Comments on: ACM 116: Stochastic Processes

By: AnonEcon

AnonEcon — Sat, 14 Oct 2006 05:19:42 +0000

After your hint I shamelessly formula-mined Knuth’s Fundamental Algorithms to find

sum_{k=1}^{N}{1/k} \approx ln(N) \plus \gamma

where \gamma is Euler’s constant: 0.5772156649…

Playing around with that a bit I got

sum_{k=1}^{N}{1/(2k-1)} \approx
(1/2) ln(N) \plus ln(2) \plus \gamma/2

Even for N as low as 10 the error in this approximation is less than one part in a thousand.

By: Alex

Alex — Fri, 13 Oct 2006 16:23:32 +0000

That’s isomorphic to the method I used. As for simplification, for some reason, the prof said that the answer is proportional to ln N as N gets large, but I don’t see that from the result.

By: AnonEcon

AnonEcon — Thu, 12 Oct 2006 11:17:57 +0000

Hey, Wordpress has eaten up all my plus signs. It should have been
r(k) = p_k[1 \plus r(k-1)] \plus (1-p_k)[r(k-1)]
or r(k) = p_k \plus r(k-1)
or r(k)-r(k-1) = p_k =1/(2k-1)

By: AnonEcon

AnonEcon — Thu, 12 Oct 2006 08:57:30 +0000

Let r(k) be the expected number of loops if you begin with k ropes. If you pick two ends at random from k ropes the probability of getting both ends from the same rope is p_k=1/(2k-1). If you do form the loop you have 1 loop and k-1 ropes from which we can expect r(k-1) more loops and otherwise we have k-1 ropes from which we can expect r(k-1) loops. So, we have the recursion:

r(k) = p_k[1 r(k-1)] (1-p_k)[r(k-1)]
or r(k) = p_k r(k-1)
or r(k)-r(k-1) = p_k =1/(2k-1)

We also know that r(1)=1. From this we get the answer by summing.

What method did you use? I was wondering if the result can be further simplified?

By: Alex

Alex — Thu, 12 Oct 2006 06:38:29 +0000

I didn’t write down the soln in class, so I could attempt recalculating it myself. I just did it, and got the same as you… so we’re both right or both wrong what method did you use?

By: AnonEcon

AnonEcon — Thu, 12 Oct 2006 02:48:08 +0000

Is it

sum_{k=1}^{k=N}{1/(2k-1)}