Wikipedia:Reference desk/Archives/Mathematics/2011 April 26

= April 26 =

Integration
Suppose you have to evaluate the integral of f(s) wrt s around the closed contour C numerically, and you don't actually have the entire curve C, merely a, rather large, number of discrete points on C. how do you proceed? Just for the record, I can evaluate a contour integral analytically but this has to be done with a computer, a realm in which my skills do not lie. Thanks. 131.111.55.91 (talk) 09:17, 26 April 2011 (UTC)
 * Of the top of my head: Let $$z_1,\cdots z_n$$ be your points where $$z_1$$ is the least according to the lexicographical order. Starting from i=1 do the following: Consider a circle of sufficiently big radius centered around $$z_i$$ which definitely is big enough to enclose (or contain on the boundary) some other $$z_j$$. Consider the circle of radius $$|z_i-z_j|$$ now. Does it contain some other $$z_t$$? If so consider the circle of radius $$|z_i-z_t|$$. Iteratively, consider circles inside previously considered circles such that all of them are centered around $$z_i$$ until you have found the $$z_p$$ which is closest to $$z_i$$. Put $$z_i$$ ,$$z_p$$ in an array. Go on to $$z_p$$ and strike i out of the list of yet to be considered values amongst 1,2,...n . Keep on repeating until the list is empty and in this new reordering call the values $$Z_1,\cdots Z_n$$. This procedure would give you $$Z_i$$'s arranged such that each point is followed by the point closest to it. Now you can use the sum $$\sum_{i=1}^{n-1}f(z_i)|Z_i-Z_{i+1}|$$ as an approximation to the integral.-Shahab (talk) 11:13, 26 April 2011 (UTC)
 * I think it should be $$Z_{i+1}-Z_i$$, not $$|Z_i-Z_{i+1}|$$. Looie496 (talk) 16:35, 26 April 2011 (UTC)
 * It depends on whether it's a line integral wrt arc length, or a proper contour integral in the sense of complex variables. I don't feel like the OP gives enough context to decide which. In either case, I don't see why there is any fussing over the order of the points: they need to be given in the order they appear on the curve. Sławomir Biały  (talk) 16:46, 26 April 2011 (UTC)
 * And it's suspicious that the value $$f(z_n)$$ isn't used. To estimate a complex line integral I think a better formula is $$\sum_{i=1}^{n}\frac{1}{2}(f(Z_i)+f(Z_{i+1}))(Z_{i+1}-Z_i)$$ where $$Z_{n+1} = Z_1$$ (wrap around). That is, compute the integral over a closed polygon, and approximate the value along each segment by the average of its endpoint values. 98.248.42.252 (talk) 15:32, 27 April 2011 (UTC)

Showing $$\ell_p$$ is of first category in $$\ell_r$$ for r > p (and a second query on weak convergence)
Hi all,

I want to show that the sequence space $$\ell_p \subseteq \ell_r$$ for $$1 \le p \le \infty,\, p < r $$ is of first category in $$\ell_r$$ under the standard $$\ell_r$$ norm. My thought was to write $$\ell_p = \bigcup_{n} \{x \in \ell_p: \, \|x\|_p \le n\}$$ and then find for a given $$x,\, \|x\|_p \le n$$, some $$y$$ with $$\|y\|_p > n$$, $$\|x-y\|_r < \epsilon$$, i.e. given any arbitrary ball in one of the closed sets $$\{x \in \ell_p: \, \|x\|_p \le n\}$$, we have some y not in that set contained in the ball. However, I'm failing to find such a y for a given x: if this is the right method, could anyone suggest anything? Otherwise, what am I doing wrong? I also wanted to find out what happened in the case r = p: is there anything different going on then? Delaypoems101 (talk) 16:49, 26 April 2011 (UTC)


 * This should work. The trick is to modify the tail of $$x\in\ell_p$$ so that $$\|y\|_p>n$$ but $$\|x-y\|_r<\epsilon$$.   Sławomir Biały  (talk) 14:49, 28 April 2011 (UTC)


 * I'm afraid I've also come across a second question I'm having trouble on in the meantime, it's not going very well! I wasn't sure whether to ask in a separate question or not, given it's on a similar topic: I'm very sorry if this is all too much. If X is a complex Banach space, we say a sequence $$x^i$$ ∈ X converges to $$x$$ ∈ X weakly if $$\phi(x^i) \to \phi(x)\forall \phi \in X^{*}$$, the dual space of X. I want to show that if X is $$\ell_2$$, then if $$\|x^i\| \leq 1 \forall i$$, then there is a subsequence converging weakly to $$x$$ ∈ X with $$\|x\| \leq 1$$, and if X is $$\ell_1$$, $$x^i$$ converges to some x ∈ X weakly iff it converges to x in the usual sense. For the first part of this, I tried using the fact that every coordinate is bounded and thus has a convergent subsequence (is a bounded sequence), so find a convergent subsequence in the first coordinate, then find a convergent subsequence of that sequence in the second coordinate, and then a convergent subsequence of that sequence in the third coordinate, and so on. Then take the n-th term from the n-th subsequence, and this gives you a sequence converging eventually in all coordinates: this seemed like the 'logical' subsequence. However, it isn't obvious to me whether or not it converges weakly, and if so why it only converges weakly. Similarly in the second part, it's obvious if it converges it converges weakly, but why is it that if it converges weakly it must converge, when this doesn't hold in $$\ell_2$$? Why do these behave differently? Many, many thanks for your help Delaypoems101 (talk) 17:51, 26 April 2011 (UTC)


 * In $$\ell^2$$, this argument doesn't work basically for the reason you have stated. A modification will work, though: produce convergence on each element of a countable dense subset of $$\ell^2$$ by using this kind of diagonalization argument.  (This is the sequential Banach-Alaoglu theorem.)  This technique fails in $$\ell^1$$ because the dual space, $$\ell^\infty$$, is not separable.   Sławomir Biały  (talk) 14:49, 28 April 2011 (UTC)