@UserX I'm afraid I don't understand what you mean.

How do we get a circle?

Which line?

the first equation represents a circle for a\neq 0. The second a line.

@MikeMiller If I parse that correctly, yes.

But my explanation on why we dont want a=0 is nonsensical to me too

@UserX Aha. More or less. Except - there is something about the first that you should notice.

We can't ensure the circle has a positive radius!

Forget about the $r$ and focus on the function $(x,y) \mapsto a(x^2+y^2) + 2x + y$.

@UserX Yes, one can put it that way.

Okay, I got no other explanation. What's yours?

For $a \neq 0$, the function is never surjective.

Since it's $a(x-\xi)^2 + a(y-\eta)^2 + \gamma$.

So $\gamma$ is the minimum (for $a > 0$) or the maximum (for $a < 0$) attained value.

@DanielF To me, fixed point free would mean that no point is fixed by the action of every element of $V$.

Wait what. You lost me on that last sentence

@MikeMiller Hm, that could also be. Let me have a look.

In fact now you lost me step 1. How is this a function?

My free certainly implies fixed point free.

Ohhh

I can only see this as a function if it has two variables

@MikeMiller Do you understand German? Otherwise I'd have to translate.

Not a chance, @DanielF. What were you going to link?

@UserX Right. And the graph is - for $a\neq 0$ - a paraboloid.

And the other graph is a plane. Now I got lost again, why can't we have a plane and a paraboloid?

@MikeMiller Not linking, the definition of free (more precisely fixed point free) actions of groups on sets in Stöcker/Zieschang (Algebraische Topologie). Which is $\bigl(\forall g\in G\setminus\{1\}\bigr) \bigl(\forall x\in X\bigr)\bigl(gx \neq x\bigr)$.

Quick opinion: someone's (relatively easy) limit question led me to a separate question about convergence of a sequence that I don't know the answer to or even a reasonable approach to; my interest is purely curiosity, but I think it might be a curiosity to others too. Should I go ahead and ask the Q, or (since it's not related to an 'actual problem I currently face') should I just leave it in the back of my brain for now?

So, @Mike, "free = fixed point free". At least by that definition. And "fixed point free" is the more precise (genauer) term according to Stöcker and Zieschang.

(The question, for the record: what is the limiting and/or distribution behavior of the sequence $s_n=\dfrac{n^n}{n!}\bmod 1$? A quick dip into Alpha suggests that it's all over the place but might have some bias toward the endpoints of the interval)

@StevenStadnicki If you can make it a good question, ask.

Okay, did I do this correctly? An 8 character password contains 3 of one letter, 3 of another, 1 or another, and 1 of another letter (8 total, 4 unique). So it would be $26{8 \choose 3} + 25{8 \choose 3} + 24{8 \choose 1} + 23{8 \choose 1} = 3232$ ?

@David WAT.

You want to find all possible combinations?

I guess.

@PedroTamaroff Yes / how many possible passwords exist. Assuming all lowercase.

I take it order matters.

@David Then there are $(3+3+1+1)!/(3!3!1!1!)$ unless I am misreading what you wrote.

That's $8!/3!3!$.

Pedro is right.

@David Maybe you can explain your line of thought.

I was thinking to begin I had 3 of 8 positions that needed to be filled, and there are 26 letters to try. Then I have another 3 positions to fill, but now I only have 25 letters let, then 1 position to fill, but only 24 letters left, and so on.

@David Oh, wait.

I had used Pedro's method when I knew which letters I had. As in, if I knew I had 3 a's, 3 b's, 1 c, and 1 d...

So first you have to choose the letters you'll use.

Right.

What you can do is break the process in two.

First, you must choose four letters, yes?

Out of 26.

This can be done in $\binom {26} 4 $ ways.

Now you must choose the two special letters.

Four letters, two of which have three of each.

This can be done in $\binom 42$ ways.

And now you do $\binom{8}{3,3,1,1}$

So the final answer is the product of those.

Do you agree?

Let me know if I'm missing something.

@Daniel, Your free is my free. My "fixed point free" is $(\forall a \in X)(\exists g \in G)(gx \neq x)$.

Pedro, Should that last one be $\frac{8!}{3!3!1!1!}$?

@David Yes, that's what $\binom{8}{3,3,1,1}$ stands for.

@DanielFischer I just understood what you were talking about...

So I get there are $26 \choose 4$ ways to pick four letters. And I understand that two are special because there are three of two of them, but I'm not understanding how $4 \choose 2$ encapsulates that part.

@David It suffices you choose what the two "special" letters are, i.e. the ones you're going to repeat.

Once you've chosen four letters, you must choose 2 out of those 4, right?

So there are $26 \choose 4$ ways to pick four letters, of those four, $4 \choose 2$ are special, and then you multiply that all by how many possibilities there are once you know what you have?

Right.

Giving 100,464,000 possible combinations?

DAAAAAAAAAAAAYUM. Really?

Yep, that's what it came out to. Of course taking say 10ms to try one password, you get 100 per second and it would take 11.627 days to try them all.

In reality, I think it should be ns though, so 10ns gives 100,000,000 per second, which means your password just got cracked in 1 second. ;)

I'm curious how it would change if we knew one letter, but didn't know how many of that one. Just changed 26 -> 25?

And 4 -> 3

What does this mean; $]0,1[$?

@UserX Umm, somebody needs to learn their left and rights?

I.e. typo in the book?

No, no typo. In Europe they write $]a,b[$ for the open interval.

No confusion with ordered pairs.

Hi @DanielF @Pedro

@TedShifrin isn't $(a,b)$ the open interval?

@TedShifrin France.

Sometimes, often ordered pair. Hence, American notation confuzling.

Not Britain, @Alizter?

If you can't tell from context which is which your paper sucks.

@TedShifrin We very much use (a, b). Same in Germany I think. Ask Daniel.

One reason I use vertical vectors :)

Because I might have a point on a graph at interval 0-1? Or maybe x can only be within point (0,1)? Context is everything. I agree with Alizter.

@TedShifrin Oh I thought you were talking about binomial coefficients ;D

Sometimes context is not 100% clear.

Conte t is 85% clear

smacks @Alizter

(sad music plays as Alizter is wheeled into hospital)

Melodramatic Alizter ...

@TedShifrin How is your combinatorics?

Barely existent.

Could you help me?

Well it is kind of a combinatorial series

Better to ask @Pedro or @Kaj?

ok

@Pedro

Can you help me?

You should post your question, @Alizter

@TedShifrin noo

Too scary

why?

It's a Halloween question?

I also like the one to one

I meant to post it in here

oh