
Re: Average length of random 1expansions
Posted:
Nov 20, 2000 4:55 AM


r3769 wrote: > >Note that f(1/(2^k n)) = n/2 + f(1/n). > > I think you wanted to write f(1/2^k n))=k/2 + f(1/n), no?
Yes, my bad.
> It is kind of interesting to note that the "coinflip 1expansion" defined > above can be generalized: > > Given any integer b>0, let amin be the smallest integer s.t. > floor(x[i]*amin)=b and amax the largest such integer. > > Now let f(b,x)=the average length of all the coinflip bexpansions of x. > > For example f(1,1/11)=16, f(2,1/11)=6, f(4,1/11)=4, f(11,1/11)=51, and > f(113,1/11)=18. > > I suppose it's not too difficult to show the sequence <f(b,1/11)> is > bounded, so what is it's sup?
I haven't had a chance to look at this yet, but I'll try to come back to it.
Hugo

