S$.

@TedShifrin I see. That's not required.

@copper.hat: thanks. I had got confused. Define $f:\mathbb R\to \mathbb R^\omega$ by $f(x)=(x,x,...)$. Then $f$ is continuous. By continuity of $f$, the set $f(\mathbb R)=S$ is connected.

But the answer is wrong :(.

Right. So is $S$ homeomorphic to $\Bbb R$?

The answer key says- disconnected.

I'm thinking about homeomorphism.

particularly about continuity of $f^{-1}$

each coordinate function is an inverse.

That is confusingly stated, plus for Koro to think, not you.

oh no, this is correct. I mistakenly saw answer key to some other question.

i thought we were in the smoking phase of the problem

I have doubts whether $f^{-1}$ is continuous. Let (a,b) be a basis element in $\mathbb R$, ($f^{-1}: \mathbb R^w\to \mathbb R$) then $(f^{-1})^{-1}(a,b)=\cup_{t\in (a,b)}\{ (t,t,...)\}$. But is this open in $\mathbb R^\omega$?

@copper.hat haha.

I never heard of smoking phase before 😅.

But I don't see why S homeomorphic to R is required here.

it is not needed, but it is true

It’s not. I’m adding it for you to learn.

ohh.

I'm still thinking about continuity of $f^{-1}$ though.

thankfully you are not paying much attention

$f^{-1}: \color{blue}S\to \mathbb R$. I earlier wrote $\mathbb R^\omega$.

Correct.

I am not sure how to simplify $\cup_{t\in (a,b)}\{ (t,t,...)\}$. I know for sure that it is not $(a,b)\times (a,b)\times \cdots$.

if $f^{-1}$ were continuous. Because that product is not open in product topology.

We’re talking open in $S$. But is this the easiest way to proceed?

This seems to be the most natural way to me as of now.

if there were finite product, this would be a line through origin.

think of $t \mapsto (t,t)$ for a simpler example.

how is the product topology characterised from a functional perspective?

it's generated by inverse of projections.

mmm, not what i meant, the coarsest topology such that...

Take any (t,t,...) in the set. Then consider: $((a,b)\times (a,b)\times \mathbb R\times \mathbb R\times...)\cap S=:T$. Then $(t,t,t,...)\in T \subset \cup_{t\in (a,b)}\{(t,t,...)\}$. Therefore the set is open.

@copper.hat ohh, I'm not sure. But I know that: basis elements of the product topology are of the form $\Pi_i U_i$, where $U_i=\mathbb R$ except for finitely many $i$.

you are overthinking this

I'm still not used to thinking using projection maps :(.

that is the key here. try thinking in terms of the simpler example above. what is the inverse of that map?

$(t,t)\mapsto t$.

and the product topology is the coarsest topology such that...

this map is continuous.

?

is this correct? @copper.hat

I said (guessed) that because I recall studying something similar while studying quotient topology. But I didn't think much about it then.

the projections (coordinates) are continuous, so each projection is an inverse.

preimage of a basis element (a,b) then is a line segment of the line y=x.

which seems open by intuition (being the intersection of S and an open set).

But I don't see how projections help here.

$\pi_k((x,x,x,...)) = x$ for each $k$ is continuous.

I like when I'm doing repeated least square solution problems and they decide to surprise me with a system that has an exact solution. Me calculating the least squares error of $0$ at the end :/

when i used to do more numerical work, an exact answer of zero was almost always an indication of a programming oversight :-)

@Akiva I made an interactive 3D Borromean / Brunnian thing. Here's a screenshot.

sagecell.sagemath.org/…

There's a circled 'i' icon in the interactive view that opens a menu which lets you save the HTML or a PNG.

Does anyone else lose motivation to write answers sometimes? I saw a really interesting geometry problem with no answers, which I know exactly how to solve, but I just can't get myself to write out an answer

Is this burn out?

Well, let me ask you this, why do you care about that anyway?

You want to help somebody, you write an answer for them.

Does attribution be included in a question, or am I thinking too "Puzzling SE terms"?

@Stevo You must attribute any material (text and images) that you post that you didn't create yourself. Please see math.stackexchange.com/help/referencing That help article is mainly focused on answers, but it applies to questions too.

And of course it also applies to chat. Everything you post on the network has a Creative Commons license, so others are free to re-post it, if they attribute it properly. But they can't give proper attribution if they mistakenly think you're the author of material you didn't create.

Hi Folks, I got a hard time trying to figure out what they did here: imgur.com/GcQ3Mdo. Theta and T are possibly dependent random variables. Could you give me some hints?

How come E(T) = E(T - E(theta|X)|X) ?!

@PM2Ring Cool!

cool indeed

what is the relationship between $(int_a^b |f(x)|dx)^2$ and $int_a^b |f(x)|^2dx? like in terms of inequalities/equalities

if you scale by the legnth of the interval, there's jensen's inequality. squaring is a convex function and jensen's inequality is generally about that.

Cauchy-Schwarz?

$\langle 1,f\rangle^2 \le \langle 1,1\rangle \langle f,f\rangle$

ooh, also that. it might even say the same thing as jensen in this case.

langle 1, 1 rangle being that scaling factor.

Yuppers

good morning, ted. day care has veterans' day off but i don't, so la gritona is with her grandmother.

Oh, I'm surprised the grandparents don't get custody more often.

they let her get away with too much.

and as you well know, i say that as someone who lets her get away with too much.

over indulgence is never a good thing, at any age

When you have a function $f(k)$ and you take the derivative with the $a$-method, would you usually write $f'(a) = \lim_\limits{k \to a} \frac{f(k) - f(a)}{k - a}$ with $k \to a$ or with the conventional $x \to a$, like $f'(a) = \lim_\limits{x \to a} \frac{f(x) - f(a)}{x - a}$?

either one, they are literally the same thing. if you'd been using k as a 'variable' elsewhere you might keep using it in writing out the limit definition

The former, right?

@leslietownes alr, thanks. You would probably use k more often since f is already in terms of it then

yeah. particularly if the variable has some kind of 'meaning' and calling it k reminds the reader what that meaning is.

What does it mean to fix a point on a curve?

don't let it move

it it fixed at one place

just to choose a point. for the purpose of some discussion or argument.

'fix' i guess also adding a reminder that the point is not going to change during that discussion or argument.

Ok, now what is an "open ball"?

a set of points less than some fixed distance away from a given point.

Sorry if these questions seem random, i'm trying to motivate the implicit function theorem for a few classmates and I can't think of simple ways to explain the meaning of these words.

the given point is kept fixed at one location

open balls are what you end up having to consider if you want to make an argument that involves small changes away from a given point, where you have no control over what direction those changes may be made in, but do have control over the size.

the epsilon-delta definition of the limit in R^1 is about open balls, although there the open balls are intervals and maybe don't look like physical balls.

there area few points in calculus where it might be pedagogically helpful to think in terms of multivariable functions from the beginning. open balls might be one of them. people always want to turn absolute value stuff in one variable into case-by-case analysis, if k > a then it's this, if k < a then it's that. that move isn't always helpful, and seems less natural with open balls.

Thanks Leslie, your example will be quite helpful!

aren't the meanings of these words explained in your class notes?

@user4539917 Yeah, I thought so too. Looks like I was overthinking things as usual. Heavy sigh...

I'm a high school student

but i've learned a fair bit of multivariable calc

I am a math ambassador at my school, meaning I help out with math related stuff.

cool

Since some of my peers wanted to learn about the Implicit function theorem, I am teaching them. Well, doing my level best to.

This is for the second part, I already did an introductory part about the inverse function theorem.

youtube.com/playlist?list=PL5I-Eyk8l9FHdJUd9UujGcvumjCFPHbrd

Ahh yes, Ted's videos

I've actually watched quite a few of them

watching them in order will take you a long way

Couldn't agree more.

after the 110th day you'll be able to teach like a pro

@leslietownes Oy vey.

@user4539917 LOL

Fact: given any open set U in $\mathbb R^n$, there exists sequence of compact sets $\{C_n\}$ such that $C_n\subset U$ for all n, and that $\cup_n C_n=A$.

except $A=U$.

yes, indeed. A=U

Have you proved said fact?

yes.

now make it so that C_n is contained in the interior of C_n+1

that's trickier

It is required in defining (improper) integral $\int_A f$, where A is open in R^n and f need not be bounded.

Thor: note that I did not say that the sequence is nested.

well, you can make it nested WLOG

but what I'm suggesting is genuinely stronger

in that case, yes we can make C_n contained in int C_{n+1}.

@leslietownes You really need to get La Gritona an instrument so she can channel that energy. Perhaps she'd be good on the trumpet, like Alba and Elsa Armengou.

Hi everyone. I have trouble understanding what a problem is asking me to find.

It asks "Find the area of the part of the hyperbolic paraboloid $z=x^2-y^2$ where $0\leq x^2+y^2 \leq R^2$". Does that mean find the surface area of the hyperbolic paraboloid?

That portion of it, yes, @rb3652.

I see, and I understand that the surface area is given by the pythagorean theorem: $\int \sqrt{1+(f'(x))^2} dx$. Would that be my strategy for working this out?

No. Better look this up. It will be a double integral.

Hm, OK. Let me try searching for a similar problem.

You probably want to do the integral in polar coordinates.

unfortunately, with climate change, polar coordinates are being used less & less.

You can do Mercator.

@TedShifrin It seems my idea is essentially correct. I just need to add an extra term for the extra third dimension: youtube.com/watch?v=OQpVDg6yvHo

Right.

As I said, switch to polar coordinates because the region is a circular disk.

Sure, so I would have $\int \int \sqrt{1+(z_x)^2+(z_y)^2} dy dx$

I'm using subscript notation for partial derivatives.

i've forgotten all my projections. the maps (sectionals) used by small airplane pilots use the Lambert conformal conic projection.

Right.

After plugging in, I have $\int \int \sqrt{1+(2x)^2+(-2y)^2} dy dx$

Now, I must convert to polar coordinates, as you suggested, @TedShifrin

It's basically undoable if you do not.

That means $dy dx = r d\theta dr$ and $x=r\cos(\theta)$ and $y=r\sin(\theta)$, which gives

$\int \int \sqrt{1+(2r\cos(\theta))^2 + (2r\sin(theta))^2} r dr d\theta= \int \int \sqrt{1+4r^2} r dr d\theta$

This calls for u-substitution, where $u = 1+4r^2$, I believe.

This gives $\int \int u^{1/2} du d\theta = \int \frac{2}{3}u^{3/2}d\theta=\frac{2}{3}*2\pi*u^{3/2}$

Now I'll add the bounds, which is from $0\leq r \leq R$

But $u=1+4r^2$, which implies that $1\leq u \leq 1+4R^2$

This ought to give $\boxed{\frac{4\pi}{3}((1+4R^2)^{3/2}-1 )}$

Is that correct -- or at least my general strategy?

It looks pretty good.

Thank you, and I presume I would execute a similar strategy for finding the area of the part of $z=\sqrt{x^2+y^2}$ where $0\leq x^2+y^2 \leq R^2$ (i.e., pythagorean + polar + u-sub)?

Interesting that you get the same area for $z=x^2-y^2$ as you do for $z=x^2+y^2$, which have very, very different shapes.

Wow, that's crazy. Hang on, let me GeoGebra this.

I can't answer your new question. The set-up is the same, but the methods of integration may be different depending on what you get.

@TedShifrin Maybe this is why -- $z=x^2+y^2$ (red) is the negative space of $z=x^2-y^2$ (blue)!

I don't know what you mean.

No, no. We're talking about the region over $x^2+y^2\le R^2$ in both cases.

You're doing something very wrong.

Hmm, then I don't why they have the same area over a circle. It's interesting.

The correct graph of $z=x^2-y^2$ is a saddle surface.

I have no idea what your software is drawing.

Yes, GeoGebra shows a saddle, but it stretches out to infinity, so it may be hard to recognize. A pringle chip.

Just draw the portion over the disk.

Leave out the paraboloid.

I do this stuff in Mathematica and I do it in polar coordinates if I want it over a circular region.

Perhaps this is a better angle:

You're not limiting the domain correctly.

But it's a circular region.

I just drew a circle of finite radius.

Here's a much better picture.

Oh wow, nice.

Now it makes sense why they have the same surface area!

It's not geometrically obvious to me.

Well on the left and right sides, the paraboloid and chip exactly coincide, and if you could imagine taking the front and back parts of the chip that are downward and reflecting them, it would exactly coincide with the paraboloid. Hence, equal surface areas. It's a hand-wavy argument, sure, but it seems intuitive.

No, I don't think it's right, although it sounds good.

Hyperbolas don't match up to circles (horizontal cross-sections).

Curvature doesn't match up: negative curvature surface is not matching up to positive curvature.

Hmm, well I'm not quite sure then.

Yeah, it's a computational fact, but I don't see it visually.

I have another question, if you don't mind.

Yes?

I'm asked to evaluate a triple integral over a tetrahedron. I've decomposed the tetrahedron to get my boundary conditions as $0 \leq x \leq 1, 0 \leq y \leq x$, but I'm stuck on my limits of integration for $z$, because the side of the tetrahedron is a plane governed by $x-y=0$, but it has no z-dependence inside of it (i.e., z can be anything, as it's a cylinder). How would I formulate my z boundaries, then?

The tetrahedron in question is bounded by (0,0,0), (0,1,0), (1,1,0), and (1,1,1).

So how do you say $z$ can be anything? I see a plane in there.

Sometimes using a computer to draw instead of your brain is a mistake.

Haha, ok. But the equation of the plane is $x-y=0$

I found it using the cross product and that whole mess, but it checks out -- it passes through my 3 desired points.

No, that equation does not come from the picture.

What do you mean?

What three desired points?

Oh, that's terrible! I see my mistake! Agh!

I chose the wrong three points. Please disregard.

LOL ... indeed.

It would be very funny ... if I didn't waste 20 minutes on it.

There is one plane in your picture that is not parallel to any of the coordinate planes.

Yup, guess who just realized.

Smarty McSmarty

I lied. There are two planes in your picgure that are not parallel to any of the coordinate planes. You have the front one: $x-y=0$. What's the back one?

I got $z=x$ and it seemed to work out fine.

Right.

Yup.

This site has gotten more annoying. I answered what I thought was one of the most interesting questions (on Lagrange multipliers) two days ago and neither the OP nor anyone else has responded with any interest.

Meanwhile, I continue to get hundreds of upvotes for the stupidest answer ever.

Haha

A real quickie: For this question ...

I must find the volume within both the cylinder and the sphere, is that right?

If so, would the limits of integration be $0 \leq r \leq b, 0 \leq \theta \leq 2\pi, -\sqrt{b-a}\leq z \leq \sqrt{b-a}$? I got the $z$ bounds by setting $r^2+z^2 = b^2$ (the sphere) equal to $r^2=a^2$ (the cylinder).

I'm reading about free products of $T_1$ topological spaces. Seemed unnatural at first, but it does resemble some kind of free construction on the second thought, with many properties being inherited from the original spaces ($T_n$ for $n = 2, 3, 4, 5$ and compactness)

the what now