@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

@psie ah ok, so what I wrote here was actually incorrect?

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

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

I guess the AI fever hasn't reached there, yet.

they were both closed on Area 51

@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

@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

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:

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

@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?

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

@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.

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?

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

@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?

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

:)

@leslietownes that is BAD.

GPT might, possibly, be better.