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08:18
@psie the binomial coefficients here are $\binom{m}{k}\binom{m-k}{n-k}$.
yeah, make some kind of substitution, $l=n-k$?
You can. This is just binomial theorem then
I get $$\sum_{l=0}^{m-k}\binom{l+k}{k}\binom{m}{l+k}p_2^kq_2^lp_1^{l+k}q_1^{m-l-k}.$$
No you don't
$\binom{m}{k}\sum_{l=0}^{m-k} \binom{m-k}{l} ...$
@psie ah ok, so what I wrote here was actually incorrect?
08:33
No...
ok, it's early, haven't slept well :D
You just want to replace binomials by the other binomials and then observe you can substitute
You could do this in different order but then we'd have to think about how to replace those binomials after substitution so why did we think of how to replace them in the first place. I just don't think this is necessary and it adds work
ok
09:01
Anyone know any chat rooms for Machine Learning stuff? Coudlnt find any on their stack exchange
The last message was posted 568 days ago.
RIP
Seems like people just keep on piling unanswered questions there
09:19
I guess the AI fever hasn't reached there, yet.
they were both closed on Area 51
Artificial Intelligenceai.stackexchange.com

Launched Q&A site for conceptual questions about life and challenges in a world where "cognitive" functions can be mimicked in purely digital environment.

 
1 hour later…
10:56
@Thorgott If $S\subseteq R$ is a subring of ring $R$, then whats the most common notation for $\{xy^{-1} : y\in S\text{ is invertible in }R, x\in S\}$?
Also @LukasHeger
theres this problem: Let E be the set of all $x \in [0,1]$ such that $x$ contains only 4 and 7 in its decimal expansion. is E closed? My argument for it being not closed is that, if i consider the set of all decimal expansions without 4 and 7, and i consider a point p in this, the counter argument would require there exist an $\varepsilon_p$ such that $(p-\varepsilon_p,p+\varepsilon_p)$ would contain no 4 and 7 in decimal expansion
but isnt it possible for large enough decimal place, changing a particular digit occurence to 4 or 7 whilst staying inside the $\varepsilon_p$ ball of p?
@nickbros123 this is the Cantor set
oh wait, i think i blundered, the complement of the set mentioned above would actually be, "not all digits are 4 and 7"
so, ig forget it
@Jakobian how?
@nickbros123 thats not entirely correct
Digit expansions aren't unique
@nickbros123 it just is
11:13
@Jakobian there is only one outlier roght?
right?
@nickbros123 huh. No
Oh wait
In this case the exapsnions will be unique
I was just being wrong
atleast for the 4 and 7 case, it is unique. yu were worrying about the continued 000 s?
No I just wrongly assumed that repeating the same digit infinitely many times always leads to two different digit expansions
@Jakobian why is this wrong? for eg, 0.3 vs 0.29999 are different no?
Anyway the homeomorphism here given some digital expansion using 4's and 7's just maps 4 to 0 and 7 to 2
In ternary expansions
Alternatively you can consider the quotient map from {0, 1, ..., 9}^ω to [0, 1]
By restricting to the preimage of numbers whose expansion contains only 4's and 7's we obtain a bijective quotient map i.e. homeomorphism between our set and {4, 7}^ω
11:22
in the definition using induction given by rudin it goes like this: find largest integer n0 $\leq$ x, then find n1 such that n0+n1\10 $\leq$ x and so on, and x shall be the supremum of the set formed by these, and the "expansion" is the terms of the series. so by this definition, 1 has a unique representation no? which would be only 1.0000 and not 0.99999 ?
@nickbros123 Rudin's definition is just a work around
By this definition certain sequences of digits don't correspond to any numbers
In reality digital expansions are given by series
They're not unique and this is reflected by the map:
8 mins ago, by Jakobian
Alternatively you can consider the quotient map from {0, 1, ..., 9}^ω to [0, 1]
Not being injective
If we name this map $f$, then digit expansion of $t$ is just an element $x$ such that $f(x) = t$
Cantor set has really nice interpretation by using products of discrete spaces
You just need to know some topology
the definition of cantor set i know is trisecting [0,1], tossing the middle one and doing the same for the resulting sets
All you need to recover standard interpretation is continuity of the map which maps the product into [0, 1]
And knowledge of some topology
@nickbros123 well I guess my definition assumes you know its the same as numbers with ternary expansion using 0's and 2's
So this is perhaps something non-trivial that this construction leads to the same thing
But other than that its all just topology facts
11:38
@Jakobian you mean through the homeomorphism or something? cuz there definitely must be numbers above 0.6 in the cantor set, but that cant be achieved using 0 and 2 decimal expansions?
ok wait
stupid question
@nickbros123 ternary expansions
Not decimal
@Jakobian yeah yeah got it , this actually makes sense.
Yeah. Removing middle third corresponds to remo ing numbers which can't be written without first digit $1$ in ternary
And so on, second, third, ...
The map {0, 1, 2}^ω of ternary digit expansions is continuous all the same
And topology argument shows itshomeomorphic to {0, 2}^ω, the same one as above
And this is homeomorphic to {0, 1}^ω by simple bijection of {0, 2} and {0, 1}
I can give you reference to all those facts, I think this is the simplest proof
Not now though
It avoids unnecessary "analysis"
Which analysis, in my opinion, is hardly, rather manipulation of strings
proof that this is the cantor set? I am already working on it!
Oh so you know products, Tychonoff theorem, topological spaces?
11:49
oh wait no
i meant that this ternary expansions without 1 is the trisection thing, nothing much
Well. I guess you need to do this "by hand" then
I wish you had all the tools I do and see how simple this is
ill get there one day
 
3 hours later…
14:24
@Jakobian I've never done something like that. I suppose $S\cdot(S\cap R^{\times})^{-1}$ works and is universally understood, but it's not all that short.
 
4 hours later…
18:52
Niemann humiliated by GothamChess wants to quit chess
Niemann played Caro-Kann, GothamChess is an expert in Caro-Kann, he has been playing it since he was a child, Niemann wanted to humiliate him, he wanted to prove that he knows how to play the Caro-Kann opening
better than him and that he is superior
What do you think ?
is the power set of any set considered a topology?
19:20
a trivial one
Niemann certainly does not engender warm fuzzy opinions.
obliv: if you mean, would it satisfy the axioms of a topology, the answer is clearly yes. but is it considered a particularly interesting topology, or a fruitful place for applying theorems about topology? in most contexts, probably not.
so many chess players in chat
20:10
@Obliv What do you mean by "considered"? A thing either is, or is not, a topology. In this case, the power set IS a topology---it satisfied the definition of a topology.
@SineoftheTime O⁠_⁠o
Has anyone watched Bungo stray dogs?
21:14
en.wikipedia.org/wiki/Discrete_space is a great example of wikipedia having some pages that are just alarmingly low quality. this is bad. even the worst textbook would not subject you to this
just random spew "about" a topic
ordinarily, obliv, i would just drop in a wikipedia link, but not today
:)
21:26
@leslietownes that is BAD.
GPT might, possibly, be better.

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