@MaryStar It seems it will work because you effectively defined a curve that trace out the same circle many times (and it slows down logarhtimatically as you increase t), but the starting point and end point never matches (because the starting point is undefined), hence the curve will remain open (in fact, it only has one end point, $t=e^{2\pi}$)
Maybe me, once I've reviewed my chem lessons, physics lessons, finished my 5 assignments that each take 3-4 hours and finished redoing all the chem exercises ._.
@Rememberme: https://dl.dropboxusercontent.com/u/2098511/FAnotes.pdf Still this one, but I might skip some chapters because I want to focus on chapters 10 and 11.
@TalenKylon If you divide one polynomail with another of lower degree and get remainder of zero, it means the polynomial id divisible by the lower degree one and thus by factor theorem the lower degree one is a factor of it. Hence the original polynomial will be reducible as you can write it in terms of the factors (lower degree polynomial)(factor from the long division)
@Rememberme: The author is the best teacher I've ever had. I just talked with some friends about him today and he doesn't seem to only have this impact on me. He has a very broad knowledge, thus always finds many different reasons to motivate new things and in person, he is really really good. He always does the whole lecture without any notes, yet doesn't lack structure and can show many examples interesting for anyone.
@Rememberme: Just check out section 1.4 for example. I've actually seen the Proposition 1.16 (and even proved it personally) before, but haven't really thought about it in the specific context of the heat equation and the wave equation. And he does exactly that and now I am actually a bit ashamed to admit I have never really understood those two fundamental equations up to the point where I read that section.
Is there some way using wolframalpha, or another tool, I can check my multiplication of two polynomials in GF(256). The two polynomials I'm multiplying are (x^8+x^6+x+1)*(x^7+x^6+x^2+1). I'm getting, x^56 + x^48 + x^36 + x^16 + x^12 + x^3 + x^2 + x + 1. I would like to double check my answer to make sure I am correct.
@Rememberme Homotopy between $[f][g]$ and $[g][f]$ doesn't make sense. You're already in the equivalence class. What you want is a homotopy between $f * g$ and $g * f$. Anyway, that bit is obvious. The nontrivial part is to find what the homotopy is.
If you know that the combination of two functions make the same kind of function, since the combination of $n$ functions yields that kind of function by your induction hypothesis, then the combination of $n+1$ functions is just the combination of (one function) and ($n$ functions), and hence is of the type you want
Let $(a_{ij})$ be a skew-symmetric $3 \times 3$ matrix (i.e., $a_{ij}=-a_{ij}$ for all $i, j$).
Let $v_1$, $v_2$ and $v_3$ be smooth functions of a parameter $s$ satisfying the differential equations
$$v'_i= \sum_{j=1}^3 a_{ij} v_j$$
for i = $1, 2$ and $3$, and suppose that for some paramete...
@user159870: I haven't read the question, but if you want to show they are orthogonal, you want to show the dot product is zero. However, if you show that the derivative of the dot product is zero, you know the dot product is constant. Therefore, you can evaluate the dot product at some point where it's easy to see it is zero. I'm assuming this is what the hint is referring to here.
and I just saw that in your question, you are given that at $s_0$ your vectors are orthogonal. so if you show that the derivative is zero, you know they are orthogonal everywhere
@AndrewG It depends on what you mean by Galois theory of covering spaces, though. Do you just mean plain old covering space theory, or the rigorous Grothendieck formality of covering spaces and Galois theory?
Describe all curves in $\mathbb{R}^3$ which have constant curvature $κ > 0$ and constant torsion $τ$.
Any ideas what we can do to describe all such curves?
Do we have to use the formulas of the curvature and of the torsion?
Not sure, but from my point of view it makes sence. I remove every number that exists in both sets from the A set. Since R contains all numbers it will have all the numbers that are in A.
It's not about symbols only. To be honest my question is: "is there any chance a student will write a proof that he just though and this proof is as good as the one in the book"?
Sure ! Believe in yourself :-) At best, your proof will be better than the book's. At worst, you'll see where your proof is lacking and do better next tiùe
@Samuel it's pretty cool. do you know about the cayley graph? if $G$ is a finitely presented group with generating set $S$, you can make a graph $\Gamma(G, S)$ out of it as follows
each vertex of $\Gamma$ is associated to an element of $G$
each edge joining vertices of $\Gamma$ correspond to relations of the form $g_1 = sg_2$, where $g_1, g_2$ are the elts the vertices correspond to and $s$ is an elt of $S$
but now note that if we choose different generating sets $S, S'$, we get different graphs $\Gamma(G, S)$ and $\Gamma(G, S')$ (i.e., nonisomorphic graphs). e.g., $G = \Bbb Z$, $S = \{\pm 1\}$, $S' = \{\pm 1, \pm 2\}$. $\Gamma(S)$ looks like a long line with vertices sticking at integer intervals.
but $\Gamma(S')$ looks like a crissed-crossed stip.
@SuperHeroY: Say you have some password consisting of $n$ characters out of an alphabet of of $m$ different characters. You want to crack the password. For this, you can choose a string of arbitrary length containing characters of given alphabet. Using this string, you can enter the password as follows: Assume for example the string is "199285" and you know $n=4$. Then, the string will first check whether "9285" works, then whether "9928" works and then whether "1992" works.
What is the shortest possible string which will crack the password?
ok. now the cool thing to note is that $\Gamma(S)$ and $\Gamma(S')$ are "coarsely" the same. informally, assume you have a very hazy vision and you're looking at things from a distance. then $\Gamma(S')$, the crissed-crossed strip will look approaximately the same as $\Gamma(S)$, right?
this idea is formalized by quasi-isometries. if $(X, d)$ and $(X, d')$ are metric spaces, then a quasi-isometric embedding is a map $f : (X, d) \to (X, d')$ such that $1/q \cdot d(x, y) - c \leq d(f(x), f(y)) \leq q \cdot d(x, y) + c$ for some constant $c > 0, q > 0$. quasi-isometries are quasi-isometries embeddings $f: (X, d) \to (X, d')$ which has an inverse which is also a quasi-isometric embedding.
approximately, this can thought off as "an isometry, but upto hazing by a factor of $q$ and stretching by $c$"
now, a quasi-isometry between groups are just quasi-isometries between their cayley graphs, the graphs getting the structure of a metric space by equipping each edge with the length of $1$.
Fact : given any finitely presented group $G$, for any two generating sets $S, S'$, $\Gamma(G, S)$ is quasi-isometric to $\Gamma(G, S')$
yep. Gromov initiated the program of classifying groups upto quasi-isometry
as you see, quasi-isometry type of the groups is an inherent nature of the group - as choice of the generating set doesn't matter
what's fun now is that hyperbolicity can be defined in certain groups (this is what Gromov did). if $G$ is a finitely presented group, then $\Gamma(G)$ (it's quasi-isometry type, note that i'm omitting the generating set) can actually be given the structure of a geodesic metric space. let me elaborate on this a bit : assume you are starting off with the free group on two gens : $G = F_2$
$\Gamma(F_2)$ then looks like a $4$-tree.
now take any three points on the $4$ tree, say the vertices corresponding to $x$, $x^{-1}$ and $yx^{-1}y$. join each pair of the three vertices with the shortest path available through the edges.
what you'll get will seem like an extremely deflated version of a triangle -- hey, that's what a hyperbolic triangle is!
indeed, there is a rigorous definition under which a notion of hyperbolicity can be attached to $F_2$. fact : groups quasi-isometric to hyperbolic groups are hyperbolic.
so, in a sense, geometric group theory is a somewhat mixture of geometry, graph theory and group theory. if you love at least two of the three, maybe you'd want to have a peek. just suggesting.
I have to wonder, this and this question both are seemingly defining what I'm used to as being called $\ell^p$ spaces. Is the notation $\ell^p$ not actually standard? And if not, are there notable examples of who does and who doesn't use that notation?
Sorry, off-topic, but is the site bugged at the moment? I can't add tags to my questions, and my profile seems to have deleted all my previous questions. :(
@UserX If you mean "is it consistent", it's as consistent as hyperbolic geometry, etc. One is consistent iff the other is. I don't think there's actually a proof that it is in fact consistent, though.
Let $(a_{ij})$ be a skew-symmetric $3 \times 3$ matrix (i.e., $a_{ij}=-a_{ij}$ for all $i, j$).
Let $v_1$, $v_2$ and $v_3$ be smooth functions of a parameter $s$ satisfying the differential equations
$$v'_i= \sum_{j=1}^3 a_{ij} v_j$$
for i = $1, 2$ and $3$, and suppose that for some paramete...
