Hmm, so I'm gonna say let's make the manifold compact and orientable. So let's say $f:M\to \partial M$ is a retraction and $\omega$ is an $n-1$ form on $\partial M$
Okay I'm guessing the cohomological interpretation of this is that if $M$ is an $n$-manifold and $f:M\to \partial M$ is a retraction, then take the inclusion $i:\partial M \to M$, then $H^n_{DR}(M) \to H^n_{DR}(\partial M) \to H^n_{DR}(M)$ where the first map is $f^*$ and the second is $i^*$
Let $f(z)=(1+z^2)^{-3/2}$ and $g(z;\epsilon)=f(z)-f(z-\epsilon)$. For small $\epsilon>0$, find the value of $z$ for which $g(z;\epsilon)$ achieves its maximum.
(The last problem did something I hate and labelled the separation between capacitor plates as $d(t)=d_0\cos \omega t$...and then ultimately wanted them to take a time-derivative. so $\partial d/\partial t$)
i had taken tschinkel my first semester at NYU, (with no mathematics background in undergrad mind you). had to drop very quickly... went back with my tail between my legs, read some algebra
I had his course last semester @JoeShmo. He definitely teaches at a rapid pace. I was lucky enough to have had the grad course in algebra at my undergrad institution.
If I imagine a little current going around each of those rectangles, then the superposition of them will just be the original two currents since the vertical edges will have cancelling currents
one thing I like is that it makes it obvious (to me) why the combined field you end up with in that limit behaves like $g(z;\epsilon\ll 1)\sim \epsilon z/(1+z^2)^{5/2}$.
A bit slow for my liking, in truth. On the flip side I had graduate alg top last year in undergrad. But, Cappell is a great lecturer, and I definitely gained some insight even if I already knew most of the basics @JoeShmo
It's not in my linear algebra course, no. It could have been, but one can't do everything. It fits a more advanced course, or more algebra-oriented course, just fine.
it just wasn't as obvious physically to me why it had to be like that. (in terms of dimensionful variables it's $g(z)\sim z/r^5$, and you can't distinguish $z$ vs. $r$ from dimensional analysis)
@Semi I've seen two proofs. One is by proving it using diagonalizable matrices and then making a density argument, another I've seen in Hoffman and Kunze just does a bunch of matrix stuff
correction: we sum across the k-th row instead. So we pick the k-th row, evaluate the first entry at e_1, etc, and then sum them up. The RHS, i.e. detB I, is still detB e_k The LHS becomes Σ_j (adjB B)_kj e_j = Σ_j Σ_i (adjB)_ki B_ij e_j = Σ_i (adjB)_ki Σ_j B_ij e_j = Σ_i (adjB)_ki (A e_i - Σ_j "A"_ij e_j) = Σ_i (adjB)_ki (A e_i - Σ_j A_ij I e_j) = Σ_i (adjB)_ki (A e_i - Σ_j A_ij e_j) = Σ_i (adjB)_ki (A e_i - A_i) = Σ_i (adjB)_ki 0 = Σ_i 0 = 0
let's just check it. Let $A = \begin{bmatrix} a&b\\c&d \end{bmatrix}$. Then, $B = \begin{bmatrix} A-aI & -bI \\ -cI & A-dI \end{bmatrix}$. If $\operatorname{adj}(B) = \begin{bmatrix} P & Q \\ R & S \end{bmatrix}$, then $B \operatorname{adj}(B) = \begin{bmatrix} PA-aP-bR & QA-aQ-bS \\ RA-cP-dR & SA-cQ-dS \end{bmatrix}$. Pick the first column, and do that black magic, and we get $(PA-aP-bR)e_1 + (RA-cP-dR)e_2 = P(Ae_1 - ae_1 - ce_2) + R(Ae_2 - be_1 - de_2) = 0$
the website is wrong since the website considered adj(B) B instead of B adj(B)
ok the website is not wrong, it just uses a different convention, i.e. somehow it multiplies row vectors on the left.
Suppose M is a compact, 2-dimensional, oriented submanifold of R^3\{(0, 0, 0)}. Let omega := x . *dx. What are the possible values of the integral of omega over M?
I have this exercise: Compute the integral $\int_{\gamma_r}(z-z_0)^n$, where $n$ is an integer not equal to $−1$ and $γ r$ is the circle $|z − z 0 | = r$ traversed once in the counterclockwise direction.
I have this exercise: Compute the integral $\int_{\gamma_r}(z-z_0)^n$, where $n$ is an integer not equal to $−1$ and $γ_r$ is the circle $|z − z_ 0 | = r$ traversed once in the counterclockwise direction.
This is rather a philosophical question. Although it uses topological notions, it isn't any precise mathematics, so maybe one cannot take it very seriously.
Sometimes I try to picture an infinite set of points. Assuming that the space $E$ in which my visual imagination takes place is a simply co...
Technically if one can do so, they would have already well ordered the reals in the process
We can instead consider the question of making a faithful visualisation of a dense linear order. If there exists a picture (a configuration of points in $\Bbb{R}^2$) that can do that, we can use that to visualise many dense linear orderings including the reals
I want to show the following.
Let $\gamma :[a,b] \to \mathbb{R}^n$ be a piecewise continuously differentiable path. Then there exists a reparametrization of $\gamma$ which is continuously differentiable.
There was a hint provided which says that you need to "slow down" at the corners to mak...
How do you mean deviation from the original parametrization? I thought this construction will be a function $[a,b] \to [a,b]$ which kind of slows down the input at certain points (where the steps of the staircase are).
the reparametrization is then the composition of this and the actual path
you might want to include a coefficient on the trig function as well, so that you can control how big of a deviation you get (it probably doesn't need to be big)
I want to show the following.
Let $\gamma :[a,b] \to \mathbb{R}^n$ be a piecewise continuously differentiable path. Then there exists a reparametrization of $\gamma$ which is continuously differentiable.
There was a hint provided which says that you need to "slow down" at the corners to mak...
