Re: [math-fun] Fracfac expansions (fractions in factorial base)
Thanks for the correction of my careless error. What I should have written is that uniqueness holds if the greedy algorithm is used. (For each n = 1,2,3,... choose c_n, 0 <= c_n <=n, to be as large as possible such that the partial sum c_1/2! + c_2/3! + . . . + c_n/(n+1)! does not exceed the real number x being represented.) Which brings up the question: Is the greedy algorithm the most efficient? (Say in terms of minimizing the number of nonzero coefficients when x is rational.) It's well-known that this is not the case for Egyptian fractions. But for factorial denominators this seems more likely. --Dan Joerg wrote: << I wrote: << Since high school I've thought factorial base is an interesting way to express real numbers in the unit interval (0,1). That is, since 1/2! + 2/3! + 3/4! + ... = (d/dx((e^x-1)/x))(1) = 1, every x in (0,1) has a unique representation of the form x = Sum_{n=1..oo} c_n/(n+1)! if the integers c_n satisfy 0 <= c_n <= n.
I do not think so: 1/2 = 0.1 = sum( n>=2, n/(n+1)! ) The problem (same as 1.0 = 0.999.. ) stays with all mixed radix representations. For uniqueness I'd should check non-adjacent forms (no two adjacent digits nonzero), e.g., digits 0,+1,-1 in binary.
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* Dan Asimov <dasimov@earthlink.net> [Jan 03. 2012 07:10]:
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For uniqueness I'd should check non-adjacent forms (no two adjacent digits nonzero), e.g., digits 0,+1,-1 in binary.
I may have been to fast here: at least when using digits +1 and +i and radix (-1+i) you get http://www.jjj.de/tmp-rd/rd-naf-m1p1-d1.png where the boundaries of the underlying IFS meet (as will (mod my error) always be the case for the regions of the digit zero and the digit(s) nearest to it). No idea for your question regarding greedy selection apart from the obvious: whenever there are two representations as in 1.0 (terminating) and 0.999... (infinitely many nines (=radix at place minus 1) then greedy gives the terminating representation.
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