Elekitu

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See, that’s what the app is perfect for.

Sounds perfect Wahhhh, I don’t wanna
patricia-taxxon
patricia-taxxon

Strawberry Jam handles difficulty in a very different fashion to Spring Collab. Like, the variance in difficulty within each lobby is so much greater, to the point where the extremities intermingle with each other a little & even overlap in some cases. The red advanced maps are basically expert and vice versa. I think this might be preferable to SC's problem where every new lobby felt like you were wandering somewhere you shouldn't be, SJ has you enter each new lobby decently well prepared for at least the easier stuff.

This comes with some new problems though, like do you remember that part in the video where I talked about how the community corelates speed tech with difficulty & how this creates artificial segregation of mechanics between lobbies? Well, in SJ, the difference between red advanced and green expert is so minimal that the only concrete difference is tech usage. The expert lobby has a lot of maps that are basically just middle of the road advanced maps with the occasional ultra or dream hyper, and also a couple maps that by all accounts should be in grandmaster but they technically never have you do more than one ultra in a row (which would make it a chained ultra) so it only uses "expert tech" i guess.

This tech-focused difficulty categorization is a years-long uncorrected error on the part of the celeste modding community, honestly. Like they must understand that tech use on its own can't denote difficulty, or else they'd have put summit down-side in the beginner lobby.

bubbloquacious
bubbloquacious

So the exponential function is given by

exp x = x^0/0! + x^1/1! + x^2/2! + x^3/3! + ...ALT

which evaluated at a real number x gives you the value , hence the name. There are various ways of extending the above definition, such as to complex numbers, or matrices, or really any structure in which you have multiplication, summation, and division by the values of the factorial function at whatever your standin for the natural numbers is.

For a set A we can do some of these quite naturally. The product of two sets is their Cartesian product, the sum of two sets is their disjoint union. Division and factorial get a little tricky, but in this case they happen to coexist naturally. Given a natural number n, a set that has n! elements may be given by Sym(n), the symmetric group on n points. This is the set of all permutations of {1,...,n}, i.e. invertible functions from {1,...,n} to itself. How do we divide Aⁿ, the set of all n-tuples of elements of A, by Sym(n) in a natural way?

Often when a division-like thing with sets is written like A/E, it is the case that E is an equivalence relation on A. The set of equivalence classes of A under E is then denoted A/E, and called the quotient set of A by E. Another common occurence is when G is a group that acts on A. In this case A/G denotes the set of orbits of elements of A under G. This is a special case of the earlier one, where the equivalence relation is given by 'having the same orbit'. It just so happens that the group Sym(n) acts on naturally on any Aⁿ.

An element of Aⁿ looks like (a[1],a[2],...,a[n]), and a permutation σ: {1,...,n} -> {1,...,n} acts on this tuple by mapping it onto (a[σ(1)],a[σ(2)],...,a[σ(n)]). That is, it changes the order of the entries according to σ. An orbit of such a tuple under the action of Sym(n) is therefore the set of all tuples that have the same elements with multiplicity. We can identify this with the multiset of those elements.

We find that Aⁿ/Sym(n) is the set of all multisubsets of A with exactly n elements with multiplicity. So,

exp A = A^0/Sym(0) + A^1/Sym(1) + A^2/Sym(2) + A^3/Sym(3) + ...ALT

is the set of all finite multisubsets of A. Interestingly, some of the identities that the exponential function satisfies in other contexts still hold. For example, exp ∅ is the set of all finite multisubsets of ∅, so it's {∅}. This is because ∅⁰ has an element, but ∅ does not for any n > 0. In other words, exp 0 = 1 for sets. Additionally, consider exp(AB). Any finite multisubset of AB can be uniquely identified with an ordered pair consisting of a multisubset of A and a multisubset of B. So, exp(A + B) = exp(A) ⨯ exp(B) holds as well.

For A = {∗} being any one point set, the set Aⁿ will always have one element: the n-tuple (∗,...,∗). Sym(n) acts trivially on this, so exp({∗}) = {∅} ⊕ {{∗}} ⊕ {{∗∗}} ⊕ {{∗∗∗}} ⊕ ... may be naturally identified with the set of natural numbers. This is the set equivalent of the real number e.

bubbloquacious

#oh very clever #is there actually an inclusion of the reals into Set that respects this operation though? #or even of the naturals?ALT

Nope! Any nonempty set A will have exp A be infinite, so it does not correspond to any numerical exponentiation operation if you pass to cardinal numbers.

the-real-numbers
the-real-numbers

Statisticians be like:

😳 what if we were two population means $\mu_1, \mu_2$ 😳

😳 and we were from two random samples of the same size $n$ 😳

😳 with sample variance $s_1, s_2$ 😳

😳and $\upsilon = \frac{(s^{2}_{1}/n + s^{2}_{2}/n)^{2}} {(s^{2}_{1}/n)^{2}/(n-1) + (s^{2}_{2}/n)^{2}/(n-1) } $😳

😳 and our means were within $t_{0.995, \upsilon} \frac{(n-1){s^{2}_{1}} + (n-1){s^{2}_{2}}} {2n - 2} \sqrt{2n} $ of each other? 😳

elekitu

What kind of monster writes s^{2}_{1} instead of s_{1}^{2} like God intended