Mathematics - 2018-08-28 (page 1 of 4)

4:13 AM

@usukidoll Hey! ;)

I need help x_X

okay ..

I'm not getting this...like I have a pattern for it but UGH! I have to provide a proof for a general case that this satisfies the definition not setting k and m at a certain number. I'll screenshot the question and type out what I have so far. So, it's gonna take 5-10 min to get it set up

http://prntscr.com/knnk9s
#11 is the problem. I already did #10 and it did satisfy the definition which is
closed under complements, unions, and intersections.

http://prntscr.com/knnkjm

But now I have this long string and I have to use m =2 and k = 1 which that was the two sets A and B and only considering one point in either A or B which is probably what the k = 1 meant

so the symmetric difference is
$A + B = ( A \backslash B) \cup (B \backslash A)$

and then by complement definition
$A + B = ( A \cap B^{c}) \cup (B \cap A^{c})$

which would mean all points in set A or all points in set B

If I take the complement of that then it's
$A + B = ( A \cap B^{c})^{c} \cup (B \cap A^{c})^{c}$

$A + B = ( A^{c} \cup B ) \cup (B^{c} \cup A)$

taking the complement back again will give me the original A+B and I think the m = 2 is the number of sets in total and k =1 the number of sets considered when taking the points?

So now like if m = 3 and k = 2 then I will have three sets A,B,C

which will become
ABC with three cases if I were to take two of any of these sets
A and B but Not C
A and C but not B
B and C but not A

$(A \cap B \cap C^{c}) \cup (A \cap B^{c} \cap C) \cup (A^{c} \cap B \cap C)$

so that would mean out of three sets only points from each of the two sets will be considered. it goes on and on like when m =4 and k =3. That's ABCD but then what I'm stuck is how to write an overall general proof of this since this has to work on every case and not when picking a specific number like what was done here

wait wait .. $(A \Delta B)^{c} = ( A \cap B^{c})^{c} \cap (B \cap A^{c})^{c} = (A^c \cup B) \cap (A \cup B^c)$

ugh did I typo? ;-; why is there an intersection on the middle?

oh wait I didn't take the .. oohh forgot to change the middle

complement of or is and .-.

but then ... k = 1,2,3... and m = 2,3,4... fdsklfjdslfjsdlfjsdal

4:27 AM

yea .. but I think there might be simpler ways of seeing this .. first perhaps we could try to show 'belongs to atleast $k$ of them' is an event ..

then 'exactly $k$' is same as saying that it's in atleast $k$ but not in $(k+1)$

that's what I'm trying to get....like a clearer picture of this. When I did the previous three exercises 8-10 they were all related to the definition.
Ohhh is it like induction

base case, k case, k+1 case?

like if k = 1 then we have to show that it belongs to at least one of them and it's it's 1 and not in 1+1=2...

it's in exactly one set is the event $\displaystyle \cup_{i=1}^{n} \left(A_i \setminus \cup_{j\neq i} A_j\right)$

maybe show that $A_{1},A_{2},...,A_{m} \in \mathscr{F}$ by the definition.
aren't those also disjoint? so... $A_{1} \cup A_{2} \cup,..., \cup A_{m}$?

r9m I confused myself again.

I was reading my other probability book and it had those A's and then wrote it into unions. Wait I'm getting way off track

4:36 AM

Here's the question imgur.com/a/7jDbGfs

wait .. usually one doesn't put disjoint assumption in a definition ..

yeah I know... I thought it was a bad idea so I got rid of it

lim_{t \to \infty} \int f_t d\mu only makes sense if the integral is the same for any sequence (t_n) \to \infty.

So let (t_n) -> \infty be given.

Then f = \lim_n f_{t_n} and |f_{t_n}| <= g a.e..

but wouldn't this now be called
$\displaystyle \cup_{i=1}^{n} \left(A_i \cap ( \cup_{j\neq i} A_j)^{c}\right)$

by complement definition

@usukidoll yes

4:38 AM

x in A_{i} but not in that huge family index set of union

s

Now apply DCT: \int f = \lim_n \int f_{t_n}.

Isn't that the entire argument though

like. I don't understand why it should need more than that...

but having that is kinda in F since it needs to be nonempty. And if A is in F then so is the complement A or $ \Omega \backslash A \in F$ and then it's closed under finite unions... hmmm

@r9m?

@DrewBrady I think what they are asking is a proof by contradiction .. suppose $\lim_{t \to \infty}$ don't agree for the limit of integral and integral of the limit .. then there is a sequence $t_n$ etc

atleast that was the essence of the C7 appendix they quoted

prove convergence by sequential criterion?

wait a sec... I think I might have some idea. Let
$\displaystyle \cup_{i=1}^{n} \left(A_i \cap ( \cup_{j\neq i} A_j)^{c}\right)$
is in F
The event space F contains both the empty set and $\Omega$ . Since F is nonempty, then $A_{i} \in F$ and the complement is also in F. So by setting $A_{i}=A$ and $A_{j} =A^{c}$ for $i \geq 2$ then F also contains the union $\Omega = A \cup A^{c}$

4:46 AM

Isn't that equivalent

@DrewBrady exactly .. I don't know if anything more is required

Like if something holds for t->\infty, isn't that the same as n -> infty for all (t_n) -> \infty.

I.e., isn't that the definition?

yes

So $\Omega = A \cup A^{c}$ then the complement must also be in F which results in $\Omega \backslash \Omega = \emptyset$ o_O

@usukidoll any point is always either in an event $A$ or not (i.e. it's complement)

4:49 AM

so it will be any point in this?
$\displaystyle \cup_{i=1}^{n} \left(A_i \cap ( \cup_{j\neq i} A_j)^{c}\right)$

Oh I see, what you're saying. It's a "hint" because this textbook does not actually define lim_{x \to a} f(x) as defined iff lim_{n to \infty} f(x_n) for all (x_n) distinct -> a.

they use an eps-delta definition which is equivalent to my characterization (which is how I learned it in analysis)

ok.

just a question of semantics really then it looks like.

@DrewBrady perhaps you are right .. they did go for the pain of adding it in appendix after all ..

if you take the eps-delta definition it is not immediate that my characterization is equivalent, which is why they added it to the appendix

I prefer my characterization (the one for any sequence) though I suspect it is less general.

i.e., may not hold outside of the metric space setting.

yes .. if we are asked to generalize in $X \times Y$ setting ..

and $Y$ is not separable say

precisely, in the case Y is only topological and may not Hausdorff

*maybe not

then it probably is not enough to use the t_n trick.

4:54 AM

$X \times Y$ product topology?

@usukidoll it reads .. point is in $A_i$ bot not in any other $A_j$ (which is why I took the union of the rest and differenced it out of $A_i$) .. then you take union on all $i$

it's 10:30 am here // bbl .. @usukidoll now it's easy to generalize to $k$-case ..

if $A_{i} \in F$ then the complement is also in F which is $ (A_{i})^{c} \in F$ so the complement is $A_{j} = (A_{i})^{c}$ ???

no .. $F$ is a collection of sets .. I don't see how you made the conclusion $A_{j} = (A_{i})^{c}$ from what I wrote ..

ooh x-x . So I have to use this definition http://prntscr.com/knnuv1 for this.
in t his case $\displaystyle \cup_{i=1}^{n} \left(A_i \cap ( \cup_{j\neq i} A_j)^{c}\right)$

denoting $[n] = \{1,2, .. , n\}$ and $I_k = \{(i_1, i_2, \cdots, i_k) \in [n]^k: i_s \neq i_t \text{ if } s \neq t\}$

all $k$-tuples with distinct entry

5:08 AM

hmmm $A_{i} = \{ A_{1},A_{2},...,A_{i} \in [n]^{i} $ $A_{i} \neq A_{j}$ if $i \neq j$

sd;fj;ldskfjsdl

What question are you all working on by the way?

wait .. @usukidoll .. I'll be back later // need to start my day // it's getting late here ..

ok

did you ask it on main? .. might take longer to explain than I thought ..

if you do end up posting, share the link here too!

5:10 AM

I could try on main...

@DrewBrady I don't get what 1-11 is talking about but it's related to 1-10 which I already did because I used the definition to prove that it's in F which is a collection of events prntscr.com/knnwiu

it said something about previous case is the symmetric difference with m = 2 and k =1

F is a sigma-algebra?

yes. I'll attach the definition

prntscr.com/knnwzr

I did exercise 8-10 using this definition

Well A \triangle B = (A \ B) \cup ( B \ A)

12 isn't required.
but 1-11 WHAT THE HECKKK

And A \ B, B \ A are in F.

So you're done.

5:12 AM

I'm talking about 1-11

1-10 is done already but my prof isn't asking for that one

I understand, just checking that you did 1-10 this way

Okay.

he's asking for 1-11 and 1-21 which i did 1-21. It was elementary stuff

but 1-11 is driving me nuts

This is not bad right.

Let I = all k multi indices.

Do you know what a multi-index is?

no :/

It is not hard.

A multi-index of 1.,...,m is just (a_1, ..., a_k) where a_i are distinct and a_i \in \{1, ..., m\}

Q: How many rectangles can be observed in the grid?

5:17 AM

1

In a $2 × 4$ rectangle grid shown below, each cell is a rectangle. How many rectangles can be observed in the grid? My attempt : I found a formula somewhere, Number of rectangles are $= m(m+1)n(n+1)/4 = 2\times4\times3\times5/4 = 30$. Can you please explain in formal way? Update :...

This question should be reopened because this post has 12 times more views than linked post.

@usukidoll does that make sense?

whoa whoa whoa whoa .. so it will go like
Suppose we have a multi-index such that $1,...,m$ is $(a_{1},...,a_{k})$ where $a_{i}$ are distinct and $a_{i} \in \{1,...,m \}$

Right.

so little $a_{i}$ is in an index and it's in 1,...,m

Actually Usukidoll, let's do this instead.

Let's let I = all cardinality k subsets of {1, ..., m}

Does that make sense to you?

5:19 AM

cardinality is the size of a set

It's a little cleaner if we do it this way.

Yes, that's right.

so... suppose we have k subsets of $\{1,...,m \}$

Hi @r9m , am I right?

No.

@usukidoll we are looking at subsets of {1,...,m} of size k.

OH!

OK!

5:20 AM

So usukidoll, how many of these subsets are there.

Suppose we have subsets of $\{1,...,m}$ of size k

skdjlskfjd

How many subsets of size k of {1,...,m} are there?

m

No.

try again.

ok cardinality is the size of the set

so it's like how many elements

5:22 AM

Yes, that's right.

there's 1,...,m elements

1,2,3,4,5,6,...m

a lot of elements

There are m elements, but how many subsets of size k of these m elements are there?

oh like 1 is a subset of 1,..., m
2 is a subset of 1,..., m ... etc etc

No.

Let's try a concrete example. when m = 5.

and k = 3.

size 3 with 5 elements

5:24 AM

No.

How many subsets of size 3 are there over 1,2,3, 4, 5?

oh...
1,2,3
1,2,4
1,2,5
1,3,4
1,3,5
2,3,4
3,4,5

I found 7 of them.

There should be 10.

You're missing 1,4,5.

You're missing, 2, 4, 5.

You're missing 2, 3, 5.

But, usukidoll, how did I know in advance there would only be 10 of them?

feels like permutations.... like with
a b c
a b
a c

kind of.

b c
a
b
c

5:28 AM

It will always be m choose k.

That's what combinations are.

m!/k! (m-k)!

They're distinct subsets on {1,...,m}.

Yes, so anyways, there m choose k distinct subsets of size k over 1,2, ..., m.

Now usukidoll, is that, in general, a finite number, a countably infinite number, an uncountable infinite number, etc?

For a given m and k, is m choose k finite or countable or uncountable?

finite there's only so many values out there

Right. m choose k is finite.

Okay.

Now, so we're letting I be all subsets of size k over the elements 1, 2, ..., m. right?

right

5:31 AM

So let's call J an element I. It looks like a set. J = {a_1, a_2, ..., a_k}, where a_i are distinct and elements of 1, 2, ..., m.

ah ok so so far it seems like we have
Set I = $\{1,...,m \}$ of size k
and then set J = $\{ a_{1}, a_{2},...,a_{k} \}$ and then the $a_{i} $ are distinct elements of set I which was 1,2,...,m

Now we have to figure out how to write the set {x : x is in A_j for j in J, but not in A_k for k not in J}.

hmmm
$ \x : x\in A_{j}$ for $j \in H$

@usukidoll any idea how to write that set using intersection, union, set minus, etc.?

-_-

5:34 AM

No.

wait wait I'm trying to use latex to understand what you're typing on the set thing

$\cap_{j \in J} A_j \setminus \cup_{k \in J^c} A_k$.

It's going to be the set above.

$\{ x : x \in A_{j}$ for $j \in J$ $\backslash$ $A_{k}$ for$ k \notin J$

oosfldjfkl

ahhh

So for each $J \in I$, let us write $B_J = \cap_{j \in J} A_j \setminus \cup_{k \in J^c} A_k$.

Then the set you're interested in is $\cup_{J \in I} B_J$.

WHOA that looks ugly if I were to take the complement it's going to be one nasty $B_{J}$

5:37 AM

It's going to be in the sigma algebra because I is finite, and B_J is in the sigma-algebra since it is finite set operations to sets in the sigma algebra.

So basically that's the argument.

$B_J = \cap_{j \in J} A_j \cap ( \cup_{k \in J^c} A_k)^{c}$

Based on our conversation, it seems unlikely that you followed the entire discussion, so my strong recommendation to you would be (1) make the question a post on math.stackexcahgne, and post the link here, so I can explain the discussion we just had in greater detail.

and (2) to read the discussion carefully.

I still got confused on some parts. It was this one huge index of two things and then I sort of see that since A_{j} was in F then so was A_{k}$ because that's the complement or something like that. x.x

yeah I think I should post the question on main but what do I call it?

I'll let you figure it out. I gotta get going.

ok

6:06 AM

@DrewBrady here it is math.stackexchange.com/questions/2896921/…

Q: How many rectangles can be observed in the grid?

7:28 AM

1

In a $2 × 4$ rectangle grid shown below, each cell is a rectangle. How many rectangles can be observed in the grid? My attempt : I found a formula somewhere, Number of rectangles are $= m(m+1)n(n+1)/4 = 2\times4\times3\times5/4 = 30$. Can you please explain in formal way? Update :...

combinatorics discrete-mathematics permutations combinations

This post should be reopened because this post has 12 times more views than linked post.

@MithleshUpadhyay Why should that matter?

7:43 AM

@TobiasKildetoft , Because people prefer my post to understand concept.

math.stackexchange.com/help/badges/28/…

Martin Sleziak

Merging probably isn't option in this case, since one of them has 4x2 grid and the other one has 4x3 grid.

@MithleshUpadhyay No, more people have viewed it, which is something else entirely. The upvote-to-view ratio is way better for the other question.

Anyway, it is customary for good reason to mark the newer question as duplicate unless there is a good reason to choose otherwise.

Martin Sleziak

Typically factors taken into account when choosing which of the two questions should be duplicate target are: age, quality of question, quality of answers.

I would consider good title which is easy to find as a part of "quality of the question" criterion. And high number of views might indicate that this one is easy to find.

Related discussion on meta concerning age of the question and duplicates: Original post marked as duplicate.

In any case, probably more suitable place to discuss this might be CRUDE or the reopen request thread.

@TobiasKildetoft , The upvote-to-view ratio is ALSO better for the ANSWER in my post. Anyway, that's depend on community to vote.

@MartinSleziak , agreed.

8:48 AM

@loch hi

loch

9:14 AM

hi @LeakyNun

9:24 AM

What just happened?

@OskarTegby Once you reach 200 rep on one site, you get a bonus of 100 on all others.

Alex Clark

(as well as on the original)

@AlexClark Ahh, I forgot you got it there too.

@TobiasKildetoft oh hej du er finally here

Oh! Okay. Haha!

9:27 AM

@LeakyNun Wow, that was quite a mix

Alex Clark

Ja er est now ere

@TobiasKildetoft er det?

Alex Clark

Oui

@TobiasKildetoft also the stod is impossible

and the blode d

@LeakyNun You could at least write the words correctly

9:28 AM

I don't have a Danish keyboard

sto/d

then switch to it via software. That is what I do (I have an American keyboard)

stød

@TobiasKildetoft Danish is a really fascinating language

@LeakyNun Not when you need to teach your kids to read

lol

@TobiasKildetoft does $\Bbb Q_p$ have all $p$-th roots of unity?

I would believe so

So many cases of spelling their way through a word only to come up with something completely different than the word really is because of 3 exceptions occurring in that word

9:34 AM

because roots of unity are always unramified, but the residue field already contains them all (they are all $1$ in the residue field)

@LeakyNun The residue field does not contain them, since as you say they are $1$

what I mean is that $x^p-1$ splits in the residue field

ahh, right

so e and f must both be 1

but to prove it directly, I seem to need to find $p$ distinct $p$-th roots of unity in $\Bbb Z/p^2\Bbb Z$

@LeakyNun Hmm, I was thinking of doing it even more directly, by finding a sequence whose $p$'th power converges to $1$, but that seems tricky.

9:38 AM

well what I said is the first step of what you said :)

ahh

Do you need to find those roots, or just show they exist?

well, you see

$|(\Bbb Z/p^2\Bbb Z)^\times| = p$

so, bingo

jackpot

congratulations

no, that is not correct

$|(\Bbb Z/p^2\Bbb Z)^\times| = p(p-1)$

right

9:41 AM

~~now apply Sylow theory~~

Just either Cauchy or the fact that it is cyclic

is it cyclic?

yes

by Sylow theory?

no, because Sylow theory says basically nothing about abelian groups

9:42 AM

why is it cyclic?

Because I said so :) (you should prove it yourself. It is a good exercise to show that this holds for all prime powers of odd primes)

For powers of $2$, we get an extra factor of order $2$ and the rest is cyclic.

To be fair, I don't actually recall the full argument, so I would need to work a bit on it. The idea is to use a generator for the prime power one below to construct one above.

I've done something like this

I ended up using lifting the exponent

$p+1$ seems to be the generator

For the powers of $2$, one can be much more explicit. One factor is generated by $-1$ and the other by $5$

right

$p+1$ has order $p$

it generates the group $\{np+1 \mid n \in \Bbb F_p\}$

$(p+1)^n \equiv np+1 \pmod {p^2}$

Is it possible to compute the partial derivative $\frac{\partial F(x,y,u,v)}{\partial x\partial y}$ implicitly?

9:50 AM

sure

How?

I have the system of equations given by $F(x,y,u,v)=x^2-y^2-u^3+v^2+4$ and $G(x,y,u,v)=2xy+y^2-2u^2+3v^4+8$.

that isn't implicit...

also $F_{xy}=0$, $G_{xy} = 1$

I have functions $u=f_1(x,y)$ and $v=f_2(x,y)$. Before, I was doing exercises where I was asked to compute $x'(z)$ and $y'(z)$ implicitly.

They didn't ask me to compute any derivatives implicitly of $f_1(x,y)$ or $f_2(x,y)$, but is it a reasonable thing to ask?

Hi, I am currently reading Munkres' Topology. I started a couple of days ago. I wanted to get an idea of the pace I should set?

10:06 AM

@AnotherJohnDoe you should finish the whole book within an hour

I'm sorry?

@TobiasKildetoft shocking news

the order-$p$ subgroup of $(\Bbb Z/p^n\Bbb Z)^\times$ is always $\{kp^{n-1}+1 \mid k \in \Bbb F_p\}$ as long as $p$ is an odd prime

so no, it can't be done

@AnotherJohnDoe you don't need to be sorry

@TobiasKildetoft but this would shock me

because it would mean that $\Bbb Q_p(\mu_p)$ is a proper extension

What's your problem??

@AnotherJohnDoe I don't have any problem

Well if you don't want to be helpful why respond and waste your time?

10:12 AM

take a joke

why so serious

@LeakyNun What can't be done?

can't find $p$-th roots of unity

ahh

Well I hope you've had your fun?

@AnotherJohnDoe Was that one comment really that bad? We try not to be too serious here, and the question you asked is very hard to answer without knowing much more about you as a mathematician

10:20 AM

Well, not really. It's my first time here so it wasn't what I was expecting, that's all

It would be my first introduction to the subject. I'm atleast somewhat familiar with the material in the first chapter

Munkres is a hard but useful book. It is best to spend more than half a year on it to ensure you fully grasp the principles as topology can be quite non intuitive for infinite dimensional spaces

i will say, unless you have good coursework related reasons, do it slowly, better to understand than to be confused

@Secret where did "infinite dimensional" come from?

@LeakyNun, I hope I didn't come across as too uptight/ rude. I do think indeed think I can take a joke :/

@TobiasKildetoft well I knew at least functional spaces will need more attention on when first came across those since there are no diagrams to guide you and the proofs will be mostly algebraic

@Secret The notion of dimension is not usually considered at all in topology, as it is often not particularly useful.

10:27 AM

Perhaps the better term in trying to describe what I mean are infinite operations and infinite sets in topology, because these get hard very quickly

@Secret, I'm definitely not in a hurry. Not course related either.

Ok then take it slow.

topology itself is pretty fun especially with rationals

That's it, I think. Thanks!

10:56 AM

Hi, I'm trying to write a proof for $(A\cup B)\cup C=A\cup(B\cup C)$. I expressed both sides with respect to definition of $\cup$ of sets. And now, it seems obvious to me because $(p\lor q)\lor r$ has the same truth value of $p\lor(q\lor r)$. But I feel like I couldn't show it clearly. Any ideas?

@AbdullahUYU truth table

@AbdullahUYU That is really the reason. To show the last part, just split into the 8 cases.

:P

So refering to this equivalance of truths is enough, you say?

yes

10:58 AM

Ok, thanks @LeakyNun @TobiasKildetoft

afterall, sets are just functions from sets to {true,false}

Hmm, this is recursive @LeakyNun. Are you sure?

don't think about it too much

I think I see what you mean.

@TobiasKildetoft what is your favourite proof that $\overline{\Bbb Q_p}$ is not complete?

11:06 AM

@LeakyNun I don't know any proofs of that

I don't do p-adics

my favourite is BCT

I thought you do LCFT

I actually read Topology by Munkres. Schuams Outline for General Topology and Topology without tears are great books

Not to mention the amount of Topologies out there is...a lot

Whoops. Late to the party

@LeakyNun I could not even tell you precisely what LCFT is all about

oh

I do characteristic $p$, which is an entirely different beast

11:32 AM

this might be another thing where I am making mistakes because I am tired

so, we have ${x}^3+x+1$

which is strictly increasing, right?

right

so, we can write this as $(x^2+ax+b)(x+c)$

where $a+c=0,ac+b=1,bc=1$

because the polynomial is strictly increasing there is exactly one root

(which just so happens to lie between -1 and 0 (I think)

which would mean that the root is exactly equal to -c, or just a right

I was trying to come up with a new polynomial in x by substituting a for x, and c for -x

I got 0

I guess that shouldn't have surprised me

basically, what I am trying to do is factorize that polynomial without looking the answer up

just checking, can we say that $b^3-b-1=0$?

actually, I think that it should be $b^3-b^2-1=0$

is that fair to say?

obviously c, and a won't give terribly helpful polynomials, because they will be like the starting point

(a note based on the second message on the starboard) I've learned how to type most equations in word just with the keyboard

perhaps I am just wasting my time:

11:55 AM

Actually, after testing the rationals from mathworld.wolfram.com/RationalZeroTheorem.html, it's likely the case that you end up with roots like that. @micsthepick

Derivation doesn't do much.

I went into this knowing that there should not be any rational roots

How so? I read what you wrote, is there an indication for rational root?