Comments on: Combinatorial approach to e http://concretenonsense.wordpress.com/2009/02/26/combinatorial-approach-to-e/ A group blog about mathematics Tue, 01 Dec 2009 16:53:01 +0000 http://wordpress.com/ hourly 1 By: Simon Spicer http://concretenonsense.wordpress.com/2009/02/26/combinatorial-approach-to-e/#comment-142 Simon Spicer Fri, 27 Feb 2009 10:19:52 +0000 http://concretenonsense.wordpress.com/?p=340#comment-142 This is a rather cool way of extracting e from the uniform [0,1] distribution. Of course, we can obtain pi by considering a related problem: what is the probability that the sum of the squares two U[0,1] random variables is less than one? (Answer: pi/4) My question is thus: is it possible to do something similar to extract the Euler-Mascheroni constant from the uniform [0,1] distribution? This is a rather cool way of extracting e from the uniform [0,1] distribution. Of course, we can obtain pi by considering a related problem: what is the probability that the sum of the squares two U[0,1] random variables is less than one? (Answer: pi/4)

My question is thus: is it possible to do something similar to extract the Euler-Mascheroni constant from the uniform [0,1] distribution?

]]> By: Top Posts « WordPress.com http://concretenonsense.wordpress.com/2009/02/26/combinatorial-approach-to-e/#comment-141 Top Posts « WordPress.com Fri, 27 Feb 2009 00:45:39 +0000 http://concretenonsense.wordpress.com/?p=340#comment-141 [...] Combinatorial approach to e I plan to continue my discussion of Boij–Söderberg theory from last time, but I’d like to make a quick detour [...] [...] [...] Combinatorial approach to e I plan to continue my discussion of Boij–Söderberg theory from last time, but I’d like to make a quick detour [...] [...]

]]> By: David Speyer http://concretenonsense.wordpress.com/2009/02/26/combinatorial-approach-to-e/#comment-140 David Speyer Thu, 26 Feb 2009 16:17:37 +0000 http://concretenonsense.wordpress.com/?p=340#comment-140 Solution 3: Let p(k) be the probability that, after choosing k numbers, the sum is still less than 1. So the probability that we will pick exactly k numbers is p(k)-p(k+1) and the expected number of picks is sum k (p(k)-p(k+1)) = sum p(k) Now, p(k) is just the volume of the pyramid {(x_1, ..., x_k) : 0 \leq x_1, .., x_k, x_1+x_2+...+x_k \leq 1} This is 1/k!.So the expected number of picks is sum 1/k! Solution 3: Let p(k) be the probability that, after choosing k numbers, the sum is still less than 1. So the probability that we will pick exactly k numbers is p(k)-p(k+1) and the expected number of picks is

sum k (p(k)-p(k+1)) = sum p(k)

Now, p(k) is just the volume of the pyramid

{(x_1, …, x_k) : 0 \leq x_1, .., x_k, x_1+x_2+…+x_k \leq 1}

This is 1/k!.So the expected number of picks is
sum 1/k!

]]> By: remyoudompheng http://concretenonsense.wordpress.com/2009/02/26/combinatorial-approach-to-e/#comment-139 remyoudompheng Thu, 26 Feb 2009 15:40:24 +0000 http://concretenonsense.wordpress.com/?p=340#comment-139 I propose the following proof : let $latex (X_i)$ be a sequence of i.i.d. uniformly distributed random variables, and $latex S_k = X_1 + \cdots + X_k$. We need the expected value of the minimal k such that $latex S_k \geq 1$, and this is easily proved to equal the sum over all positive n of the probability P(n) that k is greater than or equal to n. For each n, P(n) is the probability that the sum of n numbers in [0,1] is less than 1. Then P(n) is the volume of the simplex $latex \sum x_i \leq 1$, which is easily proved to be 1/n! (by induction on n for example). You thus obtain another equivalent definition of e, the one with power series. I propose the following proof : let (X_i) be a sequence of i.i.d. uniformly distributed random variables, and S_k = X_1 + \cdots + X_k. We need the expected value of the minimal k such that S_k \geq 1, and this is easily proved to equal the sum over all positive n of the probability P(n) that k is greater than or equal to n.

For each n, P(n) is the probability that the sum of n numbers in [0,1] is less than 1. Then P(n) is the volume of the simplex \sum x_i \leq 1, which is easily proved to be 1/n! (by induction on n for example).

You thus obtain another equivalent definition of e, the one with power series.

]]> By: Pete Bevin http://concretenonsense.wordpress.com/2009/02/26/combinatorial-approach-to-e/#comment-138 Pete Bevin Thu, 26 Feb 2009 15:19:51 +0000 http://concretenonsense.wordpress.com/?p=340#comment-138 Nice article! BTW, isn't the second inequality actually an equality? Nice article! BTW, isn’t the second inequality actually an equality?

]]> By: d http://concretenonsense.wordpress.com/2009/02/26/combinatorial-approach-to-e/#comment-137 d Thu, 26 Feb 2009 06:02:14 +0000 http://concretenonsense.wordpress.com/?p=340#comment-137 Let $latex X_j$ be iid uniform in $latex [0,1]$ random variables and $latex S_n = X_1 + ... + X_n$. Then For $latex s < 1$, assume $latex P(S_n < s) = s^{n} / n!$ Then $latex P(S_{n+1} < s) = P(S_n + X_{n+1} < s) = \int_{0}^{1} P(S_n < s - t) dt = \int_{0}^{s} P(S_n < s - t) dt = \int_{0}^{s} P(S_n = 1$. Then $latex P(N > n) = P(S_n n) = \sum_{1}^{\infty} 1/n! = e$ Let X_j be iid uniform in [0,1] random variables and S_n = X_1 + ... + X_n. Then

For s < 1, assume P(S_n < s) = s^{n} / n!

Then
P(S_{n+1} < s) = P(S_n + X_{n+1} < s) = \int_{0}^{1} P(S_n < s - t) dt = \int_{0}^{s} P(S_n < s - t) dt = \int_{0}^{s} P(S_n = 1. Then P(N > n) = P(S_n n) = \sum_{1}^{\infty} 1/n! = e

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