@user400188 P({a,b,c,d}) = ?

wrong

I am unsure what the mistake it

where is {c,d}?

3rd last

It didnt show up in the copy paste I made from the earlier question

Appart from that was there any fundamental error?

you still missed 2 sets

but let's move on

{c,d} {a,d} {b,d}

{apple, orange} \ {orange} = ?

I am not sure what the \ means

I have heard it called excluding though

so {apple}

it's called set difference and you are right

LaTeX is \setminus

you will also see {apple, orange} - {orange}

I've always wonder what set difference is

the elements in the first set removing the elements in the second set

have you done any cs?

I've just wondered how it was achived using only $\land\ \lor\ \not$ Plus whatever extra stuff is needs from first order logic.

CS?

Computer science?

yes

@anon Sorry, I was at the ER with my wife. She's okay, just had a breathing problem.

I have done a tiny bit. But not a lot

@user400188 we don't use those on sets. That's the last time I say it.

do you speak Python?

No that is not a launguage that I have covered

let's move on to set comprehension

aka set building

have you heard of it?

nope

@DHMO nope

@user400188 so we say {x|x is a fruit}

or {x:x is a fruit}

then orange is in that set

Im not sure if I can read that notation

x such that x is a fruit?

: means such that

we say "the set of x such that x is a fruit"

so {x|x is a fruit} intersect {x|x is red} = ?

{x|x is a fruit and x is red}

No,red is not a fruit

@user400188 correct

can you name some elements of the set you just built?

@Fawad you need to study English ;-)

Hello @DHMO

nice

can you use set comprehension to define A union B?

{A|A is B) would make conjuction. I am not sure hwo to do union yet

no that is not how you do it

yes

I hope he didn't read…

I read it and was about to say I agreed then DHMO answered

forgive me

WHich I know is wrong but I cant get my head around why it wont work

@user400188 what the hell did I just read

used convention that multiplication is AND

comma means and

and the last sentence is nonsense

Yes it means AND but in set theory the notation used in sets seems to mean or

depends on context

A{x:x is even} and B{x:x is prime} what is union @user400188 what is A︶B ? And A︵B?

what the hell is x(x=fruit)

︶ means union,︵ means intersection

what the fun

ROFL

@user400188 You do have a unique way to create nonsense.

I call those "elaborate nonsense"

Its how I think about everything to be honest

thats why im so worried about syntax most of the time

I have to be able to translate correctly

It actauly does work: the result of the simplification is just x AND fruit. Which acording to my nonsense is the same as x intersect fruit.

{x|x is a fruit} means nothing more than the set of x such that x is a fruit

is fruit a larger set than x? and is x possibly larger than the set x is a fruit?

if so then then the set of x such that x is a fruit could be thought of as the intersection of the two sets

well it contains all the fruits really

x is a variable

@DHMO what you are explaining is common math or related to computer science?

common math

can also be related to computer science

How?

Haskell is written using many of the conventions and concepts of set theory.

@user400188 you understand set theory?

Not really. I haven't done any courses on it. And From what I have talked about here the stuff I thought I did undertand was in fact wrong..

@user400188 here is the answer: A union B = {x|x in A or x in B}

Shouldnt that be subset as opposed to in?

Just want to confirm: if $x \in K(x)$ then $x$ is algebraic over $K$, correct?

it depends, $\pi\in\Bbb Q[\pi]$

@user400188, yes. $x \in A \cup B$ implies that $x \in A$ or $x \in B$

Oh yeah, since $\pi$ is not algebraic over $\mathbb {Z}$

Hi chat

Thanks, @AlessandroCodenotti.

however if $|K[\alpha]:K|$ is finite then every element of the extension is algebraic over $K$ (the extension is said to be an algebraic extension)

@HarryEvans If A and B were sets; and I wanted to define thier union: would x, such that x∈A or x∈B be enough? Or would that just define the union of two single elements (and not every element)?

(not every algebraic extension is of finite degree, can you find an example?)

Yes, @user400188. That's the definition of being in $A \cup B$

Remember that $x$ is an arbitrary element

it seems there should be a for all somewhere

or is the the for all part somehow described in the way we write it: $A \cup B = \{x|x \in A\ or\ x \in B\}$

@AlessandroCodenotti Is $[\mathbb {R} : \mathbb {Z}]$ an algebraic extension

not every real is algebraic over $\Bbb Z$

@user400188, definitions are "if and only if"

also $\Bbb Z$ isn't a field, are we doing field or ring extensions?

ah thank you: the = should be a $\iff$

Sorry, it should have been $[\mathbb {R} :\mathbb {Q} ]$

you already know whether that is an algebraic extension, is every element of $\Bbb R$ algebraic over $\Bbb Q$?

No @AlessandroCodenotti, because $\pi$ is a counterexample

right, so that's not an algebraic extension

What about $[\mathbb {Q} (\sqrt {2}, \sqrt {3}, \dots, \sqrt {n}, \dots : \mathbb {Q}]$?

that works

also $\bar{\Bbb Q}:\Bbb Q$

Still whacking my brain over a homework for my Complex Analysis class, though

@AlessandroCodenotti, are you someone I can consult?

I know nothing about complex analysis sorry

but if you have more abstract algebra doubts I might be able to help

Thanks, @AlessandroCodenotti!

Anyone who could help with Complex Analysis?

@DHMO Thank you for the help.

I'm glad everything is okay @robjohn

@AlessandroCodenotti is the axiom of subset a schema?

@HarryEvans Complex Analysis is a broad topic. Do you have a more specific question?

Hey there, @robjohn! Yes, I did have a few

@skillpatrol thanks

@robjohn I'm currently working on a homework and I need someone's help. First, suppose we have the power series $\sum_{n = 0}^{\infty} \frac {B_n z^n}{n!}$ which converges to $f(z) = 1$ if $z = 0$ and $f(z) = \frac {z} {e^z - 1}$ whenever $z \ne 0$. We need to show its radius of convergence is $2 \pi$

I don't know if Hadamard's formula works here. In fact, we need to also show that $B_n = - \frac {1} {n + 1} \sum_{k = 0}^{n -1} C(n+1, k) B_k$

@DHMO yes

@HarryEvans show that $f$ is holomorphic inside the open disc of radius $2\pi$ then use Cauchy's integral formula

@mercio, Unfortunately, we haven't studied Cauchy's integral formula yet

w h a t

shocking

I think I can show that $f$ is analytic/holomorphic inside an open disc of radius 2$\pi$

how did you prove that $f$ is holomorphic implies $f$ is analytic ?

iirc we have to use some kind of cauchy formula for that

In our class, we defined an analytic function as synonymous to being holomorphic

uh

maybe I meant $f$ differentiable implies $f$ analytic ?

Our last theorem is if $f(z)$ is the pointwise limit of the power series $\sum_{n = 0}^{\infty} a_n z^n$, in $D: |z| <R$, then $f$ is analytic in $D$.

but

what's your definition of analytic-holomorphic ?

So if the power series converges on a certain disc of convergence, it is analytic

because convergence of power series on an open disc is my definition of analytic

Analytic: the function is differentiable at every point in $D \subseteq \mathbb {C}$

that's holomorphic

so have you proved the converse of your last theorem ?

In our class, though, that's how it's defined, and that's why analytic and holomorphic are synonymous terms

Not yet, @mercio, but we stated it as something to prove

I see

I guess your only tool then is to get an estimation of the growth of the $B_n$

@mercio So what you mean here is to show that $f'(z)$ has radius of convergence equal to 2$\pi$?

how did you show that your sum converged to $\frac z {\exp(z)-1}$ ?

@mercio, it's a given

bwrrrrrrk

How do I upload a picture here?

put it on imgur then give a link

but if you haven't proven that $R > 0$ it's useless to know what the sum converges to

imgur.com/a/I1EnB

<img>http://imgur.com/a/I1EnB</img>

<blockquote class="imgur-embed-pub" lang="en" data-id="a/I1EnB"><a href="//imgur.com/I1EnB"></a></blockquote><script async src="//s.imgur.com/min/embed.js" charset="utf-8"></script>

You can't use markup here

The first one is easy...the succeeding ones are damn difficult

Try the Upload button next to Send

wait so you haven't even got a definition of $B_n$

Did you see the file? imgur.com/a/I1EnB

Yes, $\{B_n\} \subset \mathbb {C}$

@HarryEvans this?

Yes, @MickLH. How'd you do it?

you linked the album and not the image

I fed the URL of the raw jpeg image into the Upload dialog on here, I think just pasting the raw jpeg link does the same

But you can skip the hassle, and just upload from your PC, it goes to imgur anyways

oh wow i never noticed the upload button

I don't see an upload button on mine...maybe I'm too basic...not enough reputation?

But, @mercio, @MickLH, do you guys have any idea? I mean, I can show that the radius of convergence of $f'(z)$ is 2$\pi$, correct?

is there a reason you aretalking about $f'$and not $f$ ?

Both $f$ and $f'$ have the same radius of convergence

yeah

oh forget about that idea...$\{B_n\}$ is a complex sequence...

have you showed the identity theorem ?

so far you haven't showed any magically strong theorem about holomorphic function so really I'm not sure how you're supposed to show a)

Is this the Identity Theorem: If $f$ is analytic on $D \subseteq \mathbb {C}$, $f = 0$ on $S \subset D$ contains an accumulation point in $D$. Then, $f = 0$?

yeah

I don't see the link, though

me neither I was just wondering

well you can show that $R$ can't be greater than $2\pi$

So I just take an open disc bigger than $2\pi$, and show that the power series diverges?

if it were you would be able to evaluate $f$ at $2i\pi$

then, multiplying by $(\exp(z)-1)$ (which has infinite radius of convergence)

you get $f(z) \times 0 = 2i\pi$

contradiction

Do I show that the series converges at any disc smaller than $2\pi$?

i really don't know how to do that without cauchy integral formula

So, let $z = 2\pi i$ thus, $f(2\pi i) (e^{2\pi i} - 1) = 2\pi i$

yes, assuming $R > 2\pi$

We can simply get $f(2\pi i)$ is not defined

Thus it is divergent there

Coolio

I'd rather be outside walking pokemon now that the gen 2 update has been released...what a waste of a fine Saturday morning :(

Can any1 explain the last three lines of Donantonio's answer please?

@BAYMAX, by the Sylow theorems, since 17 is a factor of 595, we determine how many Sylow 17-subgroups there are. The possibilities are 1 or 35. Now, there cannot be 35 Sylow 17-subgroups since that would be a contradiction

@BAYMAX, all of these are consequences of the Third SYlow Theorem

@Fawad hey Ian here

*I am

Ok

@Fawad should I ping anybody while asking questions

Ask your question,if someone can help then they will help you

No,not anybody

OK

if 35 Sylow 17 subgroups are there then no of elements total in these Sylow 17 subgroups will be $35 . (17-1) = 560$ non trivial elements?

@HarryEvans

@BAYMAX, If there are 35 Sylow 17-subgroups, then if there are at least 85 Sylow 7-subgroups, then $35 \cdot (17 -1) + 85 \cdot (7 - 1)$. What is the total?

@satyatech the latter is correct

@DHMO which one the above one or below one

below

@HarryEvans 1070 elements and hence a contradiction!

@BAYMAX, yes, that's right

@DHMO ,but Sir (don't think I am going to make wrong thing correct) in my book what it did was the above one and used G.P ,there was a G.P ,which led to the answer. Let me post the image for better understanding.

Sir :p

This seems very wrong to me

@satyatech remove the second line

@Astyx that's how Indians talk.

I know, I just find it amusing

The notations are confusing in your picture

@DHMO I couldn't understand which line to remove and which notation s are confusing

What they mean is $C$ happens n times then $A$ happens

And usually one does not write this as $C\cap C \cap \dots \cap C \cap A$

because this is simply $C\cap A$

Sorry friend I am seeing only dollars and slashes

Check Chatjax on top right

tinyurl.com/hzl2955

@satyatech ^

One would rather write this as $P(X_1 = C \text{ and }X_2 = C \text{ and } \dots \text{ and } X_{n-1} = C \text{ and } X_n = A)$

Or $P((X_1 = C) \cap (X_2 = C) \cap \dots \cap (X_{n-1} = C) \cap (X_n = A))$

Friend the problem is I am using a smartphone and my desktop is damaged ....😑

Oh I see

Any site to covert latex to math

Do you see images ? ^

Yes

Cool

Where the Xi are the random variables of each event

@Astyx Thanks ,now I understand

Glad to hear it :)

Hi @Astyx what's up?

@Astyx Exercise: monstrer que l'intersection des ensembles X et Y existe

Hi @Fawad, enjoying my penultimate day of holidays, and you ?

@DHMO Montrer* Et tu veux dire avec les axiomes de ZF ?

Preparing for exams

@Astyx oui, avec les axiomes de ZF

Haha typo :p

Je ne les connais pas par coeur, tu peux me ré-envoyer le lien que tu avais hier ?

By the way, I'm still unsure about how to proceed with my homework for my complex analysis course... imgur.com/D2fNUGh

Can anyone help? Thanks again! imgur.com/D2fNUGh

@HarryEvans What have you tried so far ?

Well, #1 is simple. #2a, we show that the power series diverges for $z = 2\pi i$

But #2b and #2c are quite challenging, @Astyx IMO

@Astyx mathworld.wolfram.com/Zermelo-FraenkelAxioms.html

Merci @DHMO

@HarryEvans Wait just a sec :)

Thanks, @Astyx

I shall get breakfast. I wanted to walk outside and catch pokemon, though. Ruined, because of this homework.

@DHMO C'est pas juste une application de l'axiome des sous-ensembles $A\cap B = \{x\in A | x\in B\}$ ?

@Astyx marveilleux!

merveilleux :)

(Yeah, French is frustrating)

@Astyx, is that really French? I mean "merveilleux"

Yes it is

Whoa, coolio!

@DHMO What reason do I have to expect that the arbitrary sets X and Y must intersect and henc have nonzero intersection A?

@Secret when did I say that A is nonzero?

If A is empty, then there is no element u, and the line that follows in that sentence will break down

@Secret u has no domain

and the line that follows would be a vacuous truth

The vacuous truth will hold even if u is in only one of the sets X or Y?

because if u in X and u not in Y, then by the truth table of "and", you get "false" as a result, even if u in A where A is empty is vacuously true, thus you have true=false, and I don't know how that will be consistent

potentially trivial question: must a topological space in which every closed proper subset is compact be compact?

I don't think that's true but I don't see any easy counterexample

Maybe that's actually true, I need a space that can't be written as a finite union of closed sets, but then it also has "few" open sets

@Secret Can you give me a specific example?

Consider X={1,2,3} and Y={4,5,6}. Clearly they are disjoint sets, thus A is empty. Now let u=4, then u in Y but not in X, thus "u in Y and u in X" is false. Now for the left side of the $\equiv$ there is no element in A, thus for all u in A, (property) is vacously true. Thus we have true $\equiv$ false, which is absurd

@Secret no, A is empty, so "u in A" is false

ignore my comment about "vacuous true"

ok now it makes sense

@HarryEvans Try and compute $$\sum_{n=0}^{+\infty}{1\over n!}{-1\over n+1}\sum_{k=0}^{n-1}{n+1\choose k}B_k z^n$$

Hmmm...@Astyx, I don't know

And read the hint of the original question

I don't know $\{B_n\}$...

That's not an issue

And you might want to add $\sum_{n=0}^{+\infty}{B_n\over n!}z^n$ coming to think about it

Another hint : ${n+1\choose k} = \dots$ ?

you're trying to do part (b) in the imgur link on the starboard? if so, plugging the desired formula for $B_n$ into the power series works as verification, but it doesn't tell you how the recursion was derived in the first place - which is what you find out when you follow the directions

@Secret are you typing a long paragraph?

I see that you're trying to relate the series you wrote with the Cauchy product

not really. astyx just replaced $B_n$ with the desired formula inside the generating function for $B_n$

Oh yeah

@Astyx, sorry, I don't see how $\sum_{n=0}^{\infty} \frac {B_n z^n} {n!}$ comes into the pic...Oh, I am so frustrated!!!

Also, I don't get your hint of $\begin {pmatrix} n + 1\\ k \end {pmatrix}$

@HarryEvans You see how that $\sum$ is supposed to be a fraction right? It's supposed to be $f(z)=\frac{z}{e^z-1}$. The Cauchy product involves multiplying by another function. What other function $g(z)$ do you think we're going to multiply $f(z)$ by?

Hint: $f(z)$ is expressed as a fraction. What do we tend to multiply fractions by?

Their conjugate?

SOrry, I'm so stupid

Just throwing random things here :'(

we tend to multiply top and bottom of a fraction by a conjugate

think simpler than though

like, solving $\frac{1}{2}x+\frac{3}{2}=-\frac{5}{2}$ simple

if you don't like fractions, what would you do to that equation?

by the denominator

right

multiply $f(z)=\frac{z}{e^z-1}$ by $g(z)=e^z-1$.

So, $g(z) = e^z - 1$?

mmhmm

you end up with $f(z)g(z)=0+1z+0z^2+0z^3+\cdots$ on the one hand

use cauchy product to figure out the taylor coefficients of $f(z)g(z)$ on the other hand

But isn't $f(z)g(z) = z$?

yes

Oh, I see what you did there

yes, I explicitly said what I did there

:P

So we use a Maclaurin series representation for $e^z - 1$, is that correct?

Sorry I went away

yep

That's OK, @Astyx

Arctic is surely better than me at explaining

If events A and B are mutually exclusive, and $P(A) = 0.75, P(B) = 0.02$, then $P(A and B) = 0.77$, right?

I shall try that

But, please stay there huhuhu

Sure

@SylentNyte The values of P(A) and P(B) are irrelevant. If A and B are mutually exclusive they can't happen simultaneously, so P(A and B)=0

$e^z - 1 = \sum_{i =0}^{\infty} \frac {z^n} {n!} -1$

@SylentNyyte, no it's 0

yea i thought so..

@Astyx huh?

@SylentNyte, no, since they're mutually exclusive $A$ and $B$ don't happen together

it just seemed a bit weird that he'd (ny teacher) give different examples of P(A) and P(B), yet keep them mutually exclusive and keep asking for p(a and b)

do you mean independent, sylent nyte?

no,

i mean mutually exclusive

lmao

So there's no probability of them happening together, $P(A \cap B) =0$

ooohhh

i think he made a type

because he included saying that P(A and C), and p(a) and p(c) are independent

well, ill answer the questions assuming he made a type

typo*

thanks for helping

@SylentNyte if you answer a question assuming they worded it incorrectly, say so. another possibillity is that it's a trick question.