Harald Hanche-Olsen
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 8m comment Series converges point-wise Hint: You can compute the partial sums explicitly – it's a geometric series, after all. And as the other comments point out, your notation is bad. (But we do understand what you mean.) 13m comment Triangles - sin, cos etc. @Nehorai In many parts of the world (e.g., much of Europe) one uses the comma instead of a period as decimal separator. 16m comment What is the sum of all the natural numbers between $500$ and $1000$. Looks like a good one to emulate, yes. How about: Find the sum of all even numbers in the given range. Find the sum of all multiples of 14 in the given range. Subtract. Feb 5 answered How to prove $\lim_{s \rightarrow \infty} \zeta(s) = 1$? Feb 2 comment Idea behind the tangential vector space? To expand a little bit on the comment by @HagenvonEitzen, try to replace $U$ by the unit sphere in $\mathbb{R}^n$. You will find that the tangent space at any point is $n-1$-dimensional, and can in fact be identified with the hyperplane touching the sphere at the given point. And yet, there is no sensible way to identify all these spaces with $\mathbb{R}^{n-1}$ in a coherent way. Jan 31 comment Banach Tarski proof understanding Probably, assuming that $S$ being decomposable into the elements of $X$ means that you can write $S=P_1\cup\cdots\cup P_n$ (with the $P_i$ pairwise disjoint) and $P_i$ congruent to $Q_i$ for $i=1,\ldots,n$. Jan 31 comment Showing summation of disjoint sets can be split By that, I mean results like: The sums over disjoint sets add up to the sum over the union, and the double sums over two sets is a single sum over the cross product (as I mentioned in my first comment above). If you assume nothing known, other than the commutativity and associativity of addition, proving an inequality like this is going to be a lot more work. Jan 31 comment Showing summation of disjoint sets can be split I'd say the answer depends on context. First, are we talking finite or infinite sets? And it all depends on what related results you can rely on. One way would rely on showing that the sum on the left is a (single) sum over $(u,v)\in(X\cup Y)\times Z$, and similarly for the sums on the right – and noting that $(X\cup Y)\times Z$ is the disjoint union of $X\times Z$ and $Y\times Z$. Jan 31 comment Banach Tarski proof understanding Aren't two sets said to be equidecomposable if each can be decomposed into $n$ pieces which can then be paired together so that each piece from one set is congruent to the corresponding piece of the other? If so, congruent sets are trivially equidecomposable (with $n=1$). Jan 31 comment Banach Tarski proof understanding $SO(3)$ (I have never seen it written in blackboard bold, as you did) is the group of orthogonal $3\times3$ matrices with determinant 1. Also known as the group of rotations (in Euclidean 3-space). Jan 31 revised Banach Tarski proof understanding You can't use TeX constructs outside of math mode Jan 27 comment How do I go about proving da db/a^(-2) is a left Haar measure on the affine group? But ultimately, the proof is in the pudding: You guess it's absolutely continuous, you find out what is required of the Radon–Nikodym derivative, you find a function satisfying the requirements, and conclude (if your argument works in reverse, as it should) that the resulting measure is invariant. Jan 27 comment How do I go about proving da db/a^(-2) is a left Haar measure on the affine group? @Maximiliano I haven't got any brilliant insights, but at least it should be clear that Haar measure is nonatomic: For if it has a point mass at one point, it must have one at every point, which is absurd. Also, for a compact Lie group, you can always average all translates of Lebesgue measure wrt Haar measure (i.e., perform a convolution), to get another Haar measure, this time a.c. wrt. Lebesgue measure. By uniqueness of Haar measure, this must be the original measure (up to a constant multiplier). Jan 20 comment Why do derivatives of functions exist? – or did I misread the quesetion? It looked to me that you were asking about justification for the existence of $f'(0)$. If that is not what you meant, perhaps you could clarify the question. Jan 20 answered Why do derivatives of functions exist? Jan 20 awarded Yearling Dec 16 answered How is Big-O notation used in equalities? Nov 21 comment Every set of well-ordered sets is well-ordered @DacianBonta Ah, now I see what you mean. But that is not the right way to read the statement: It should be read as follows: “By assumption, there exists some $B\in W$. Pick any of them. Then …” (If you want to be pedantic, though, there is a small gap in my proof: You might imagine there is no initial segment of $B$ which is order isomorphic to some $A\in W$. But $B$ itself is such an initial segment, order isomorphic to itself.) Nov 20 comment Every set of well-ordered sets is well-ordered @DacianBonta Not sure if I understand the question. When $W=\emptyset$, there is certainly no $A\in W$, let alone one additionally satisfying the other requirements in the problem as stated. Nov 9 comment Is this matrix positive semi-definite? @soumitra Yes, that is another, perfectly valid, way of doing it.