there is an eventual transition over to lesliecoin in a promised future, but for complicated reasons, campaign contributions will only be accepted in US dollars.
Consider $M \times \Bbb R$ where $M:=\big\lbrace \log^2 x +\log^2 y=1 : (x,y)\in \Bbb R^2_{\gt 0}\big\rbrace.$ Consider intersecting two of these surfaces at right angles. Then I want to show that the intersection curves either lie on a plane or on a hyperbolic sheet $N\times \Bbb R$ where $N:=\big\lbrace xy=1: (x,y) \in \Bbb R^2_{\gt 0}\big\rbrace.$
By this point, you should be more skilled at asking questions that are well-defined.
So you have a "logarithmic cylinder" parallel to the $z$-axis in $\Bbb R^3$ and another parallel to the $x$-axis. ....
Your hyperbolic sheet is indeed a hyperbolic cylinder.
So why are you asking this question instead of solving it yourself? Also, your hyperbolic cylinder should be parallel to the $y$-axis, given your two equations, should it not?
$S_1=S^1\times (-1,1), S_2= S^2-\{N,S\}$. Consider the diffeomorphism $\phi: S_1\to S_2, \phi(x,y,z)= (x\sqrt{1-z^2}, y\sqrt{1-z^2},z)$. How to orient $S_2$ such that $\phi$ is orientation preserving?
so if $N_1$ is the orientation on S_1, then I think I must find orientation N_2 on S_2 such that $N_1(p)= N_2(\phi(p)$ for all $p\in S_1$, right?
by orientation, I mean a smooth choice of normal vector field N(p).
**N(p)**= (p, N(p))
@SoumikMukherjee $\int_A f=\int_A\sum_n \frac{\{nx\}}{n^2}\le \int_A \sum_n \frac 1{n^2}=\int_A \frac{\pi^2}6= \frac{\pi^2}6 |A|,|,|$ is Lebesgue measure. |A| is finite as A is closed and bounded.
there are plenty and I don't think there's a best one, just this specific one is very bad cause it makes it seem like orientation is an extrinsic structure instead of an intrinsic one
@Koro You didn't say how you were orienting $S_1$.
But, yeah, Thor is right. You can't explicitly involve $N_i$ in a computation with the map, other than to invoke the right-hand rule to say what an ordered basis on the tangent space is in each case.
Ultimately, you need to consider the derivative of $\phi$ ... and you need it in local coordinates, not $\Bbb R^3$ coordinates. But you can do it pictorially by seeing where $D\phi$ maps the tangent vectors to curves in the cylinder. It's very easy with pictures.
Now pick a random point on the cylinder and follow where $\phi$ maps the circle through that point and the vertical (through that point). Take tangent vectors appropriately.
Several people downvoted me in recent days. Just because mick asked a bad question (so he couldn't define what one function "smaller than" another meant), don't punish me!
@Shaun This OP makes it clear he's just beginning his study of algebra. You write fancy stuff, including Tietze operations, which I've never heard of, and I've written a textbook on algebra. Come on! Surely you're intelligent enough to understand why you're getting these downvotes.
I did not even read to see if it's correct or incorrect. That's not relevant to my observation.
I'm aware that nobody can possibly know why somebody has downvoted, except the downvoter, when there's no comment explaining the situation; I'm aware of - and, in fact, am the guy who started - the chat room "Helpful Commentary" on the main site; but I keep getting downvotes on various posts of m...
I know, by definition, $$e^A=\sum_{k=0}^\infty \frac{A^k}{k!},$$ and that this always converges, since the power series of $e^z$ converges for all $z$, and you can define functions of matrices in terms of entire functions. But what about $$e^{tA}=\sum_{k=0}^\infty \frac{(tA)^k}{k!}?$$ Can the $t$ somehow mess with the convergence here? I'm supposed to motivate why this series converges.
psie: it might help to fix what notion of 'convergence' you are talking about. e.g. if you are working in a normed algebra where A has some finite norm, tA also has some finite norm (indeed, t * ||A||) and if you stare at the proofs of convergence for e^z even in the case of complex z that is probably all you need to prove convergence in the norm.
you might not have convergence that is 'uniform' in t, but you don't have that even in the real or complex case.
if A is something slightly more than a matrix (e.g. an operator, not necessarily bounded, on a hilbert space) you might not have a sense of norm convergence for the operators, but you might have norm convergence in the hilbert space.
ok, I guess absolute convergence would probably be the most sensible here, so we have $\lVert t^kA^k\rVert \leq |t|^k\lVert A\rVert^k$ (using $\lVert AB\rVert\leq\lVert A\rVert\cdot\lVert B\rVert$), then apply comparison test I guess
