Not aware of one. Also, the notation above is different to what I am used to, I would write the derivative of $x \mapsto a^T x$ as the mapping $h \mapsto a^T h$, or more usually in matrix notation, $a^T$.
Let $f:\Bbb C\to\overline{\Bbb C}$ be a meromorphic function, such that all its poles are simple with integral residues. Then there exists a meromorpihc function $h:\Bbb C\to\overline{\Bbb C}$ with $f(z) = h'(z)/h(z)$. This problem is really neat.
@dezdichado gives the only if direction. If $\prod_{n=1}^\infty \left(1-{z\over a_n}\right) = :f$ is entire, then taking logarithmic derivative we get
$${f'(z)\over f(z)} = \sum_{n=1}^\infty{-{1\over a_n}\over \left(1-{z\over a_n}\right)} = \sum_{n=1}^\infty {1\over z-a_n}.$$
Since the LHS is mer...
I wrote the solution. Can someone please check if it's a correct arguement?
@Jakobian Actually that's the part I'm worring about. First, if $z = 0$ is a pole then it implies $a_n =0$ is allowed but then the infinite product doesn't make sense because it contains $z/a_n$. For the absolute and uniform convergence, since the given meromorphic function is analytic at $0$, I can choose a small nbd near $0$. I think maximum modulus principle implies the convergence.
i am allergic to watching videos. ubpdqnmathematica.wordpress.com/2021/12/05/… relates the problem to a pell equation and gives a continued fraction without saying too much how its convergents have anything to do with the problem.
Prove that for any given prime $p \ge 5$, there doesn't exist any matrix of size $p\times p : [x_{i,j} | i,j\in \mathbb{Z}, 0\le i,j \le p-1]$, such that
$\forall i_0, \exists j_1, j_2 (x_{i_0,j_1} = x_{i_0,j_2})$, i.e., for any row $i_0$, there are $\le p-1$ distinct elements in it
$\forall j_0...
For $g$ analytic near $0$ we have a linear map between $g_t(x)$ and $\sum_{n\ge 2} (g_t(n^{-s})-g(0))$ and you can find one from the other, ie. (restricted to the correct space of Dirichlet series) there is an inverse which is linear as well.
Since we can essentially recover a zeta function from $g_t(x),$ can we generalise $g_t(x)$ itself and then try to connect its generalization to a higher dim. zeta function?
and by generalizing $g_t(x)$ I mean generalizing it from a collection of curves to a collection of surfaces, and maybe having a linear map from each surface to a higher dim. zeta function
and maybe if the collection of surfaces, the disjoint union, forms a geometric structure like a riemannian manifold, you could use techniques from geometry and prove things about the manifold, that may translate to proving things about zeta functions
@XanderHenderson Thanks Xander. I click on the link but how to I make it active? I did create a bookmark and click on startChatJax, but I don't see things being rendered on this page (on my end)
I have read a few many decades ago but it was mostly equations and I knew the general area. I really liked the old style of Soviet writing in the continuous parameter optimisation arena.
@copper.hat ah but does this proof for the first statement makes sence? $E(X_kZ)=E(X_1Z)$ is always true since the $X_i$'s are identically distributed. But since this holds for all $Z$ measurable the equality I wrote tells me exaclty that $E(X_k|X_1)=X_1$. Is this ture?
But Is my proof above about this statement $E(X_k|X_1)=X_k$ wrong? I use the the definition that if $B\subset A$ is a subalgebra and $X\in L^1$ then $E(X|B)$ is the unique random variable such that for all $Z\in L^\infty$ $E(XZ)=E(E(X|B)Z)$
@Overtherainbow No, note that $E X_k$ is $\sigma(X_1)$ measurable and $\int_A (E X_k) dP = \int_A X_k dP$ for all $A$ that are $\sigma(X_1)$ measurable (the last is the catch).
That is what in probability one means by saying this process is a version of a specific random variable $Y$. Sometimes one uses version to denote a process or r.v. that has certain properties, i.e. a version of Brownian motion. In the context of $E[X|\mathcal{A}]$ usually one means a random variable $Z$ (defined in the same space as $X$) that has the properties that you quoted above. Of course $Y$ is not unique but all versions differ only by a set of measure $0$. You know this any way...
@copper.hat is this because $E(X_k)$ is a constant (so \sigma(X_1) measurable) and $X_k$ not, in particular you can't find a measurable function such that $X_k=f(X_1)$?
@Overtherainbow A nice characterisation is that if $X,Y$ are rvs. then $Y$ can be written as a Borel function of $X$ (that is, $Y=f(X)$) iff $\sigma(Y) \subset \sigma(X)$.
@Overtherainbow I would write Suppose $A \in \sigma(X_1)$, then $X_k, 1_A$ are independent and so $\int_A X_k dP = E[X_k \cdot 1_A] = EX_k\ E 1_A= \int_A (E X_K) dP$.
Yep, I think Radon Nikodym should be taught in Kindergarten.
@Overtherainbow I am not a huge fan of Durrett's writing in PT&E (a little too concise for my liking), but the part on Conditional Expectation in the Martingales section is a nice summary.
A colleague from a long time ago argued that Radon Nikodym should be taught in undergrad as the definition of conditional expectation. It came up in one of my many rants to @leslietownes
@Overtherainbow There is a 'geometric' (meaning $L^2$) interpretation of all of this stuff that may help with intuition, personally I found it unsatisfactory.
@Overtherainbow Personally it seemed a bit contrived. I think that looking at the 'normal' notion of conditional expectation is a more fruitful perspective.
@copper.hat Yes I had the same thoughts, it's nice to see that there is such a way to think about it but I mean one really needs to understand this definition and work properly with all this "ugly" stuff
I do not understand why lecturers & books do not focus more on what it means to be measurable with respect to a function. Doob is the only well known text I am aware of that deals with it.
I'm back. Having read a bit on approximation theory just now, I can say that I'm looking for a way to represent a comonotone approximation with the least information as can be possible according to some arbitrary general representation specification capable of representing more than one object.
