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Re: MX2 2016 Integration Marathon $A guided evaluation of \\ \\ $\lim_{R\rightarrow\infty}\int_{-R}^R \frac{\sin{x}}{x}\, dx$\\ \\ that I don't think anyone answered in the end. Not overly difficult. $

Yep x-1. Will edit, thanks.

A rather dry problem but I was a little bored. Here goes: 64C2=2016 24C3=2024 17C4=2380 15C5=3003 14C6=3003 The above binomial coefficients xCy are each those with the least x for a given y such that the binomial coefficient is at least equal to 2016 (v easy to bruteforce with a calc by...

No worries :).

Because |1+\alpha|=1+\alpha=1+|\alpha| noting that taking the modulus does not change a positive number, and also using the fact that alpha is positive by assumption.

1. Yes, this is a pretty common convention. Another one is e_i to denote the coordinate vector whose only nonzero entry is it's i-th one, which is equal to 1. 2. Yep. In fact the dot product is actually a more fundamental notion than that of an angle. We use the dot product (or more generally...

Comes from the absolute homogeneity of norms. (\|cx\|=|c|\|x\|) So: \|a+\alpha a\|=\|(1+\alpha)a\|=|1+\alpha|\|a\|=(1+|\alpha|)\| a \|=\|a\|+\|\alpha a\|.

A bit on the easier side, but just to get the ball rolling in this thread again: A function g(n) defined on the non-negative integers satisfies: i) g(0)=g(1)=0. ii) g(p)=1 if p is prime. iii) g(mn)=m*g(n)+n*g(m) for all non-negative integers m and n. Find all n such that g(n)=n...

Re: Extracurricular Integration Marathon I=\int_0^\infty \frac{2x^2}{x^4+2x^2+5}\, dx=\int_{\mathbb{R}}\frac{x^2}{x^4+2x^2+5}\, dx This integral is clearly a prime target for contour integration. (We could factorise into its real quadratic factors, use partial fractions, and integrate the...

(And yes, this is strictly more work than one needs to do to answer your original question, where a construction using a factorial explicitly gives us prime gaps of length blah. I just did it this way because I personally wanted to see how "low-tech" I could make a proof of the fact that the...

$Let $S_p$ be the set of natural numbers indivisible by $p$. \\ \\ It is trivial that $\frac{|S_p\cap [1,n]|}{n}\rightarrow 1-\frac{1}{p}$ as $n\rightarrow\infty.$\\ \\ Moreover, since primes are pairwise coprime, we have $\frac{|(\cap_{k=1}^m S_{p_k})\cap [1,n]|}{n}\rightarrow \prod_{k=1}^m...

(It suffices to show that pi(x)=o(x), which is a quite weak statement about the distribution of the primes that can be proved by a completely elementary sieving process. Am just thinking of the cleanest way to write this. Will post shortly.)

$First, we look at the second assertion. The number of primes in the set $I_n(x)=\{x+1,x+2,\ldots,x+n\}$ is given by $g_n(x):=\pi(x+n)-\pi(x)$ for any non-negative integer x.\\ \\ If we look at the shifted interval $I_n(x+1)$ instead, we might lose a prime from no longer having $x+1$ in our set...

Re: Extracurricular Integration Marathon The integrand has its modulus squared equal to (x^2 + (1-x^2)cos^2(theta))^n =< (x^2 + (1-x^2))^n=1. The triangle inequality completes the proof.

Of course. Fun is most of the reason I play the game myself. (Not that I don't want to improve, just it isn't a big priority right now.) It seems that we won't have enough people for a tournament, but I will still be down for some chill games from 8ish.

Let's say 10 mins for the sake of speed. All depends on the number of people interested though. If it's small enough for a round robin (like 4-5 people) then we can probably play two games per pairing so the colours are fair.

Am thinking of starting at like 8-8:30pm. Whoever can make it then let me know, otherwise let me know a more preferable time.

Anyone up for a mini tournament this evening?

Still around?

http://lichess.org/jXLXattU