25 Sep
2016
25 Sep
'16
1:10 p.m.
As for periodic expansions in the most obvious sense: Let S = (2/3)^n + (2/3)^(n+k) + (2/3)^(n+2k) + ... = (2/3)^n / (1 - (2/3)^k) = 2^n / (3^k - 2^k). Not much thought went into this, but can S = 1/3 be achieved by some choice of n and k positive integral? We'd need 3^k = 2^k + 3*(2^n) which is impossible with n, k in Z+. —Dan Allan wrote: ----- ... It is entirely conceivable that there exists a *non-canonical* but *periodic* expansion. Jim provided one for 4/5 -- I'm sure the canonical expansion is completely chaotic in this case too. Jim was asking whether there was *any* periodic expansion. -----