Hey guys , I have a question

The method I devised for this was to use the standard $T^2=S \cdot S_1$

Using that , I get the combined equation of the pair of tangents as $9y^2+y-34x+4=0$

the parametrization is (x,y) = (4t^2,4t) right

Now , I read in straight lines that angle between any 2 homogeneous lines can be given by $\tan \theta = 2 \dfrac{\sqrt{h^2-ab}}{a+b}$

maybe that isn't necessary

@LeakyNun That is fine , but I want to know why the combined equation of the tangents I get is not homogeneous

no idea

The $\Delta=abc+2fgh-af^2-bg^2-ch^2$ is not coming out to be $0$ for this.

If $Y \subseteq X$ ist a subset of a metric space $X$ and a set $O \subseteq Y$ is open in $X$, is then also $O$ open in $Y$?

$Y$ having the induced topology

Wow I just hit Start Chatjax for the first time and it's like a whole new World in here

lol

@philmcole O is open in Y iff $O = Y \cap B$ for some B open in X

@ProducerofBS like a colour-blind seeing colours?

(sorry for the misinformation earlier)

right I know that

@philmcole I think that's enough hint for you to solve it

:p

So $B$ is just $O$ right?

since $O \subseteq Y$ the intersection gives me $Y \cap O=O$

bingo

jackpot

@LeakyNun yeah

yeah

back to first semester set theory

lol

@LeakyNun Now I can at least read what I don't understand

lol

@ProducerofBS that's the first step toward understanding

@AlessandroCodenotti hi

now I have the recipe to answer mercio's question

Given a generic ellipse positioned at the origin:

$$\frac{z^2}{a^2}+\frac{w^2}{b^2}=c^2$$

and given a pair of slopes of tangents $\pm m$ where $m$ can be complex

Then, you can use $\pm m = -\left(\frac{b}{a}\right)^2\frac{z}{w}$ to solve for the pair $(z_1,w_1)$

And this is all you need as after solving all 4 pairs of simutaneous linear equations, the intersections are given by:

$$(\pm x, 0), (0,\pm y)$$ where

$x = \frac{w_1^2 \left(\frac{a}{b}\right)^2}{z_1}$ and $y = w_1$

Thus, to find the points such that the intersections are all real, you need $\text{Im} (w_1) = 0$ and $\text{Im} (x) = 0$

@LeakyNun hi

@AlessandroCodenotti All too often it seems like the first step towards misunderstanding :(

and $\text{Im}(x) = 0$ simplifies to $w_1^2 \left(\frac{a}{b}\right)^2 = \bar{w_1^2} \bar{\left(\frac{a}{b}\right)^2}$

So plugging $m = \pm i$ we get:

$\frac{z_1}{w_1} = \mp \left(\frac{a}{b}\right)^2 i$

and thus using the above conditions, we have:

hello

if somone have an idea : math.stackexchange.com/questions/2756351/…

$y = \text{Im}(w_1) \implies \mp \left(\frac{b}{a}\right)^2 z_1 = \pm \bar{\left(\frac{b}{a}\right)^2} \bar{z_1} \implies z_1 = 0 \lor \binom{z_1}{|z_1|}^2 = - \bar{\left(\frac{b}{a}\right)}^2 \left(\frac{a}{b}\right)^2$

$\text{Im}(x) = 0 \implies \text{Im}(\pm i w_1) = 0 \implies w_1 = 0$

In other words, for tangent slopes $\pm i$ and any ellipse at the origin, we have the following points of intersection being real: $(0,0),(\sqrt{-\bar{\left(\frac{b}{a}\right)}^2\left(\frac{a}{b}\right)^2}|z_1|,‌​0)$

Now as for arbitrary ellipse, note how the coordinates of the intersections and the tangent points form a nice basis set of vectors (thus allowing you to compute the entries of the matrix under the standard basis), which can be transformed by any linear map to their new positions. We can thus transform a warped ellipse to the standard form shown above, deduce the equation such that the imaginary parts disappear, and then transform the points back to get the correct relation

@mercio Let me know if you find any problems

depends on what question

Whats the secret of mastering algebra, anyone?

the millenium puzzle

The millenium puzzle?

ok I am not confident with metrics yet so I cannot help you. Wait for the metric guys to return shortly

thank you , please vote for this question

hi. i have a question. if f(x) and g(x) are not continuous in a, can (f.g)(x) be continuous in a? if yes, what is the example?

Is there a name for a function in a topological space which has identical domain and codomain, i.e. $f:X \to X$? In Linear Algebra it's called endomorphism but is this name also used for more abstract functions not on vector spaces?

@philmcole that's sometimes called a "transformation", but unfortunately the word "transformation" sometimes is just a synonym for a map

that's a definition in any context, not just topology

I think a transformation can have different domain and codomain though?

it sometimes means a map like you described, and sometimes it's just a synonym for "map"

ok

@philmcole If you work in the category of topological spaces, it would also be called an endomorphism

thanks Tobias

doesn't "endomorphism" imply some operation on $X^2$ though?

@GFauxPas No, it just means a morphism from an object to itself

oh im thinking of "homomorphism"

Somehow endomorphism sounds unusual to me when used for a topological space, while automorphism doesn't, weird

@GFauxPas homomorphism is a morphism

but im talking about the property

$f(a \circ b) = f(a) * f(b)$

Hi! can someone let me know if I am moving in the right direction here .. if I argue that the map $u \mapsto \left< f'(u),\cdot \right>_{L^2}$ is continuous map from $L^p(\Omega)$ into $L^q(\Omega)$ my claim should follow ..

@GFauxPas aha, the generalization of that is from model theory

not that it is relevant

using sequential continuity followed by Egorov's theorem and the $L^q$ boundedness & continuity of $f'$ seems to settle the matter ...

seems to me $H^1$ is doing nothing here .. it seems to be frechet differentiable in $L^p$ itself .. -_-

howdy @Leaky, @r9m, @GFauxPas, demonic @Alessandro, @Tobias

hi ted

hi

hello professor @TedShifrin :)

Hi @Ted!

@Farhad: Have you found an example yet?

@TedShifrin Yeah, ignore me, I won't say anything !

LOL, je ne t'ai pas vu, cher @Astyx :P

Ça va ? :)

Bien sûr, et toi? Quel espèce de maths apprends-tu ces jours-ci?

On fait de la théorie de la mesure déguisée en probabilité et de la topologie (principalement des espaces métriques pour le moment)

Et de la mécanique quantique

Formidable. Beaucoup d'analyse!

Oui, j'aurais apprécié un peu d'algèbre mais pas avant la 3eme année !

Quel dommage!

C'est trop réglé, à mon avis.

Oui, j'avoue être un peu déçu

Réglé ?

Régimenté? :)

Trop de règles?

Reglementé

aha, merci :)

Comme tu sais, je ne parle guère plus le français ...

Oui, enfin pas tant que ça en fait

Cette année (ces trois mois) on a le tronc commun donc on ne choisit rien

Aha ...

hi Nicholas

Mais dès Septembre prochain, on devra choisir trois matières (entre maths, physique, économie, etc)

ça va

Mais malheureusement tous les cours de maths proposés sont des cours d'analyse iirc

hmm ...

pas d'algèbre, pas de géometrie :(

La raison étant que c'est des maths qui s'appliquent aux autres matières (ex les séries de Fourier pour la physique)

Après rien ne m’empêche d'apprendre des maths de mon coté !

Si on s'intéressait à l'informatique, il faudrait plutôt de l'algebre ...

C'est sûr

Après on a aussi des cours d'informatique qui s'occupent de ça

aha

Du genre Data science, etc

Pour ça il faut aussi la probabilité, etc.

Justement de la proba on en fait en ce moment !

Oui, je m'en suis rappelé. :)

@Ankit practice

Qu'est-ce que se passe ici?

j'ai jamais fait de mécanique quantique ;w;

Hello

Olleh

Actually eyB as well because I am going to sleep now

Nice

good evening crack exchange can someone please tell me what the latex is for an interation of unions or intersections over a family of sets or a parition

iteration*

\bigcup or \bigcap if I understand the question correctly

detexify.kirelabs.org/classify.html

hey isn't the union of non empty members of X just X?

i'm reading a definition of the axiom of choice on wiki and it goes, X is a set of non empty members, then there exists a function f, the choice function, that maps X to the union of its members, such that for all $Y \in X$ there is $f(Y) \in Y$

yea as far as I know but in the book I'm learning from for AA it asks questions regarding showing that there is a partition for every possible equivalence relation and things like that so i still need to know it. but yes it is rarely used. The union of the elements of a partition won't be the partition itself no, it will be all unique elements in all of the parition's elements, which are sets themselves

@mercio Tu en es où dans tes études ?

je lis des livres de temps en temps

i think you mean the union of the elements of a partition is the original set from which the partition was contructed\

@user548331 OH true.. so that's why it's considered a choice function because you choose the partition

thanks

Mais en terme de formation ?

j'ai fini il y a longtemps :(

Il y a pas mal de bouquin sur la méca quantique

Et quand je dis pas mal, je veux dire énormément

no strict English rule that's nice these rooms are cultured

oh that's why the axiom of choice was controversial. When you involve the concept of infinity, you have to be more rigorous with the axiom

y'a "mécanique quantique pour les mathématiciens" ?

Haha la mécanique quantique c'est quasiment que des maths

Des probas

Hi Sha

Wassup ?

yoos Astyx

nothing much, studying complex right now

you?

Same

Except I'm not studying complex

hahahaa, I was confused for a sec:p

:p

anyhow, Imma drop a Q

jfc why did no one show me detexify.kirelabs.org/classify.html until now

@user548331 lmaoooo

rip

@user548331 yeah whatever lets you talk about math works.

yeah detexify is a life saver. i'd pay for its service tbh

@Sha The derivative of the integral is the function itself

Modulo some hypothesis

right, but I thought we weren't taking the derivative of the integral, but just the $1/h$ limit

ohh wait

For instance if $f$ is continuous it's cool

yea then it's even differentiable

@Obliv If $X=\{\{3\},\{2\}\}$ then $\bigcup X=\{2,3\}$

There you go

right I have to reread it, I think I see it indeed

@Astyx o lols

can a surjective function that maps $\mathbb{R} \to \mathbb{Z}$ exist theoretically?

Why theorically ?

Take the integral part for instance

$x\mapsto \lfloor x\rfloor$

you can then compose that with any bijection on the integers to get another such surjection

@astyx sorry i've never seen/studied ceiling/floor functions. I'll look up the definition

e.g. $f(x)=\lfloor x r\floor$ is a surjection and $g(x)=x+1$ is a bijection, so $g(f(x))$ is also a surjection

it just means to round down to the next lowest integer

so 2.5->2, 1.3->1, -1.2->-2

Ooh i see. thanks for the example

man, there are some days when afternoon coffee is just great

Are there days when it's not ?

yeah, the days when I don't get afternoon coffee

Fair point

Astyx, mind answering one more follow-up question?

the most recent XKCD isn't that remarkable, but the alt text...the puns, they hurt so good: xkcd.com/1986

@Sha A lot

Go on

haha:P

alrighty

any idea why $\dfrac{\partial U}{\partial y}=-\operatorname{Im}f$?

because I would think that since we still work with $U$, we still get $\operatorname{Re}$

I don't see what goes different, except that now we integrate along the vertical axis (but the integrals still seems the same)

What is F ? $\Bbb R^2 \to \Bbb C$ ?

yes

I want to say it's because $x+i (y+h)-(x+i y)=i h$ not $h$

here is more context, if that helps

So U is both a set and a function

Good job editor/author

hahaha

Note that, if you change the imaginary part of $z$ by $h$, then you change $z$ by $ih$.

I strongly suspect that's where it's coming from.

ahh

that makes sense

So for instance I wouldn't be surprised if the integral of interest is initially $\int_0^{i h}$

that would explain the minus sign, in any case

I'd be lying if I said I was willing to work out the details myself, though

$ih$ in the upper limit is not cool

I would still keep it as a 'real' integral

eh, think of it like a contour integral

is there a study of functions that are neither surjective nor bijective but have surjective inverses?

I agree with Semi

(as usual)

you're doing the contour integral from 0 to ih

is a contour integral the same as a line integral?

if you parametrize it, yes

so that's just $z=i t$ with $t=0$ to $h$

what the difference then? a contour integral doesn't have an orientation yet?

you'll have $dz=i \,dt$ as well, of course

well, keep in mind that a line integral is usually written as $\int \vec{f}\cdot d\vec{l}$

not in my book

it's written as:

Here we don't have vectors and therefore don't have dot products

$\oint_\gamma f(z)dz$

To me, that's just a (closed) contour integral.

oh okay then

yea I don't know why they chose the circle while they don't even mean closed

:/

yeah

the problem for me with your approach is that we only defined complex line integrals in terms of a real integral ($\int_a^b$), but I'll think about how to avoid putting $ih$ in the upper limit

ohh I see it

I had to write down $\int_0^h f(P+it)i\,dt$, and then remember that $f=u+iv$