Michael Lugo
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 Apr10 comment Compute $\sqrt[7]{0.999}$ to three decimal places.(From Gelfand's Algebra text.) I don't agree with your assumption; my assumption is that Gelfand was looking for a smarter answer like mine, given that the numbers involved are so close to 1. (If he'd asked, say, to compute the cube root of 3.257, then I'd expect that he was looking for the long-multiplication answer.) But I can't prove this. Feb24 comment Is $\int_0^\infty x^{a-1} (1-x)^{b-1} e^{t-cx} dx$ integrable? Thanks! I hadn't realized that your context was a beta distribution, and this was the first thing that came to mind. Jan30 comment Explain why catastrophic cancellation happens I'm seeing the same thing in R. This supports the idea that this is an issue with floating-point numbers. Jan29 comment Why is Multiplicative Notation Used for Groups (Instead of Additive)? I'm not sure I'd call this "multiplicative" notation - we don't explicitly write out the multiplication sign. I think that this is a case of juxtaposition having two meanings, multiplication (when used for numeric variables) and an arbitrary group operation. Jan20 comment Mathmatical notation for a term of a polynomial I think that's the closest there is to a standard notation. (But I honestly wouldn't use it without definition unless speaking to an audience of people who deal with generating functions all the time.) Jan15 comment Find a quartic (degree 4) polynomial with integer coefficients whose roots are the primitive 12th roots of unity The polynomial $(z-\alpha_1) (z-\alpha_2) \cdots (z-\alpha_n)$ has roots $\alpha_1, \alpha_2, \cdots, \alpha_n$. Jan14 comment Expectation of maximum of Binomial RVs possible duplicate of Bounds for the maximum of binomial random variables Jan9 comment Visually stunning math concepts which are easy to explain The basic idea is pretty simple: ${i \choose k} = {i \choose k-1} + {i-1 \choose k-1}$, and this recurrence holds $\mod p$ as well. Jan9 comment Visually stunning math concepts which are easy to explain I'd also note that it's possible to compute ${i \choose k} \mod p$ without computing $i \choose k$. For large $i$ this would matter. Dec15 comment Taylor approximation for $\ln(1.3)$ Your book probably should say $0.255$. Dec15 comment Expected value of this deceptively simple variable To find $E(Z)$: note that $Z$ is either 1, 0, or -1. With what probability does it take each of those values? Dec10 comment What is the probability that a Poisson random variable is prime? Asymptotically I'd expect $Q(\lambda, 0)$ to be "the probability that a number near $\lambda$ is prime", i. e. about $1/\log \lambda$. Dec9 comment Can you select random entry from unknown number of entries? If I recall correctly this is called "reservoir sampling", also worth googling. Dec8 comment Why is $\frac {1\cdot2\cdot3\cdot…\cdot n}{(n+1)(n+2)…(2n)}\le \frac 1 {n+1}$ See my edits to answer that question. Dec5 comment Chance on pairs when picking 'Sinterklaas tickets' In the limit of large $n$, the number of 2-cycles of a permutation on $n$ elements is Poisson-distributed with mean $1/2$. (For $k$-cycles, it's $1/k$.) In particular, it's zero with probability $e^{-1/2}$. Restricting to derangements doesn't change this too much. This is a high-level explanation for the asymptotic result quoted in OEIS. Nov24 comment Will it become impossible to learn math? I don't know this book, but I am intrigued. I think this may shed some light on the original question: there could be an upper limit on the depth of possible research imposed by the human lifespan. Nov19 comment Numerical value of $\sum_{p \in \mathcal P} \frac1{p\ln p}$ Your difference should be $1/(2 \log 2)$, not $1/2$. Nov19 comment How to draw greek letters on paper / blackboard? I have seen $\phi$ and $\varphi$ used by the same lecturer in the same lecture (in physical chemistry). One was pronounced "fee" and the other "fie". I found this lecture essentially impossible to follow because I thought of those two glyphs as interchangeable. Nov19 comment Prove that $\sum_{n=0}^{\infty }\frac{(2n+1)!!}{(n+1)!}2^{-(2n+4)}=\frac{3-2\sqrt{2}}{4}$ Yes - that numerical value corresponds to your analytical value. Nov19 comment Prove that $\sum_{n=0}^{\infty }\frac{(2n+1)!!}{(n+1)!}2^{-(2n+4)}=\frac{3-2\sqrt{2}}{4}$ As a check on this value, when the sum is computed numerically one gets 0.1035533906.