Marek
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One way to see this is by noting that $$\arctan(x)' = \cos^2(\arctan(x)) = {1 \over 1 + \tan^2(\arctan(x))} = {1 \over 1 + x^2}$$ where we used that $\tan(x)' = {1 \over \cos^2(x)}$, the rule ...

Just to complete the details of Dan's hint. The top homology of a non-compact manifold vanishes, e.g. by Poincare duality $H_n(M) \cong H^0_c(M)$ where $H^*_c(M)$ is cohomology with compact support. ...

In both information theory and physics there is a fundamental quantity called entropy associated with every measure. For the sake of simplicity, let me assume that the number of possible events is ...

This is a partial answer but too long for comment. First plug in $x = 1$ to get $$f(y + {1 \over y}) = yf(1 + y^2)$$ and $y = 1$ to get $$f(f(x) + x) = xf(x^2 + 1)$$ Observe that the RHS of both ...

Use the relation $$T^{-1} \circ T = {\rm Id}$$ and linearity of $T$ and $\rm Id$ to obtain $$T^{-1} (a T(v) + b T(w)) = av + bw.$$ Now write $v' = T(v)$ and $w' = T(w)$. We get $$T^{-1} (a v' + b w') =... View answer Accepted answer 10 votes Suppose 1 < 0. Adding (-1) to both sides we'd also have 0 < -1 (addition axiom). But if 0 < a then it must also hold that 0 < a^2 (multiplication axiom). For a = -1 this means ... View answer Accepted answer 9 votes$$\int {x^{n-2} \over (1+x)^n} {\rm d} x = \int (1+x)^{-2} \left({x \over 1+x}\right)^{n-2} {\rm d} x =\int y^{n-2} {\rm d} y$$using the substitution y = {x \over 1 + x}. View answer 9 votes I think operations other than \times are irrelevant here because bundles are locally product structures. So let's restrict just to this case. Then you are in fact asking whether the tangent bundle ... View answer Accepted answer 8 votes EDIT 2: Oh, I think I finally understood your reasoning. You argued as follows: assume \eta_S vanishes. Then the integral also has to always vanish. But for your \omega it does not. The problem ... View answer 8 votes$$b^{49} = (b^7)^7 = (ab^4a)^7 = a (b^7)^4 a = a (a b^4 a)^4 a = b^{16}.$$View answer Accepted answer 8 votes First suppose that n=0 on a genus g \geq 2 surface. Then the surface can be endowed with a hyperbolic metric (this is essentially the uniformization theorem). Note that g \geq 2 is necessary, e.... View answer Accepted answer 7 votes Well, projections are open since basic open sets in the product topology are U \times V with U and V open, so projecting down leaves you with U and you're done. Now, since submersion is ... View answer Accepted answer 7 votes They are not uninteresting. Just look up the term n-category (e.g. Baez's introduction). Nevertheless, it's true that functors and natural transformations already suffice for most common ideas and ... View answer 6 votes Consider the map z \mapsto z^2 of the unit circle in the complex plane. This gives us a fibration S^1 \to S^1 of degree two. If you travel around the target circle once then this will take you to ... View answer Accepted answer 6 votes No, this is not possible because the integration depends heavily on the representative you choose. Here's an illustration: take a circle S^1 and a bump form b_1 around 1 (I am thinking of the ... View answer 6 votes The short answer is no. Take constant function f(x) \equiv C \in \mathbb{C}. Disregarding normalization, we have \hat{f} = C \delta (in the sense of distributions). Clearly, there is no way to ... View answer Accepted answer 5 votes B(x,r) is not open in uniform topology. For example take y = (1, 1/2, 1/3, \dots). Then z := (1, 1, 1, \dots) \in B(y, 1) but clearly B(z, \epsilon) \not \subset B(y, 1) for every \epsilon &... View answer 5 votes All of these operations are only up to sign. Once you pick the sign though, everything is fixed. It is similar to integration: you pick a generator \omega \in H^n(M, \mathbb R) and this gives you a ... View answer Accepted answer 5 votes I think you are just mixing up two definitions of Gauss map. One is intrinsic to the manifold (and is equal half of the Euler characteristic) while the other depends on the embedding into a Euclidean ... View answer Accepted answer 5 votes EDIT: upon your request, I'll add a further discussion. I'll start with (B), which is both true and also harder than the rest of the cases. Note that after the substitution I mentioned we have$$n \...
No, we can't. Take for example the modular group. This group can be generated by elements $S^2 = 1$ and $(ST)^3 = 1$. Yet $S(ST) = S^2 T = T$ is a translation which generates (a group isomorphic to) $\... View answer 5 votes Calabi conjecture stated that on compact Kähler manifold there exists a unique Kähler metric with prescribed Ricci form. This problem can be reformulated in terms of a (non-linear) PDE and has been ... View answer 5 votes I enjoyed Fecko's Differential Geometry and Lie Groups for Physicists. It doesn't contain complete bottom-up theory building and omits hard proofs but it is a very neat general introduction to the ... View answer 5 votes On the level of Lie algebras we have that${\mathfrak {so}}(n)$are just antisymmetric matrices$n \times n$. It turns out that the six-dimensional space of such$4\times 4$matrices decomposes into ... View answer Accepted answer 5 votes If you know what Taylor expansion is, then you should know that$S_n(s)$in e.g.f. is just its$n$th derivative at zero. For the ordinary g.f., it's almost the same, except that you have to divide the ... View answer 4 votes If a measurable set$A$is a subset of a measurable set$B$then$B \setminus A$is also measurable (by axioms of the$\sigma$-algebras) and since$A$and$B\setminus A$are disjoint, by additivity of ... View answer Accepted answer 4 votes It's hard to say what Terry meant exactly unless you ask him. Nevertheless, it seems quite probable that he is talking about flows on manifolds (or more general kind of spaces but let's stick with ... View answer Accepted answer 4 votes As Daniel mentioned in comments, there is an intimate relation between flows, integral curves and vector fields (a collection of arrows at every point of the sphere). Given a flow, one can follow ... View answer Accepted answer 4 votes It's not that hard to prove that$S^2/X$is homeomorphic to$S^2 \vee S^2 \vee S^2$directly. If you can draw the picture with the three disks then you know where to send every point (think about the ... View answer 4 votes Consider all the possible results of drawing the balls. Each result can be represented by a binary number having 1 when we draw a red ball, so that the number of 1s in the number is$r\$. Your question ...