12:11 AM
@DanielFischer After having shown by induction that $h(k) \geq k$ we have that $k \leq h(k)<n \rightarrow k<n \rightarrow k \in n=\{0,1,2, \dots, n-1 \}$.. But what do we conclude from that?

@evinda That the domain of $h$ is a natural number not larger than $n$. And since by construction $h$ is a bijection between its domain and $f(B)$, that shows that $B$ is a finite set, whose cardinality is less than or equal to the cardinality of $A$ (which is $n$).

We didn't say before that k is the domain of h... So, is k the same as m? @DanielFischer

@evinda $k$ is an element of the domain of $h$.
$k\in m$

@DanielFischer But why does this imply that it is the greatest possible?

What?

12:30 AM
@DanielFischer So since $k$ is an arbitrary element of the domain, we know that each element must be smaller than n and that means that the greatest natural number that can be the domain is n, right?

Yes.

159870

@DanielFischer So then do we say that $f(B) \sim l$ where $l \in \omega$ and $l \leq n$ ?
@111 What do you mean?

What speed covers 220 miles in 8 hours and 48 mins?

s=d/t
= 200 / (8 48/60)
im not sure how to progress to i make that into a top heavy fraction ?

12:50 AM
@Alex $s = \dfrac{d}{t} = \dfrac{220~\text{miles}}{8+\frac{48}{60}~\text{hours}} = \dfrac{220}{\frac{528}{60}} = \dfrac{220}{\frac{528}{60}}\cdot \dfrac{\frac{60}{1}}{\frac{60}{1}} = \frac{13200}{528}=\dots$

why is 528 still kept over 60 even tho its in minute form now ?

528/60 hours
the final answer as i have it would put it in miles per hour
if preferred, you could do it as $\frac{220~\text{miles}}{528~\text{minutes}}$, but miles per minute is not a common set of units

i see thank you
im trying a question similar

Good night!!!

1:06 AM
is there a quick way to tell if a number is divisible by 4 ?

A number is divisible by 4 if the final two digits are divisible by 4

@Alex The first thing to do is check if it's odd. That'll catch 50% of the numbers you're testing.

does the same rule with for all the final two digits?

Testing the final digit by itself works for divisible by 2, testing the final two digits for divisible by 4, you need to test final three digits for divisibility by 8

wow thats gonna save me so much time
6?

1:16 AM
To test divisibility by 3, add together all of the digits. If the number is larger than 9, add the digits of the newly formed number. Repeat this process until you get a number between 1 and 9

For 3, add the digits together; if that number's divisible by 3, the original number's divisible by 3.

If at the end you get a 3,6, or 9 then it is divisible by 3
for divisibility by 6., first test if divisible by 3, and then test if divisible by 2. If it is divisible by 6 it must be both divisible by 3 and 2 simultaneously
Example. $12345678\mapsto (1+2+3+4+5+6+7+8)=36 \mapsto (3+6)=9$ and 9 is divisible by 3, so 12345678 is in fact divisible by 3
Furthermore, 12345678 is even, so it is also divisible by 2. Since divisible by both, it is divisible by 6
You can do similarly for divisibility by 9, but instead of getting a 3,6, or 9 at the end of repeated summation of digits, you want only to get a 9 (or 0)

is there a name for all these rules? i never knew they exsisted. im laughing at this now lol its like the cave man just discovered fire

It falls from what is called "elementary number theory" (though most people don't learn it until they are older)
divisibility by 7 and by 11 are a bit harder
I don't remember the algorithm exactly, but for 11, it requires alternating summation/subtraction of digits i.i.r.c.
mathsisfun.com/divisibility-rules.html has the list for small integers (1 through 12)

its a really good website was learning fractions yesterday. thanks for the help :) i never knew about those rules. really cool tbh

1:26 AM
Anyone in the room who might have an idea for a proof regarding Galois theory? @Emrakul perhaps?

Heh, I wish. I'm still chewing my way through the fundamentals of abstract algebra. Sorry!

Trying to reach a contradiction, but all I got to was a nonlinear system of three equations and a finite (but potentially large) number of unknowns. I'm thinking I should reverse direction away from proof by contradiction and look for some result which can go more directly

@JMoravitz what's the problem

Let $K = \mathbb{Q}(\sqrt[3]{2}e^{2\pi i/3})$. Claim: $x_1^2+\dots+x_k^2 = -1$ has no solution with $x_i\in K$
from M. Artin 2nd ed

the fields which have -1 equal to a sum of squares has been studied specifically, but I don't remember what the results were
ah
I think I know how to prove it for that field :-)
think purely in terms of its isomorphism type, don't be distracted by the fact it has nonreal elements in it

1:34 AM
it winds up being, whatever $x_i$ is equal to specifically, an equation of $y_0 + y_1 \beta + y_2 \beta^2$
where $\beta$ is the messy cube root with root of unity, and $y_i$ is some element of $\mathbb{Q}$

ugh, writing things out like that would surely yield an ugly proof
see if you can determine other descriptions of the field
for instance, if X is algebraic over L, then L(X) is isomorphic to L[x]/(f(x)) where f is X's minpoly
what is the minpoly of $\sqrt[3]{2}e^{2\pi i/3}$ over $\Bbb Q$?

$x^3 + 2 = 0$ i.i.r.c

so if I cube $\sqrt[3]{2}e^{2\pi i/3}$ I should get $-2$ you're saying?

mm, i think not,. Yea, my mistake there. cubing $\beta$ would give you positive 2

so it's a rupture field for $X^3-2$. all rupture fields are isomorphic; what's another rupture field for that polynomial?
(rupture field for a polynomial means field obtained by adjoining a single root)

1:42 AM
(potentially lagging out)
refreshed.
I expect you are pointing me towards $\mathbb{Q}(\sqrt[3]{2})$ (without the root of unity)

yes
and why would I be pointing you to that?

anon's here

yup

In $\mathbb{Q(r)}$ where $r$ is any real number, a sum of squares is always nonnegative, and could never equal -1

right.

1:47 AM
I suppose I need to think further to understand why the rupture fields for a splitting polynomial are isomorphic and how we can apply results from one to the other. It seems an incredibly obvious statement when considering $\mathbb{Q}(\sqrt[3]{2})$, but it is still not immediately apparent how to transfer results from the one to the other

If $x$ and $y$ are roots of an irreducible $f\in K[T]$ then $K(x)\cong K[T]/(f(T))\cong K(y)$
since $\sqrt[3]{2}$ and $\sqrt[3]{2}e^{2\pi i/3}$ are both roots of $T^3-2\in\Bbb Q[T]$, irreducible, then $\Bbb Q(\sqrt[3]{2}e^{2\pi i/3})\cong\Bbb Q[T]/(T^3-2)\cong\Bbb Q(\sqrt[3]{2})$. this isomorphism preserves $-1$, squares and sums of squares, so any solution in the first field would be transported by the isomorphism to a solution in the last field.
@JMoravitz I said minimal (=> irreducible) polynomial, not splitting polynomial

my mistake, confusing terms again.
Well, thank you again for taking the time to explain. It is nice that there are users like you who can answer upper level math questions (though still early grad level, probably not upper level to you).
I get to contribute by helping the undergrads at least. In another few more years hopefully I'll get more algebra under my belt enough to return the favor to the next early grad students down the line.

2:31 AM
it seems to differ from place to place, what generally counts as undergraduate vs. graduate material? where do most people draw the line in various subjects?

2:49 AM
@SamuelYusim I personally don't care about it much.
It seems better to pick the book, and decide if it suits you or not.
@anon Hello and goodbye mister.

/shrug, I referred to it as "graduate" since it is a problem I received while being a graduate student
though still well below the level of difficulty of comprehensive/qualifying exams
Its kind of a fuzzy-logic scenario. At what point is something hot or cold, at what level of difficulty is something graduate or undergrad.,

3:03 AM
Is there a reason the epsilon-delta definition starts with $(\forall\epsilon>0)(\exists\delta>0)$ instead of $(\exists\delta>0)(\forall\epsilon>0)$?
(I ask, because it only made sense to me given the rest of the proposition once I reversed the two conditions.)

oh, I wasn't actually asking because of that question, the comment after it just got me thinking.
lambda? they call it epsilon-delta for a reason!

er

I'm just fooling around, it's no big deal
the other way means that the same delta works for all epsilon

If you change all deltas to lambdas, the proof no longer holds!

which isn't necessarily true
whereas for all epsilon > 0 there exists delta > 0 means that you're allowed to choose the delta after you know which epsilon you want it to work for
which isn't necessarily true of continuity*

3:07 AM
Hm. That makes sense, yeah.

when you have it the other way around it's still a reasonable concept though, I think some people call it absolute continuity
because, as you can hopefully see, it's an even stronger condition than regular continuity
i.e., absolute continuity implies continuity but not the other way around

If you have it the other way around, then let $\epsilon\to 0$, I'm pretty sure you get that within a neighborhood $(x-\delta, x+\delta)$ there is essentially no change
implying that the function is constant
Continuity is defined for a specific point (continuous at $c$). Absolute continuity is when the same $\delta$ works for the epsilon regardless which $c$ it is in question

...now I need to look up absolute continuity.

whoops! look up uniform continuity, I think that's the one you want
actually even then I think I was quite wrong. see: @jmoravitz or the definition of uniform continuity on wikipedia. shows how much of this stuff I remember

Thanks, both! :]

3:18 AM
Well, I was forgetful as well. Absolute continuity gets into measure theory, which says that for each $\epsilon$ there exists a $\delta$ such that breaking the domain into several disjoint subintervals of total measure less than $\delta$ the total variation on those intervals is less than $\epsilon$
but yea., wikipedia is quite helpful for remembering definitions instead of having to flip through the index of a textbook
(or relying on the faulty memory of others)
But yea, to summarize., for function $f:X\to Y$: everywhere Continuous: $(\forall c\in X)(\forall \epsilon > 0)(\exists \delta >0)~ s.t.~ ((\|x-c\|_X<\delta)\Rightarrow (\|f(x)-f(c)\|_Y<\epsilon)$. Uniformly continuous $(\forall \epsilon > 0)(\exists \delta >0)(\forall c\in X)~ s.t.~ ((\|x-c\|_X<\delta)\Rightarrow (\|f(x)-f(c)\|_Y<\epsilon)$

Hello @user130018

Hi @JasperLoy

3:39 AM

4 hours later…
7:20 AM
If some mod will be around, they can respond to this, in case I have omitted something important on give an incorrect advice there.
in Mathematics Meta, 3 hours ago, by Thorsten
In a comment, I asked for something which was clearly in the OPs message. Should I delete my comment(s)? http://math.stackexchange.com/questions/1102127/equivalent-dfn-of-filtered-categ‌​ories
in Mathematics Meta, 6 mins ago, by Martin Sleziak
Your comment might help other users which have the same problem as you have. (I personally would prefer to leave it. If it is the comment I linked to at all.)
in Mathematics Meta, 2 mins ago, by Martin Sleziak
But it's you're comment and you are the one to decide what to do with it. But if you decide to delete it, I think that it is useful to wait some time whether the OP deletes comment reacting to yours. If not, you should flag it as obsolete. One of the moderators will see your flag, they will see that it is a comment reacting to a deleted comment and remove the flagged comment, too. But occasionally it might happen that such flags are declined.

2 hours later…
8:59 AM
The title of this question contained $\lim_{n \rightarrow \infty} \dfrac{1-\left(1-\dfrac{1}{n}\right)^4}{1-\left(1-\dfrac{1}{n}\right)^3}$. I have change it to $\lim_{n \rightarrow \infty} \frac{1-(1-1/n)^4}{1-(1-1/n)^3}$.
It the way it looks now acceptable? Or would it be better just to change dfrac to frac?
Or some completely different suggestion?
In fact, it seems that it was one of the editors who put that thing into the title.

9:42 AM
I hope things are improving @Jasper!

10:23 AM
@MartinSleziak $\lim\limits_{n\to\infty}\frac{1-(1-1/n)^4}{1-(1-1/n)^3}$ is good, too.

10:35 AM
@Committingtoachallenge I am not sure how things will unfold for me. My mental condition declined since Dec, but today I don't feel too bad. I might try to do some part time job this year and the next before studying, we will see.

10:49 AM
Moin @DanielFischer !!!
I am looking at the following:
**Propostition**:

The Cartesian product of finite sets is a finite set.

**Proof**:

Let $X,Y$ finite sets and therefore there are $n,m \in \omega$ such that $X \sim n$ and $Y \sim n$, i.e. there are $f: X \overset{\text{1-1 & surjective}}{\longrightarrow} n$, $g: Y \overset{\text{1-1 & surjective}}{\longrightarrow} m$.

We define the function $h: X \times Y \to n \times n \\ \langle x,y \rangle \to \langle f(x), g(y) \rangle$

It is obvious that $h$ is well-defined and it is $1-1$ and surjective.
Hey @Huy

Morning, @evinda

11:01 AM
Good day

Hi @ValerySaharov

Anyone familiar with categroy theory here guys?

@JasperLoy Part time work could be beneficial. Have you tried doing a page of a textbook a day? It is a start

@Committingtoachallenge I have not. You must understand that I cannot change certain obsessions of mine, to start only in certain ways.
@Committingtoachallenge Severe OCD is a terrible condition. Most people who get it are mostly screwed for life.
@Committingtoachallenge I will figure out something for myself. I am 34 this year. I will do my best to get well and enter grad school by 40. That is the deadline I set for myself...

@JasperLoy How do you spend time?

11:09 AM
If anyone asks my location, just tell them I am in Antarctica, thanks.
@Oracle I spend a lot of time sorting out my confused thoughts every day. I also go for long walks, read things for fun (non-math) and chat with people online.

@JasperLoy Can't you turn your obsessiveness into something more productive, such as studying?

@Oracle No, I cannot.

@JasperLoy Maybe you can. Remember, RH is waiting!

@Oracle You will not understand OCD or PTSD if you don't have them. I have both.
Hi @skullpatrol
@skullpatrol I have watched the documentary. It was so short.

11:42 AM
Is this: $f=\{ (x,f(x)): x \in dom(f) \wedge f(x) \in rng(f)\}$ a right definition of a function?

@evinda Seems right to me. It is the set of ordered pairs basically.

@JasperLoy That's why when we have the function $f: \{a\} \to 1$ with $f(a)=0$ we can symbolize it as $\{ \langle a,0 \rangle \}$ ?

@evinda That function would just have one ordered pair, yes.

A ok @JasperLoy Thanks!!!

@evinda It helps to translate symbols into words to understand. In words, a function is defined as the set of ordered pairs making up the points on its graph.

11:53 AM
@JasperLoy Yes, this makes sense!

12:09 PM
@evinda Everything I say makes sense, lol.

@JasperLoy :D

1 hour later…
1:26 PM
Hi @skullpatrol

Hi pal @JasperLoy

Hi

Hello.

if a and b are real numbers >0
if b >> a, can I say the following statement
b=inf ?

Is inf a real number?

1:34 PM
@barznjy What is b>>a?
@barznjy What is inf?

>> means much greater than
inf means infinity

@barznjy What is much greater than?
@barznjy What is infinity?

Thanks

@barznjy A real number is a real number, it is not infinity, end of story.

Heya

1:41 PM
Yo

Hello @user130018

Hi @JasperLoy

in Committingtoachallenge's room, 20 secs ago, by Committing to a challenge
The regression equation is
(Hours of Study) = 2.63 + 0.711 (Coffees) - 0.273 (Hours Sleep)

@user130018 Have you thought of which grad school you want to apply?

Minitab gives very strong evidence against the null hypothesis in terms of coffee consumption. E.g. There is very strong evidence that cups of coffee consumed is correlated with amount of study done.

1:51 PM
@JasperLoy All of them

@user130018 OK. I will try to get into grad school in at most 6 years from now.

@JasperLoy What are you going to do until then

@user130018 I will do my utmost to get well and then study math and then apply.

Today I got 11 hours of sleep and had 2 coffees, so by that equation I should have done 1.04 hours, but I have done 1.5333 whoops, I better have a coffee ;P

@Committingtoachallenge How did you sleep for 11 hours?

2:01 PM
@Committingtoachallenge: I've had about 10 hours but no coffees, so I should start unlearning stuff really soon, right?

@JasperLoy Girlfriend left for her next week at work, and since she normally wakes me at 5am, I slept in until 9

@Committingtoachallenge: That doesn't make any sense.

@Huy That regression was from 33 days of my data
@Huy What doesn't? My regression or sleeping in?

@Huy Heya, do you have a minute?

@Committingtoachallenge: If she normally wakes you at 5am, your body should be used to it and you should wake up at 5am automatically, feeling normal.

2:08 PM
@Huy Except I have a sleep disorder

I was trying to show that $$\lim_{n\to\infty} z_n = z \ \Longrightarrow \ \lim_{n\to\infty} \frac{z_1 + z_2 + z_3 + \cdots + z_n}{n} = z$$

What is $z$?

@Huy Clearer?

Yes.

@DanielFischer Hello!!!

I am looking at the following:
**Propostition**:

The Cartesian product of finite sets is a finite set.

**Proof**:

Let $X,Y$ finite sets and therefore there are $n,m \in \omega$ such that $X \sim n$ and $Y \sim n$, i.e. there are $f: X \overset{\text{1-1 & surjective}}{\longrightarrow} n$, $g: Y \overset{\text{1-1 & surjective}}{\longrightarrow} m$.

We define the function $h: X \times Y \to n \times n \\ \langle x,y \rangle \to \langle f(x), g(y) \rangle$

It is obvious that $h$ is well-defined and it is $1-1$ and surjective.

2:12 PM

@Huy Why is it correct?

@N3buchadnezzar: Urm. I'm not sure what you mean. You can prove it using the epsilon-definition of the limit.

@Huy Uuugh

@N3buchadnezzar: The idea is to split up the sum into two, the first is finite and thus can be bounded easily, the second you can bound using the fact that $z_n$ converges.

@Huy Just pick an $k\in \mathbb{N}$ ?

2:21 PM
@N3buchadnezzar: I think it's a bit easier if you pick a very certain element of the sequence at which you split the sum into two. Remember that since $z_n$ converges to $z$, for all $\varepsilon > 0$ there exists some $N \in \mathbb{N}$ such that for all $n \geq N$, we have $|z_n - z| < \varepsilon$. What happens if you split up the sum at this $N$, for an arbitrary $\varepsilon > 0$ given?

@evinda Typo: It ought to be $h \colon X\times Y \to n \times m$. One takes $h(\langle x,y\rangle) = \langle f(x), g(y)\rangle$ because given two bijections $f \colon X \to n$ and $g\colon Y \to m$, that is the most straighforward and obvious bijection $X\times Y \to n\times m$.
@N3buchadnezzar Look at $$\frac{1}{n}\sum_{k=1}^n (z_k -z).$$ You have a couple of potentially large [in absolute value] terms at the start of the sum (where "a couple" can be a huge number), and then a lot of small terms. If you let $n$ grow, the number of large terms remains constant, so the average is dominated by the small terms, and hence itself small. Making that precise is the $\varepsilon$-$N$ argument.

2:37 PM
$$\lim_{n\to \infty} \frac{1}{n}\left( \sum_{k=1}^N z_k + \sum_{k=N+1}^n z_k \right)$$
The first term should go to zero since it is a finite amount of terms, right?

@N3buchadnezzar: I didn't mean you should look at the sum itself, rather at the sum minus its (supposed) limit. I thought that was clear, but I see now it wasn't.

@N3buchadnezzar Since it is a fixed finite amount of terms.

@Huy Right so you want to prove that for every epsilon we have $$\left| z - \lim_{n\to \infty} \frac{1}{n}\sum_{k=1}^n z_k \right| < \varepsilon$$

@N3buchadnezzar: No, I was going to bound $\frac{1}{n}\sum_{k=1}^n z_n-z$.
@N3buchadnezzar: I should continue with my work. If it's still unclear, there are a number of questions about that statement on MSE, e.g. math.stackexchange.com/questions/533626/…

@Huy Yeah go back =)
Thanks a bunch!
@Huy I know, but I wanted to try myself before looking up any cesaro limit spoilers :p

2:46 PM
@N3buchadnezzar: Sure, try yourself. I think it's a good exercise if you're not used to proving convergence with epsilons, or just revising the method.

We want to show that the following expression is less
than epsilon, for all epsilon
%
\begin{align*}
\left| z - \frac{z_1+z_2+\cdots z_n}{n} \right|
= \left| \frac{1}{n} \sum_{k=1}^n z-z_n \right|
\end{align*}
%
Since the limit $\lim_{n\to\infty} z_n$ exists,
we know that there exists some number $N\in\mathbb{N}$ such that for all $n\geq N$ we have $\left| z_n-z\right|\leq \varepsilon$. Hence
%
\begin{align*}
\left| \frac{1}{n} \sum_{k=1}^n z-z_n \right|
\leq \frac{1}{n} \sum_{k=1}^n \left| z-z_n \right|

@MikeMiller yo mr mike
got a question for you
i'm trying to understand why the map $q:S^2 \to \mathbb{R}P^2$ that sends each point on the sphere to the line through the origin and the point, is a covering map
it's clearly a continuous surjection

@iwriteonbananas Can you see that it is a local homeomorphism?

@DanielFischer hmm yeah i think so
@DanielFischer i mean...if we think of $\mathbb{R}P^2$ as $S^2/\sim$ with $x\sim -x$, it seems obvious that its a covering map

@iwriteonbananas Right. Think of it that way.

2:59 PM
if we have a point in the projective plane, can't we just take a small intervall around that point and then the preimage of this intervall will be exactly the union of this interval on the upper hemisphere and on the lower hemisphere
which is a disjoint union of connected open subsets

@iwriteonbananas If by "interval" you mean "disk" or some other small open neighbourhood, then yes, that it exactly the argument.

@DanielFischer yeah thats what i meant. but how exactly do we choose this disk?

Where by "disk" I obviously meant "spherical cap" ;)

@iwriteonbananas Well, you have an $x\in S^2$ and an open neighbourhood $U$ of $x$. You want $U$ and $-U$ to be disjoint ...

3:07 PM
Yeah...dont we get a problem if $x=(-1,0)$ or something?

@iwriteonbananas No. What do you think might be a problem there? (But a point in $S^2$ has three components.)

oh yeah im thinking about S^1 for some reason
anyways, i'll think about it more, i'll figure it out from here

@DanielFischer Do we define the function h to be the one that maps $\langle x,y \rangle$ to $\langle f(x), g(y) \rangle$ ?

@evinda Yes.

@DanielFischer Because we know that $f(x) \in n$ and $g(y) \in m$, right?

3:10 PM
Yes.

@DanielFischer Nice... Can we show as follows that $h$ is bijective?

Let $x_1, x_2 \in X$ with $x_1 \neq x_2$ and $y_1, y_2 \in Y$ with $y_1 \neq y_2$.
Then $f(x_1) \neq f(x_1)$ since $f$ is 1-1 and $g(y_1) \neq g(y_2)$ since $g$ is 1-1.

Since $x_1 \neq y_2 \wedge y_1 \neq y_2 \rightarrow \langle x_1, y_1 \rangle \neq \langle x_2, y_2 \rangle$.

Since $f(x_1) \neq f(x_2)$ and $g(x_1) \neq g(x_2) \Rightarrow \langle f(x_1), g(x_1) \rangle \neq \langle f(x_2), g(x_2) \rangle \Rightarrow h(x_1) \neq h(x_2)$

how can you cross out text in mathjax
?
Like, but a line through text and then write something after

@evinda You have one typo, you typed $f(x_1) \neq f(x_1)$ where you meant $f(x_1) \neq f(x_2)$. Just for the record. You have a conceptual problem, because $\langle x_1,y_1\rangle \neq \langle x_2,y_2\rangle$ means $(x_1 \neq x_2) \lor (y_1 \neq y_2)$, not $(x_1\neq x_2) \land (y_1\neq y_2)$. So your argument for injectivity is incomplete. It's fixable.
The surjectivity part is okay, but not written up well. You need equivalences instead of $\Rightarrow$, and you should say something like "since $f,g$ are surjective, such $e$ and $k$ exist".

@DanielFischer At which point did I use that $\langle x_1,y_1\rangle \neq \langle x_2,y_2\rangle$ means that $(x_1\neq x_2) \land (y_1\neq y_2)$? :/

3:26 PM
@DanielFischer Thanks for checking my proposed proof in comments.
@DanielFischer If you or anyone else has a chance please look at my post which is a question using Mac Shane's extension.

@evinda You start by saying "Let $x_1,x_2 \in X$ with $x_1 \neq x_2$ and $y_1,y_2 \in Y$ with $y_1 \neq y_2$". So you never consider the case where one of the components is equal and the other different.
@JohnDoe You're welcome.
@JohnDoe Do you know that Lipschitz functions are almost everywhere differentiable (Rademacher)? And the distributional derivative is given by the a.e. defined gradient? So $\lVert \nabla \overline{v}_m\rVert_{L^\infty(I)} \leqslant 2\sigma$ is just Mac Shane.

@DanielFischer So it should be like that, right?

Let $x_1, x_2 \in X, y_1, y_2 \in Y$ with $x_1 \neq x_2 \lor y_1 \neq y_2$. Then $\langle x_1, y_1 \rangle \neq \langle x_2, y_2\rangle$.

Then because of the injectivity of $f,g$ it holds $f(x_1) \neq f(x_2) \lor g(y_1) \neq g(y_2)$.

Therefore $\langle f(x_1),g(x_1) \rangle \neq \langle f(x_2), g(x_2) \rangle \Rightarrow h(\langle x_1,y_1 \rangle) \neq h(\langle x_2, y_2 \rangle)$.

@evinda For example. But typically, it is more convenient to show injectivity the other way round, suppose $h(\langle x_1,y_1\rangle) = h(\langle x_2,y_2\rangle)$ and deduce $\langle x_1,y_1\rangle = \langle x_2,y_2\rangle$.

3:52 PM
@DanielFischer Is it better like that?

Suppose $x_1, x_2 \in X, y_1, y_2 \in Y$ such that $h(\langle x_1, y_1 \rangle) =h(\langle x_2, y_2 \rangle) \Rightarrow \langle f(x_1), g(y_1) \rangle =\langle f(x_2), g(y_2) \rangle \Rightarrow f(x_1)=f(x_2) \wedge g(y_1)=g(y_2) \overset{\text{injectivity of f,g}}{\Rightarrow} x_1=x_2 \wedge y_1=y_2 \Rightarrow \langle x_1,y_1 \rangle=\langle x_2, y_2 \rangle.$

And can we show like that that the function $h$ is well-defined?

$\langle f(x_1), g(y_1) \rangle \neq \langle f(x_2), g(y_2)\rangle \rightarrow f(x_1) \neq f(x_2) \lor g(y_1) \neq g(y_2) @DanielFischer And the distributional derivative is given by the a.e. defined gradient? By distributional derivative are you referring to 'weak derivative'? @evinda The first part is fine. I don't know what the purpose of the second ($h$is well-defined) part is. @DanielFischer In my lecture notes, it says that$h$is obviously well-defined and it is 1-1 and surjective. And I was sondering how it could be shown. @DanielFischer If we want to show that a set is finite and we find a bijective function, do we have to show that it is well-defined? @JohnDoe Almost certainly "Yes" or "Yes, but". Depends on what you call "weak derivative". The derivative in the sense of distributions is know to exist, and is sufficiently nice, hence it coincides with the weak derivative - of which we may not know a priori that it exists - in all reasonable definitions of "weak derivative". @evinda I don't even know how well-definedness of$h$could be an issue. Maybe it would make sense in context. This chat is boring with Teddy. 4:05 PM @DanielFischer So a kind of generalized definition of weak derivative. Is the following in line with what you mentioned:$\frac{|v_{m}(x)-v_{m}(y)|}{|x-y|} \leq 2\sigma$for all$x,y$, it follows then that$\nabla v_{m}(x) = \lim\limits_{x \rightarrow y}\frac{|v_{m}(x)-v_{m}(y)|}{|x-y|} \leq 2\sigma$. Therefore$\| v_{m} \|_{L^{\infty}(I)} \leq 2\sigma$. Because$v_{m}$is Lipschitz it is almost everywhere classically differentiable as you stated. @DanielFischer It is just said that it$h$is well-defined. Anyway... :) How could we show that the set$n \times m$has the same cardinality as the natural number$n \cdot m$and thus$n \times m \sim n \cdot m$, i.e.$Card(X \times Y)=Card(n \times m)=n \cdot m$? Do we consider the function$f: n \times m \rightarrow n \cdot m$? @JohnDoe It's not$\nabla v_m(x) = \lim\limits_{y\to x} \dotsc$, you must look at each partial derivative for itself. But basically, that's it, $$\lvert \partial_k v_m(x)\rvert = \lim_{h\to 0} \left\lvert\frac{v_m(x+h\cdot e_k) - v_m(x)}{h}\right\rvert \leqslant 2\sigma$$ for$1 \leqslant k \leqslant n$, hence$\lVert \nabla v_m\rVert_{L^\infty} \leqslant 2\sigma$. @evinda You construct a bijection.$f(\langle a,b\rangle) = n\cdot b + a$is one obvious choice. Today I saw someone who looks like Teddy. @DanielFischer Yeah that makes more sense. You are of course using that$|v_{m}(x+h\cdot e_{k}) - v_{m}(x)| \leq 2\sigma |(x+h\cdot e_{k}) - x|$? @DanielFischer Where$a,b \in \omega$? 4:17 PM @JohnDoe Yes, where$\lvert (x+h\cdot e_k) - x\rvert = \lvert h\cdot e_k\rvert = \lvert h\rvert$. @evinda$a\in n$and$b\in m$. @DanielFischer Kewl thanks, makes sense. It feels rather easy, but I don't see how one would go from the second to the third equation here: i.imgur.com/E93HvSC.jpg I should add that the alphas are vectors of length z, so are the omegas, and the gammas are square matrices @DanielFischer Why do we take this function? I wanted to prove that it is bijective but I don't know how to continue. Let$a_1, a_2 \in n, b_1, b_2 \in m$. Then$f(\langle a_1, b_1 \rangle)=f(\langle a_2, b_2 \rangle) \rightarrow n \cdot b_1+a_1=n \cdot b_2+a_2 \rightarrow n \cdot (b_1-b_2)-a_2-a_1$. How can we deduce from this that$\langle a_1, b_1 \rangle=\langle a-2, b_2 \rangle$? 4:33 PM I've answered my own question, so please ignore it! hey, if I have a function f with ||f||_Lp(X) = c < infty, L(X) < infty (Lp stands for Lp norm, L for lebesgue measurement). is it true, that when I integrate f over C_eps with L(C_eps) = eps: ||f||_Lp(C_eps) --> 0 for eps --> 0 4:45 PM Hello! Can someone tell me where can I find the geometric inequalities by Kang Ying Liu ? @DanielFischer How can we show that if$f: X \to \omega$is bijective that then$f^{-1}: \omega \to X$is also bijective? @evinda Pretty much by definition. Just check that it is injective and surjective. @DanielFischer So we pick$y_1, y_2 \in \omega$with$f^{-1}(y_1)=f^{-1}(y_2)$, right? And can we take now the compostition with the function$f$? @DanielFischer Or do we have to use the injectivity and surjectivity of f? 5:23 PM The set of permutations having$n-k$elements in their original place$\subseteq$set of permutations having$n$elements in their original place, right? Hi @Ted Or I guess I should say morning Good afternoon, @Mike. 6:16 PM Good afternoon :3 SamuraiSamurai 22:15 Can someone please tell me where can I find the geometric inequalities by Kang Ying Liu ? Here is a link describing about those inequalities. 6:41 PM @Samurai I can't find them anywhere either :/ Greetings hi @Chris'ssis Good day... @robjohn Hello :-) @Chris'ssis Good evening, Chris 6:47 PM @Hippalectryon I have prepared something for you @Chris'ssis How does the book ? :-) @Mircea Hello. How is it going? @Chris'ssis Oh, what is it ? @Hippalectryon $$\int_0^{\infty } \log \left(\frac{1}{1-e^{-x}}\right) \cos \left(\log \left(\text{Li}_2\left(e^{-x}\right)\right)\right) \, dx=\frac{\zeta(2) }{2} \left(\sin \left(\log \left(\zeta(2)\right)\right)+\cos \left(\log \left(\zeta(2)\right)\right)\right)$$ @robjohn He's getting bad grades ... ;) @Chris'ssis thanks 6:49 PM @Hippalectryon Oh, sorry I asked... @Chris'ssis I'm pretty good, doing my math homeowrk @robjohn I was joking :o Don't ever take anything I say seriously @Mircea Great! Except this last sentence @Hippalectryon Oh, great... I deleted my comment :-p 6:49 PM lol Troll powers 110% @TedShifrin: Hey, Ted! @Chris'ssis I calculated an integral but I got a wrong answer and I can't see where I did wrong... @Mircea What integral? integral of tan(x)^4 from 0 to Pi/3 I got -Pi/3 @Chris'ssis forgot to PM @Mircea It should be$\pi/3$6:53 PM @Chris'ssis i know @Chris'ssis if i show you how i did it can you tell me where i did wrong? @Chris'ssis if you're not too busy @Mircea The integrand is positive, so a negative answer seems unlikely. @Mircea Just a second ... @robjohn i know... @DanielFischer Could you explain me why we can use the function$f(a)=a+nb$in order to show that$n \times m \sim n \cdot m$? @Mircea how about writing things like that? $$\tan(x)^4= \tan(x)^2(\tan(x)^2+1-1)= \tan(x)^2(\tan(x)^2+1)-\tan(x)^2$$ 6:57 PM @Chris'ssis oh wait, I just discovered my mistake @Chris'ssis i wrote the derivativeof tan as -1/cos^2 @MikeMiller Mikey McMike i got a question for u @Mircea I see. How about my way suggested above? @Chris'ssis hmm, that's pretty interesting @DanielFischer I wrote it wrong.. We take the function$f(\langle a,b\rangle) = n\cdot b + a$.. But why do we know that$n \cdot b+a \in n \cdot m\$ ?