Hah, indeed

(It's apparently from Purgatorio, canto 26)

@Fargle For real, I know I need to stop, I don't know how, I want to know how to stop, I want to play, I don't want to waste time, I want to do Math, I don't want to be gamer-forever, I want to live outside the flat land of Minecraft, I don't want to live without some recreation time, I want Minecraft to die, I don't want to see it dying. Holy crap, that's .. complicated.

uninstall n00b

Self-control is tough.

delete game files

break disk

etc

If I had good advice I'd give it.

@Semiclassical I feel better now, in some sense.

hack so your acct gets banned

lol

lol

just find an rp server without a whitelist and grief like you've never griefed

Then it'll become boring .. and I'll quit genius.

self control is very hard

I've never played Minecraft, but I had my own over-commitment to a certain online game.

in fact i would say self control is probably NP-hard, though i lack the proof

And I eventually had to take a hard break from that. Changed my password and stepped away for a few months.

@Fargle Problem is I'm op on a server, and I got a lot of friends, even from real life.

semi what game

@Semi was this a game where you're in a world, and people are crafting for the purpose of war?

i can suggest replacing playing games with listening songs, watching films etc. but it's true that the latter won't feel as engaging as playing games if you like taking active part in an environment rather than observing

Nah.

It was/is an old school text RPG. (I still log in from time to time but there's not many people left who I know anymore).

Definitely doing anything to make something less easy to access will help (e.g. uninstalling).

Ooh, that sounds like fun.

lol

I've had my character on there since about the end of my freshman year in high school.

@BalarkaSen That's really good, sadly you can't replace gaming with doing Math, no matter how much you love Math $:($

Which would make it spring of 2003, I think? (Fall of eighth grade was 9/11, so that's how I have to track it back.)

that's because the you need to get your mental faculty to work to find math entertaining

So yeah. I've got a 14-year-old character on a text RPG. :)

i mean, so would you in games or watching movies but you don't need to work so hard

@PVAL-inactive It fails, once I go on holidays, I can't help myself anymore, and I go ahead and install it again, that's the fourth time, lol.

Since all proofs are programs and vice versa, it'd be neat (if 100000% infeasible) to see someone create a first-person shooter that does complicated proofs in disguise.

Well, any first-person shooter with a physics engine is doing a lot of computations.

Not proofs in the math sense, but definitely a whole heck of a lot of math going on.

I just want to see an arXiv paper that just shows some CoD gameplay as an alternative proof to FLT.

first person shooter game which computes homotopy group of spheres? sounds like something the HoTT guys would find entertaining

For my part, I'd like to see a proof by crochet :)

@Semi That sounds way more feasible than my idea, funnily enough.

oh wait, that's already (kinda) a thing: math.cornell.edu/~dwh/papers/crochet/crochet.html

there's a paper analyzing the computational complexity of some final fantasy minigames out there

"shoot that ugly alien, man, that computes $\pi_{137}(S^{68})$"

much more hyperbolic crochet here: pinterest.com/haakaan/crochet-and-math

Hello @Balarka long time no see. How is your sleeping?

'sokay

We solved a problem in class today, and I was a little confused about something the teacher didn't know/want to clarify :

isn't it 3AM right now?

@BalarkaSen Wow, you just used a very hard word not in my dictionary.

I took a mini-nap for an hour

this is a nice one: Two crocheted klein bottles. The right hand one is made by zipping together two Mobius strips

Hey everyone!

Hi, I am not human though.

@Semiclassical I demonstrated that gluing Mobius strips along the boundary gives a Klein bottle using newspaper once

neat.

@BalarkaSen That sounds hard.

Had to sellotape a lot to not turn it into a crumpled mess

Yeah.

$f$ is a function defined on $\Bbb R$, such that $$\forall (x,y)\in \Bbb R^2 : |f(x)-f(y)|\le k |x-y|,$$

Now why on Earth did this lead to proving that $f$ is continuous on $\Bbb R$ ?

$k$ has to be a uniform constant there, which does not depend on $x$ or $y$. Have you tried an epsilon-delta proof?

@BalarkaSen I believe it is cello and not sello?

'swhat i said

:)

I stopped distinguishing between the constants and the variables @BalarkaSen

That's known as Lipschitz continuity, for reference.

With the point being that every Lipschitz-continuous function is continuous in the usual sense (but not vice-versa).

But yes we used the epsilon-delta proof, I want to explore the intuitive meaning behind that.

The picture on the Wikipedia page may help: en.wikipedia.org/wiki/Lipschitz_continuity

The point is if you choose $|x - y|$ small, then $|f(x) - f(y)|$ is small too

Because it's bounded by a constant (uniform!) multiple of $|x - y|$

Uniform ?

$k$ does not depend on $x$ or $y$. For whatever $x$ and $y$ you choose, the $k$ is the same.

That's part of the definition of Lipschitz functions.

@BalarkaSen Thanks $:)$

@Semiclassical Thanks $:)$

That picture makes it fairly intuitive, I think.

@Mahmoud Why did you say your teacher didn't know or clarify if you used the proof?

The teacher told me that it is the case, because it makes the limits exists @JasperLoy :/

@BalarkaSen Ah, I didn't realize that, myself. So it's "$\exists k : \forall x,y$" and not $\forall x,y\;\exists k$?

@Mahmoud Then you should ask further why, why why.

Right-o, @Fargle.

there are also a lot of papers proving that various videogames are turing complete

Thanks. That's a perfect example of something a classmate was asking me about the other day with quantifier order.

I'll have to remember that.

@JasperLoy He didn't want to answer/know what I was asking .. so I gave up

@AlessandroCodenotti Why mathematicians have so much time to spend on games?

@AlessandroCodenotti Magic: the Gathering is a card game that's Turing complete. (And is also my current non-electronic addiction.)

One approach if someone isn't able to tell you more is to ask if there's a particular name for the concept/method.

For instance, the fact that I recognized what you were describing as Lipschitz-continuity meant I could quickly cite Wikipedia.

@Fargle I used to play semi-competitively, now my wallet is happier and so am I

@Fargle You can weaken it to $k$ being constant on an interval around any point you choose. That's called "locally Lipschitz" and those are still continuous.

@JasperLoy There's a certain school of thought in which mathematics itself is a symbol game with no inherent meaning...

@Fargle for a list of weird turing complete things

@AlessandroCodenotti Thanks for the list. What was the last set you played in?

And if your prof had been able to tell you the name of it, you'd have been able to do the same.

@Fargle Sometimes I think so too, all of life is a game

@JasperLoy How very Wittgenstein of you :)

I like that

There are two kinds of people in this chat who criticise their teachers. One kind really knows stuff and the teacher is really stupid. The other kind doesn't know stuff. End of story.

@Semiclassical Yes, but I cannot dream of such stuff, my education system is a translated version of the french system, with no existence of English resources, so asking that wouldn't make sense, until I get here of course $:)$

Well, you can still be strategic. For instance, you did know it was related to continuity.

And if you look at the Wikipedia page for continuity, you will find a discussion of Lipschitz-continuity there.

@Mahmoud French mathematics is very very powerful, like the French loaf.

@JasperLoy The teacher isn't stupid, it's just the education system.

(albeit not very clearly there, alas. so there are limitations on that approach.)

@Fargle innistrad I think? I played legacy though so I didn't pay much attention to the newer expansions

Institutional problems with education are a pain.

@JasperLoy I mean compare French and English Wikipedia articles.

@Mahmoud Hmm best to consult books first if possible.

If at all possible, try to distill the concept down to some key points. That'll make it easier to track down if you look through books/references/google

Admittedly, though, it can be hard to find something when it's just a name associated to it e.g. Lipschitz

@AlessandroCodenotti Oh man, that was a while ago. They recently returned to Innistrad, right after they returned to Zendikar.

There were Eldrazi, and there was much salt.

They should see if Resident Evil is Turing complete, lol.

Ah, Magic.

I never really played myself, but I loved all the lore stuff.

@Semiclassical How advanced is L'Hôsipatal's rule compared to stuff like multiplying by the conjugate, factorizing ..

Well, the latter is algebra and the former is calculus.

@Mahmoud It is very hard to prove, compared to similar results.

So I'd count L'Hopital as a step above the latter.

Okay, I'm lingering too long. later

Rudin has a very short proof that combines several cases all into one.

What's the difference between french and english mathematics ?

@Astyx Language, and whether $0\in \Bbb N$ or not, come on it's Mathematics !

But globally not much differs

oops lol.

There is this problem too : $$f : \begin{cases} \frac{3x^2-x-a}{x-1} & , x\not= 1 \\ b , x=1 \end{cases}$$ Prove that $a=2$ and find $b$.

Now something strange happened in the ''proof'' we did,

$\lim_{x\to 1} \frac{3x^2-x-a}{x-1}=''\frac{2-a}{0}''$

If $a\not= 2$ then $\lim_{x\to 1}=\infty$, but that's a contradiction because that limit equals $b$, so we must have $a=2$.

But how did we get to talk about an undefined form ?

Even putting $"$ around it, doesn't fix the nonsense.

The fact that the proof actually worked is what I like to describe as Mathematical Black Magic, no ; proof is not a synonym of truth.

To me the visual "multivariate calc / diff geo." proof of L'hosp. is the most illuminating.

If one looks at the the parameterization$ t \mapsto ((f(t),g(t))$ (say with f and g approaching 0 as t approaching 0 though by substituting 1/t for t you can do all the other cases). Then g(t)/f(t) is the ratio of y/x and g

g'(t)/f'(t) is just an approximation of that using the derivative

i guess you are not talking to me?

i literally have no idea what to do.... would be glad if someone could help

(You will also need to write $\frac{1}{x^2\sqrt{1-x^2}}$ in terms of $t$)

uhm.. $t$ in terms of $x$ is $1/x$ :D

not what I asked

compute $\frac{dx}{dt}=~?$

i think i dont know important things. i thought, dx is just the statement at the end of the integral which says we have to integrate about x...

@T_01 If $x=\frac{1}{t}$ then what is $\frac{dx}{dt}$?

$(d/t)/(dt)$ ?...

what is the derivative of 1/t with respect to t

ln(t)

no

no, that's an antiderivative

awit

okay so what does dx/dt actually mean?..

yes. $\frac{dx}{dt}=-\frac{1}{t^2}$

which means $dx=-\frac{dt}{t^2}$.

what does this "d[something]" mean? i thought we could'nt "do something" with it

sounds like Leibniz notation isn't something you've been taught to use

nope...

It's kind of a "differential form"

easier to explain what dx/dt means than dx or dt alone

okay.. thank you

consider the following (await my typing)

In that case, try writing the substitution rule in the specific case of $u=g(x)=1/x$.

Once arctic finishes his spiel, you should be able to see how the two notations are related

but what is $f(x)$ then?.. $\frac{1}{x\sqrt{1+x^2}}$ or what?

Consider a function $u:[0,1]\to[0,1]$. You can think of this as a kind of transformation of the unit interval that stretches it and shrinks it in various places. The effect on the graph, $y=f(t)$ versus $y=f(u(t))$ is that it stretches and shrinks the graph horizontally in various places.

So integrating $f(t)$ and $f(u(t))$ will give different areas, because the area in the second region has been augmented. To fix this, we need to "weight" the heights in the graph that we're integrating according to just how much $u(t)$ is stretching and compressing horizontally, namely by $u'(t)$, which is intuitively why $$\int_0^1 f(t)\,{\rm d}t=\int_0^1 f(u(t))\, u'(t){\rm d}t$$

It'll be easier to see what $f(x)$ has to be once you've done the g(x) part.

@T_01 I thought you said you were integrating $\frac{1}{x^2\sqrt{1-x^2}}$?

yes

So now rewrite that in terms of $t$.

t is 1/x

not what I asked

yeah wait

Rewrite $\frac{1}{x^2\sqrt{1-x^2}}$ in terms of $t$.

u want a term with x's or with t's ?

so a term with t's okay, wait a moment

sorry

$t^2\frac{1}{\sqrt{1-\frac{1}{t^2}}}$ maybe.. like this? forgot the root wait

yes

what did i do formally?

okay i just put 1/t for x, or not?

so now you can use substitution to replace $\frac{1}{x^2\sqrt{1-x^2}}$ with an expression involving $t$ and replace $dx$ with an expression involving $t$ and $dt$. do this inside $\int \frac{1}{x^2\sqrt{1-x^2}}dx$

Hi tern and @T_01.

hello.. ^^

hello

Hi @Ted!

Heya @Fargle!

How goes it?

I'm limping along, thanks :) How're you?

Hi @TedShifrin

Hi @Semiclassic

i dont understand... how i get from dx to dt...

@TedShifrin I'm back from the dead.

Wow, @Julian. I actually wondered what had become of you!

@TedShifrin Alright. Bored. I had some proofs HW for real analysis, but it was just "prove that every non-empty set of integers which is bounded above has a maximal member", plus the usual induction proofs (i.e. $(1+r)^n \leq 1+nr$, Gauss's sum, etc.)

There are many things going on in my life

May I pm you?

Blah @Fargle

Sure, @Julian.

My thoughts exactly. I fear measure theory is gonna hit my classmates like a truck.

And maybe me too.

There's a long conversation I had on here re: using the substitution rule. Lemme find it

Hey @TedShifrin relatively long time no see $:)$

Hi @Mahmoud ... not that long.

Anyway, I'm off to play games.

Hey @Ted!

i think im just too stupid for substitution.

Bye, @Fargle. Hi @Daminark.

How's it going?

@TedShifrin Please how would you prove that $a=2$, for this function :

You're missing something, @Mahmoud. What condition do you want on $f$?

Knowing that $\lim_{x\to } f(x)=f(1)$

@T_01 You know $\frac{1}{x^2\sqrt{1-x^2}}=\frac{t^2}{\sqrt{1-1/t^2}}$ and $dx=-\frac{dt}{t^2}$. Do the appropriate replacements inside $\int \frac{1}{x^2\sqrt{1-x^2}}dx$. It literally requires no math ability: you just replace one thing with another thing mindlessly.

Ohhhh, you need to say that, @Mahmoud.

Or continuous at 1 @TedShifrin using words $;)$

Rehi @Ted

So what do you need $a$ to be in order for the quotient to have a limit at $1$, @Mahmoud?

then we have $\int{\frac{t^2}{\sqrt{1-1/t^2}}}(-\frac{dt}{t^2})$

perfect

now work that out

but i didnt understand this thing with dt and dx

Rehi @Alessandro

What are you going to do in measure theory @Fargle? I should start studying it again too

we have dx and x = 1/t, and we want to know.. dx with t or what?

A number that'll allow me to factorize and divide by $(x-1)$ ? @TedShifrin

so we want to "fix this deriative" at the end so we dont mess up the integration, as you explained, right?

@T_01 we want to know the relationship between dx and dt, which requires calculating dx/dt

does my sentence go in the right direction?

basically you have $x(t)$ is a function of $t$, and $\int f(x)dx=\int f(x(t))x'(t)dt$, where you replace $f(x)$ by $f(x(t))$ (which will be a function of $t$), and replace $dx$ with $x'(t)dt$.

@Mahmoud: Algebraically, what's the easiest way to figure that out?

hm... okay...

so we had dx there

and now you say we need dx / dt

in order to replace $dx$ with $x'(t)dt$, you need to know $x'(t)$, yes

that's why I told you to calculate $x'(t)$

@TedShifrin I have no idea.

x'(t) is -1/t^2

yep

so we say dx = (-1/t^2)dt ?

What must the value of the numerator be at $x=1$, @Mahmoud?

@T_01 yes

How can we know what it should be ? Does it have to be $0\over 0$ ? @TedShifrin

@arctictern is this correct?

@T_01 yes

Yes, @Mahmoud, or else you couldn't possibly have a limit!

hm.. okay.. and.. now?

@TedShifrin Makes sense, but how do we know that it has to be one of the four undetermined forms ?

I mean it could be some number over $0$ too.

@T_01 now do that integral. simplify the integrand first.

Then you would not have a limit, @Mahmoud.

Why ?

I must now leave.

hm okay..

thank you very much

It will be $\infty$, which won't go well with the fact that $f(1)=b \in R$

@Mahmoud. No, it won't even be $\infty$ (because of left- and right- issues).

Ok @TedShifrin, then we know that $3x^2-x-a=0$ and $x=1$ hence $a=2$.

OK, now finish.

But @TedShifrin My teacher did this,

And quoted the undetermined form, which I've never seen before

Blah.

@TedShifrin Is it riogorous to consider that Math ?

It's a sloppy way of saying what I did — that in order for the quotient to have a limit, you need the numerator to approach 0 at $x=1$.

Thanks sir $:)$