are you in a hotel, or just buy a microwave or something a cheapo one will do
actually i steam the food in a rice cooker which heats more evenly than a microwave but is slower amd introduces moisture but a microwave is clearly the easy option
If $\pi: \mathbb{R}^n\rightarrow \mathbb{R}^k$ is the projection map, and $A\subseteq \mathbb{R}^n$ is a k-slice, is it true that $\pi|_A: A\rightarrow \mathbb{R}^k$ is a diffeomorphism?
$G, \mu, c $ Are constants and we are given that $u(a) = u(-a)=0$
So, we can solve that equation using indefinite integrals and then using Algebra to solve for constants or we can use the definite integrals. But both of these methods are giving the answer which differ by a minus sign.
Another thing you could do, you're given $u'$ is linear, this means that $u(y)$ is a quadratic, you know that it vanishes at $\pm a$, so $u(y) = C(y-a)(y+a) = Cy^2 - Ca^2$ for some constant $C$, differentiating you get $u' = 2Cy$. Compare this with $\frac{Gy}\mu + c $ you get $c=0$ and $2C =\frac{G}{\mu}$.
whenever you're talking about the flow, and you do a lot, I think you should make that more explicit, e.g. I don't think anyone would guess that for the "solution space" for $Ax$, that sounds like the null space to me instead
Also I'm not convinced, to shrink, you need a negative eigenvalue
right?
like for example solutions to $x' = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix} x$ trace out hyperbolas
oh, it would seem you're talking about things like $x(t) = e^{t \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix} }x_0$? but there's the $t$ there that you didn't include. Not sure why I tried to decipher what you wrote
@Manolis What matters for a loop in the plane minus two points is only how it turns around the two removed points. If you have a loop that only turns around one of these, the situation will be the same as with a circle, so you won't find two non-commuting elements. So try looking at simple loops that turn around both of the removed points.
@ManolisLyviakis Let a be the loop which circles about the red point and b be the loop which circles about the green point here. I claim aba^-1b^-1 is homotopic to that loop.
Also yes, punctured torus indeed does deformation retract to S^1 V S^1
So the statement is : The fundamental group of the 2 point punctured C is the free group with 2 generators and if the loops around the 2 points where commutative on the operation of concatenation then i would only count the winding number on the first point and the second to get the equivalnce classes and the fundamnetal group would be ZxZ
@BalarkaSen I have always liked the handle-theoretic picture of classification of surfaces, which comes to mind because your two manifolds equivalent to S^1 v S^1
If you assume a single 0-cell (and as many 2-cells as necessary to close up the surface), the picture is: a circle, equipped with n pairs of points, where each point is labeled with a normal orientation
Your points should be thought of as teleporters and the arrows tell you which way you'll come out
can also draw them as + and - so long as you understand clearly the rules for how to go from one teleporter to the other
Then the handleslide move is just moving a point across a teleporter
And birth/death is just adding/deleting pairs of adjacent cancelling teleporters (where the teleportation is just "continue along the circle as expected", more or less)
Let $j,k,l,n$ be positive integers such that $j,k,l \leq n$, $j-l \leq k$ and $l \leq k$. Is there any way to simplify the product
$$
\binom{n-j-k-l}{k-j-l}\binom{n-j-k+l}{l}^2,
$$
perhaps as a product of just two binomial coefficients (or just as one such coefficient)? I've been staring at this ...
When someone talks about compact subsets, what do you think of? What's the intuition behind a compact subset and a non-compact subset? What's the importance? Why do we care if a subset is compact or not?
In the 19th century, several disparate mathematical properties were understood that would later be seen as consequences of compactness. On the one hand, Bernard Bolzano (1817) had been aware that any bounded sequence of points (in the line or plane, for instance) has a subsequence that must eventually get arbitrarily close to some other point, called a limit point.
Consider the the projection $\Bbb R' \times [0, 1] \to \Bbb R$, where $\Bbb R'$ is discrete. Each fiber is homeomorphic to $[0, 1]$ hence compact. But, the inverse image of $[0, 1]$ is $[0, 1]' \times [0,1]$, which is not compact
@Thorgott So there must be $4$ Sylow 3-subgroups on which the group acts transitively. I assume that this will give the injection to $S_4$, but I don't have an argument for why it is injective yet
Should I send some images of my book or write down? (the printing have some missing letters in between but image will be a quick thing, what you demand sir?)
yes I agree with that. But in this case if we take partition $P=\{z_0, z_1, \cdots z_k\}$ such in each integral $[z_{I-1}, z_i]$ the value of the function is constant
thats P*, a fixed partition that comes from the funcction. Its not the same partition as the partitions P_n, which are the ones you're testing integrability with
@Tobias the source I'm reading argues by looking at the core of a normalizer of such a $3$-sylow and shows it is trivial, then we get the injection by having the group act on the cosets of that normalizer
@BalarkaSen I see. So, you would take a inverse image of a compact set containing $1$, and that would be something like $(0, a] \cup [b, 1)$ (not compact).
@feynhat Yeah I mean just think sequentially. A map of metric spaces is proper if every sequence escaping to infinity is mapped to some sequence escaping to infinity
