If $\omega$ is a nowhere vanishing closed 1-form on a smooth manifold $M$, then can I always find a flow $\psi$ on $M$ such that $\omega({d\psi\over dt}) = 1$?
suppose f: [a,b] -> R be a differentiable function. let a <c<d<b.if f'(c) <0 and f'(d)> 0, then show that there exists a point p in (c, d) such that f'(p)=0.
Any ideas on how to solve this?
@robjohn and @leslietownes Can you please help me with this?
@SoumikMukherjee and @TedShifrin I have not encountered the notion of connected sets but yes, I was able to conclude it because the continuous image of an interval is an interval. Thanks!
connectedness is subtle in general. the real line is about as simple as it gets as the only connected sets are the intervals (taking points, the whole line and the empty set as intervals).
in the context of the principle of inclusion and exclusion, where $p_1=\sum_{i=1}^n P(A_i), p_2=\sum_{i \leq j} P(A_i \cap A_j) ...$ and so on, we have $P(\cup_{i=1}^n A_i)=p_1-p_2+p_3 ..... +(-1)^{(n+1)}p_n$. my question is, my book says it is clear to see that $p_1 \geq p_2 \geq ....$, but im unable to even prove $p_1 \geq p_2$. could anyone help
this question is wrt probability, not cardinality, btw
nick: you maybe want that sum defining p_2 to be over i < j [and similarly for the others]? is there some assumption about the relation between the sets A_1, ..., A_n? there is no general reason why p_2 could not be larger than p_1
If $\omega$ is a nowhere vanishing closed 1-form on a smooth manifold $M$, then can I always find a flow $\psi$ on $M$ such that $\omega({d\psi\over dt}) = 1$?
this a part of an argument to show that $\omega$ makes $M$ a fibered 3-manifold over $S^1$ by map $f:M\to S^1$ by $x\mapsto \int_{x_0}^x\omega$.
the book is Hogg, McKean, introduction to statistics
@leslietownes yeah, that i<j is there, sorry for writing it as $i \leq j$. and yes, it is implied for the others as well. the book doesnt speak about any conditions on the sets, but still asserts this inequality...(see above screenshot)\
nah, that can't be right. if the C_i are all equal to the same set C having nonzero probability, you get scaled binomial coefficients for these p_j's, and there is obvious trouble when k = 4, where p_1 is 4 p(C) while p_2 is 6 p(C).
sometimes authors don't pay as close attention when "remarking" on stuff as they might when stating things that they actually intend to prove in the text.
i found a copy of the book online. "as shown in theorem 1.3.7" is the beginning of an observation that is unrelated to the false statement about the order relation between the p_j. it's just an oddly typeset further remark.
mckean maintains an errata sheet for the most recent version of this text [which contains the error] and this isn't on it. maybe someone should tell him
ooh, for counting measure someone even asked this on main.
If $|A_1 \cup A_2 \cup \ldots \cup A_n| = c_1 - c_2 + \ldots + (-1)^n c_n$,
where $c_i$ is the sum of the sizes of all of the intersections of $i$ sets at a time (inclusion-exclusion principle);
i.e
$c_1 = |A_1| + |A_2| + \ldots + |A_n|$,
$c_2 = |A_1 \cap A_2| + |A_1 \cap A_3| + |A_2 \cap A_3| +...
@onepotatotwopotato given a point $p$, you can find coordinates $x_1,\dotsc,x_n$ on a nbhd of $p$ s.t. $\omega=\sum_{i=1}^na^idx_i$ for smooth functions $a^1,\dotsc,a^n$ on that nbhd and there is an index $i$ s.t. $a^i$ is nowhere-vanishing on that nbhd. then, take the vector field $\frac{1}{a^i}\frac{\partial}{\partial x_i}$.
what Ted said works too without working locally and in coordinates: pick a Riemannian metric and then this is one of the musical isomorphisms (I always forget which is which)
@onepotatotwopotato that's true for all compact manifolds
@leslietownes Consider $C(X, R)$ with $X$ a topological space, $R$ a topological ring. How much can we extend Stone-Weierstrass theorem? As in, results generalizing choice of $R$, say $R = \mathbb{R}^n$.
I've noticed I only really saw results for $R = \mathbb{R}, \mathbb{C}$ or $\mathbb{H}$ (quaternions)
I don't mean generalizations to very general $R$, just all nice enough $R$, perhaps
say that a topological ring $R$ has Stone-Weierstrass property if for any compact space $X$, any separating (unital) subalgebra $\mathcal{A}\subseteq C(X, R)$ where $C(X, R)$ is equipped with compact-open topology, is dense in $C(X, R)$
$X$ compact Hausdorff, maybe
Or maybe $X$ Hausdorff since I'm using compact-open topology
If $\mathcal{A} = \langle f\rangle \subseteq C(R, R^2)$ where $f(x) = (x, x)$ then $\mathcal{A}$ is a separating subalgebra of $C(R, R^2)$. If $g\in\mathcal{A}$, then $g(R)\subseteq \Delta_R$. Since $\{g\in C(R, R^2) : g(R)\subseteq \Delta_R\}$ is closed in $C(R, R^2)$, for $\mathcal{A}$ to be dense we need to have $R = 0$
So if $R\neq 0$ has Stone-Weierstrass property then $R^2$ doesn't
(with point-wise multiplication)
clearly I only prove trivial theorems
But this helps, because it shows that multi-variable Stone-Weierstrass theorem needs to be stated carefully, as a corollary of the one-dimensional case
Is there an alternative to a Cayley Table which can represent an operation whose image grows (with some known maximum bound) the more it is successively applied? So A operation B operation C has a larger image than D operation E?
Best I could think of so far is a Cayley Hybercube.
But that's maybe of limited utility in > 3 dimensions.
I guess maybe one should abandon the concept of applying each quasi-operation in "A quasi-operation B quasi-operation C" independently of the other?
The example I'm thinking of here is digital sum in base b (en.wikipedia.org/wiki/Digital_sum_in_base_b) operating on inputs in base < b. So if you take a digital sum in base 4 of binary numbers, digits in A quasi-operation B quasi-operation C can go up to 3, but digits in A quasi-operation B can only go up to 2.
That is a poorly written Wiki entry. Inputs in base $<b$? Bizarre. You're definitely mapping one set to another, and repetition doesn't seem to be continuing to use the same function at all.
Yeah, this actually comes up in combinatorial game theory in the optimal strategy for Moore's Nim$_k$ (math.stackexchange.com/questions/4831841/…). The answer is long but the general idea is as I just described in the above message.
I think a multidimensional Cayley table might handle it, but that's obviously not super useful in 4+ dimensions.
@Jakobian interesting, i have not thought about this before. one thing to keep in mind for arbitrary R is that even when R = complex numbers, the "separating subalgebra" hypothesis alone is not enough for density in general (e.g. X = S^1 and 1/z not being a uniform limit of polynomials in z). i don't know what would replace "self adjointness"-like conditions for arbitrary R, but maybe people doing nonselfadjoint stuff over C is already a place to look for inspiration
@leslietownes I've asked a question about example of a topological field with the SW property, and there I defined the subalgebra to additionally be closed under topological field automorphisms
So, for example, in A operation B operation C, the two operations have different Cayley tables, and the Cayley tables also vary depending on grouping even if associative?
Cayley table is used for a group operation. You're just representing a function by an array. But in your case, the array will no longer be square, and then you have to decide how your arranging the rectangular array of outputs linearly to proceed to the next level.
Think of this as a graph of a multivariable function (in some high dimension). The only difference is that your domains are finite and discrete instead of regions in $\Bbb R^k$.
jakobian: have you seen ime.unicamp.br/sites/default/files/pesquisa/relatorios/… and its references? i encountered it through the paper of chernoff et al on stone-weierstrass for "valuable" topological fields [in the world to come, every field is valuable]
Is there some know ways to solve diophantine equations involving terms of the Fibonacci sequence? Eg. $13F(n-1)-8F(n)=0$.
WolframAlpha gives one solution viz n=7 but is there some way to calculate that efficiently on pen and paper? I don't think I can go around putting the value $\frac{phi^n - \psi^n}{\sqrt 5}$ for $F(n)$ and expect to be able to find $n$ from the equation.
after the first few terms, the sequence of fibonacci ratios c_n = f_n/f_{n-1} is strictly decreasing to phi [and it satisfies a recursion, namely c_n = 1 + 1/c_{n-1}], suggesting an approach that is slightly more enlightened than complete brute force
namely, if someone asks me you to solve a f_{n-1} + b f_n = 0 (i.e. c_n = -a/b), first look where -a/b is in relation to phi, and if it's in the realm of possibility (i.e. one of the first few values of c_n, or slightly above phi) compute values of c_n using the recursion until you either hit it, or go below it
kaplansky offers "any non-completable division ring", such as the rationals in the 6-adic topology, as a nonexample
i wasn't directly quoting or expecting to help as much as saying "the author of a person on a paper like this suggested that most topological fields people use are valuable"
he cites kaplansky who gives that example, he also cites bourbaki who may give nonexamples
@TedShifrin So for associativity in a non-trivial sense you'd need at least three input dimensions I suppose, so the Cayley Table discrete surface wouldn't be expressive enough. To test whether Z = A operation (B operation C) = (A operation B) operation C maybe you could express Z in color and A, B, and C in spacial dimensions. The only test I could think of would be to start with the line Z(B) with A and C constant, and then extend the line into a surface in two branches
first by varying A and then varying C on that output, and second by varying C and then varying A on that output. Maybe you get the same colored surface in both branches iff the opration is associative. Or maybe you just always get the same colored surface in both branches I can't really visualize it without simulation.
Hmm, I guess technically my method doesn't produce arbitrary colored surfaces, but colors every unit cube in the bounded input space R^3 cube. I wonder if maybe there is some massaging that allows associativity to be expressed/generalized in terms of Rubik's Cube operations, or maybe some extension of a Rubik's Cube from a hollow into a solid cube. I don't know anything about Rubik's Cube math but it looks like group theory is indeed relevant.
i haven't been following the chat, but "seeing" associativity of a binary operation from its list of outputs/cayley table is informally known to be a difficult and probably unsolved problem. people sometimes publish "shortcuts" or partial results in recreational math journals.
i say "shortcut" in quotes because it'll be stuff like, "print out 8 copies of the cayley table and cut them into strips, now glue the strips in the following way..."
that's one for checking whether a commutative binary operation is associative by hand. here A is a set having n elements, and O is a cayley table.
If a function is differentiable does $ \lim \limits_{\mathbf h \to \mathbf 0} \frac{1}{|h|} (f( \mathbf a+ \mathbf h)-f(\mathbf a)) = (h_1(g(x,y) + h_2(s(x,y)$ $+$ remainder) for some function $g(x,y)$ and $s(x,y)$.
oops I dropped the $\frac{1}{\mathbf h} $ before $(h_1(g(x,y) + h_2(s(x,y)$ $+$ remainder)
You can't write limit on the left and not on the right. If $f$ is differentiable, then, yes the partial derivatives exist, so $$f(a+h)-f(a) = (\partial f/\partial x)(a)h_1 + (\partial f/\partial y)(a)h_2 + \text{remainder},$$ where $\dfrac{\text{remainder}}{|h|} \to 0$.
This is a vague question but consider a function $f: R^n \to R^m$ which is differentiable and its derivative is a $mxn$ matrix $[Df(x)]$. Then when you take the derivative of that $[Df(x)]$ do you think about it like some sort of 3 dimensional $mxnxn$ array kind of like a rank 3 tensor?
@User13114 The derivative at a point is a linear map from $\mathbb{R}^n$ to $\mathbb{R}^m$. The $k$th derivative is a $k$-linear map from $\mathbb{R}^n$ to $\mathbb{R}^m$
so, in essence, we can think of it as a linear map from $\bigotimes_{i=1}^k \mathbb{R}^n$ to $\mathbb{R}^m$
I think you're trying to do something that physicists do, putting that $\mathbb{R}^m$ into the tensor product?
perhaps that's not exactly what a physicist would do..
at least, I doubt they'd put it the same way as me
@leslietownes I've looked at Cauchy-Schwartz recently, and I think $|u\cdot v| = \|u\|\cdot \|v\|$ iff $u, v$ are linearly dependent has some interpretation in terms of convexity of the unit ball?
