We had that as an exercise in the probability course

hey dudes

whatsup?!

I need your "opinion" regarding a proof

my it's a proof in the field of theory of computation

I hope some of you has studied theory of computation

anyway

@AkivaWeinberger Assuming uniform distribution?

the problem is to show that $ALL_{DFA} = \{ \langle A \rangle \mid A \text{ is a DFA and } L(A) = \Sigma^* \}$ is decidable

@Lozansky Yeah

I simply proved it by reduction

We want to show that the language $ALL_{DFA} = \{ \langle A \rangle \mid A \text{ is a DFA and } L(A) = \Sigma^* \}$ is decidable. We show this by reducing this language to the language $A_{DFA} = \{\langle B, w \rangle \mid B \text{ is a DFA that accepts input string } w \}$, which we know to be decidable (Sipser, $3$\textsuperscript{rd} ed., $4.1$).

To be more specific, for each string $w \in \Sigma^*$, where $\Sigma$ is the alphabet for the DFA $A$ in the language $ALL_{DFA}$, we can run the Turing machine decider $TM(A_{DFA})$ by giving to it as input $\langle A, w \rangle$, where $A$ is the DFA from language $ALL_{DFA}$.

I think it's correct

I simply don't know if it's enough

@AkivaWeinberger Actually, that doesn't make sense

I think

@AkivaWeinberger It is true that $C[x,y] / (x - y) \cong C[x]$ right ?

There is no uniform probability distribution over the positive real numbers

just want some sanity check

It is (write down an isomorphism).

did you see? huy is back

Oh hi @Huy

welcome back pal

no computer scientist here?

wtf

@Lozansky it's uniform in (0,1)

Hi @Balarka

hey

@nbro I guess they want you to be the one. What a mindjob.

sup @BalarkaSen

long time

@Alucard what?

has Jasper finally started to study?

i dunno, he doesn't come here a lot. i don't think so

=(

I saw that you, Ted and many other similar names are still around

you must be like 15 already now, no?

something like that

very good

are you still teaching high schoolers and studying geometric topology

no I'm done studying and just teaching

gotcha. not a bad decision

so only aspirant mathematicians here

so bad

:D

:D

kidding ;) :-*

by "I'm done" I meant "I finished my degree" btw. :P

Hello, I hope everyone is doing good, I have a simple question about geometry! Is it possible that we have 1-dimensional and two faces at the same time

Oh. Congrats.

@Huy you got your phd ?

no, just my MSc

don't need a PhD for teaching high school

oh okay awesome

congrats

thanks !

in MathEd?

no, there is no "math ed" here

orly?

@Lozansky There is a uniform distribution over $[0,1]$, though.

you need a MSc in maths + some education degree to teach at high school

You're just choosing numbers between $0$ and $1$ each time.

and I'm working on my education degree also

cool, congrats :-)

@Adeek You're just insane for different reasons

So $f_X(x) = 1, 0 \leq x \leq 1$ and $0$ otherwise

Probably

Yeah that's the density

@AkivaWeinberger yeah

anyway it is yeah

@BalarkaSen: what about you? are you done studying analysis yet?

Yeah

@Huy Yep. Doing the good stuff now.

what would that be?

pictures of different sorts

foliations etc

Is there any one is working in geometry

It is trivial that $C[x,y] \cong C[x,x - y]$ and rest follows readily

Sure

@AkivaWeinberger do you have facebook btw ?

Yeah

can I add you ?

Sure

kids these days looking for new FB friends in maths chatrooms

lol

@Huy I like math friends :D

(i don't have fb)

Didn't your mum tell you to beware of strangers on the internet?

^

@AkivaWeinberger what is your email ?

not if they have cute puppies as their avatar

got it

/jk

@BalarkaSen: any interesting drama happening around here lately?

@Huy none here does though

@Huy the drama queen retired

NO

@AkivaWeinberger sent you email

She did a couple attempts not too long ago though if we're thinking about the same person

oh the calculus person ?

"analysis", not calculus

@Alessandro We probably have the same person in mind.

yeah

@Adeek Weird, the last name given on your email is different from the last name on your Facebook account

I'm not

nor ted

the two anti-fun people

the two productive people

Ted isn't anti-fun

Me and productivity are completely contradictory

What are you talking about

@AkivaWeinberger they messed up in my name when I came here

@AkivaWeinberger when I came to Canada people issuing passport messed up my name

I gotta go, bye

cya

cya

@Adeek The two names aren't even remotely similar, though

@AkivaWeinberger That name is my great great father name they shouldn't have included it

Arright

they should only include the name of my grand father which is the same as I have in fb

Bye

cya

cya

I am making myself a new facebook profile

because they want my id card and to be honest i am too lazy to go to the lost property office to get it back so here we go

Hello. Let f : D -> R be a differentiable function. I am trying to show that f is an increasing, function if and only if f'(x) >= 0 for all x in D. For the one direction, suppose that f is increasing, and let x be arbitrary. Then x < x+h for all h > 0. Hence, f(x) <= f(x+h). This implies that the ratio (f(x+h) - f(x))/h is positive for all h > 0. Since f is differentiable, the limit of this ratio, which is the derivative, is equal to the right-hand limit, which is a limit of positive numbers.

Hence, f'(x) => 0. Does this seem right for the one direction?

I believe I am implicitly using the fact that a limit of positive numbers converges to a positive number, although I am not sure if that would be a precise statement of the theorem I have in mind.

@user193319 you are trying to prove f(x) is increasing function if f'(x)≥0 ?

@user193319 To be truly pedantic, the limit of positive numbers converges to a non-negative number.

As I believe zero is not considered a positive number.

No, first I am trying to show that f increasing implies f'(x) => 0.

I think he means increasing over some interval.

@TimTheEnchanter Yes, that is what I mean.

And a limit of positive numbers does indeed have to be non-negative as if the limit of a sequence of positive numbers is -a, setting $\epsilon = a/2$ in the epsilon delta definition of a limit throws a contradiction.

I am not aware of any name for the theorem (if any)

@TimTheEnchanter So, we would need something like: If lim_{x \to c} f(x) exists and there exists an interval (a,b) containing c such that f(x) => 0 for every x in (a,b), then the limit is nonnegative. Is this wrong, or perhaps more than we need for the present problem?

Comma, comma, period. Question mark?

,,.?

Your wish is my command

gotta get stuff done

Somehow, posting that on the chat doesn't seem like a necessary step in "getting stuff done"

@user193319 Yes, epsilon-delta my friend. Works every time.

@Akiva trying to say it out loud so i can motivate myself to do it

@BalarkaSen

@TimTheEnchanter Hmm...Okay, I am having one issue then. Going back to my proof, what is the interval (a,b)? In order to use the theorem in my proof, I need to find an interval. All I have is h > 0, but not interval. I don't think (x-h,x+h) would work.

@user193319 Why not?

When you say a derivative exists at a point x, that means the interval (x-h,x+h) has to be within the domain for sufficiently small h.

@TimTheEnchanter And that ratio is always nonnegative on (x-h,x+h), right?

hello, i have $||u_n||\rightarrow\infty$ when $n\rightarrow\infty$ what i can say about this limit $$\limsup_{n\rightarrow\infty}\int_{\mathbb{R}^N}\dfrac{1}{||u_n||}dx$$

@user193319 Yes, as long as you account for the sign difference in right and left derivatives.

@TimTheEnchanter Okay. Thanks for your help. I'll take some time to think about this more carefully before I prove the other direction.

@user193319 No problem, good luck.

@TimTheEnchanter that was what i needed, thank you Sir.

someone can help me?

@TimTheEnchanter Okay, to prove the other direction, that f'(x) => 0 implies f is increasing, it seems that the mean value theorem yields a very nice proof. Are there any simpler or more elementary ways, or is this as simple as it gets?

Heya =)

Oh

Hallelulha

Heya :-)

writing up some revision notes for a course i am TA'ing. I have a few theorems and results which are identical for cosine and sine. Any ideas how to best present the results to minimize confusion?

what year?

Yes, make with your students sport and tell them about it in nature.

I dunno 19-20 year olds? It is about integrating trigonometric functions, as a sidenote in multivariable calculus.

@N3buchadnezzar teach them, that they can use it for something, i guess otherwise noone will even try to understand.

I am just unsure about which of the three options above is best for writing down the main results.

To be perfectly clear you should use option 2 and then point out option number 3.

Option number 1 is the briefest.

@Daminark helo, can you help me please ?

f(x) = (cos or sin)(x)

Actually if i had 3 options i would ask a RNG and be done with it

what's an "RNG"?

random number generator

@N3buchadnezzar What theorems?

@user193319 The mean value theorem is quite direct, but you could also try proving the contrapositive using the same means as you used earlier (A decreasing function has f'(x) $\le$ 0).

@AkivaWeinberger theorems is stretching it. Just some simple results about integrating sine and cosine.

$$ \int_K (\sin x)^{2k} \mathrm{d}x = \left( 1 - \frac{1}{2k} \right) \int_K (\sin x)^{2k-2}\mathrm{d}x$$

Where $K$ is any interval which is a multiple of $\pi/2$. Eg $K = [\pi n/2,\pi m/2]$, given $m\geq n$, and $m,n \in \mathbb{Z}$

Ah, OK

I would give both equations but just prove the first one, saying "the proof for (2) is similar"

Then again, I am not a teacher

Yeah, I ended up doing that. I just feel it is useless stating the same result for sine and cosine every time

If anyone has been keeping up with my FAIL system, I believe the current version goes up to the small Veblen ordinal in the fast growing hierarchy

So you're saying that it fails to reach the large Veblen ordinal

I am unsubtle

If it reaches the small veblen ordinal in its current state, then it will reach the LVO no problem with one augmentation

And hopefully with two augmentations, it will surpass the Bachmann Howard ordinal

@AkivaWeinberger chat.stackexchange.com/transcript/message/36530523#36530523

I haven't worked out all the details yet, but the above link is a basic guide up to basic @ symbols

XD

@AkivaWeinberger Ugh any hints on the expected value question?

@AkivaWeinberger btw, if you haven't noticed, I did it all without any ordinals

How can i model a selection of an object from a collection such that the object that has been selected previously many times has least probability!

suppose object x1,x2,..xn have been selected a1,a2,..an times and then what should be the probability that an object xi should be selected?

$\frac{a_i}{\sum a_j}$?

@AkivaWeinberger funny question, I just added an answer

@fluffy_muffin :-) You good?

@N3buchadnezzar Oh? I didn't know you were talking about the proofs. In that case, your choice is the best.

I thought you meant just the results.

sweet, throw away email adresses :D

Hi all

Ya @TedShifrin , if you are free we can talk about continuous extension!

@Lozansky There's a 100% chance that there are at least two numbers. What are the odds that it's at least three numbers? four?

A bit of geometry might help

Here is a cool question: how do you invite someone to something in a public place but you don't want that anyone else notices. Hint: Overflow

I like how $\bigcap\emptyset=V$

'Cause everything is in every element of the empty set

The set with no elements.

Hi, could anyone point me in the right direction/provide keywords for finding material on describing sets generativeley. I'm trying to write something like (P = \{\forall n \in \{0,1\}^4, p_n \leftarrow \{n\}^{16}\}).

which is supposed to read: a set P, containing the elements (0,0,0,0,0,0,0,0), (1,1,1,1,1,1,1,1), ..., (15,15,15,15,15,15,15,15) (each referred to as p_n where n = 0,1,...,15)

Set builder notation

thank you :)

np

I go to church, see you later :D

cya

@AkivaWeinberger Monster group is the only true example among all those listed, tbh

the field with one element is not a field

oh, I see rschwieb already said that in a comment

Hello. I am reading this: math.stackexchange.com/questions/1266906/…

I am wondering, aren't the two cases in the first proof unnecessary?

The cases I am referring to are x_1 = x_2 and x_1 < x_2. In either case, couldn't we choose the second coordinate to be the midpoint of r_1 and r_2 and let the first coordinate be x_1?

Hi chat

Hey @Astyx!

Oh @Balarka regarding the problem about the Gauss map from some time back

Last night when I was doing it I realized that just using Lagrange multipliers should work