Michael Lugo
• Member for 11 years, 6 months
• Last seen this week
• Atlanta, GA

So I think you want to know "why" $\pi$ and $e$ appear here based on an explanation that goes back to circles and natural logarithms, which are the usual contexts in which one first sees these. If ...

This has nothing to do with the Riemann zeta function, but is rather a property of random sequences of digits. (As a rule there's no reason to assume that there's anything significant about the ...

Assuming Arnold means to find an approximate value -- I'd do this as follows: first, we may as well find the average of $\cos^{100} x$. I'll do this over a half-period, $-\pi/2 \le x \le \pi/2$. But $\... View answer 47 votes See Durrett, Probability: Theory and Examples (link goes to online copy of the fourth edition; original defunct link). On p. 164 Durrett gives a proof that simple random walk is recurrent in two ... View answer 37 votes An exact formula can be found in Anirban DasGupta, The matching, birthday and the strong birthday problem: a contemporary review, Journal of Statistical Planning and Inference 130 (2005), 377-389. ... View answer 36 votes From Srivatsan Narayanan's comment: there are on the order of$n^7$numbers satisfying the digit constraint, with$n$digits. The probability that a random$n$-digit number is prime is of order$1/n$. ... View answer 33 votes A lot of people have mentioned Binet's formula. But I suspect this is not the most practical way to compute the nth Fibonacci number for large n, because it requires either having a very accurate ... View answer 27 votes As has been pointed out in the comments, this is not a special property of the primes. Rather it's true that whenever you reverse a number and append the result to itself, the result is always ... View answer 25 votes I've long used Simon Plouffe's inverse symbolic calculator for this purpose. It is essentially a searchable list of "interesting" numbers. View answer 24 votes This is the secretary problem. If there are$n$suitors, the optimal strategy is to reject the first$r-1$of them and then accept the first one that is better than all of those$r-1$. In general$r \...

Cathy O'Neil gave a talk at MIT entitled "Math in Business" last week; she summarizes that talk in this blog post. There may be some ideas here.

Look at the logarithm. More specifically, consider $f(x)^{g(x)}$ as $x \to \infty$, where $\lim_{x \to \infty} g(x) = \infty$ and $\lim_{x \to \infty} f(x) = 1$. (This is something of form $1^\infty$....

A probabilistic method that I learned from Jim Pitman's book Probability (exercise 3.3.10) is as follows. Let $X$ be uniformly distributed on the set $\{ 1, 2, \ldots, n \}$. Then $$E(X^3) = (1^3 + 2^... View answer 19 votes Yes. We only need to know how many powers of two appear in the prime factorization of C(n,p). The number of powers of two that appear in the prime factorization of n! is \lfloor n/2 \rfloor + \... View answer 19 votes Somewhat informally: if you take a matrix, and change the entries of it a little bit, then the determinant also changes by a little bit. (Formally, the determinant is a continuous function of the ... View answer 19 votes The book "The Mathematical Mechanic" by Mark Levi is a very good source of such examples, which Levi has been collecting for some time. The first two here are in the book, if I recall correctly. View answer Accepted answer 18 votes Your original answer of \dfrac{3 \times 10^{14}}{52!} is not far from being right. That is in fact the expected number of times any ordering of the cards has occurred. The probability that any ... View answer 18 votes The usual reason I've heard is that dividing by integers is computationally easier -- it's easier to find, say, (5\sqrt{3})/3 by computing 5 \times \sqrt{3} \approx 8.66  and then dividing by 3 ... View answer 17 votes Another proof is by comparison: note that$$ {1 \over k^2} < {1 \over (k-1)k} $$for all integers k \ge 2. Therefore$$ {1 \over 2^2} + {1 \over 3^2} + \cdots + {1 \over n^2} < {1 \over 1 ...

I'd say that each of them arrives at some time uniformly distributed on [12:00, 12:45], where the times are independent. They meet if their arrival times differ by less than fifteen minutes. In the ...

Another hint: can you factor $a^{15}-1$? In more than one way? Six years later, for the record, let's spell out what I meant by this. What I had intended was to observe $$a^{15}-1 = (a^3)^5 - 1 = (... View answer 16 votes If you know about the Euclidean algorithm, you can see that \gcd(2k+3, 2k+1) = \gcd(2k+1, 2) = \gcd(2, 1) = 1. As for making a linear combination: try small values of k instead of diving right ... View answer 16 votes You can write this as$$ S(n) = n! \left( {1 \over 0!} + {1 \over 1!} + \cdots + {1 \over n!} \right) $$and now recall that e = 1/0! + 1/1! + 1/2! + \cdots . So in fact$$ S(n) = n! \left( e - \...

One-line proof (some details omitted): ${m+n \choose m} > 1$ if $0 < m < n$.

Take a look at the Wikipedia article on the Dirichlet distribution. In particular the Dirichlet distribution with $\alpha_i = 1$ for all $i$ is the uniform distribution on the simplex. Furthermore, ...

Say $x^y = y^x$, and $x > y > 0$. Taking logs, $y \log x = x \log y$; rearranging, $(\log x)/x = (\log y)/y$. Let $f(x) = (\log x)/x$; then this is $f(x) = f(y)$. Now, $f^\prime(x) = (1-\log x)/... View answer 15 votes I don't really have the intuition for this, but I know the formal proof. This is equivalent to showing that the probability that in a permutation of$[n]$chosen uniformly at random, two elements ... View answer 15 votes They're interchangeable. I think a lot of people tend to use parentheses just because they're easier to write by hand. View answer 14 votes This sum does not converge. Consider for example the sum of the reciprocals of all the even numbers greater than 2:$1/4 + 1/6 + 1/8 + 1/10 + \ldots\$. This diverges, the proof being analogous to the ...