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 May 31 comment For $x \, \epsilon \,R^n$, prove $\int_{|x| \leq 1} x_i x_j dx = 0$ Thanks @EthanBolker, so simple :) 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 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 comment Halting probability of random tree-generating algorithm @Vortico, I rephrased all for equal probability of stopping or not stopping. 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 comment Halting probability of random tree-generating algorithm Hum, it does not, I understood that 3 new nodes and no node were all equiprobable, not as you say. But the outcome is the same, $(1/2) \frac{1}{1-(1/2)}$ = 1. Nov 28 comment Halting probability of random tree-generating algorithm I have expanded the answer to your request, if I understood the problem correctly. Hope it helps. Nov 23 comment $x,y,z$ are positive real numbers and $x+y+z=1$ $\implies$ $\bigg(1+\dfrac 1x\bigg)\bigg(1+\dfrac 1y \bigg)\bigg(1+\dfrac 1z \bigg)\ge 64$? Hum, Jensen's inequality seems to give an smaller bound ... sorry. Nov 23 comment $x,y,z$ are positive real numbers and $x+y+z=1$ $\implies$ $\bigg(1+\dfrac 1x\bigg)\bigg(1+\dfrac 1y \bigg)\bigg(1+\dfrac 1z \bigg)\ge 64$? It smells like Jensen's inequality. Nov 6 comment Explain proof of irreducibility of $x^{p-1} + 2x^{p-2} \dots (p-1)x + p$ Isn't $Q(0) = p$? as if all zeros are there, then $Q$ = original $P$. Nov 3 comment Prove that the set of limit points of a function is all $R$ A limit cannot be whole real line. I bet you need to restate what you want to prove. Oct 16 comment Propositional Calculus: Stating and proving the unique readability theorem in Polish notation "an initial segment of one another, which we know is not possible" I bet, that had to be proved as a lemma, ah? Oct 14 comment Is this true: $\lim_{\lambda \rightarrow 0}E\left[ e^{-\lambda X} \right] = P\left(X < \infty \right)$? I bet the OP meant $\lim_{\lambda \to 0}$ Oct 12 comment $(x+1)/x = \sqrt{3}$ in form $a+b \sqrt{3}$ @columbus8myhw, thanks. Yes, even simple questions in maths can have deeper answers. Oct 2 comment Formula for the terms of the sequence defined by $a_0 = 1$, $a_1 = -2$ and $a_{n}=-4 a_{n-1}-4 a_{n-2}$ @Mark, induction is: show that the first domino falls, and that every domino that falls knocks the next one; then you can be sure all dominoes fall. Strong induction is, show that the first domino falls, and that if all dominoes up to a certain point fell, then the next one falls too. So, strong induction -although equivalent- allows you to use as argument a stronger assumption ("all dominoes up to a point") to prove that the next one also falls. Stronger assumptions make proofs easier.