That's better, @Studentmath ... Now prove it :)

@PedroTamaroff LOWUT I'm 5 ft.

I needed to refresh my memory of exponential RV :P

Of course, the way you just said it, @Studentmath, it was tautology.

Told ya, @Pedro. Argentines are not tall.

and Indians are?

50-50

the world is 50-50

They are showing 3 James Bond movies on TV the next 3 nights.

the old ones are ok

I should start a new movie series. I am Loy, Jasper Loy, LOL.

So the movie would be "I am Loy, Japser Loy, LOL"?

LOL

licenced to LOL

I am Parker, Bart Parker, LOL.

@anon: More promising: hyperbolic lengths of $|AB|$, $|AC|$, $|BC|$

That did not come out as expected.

I tried that.

@Ted this is great practice, as my automatic approach to use indexes won't work here, as we are working over continous range

well, the analog works

@ccorn SL2(R) acts 2-transitively on H doesn't it? so we can determine when {i,2i,u} and {i,2i,v} are mapped to each other by a mobius transformation

hmm, not sure if it's 2-transitive actually

think we'll have to do {i,ir,u} and {i,is,v}

@anon You can map any point to $i$, but then there is only one real degree of freedom left. The mapping to i fixes $a^2+c^2$ and $ab+cd$, and we have $ad-bc=1$ anyways.

I suppose one can map any point to $i$ and another one somewhere on the imaginary axis.

@BalarkaSen You can call something bullshit and still study it.

@BalarkaSen That's not true.

LOL

@TedShifrin You totally thought about probability.

huh? @Pedro

@TedShifrin Dang.

@anon Given that the hyperbolic lengths are preserved, once a vertex has been mapped to $i$, there are only two choices (above and below $i$ at an invariant distance) for the next vertex. After that choice, the image of the third point is then determined uniquely.

One could fix the choice for the next vertex to always be above $i$. Then only three possible image sets remain, corresponding to cyclic shifts of $(A,B,C)$ (assuming you wish to keep the orientation of the triangle).

yeah I want to keep the orientation

So it seems $(|AB|,|BC|,|CA|)$, up to rotation, is a suitable encoding. One representative would then be $A=i$, $B$ the hyperbolic length |AB| above $i$, and $C$ to the left (that is, negative real part) where $|BC|$ and $|CA|$ match. (Assuming $A,B,C$ are in counterclockwise order)

@Ted sigh Poisson :)

@Studentmath Poisson = way cool

Indeed it is

I am ashamed it didn't pop up immediately to me...

It actually took me almost 15 minutes to figure that out..

I better catch some sleep. Night!

@Studentmath Good night

This question...seriously?

What's a reasonable time frame to work through Atiyah-MacDonald's Introduction to Commutative Algebra? For those who did work through it, how long did it take?

By "work through" I mean do all the exercises as well.

about a zoggle and a half

math.stackexchange.com/questions/977733/… stupid book!

Math.SE is honored to be the first SE site with a warning about all-caps titles! (At least I haven't gotten it on any other, including SO)

This was the final straw.

@CareBear you're good at pdes right?

@CareBear "Here too" seems to imply it was somewhere else before.

there is a big difference between shouting and a desperate cry for help

also in real life, you can't stop a person from shouting by shouting back at them

"DON'T SHOUT!"

@MikeMiller I tested on SO, SU, AU, and some other site -- no such warning. I think Shog meant "too" as "in addition to the useless words filter".

@CareBear That makes sense - is the useless words filter a new addition?

Yes, October 8

What's next, a filter for WebAssign screenshots?

a "word filter" sounds alot like a step towards censorship

obligatory xkcd 1357

Speaking of bull..., there's a neat web app for monitoring the new answers posted on the site, Blaze. [The API site name is "math", naturally]. I dropped a couple more flags today because of it.

@MikeMiller mi lad

@CareBear Thanks for sharing that :-)

you really do care

Hey bros can someone tell how to do this integral

$\int 3t \sqrt {t^2+1} dt$$

@XNova: Use the substitution $u = t^2 + 1$

You'll get $\frac{3}{2}\int \sqrt{u}\, du$

@XNova: Did you get it?

@Chris'ssis I did get your point but I couldn't continue it since I was sleepy. I've just come from school

@Nick @BalarkaS Hey!

@PedroTamaroff So the open spheres in $(X, d)$ are not open spheres in $(X, \bar{d})$? I was under a different impression.

Oh I see. The open spheres $S_r(x)$ in $(X, d)$ are just $S_r(x)$ in $(X, \bar{d})$ and the open spheres $S_r(x)$ in $(X, \bar{d})$ for $r < 1$ are also $S_r(x)$ in $(X, d)$. My bad.

But potato pohtato. Nothing but a bad terminology.

@PedroTamaroff Hahaha

@Pedro Somewhat similar I found on Simmons : $\prod X_i$ be your space, $(X_i, d_i)$ be metric spaces and define two metric on the product space as $d(\mathbf{x}, \mathbf{y}) = \max \, d_i (x_i, y_i)$ and $\bar{d}(\mathbf{x}, \mathbf{y}) = \sum_{i = 0}^n d_i(x_i, y_i)$. Prove that these are indeed metrics and that $(X, d)$ and $(X, \bar{d})$ are topologically equivalent.

Fun stuff.

@BalarkaSen You didn't reply to a hello! I must inform you can be sentenced to 2 years in prison for this evil act.

$S_r(x) \subset X^\bar{d}$ goes to $S_r(x) \subset X^d$ while $S_r(x) \subset X^d$ goes to $S_{r/n}(x) \subset X^{\bar{d}}$

Oops, sorry, it should be the other way.

@Sawarnik evil & rude

The $\bar{d}$ sphere $S_r(x)$ is in the $d$-sphere $S_r(x)$ as $d(x, y) < \bar{d}(x, y) < r$ and the $d$-sphere $S_r(x)$ is in the $\bar{d}$-sphere $S_{r/n}(x)$ as $\bar{d}(x, y) < n \cdot d(x, y) < n \cdot k$, forcing $k = r/n$

Right. I need more practice.

@IceBoy what evilry did Sawarnik do now?

Oh I see the starred message. Nah, I have him on ignore.

icic

That is an important part of our freedom of choice.

To ignore or not to ignore.

LOL ignoring people?

yep :-)

@Sawarnik: Hi :D

@Nick Hi

@Nick Hey!

@Nick The What If column is quite awesome actually :)

@Sawarnik: I knew that ever since the post about the one mole of moles

@Nick Have you read all of them?

I have a physics question @Nick.

Whatever comes our way, whatever battle we have raging inside us, we always have a choice. My friend @Iceboy taught me that. He chose to be the best of himself. It's the choices that make us who we are, and we can always choose to do what's right.

Well, chemistry.

@BalarkaSen: Potato, Tomahto. Just Shoot!

@Sawarnik: Most yeah. There are a few gems in there but overall most posts are just filler content.

@Nick True :)

but its still fun to read

@BalarkaSen You put me on ignore huh?

:D True dat, Homie. Word to the Hissay on the streets, dwg.

@BakralaSen: Wait... a clone!?

@Nick My book says that copper doesn't react with H2SO4. I can't believe that as CuSO4 copper sulfate is a well-known object and I'd think it forms by H getting replaced bu Cu, similar to the reaction with Zn.

@Nick A duplicate [not exactly] :D

In fact blue vitriol is a a crystal CuSO4, 5H2O

Whaaaaaaa, Balarka is studying chem, with such attention!

@BakralaSen: Do I know you?

@Nick Yes, that's @Sawarnik

@Nick Not face to face though.

@BakralaSen Chemistry is just as much a science as physics.

@Nick delete that please.

I don't wish to be an idol. I barely know any mathematics.

@IceBoy -.-

@BalarkaSen: Copper sulfate is produced industrially by treating copper metal with hot concentrated sulfuric acid or its oxides with dilute sulfuric acid. Your textbook is hence correct.

Guess why.

lol Balarka!

:P :P :P

@Nick What stops that from Cu reacting with H2SO4

OH WAIT. I think I can explain it.

@BalarkaSen: exactly what stops me from exercising. Lack of energy.

H2SO4 splits in ions in diluted form as H+ and SO4--

So for Cu to bond with SO4--, that guy needs to give out some electrons, right?

@Nick :D

But consider the usual battery now. It contains a Cu sheet, a Zn sheet and diluted H2SO4.

if I recall correctly, Cu takes electrons in that process. so we can say that Cu "doesn't want to give its electrons"

@Nick ^

galvanic cell is the word.

i am not good with names, sorry.

@BalarkaSen: You can also call it a voltaic cell

@Nick so I trust that Cu has a strong electronic structure which can be probably explained by bonding and ninja stuffs I don't know?

@BalarkaSen: There is no ninja stuff. What was your doubt again? I think you have a clear grasp of ionization ;/

"Why doesn't Cu react with H2SO4?"

Yes, I am aware of Arrhenius.

this room is empty :-)

@BalarkaSen: he's right. Let's jump in there

OK.

@Nick Oops, I think I have to go so I can't discuss right at this moment. I'll love to say in a couple hours.

Sorry.

:D That's ok but from my knowledge on metal and base reactions. The following should happen

$$Cu + H2SO_4 \to CuSO_4 + SO_2 + H_2 O$$

Why do I think so? Well,

As I said earlier, we need some heat/electricity/energy for this happen.

But again, I would like to emphasize that your textbook was right.

This reaction does not occur in aqueous solutions of dilute sulfuric acid.

So, you can't do this reaction in your lab

The correct reaction with the state symbols included is: $$Cu_{(s)} + H_2SO_{4(l)} \to CuSO_{4(s)} + SO_{2(g)} + H_2O_{(g)} $$

You cannot use $H_2SO_{4(aq)}$ . You need pure concentrated sulfuric acid.

But then if this reaction has requirements of both pure sulfuric acid and of heat energy,how is it economical to produce $CuSO_4$ ?

We simply use $CuO$ instead of $Cu$ metal.

$$CuO_{(s)} + H_2SO_{4 (aq)} \to CuSO_{4 (aq)} + H_2O_{(l)}$$

Also, when you said:

That ionization isn't a one step process.

@BalarkaSen: Think about it. Hopefully, that was helpful.

@IceBoy: Now, can we get onto mathematics? lol

I'm trying to explain ellipses to my little brother.

He has understood an ellipse to be a "wide" circle

@Balara, @Nick is right nod

I'm still trying to figure out if what I manage to prove is significant at all..

At first he couldn't distinguish between a circle and an ellipse but after we got through that, he's now stuck at differentiating between ovals and ellipses.

I'm a horrible teacher.

@Studentmath: What did you manage to prove?

@IceBoy: Any advice for me?

I am trying to prove in an elegant way that a certain component in a graph is of size $\Theta(n^{\frac23})$. I managed to prove it is certainly smaller than $n^{\frac23+\epsilon}$ and larger than $n^{\frac23-\epsilon}$ where $\epsilon$ is any positive constant, as small as you would like as long as it is a constant.

Now obviously if I pick $\epsilon$ to be dependant on $n$ ($1/n$ or $1/n^{\frac{1}{100}}$ for examples), I still ain't where I want to be, but on the other hand $n\to \infty$ so maybe I am really close

@Studentmath: If you are, then always remember: The last lap is the longest lap.

Yeah, I won't manage to prove that small range with my current methods, I believe

That always set my mind right

@Chris'ssis I am able to integrate the last integral but it's very lengthy. I am unwilling to post my own solution

@Studentmath: Yeah, I've already told him that a circle is an ellipse which is an oval but not necessarily the other way around. But he was still confused. Just now, I tried explaining that an oval need not be symmetric about the y-axis and the idea simply clicked :D

an ellipse is to a circle as a rectangle is to a square

lol, funny are the things one knows but never phrases.

I'm going to put that line in a textbook.

:D

Gosh, now, he's contemplating on the relative obesity of ellipses. He doesn't know how to divide yet, so I'm not going to explain eccentricity to him. Hopefully, squeezing an orange will be enough of an explanation.

Catch you on the citrus side, folks.

:D

later pal

@nick

@Nick Ow, age?

@Nick To be honest, I have very much the same idea.

@Sawarnik: A standing $\infty$

So, He's gone to watch Digimon on TV now. Enough geometry for today :D

DIGIMON DIGITAL MONSTERS DIGIMON ARE THE CHAMPIONS!

@MickLH: I only like those Human + Digimon crosses. Which show was that? Do you remember?

Code Lyoko?

@MickLH: lol, no. It was some Digimon: ... show.

@Nick Maybe he is too young to differentiate between them

@Nick :P Why not Doraemon?

coming in 5 mins...

ew. tmi

digimon are cool, but sheesh

@Sawarnik: He's growing. I remember when I moved from my Dora the Explorer to Dragon Tales phase. Then, I was all like CSI and stuff.

@Nick I'm sorry, I believe I am failing you here. My knowledge on the topic is severely abridged.

@MickLH: I'll just Wikipedia it :D

It seems you've got me beat with calculus and chemistry too, can I offer you some electrical engineering or computer programming?

@MickLH: You wouldn't happen to know how to deal with Constructor Overloading, would you now?

Also, I found it. I was thinking of Digimon Frontier.

@Nick Ask him what a circle is.

Good ol' digimon.

@MickLH: Ah, then, is it useful?

@Nick I find it can give you some very pretty APIs

As long as you obey RAII principles it's useful

@Nick Convince him that an ellipse cannot be drawn by a compass.

He is kind of right though.

Ellipse is projectively "a wide circle", yes.

@BalarkaSen: Ah, well, your abv's are XYZ to me. But I think I'll learn off of youtube, thank you very much :D

Abv?

@BalarkaSen: abbreviations. Also, I did the whole thread and pin thing.

good idea^

You need convince him what a circle is, first.

move the pins closer and closer together and observe the shape

Convincing him that the (sum of green to pen and pen to yellow) is constant is hard. Infact, explaining $\pi$ to him is hard.

I should wait till he can divide.

or comprehend ratios.

@IceBoy: That's not me and yes, I did try to illustrate that when the pins coincide, a circle is formed.

@Nick Can he equate areas of squares and stuffs?

@BalarkaSen: Nope. I'll get that eventually.

in eventuality... lol

early exposure is good for him

But you claim he can multiply?

If he just digested a bunch of multiplication tables, he doesn't really know what multiplication is. Try to explain him the geometric aspect of multiplication (i.e., lattice counting on a rectangle)

how old are we talking about?

That's how I taught my cousin to multiply.

First the motivation, then the digestion.

He's a standing $\infty$

@Nick I watched Doraemon religiously till 12 perhaps.

@Sawarnik WAT

[Yay, I m unignored!]

@Sawarnik: I watched FRIENDS monotonously till 2004! I win.

Oh :P

Friends is different than Doraemon!

I was addicted to DBZ. 5-9 perhaps.

@Nic Pedagogy is fun stuff. Some guy asked me in school how to realize discontinuity physically.

@BalarkaSen: I lost track of DBZ the moment they stopped airing episodes of it on CN India. Last I saw, goku was on a holiday from heaven to fight a tournament on earth during which some wild adventure occured.

After a day of thought, I came up with this : Take a stick. Revolve it fixing one end. A point on the stick is displaced that way. Let it be displaced some angle $\theta$. Any point on the stick is displaced angle $\theta$, except the endpoint.

I forget details. To think of it, there weren't many to begin with.

The endpoint is left fixed, i.e., displaced $0^\circ$.

That is, a discontinuity.

@Nick You mean the Majiin Buu saga?

I liked the cell games saga.

@BalarkaSen: No, I remember he tuned into kid buu. Then.. it's a blur.

After kid buu there is nothing much in there.

@BalarkaSen How would you realize division by zero physically?

Goku fights Buu while in SS3 but it turns out that both have almost the same ki. Then he makes up a spirit bomb and destroys Buu forever.

@BalarkaSen: Yeah, I never saw or heard of DBZ after that.

@IceBoy How would you define "division by zero"

1/0? It's not a real number.

@IceBoy: Physically, I would defined the result as NaN. Although, I know mathematically, you like "indeterminate"

You can make sense out of it if you adjoin 1/0 to your real line though. That is, the Riemann sphere.

@BalarkaSen: Does $\infty \in \mathbb R$ ?

I thought the whole point of "undefined" is that is depends on the context and can't be reasoned on by math alone

Yes, but there is no "physical realization".

@Nick Define "$\infty$".

@IceBoy If you define it, there is.

not "physically"

Define it, I'll give you a physical realization.

Nothing / 0 = Universe

1/0 = no thing

@BalarkaSen: Infinity is something over nothing.

@IceBoy: $\lim_{x \to 0+} \frac{1}{x} = \infty$

@IceBoy in the classical sense, m/n is the number of objects every person has after distributing m objects into n people.

ok

@IceBoy If you distribute 10 objects into 10 people, each will get 1 object.

distributing 10 objects into 5 people, each will get 2 objects.

distributing 10 objects into 2 people, each will get 5 objects.

distributing 10 objects into 1 people, each will get 10 objects.

Distribute all your possessions to zero people, you will know you're selfish.

So, by continuous extensions, distributing 10 objects into 0 people, each will get "something very large" object.

But by common sense: 0 objects per person, because the number of objects given out to people is 0

Now, now, you can't confuse common sense with mathematics ;)

@MickLH: But $\require{cancel} 0 \cancel\in \mathbb N$

@BalarkaSen Well I would argue that it's surely possible ;)

Sorry, I assumed we were counting things

But wait. Look at your unit!

$0$ objects per $0$ person

You're saying $\frac{0}{0} = 1$

yeah

I'm distributing 10 object into 0 people, not 0 objects to 10 people.

@Mick ;)

I've thought this whole thing through man get off my case honestly

I see the glitches, do you want a paper on it?

:p

We were discussing pedagogy not actual mathematics.

^Actually you can prove the above

@BalarkaSen: Isn't that right?

@Nick how are you going to prove that?

@BalarkaSen: Ah wait, can you somehow manipulate $\frac{0}{0}$ into $0^0$

I have a proof for the latter. lol

OK, prove that $0^0 = 1$.

how are you going to do it?

@BalarkaSen: $(0+0)^0 = 1$ Hence proved

How is $(0 + 0)^0 = 1$?

Whoops, I meant to say something else entirely. Forget I said that.

@BalarkaSen: Using Binomial theorem, expand $(0+b)^x$

What do you get?

$b^x$.

The logic is that counting the number of objects received by each person during the transaction will show that when distributing to zero people, you have given zero objects to each member of your group of receivers.

@BalarkaSen: :( No. Ok, expand $(a+b)^x$

OK, expanded.

Put a = 0

You'll get $0^0 b^x$

Huh? You have $x$ as an exponent.

$$(a + b)^n = a^n + \binom{n}{1} a^{n-1}b + \binom{n}{2}a^{n-2} b^2 + \cdots + \binom{n}{n-2} a^2 b^{n-2} + \binom{n}{n-1} a b^{n-1} + b^n$$

Set $a = 0$. You'll get $b^n$.

Like I said.

lol, small $x$ and $n$ are horribly confused in $\LaTeX$ sometimes

You are going nowhere with your proof.

It's BS.

$0/0$ is not defined. This is more clearly explained if you look at $\lim_{(x, y) \to (0, 0)} x/y$.

Approach the limit through the parabola $y = x^2$, it's $0$. Approach it through the line $y = x$, it's $1$. Undefined.

$$b^x = (0+b)^x = \sum_{k = 0}^{\infty} \binom{x}{k}0^k b^{x-k} = \binom{x}{0} 0^0 b^x \implies 0^0 = 1$$

No, @Nick. The $\infty$ limit above is wrong.

And how come the $0^0$ appear?

@Nick Utter BS.

What about the rest of the expansion?

Utter bachelor of science?

@BalarkaSen: The rest are defined $0^n = 0 \, , \forall n \in \mathbb N$

@Nick Oh right. But it's still nonsensical.

@BalarkaSen: Ofcourse it's BS but it's playful BS. I like to think about it and why it's wrong.

There is nothing wrong about it. $0^0$ can also be shown to be $0$.

the $0^0 = 1$ definition is useful: easy life with power series

But it's better if it's one

betterexplained.com/articles/…

That's why it's called "indeterminate".

$1/0 = 2/0 = 3/0$ ... So, division by zero also yields indeterminates?

Bah this is nor math.

@Nick No, indefinites.

No, it's blasphemy.

@BalarkaSen: Terminology keeps us sane.

@Chris'ssis theoatmeal.com/comics/running4

@MickLH: You're right but seriously, it is difficult to comprehend how one can take two empty barrels and through mathematics come up with one full one.

lol I was lucky enough to screw off school

so by the time I decided to learn math I skipped all the metaphors and real world imagery

Hi MSE people! I have started a bounty on a question about estimating a large covariance matrix given only few observations of a multivariate normal distribution: Here it is in case you are interested: math.stackexchange.com/questions/945690/…

Anyone can make their imagination into a reality by climbing up the right branch: $$ i = \sqrt{-1} = (-1)^{\frac{1}{2}}= (-1)^{\frac{2}{4}} = \left((-1)^2\right)^{\frac{1}{4}} = 1^\frac{1}{4} = 1$$

Sorry for spamming those who are not interested... Please take a look and help me!

@BalarkaSen: ^ I can prove it using Freshman's dream and a common proof shared among school students.

@BalarkaSen: I can throw no more BS at you.

@Anastasiya-Romanova Very lengthy? Is this a joke? It's pretty fast.

(actually)

:D

Hallo

Hey @PedroTamaroff

hi

@PedroTamaroff: 1-1 3 1_ 1_ 0 !

Huh?

H3LL0 :D !

Oh. Hello

Here's riddle for you from under the hat of the the Mad Hatter: $$\text{" Why is a raven like a writing desk? "}$$

hello, Pedro

Hello @Mike, @IceBoy, @AlecTeal and @Nick.

@Nick hi pal :-)

@Chris'ssis No, it's not. But perhaps I'm missing something. Could you post your own solution to enlighten me, please? Thank you

@Chris'ssis IMO, the better way approach to tackle that integral is using Feynman's trick twice. I tried to solve it using Feynman's trick once, but it turns out the approach becomes too lengthy