@Obliv Yes. That is the argument. "Suppose that you have any function $f:\{0,1\}^{\mathbb{N}}$. That function cannot possibly be surjective for [reasons]."
@XanderHenderson Do you happen to know an outline for the incompleteness theorems? Do we define some sort of function or relation that 'connects' axioms to theorems or something
i guess that's for the first incompleteness theorem, idk about the consistency one
Hi, I am unable to post on MathStackExchange. I keep getting the error "Your post cannot be submitted at this time due to the volume of spam and abuse originating from your network. We apologize for any inconvenience.". I haven't tried to post on this forum previously, so I am not sure how did the spam originate. Could someone help pls?
@Obliv Peano Arithmetic (PA) is a formal proof system. Statements and proofs must follow a very strict syntax.
Step 1: Encode sentences in PA as numbers (regardless of if they have a free variable or not). Step 2: Encode proofs in PA as numbers. (These two steps are called "arithmetization.")
Step 3: If x encodes a sentence F with a free variable and y is a number, let sub(x,y) be the number encoding F(y) (i.e. y substituted into the free variable of F). Step 4: Let P(x) be the sentence "the sentence encoded by x (has no free variables and) has no proof."
Step 5: Let n encode the sentence (with a free variable) P(sub(x,x)). Then n is simply a number.
[Small caveat. PA's syntax can't handle functions directly, so we need a workaround: instead of a function sub(x,y), you're going to want a sentence Sub(x,y,z) that's equivalent to sub(x,y)=z. Thus instead of n encoding P(sub(x,x)), it should really encode ∃y,Sub(x,x,y)∧P(y).]
Step 6: Note that sub(n,n) encodes P(sub(n,n)). (The construction of sub(n,n) - which is simply some integer - and the sentence it encodes is called "diagonalization.") Step 7: P(sub(n,n)) is true but unprovable.
(You can also get it indirectly from the halting problem, by considering the program that tests if programs halt by searching for proofs or disproofs of their haltingness.)
Let $X_{(k)}$ denote the $k$th order variable of sample of size $n$, where $X_1,\ldots,X_n$ are the samples with distribution function $F$. I'm working the following problem;
> Problem: Suppose that $F$ is continuous. Compute $P(X_k=X_{(k)},k=1,2,\ldots,n)$, that is, the probability that the orginal, unordered sample is in fact (already) ordered.
If the entries are distinct , this is correct. But duplicates are not explicitely ruled out. In particular , if all entries are equal, the original sample is ordered with probability $1$.
@Peter ok, thanks. I guess I have things left to compute then. Do you see why we require $X_1,\ldots, X_n$ to come from a continuous distribution? I don't see why this is necessary.
@Peter I think it does rule out that $P(X_i=X_j)=0$ for $i\neq j$, but maybe I'm wrong. The way I'd argue is that $$P((X_1,X_2) \in A) = \int_A f(x,y)\, dxdy,$$i.e. $P(X_1=X_2) = \int_{y=x}f(x,y)\, dxdy$. But the set $A=\{(x,y):y=x\}$ is of Lebesgue measure $0$ in $\mathbb R^2$.
Yes, you need to use the Axiom of Choice for this proof. Or rather the Partition Principle, which says that if $A$ can be mapped onto $B$, then $B$ can be injected into $A$. The range of such an injection is this set $C$ you're looking for.
It is currently open whether or not the Partition Princi...
Has activity on math.se declined? My questions never seem to get answers these days, even when I post a bounty... This one seems answerable, but only 58 views and 2 upvotes in a week of bounty. math.stackexchange.com/questions/4947604/…
@psie Here's a follow-up problem. Denote now by $X_{k;n}$ the $k$th order variable from a sample of size $n$, where $X_1,\ldots,X_n$ come from a continuous distribution $F$. Compute $P(X_{k;n}=X_{k;n+1})$, that is, the probability that the $k$th smallest observation is still the $k$th smallest one. Is this simply $1/n$?
Sequences and series has 1500 watchers, and Functional Analysis has 1800. They should have seen it too right? There's nothing special about the last tag is there?
@SoumikMukherjee The above was in reply to Soumik sorry.
Wow, someone went through and down voted every question I had asked on MSE. it was reversed automatically, but I'm really curious who I triggered recently!
@Thorgott yeah I see this. So all the maps preserve $X\sqcup Y$ in this case (or, well, $X\times \{1\}\sqcup Y$ which is the same thing as we identify $X$ with $X\times \{1\}$ in the mapping cylinder)
For a finite dimensional real vector space defining an inner product can be done by just choosing a basis and inner product is then just the sums of corresponding "coefficients" of a representation of vectors in that basis. I figure this cannot really be done in an infinite dimensional vector space (because of infinite sum issue) but still I'm curious whether there is some sort of similar statement at least for function spaces. For example let C[0,1] denote the vector space of continuous func
defined on [0,1]. We have the integral product well defined on this vector space. Is there some sort of basis for C[0,1] which gives rise to this inner product analogously to the finite dimensional case
I think he meant if $\{v_i\}_{i\in I}$ is a basis of $C([0,1])$ and $f(x)=\sum_i a_i v_i$, $g(x)=\sum_j b_j v_j$, then $\langle f,g \rangle =\sum_i a_i b_i$
@ephe The "usual" inner product on $\mathbb{R}^n$ is something like $$ \langle x, y \rangle = \sum_{j=1}^{n} x_j y_j, $$ where the $x_j$ and $y_j$ are the "coefficients" of the basis vectors.
In the way @SineoftheTime indicated, while I was typing.
@ephe for an inner product space $X$, we can find an orthonormal basis $e_i$ so that any $x\in X$ is of the form $x = \sum_{i\in I} x_ie_i$ where the sum is meant to be convergent in the sense of being summable iirc
I was just expecting the inner product to come about in a natural manner from summing these coefficients "in a continuous manner" whatever that means. Well thanks anyway!
@ephe I mean, there are, perhaps, ways of seeing the inner product as a sum of coefficients, e.g. by pfaffing about with the Fourier transform (which allows you to write functions on closed intervals as sums of the basis functions $\sin(nx)$ and $\cos(nx)$, but I am not sure that this is the most "natural" point of view.
as long as you are able to write $x, y$ in terms of this orthonormal basis, this can be e.g. $\sin(nx), \cos(nx)$ times appropriate constants, then you have their inner product
The thing is, I don't think that most people think about summing coefficients in the way you are trying to think about it. An inner product is a function which has certain properties. That it. It is often possible to define an inner product on a vector space without having any idea about what basis might represent that vector space.
Of course, once you have defined an inner product, you can Gram-Schmidt yourself to a basis, and they, I suppose, rewrite the inner product in terms of summing up coefficients over that basis, but it seems kind of backward...
that's why if $x = \sum x_ie_i$ then $x_i$ are sometimes called the Fourier coefficients of $x$, even though they're just some numbers and this is an abstract inner product space
@XanderHenderson Oh no it's not really the way I think about inner product spaces really. My linear algebra course was way way more abstract than this. I was just curious if the coefficients part could somehow be molded into the sum <-> integral heuristic
@ephe If you have a countable basis (e.g. $C[0,1]$ has basis $\{ \sin(nx), \cos(nx)\}$), then it is still possible to write the inner product as an (infinite, countable) sum of coefficients. This is, essentially, the idea behind Parseval (or is it Plancherel...? I can never remember which is which...)).
my question is simple: is there a derivation for the formula for multiple matrix multiplications: ABCDEF... that would be similar to (1) (images above.)
@MahNeh to further emphasize, this is not sufficient context to be at all useful. in particular, you haven't said what $\sigma$ represents nor how matrix elements $A_{i,j}$ are showing up
i can maybe guess what the w_{i,j}^(k)'s and X_{i,j}'s are supposed to be but you can't just give two formulas containing entirely different elements and expect us to guess how they're related
it's pretty common in studying "dynamics" to want to consider long stacks of linear maps composed one after the other, sometimes in a 'time dependent' way (i.e. different maps at each multiplication) and sometimes when applying other maps on top of the linear maps (maybe what sigma is) and also not for that kind of stuff to simplify very much
or be at all easy to analyze
so if something nice is going on here it is because these maps are somehow 'choice' and related in ways that we cannot infer from the symbols used for them alone
i and a lot of other people used to use sigma for something that took values in the scalars or at least the center of an operator algebra of interest, of which tr would be an example
if you go to the H=2 case, tho, this becomes (discarding q) $Y=\sigma(W_2^\top \sigma(W_1^\top X))$
which only makes sense if $\sigma(W_1^\top X)$ is also a matrix
so you'd need $\sigma(W_1^\top X)=\sum_{ij} X_{ij}A_{ij}w^{(1)}_{ij}$ and also $$\sigma(W_2^\top \sigma(W_1^\top X))=\sum_{ij} X_{ij}A_{ij}w^{(1)}_{ij}w^{(2)}_{ij}$$
Problem: Consider $X$, a metric space and $I=[0,1]$. Let $C(I,X)$ denote the set of continuous paths in $X$ with the compact open topology. Assume $A\subseteq C(I,Y)$ and that are given a continuous function $\delta:X\rightarrow (0,\infty)$ with:
For each $f\in A$, there exists a constant $r_{f}>...
