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 Mar7 comment Bivariate normal distribution: showing that linear combinations of joint Gaussians are Gaussian This is merely equation (21) with $\rho$ used in place of that ratio of covariances. Feb19 answered Prove that $a_i\leq 0$ for $i=1,2,…,N-1$? Feb18 comment Prove that $a_i\leq 0$ for $i=1,2,…,N-1$? You might have the wrong sign in the equation you obtained: check it against the sequence $(0,-2,0)$ for $N=2$, for instance. As a hint, relax the conditions and suppose only that $a_{i+1}-2a_i+a_{i-1}\ge 0$ for $i=1,2,\ldots,N-1$. Can you still draw the desired conclusion? Feb13 revised Must any set of positive Lebesgue measure contain a bounded set of positive measure? edited body; edited tags; edited title Feb13 awarded Good Answer Feb11 comment Difference between Real Analysis and Probability Theory? Nice little glitch in the SE technology: I have been able to upvote this answer twice--once on CV and once again here :-). Feb11 comment Difference between Real Analysis and Probability Theory? +1 This answer gets to the heart of the matter. Analysis and probability have different interests: although they use similar sets of tools, they ask different questions and pursue almost completely different avenues of investigation. The two disciplines will nevertheless remain closely intertwined because insights from one can lead to progress in the other, much as (say) investigations of general relativity and quantum mechanics in physics have inspired advances in low-dimensional topology and operator theory, respectively. Jan30 comment What's a proof that the angles of a triangle add up to 180°? @JoeZ. Or $-2\pi$: the complex arithmetic keeps track of angle orientation, too. Dec29 comment Interesting and unexpected applications of $\pi$ This is David H's answer in disguise: the ARE is algebraically related to (a) $e^0=1$ and (b) $\int_\mathbb{R}e^{-x^2}dx=\Gamma(1/2)=\sqrt{\pi}$. The appearance of $\pi$ is due to the duplication formula for the Gamma function, $\Gamma(z)\Gamma(1-z)=\pi\csc(\pi z)$, exhibiting $\Gamma$ as a kind of "square root" of a trig function. This is because $\Gamma(z)=\Gamma(z+1)/z$ implies $\Gamma$ has (simple) poles at $0,-1,-2,\ldots$, whence $\Gamma(z)\Gamma(1-z)$ has poles at $\mathbb Z$, strongly suggesting periodicity--which is why $\pi$ should show up! Dec28 revised Prove without induction $2^n \mid (b+\sqrt{b^2-4c})^n + (b-\sqrt{b^2-4c})^n$ Appended some more discussion. Dec28 revised Prove without induction $2^n \mid (b+\sqrt{b^2-4c})^n + (b-\sqrt{b^2-4c})^n$ [Edit removed during grace period] Dec28 revised Prove without induction $2^n \mid (b+\sqrt{b^2-4c})^n + (b-\sqrt{b^2-4c})^n$ deleted 4 characters in body Dec28 comment Prove without induction $2^n \mid (b+\sqrt{b^2-4c})^n + (b-\sqrt{b^2-4c})^n$ @Thank you for noticing that, Thomas. That is easily fixed by replacing one of the $c$'s by $1$; I will make the change. Dec24 comment Probability that there is sub-sequence of exact length There is a lot of literature on this subject. The asymptotic distribution is known. I don't believe exact formulas for the case of finite $N$ are available. Dec24 revised Urn problem with balls added 12 characters in body Dec24 comment Urn problem with balls @Brian Thank you for that observation. It's very much in the spirit of this answer, which is an attempt to find as simple a solution as possible. Dec24 revised Urn problem with balls added 2 characters in body Dec24 comment Using triangle inequality for upper limit of integral. Why not just compute the integral? It's dead easy--the exponential part, being analytic throughout the triangle, integrates to zero; the remaining contribution from $\bar z$ will give $-2i$ times the area of the triangle. Dec24 comment The point of contact of between two circles and common tangent at this point. +1 for the simplified elementary approach. But note that the solution is incomplete: it has not (yet) addressed whether the circles are internally or externally tangent. Dec24 comment Urn problem with balls It might be even simpler just to count the ways of choosing one ball in one urn and the other in the other: $(n+1)^2$. Divided by $\binom{2n+2}{2}$, this gives the answer directly.