carlosayam
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 Mar 25 awarded Tumbleweed Mar 18 asked Different positive and negative bounds on Lipschitz condition Jun 5 asked Integral over balls in $\mathbb{R}^n$ in different norms and measures May 31 answered For $x \, \epsilon \,R^n$, prove $\int_{|x| \leq 1} x_i x_j dx = 0$ May 31 comment For $x \, \epsilon \,R^n$, prove $\int_{|x| \leq 1} x_i x_j dx = 0$ Thanks @EthanBolker, so simple :) May 31 asked For $x \, \epsilon \,R^n$, prove $\int_{|x| \leq 1} x_i x_j dx = 0$ Dec 1 comment How can I find the lenght of the third side of any triangle Sorry, apart from moving I don't see another option. If you can move to another position, please ensure you get reasonable measurements; this is, they need to be reasonably different from the ones you have (or put other way, you have to move at least 40 or 50 yards from your original point if your measurements are only precise +- 1 yards, I believe). Nov 30 answered How can I find the lenght of the third side of any triangle Nov 29 revised Halting probability of random tree-generating algorithm i stand corrected Nov 29 comment Halting probability of random tree-generating algorithm @goos, so your understanding of the problem is that although it halts on a given node, the machine could still work on any other node that is still "open". Like a deep-first algorithm, right? Nov 29 comment Halting probability of random tree-generating algorithm @goos, yes I understood it differently. Not sure of any edits by the OP now :( Nov 28 comment Halting probability of random tree-generating algorithm @Vortico, although the algorithm seems to produce a tree, it is actually a list as it never recurses on previous nodes to generate new ones there - as I understood it. But even in that case, you could apply the very same reasoning to the following: either it creates a node in any place on the already generated tree with probability 1/2 or it stops with probability 1/2. The answer would be the same. Nov 28 comment Halting probability of random tree-generating algorithm @Leo163, your first claim seems wrong. Please see my answer on how to approach this recursively. Nov 28 comment Halting probability of random tree-generating algorithm @Vortico, if you prefer a recursive approach, I just added that. Nov 28 revised Halting probability of random tree-generating algorithm recursive approach Nov 28 comment Halting probability of random tree-generating algorithm @Vortico, I rephrased all for equal probability of stopping or not stopping. Nov 28 revised Halting probability of random tree-generating algorithm modified to prob = 1/2 Nov 28 comment Halting probability of random tree-generating algorithm @Vortico, the essential part of the argument is considering any past history and the current event as independent events; hence one can multiply their probabilities. Nov 28 comment Halting probability of random tree-generating algorithm Probability of halting in step n is $(1/2)^{n-1} (1/2)$ = $(1/2)^n$ for n=1,..; which becomes $\sum_1^\infty{(1/2)^{k}}$ = 1. Even easier :) Nov 28 revised Halting probability of random tree-generating algorithm added 228 characters in body