Mathematics - 2017-02-20 (page 1 of 4)

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12:30 AM

Hello algebra people.

Esp @arctictern

hi @Jessy

How do I know that the inverse image under a ring homomorphism of an ideal is nonempty?

Hey @ZachHauk

snakeplissken.wikia.com/wiki/Bob_Hauk

Best movie EVER

what the hell

that's my dad's name

I always think about that when I hear your name

Apparently, he's the guy who sent Snake Plissken in to rescue the President.

Hello guys. My answer here was wrong, and i dont understand why: imgur.com/a/6Yva9

12:33 AM

@Will all I'm seeing is pictures of food from Chipotle

@JessyunBourne uhh... you sure? it just took me to the image of the equation

Er...$(-\infty,4)$ is correct b/c the denominator is not defined there

@Will it is now. I accidentally clicked on "Next Post"

@Will notice that it's not continuous at $-4$...

oh ok lol.

3

Q: Prove: The pre-image of an ideal is an ideal.

Let $\phi : R \to S$ be a homomorphism. If $N$ is an ideal of $S$, then $\phi ^{-1} (N)$ is an ideal of $R$.

abstract-algebra ring-theory ideals

12:35 AM

because you know

that would make it $x / 0$ which math doesn't like

I tried (-infinity, -4] and it was invalid. I don tunderstand

The guy in the comment said clearly, $\phi^{-1}(N)$ is nonempty. How do we know that?

@ZachHauk that's how the zombie apocalypse happened.

What i mean is it highlighted ( and ] in red. That usually means you are entering invalid input. I thought i was doing something wrong

@WillNjundong try (-infty, -4(? idk

or ]-infty, -4]?

whatever

@JessyunBourne well ideals contain the additive identity... right?

and homomorphisms map identities to identities

@ZachHauk oh got it to accept my answer :D

12:38 AM

@Zach, I thought that was isomorphisms that map identities to identities

no, homomorphisms do

isomorphisms are more special-er though, because they preserve the whole structure

Oh, right. Isomorphisms are ones where the only point htat maps to 0 is 0.

or monomorphisms

Anyway, let's say we had no idea what the inverse image of 0 was.

In case you don't believe, notice that $\phi(0)+\phi(a) = \phi(0+a) = \phi(a)$

0 is in N, so, its inverse must be in $\phi^{-1}(N)$, yeah?

yeah

12:40 AM

I straight up trippin, boo.

we all have moments, it's ok

12:52 AM

hi, everyone!

quick question

what is wrong here: $e^{-2*pi} = (e^{2*i*pi})^i=1^i=1$ ?

looks like the power rule doesn't hold for all complex numberes

@JessyunBourne all homomorphisms map 0 to 0

damn, that makes everything more complicated

@Mircea See where it talks about $e^{xy}=(e^{x})^{y}$ en.wikipedia.org/wiki/…

@Brody thanks

oh my, that looks quite disastrous; most of the exponential identities are not valid for complex numbers

1:12 AM

is A the answer here? imgur.com/a/5CpMP

Hence why you see weird things like $i^i=(e^{i \pi/2})^i=e^{-\pi/2}$.

@Akiva my parents said no unless it's absolutely free, which it won't be unless i negotiate with the people

@WillNjundong no, it's C; f(-9) is undefined

@Mircea but isnt c only partitally true?

oh you know what, i get it now. nvm. Thanks :D

@WillNjundong no problem :)

@Brody so exp(a+b)=exp(a)exp(b) is still true for complex's, right?

1:22 AM

yeah

LegionMammal978

Silly idea I just had: A Rubik's cube solver is just a partial function $s:(6\times3\times3\times6)\rightarrow\mathscr P(6\times3)$

(With standard PA numbers here)

@LegionMammal978 where do those numbers come from?

LegionMammal978

@Mircea [0..5] colors, [0..2] x [0..2] side positions, [0..5] sides; [0..5] sides, [0..2] rotations (90° CW, 180°, 90° CCW)

Wait, no

@LegionMammal978 what exactly do you mean by side position?

@LegionMammal978 I'd say it's a partial function from the set of permutations of the 20 cubes on the sides to an ordered set of (6 x 2) moves (you don't need the 180 deg one)

LegionMammal978

@Mircea 20 cubes?

But yeah, most solvers include a 180° turn

If you're being real pedantic, it would only include 90° CW turns

1:37 AM

@LegionMammal978 yes, all the cubes that are not a face center cube or the middle cube in a 3 x 3 x 3 cube

LegionMammal978

idk what I'm talking about

2:10 AM

well that

is depressing

whatever, i can dream

:[

2:26 AM

Hey guys! Was wondering if anyone knew what it's called(how to google) when solving for optimization with a subspace constraint

i.e. optimize h(x) given Ax = b where A,b are constant.

I tried googling it up and found no definitive results.

3:27 AM

should i just apply anyways

To where?

mathcamp

my mom said she wouldnt let me go

@Mircea Yeah. The fact is complex exponentiation is itself very weird. It's a richer, more alien landscape compared to the positive reals; it's quite neat. I'm still learning the ropes.

That sucks, @Zach. Sorry to hear. :(

Just apply to see if they accept you, then if they do show her the acceptance letter. I'm sure she'll change her mind.

If they don't; no big deal, right?

3:43 AM

How expensive is the problem? Money does often get in the way of extracurricular excursions like these, especially if parents don't think them too important or particularly rewarding. @Zach

*program

But the more your mom can be convinced of its prestige/selectivity and its several benefits to you, the more likely she'll entertain the thought of sending you off.

4:19 AM

@Zach: Check into the scholarship options at PROMYS and Ross, too.

Hi

Yeah I think my mom won't let me go unless she realizes what it is. All I told her was a summer math program and she said no

It's about $500 given my family's income

however we're in a lot of debt, too

because you know, 5 siblings and all their student debt

Damn it Zack if i were rich I would sent you that money

idk. I'm going to regret it if I don't go

Is it there any other option ?

@Kasmir I'm turning legal age to work next month

4:30 AM

Like you could find some online course ?

eh, it's not the sam. the emphasis is on collaboration

its still 3 months + to summer

So I hope you can get that amount you need by that time

Well now I have an ocular migraine

Goodnight math.se...

And thanks Kasmir

@AkivaWeinberger They do. In C everything is connected.

5:17 AM

Nice to see your dual presence @anon @arctictern

Or should I say one of a multiple combination of presences?

:-)

3 hours later…

8:32 AM

it's so quiet here

8:44 AM

SHOUT! SHOUT! LET IT ALL OUT!

Better?

yes

$\mathbb{YELLS IN MATH BB AND GOES TO SLEEP}$

9:01 AM

internet's hella slow.

gah, so tired. i guess i'll plug in the headphones and listen to stuff

9:54 AM

Hey, I am plotting a spirograph using a program, something like epicycloid if I am correct. But it is drawn with many short lines although it is a curve. Wondering if a bezier curve can be generated instead of plotting each point with paramedic equation?

sabithpocker.github.io This is what I have. Problem is the huge number of points. The shape suggests that this can be drawn with lesser points using a bezier curve? Correct me if I am wrong!

function x(t) {
return ((R + r) * Math.cos(t)) + (p * Math.cos((R + r) * (t / r)))
}

//y(t)=(R+r)sin(t) + p*sin((R+r)t/r)
function y(t) {
return ((R + r) * Math.sin(t)) + (p * Math.sin((R + r) * (t / r)))
}

10:19 AM

hello, can someone explaine me how we get that $f([0,+\infty))=]-\pi/2,\pi/2[$ where

$$f =\begin{cases} \arctan(x)\sin(x) \text{ for } x\geq 0 \\ x\sin(x) \text{ for } x<0. \end{cases} $$

10:39 AM

can someone help me ?

heyyy

sooo

in a circle

a lone passing through 1 point of a circle is called a tangent

is there a line passing through 2 points of a circle just like the tangent but not entering a circle?

not like the secant

where it goes through the circle

but one where it passes through 2 points from the outside

like the tangent

@DHMO have you an idea please ?

10:55 AM

@Vrouvrou $-\arctan(x) \le \arctan(x)\sin(x) \le \arctan(x)$

right

thank you very much

@Vrouvrou tu es bien venu

@DHMO can you help me?

@MartianCactus A line cannot pass through two points on a circle from the outside

oh...

so is there any proof?

oh wait

yeah it goes from inside the arc

10:58 AM

"Try to draw it"

yeah i did and then rechecked

thanks!!

Lovro Sindičić

11:20 AM

hy

Is anyone here?

@LovroSindičić yes

Lovro Sindičić

Can you please go on web.math.pmf.unizg.hr/nastava/difraf and check is the web site work. (my college web site)?

yes

Lovro Sindičić

for me web site doesn't work :(

Diferencijalni račun

Integrali

funkcija više varijabli

E-mail adrese nastavnika i asistenata su oblika ime.prezime@math.hr.

20.02.2017.: Popravni usmeni ispit za sve studente odrzat ce se u cetvrtak 2.3. u 11 sati.
20.02.2017.: Usmeni ispiti za studente koji su ostvarili 35 bodova na popravnom ispitu kod doc. Gogica odrzat ce se u cetvrtak 23.2. u 11 (prezimena M-P), te u 12 (prezimena R-Z). Raspored usmenih ispita kod prof. Tambace bit ce oglasen veceras.
20.02.2017.:Uvidi u zadace kod asistenta Zunica (prva cetiri zadatka) u utorak u 14:30 u njegovom uredu (soba 33, prizemlje)…

Lovro Sindičić

11:24 AM

Can you please send me link of Ukupni rezultati nakon popravnog ispita and send to my mail lovro.sindicic@gmail.com

web.math.pmf.unizg.hr/nastava/difraf/dif/2016-17/…

Lovro Sindičić

No I couldn't download it :/ can you please download it and send me .pdf

user image

@LovroSindičić ^

Lovro Sindičić

numeber 1191236687 it write završni?

in last row

It is zavrsni even in the original pdf

Lovro Sindičić

11:32 AM

@DHMO Thanks you :)

you save me

you are welcome

12:12 PM

hi guys

when dealing with quaternions

we define $ij=k$

but is it a definition that $ji=-k$

or can we deduce that?

It is a definition.

You cannot deduce $ji = -k$ from $ij = k$. Indeed, it is a false conclusion.

ah thanks

what is the proof that only 1 tangent cam be drawn through a particular point on the cirlcle

@MartianCactus "try to draw another"

12:28 PM

Hi guys , everybody relax as i am BAYMAX !!!

@MartianCactus a tangent must form a right angle with the radius

@MartianCactus Or else, you can find another point that the line touches.

@BalarkaSen Can we make $ij = ji = k$?

Sure.

is multiplication associative?

mhm

so -1 = k^2 = (ij)(ji) = i(-1)i = 1?

12:38 PM

Of course you can't make k^2 = -1! You have to let k^2 = 1.

and so on

I thought k^2 must be -1

In quaternions, yes. Not in this world.

It's unclear what you want to do

You seem to imply that ji=k is consistent with quarternions

that ji=k is independent with ij=k

Why does firefox give me a security warning for this page?

Per se, it is. With no axioms whatsoever, ij = k does not imply ji = -k...

hey, is someone familiar hier with prooving injective, surjective and bijective?

12:44 PM

@jublikon just ask; don't ask to ask

okay, sorry @DHMO

Let $f: X \rightarrow Y$ and $g: Y \rightarrow X$

If $f \circ g$ is surjective, then f is surjective, too.

**Question**: Is my proof okay?

$\forall x \in X \quad \exists y \in Y \quad g(y) \rightarrow g$ is surjective

There has also to be true:
$\forall y \in Y \quad \exists y \in Y \quad f \circ g(y)= y \rightarrow f \circ g$ is surjective

@jublikon the first line of your proof makes no sense

$g(y)$ is an object

"for any x in X, there exists a y in Y, such that g(y)" doesn't make sense

I thought that that means that every X is hit by an Y

no that doesn't

hm...

12:48 PM

you meant $\forall x \in X: \exists y \in Y: x = g(y)$

oh yes, I see. My mistake. Was like sleeping while typing

sorry

Let $f: X \rightarrow Y$ and $g: Y \rightarrow X$

If $f \circ g$ is surjective, then f is surjective, too.

**Question**: Is my proof okay?

$\forall x \in X \quad \exists y \in Y \quad g(y)= x\rightarrow g $ is surjective

There has also to be true:
$\forall x \in X: \exists y \in Y: \quad f \circ g(y)= x \rightarrow f \circ g$ is surjective

it's the same

hope its better now

I don't understand the logic of your proof

you are given that $f \circ g$ is surjective

yes

12:52 PM

and you start with $g$ is surjective

yes

should it be the other way round?

I don't understand.

Can you tell me what you are trying to do?

starting with $f \circ g$ is surjective?

As you like.

I have to decide that if $f \circ g$ is surjective, what $f $ itself will be. So if it is surjective, I have to proove that

12:53 PM

I know

can you give me a hint how I could do it better?

start from the premise and work step by step towards the conclusion

ok, so the premise is that $f \circ g$ is surjective

yes

so because $f \circ g $ is surjective there is true that

$\forall x \in X: \exists y \in Y: \quad f \circ g(y)= x \rightarrow f \circ g$ is surjective

correct?

12:57 PM

no

firstly, why is your arrow pointing in that direction?

secondly, what is the co-domain of $f \circ g$?

firstly: I do not know..
secondly: the domain of $f \circ g$ is $Y$ and the co-domain also is

but you made that arrow

just to say, I make a conclusion from that

from what?

from the stuff before: $\forall x \in X: \exists y \in Y: \quad f \circ g(y)= x $

1:00 PM

that is not your premise

your premise is that $f \circ g$ is surjective

and as you said, the co-domain is $Y$, so please correct your formula.

yes. and I have found on the internet that this is the condition for that...

I am sorry, I am quite a beginner..

okay

that is not any condition; they are equivalent

and you should start from your premise

the arrow should turn to the other way

$\forall x \in X: \exists y \in Y: \quad f \circ g(y)= y$

like that?

no

do you understand every symbol in your formula?

yes

1:02 PM

show me

For all x in X there is a y in Y with f composed with g with the variable y equals y

alright, I'll give you the answer this time

okay, thanks

It should be $\forall x \in Y: \exists y \in Y: f \circ g(y)= x$

okay, I understand.

1:05 PM

Would you be more comfortable speaking in German or in English?

german

then go ahead

LegionMammal978

Ah dang it, how do I do a change of basis in linear algebra again?

@LegionMammal978 khanacademy.org/math/linear-algebra/alternate-bases/…

okay, also ich versuche jetzt zu zeigen, dass ausgehend von der Bedingung, die wir gerade aufgestellt haben, die variablen $x$ und $y$ die Bedingung für surjektivität für f erfüllen

LegionMammal978

1:07 PM

@DHMO Thanks, had an idea of using linear algebra for dimensional analysis

also quasi das gleiche noch einmal zeigen, nur halt mit den selben variablen und für f

damit müsste die surjektivität doch gezeigt sein, oder?

@jublikon ja...

@jublikon ich verstehe nicht, warum du "mit den selben variablen" sagst...

@DHMO damit meine ich nur, dass ich das definierte x und y wiederverwenden möchte, um es auf f anzuwenden

@jublikon aber $x \in Y$...

ja, sehe ich auch gerade

sekunde

1:17 PM

du musst zeigen, dass $\forall y \in Y: \exists x \in X: f(x) = y$

gut, ich versuche es mal

@jublikon es könnte einfacher sein, seine kontrapositive zu betrachten.

wie meinst du das?

also gegenbeweis finden?

ich meine seine Kontraposition

$f \text{ ist nicht surjektive} \implies f \circ g \text{ ist nicht surjektive}$

lass uns das bitte versuchen, wie du zuerst vorgeschlagen hast - ich würde den direkten beweis gern üben / lernen

1:25 PM

Danach wird es ganz offensichtlich.

also einfach nur die aussage für f(x) verneinen?

ja

zeige:

2 mins ago, by DHMO

$f \text{ ist nicht surjektive} \implies f \circ g \text{ ist nicht surjektive}$

Eine Aussage ist äquivalent mit ihrer Kontraposition.

@Alessandro Hi

Alessandro Codenotti

Hi @Balarka

anything fun up lately?

Alessandro Codenotti

1:32 PM

I'm in an algebraic topology lecture right now

ahh, cool

Alessandro Codenotti

So definetely fun (and interesting too)

@DHMO hoffentlich koste ich Dich nicht die Nerven heute...

Also:
$f$ ist nicht surjektiv $\Rightarrow \quad f\circ g$ ist nicht surjektiv

$\forall y \in Y: ! \exists x \in X: f(x) = y$

also:
$\forall x \in Y: ! \exists y \in Y: f \circ g (y) = x$

i shouldn't distract you then :)

Alessandro Codenotti

Yeah, maybe I should be paying attention! I'll be back later

1:37 PM

@jublikon sorry ich habe nicht es gelesen

kein ding

@BalarkaSen why does a power series only have countable amount of zeros?

@jublikon "$f$ ist nicht surjektiv" ist nicht $\forall y \in Y: ! \exists x \in X: f(x) = y$

"$f$ ist surjektiv" ist $\forall y \in Y: \exists x \in X: f(x) = y$

Wie verneinst es?

das für alle verneinen?

versuche

ich finde das durchgestrichene Symbol nicht...

$! \forall y \in Y: \exists x \in X: f(x) = y$

es gibt nicht für alle y ein x sodass gilt, wenn ich das x in f einfüge, ergibt das y

1:44 PM

@jublikon Wie vereinfachst es?

@DHMO $y \in Y : x \in X : f(x) \neq y$?

@jublikon du kannst die Quantifizierer entfernen nicht...

dann habe ich keine ahnung, wie ich das weiter vereinfachen kann.

@DHMO kannst du versuchen, mir so zu "helfen" dass ich ersti das step by step machen kann? Das ist schon echt nett, wie viel Mühe du dir gibst grade, nur verstehe ich das auf die Art leider immer noch nicht alles

das wäre wirklich prima

@jublikon $\neg(\forall x: P(x)) \equiv \exists x: \neg P(x)$

0

Q: Are $F_{\sigma}$ and $G_{\delta}$ sets are related?

Friends I was just curious while reading about $G_{\delta}$ and $F_{\sigma}$ sets, Where $G_{\delta}$ set is defined as countable intersection of Open sets and $F_{\sigma}$ set is defined as countable union of closed sets Just by seeing the definition i concluded $(F_{\sigma})^{c} = G_{\delta...

1:55 PM

@jublikon verstehst du?

sorry, leider gar nicht

"$f$ ist surjektiv" ist $\forall y \in Y: \exists x \in X: f(x) = y$

Wenn ich verneine es:

$\neg (\forall y \in Y: \exists x \in X: f(x) = y)$
$\equiv \exists y \in Y: \neg(\exists x \in X: f(x) = y)$
$\equiv \exists y \in Y: \forall x \in X: \neg(f(x) = y)$
$\equiv \exists y \in Y: \forall x \in X: f(x) \ne y$

@DHMO $\lnot( \forall x \in X : \exists y \in Y : f \circ g (y)= x) \equiv \exists x \in Y : \forall y \in Y : f \circ g(y) \neq x$

@jublikon ja

Alessandro Codenotti

2:11 PM

@BAYMAX the complement of a $G_\delta$ is an $F_\sigma$ (and viceversa), not sure which kind of relationship you're looking for

Hey guys I'm working with quaternions. Let $\alpha=c+di$ and $\beta=c'+d'i$. Is it true that $\alpha j\cdot\beta j=\alpha\cdot \overline \beta$?

@AlessandroCodenotti why does a power series have countably many zeros?

oh wait I think I get it

Alessandro Codenotti

@DHMO I know nothing about power series sorry

alright

2:20 PM

@Alessandro So, what did you learn about AT?

@AlessandroCodenotti yes i was looking for complement relationship!!

Guys any reference to problems for doing and understanding Measure theory in a problematic approach?? I will be very thankful !!!

2:44 PM

"problematic approach" ?

Probabilistic, you mean?

Let $x,y$ be quaternions. How can I prove that $\overline{xy}=\overline x\cdot\overline y$? Is the only way to go to write $x=\alpha+\beta i$ and $y=\alpha'+\beta' i$, and then completely write it out?

Where $\alpha=a+bi$ and $\beta=c+di$

wait, I'll write out what I have

what's the bar for quaternions?

does it only affect $i$, or $j$ and $k$ too?

it affects all of them

but I got it now!

2:59 PM

Hello.

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