Mathematics - 2019-01-10 (page 1 of 3)

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Secret

00:00

1 hour ago, by user193319

Not sure. That's why I came to the chatroom. Does the answer involve sophisticated algebraic topology?

Algebraic topology has reputation issues

Too many claims it will do whatever it is not

Ultradark

can entropy be measured with $H(x)=\ln(x) \ln(1-x)$

Secret

00:16

Depends on what you know about entropy

Also misread the $\not\exists $ page as "the question was violently removed by the user" and I go O_O

Ultradark

I know some things about entropy

entropy functions are concave

and entropy functions have the property of additivity

Math geek

01:14

0

Q: If $\vec{\nabla} \times \langle P,Q,Q \rangle=\vec{0}$ iff $Pdx+Qdy+Rdz$ is exact differential form.

If $\vec{\nabla} \times \langle P,Q,R\rangle=\vec{0}$ iff $Pdx+Qdy+Rdz$ is an exact differential form. My attempt:- If $Pdx+Qdy+Rdz$ is an exact differential form. Then there exists $U(x,y,z): dU=Pdx+Qdy+Rdz$. From the mixed partial theorem, It is easy to prove that $curl \langle P,Q, R\rangle=...

differential-equations pde scalar-fields

I don't understand what hello world do in his eighth line.

$$\phi_y =Q - \frac{\partial }{\partial y}\int Pdx.$$

Then he integrated with respect to $y$

$$\phi (y,z) = \int Q(x,y,z)dy-\int \frac{\partial }{\partial t}\left(\int P(x,y,z)dx\right)dt + h(z).$$

How this term $\frac{\partial }{\partial t}$ comes in the expression?

When we partially differentiate w.r.t $y$ again integrating. the answer should be like $$\phi_y =\int Q dy - \int Pdx+h(z).$$. Right?

Why did he use variable $t$?

3 hours later…

CaptainAmerica16

04:43

ted.com/talks/benoit_mandelbrot_fractals_the_art_of_roughness

This is really interesting. Fractals.

vzn

05:19

@CaptainAmerica16 fractals rock! =D coincidentally was just musing on them myself! vzn1.wordpress.com/2019/01/08/collatz-mod3

CaptainAmerica16

@vzn Nice :D I don't know much about fractals in terms of computer science.

Dair

05:46

@vzn lol looks like i'm not the only one doing fractals and collatz. (see: codereview.stackexchange.com/questions/203764/… )

CaptainAmerica16

Oh wow, lol

Derso

06:19

When are two cycles (in the same homology class) isotopic?

Is that simple to answer this?

famesyasd

06:36

suppose I have a function f: R->R and I have worked out a taylor serie $\sum_0^\infty \frac{f^{(k)}(a)}{k!} (x-a)^k$ for it. And I also figured out (for example by the root test) that this serie converges for all x in (a-R,a+R) does this immediately imply that this serie converges to f in (a-R,a+R) or do I still need to check that directly?

Akiva Weinberger

@Derso If there were, wouldn't that mean there's a simple description of $\pi_n(X)$ for all $n$ and $X$?

Finding homotopy groups, even of spheres, is notoriously hard

Balarka Sen

@AkivaWeinberger How are you directly inferring that? Two nonhomotopic maps $S^n \to X$ need not represent the same homology class.

Maps from higher dimensional spheres to lower dimensional spheres do, though, so that is a valid point.

Akiva Weinberger

06:52

@BalarkaSen All maps $S^n\to S^m$, $n\ne m$, are in the same homology class

Balarka Sen

Your comment about homotopy groups of spheres is correct. The previous one is false.

If $X$ is $(n-1)$-connected, eg, two maps $S^n \to X$ are homotopic iff homologous.

In any case Dereso seem to be asking a much stronger question: If you have two cycles with embedded manifold representative which are in the same homology class, when are they isotopic?

First of all you want the embedded representatives to be diffeomorphic submanifolds.

Akiva Weinberger

Use $S^n\times\Bbb R^k$ for sufficiently large $k$

so that our nontrivial elements of $\pi_m(S^n)$ can be represented by submanifolds

(Incidentally - do homotopy groups of spheres change when we go from the topological category to the smooth one? I'm guessing/hoping no)

Balarka Sen

It does not.

Continuous maps can be approximated by smooth maps

Akiva Weinberger

Good

Balarka Sen

@AkivaWeinberger Careful. Why need two nontrivial elements be represented by diffeomorphic submanifolds?

In general you'll get a immersed $S^m$ in $S^n \times \Bbb R^k$ representing your homotopy class

To make it an embedding you have to desingularize it: that changes topology

Maybe if $k$ is large then you can make it an embedded $S^m$ by "lifting the branches at the immersion points in different directions"

Akiva Weinberger

07:06

Yeah you should be able to

Balarka Sen

Seems right.

Cool

Akiva Weinberger

Throw Rs at it until it yields

I already know that the nontrivial class of $S^3\to S^2\times\Bbb R^3$ can be embedded.

I think that's right, anyway

Yeah

while $S^3\to S^2\times\Bbb R^2$ can't

Balarka Sen

Look at $\Bbb{CP}^1 \subset \Bbb{CP}^2$. The boundary of it's tubular neighborhood is an $S^3$ which projects to $\Bbb{CP}^1$ by the Hopf map. The tubular neighborhood is $O(1)$ over $\Bbb{CP}^1$, and $O(1) \oplus O(1)$ is the tangent bundle of $S^2$, which is stably trivial. So I can definitely write down an embedding $S^3 \to S^2 \times \Bbb R^k$ for some $k$ representing the Hopf map.

Not sure about 3

Ah, got it. Look at the unit tangent bundle in $TS^2$, which is an $\Bbb{RP}^3$ projecting to $S^2$. This map $\Bbb{RP}^3 \to S^2$ is the Hopf map quotiented by $\Bbb Z_2$. But $TS^2 \oplus \underline{\Bbb R} = S^2 \times \Bbb R^3$.

So you get an embedded $\Bbb{RP}^3$ in $S^2 \times \Bbb R^3$ which projects to $S^2$ by quotient of the Hopf map by $\Bbb Z_2$.

Some jugglary now probably gives the Hopf map

2lazy2carryitout

Akiva Weinberger

08:06

What's $TS^2 \oplus \underline{\Bbb R}$?

@BalarkaSen

Leaky Nun

@AkivaWeinberger that would be $S^2 \times \Bbb R^3$ :p

Akiva Weinberger

*visible anger*

By the way, fun fact, when you go past 100 tabs, Chrome for mobile stops displaying the number of tabs and just has ":)"

user image

(Bottom right)

(I'll go charge my phone)

Alessandro Codenotti

08:23

How did you get latex in chat to work on the mobile version of Chrome? @AkivaWeinberger

Akiva Weinberger

pastebin.com/CRSYEicL

@AlessandroCodenotti I used pastebin 'cause it's too long to fit in a message, but you make a bookmark with that^ as the link

(The text in the pastebin, not the pastebin link itself)

Kasmir Khaan

Hi chat

@LeakyNun Yo leaky !

can you take a look at this question

Leaky Nun

08:40

hi

Kasmir Khaan

Let H be a subgroup of G, let f : G--> H be a hom whose restriction to H is the identity map

my question is , how does this even work ? G---> H ?

Tobias Kildetoft

@KasmirKhaan The same way as usual

Kasmir Khaan

@TobiasKildetoft Hi Tobias !

is H is a subgroup of G

how can we embedd G in H

Tobias Kildetoft

nobody said embed

Kasmir Khaan

and the map is the identity map, could this be a typo in the question ?

I meant, how can we get a map from G--> H , being the identity map

if G is not equal to H

elements of G will make no where

i mean element of G \ H

G minus H

Tobias Kildetoft

08:44

it is only meant to be the identity map on elements of $H$

Kasmir Khaan

aha

and where do the other element map to ?

Tobias Kildetoft

Try for example the map $C_2\times C_2\to C_2$ given by $(x,y)\mapsto x$.

Leaky Nun

@TobiasKildetoft well...

$C_2$ is not really a subgroup of $C_2 \times C_2$ is it

Kasmir Khaan

exactly =p

Tobias Kildetoft

Sure it is, as long as we agree to abuse notation in useful ways

Leaky Nun

08:46

well he meant $C_2 \times C_2 \to C_2 \times \{e\}$ sending $(x,y) \mapsto (x,e)$ @KasmirKhaan

Kasmir Khaan

hmm okay make sense that way

the question is phrased very bad

Leaky Nun

no it isn't

Alessandro Codenotti

@AkivaWeinberger I tried that already but it's not working for me

Akiva Weinberger

And you tried clicking on the bookmark while you're viewing this page?

Tobias Kildetoft

@AkivaWeinberger is it just the usual bookmark?

Akiva Weinberger

08:52

(You can get to bookmarks by the "…" button on the bottom-right of the screen)

@TobiasKildetoft Yeah

Tobias Kildetoft

Doesn't work on my mobile either

Daminark

Even at 3AM there are nerds

How's it going guys?

Tobias Kildetoft

(also, your chrome looks really weird compared to mine)

Leaky Nun

@Daminark it's 09:53 here

Akiva Weinberger

There's an update that happened not too long ago

(and I'm not fond of it but it is what it is)

Daminark

08:53

Leaky ples we all know that CST is the correct time zone

Tobias Kildetoft

But that seems to have moved all the menus and stuff to the bottom?

Daminark

:P

Akiva Weinberger

I think to make a bookmark you have to make a bookmark to a regular page first, and then edit it

@TobiasKildetoft Yeah

Daminark

Yeah my phone has stuff on the bottom as well, and it's fairly new

Tobias Kildetoft

Well, I know an update I will be postponing for as long as I can

Akiva Weinberger

08:54

@Daminark Above "find in page" does it say "bookmark" with an icon of a star with a plus sign in it next to it?

That should make the current page into a bookmark

and then it lets you edit the bookmark

Kasmir Khaan

@TobiasKildetoft btw Tobias, how is S_3 the product of 2 subgroups of order 2 and 3 , and only one of them is normal ?

Leaky Nun

@KasmirKhaan it isn't

Daminark

user image

Kasmir Khaan

the theorem in my book said that they both need to be normal

@LeakyNun why it isint ?

Daminark

That's what it looks like for me

Akiva Weinberger

08:57

Oh never mind yours looks different than mine

How do you make the current page into a bookmark on that?

Daminark

At the bottom of that menu there's a star

Leaky Nun

@KasmirKhaan because any group of order 2 is C2 and any group of order 3 is C3 and C2xC3 is C6 not S3

Akiva Weinberger

@Daminark What happens when you press it

Makes a bookmark?

Kasmir Khaan

@LeakyNun imeant this , HK not HxK

H= < (12)> and K = <(123)> for example

Leaky Nun

well by counting you can establish it

the theorem is that if H and K are normal then HK is a subgroup

there's no converse to it

Daminark

08:59

Yup, and for you, if you download a pdf or something, do you see that little thing at the bottom of the screen saying "Downloaded -file-" and with a button to the right allowing you to open it? Well, when I tap the star it says "Bookmark created" at the bottom and there's a button to edit

(It only displays for a few seconds)

Kasmir Khaan

@LeakyNun we only need one of them to be normal to be subgroup

@LeakyNun if both of them are normal we have a normal subgroup

Daminark

Also if you are on a bookmarked page that star becomes blue, tapping that will allow you to edit the bookmark

Leaky Nun

your point being?

Akiva Weinberger

Weird, that's different from how it works on iOS

So edit it to the link from before

Kasmir Khaan

my point is , it might be nessarly condition not sufficent

that if H and K are subgroups

Akiva Weinberger

09:00

@Daminark When I open a PDF it gives me the option to open it in the iBooks app

Kasmir Khaan

st their intersection is trivial and HK = G

Akiva Weinberger

By the way, I heard that $\langle a,b\mid a^4=b^5,aba=bab\rangle$ is the trivial group. Why?

Kasmir Khaan

and HK= KH

then we have an isomorphism HK with G

it does seem to work in S_3 without satisfying normality of both subgroups

Leaky Nun

@AkivaWeinberger because some reduction reduces a and b to the identity?

I mean, I don't think that theorem could carry much value

@KasmirKhaan nobody said it's an iff

Kasmir Khaan

okay thanks @LeakyNun

Akiva Weinberger

09:04

@LeakyNun Right yes but what reduction

Oh wait I think I see it

Alessandro Codenotti

@AkivaWeinberger yes I had to do that

And I clicked the bookmark while on this page

Akiva Weinberger

$(ba)b(ba)^{-1}=baba^{-1}b^{-1}=abaa^{-1}b^{-1}=a$

Daminark

Ah so two conjugate elements have orders dividing 4 and 5

Akiva Weinberger

So $a$ and $b$ are conjugates

Daminark

Clever

Akiva Weinberger

09:08

Hm wait actually we don't have $a^4=b^5=e$, we just have $a^4=b^5$

Daminark

Oh right

Uh I mean

$a^4 = (ba)b^4(ba)^{-1} = b^5$, so $b^4$ and $b^5$ are conjugate

Kasmir Khaan

ab^4 = b^5 *a

that is all what you need

Daminark

If we know the group is finite I think that should take care of things because of order coprimality I think?

Kasmir Khaan

a^4 = b^4 !

from that we get that a and b can only have order 1

ie they are both the identity

Daminark

Wait I don't see why what you said is true

First off, where does $ab^4 = b^5a$ come from?

Also @Akiva was there a particular context for this group or was it just lol memes?

Akiva Weinberger

09:19

It was something about a group that was hard to prove was trivial

Like, there's a certain set of operations you can do to a group presentation that don't change the group, and it's conjectured that you can't get from this presentation to the normal presentation of the identity group using those operations

Kasmir Khaan

a^4ba = b^6a

multiplying by ba on the right

then reducing the left side using the rule aba = bab

replace each aba on the left with bab

bab^4 = b^6 a

cancel b on the both sides

ab^4 = b^5 a

but b^5 = a^4

ab^4 = a^5

hence b^4 = a^4

now a^4 = b^4 = b^5

Leaky Nun

@AkivaWeinberger It's well-known that the word problem is undecidable

Akiva Weinberger

@KasmirKhaan How did you get from a^4ba to bab^4?

2

@LeakyNun Yes but this is only half of the word problem

If you could get from any presentation of the trivial group to the usual one using those operations, then it doesn't mean the word problem is decidable, because

you could have an algorithm that methodically does all of the operations until it reduces it,

and if it's a trivial group then it would eventually halt,

but if it's not a trivial group, you'd never know when to stop

Leaky Nun

I see

it's Δ1 vs Σ1

Akiva Weinberger

Oh I get it @KasmirKhaan

2

aaaaba=aaabab=aababb=ababbb=babbbb

@LeakyNun I don't remember exactly what those operations are though

Kasmir Khaan

09:34

@AkivaWeinberger yes !

Akiva Weinberger

Oh I think it was specifically about "balanced" presentations where there's the same number of generators as relations

Found it

https://en.wikipedia.org/wiki/Andrews–Curtis_conjecture

@KasmirKhaan Well done

3

Kasmir Khaan

@AkivaWeinberger thanks Akiva ! I was just lucky =p

Akiva Weinberger

Just say "Thanks"

2

Kasmir Khaan

okay thanks

Akiva Weinberger

@Daminark I'm very curious what sort of meme would involve a group theory problem

Alessandro Codenotti

09:54

user image

2

Do geometric group theory memes count?

Learner

10:10

2

Q: Have I correctly proved that $\lim_{||(x,y)||\to\infty}\frac1{y-x}\int_x^y\exp(-1/|t|)dt$ equals $1$?

I should prove that, as long as $y\ne x$, $$f(x,y)=\frac1{y-x}\int_x^y\exp(-1/|t|)dt\longrightarrow1 \ \text{as $||(x,y)||\to\infty$}$$ and I would like to do it without $\varepsilon-\delta$ reasoning. My idea was to let $h=y-x$, and then separately consider the cases $|h|\to\infty$ and $|h|\not\...

real-analysis integration multivariable-calculus proof-verification alternative-proof

Can anyone tell me if my proof is correct?

famesyasd

10:31

user image

does this follow directly from Mertens' theorem?

i.e. if two series converge to A and B respectively and either one of them absolutely converges then their Cauchy product converges to AB

Akiva Weinberger

@famesyasd I think so

famesyasd

nice

Learner

@AkivaWeinberger Sir?

Akiva Weinberger

10:54

Yes?

Oh hold on

What's $\|(x,y)\|$? @Learner

@Learner I think when you do L'Hopital, you treat $x$ as if it's a constant with respect to $h$, even though $h$ depends on $x$

Learner

11:15

@AkivaWeinberger Thank you for your feedback! I see, you're right... is there a way I can salvage that approach? Some helpful theorem?

As in, something that guarantees that in this case no harm is done when one uses L'Hopital

@Akiva $||(x,y)||$ is the norm of $(x,y)$

Akiva Weinberger

11:54

https://transcendentcode.files.wordpress.com/2014/03/truchet.gif

Learner

@Akiva ?

Akiva Weinberger

I dunno

Hm, what happens if $x=0$ and $y\to\infty$?

$\displaystyle\frac1y\int_0^ye^{-\frac1{|t|}}dt$

Why does this go to 1?

Learner

In that case you can apply L'Hopital

So that $\displaystyle \lim_{y\to\infty}\frac1y\int_0^ye^{-\frac1{|t|}}dt=\lim_{y\to\infty}e^{-\frac1{|‌y|}}=1$

Silent

12:15

Suppose $Q\in M_{3\times3}(\Bbb R)$ is a matrix of rank 2. Let $T:M_{3\times3}(\Bbb R)\to M_{3\times3}(\Bbb R)$ be the linear transformation defined by $T(P)=QP$. The the rank of $T$ is .....

I thought of forming matrix but it seems to be of 9*9 dimension

and question is of two marks only.

Please help

user193319

How do I compute the projection/closet vector to a subset? If it helps, I am working in $\Bbb{R}^2$, but I would like formalue in terms of norms and inner products, if possible.

For context, I am trying to prove that the figure eight is a deformation retract of the doubly punctured plane. And it is annoying that everything hinges on these annoyingly simple question.

user629353

How to divide the polynomial $x^2+xy+y^2$ by $x+y$? Since it has two variables it find it confusing. Should I start multiplying the divisor by x or by y?

Somebody please help

Leaky Nun

13:11

@user629353 I don't think you can

afterall, $\Bbb C[X,Y]$ isn't a Euclidean domain

Tobias Kildetoft

@LeakyNun Well, once you fix a term order, you can do it

Leaky Nun

but...

Learner

@user629353 At most you can say $\tfrac{x^2+xy+y^2}{x+y}=x+y-\tfrac{xy}{x+y}$

Tobias Kildetoft

@LeakyNun You then get a remainder that is smaller in the term order (but the term order need not give you an euclidean function)

user629353

Actually this (ibb.co/mqbkjDY) is my working. I am getting a non-terminating expression. Is it wrong?

Tobias Kildetoft

13:22

@user629353 Are you familiar with term orderings?

user629353

What is that?

Tobias Kildetoft

more or less what you need to choose to get a well-defined quotient and remainder when doing division of polynomials with more than one variable

(technically you also use one for a single variable, but there you have just one choice)

Why do you need to do this division?

user629353

@TobiasKildetoft what does "more or less what you need to choose" mean

Tobias Kildetoft

@user629353 It means that this thing is (more or less) the thing you need. I.e. there might also be some other thing you could use, but it would be essentially the same.

user629353

Is there any way to do

@TobiasKildetoft can you do it in a proper way

Abdullah UYU

13:38

Hi, if a fix an element of a set by saying let's $a\in A$, can I use it later when defining a set like $B=\{c|P(c)=a\}$? My concern is that $a$ is probably not in the scope of the set generator.

I mean, $a$ is a free variable in that definition, so can be changed to anything.

user629353

@TobiasKildetoft "two numbers divided to yeild irrational quantity, then one of them must be irrational" is the case also with polynomials

Lucas Henrique

Hi chat.

Abdullah UYU

13:59

Hi @LucasHenrique

Lucas Henrique

@AbdullahUYU Informally, yes.

user629353

Can somebody answer to mine

Lucas Henrique

Usually when I pick some fixed, generic $a$, either I construct something based on this generic $a$ or I say a general fact about any element. So if you were to state this more rigorously, either you construct a set of objects based on every $a$ or you start a formula like $\forall a(a \in A \implies \dots)$

@user629353 The question was already answered. You can't divide like in integers or 1-variable polynomials.

This because $\Bbb C[X, Y]$ is not an Euclidean domain, as @Leaky pointed out.

And, the same way he told you, you can say at most $\tfrac{x^2+xy+y^2}{x+y}=x+y-\tfrac{xy}{x+y}$.

3

A: Is $F[x,y]$ a Euclidean Domain?

Every Euclidean domain is a principal ideal domain, but the ideal $(x,y)$ in $F[x,y]$ is not principal.

Abdullah UYU

Yes, I think that's fair. There has to be a set theoritic way of clearly stating it.

Lucas Henrique

@user629353 Informally speaking, the main reason is that we can't establish a criterion to say where the division algorithm must stop. Your own example illustrates this specific case. How do we divide $xy$ by $x+y$? The degree of the "remainder" is, in fact, greater than the dividend's!

user629353

14:16

@LucasHenrique sorry, but I am unfamiliar with abstract algebra

or it's related topics

As you send in above Euclidean domain example

Lucas Henrique

@AbdullahUYU, yup. Usually on metric spaces, we write stuff like "fix some $a$ neighboring $x$. Then either ... blah blah blah *... so the possibilities are ... *blah blah ... Now pick every neighbor of $x$. Then at least one is ...". If you were to do this rigorously, you'd have a universal quantifier and then you'd pick the neighborhood of $x$ and use the general fact to get a contradiction, etc...

By the same way, you can get a generic structure based on each arbitrary $a$, like the set of all the balls around each element of a metric space. You could state a lot of things about this "specific" but what you're really doing is creating a set (or a class) with such structure for each of the elements.

@user629353 to understand deeply (with a proof), you must know some ring theory, but you may trust me: unfortunately (well, at least I like Euclidean domains :p), $\Bbb C[x,y]$ doesn't have a well-defined division algorithm for all its elements. The case you sent us is such example. You can't divide "more" because you can't really tell when to stop.

You might want to read my first reply again. The last part says it (the degree part)

Also, try to write $x^2 + xy + y^2 = (x+y)q(x,y) + r(x,y)$, with $deg r < deg (x+y) = 1$.

user629353

14:32

@LucasHenrique so the polynomial long division process is for single variable polynomials only.

Lucas Henrique

Yes.

user629353

Thanks

Lucas Henrique

You're welcome. :)

Semiclassical

15:06

just posted the following question on main:

0

Q: What is the surface area of the 3-dimensional elliptope?

The $n$-elliptope is defined as the set of $n$-by-$n$ correlation matrices; that is, the set of $n$-by-$n$ symmetric positive-defined matrices with ones on the diagonal. Such matrices are parametrized by their $n(n-1)/2$ upper off-diagonal elements. In the case of $n=3$, this yields the 3-ellipto...

calculus geometry statistics multivariable-calculus correlation

working on adding my own attempt at the moment as well

Balarka Sen

@AkivaWeinberger Direct sum of tangent bundle of $S^2$ with the trivial line bundle on $S^2$.

Abdullah UYU

15:19

@Semi Dont't know much about you asked but is it normal that the volume of somthing can be expressed as some multiple of $\pi^2$?

Semiclassical

That is a bit funny, isn't it.

@AbdullahUYU the linked question has the calculations for that, so you can verify it for yourself

Abdullah UYU

Yeah, coming from there.

Semiclassical

But nonetheless it is funny that the answer comes out so simply.

Kasmir Khaan

Hi guys

is there something that can be done to a 3x3 upper triangular matrix

that sends the entry 12 and 23 to 0

leaving 13 unchanged?

i have that the kernel = the 3x3 matrix identity exept that the entry 13 can be anything

trying to find a map with such kernel from the upper triangular matrices to G' with that as a kernel

Rithaniel

15:35

That should be relatively easy with row operations, right?

R2-R3(23)/(33) followed by R1-R2(12)/(22)? If that makes sense.

Lucas Henrique

15:50

@Ted: here you use $f(t) = ||x - ty||$, and its motivation is clear. However, IIRC in your book you work with $\frac{x}{||x||}$ and $\frac{y}{||y||}$ and the module of their difference (you've used that they're greater of equal to 0). What's the geometric interpretation of this construction? I can understand why you'd use same norms, i.e, scaling them to the same circumference... but it looks like you don't use this property. Where this idea this come from?

quallenjäger

16:40

For a cointinuous map, we know it is for every $\epsilon>0,\text{there is a \math{\delta}},\mid f(x)-f(c) \mid<\epsilon \forall x, \mid x-c \mid<\delta$. If $\epsilon \rightarrow 0$, does it has to be $\delta \rightarrow 0$?

No right? because I can take a constant function

Pig

you are right

continuity only requires existence of such $\delta$ - in many cases they are supposed to be "small", but it doesn't have to be, e.g. the constant function you mentioned

and so in particular even if $\epsilon \to 0$, it's entirely possible for $\delta$ to stay pretty wide. Strictly speaking though, constant function is the only counterexample to this

Ted Shifrin

@LucasHenrique In the lectures, I was trying to tie in calculus, of course, so minimizing $f(t)$ was natural (of course, we don't really need calculus for that). The idea for the other proof is simple: We know that $\|z\|\ge 0$ for any vector $z$ and the norm is $0$ ONLY for $z=0$. For Cauchy-Schwarz we know that equality will hold only when $x$ and $y$ are positive scalar multiples of one another, which means that $x/\|x\|=y/\|y\|$. So we consider $z$ to be their difference.

Hi @Piggy.

Pig

hey @Ted

Érico Melo Silva

hlo nerds

Ted Shifrin

Hello @Eric @Leaky

Pig

16:53

heya

Math geek

0

Q: If $\vec{\nabla} \times \langle P,Q,Q \rangle=\vec{0}$ iff $Pdx+Qdy+Rdz$ is exact differential form.

If $\vec{\nabla} \times \langle P,Q,R\rangle=\vec{0}$ iff $Pdx+Qdy+Rdz$ is an exact differential form. My attempt:- If $Pdx+Qdy+Rdz$ is an exact differential form. Then there exists $U(x,y,z): dU=Pdx+Qdy+Rdz$. From the mixed partial theorem, It is easy to prove that $curl \langle P,Q, R\rangle=...

differential-equations pde scalar-fields

I don't understand what hello world do in his eighth line.
$$\phi_y =Q - \frac{\partial }{\partial y}\int Pdx.$$
Then he integrated with respect to $y$
$$\phi (y,z) = \int Q(x,y,z)dy-\int \frac{\partial }{\partial t}\left(\int P(x,y,z)dx\right)dt + h(z).$$
How this term $\frac{\partial }{\partial t}$ comes in the expression?
When we partially differentiate w.r.t $y$ again integrating. the answer should be like $$\phi_y =\int Q dy - \int Pdx+h(z).$$. Right?
Why did he use variable $t$?

did he do wrong calculation?

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