3 ms·

Thanks a lot for this clear analysis. I had gotten till the “difficult combinatorial question” and was stuck. Didn't think of just going into continuous space.

Thanks a lot for this clear analysis. I had gotten till the “difficult combinatorial question” and was stuck. Didn't think of just going into continuous space. The “crappy approximation” is actually what I used up until now and it's good to see that I wasn't wrong in assuming it would overestimate the loss. You made one small mistake at the end: The approximation says you loose 2.3 nats = 3.32 bits, so they actually differ by over a bit. Btw: What did you ask wolfram alpha to get the -log m + 1 - 1/m? It was out of computation time when I asked :) http://www.wolframalpha.com/input/?i=-+int_0^1+%28m*%281-x%29^%28m-1%29%29+*+log%28%28m*%281-x%29^%28m-1%29%29%29++dx http://www.wolframalpha.com/input/?i=-+int_0^1+%28m*%281-x%2... Or do you have a subscription?

gjm11 11y ago
> You made one small mistake at the end Oh yes, so I did. Sorry about that. > What did you ask wolfram alpha to get the -log m + 1 - 1/m? This, I think: http://www.wolframalpha.com/input/?i=Integrate[m+t^%28m-1%29+Log[m+t^%28m-1%29],{t,0,1} http://www.wolframalpha.com/input/?i=Integrate[m+t^%28m-1%29...] I get a message about computation time exceeded, but I also get the answer (both the indefinite integral and the definite). I don't have a subscription.
- chronial 11y ago
  Fascinating ^^. Removing the minus before the integral and changing 1-x to x is the difference that matters. I would have though that wa is so quick with those transformations that it doesn’t make a difference. Well, I was wrong :).