Right, if I had been smarted I'd probably have set $\vec{g}(t) = (x(t), y(t))$, and I'd write $f'_t(x, y, t)$, which would cause no problem. But oh well.
This was an issue in my DE class I ta'd for. Namely for exact equations you look for a solution to $M+Ny'(t)=0$ By looking for a 2-variable function f with partials M and N and then considering $f(t,y(t))$ and integrating both sides.
@BalarkaSen The issue you have is that the expression you wrote is a compostion of functions, so you want to differentiate that and not take a partial of anything.
@Balarka: There is no problem. $f$ is a function of $(x,y,t)$, and $\partial f/\partial t$ makes perfectly good sense. The problem comes from the sloppiness in Leibniz notation with single-variable calculus.
@TedShifrin granted, $\partial f(g(t), t)/\partial t$ makes it clear that we're diffing wrt $t$ of the third coordinate. But still, can be confusing sometimes.
So writing $\partial f/\partial t$ evaluated at $(g(t),t)$ is very different from differentiating the composite function. For the latter, it's a non-partial derivative.
Er, what? I thought I was differentiating the single variable function $f \circ \vec{g} : \Bbb R \to \Bbb R$ at $t$? That's $(f \circ \vec{g})'(t)$, no?
I haven't found either the algebra book or the multivariable book. I did find the linear algebra book, although I don't know what the publishers did about it.
In case @Akiva or anyone else wants to ponder it: Prove or give a counterexample. Suppose $f\colon X\to Y$ is a surjective local homeomorphism. Prove that it is a covering map if $X$ is compact and $Y$ is Hausdorff.
@BalarkaSen Here's an exercise (maybe its in @Ted's book) two 2-variable functions $M$ and $N$ are partial derivatives of a 2-variable function if and only if $M_y=N_x$ or vice versa.
@Ted: You can define differential forms with eg $H^{-1}$ coefficients. Then consider the sequence $\Omega^0_1(U) \to \Omega^1_0(U) \to \Omega^2_{-1}(U)$. The same proof as usual shows that this still calculates (first) de Rham cohomology, hence every $L^2$ function with equal (distribution, ie in $H^{-1}$) derivatives is given as the differential of an $L^2_1$ function.
@TedShifrin $f_t : \Bbb R^2 \to \Bbb R$ is a bunch of $C^1$ functions, $t$ running in $(a, b)$, $C_t$ level sets of $f_t(x) = 0$ and $F(s, t) = f_t(s)$ is $C^2$. Assume envelope $C$ is parametrized by $g : (a, b) \to \Bbb R^2$. Then, as normal of $C_t$ is also normal to $C$, $\nabla f_t(g(t)) \cdot Dg(t) = 0$. But we also know $F(g(t), t) = 0$, so differentiating gives $DF(g(t), t) = \nabla f_t(g(t)) \cdot Dg(t) + \partial F(g(t), t)/\partial t = 0$.
Comparing these two, I have $\partial F(g(t), t)/\partial t = 0$. Thus, the envelope is determined by the equation $F(x, t) = \partial F(x, t)/\partial t = 0$, right?
@Ted: I am unconvinced this can't be made to work, but I see your objection to what I've written above. You probably want $L^1_1$, say. But my original claim is still valid: $L^2$ functions are $L^1_1$ on $\Bbb R^2$.
It's just that my antiderivative is now only guaranteed to be $L^1_2$ instead of something better.
Hi Guys, hope someone can help me with a 1st/2nd year problem.
Consider the linear homogeneious system $x' = A(t) x$ with $A(t)$ continuous on some domain. By our general uniqueness/existence theorem, we know there exists a unique solution to the initial value problem $x' = A(t)x$ with $x(t_0) = x_0$ on an open interval containing $t_0$ (In fact, this open interval is the domain of $A(t)$ but I digress).
Now I get confused when we start talking about fundamental solution sets and linear independence of solutions. The next part in the textbook talks about $n$ solutions of the system $x' = …
In case @Akiva or anyone else wants to ponder it: Prove or give a counterexample. Suppose $f\colon X\to Y$ is a surjective local homeomorphism. Prove that it is a covering map if $X$ is compact and $Y$ is Hausdorff.
@TedShifrin I suppose the compactness of $X$ is required because $X=(0,2)\times\{0\}\cup(1,3)\times\{1\}$ and $Y=(0,3)$, with $f$ being the projection onto the first coordinate, satisfies all other hypotheses but isn't a covering space?
Where $(a,b)$ is the interval $]a,b[$ rather than the point $\langle a,b\rangle$.
@TedShifrin And as for Hausdorffness, let $X$ be $[0,1]$, and $Y$ be the Sierpiński space, and let $f:\begin{cases}x\mapsto0,&x=0\\x\mapsto1,&x\in(0,1]\end{cases}$. (Is there a better way to notate that?)
@JulianRachman Care to share?
(The Sierpiński space is the one whose open sets are $\varnothing$, $\{1\}$, and $\{0,1\}$.)
And again there is no open $U_1$ where $1\in U_1$ and the components of $f^{-1}(U_1)$ map homeomorphically onto $U_1$.
Yes. To put it simply, I got the concept of the proof for my main theorem of my paper verified and now I just need to put everything into words and the main goal of my paper would be completed. If you want me to go into detail of the mathematics, I would not mind.
See that is what I was in the middle of writing up:
I just put in combinatorics because I thought that I would also be considering where my theorem is coming from. However that does not see like the case because now after my generalization of this theorem, I is now widely applicable to so many more areas of mathematics that instead of including all of its applications, I can just put the area solely under category theory
So I mean I will probably go category theory solely but I feel like there are other areas that would be important to consider; however I don't know what those areas are.
I'm assuming "local homeomorphism" means that for every $x\in X$, there is an open neighborhood $U_x$ such that $f|_{U_x}$ is a homeomorphism onto its image.
@Julian: Volunteering at a charter high school (and working with middle-schoolers at the library another evening a week). ... At any rate, look up Newton's law of cooling. Basic example of exponential behavior. :)
@Ted: I can modify my previous argument to show that indeed any distributional differential form - including those in $H^{p,s}_{loc}$ for any $p,s$ - has a distributional antiderivative. The point is that distributional forms (currents) still calculate de Rham cohomology. The real worry is what level of smoothness I can get on this distributional antiderivative. But I think care shows I can always get it in $H^{p,k+1}_{loc}$.
I didn't know you were volunteering now, I thought you got blocked everywhwre. That's cool!
Also that hint is like hella understatement: "In proving that there is a function between a set of functions and a set of functions whose outputs are functions, it may be useful to know when two functions are equal" XD XD
Not wrong, but incomplete. It is possible to draw a polar diagram to identify in which parts of the 4 quadrants function of $\varphi $ is >0 and in which parts <0. — Narasimham19 hours ago
I am having problem solving the following question-
Differenciate $\tan^{(-1)}{(\sqrt{1-x^2}/x)}$ with respect to $\cos^{(-1)}{(2x\sqrt {1-x^2})}$, where $x$ is not $0$.
My attempt -
I took the tan function as $a$ and cos one as $b$.
Now we need ${(da/dx)/(db/dx)} $
So here if I substitute $...
@TedShifrin @AkivaWeinberger Here's a proof that if $p : X \to Y$ is a local homeo, $X$ compact _and Hausdorff_ and $Y$ connected Hausdorff then $p$ is a covering map. Take some $y \in Y$. $p^{-1}(y)$ is a closed discrete set of $X$, which is compact, hence is finite. Let $x_1, \cdots, x_n$ be the preimage points. Choose nbhds $\mathcal{U}_1, \cdots, \mathcal{U}_n$ of $x_i$ so that $p|\mathcal{U}_i$ are homeomorphisms. $\mathcal{U}_i$'s can pairwise intersect, so choose open subsets $U_i \subset \mathcal{U}_i$ which are pairwise disjoint (this is where you need Hausdorff). $p|{U}_i$ are sti…
I do not know what happens when you lift the Hausdorffness condition from $X$. Something bad should happen, as non Hausdorff spaces are dumb.
@AkivaWeinberger Easier, just restrict the universal cover $p : \Bbb R \to S^1$ to $\Bbb R_+ \subset \Bbb R$
@TedShifrin @AkivaWeinberger Here's a counterexample which I think works. Let $X = [-1, 1] \times \{0, 1\}/\sim$ where $(x, 0) \sim (x, 1)$ iff $x \neq 0$ be the interval with two origins. Consider the projection map $p : [-1, 1] \times \{0, 1\} \to [-1, 1]$ defined by $p(x, \{0, 1\}) = x$. It is clear that $x \sim y$ in $[-1, 1] \times \{0, 1\}$ implies $p(x) = p(y)$, so universal property gives me a map $f : X \to [-1, 1]$ (namely, just identify the two origins). $X$ is a quotient of a compact space, hence is compact. $[-1, 1]$ is clearly Hausdorff. I think it is also visually clear than …
@Danu Yikes, you ask troublesome questions. I can't think of one off the top of my head. It is true that if a covering is finite, top space is compact Hausdorff if and only if base space is compact Hausdorff too.
So again, one needs to look for non-Hausdorff examples. NOOO.
I am having problem solving the following question-
Differenciate $\tan^{(-1)}{(\sqrt{1-x^2}/x)}$ with respect to $\cos^{(-1)}{(2x\sqrt {1-x^2})}$, where $x$ is not $0$.
My attempt -
I took the tan function as $a$ and cos one as $b$.
Now we need ${(da/dx)/(db/dx)} $
So here if I substitute $...
