oh ok @arctictern I am just solving AM, so I was wondering I want another text in commutative algebra which is maybe more dense and covers more stuff. So, I was thinking of Eisenbudd.
Hi. Kolmogorov extension theorem (KET) says that finite-dimensional distributions (fdd's) determine the distribution of the process.. now what I'd like to do is to specify conditional (on a discrete random element) fdd's and to say something about existence of the corresponding process. I suppose that the problem itself makes sense and the KET is not directly applicable, right?
$a_{b}(c,d)$, $b \in \textrm{On}$, $c \in K$ where $K$ is anything constructed by this procedure (thus it can take the function a as argument), and $d\in \Bbb{N} \cup \omega \cup S$
$S's$ existence is so that I can step up the right argument d without defining another function. Think of it like -1 in programming
A bit of a problem: given vectors $\mathbf v_1,\mathbf v_2,\dots,\mathbf v_n,\mathbf w\in\Bbb Z^6$, I'm trying to generate $c_1,c_2,\dots,c_n\in\Bbb Z$ with $c_1\mathbf v_1+\dots+c_n\mathbf v_n=\mathbf w$ such that any satisfying combination of the $c_i$ has non-zero probability of being selected. Any ideas on how I'd do this?
@LegionMammal978 Let's say I am actually defining some kind of well ordering to $\Bbb{N}$, so theoretically I can get a bijective mapping to map the smallest element (by set membership) to 0, then next smallest to 1 and so on, hence will be an order preserving bijection
I want $a_0(0,n)$ to be bijected to $n$ where $n \in \Bbb{N}$
So theoretically I can do that by $a_0(0,n+1)=a_0(0,n)^+$ from the base case, where $^+$ is defined to be the successor operation. however I cannot generalise that to a_m for any $m$ as the base case will mean I will have one elemen with multiple successors, which is forbidden by the definition of successors
We saw in class that giving initial conditions to a PDE on a characteristic surface can have disastrous consequences (both uniqueness and existence of a solution can fail), but it should be safe to give them on a surface intersecting a characteristic one, right? (maybe intersecting without ever being tangent or being equal on an open set)
@Semiclassical maybe you can help with the question above?
I really don't get why people say equalities with divergent series have a "nonstandard use of the equals sign" - the equals sign means the same as it always does, it's the sum that's being given a definition different from the usual one (limit of partial sums)
Hello guys could someone please help me with a calculus question? at least give me a hint?
I have an assignment, I'm having difficulties with. Ive tried searching, and asking elsewhere to no avail.
Your help would be much appreciated.
Let the function $f(x)$ be continuous in the interval $[0,1]$. Define $f_n(x) = f(x^n)$ for every natural $n \in \mathbb N$ and real $x \in [0,1]$.
Prove that for every choice of $\alpha \in (0, 1)$, the sequence of functions $f_n$ converges uniformly to the function $f(0)$ after restriction to $[0, \alpha]$.
Let $\Bbb{R}_K$ denote the reals endowed with the $K$-topology; i.e., the topology whose basis is composed of open intervals $(a,b)$ and $(a,b) - K$, where $K = \{\frac{1}{n} ~|~ n \in \Bbb{N}\}$. My question is, does $\Bbb{R}_K$ have the intermediate value property? It seems that it would, since if $f: X \to \Bbb{R}_K$ is continuous, where $X$ is a connected topological space, then $f$ must also be continuous considered as a function from $X$ to $\Bbb{R}$ (reals with standard topology)...
since the $K$-topology is strictly finer than the standard topology. Does this seem right?
A bit of a problem: given vectors $\mathbf v_1,\mathbf v_2,\dots,\mathbf v_n,\mathbf w\in\Bbb Z^6$, I'm trying to generate $c_1,c_2,\dots,c_n\in\Bbb Z$ with $c_1\mathbf v_1+\dots+c_n\mathbf v_n=\mathbf w$ such that any satisfying combination of the $c_i$ has non-zero probability of being selected. Any ideas on how I'd do this?
@user193319 the identity map from your $\Bbb R_k$ to $\Bbb R$ is continuous
so if you have any map $X\to \Bvb R_k$, composing with idenitty will give you a map $X\to\Bbb R$ and mean value theorem is given
@LegionMammal978 that depends on what the $v_1,...,v_n$ are. It may be that there are no solutions to your equation $c_1v_1+...+c_n v_n =w$. However if there are any solutions there are at most countably many. If you can characterise the space of your solutions there are many probability measures on a countable set so that every point has non-zero propability, for example $p(n)=2^{-n}$ on $\Bbb N$.
