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Apr
10
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.
Feb
24
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.
Jan
30
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.
Jan
29
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.
Jan
20
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.)
Jan
15
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 $\alpha_1, \alpha_2, \cdots, \alpha_n$.
Jan
14
comment Expectation of maximum of Binomial RVs
possible duplicate of Bounds for the maximum of binomial random variables
Jan
9
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.
Jan
9
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.
Dec
15
comment Taylor approximation for $\ln(1.3)$
Your book probably should say $0.255$.
Dec
15
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?
Dec
10
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 $1/\log \lambda$.
Dec
9
comment Can you select random entry from unknown number of entries?
If I recall correctly this is called "reservoir sampling", also worth googling.
Dec
8
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.
Dec
5
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.
Nov
24
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.
Nov
19
comment Numerical value of $\sum_{p \in \mathcal P} \frac1{p\ln p}$
Your difference should be $1/(2 \log 2)$, not $1/2$.
Nov
19
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.
Nov
19
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.
Nov
19
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.