In my lecture notes, "locally uniform convergence in probability" is mentioned for discrete time stochastic processes that converge towards an ito integral. I'm not sure if this is just convergence in L^2 with respect to the product of the given probability measure with the lebesgue measure
A quick google search didn't bring up useful definitions
@geocalc33 it actually isn't! I just found the relevant definition. It's defined as "X^n converges locally uniformly in probability if for all t>0, sup_{s in [0,t]} |(X^n)_s - X_s| converges to 0 in probability". The "popular" term for it seems to be "uniformly on compact sets in probability"
Maybe this is complete unrelated - but is there any way to prove it (that $\pi_1(\Sigma)$ acts isometrically on the hyperbolic plane) with using uniformization theorem, and the fact that $\Sigma_i$ is a Riemann surface ?
https://math.stackexchange.com/a/2684915/586524 My notes emphasise that the partials must be continuous. I know that if a real function is differentiable, it is continuous. That being said, is: "If continuous partial derivatives exist at some point a" less redundantly stated as "If partial derivatives exist ..."?
"Assume that u and v have continuous first-order partial derivatives in D and that they satisfy the Cauchy-Riemann equations at some point a in D. Then f is (complex) differentiable at a"
Is continuous here redundant? If they're (u and v) partially differentiable they must be continuous, or does this not follow for partial derivatives?
@Threnody For a more elementary example from real analysis, it's possible to have a function f : R^2 -> R whose partial derivatives exist at a point, but aren't continuous at that point, and this can in fact cause f to fail to be differentiable
Multivariable is a weird case even in real analysis, because even if partial derivatives exist, you don't have to have differentiability---in fact, it's possible for both partial derivatives of f to exist at a point, and yet f is discontinuous at that point
idk there's just a course on p-adic Hodge theory this semester (I probs won't take it, looks like I don't have the prereqs) but idek what Hodge theory is
@Lelouch Fuchsian groups were studied in the late 1800s, certainly before uniformization was proved!! (It was conjectured around the same time, though)
say $f : X \to X$ is topologically transitive, look at the $\Bbb R$-action on $T_f$. Pick any two open subsets $U, V \subset T_f$ intersecting the meridian nontrivially say, so $U \cap X, V \cap X$ are open subsets of $X$, we know by transitivity there exists some $n$ such that $f^n(U \cap X) \cap (V \cap X) \neq \emptyset$.
so there is some $p \in U \cap X$ and $q \in V \cap X$ such that $f^n(p) = q$. That means the $\Bbb R$-orbit of $p$ in $T_f$ contains $q$ as well
so the $\Bbb R$-orbit of $U$ intersects $V$, no? what am I missing
Is an infinite ``descend" of radicals possible in a nontrivial number field? To be more clear: If $\alpha$ is algebraic (and not equal to $0, 1$) over $\mathbb{Q}$, then does there exists infinitely many integers $n$ for which there's $\beta$ algebraic with $\beta^n =\alpha$, and $\mathbb{Q}(\beta) = \mathbb{Q}(\alpha)$ ?
@Alessandro Take the $\Bbb Z$-action on $\{0, 1\}^{\Bbb Z}$ by shift, which is topologically transitive, and consider the subspace of all functions $\Bbb Z \to \{0, 1\}$ which is $n\Bbb Z$-invariant for some $n$. This is dense in $\{0, 1\}^{\Bbb Z}$ so the restriction is also topologically transitive
Take the natural $\Bbb Z$-action on $\{0, 1\}^{\Bbb Z}$ by shift, and consider the subspace $X$ of periodic $2$-sided sequences of some unspecified period length. Clearly the original action is topologically transitive, and $X \subset \{0, 1\}^{\Bbb Z}$ is dense. Thus, the $\Bbb Z$-action on $X$ ...
@BalarkaSen uhh.... Let me be clear: Fix any permutation $\pi$ of $\{0, \cdots, 2^m-1 \}$, and for $x \in (\frac{l}{2^m}, \frac{l+1}{2^m})$, send it to $x - \frac{l + \pi(l)}{2^m}$
that's one function. Now, do it for all permutations and all $m$
@Lelouch After some thought I agree what you say should work in principle. Your group is something horrifying like the rooted automorphism group of the binary $2$-tree, but it should be topologically transitive + finite orbit.
The point is you can do it with just 1 self-homeomorphism of a space, which is what a $\Bbb Z$-action is.
small question: lim n/n = 1 as n approaches \infty, and lim (n/n)= lim (1/n + 1/n + .... + 1/n) = 0 + 0 + 0 + .... + 0 as n approaches \infty. Is this possible?
@user777 you can't swap limits and infinite sum without getting weird results.
essentially what you're doing is writing $1 = \sum_{i \geq 1} a_{j_i}$, where $a_{j_i} = \frac{1}{j}$ if $i \leq j$, otherwise $0$, and taking the limit as $j$ goes to infnity. Swapping inifinite sum and limit can be done in some cases, but not always.
$x^p-x$ has p distinct zeros in $\mathbb{Z}_p$ for p prime. I know this follows easily from Fermat's, but I want to show it using the fact that the ideal generated by a polynomial $p(x)$ in a field is maximal iff $p(x)$ is irreducible. This should be possible, yes?
I understand for all irrational $x$, $f(x)$ will be zero and hence the inequality $|f(x)| \lt \epsilon$ will be satisfied. It's only the rational number that will cause the problems
And he is representing the rational numbers in the form of $p/q$ as $$ 1/2, 1/3, 2/3, 1/4, 3/4 .... $$ But what's the role of $n$ here?
Yes, and for a rational number $p/q$ (in lowest terms), $f(p/q)=1/q$, hence $|f(x)-0|=1/q$ and thus we want to know for which natural numbers $q$, $1/q<\epsilon$
Ok, so the point now is that $1/q<\epsilon$ for all large enough $q$, hence there are only finitely many $q$ with $1/q\ge\epsilon$ and consequently only finitely many $x$ in the interval $[0,1]$ for which $|f(x)-0|\ge\epsilon$. So, in any small enough neighborhood of $a$, there won't be any of these $x$ except possibly $a$ itself.
Not at all. It's perfectly correct to say it's his book.
If you read the preface to the third or fourth editions, I'm in there. I contributed hundreds of problems and (to the fourth edition) some substantial changes to the text.
Yes, I know Strang quite well. Interestingly, when I taught at MIT for two years, I lectured the 350-person multivariable class two times, and one time he was a TA for me :)
I also credited him in my linear algebra book for influencing the way I taught the material and wrote the book.
I think Spivak's book is exceptionally clear and well-written. Which edition do you have? The chapter on limits and $\delta$-$\epsilon$ proofs changed noticeably in the fourth because of my complaints to him.