Mathematics - 2017-02-14 (page 4 of 6)

Felix.C

16:01

When we speak about a topology on a set. Is this set supposed to be finite?

Alessandro Codenotti

not necessarily

DHMO

@Felix.C not really

but finite cases are there to help you understand

Like this collection of topologies on sets containing three elements

Semiclassical

I presume there's some obvious combinatorial meaning to that.

DHMO

topologies on infinite sets include the order topology on R and the discrete topology on N

@Semiclassical collections of subsets of {a,b,c} closed under countable union and finite intersection

which includes {} and {a,b,c}

Semiclassical

Sure, I meant in terms of how many such pictures one can draw.

DHMO

16:03

should have some combinatorial meaning

I think the 9 topologies are unique up to isomorphism

Semiclassical

Hmm.

DHMO

but whether there's another is beyond me

Semiclassical

Well, on one element there'd be two topologies (the element itself and the empty set).

Felix.C

Ok, thank you guys!

Semiclassical

How many topologies on two elements?

(I'm having a hard time thinking clearly at the moment, which I blame on the above neck pain)

Alessandro Codenotti

16:05

iirc that's the same as counting preorders and there's no known closed formula

Semiclassical

huh, neat. I imagine there's an OEIS entry though.

DHMO

@Semiclassical
{{},{a,b}}
{{},{a},{a,b}}
{{},{a},{b},{a,b}}

you don't really have many choices

this?

Semiclassical

{{},{b},{a,b}}?

DHMO

@Semiclassical the OEIS counts this, but I count it as duplicate of my second one

Semiclassical

ah.

DHMO

16:07

here is the isomorphism one

Semiclassical

This one, then? oeis.org/A001930

Ah, timing.

Alessandro Codenotti

this one if you count homeomorphic ones as distinct

DHMO

homeomorphic vs isomorphic?

Semiclassical

None of which seem to offer any formulae, which bears out @Alessandro's remark.

Secret

16:22

{{},{a},{a,b}}
Considering {a}, a is separated from b because the open set {a} containing a (i.e. a neighbourhood of a) and the open set of b (emptyset???) are disjoint
Considering {a,b}, a is connected to b because the open set {a,b} containing a (i.e. a neightbourhood of a) and the open set {a,b} containing b (i.e. a neightbourhood of b) intersects (and the intersection is itself, {a,b}).

Therefore in this topology, a is separated AND connected to b at the same time??...???

Balarka Sen

Hi @MikeMiller

DHMO

16:41

@Secret two points are connected if any open set containing one must contain the other, and vice versa

one point is isolated if there is an open set containing only that point

so a is isolated

but a and b are not connected

@BalarkaSen are you here?

Balarka Sen

da

DHMO

@BalarkaSen is "connected" a real term? I seem to have invented it myself

I can't find any reference online

Balarka Sen

maybe you forgot to check wikipedia?

Oh I see you are using it differently

No, that's not the right term for saying two points are not separable by disjoint neighborhoods

DHMO

what's the right term?

Balarka Sen

dunno. it's slightly stronger than not being separated by neighborhoods.

DHMO

16:48

@Secret I've successfully created a term and passed on to you

@BalarkaSen how do you describe the element (1,0) in the order topology of {0,1} x N under the lexicographical order?

Balarka Sen

Do you mean, describe a neighborhood of (1, 0)?

DHMO

I mean, how do you say that any open set containing it must contain other elements lol

"not isolated" maybe

I just answered myself

Balarka Sen

Ah you were asking for terminology.

Fawad

@BalarkaSen in triangle ABC ,$ SinA:SinB:SinC= \frac{b×c}{∣b∣∣c∣) : \frac{c×a}{∣c∣∣a∣) : \frac{a×b}{∣a∣∣b∣)$

Then how to proceed?

Astyx

@Fawad you're missing a $

Balarka Sen

16:56

You should work with sin A : sin B first.

Fawad

@Astyx thanks,is it correct now?

Balarka Sen

And figure the constant of proportion

Anyhow, I am off for now

Astyx

It doesn't compute for me

Fawad

@DHMO ready for help

Hi @TedShifrin sir

DHMO

@Astyx je suis desole si je t'ai demande ca avant: quel sujet de mathematics tu preferes le plus?

JeSuis

16:58

Joyeux anniversaire en retard @TedShifrin

Ted Shifrin

Salut, @Astyx @JeSuis

Astyx

Salut !

Ted Shifrin

Merci bien, M JeSuis

hi Fawad

DHMO

3 hours ago, by Kasmir Khaan

Guys plz send me a message when Ted is here

@KasmirKhaan god dag

@Fawad I said I have no idea

maybe Theodore can help

Astyx

@DHMO Tu ne me l'as pas déjà demandé, mais je t'avoue ne pas trop savoir

JeSuis

16:58

(Alexandre est mon prénom)

Fawad

@DHMO god ?

DHMO

@Astyx alors, que fais-tu maintenant, in regard to mathematique?

@Fawad "good" in Swedish

Astyx

En ce moment ? On a attaqué les séries entières, mais c'est pas ce que je préfère clairement

Ted Shifrin

Alors, @Astyx, que préfères-tu?

JeSuis

@TedShifrin I have a question, I tried to find a holomorphic function $f$on $U:=\{z\in \Bbb{C}: 1/2<\vert z\vert<1\}$ such that for all $z\in U,\quad f(z)=f(e^{2i\pi \alpha}z)$ where $\alpha$ is irrational. By density we get $f(z)=f(\exp(i\theta)z)$ for all $\theta$

Astyx

17:01

@Ted C'est la question initiale, et c'est assez délicat d'y répondre :p

DHMO

@TedShifrin y a-t-il un terme pour cela?

20 mins ago, by DHMO

@Secret two points are connected if any open set containing one must contain the other, and vice versa

JeSuis

Salut @Astyx

Astyx

Salut @JeSuis

DHMO

c'est claire que "connexe" n'est pas le terme correcte @TedShifrin

@JeSuis bonjour

JeSuis

@DHMO are you french too ? ^^

DHMO

17:02

@JeSuis non, mais je parle francais un peu

Secret

Well give what Barlarka said, it seems neighbourhood is somehow different from a basis element, but...

DHMO

ou comme on dit, un p'tit peu

JeSuis

@DHMO Ok, I was thinking that we colonized the chat

DHMO

@Secret neighbourhood is quite different from basis element

Ted Shifrin

@DHMO: It means the topology is very non-Hausdorff.

DHMO

17:03

@TedShifrin I mean, do we have a term for such two points

Ted Shifrin

@JeSuis: If you write down Cauchy-Riemann in polar coordinates, I bet you'll see the function is constant.

Fawad

@TedShifrin can you help me in vectors and trigonometry related question?

Ted Shifrin

(Or use open mapping.)

I don't know, @Fawad. What's the question?

Mike Miller

Hi @BalarkaSen. I'm still not sure what your objections were.

Ted Shifrin

G'night, @MikeM.

DHMO

17:04

wow, two gods are here at the same time

Fawad

Ted Shifrin

@Fawad: OK, so use the vector product in two different ways to compute the area of the triangle.

Astyx

J'aime bien la combinatoire, la topologie, l'algèbre linéaire, la théorie des groupes, la théorie des graphes principalement @DHMO

Ted Shifrin

@Astyx: Mais il faudra que tu aimes la géométrie aussi! :P

Astyx

Et l'études des suites

Fawad

17:06

@TedShifrin very nice!

Astyx

On fait tellement peu de géométrie en France (à mon niveau) que je ne peux pas m'exprimmer sur ce sujet :p @Ted

Ted Shifrin

Now you can do it, @Fawad?

DHMO

@Astyx pourquoi ne dis-tu pas "tout de mathematique" lol

Ted Shifrin

Ça ne me convient pas du tout, @Astyx.

Astyx

Bah moins l'analyse par exemple @DHMO

Moi non plus

Ted Shifrin

17:07

Mais, @Astyx, l'étude des suites c'est bien l'analyse.

Fawad

1/2(a×b)=1/2(b×c)=1/2(c×a)

Astyx

Oui par analyse j'entends le calcul différentiel

Fawad

ab sin c = bc sin a = ca sin b

DHMO

@Astyx tu mesures?

Ted Shifrin

mais ce n'est pas juste, ça, @Astyx.

Secret

17:08

I think, the really key question will be, what is the difference the open set {a,b} vs {a} did to the topology of the 3 element set X

e.g.
{{},X,{a},{b},{a,b}}
{{},X,{a,b},{b,c},{b}}

From DHMO and barlarka have said, a neighbourhood of some point x is very different from a basis element containing x, thus we cannot call a basis element containing x a neighbourhood of x. So, what's the different between the open set {a} and {a,b} in terms of topological properties?

Ted Shifrin

OK, @Fawad, when you take magnitudes, yes.

Astyx

@Ted Les deux termes se mélangent dans ma tête :p

JeSuis

@TedShifrin you mean using the fact that $f(U)$ is open ?

Astyx

Je n'ai pas compris ce que tu veux dire @DHMO

DHMO

@Secret {a} contains 1 term while {a,b} contains 2 terms

Ted Shifrin

17:09

@JeSuis: If the function is constant on circles centered at the origin, I think it will be difficult for the image to be open.

DHMO

@Astyx tu etudies le theorie de mesures?

Ted Shifrin

But you should be able to write down C-R directly in polar coordinates, @JeSuis.

Astyx

Un petit peu oui @DHMO

Mais trop peu pour pouvoir donner mon avis je pense

DHMO

@Astyx j'etudie le theorie maintenant par moi meme

Astyx

Ça m'a l'air d'être une théorie élégante

Secret

17:10

I have worked through Munkres all the way to continous functions, and I STILL don't get what these open sets themselves are doing (I only knew they help me to define continuity and all those topological properties, but what are these open sets "measuring" wrt the points they contain?

DHMO

@Astyx c'est interessant

@Secret what do you mean? (or in Cantonese, nei gong mut gau?)

Astyx

Tu l'apprends comment ?

Ted Shifrin

@Secret: "Nearness" is defined by open sets (presumably more by smaller ones than by larger ones).

JeSuis

Ok I am trying

Fawad

ab sin c = bc sin a = ca sin b =t then sin a = at/abc and sin b = bt/abc and sin c = ct/abc then abc= sin a/a + sin b/b + sin c/c thanks a lot @TedShifrin nice catch to question

Astyx

17:12

@TedShifrin Sinon quoi de neuf ? :)

DHMO

@Astyx de l'internet, comme j'apprends les autres sujets

Astyx

@DHMO Oui mais tu suis un cours ou ..?

Simply Beautiful Art

MSE and I wish you a Happy Valentine's day!

DHMO

@Astyx wikipedia est ton ami

@Secret what do you mean by "measure"?

Astyx

@Simply You're trying too hard :p

Secret

17:13

@TedShifrin @DHMO Consider the topology {{},{a},{a,b},{a,b,c}}. So each open set define a different type of "nearness" as in a near b in one "resolution", and a near b,c in another resolution and a not near anything in yet another resolution?

as if the set is some object placed under a microscope, and the topology is how far we zoomed in?

Simply Beautiful Art

@Astyx I'll only say it took me a few hours...

DHMO

@Secret As I have said the billionth time, firstly you have to define nearness

Astyx

@DHMO Je doute que Wikipedia soit le meilleur moyen d'apprendre

DHMO

@Secret open sets are not merely nearness. The set (0,1) U (100,101) is open for example, in the order topology of R

Ted Shifrin

I honestly don't like thinking about finite sets that much, @Secret, but you can if you insist. This is not a metric topology at all, so it's not symmetric. $a$ is a limit point of $b$ and of $c$, but $a$ is nevertheless far away from $b$ and $c$ (there is a neighborhood of $a$ disjoint from them).

BAYMAX

17:15

Hey guys can you give an example of a non measurable set?

DHMO

@Astyx as-tu autres sources?

@BAYMAX vitali, if you're talking about Lebesgue measure

Ted Shifrin

Only by using the axiom of choice, @BAYMAX. You can find this on Wiki.

BAYMAX

ok thanks

DHMO

@Secret right, talking about limit points is fun. Ted just reminded me this

@TedShifrin what do you mean by limit point of a point?

Astyx

@DHMO Moi personnellement non mais je suis certain que les gens sur ce chat en ont, il suffit de demander

DHMO

17:16

I thought limit point is of a set

Ted Shifrin

It is; I was too lazy to type set brackets.

Every neighborhood of $b$ contains $a$. Every neighborhood of $c$ contains $a$ and $b$.

Astyx

Voire même sur le main, la question a dû être posée

DHMO

@Secret [A] limit point of a set S in a topological space X is a point x (which is in X, but not necessarily in S) [...] that every neighbourhood of x with respect to the topology on X also contains a point of S other than x itself.

From, you know, the Wikipedia

Ted Shifrin

Huh? @JeSuis.

What are these $\theta'$ things?

JeSuis

J'ai écrit n'importe quoi

Alessandro Codenotti

17:20

Hi @Ted

Ted Shifrin

Hi @Alessandro :)

JeSuis

I don't have an expression for the real part and imaginary @TedShifrin

Ted Shifrin

You should, @JeSuis. It's just chain rule with the regular C-R equations, or you can figure something out using differential forms (my favorite).

Secret

Ted seemed to understand what I mean there. He knew I am talking about an imprecise term, I think he knew how to translate that to you, because this "nearness" is the key to my question (which he had translated as it seems)

@Ted, so you mean for {{},{a},{a,b},{a,b,c}}, a is a limit point because there are two neightbourhoods of a ({a,b} and {a,b,c} respectively) that contains, respectively, b and b,c other than a, yet a is disjoint from b and c because the neightbourhood (which is the open set {a}) contains a only and not b and c

DHMO

@TedShifrin wait, using the definition I quoted above, it seems that a is not a limit point of {b}.

Secret

17:25

(As for some background. The primary reason I want to learn topology proper is because I had some modelling that involve highly non hausedoff sets, thus I need to know how to handle general topology, both hausedoff (which munkres has done well) and non hasedoff)

Ted Shifrin

OF course "near" is a metric intuition, and many topologies (like this) have no metric underlying them.

@DHMO: Does $b$ have a neighborhood that does not contain $a$?

DHMO

@TedShifrin but $a$ has a neighbourhood that does not contain $b$

7 mins ago, by DHMO

@Secret [A] limit point of a set S in a topological space X is a point x (which is in X, but not necessarily in S) [...] that every neighbourhood of x with respect to the topology on X also contains a point of S other than x itself.

you claim the limit point is $a$, which is x in the definition above

we're talking about "every neighbourhood of x"

Ted Shifrin

That is irrelevant, @DHMO. If we're talking about $a$ being a limit point of $\{b\}$, we look at neighborhoods of $b$. This is not something symmetric.

DHMO

so the above definition is wrong?

JeSuis

hum I am bit lost, ok so C-R in polar coordinates it's for $f=u+iv$, we have $\frac{\partial u}{\partial r}=\frac{1}{r}\frac{\partial v}{\partial \theta}$ and $\frac{\partial u}{\partial \theta}=-r\frac{\partial v}{\partial r}$

Ted Shifrin

17:26

No, @DHMO, I'm not even reading it.

OK, @JeSuis. You found that somewhere, but you should be able to prove it. What does it tell you, at any rate?

Secret

@Ted If I understood correctly, a has a neightbourhood containing b namely the open sets {a,b} and {a,b,c}, but a also has a nieghtbourhood not containg b, namely the open set {a}?

DHMO

@TedShifrin mathworld agrees with me...

> The topological definition of limit point P of A is that P is a point such that every open set around it contains at least one point of A different from P.

"every open set around it [the limit point]"

Alessandro Codenotti

I'm kinda stuck on something that should be solved just by applying definitions... The complement of a dense open set is a nowhere dense closed set, right?

Ted Shifrin

Oh, right, @DHMO. Sorry, I'm trying to do 3 things here.

I can't do that any more.

DHMO

@TedShifrin alright

Alessandro Codenotti

17:28

Q is not open

DHMO

@AlessandroCodenotti dense in where?

Ted Shifrin

Yes, @Alessandro, that's right.

Alessandro Codenotti

in some topological space $X$

JeSuis

Je ne comprends pas la question! @TedShifrin

Alessandro Codenotti

@TedShifrin ah, of course, if a point is in the interior of $A^c$ it can't be a limit point of $A$... I should have noticed that

DHMO

17:32

@AlessandroCodenotti "closed" comes from definition.
Let the topology be T on a set S.
Let the open set be X and its complement be Y.
Assume that its complement is dense in S for contradiction.
So, every point in X is a limit point of Y and every point in Y is a limit point X.
Let x be a point in X.
Every open set containing x also contains a point in Y by definition of limit point.
I'm not seeing any contradiction lol

Ted Shifrin

@JeSuis: You said that you had a holomorphic function $f$ that didn't depend on $\theta$. So ... ?

Secret

DHMO

For example, in the trivial topology of {a,b}, {a} is dense and {b} is also dense? @AlessandroCodenotti or am I missing something trivial?

Secret

(ok to test my understanding):

Alessandro Codenotti

@DHMO yes and neither is open

DHMO

17:33

@AlessandroCodenotti oh god

@AlessandroCodenotti "closed" comes from definition.
Let the topology be T on a set S.
Let the open set be X and its complement be Y.
Assume that its complement is dense in S for contradiction.
So, every point in X is a limit point of Y and every point in Y is a limit point X.
Let x be a point in X.
Every open set containing x also contains a point in Y by definition of limit point.
However, X is an open set containing x that does not contain any point in Y, by definition of complement.
Hence a contradiction is found.

@Secret {{},{a},{b},{a,b},{a,b,c}}

JeSuis

@TedShifrin By density I have $f(z)=f(\exp(i\theta)z)$ for all $\theta$, why $f$ doesn't depend on $\theta$

user3185856

Anyone with experience of cubic equations? :)

Ted Shifrin

Because what you just wrote says $f$ is constant on circles centered at the origin.

DHMO

@user3185856 depends on your following question

Kasmir Khaan

Hi Ted I have a small geometry question for you

user3185856

17:37

I have this cubic equation where I need to "depress" it to the form of y^3 + py + q. But when I solve it in TI I get different results? Can you take a look at the calculations? :)

http://i.imgur.com/8JAa3rz.png

JeSuis

@TedShifrin OKK, arrf I should have seen this

thanks

Ted Shifrin

Sure, @JeSuis.

DHMO

@user3185856 you forgot to add back 1/2

Ted Shifrin

what, @Kasmir?

user3185856

I substitute x = (t - a/3)

add back where? I might be missing a step here :)

DHMO

17:39

@user3185856 I thought you made t-1/2

user3185856

My a value is 1.5 = (t - 1/2)

DHMO

@user3185856 you transformed x^3+1.5x^2+... into t^3-3.25x+3.5=0 by t=(x-1/2) right

Kasmir Khaan

@TedShifrin I think you can help me without me putting all the question , i have to calculate line intragral around the unit circle ( top half say) but the field is not defined at the origin , so i have to take a elliptical path below the circle to avoid the singularity at the origin , from the picture , the angle should be pi /4 but they claim it to be pi /3

DHMO

@user3185856 and then you solved the latter, obtaining t=2.20

so x=2.70

you shouldn't use x in line 3

Ted Shifrin

I have no idea what you're talking about, @Kasmir.

Kasmir Khaan

17:42

I know that is not alot of info , it is just if you seen something like this before @TedShifrin I will post full problem if that is not clear

Ok ill post it in a min

DHMO

@Secret it's been 9 minutes lol, can't you break up your long post

user3185856

@dhmo true, but if I use the p and q values I optain from the depressed equation, i get a different root than if I solve it as in 1)

Kasmir Khaan

@TedShifrin F = ( -y, x) / ( x^2+3y^2) , I have to calculate line integral thru the circle x^2+y^2 = 1 , y>= 0 , y>= x

Adeek

hi @TedShifrin

user3185856

@dhmo

Secret

17:44

For this topology:
{{},{a},{b},{a,b},{a,b,c}}

a and b are limit points of a,b,c respectively, because for each point there are neighbourhoods containing:

For a: neightbohourhood {a,b} contains b and neightbourhood {a,b,c} containing ,b,c => a converges to b and c
For b: neightboourhood {a,b} contains a and neightbohourhood {a,b,c} contains
=> b converges to a
For c: neightbourhood {a,b,c} contains a,b= >c converges to a,b

Meanwhile
For a: a is isolated from b because {b} does not contain a; isolated from c because {a,b} and {a} does not contain c; not isolated from c because {a,b,c} cont…

user3185856

@DHMO p and q values when i insert them into the cardanos' formula :)

Adeek

If (X,A) is a CW pair consisting of a cell complex X and subcomplex A, then X/A ~= X U CA

Kasmir Khaan

@TedShifrin because of the singularity at the origin and the nature of the field , It is wise to take an elliptical path , r(t) = cost , sint / sqrt3 )

Adeek

where CA is the mapping cylinder

Can then one conclude that X/A ~= X ?

@TedShifrin ?

Ted Shifrin

@Kasmir: I do not understand. What precisely is the path?

Secret

17:45

Secret knew the fact it is not symmetric and try to list all possible combinations in order to analyse the open sets for each point

Adeek

I am just trying to understand this comment from hatcher.

Ted Shifrin

Karim: I cannot do so many things at once.

Adeek

oh oke

Mike Miller

No, @Adeek. That's more or less never true.

Kasmir Khaan

Karim plz one second let me finish my question

Ted the path is x^2 +y^2 = 1 , y>= 0 and y >= x

Adeek

17:46

@MikeMiller I see so we just get the fact that X/A~= X U CA

we don't get anything more

Kasmir Khaan

so from the point (1/sqrt(2) , 1 / sqrt2) to the point ( -1 ,0 )

DHMO

@Secret You have been heavily misled. We say that a point is a limit point of a set. You can also say that c is a limit point of {a,b}. Also, you need to check every neighbourhood of the limit point.

deostroll

Anyone in the mood of finding the answer to a small variation of the basel problem?

Kasmir Khaan

@TedShifrin the field is conservative to we dont have to worry about the double integral

Ted Shifrin

This is an absurd problem. No, the field is not conservative.

Mike Miller

17:47

Yup

Kasmir Khaan

i mean

JeSuis

@MikeMiller What topology we put on $X:=\{0,1\}^{\Bbb{N}}\times [0,1]$ under $(x,0)=(y,0).$ ?

Kasmir Khaan

del Q del x - del P del y = 0

we have singularity at the origin

Ted Shifrin

I don't even believe that's true, unless you have a typo, @Kasmir.

Mike Miller

Product topology for the first term, product topology for the product with the interval, quotient topology after you take the quotient.

Kasmir Khaan

17:48

I meant partial Q wrt x = partial P wrt y

Secret

Ok does not matter, I failed to solve this many weeks question and I must swtich back to chemistry tmr, so this problem will be not solved for some time, thanks for helping guys

DHMO

@Secret "Isolated" isn't the word you mean. We say that a point is isolated, full-stop. Two sets A and B are separated by neighbourhoods if there are neighbourhoods U of A and V of B such that U and V are disjoint.

Ted Shifrin

You have a $3$ in the denominator of your force field, @Kasmir. That messes things up.

JeSuis

hum so what you will be a curve into X ?@MikeMiller a continuous function from $[0,1]\to X$, but what will be a continuous function here?...

Secret

@DHMO yes I knew that, anyway I gotta sleep gnight

Kasmir Khaan

17:49

F = ( -y / ( x^2+3y^2) , x / ( x^2+3y^2)

DHMO

@Secret It is near sun-rise now, isn't it

good night

Kasmir Khaan

@TedShifrin is it clear now ?

Ted Shifrin

@Kasmir: No clearer than before

What makes you think that the curl of that force field is 0?

curl = rot

Kasmir Khaan

I did the computation is it 0

Adeek

@MikeMiller It is kinda of cool how we can play with different parts of the space to get a homotopy equivalence

Kasmir Khaan

17:51

But Ted that is not really the problem

lets assume it is 0

We gonna have to take an elliptical path to avoid the origin

r (t) = ( cos t , sin t / ( sqrt3) )

they claim that here t is between pi /3 to pi

the way is see it is from pi /4 to pi

because of the line y =x

Astyx

What is an efficient way of proving $$\zeta(s) = \prod_{p\in \Bbb P}{1\over 1-p^{-s}}$$ ?

JeSuis

Where can I find a proof of minkowski theorem using systolic geometry ?

Ted Shifrin

@Kasmir: Because of the $\sqrt 3$ this is not the usual polar coordinate.

deostroll

I am getting infinite product form of the function $e^{sin x} - 1 = \prod_{n=1}^{\infty}(1-\frac{x^2}{n^2\pi^2})$ ... why is this the same as that of $sin (x)$?

Kasmir Khaan

@TedShifrin Can you explain pls more? how to get the bounds for the angle ?

Ted Shifrin

17:54

Where is this ellipse supposed to go?

Kasmir Khaan

well it should start from the same point as the circle intersect the line

x^2 +y^2 = 1 and y =x

so the point is ( 1/sqrt2 , 1 / sqrt2 )

Ted Shifrin

That ellipse doesn't go through that point.

Kasmir Khaan

in the correction i have , it said the point is 1/2, 1/2

but i dont see how

how can you tell ?

and if they dont start at same point , wouldnt we have to calculate the work from those two points?

Ted Shifrin

Yes, you would.

Kasmir Khaan

the region must be closed in order to use greens

but they just set the line integral on the circle = the line integral on the ellips

with orientation ofc

Maks

17:57

@TedShifrin ! Can I bother you with my stupid questions ?

Ted Shifrin

No, @Maks.

Maks

:c @TedShifrin pretty please ?

Kasmir Khaan

maks please wait he is helping me

Maks

Sure

Ted Shifrin

I'm trying to be done with Kasmir, and then I'm leaving.

Maks

17:58

Will you be here at night ?

Ted Shifrin

@Kasmir: If you draw a picture of your ellipse, it touches the circle at $(\pm 1,0)$ but nowhere else.

The parameter $t$ is not the polar coordinate.

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$Mathematics$ Mathematics