I've been able to convince myself that the idea behind singular homology is, "compact spaces are nice, Euclidean spaces are nice, let's map Euclidean compact spaces into our space so that we can take the preimages of stuff and then they'll be nice too"
Yeah. Maybe I should be explicit about what a section is. If $p : E \to B$ is a covering map, a section $s : B \to E$ is a map such that $p \circ s = \text{id}_B$
Ah, ok. You can take the double $M \cup_{\partial M} M$ which is an orientable manifold with a nonorientable codimension 1 submanifold, and that's impossible
That step in the second paragraph (with the functions) is weird, it took me a bit to justify it
but I guess you take a neighborhood that looks like a half-space, and then in a half-space you definitely have such a function that's 1 near your favorite point and 0 far away enough
So my understanding is that they are constructing an exhaustion $U_1 \subset U_2 \subset \cdots$ of $\partial M$ such that each $U_k$ has a collar extension $U_k \times [0, 1)$
You can do that for the center circle of the moebius strip too...
Keep exhausting closer and closer to the twist
But never hitting the twist
So what fails?
Well it's a finite exhaustion...
Oh I see
It's the one-sidedness.
Oh duh
@AkivaWeinberger Choose boundary charts $V_1, \cdots, V_n$ near $\partial M$ like you said. Choose a smaller cover by $U_i$'s such that each $\partial U_i$ has a collar extension $\partial U_i \times [0, \epsilon) \subset V_i$, say.
If you "transit" from one local product $U \times I$ to another $V \times I$, the map $\varphi : (U \cap V) \times I \to (U \cap V) \times I$ doesn't do anything weird
@Slereah are you talking to me? I don't know much about number theory. I originally considered number theory has no application in physics until the recent, when I read from quanta magazine number theory has application in scatter amplitude.
got a complex analysis question http://prntscr.com/i3fpxl here's the graph on desmos http://prntscr.com/i3fpib is what I got so far and the boundary points are when x = 2 and y = 4 Re(z) = 2 and Im(z) = 4
The interior points should be in that purple box but not sure what else to put for this
If I plot (-1,-2) then it's an interior point for the purple box... oh so all the points involved are in the purple box . Like (-2,-2) could be another interior point of that purple box
Imagine someone bullying you for your and your family's name. Wouldn't be very funny, would it? (I am not even sure if this is genuine or troll in which this guy takes part of, I just don't like the way it's depicted) The original Osas video was funny because it's such a long name. This one's not
but how can i know if the other elements are 0 or not? if the trace is not zero, then it's not a nilpotent matrix and the laim is false, from whati understand
and then i can define vector that is perpendicular to the gradient which would be any vector that satisfies a=-b when $\begin{bmatrix} a \\ b \end{bmatrix}$
so $$ \hat{a}=\begin{bmatrix} 1 \\ 1 \\ 12\sqrt{2} \end{bmatrix}, \hat{b}=\begin{bmatrix} 1 \\ -1 \\ 12\sqrt{2} \end{bmatrix} $$ plane between these should also be considered as the tangent plane ?
or mayby not
@Narcissusjewel what you mean with $f(2,1)+f_x(2,1)(x-2)+f_y(2,1)(y-1)$ ?
@Tuki Hold $y$ constant, and see how the function varies with $x$, you obtain some slope. You want a linear approximation of the surface at a point, so you can take a plane given by the $x$ and $y$ slopes, but first you must translate this plane to the correct place on the surface
One can think of it like this
You have (x,y,z). Where z is a function of x and y
hey, I'm looking at an exercise about solving a differential equation using the Laplace transform. I get the (correct) result, that $$\mathcal{L}[y] = \frac{1}{(1+s^2)^2} e^{-\pi s} + \frac{1}{(1+s^2)^2}.$$ now, the standard solution suggests to "know" the value of $$\mathcal{L}^{-1} \left[\frac{1}{(1+s^2)^2}\right]$$ which I don't. Instead, I rewrote my result as $$\mathcal{L}[\sin(t)] \cdot \mathcal{L}[\sin(t-\pi)] + \mathcal{L}[\sin(t)]^2,$$ which is $$\mathcal{L}[\sin(t)] (\mathcal{L}[\sin(t-\pi)] + \mathcal{L}[\sin(t)]).$$
Hi there! Could somebody give me hint on evaluating $\cos(\alpha)\cos(\alpha/2)\cos(\alpha/4)...\cos(\alpha/124)$? I see that there is a geometric series, but can't find an approach to solve it.
there should be 128 not 124 in the argument of the last cos.
