Re: [math-fun] Question about base 3/2
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. -----
I think that Dan is assuming that the repeat period is of the form 000...001, and I agree that with that restriction it can't be done. I have an intuition that it can't be done *anyway,* but the proof has to rule out periods like 00101101000001 also. On Sun, Sep 25, 2016 at 3:10 PM, Dan Asimov <dasimov@earthlink.net> wrote:
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. -----
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Dan Asimov