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7:07 PM
Thanks, i posted the prettified question now. math.stackexchange.com/questions/998741/…
 
7:29 PM
What does $(a,b)\cdot [a,b]=|ab|$ mean in number theory?
Can someone spell it and explain me the notation?
Also, I get what modulo means but I don't understand how someone comes up with it in the solved exercises I saw
 
@UserX I'm not sure what that means
@UserX What was the modulo problem?
 
A very nice question to upvote
0
Q: Compute the series $\sum_{n=1}^{\infty} \frac{(H_n-\gamma)^2-\log^2(n)}{n}$

Chris's sisThe nice question by Mr. Oloa A closed form for the series $\sum_{n=1}^{\infty} \frac{H_n^2-(\gamma + \ln n)^2}{n}$ made me think of another question that is similar to that one , namely $$\sum_{n=1}^{\infty} \frac{(H_n-\gamma)^2-\log^2(n)}{n}$$ As in the presented proof there I might think o...

 
@datalava THAT was the first problem. Prove that equality.
 
@UserX I don't understand what that notation is supposed to mean
 
The modulo prob i dont get is prove $2222^{5555}+5555^{2222}\equiv 0(mod 7)$$
 
7:40 PM
@UserX hmm i'm rusty, lemme think about it
 
The solution says; It's true that; $2222\equiv 3(mod7), 5555\equiv 4(mod7)$
I have literally no idea where he found those numbers
 
@UserX $(a,b) = \gcd(a,b)$, and $[a,b] = \operatorname{lcm}(a,b)$.
 
Ohhh that makes sense. Thanks @DanielFischer
 
How is geometry going?
 
I wasn't so rusty at it after all. Checked some past exams, they're hard(really hard tbh) but half are doable
I realised the prob was number theory, which is impossible to cram and understand in a night
 
7:47 PM
@UserX pft I'm having problems
 
@DanielFischer is |ab| the absolute value? Does this equality hold?
@datalava you tell me :P
 
yes and yes
aren't you reading a number theory book? this should all be in there
 
I'm reading preparation notes
I don't have the time to read a whole book
@MikeMiller Can you help me understand that 2222 5555 prob above?
 
@UserX ok i made a calc error
 
@UserX Yes. If $a$ or $b$ are negative, it is not unusual to nevertheless consider only the non-negative representative of the set of all greatest common divisors resp. least common multiples.
 
7:50 PM
$\Large \text{WOW, NO ONE UPVOTED MY QUESTION (YET)!!!}$
 
@UserX There may be a more simple way to do this, but $2222=(2)(11)(101)$ then take each factor modulo $7$ and get $2222 \equiv 24 (mod 7)$ and $24 \equiv 3 (mod 7)$
 
That series is exceptionally amazing!
 
(sorry if my tex is wrong i don't have the plugin)
 
@datalava me neither, I read codes. My question was how did the answerer come up with that congruence relation, not why it's true
 
@UserX oh 'cause if you have $a \equiv b (mod n)$ then $a^{k} \equiv b$ for any int. $k$
@UserX and the congruence of a sum is the sum of the congruences
 
7:57 PM
I deleted my question, I suppose all know to compute $$\sum_{n=1}^{\infty} \frac{(H_n-\gamma)^2-\log^2(n)}{n}$$
 
@UserX The proofs of either of those statements shouldn't be too challenging
 
I dont care about proofs right now
 
@UserX Well then you have the statements. I think they're correct, but as I say I'm rusty
@Sawarnik @ me?
 
How do I find the last two digits of $7^{7^{7^\dots^7}}$(1001 sevens)
Why do these questions go from hard->really hard>what's going on->who am I?
 
@UserX I don't recall..
 
8:01 PM
@UserX Do you know Euler's generalisation of Fermat's theorem?
 
@DanielFischer just reached the chapter
However this exercise is before that chapter
 
@UserX Okay, then you use that $7^4 \equiv 1\pmod{100}$. So you need to find $7^{7^{\dots^7}} \mod 4$ with one $7$ fewer.
Which is easy, since $7 \equiv -1\pmod{4}$, and $7^x$ is always odd. So the exponent is $\equiv 3 \pmod{4}$, and finding the last two digits of $7^3$ is not difficult.
 
How did you conclude on the exponent being 3mod4? @DanielFischer
 
@UserX $7^{\text{odd number}}$ is always $\equiv 3 \pmod{4}$.
 
@DanielFischer How do I know that/conclude similar results for other bases?
 
8:17 PM
Question about probability: Given that player A makes 80% of their shots, and player B makes 50% of there shots, what is the likelihood that player A makes exactly 3 shots, player B? I used $5 \choose 3 (.8)^3(.2)^2$ and $5 \choose 3 (.5)^3(.5)^2$. I'm getting that it's more likely that player B makes exactly 3 shots. Is that because player A is more likely to make more than three, or am I doing something wrong?
 
@UserX Experience.
 
@DanielFischer no tricks at all?
 
@UserX Well, you know that you should reduce the exponent modulo something, and the trick is to find the best "something". Generally, if base and modulus are coprime, you need to reduce the exponent modulo $\lambda(m)$, where $\lambda$ is the Carmichael function [$\varphi(m)$, the totient is what you get by Euler's theorem, but $\lambda(m)$ is better, since $\lambda(m) \mid \varphi(m)$], but if you see that the base has a small period modulo the modulus, you can do even better.
 
Apologies, should be ${5 \choose 3}(.8)^3(.2)^2$ and ${5 \choose 3}(.5)^3(.5)^2$.
 
Anyway, new prob I'm stuck. Let the sequence $a_n=2^n+3^n+6^n-1\forall n\in\Bbb_{n> 0}$. Find all positive integers that are prime to all terms of the sequence. Can you tell me how you think each step? Am I asking too much?
Also what level are these two last problems? High school? @DanielFischer
 
8:26 PM
Is that $2^n + 3^n + 6^n - 1$ or $2^n + 3^n + 6^{n-1}$?
 
@UserX I don't know of any high school that does modular arithmetic
 
@DanielFischer first.
 
@David "what is the likelihood that player A makes exactly 3 shots, player B?" This makes no sense
 
@datalava Player A makes exactly 3 shots. Player B makes exactly three shots?
 
Sorry, only noticed the second
 
8:30 PM
@David Exactly three out of how many?
 
@datalava we theoritically do but every teacher skips it. Mathematical contests include it(they include everything on the book)so now I have to study from scratch to have the slightest chance.
 
@datalava, sorry. Out of five.
 
@UserX If it were $2^n + 3^n + 6^n +1$ rather than $\dotsc 6^n - 1$, it would be easier (for me at least). I don't see the answer to the given one right now.
 
@David ${5 \choose 3}(.8)^3(.2)^2$ and ${5 \choose 3}(.5)^3(.5)^2$ are correctly describing the situation
 
@DanielFischer should I ask it as a question?
Damn this textbook is hard
 
8:35 PM
@datalava, okay, so is my assumption as to why player A is less likely than player B correct? Just want to make sure I'm making correct sense out of it.
 
@David I think your justification makes sense. Think about the shape of the binomial distribution of each situation, if you're familiar with that.
 
@datalava, thanks!
 
@David :)
 
@UserX I don't know. If you really want an answer, that's not a bad idea I guess.
 
I really want an explanation. I'll solve similar questions in 1 day(this is on a preparation textbook for a math contest, maybe I grabbed the wrong level as these questions are way too hard for me)
 
8:59 PM
Oh, I've changed my mind (a bit). I didn't find yet a high school level proof ...
 
@DanielFischer Apparently it was an IMO question
More apparently, I definitely have the wrong textbook, I'm preparing for my locals, IMO is after 3 qualifications given that I pass this one.
 
@robjohn If I am not mistaken, I believe that the weights of the geometric series is $\binom{k+n-1}{n-1}$. This is the number of weak compositions of length n of k.
 
@UserX Then I suggest starting with the Prophitis Ilias before attacking the Mont Blanc.
 
Another awesome question
0
Q: Compute the B-series $\sum_{n=1}^{\infty} \frac{(H_n-\gamma-\log(n))^2}{n}$

Chris's sisHere is a slightly modified version of the series in the previous post Compute the A-series $\sum_{n=1}^{\infty} \frac{(H_n-\gamma)^2-\log^2(n)}{n}$ Find the closed form of $$\sum_{n=1}^{\infty} \frac{(H_n-\gamma-\log(n))^2}{n}$$ Again, I'm looking for an easy solution without using Multiple Ha...

@DanielFischer ^^ (maybe you like that last one and want to answer it)
 
@DanielFischer wait a second how could you ever know that?
 
9:11 PM
@UserX What?
 
Prophitis Ilias
 
@UserX Been there. Started in Kardamili.
 
I live 3km away from that mountain...
 
@UserX Cool, where?
 
Elassona. What were you(a tourist I presume) doing in small Greek villages?
 
9:17 PM
@UserX Don't know that one. An acquaintance of mine has a place in Marathopolis, and he used to have a group of youths there in the summer, which I helped supervise a couple of times. Occasionally, you want to get away and have some fun. So off to Taygetos.
 
Ayo.
@DanielFischer Could you help me with a bit of topology?
 
@Anthony Maybe. Which bit?
 
@DanielFischer You can't have been here... at least not by foot. That's 611 km away.
 
So we have a norm on the bounded Lipschitz functions from some Top. space X into $\mathbb{R}$.
 
@Chris'ssis Upvoted
 
9:21 PM
It's defined as $N(f) = |f| +L(f)$.
 
@UserX No, Marathopolis opposite of Proti, it's near.
 
On some metric space, @Anthony.
 
@Hippalectryon I deleted them. Things that are not appreciated I won't let them on main. I better add them to my book.
 
Crap did I already misspeak.
 
@UserX But we took a car to Kardamili.
 
9:22 PM
I guess I could just link the assignment.
 
Topological spaces are more general, is all.
 
Crap.
Yeah.
Well anyway, $|f|$ is the supremum norm of the function, and $L(f)$ is the Lipschitz constant of f.
 
@Hippalectryon besides that, I think I posted too many things on main in the last period of time.
 
@Hippalectryon LOL, I might write tons of book, $\Large \text{I'm full of ideas!!!:-)}$
 
9:25 PM
Now I need to show that the set $\{ f \in \mathcal{L}_b(X) : N(f) \leq 1 \}$ is totally bounded for the metric on $C(X)$ from the supremum norm when $X$ is compact.
I'm sure I misspoke somewhere.
 
@Anthony Do you know Ascoli?
 
$N$ is the norm?
 
Where $N(f) $ is the norm I described above.
@DanielFischer We learned it, I'm unfamiliar with it though. I am suppose to use it on this assignment.
Should I stare at it?
 
this is analysis rather than topology, and thus much too hard for me.
 
@Chris'ssis Please do :)
 
9:27 PM
@MikeMiller :P
 
But don't forget to tell me as soon as I can get them !!! @Chris'ssis
 
@Hippalectryon Just look at this question
8
Q: A closed form for the series $\sum_{n=1}^{\infty} \frac{H_n^2-(\gamma + \ln n)^2}{n}$

Olivier OloaI have found a closed form for the following new series involving non-linear harmonic numbers. Proposition. $$\sum_{n=1}^{\infty} \dfrac{H_n^2-(\gamma + \ln n)^2}{n} = \dfrac{5}{3}\zeta(3)-\dfrac{2}{3}\gamma^3-2\gamma \gamma_{1}-\gamma_{2} $$ where \begin{align} & H_{n}: =\sum_{k=1}...

@Hippalectryon Do we reallly need so much work to get the asymptotic of $$\sum_{n=1}^N \frac{H_n^2}{n}$$?
 
Hi @DanielFischer do you know why it is that you can only subtract vectors in an affine space and not add them?
 
@Anthony You should read it. And look at the problem statement. And look to and fro a couple of times until you see that Ascoli's theorem solves the problem with one move.
 
I didn't consider using Arzela Ascoli because it's asking for total boundedness, and I don't know how total boundedness relates to uniformly convergent subsequences. Or if that even makes sense to say.
But I'll continue to look.
 
9:30 PM
@Anthony Do the terms compact and sequentially compact ring a bell?
 
@Anthony You don't know how total boundedness relates to compactness?
awww, i'm late
 
@Hippalectryon all we need is the summation by parts and then we're done in one line.
 
@Alex In an affine space, you don't have vectors, you have points. You can add vectors to points, and you can "subtract" points to get a vector. But points and vectors are to be distinguished.
 
Oh I suppose every compact space is totally bounded, and if every sequence has a convergent subsequence it's compact?
Is uniformness required?
 
@Anthony Think s'more and you'll get it. You have the clues you need.
 
9:34 PM
Will do.
Gracias, all.
 
@DanielFischer Okay if you subtract points to get a vector then this vector should not be in the affine space?
 
@Alex Yes. To every affine space $A$, there belongs the vecor space $V$ of translations of $A$. You have the addition $A\times V \to A$, and the subtraction $A\times A \to V$.
 
@MikeM I bet you that there is some strong connection between profinite closures and universal covering spaces.
For example think about the $p$-adics.
 
That statement doesn't make sense to me.
 
@robjohn I do not think my sum is right actually.
 
9:40 PM
You're taking the inverse limit of the inverse system of $\Bbb Z/p^i$s with the pullback maps
 
@BalarkaSen How good is your combinatorics?
 
@DanielFischer thanks
 
But this can also be visualized as taking the inverse limit of the inverse system of cayley graphs of $\Bbb Z/p^i$s with the covering maps!
 
especially weak compositions
 
@Alizter no time now, shoo
 
9:41 PM
shooed
 
Maybe.
I told you I dunno any geometric group theory.
 
@MikeMiller I am doing fundamentals. How does the Cayley graphs of $\Bbb Z/p^i\Bbb Z$ look like?
A circle with $p^i$ equidistributed points. In general isometric to the $p^i$-roots of unities on the unit circle.
 
"The regular $p^i$-gon", but sure.
 
Then realize the covering maps $\Gamma(\Bbb Z/p^{i+1}) \to \Gamma(\Bbb Z/p^i)$ as the maps $x \mapsto x^p$.
 
Punchline?
 
9:46 PM
Take the inverse limit of the inverse system of $\Gamma(\Bbb Z/p^i)$ with the morphisms being pullback of these covering maps. What d'you get?
That's $\mathbf{Z}_p$
There is some strong connection between universal covers and profinite completions.
I tell you, strong connections.
 
Nothing you said has anything to do with universal covers.
As far as I can tell, at least.
 
No, well, I haven't.
But I have something in mind. Too vague. Afraid it'd better be rigorized.
 
Also, what category are you taking the inverse limit in?
 
@MikeMiller pardon? geodesic metric spaces?
the category defined earlier?
 
OK, the fact that you didn't put a $\Gamma$ in front of the p-adics was confusing.
 
9:51 PM
right. well it's a vague idea nonetheless.
i was thinking something along the lines of : Let $(X_i, d_i)$ be a bunch of geodesic metric space,$f_i : X_{i+1} \to X_i$ covering maps you know of. Then take the inverse limit : is it interesting? fun? worth thinking about?
 
@DanielFischer One last thing, do you know what property of topological spaces results in limit points of sets being equivalent to accumulation points of sets?
 
@Alex $T_1$.
 
heh, the mathematics I do at the moment is far too easy ...
 
@Chris'ssis I wish I had that problem.
 
@BalarkaSen What are the morphisms in your category again? I forget.
 
10:00 PM
@Alizter Mathematics is not easy, but the part of mathematics I do ...
 
@Chris'ssis @Chris'ssis Do you remember that combinatoricy sum?
 
@Alizter aha ...
 
@Chris'ssis I managed to evaluate it but I keep getting mistake.
I think I am close though.
It is a weighted geometric series
or a generating function of the weights
 
@MikeMiller OK. $(X, d_X)$ and $(Y, d_Y)$ be geodesic metric spaces. The morphism $f: (X, d_X) \to (Y, d_Y)$ is called a $(K, e)$-qi-embedding for some $K, e \geq 0$ if for all $x_1, x_2 \in X$, $1/K \cdot d_X(x_1, x_2) - e \leq d_Y(f(x_1), f(x_2)) \leq K \cdot d_X(x_1, x_2) + e$.
 
trying to work out the weights though
 
10:02 PM
@Mike The idea being that you ignore finer details and closer looks.
i.e., you vision is "hazy".
 
they are supposed to be the number of ways to write k as a sum of n integers including 0
my head hurts.
 
quasi isometries are qi-embedding with existing inverse which are also qi-embeddings.
 
Howdy
 
For example think of $\Bbb R$ and $\Bbb Z$ with the inherited archimedean metric @Mike
 
In the context of combinatorial mathematics, stars and bars is a graphical aid for deriving certain combinatorial theorems. It was popularized by William Feller in his classic book on probability. It can be used to solve many simple counting problems, such as how many ways there are to put n indistinguishable balls into k distinguishable bins. == Statements of theorems == === Theorem one === For any pair of positive integers n and k, the number of distinct k-tuples of positive integers whose sum is n is given by the binomial coefficient . === Theorem two === For any pair of natural nu...
 
10:04 PM
@BalarkaSen And it's a morphism if your category if there exist appropriate $K,e$?
 
$\Bbb Z$ just sits inside $\Bbb R$ by isometry, so that is one-sided $(1, 0)$-qi embedding.
@MikeMiller Huh?
 
Also, that doesn't work, because $\Bbb Z$ is not a geodesic metric space. :)
 
no it is. with the word metric.
 
@Chris'ssis that contradicts with another wiki article :P
or I read wrong
 
@BalarkaSen Are you conflating $\Bbb Z$ with $\Gamma(\Bbb Z)$ again?
 
10:06 PM
@Alizter I told you how to do it: start with the small cases ... (like 2 variables) You cannot fail!
 
@Chris'ssis I did two and 3 but not this way :P
or maybe that way I was just blind
 
I am browsing and found something in a post that seems interesting to me. I wrote a followup question: math.stackexchange.com/questions/999007/…
 
@Alizter Work on it, don't give up!:-)
 
@MikeMiller Am not.
 
@Chris'ssis I won't give up. This is too interesting.
 
10:07 PM
@BalarkaSen A geodesic metric space is one in which there's a geodesic connecting any two points. $\Bbb Z$ is discrete. All paths are constant.
 
Wait @MikeMiller for that problem I was talking about, I know that I can use Ascoli, but is the uniformity given by Ascoli necessary?
 
@Anthony It is for the right space to be compact.
 
If I have an exponential distribution, and I need to find out how many items fail in a given period of time, how do I go about that? I know how to use it to find out probabilities, but not a probable number.
 
Ah.
Thanks.
 
10:09 PM
Uniform convergence is just convergence in the right space.
 
@Mike ... I confused $\Bbb Z$ and $\Gamma(\Bbb Z)$
 
@MikeMiller But no matter, no matter. qi embeddings work for metric spaces in general.
 
Sure.
 
$\Bbb Z$ and $\Bbb R$ with the inherited metric are quasi isometric.
 
10:11 PM
@Chris'ssis Actually no $\displaystyle \binom{k+n-1}{k}=\binom{k+n-1}{n-1}$
 
One is the map $x \mapsto x$ and another being $x \mapsto \text{int}(x)$.
 
silly me
 
the former is (1, 0)-qi embedding the other is (1, 1/2)-qi embedding
 
Anyway, @BalarkaSen, I'm not convinced by your inverse limit thing. I don't see any reason to believe the limit of $\dots \xrightarrow{x^p} S^1$ should be $\Gamma(\Bbb Z_p)$. It also doesn't help that Cayley graphs are inherently defined with respect to some generating set; what generating set are you using for the group $\Bbb Z_p$?
 
@Chris'ssis but still it should change nothing. Why do I get the wrong answer after using $\sum_{k=n}^\infty \binom{k}{n}x^k=\frac {x^k}{(1-x)^{k-1}}$
which gives me wrong results D: tears hair out
 
10:14 PM
@Alizter I have no idea what you tried to do. How do you tackle the elementary case with 2 variables?
 
double sum ignoring distinct - sum of same
 
@MikeMiller 1) we are not covering $\Bbb S^1$. we are talking of covering spaces of graphs, not circles. it is $\Gamma(\Bbb Z/p \Bbb Z)$ we are covering 2) i am not sure if we get $\Gamma(\mathbf{Z}_p)$. in fact i don't even know if the cayley graph is defined, as the group is uncountable (that is my project at the moment, realizing uncountable groups graph theoretically). but we are getting something while inverse limiting. and it is close to $\mathbf{Z}_p$. its just an idea, a topic, not a claim
 
If I have an exponential distribution, and I need to find out how many items fail in a given period of time, how do I go about that? I know how to use it to find out probabilities, but not an "exact" number.
 
@David What is your sample size?
For example if the probability of flipping a heads is 1/2 then 100 coin flips result in 100 x 1/2 = 50 heads
 
100 elements. $\lambda = \frac{1}{700}$. How many fail in $t = 365$?
 
10:19 PM
@David Lambda is the probability they will fail?
or $\lambda e^{t}$?
or $e^{\lambda t}$?
 
$\lambda e^{-\lambda t}$
 
@Balarka In the category of geodesic metric spaces, your graphs are isomorphic to circles with the appropriate metric. It was shorthand. If you're taking the limit in the category of graphs I don't even know why limits should exist there.
 
@Alizter, sorry, not very fast typing LaTeX.
 
Ok. Just sum over $t, 0, 100$
 
oh ok. so we are really covering $S^1$s by stacks of $S^1$s by maps $x \mapsto x^{p^n}$
 
10:21 PM
and multiply by your sample size no?
 
@MikeMiller I don't know. I believe it exists. I can see it.
 
@Balarka In the category of geodesic metric spaces, your graphs are isomorphic to circles with the appropriate metric. It was shorthand. If you're taking the limit in the category of graphs I don't even know why limits should exist there.
 
or am I talking sillyness
@David is t continuous or discrete should be asked
ignore me, I know nothing
 
@Alizter Well I thought of that, but if the function gives the probability, wouldn't that still just be a probability? And I believe this should fall under discrete.
 
thanks for looking into it however @Mike. i will think about it, though you seem to be at the opinion of getting nothing out of it. maybe, maybe.
 
10:24 PM
Am I missing any factoring possible in $$(x+1)^4+(x-1)^4=16$$?
 
@UserX yes. you are.
 
After failing to complete the square to both sides(RHS didn't want to be a square apparently) I just expanded
@BalarkaSen how can I factor that?
 
$((x+1)^2 + (x-1)^2 - \sqrt{2}(x+1)(x-1))((x+1)^2 + (x-1)^2 + \sqrt{2}(x+1)(x-1))$
:P
 
@UserX $x^4+4x^3+6x^2+4x+1+x^4-4x^3+6x^2-4x+1-16=0\implies x^4+6x^2-7=0$
 
@Alizter that's wrong
 
10:28 PM
@Alizter wrong
ah you beat me to it
 
How many times did that stupid message send?
 
@MikeMiller Twice.
 
Balarka how did you factor it like that lol
 
oops sorry
 
@UserX $a^4 + b^4 = (a^2 + b^2)^2 - (\sqrt{2}ab)^2$
 
10:30 PM
The simplest way should be(now that I already have the solution) $2(x^2+7)(x-1)(x+1)=0$
 
yes
 
which gets you $(x^2+7)(x^2-1)$
Thanks for acknowledging my help @UserX have a good day.
Just simple algebra. No need for fancy tricks. Just expand the brackets.
 
@BalarkaSen I didn't even know an identity for that :P
 
Expand the brackets. You should see that they are alternating. Why does that not ring a bell. You are so brilliant at everything anyway.
 
@UserX you can always make those up
 
10:32 PM
I can see that
 
anything can be factored if you "enlarge" the base field enough
 
Anyway, off to next Q
 
shn
Hi
Does anyone have a solution to this problem
 
It might work in the category of graphs, @Balarka. The functor that sends a graph to its vertex set preserves limits, and the limit of that sequence of sets is uncountable. So the limiting graph will have undoubtably many vertices. I think you'll want a functional version of the Cayley graph to do whatever you want to do, though, which I don't think is possible.
 
shn
if we consider that d_x^S is the mean distance from x to its q nearest pionts in S
?
 
r9m
10:35 PM
@BalarkaSen and let us not forget $a^4+b^4 = a^4+4\left(\frac{b}{\sqrt[4]{4}}\right)^4 = ... $ :P
 
mean distances are very mean
 
@shn If it does not terminate it is not an algorithm no? So trivially it must terminate or else it is not an algorithm.
 
@r9m just a fancy way of writing 1
well yeah
 
shn
Ha Ha Ha. and more seriously ? @Alizter
 
$a^4 + 4b^4$ actually factors
 
r9m
10:36 PM
@BalarkaSen yes !! :P LOL
 
@Mike "So the limiting graph will have undoubtably many vertices" no doubt about that. but you might as well want to add that there would be uncountably many vertices :P
well let's see.
thanks for all your suggestions and help @Mike
 
I don't know that the limit should even exist, I shoild say... but if it does it has $|\Bbb Z_p|$ vertices
 
Right.
 
Can someone check my train of thought on this one? I might be very wrong
Find $a\in \Bbb R$ such that $a(x^2+y^2)+2x+y=r, 2x-y=-r$ has always a soluyion in $\Bbb R$ for all $r \in Bbb R$
So $ax^2+ay^2+4x=0$ but that can't happen for all $x$ no matter the $a$ we choose so there is no $a$ satisfying those conditions?
The "find" is making me nervous, I think that implies I should have found an $a$
 
@UserX For every $r$, you need to find one pair $(x,y)$ such that ...
There is exactly one value of $a$ that does what is wanted.
 
10:50 PM
I can think of $0$ working but I'm not sure
 
Check it.
 
It's always $x=0,y=r$...
 
@DanielFischer You said members of affine spaces are 'points'. How would you define what a 'point' is?
 
@Alex An element of an affine space.
 
@Alex Preferably not. Better specify how the beasts behave.
 
10:53 PM
An affine space is a set along with a transitive free action of a vector space. We call the elements of the set the points of the affine spaxe, iirc.
 
I'm not sure what I'm looking for. I can't think of any good reason why I finished or not the question, let alone correctness
 
@MikeMiller Is "free = fixed-point free"?
 
I see @mike What do you mean by 'transitive free action'?
 
@DanielF Free = no element acts as the identity except $0$. Is that not required?
 
@UserX You see that $a = 0$ works, with $x = 0, y = r$. Now you're done with the problem as stated, but I'd like you to think a little about why $a \neq 0$ cannot work.
@MikeMiller For an affine space, we even have fixed-point free (no element except the identity has a fixed point). I don't know whether that is in this scenario already implied by free.
 
10:58 PM
What I mean by free is $\forall v \forall a\left(R_v a = a$ iff $v=0\right)$.
Is this what you mean by fixed point free?
 
@DanielFischer because we'll get a circle and the line is constructed in such a way we'll always have two points they meet?
 

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