## Tuesday, May 27, 2014

### Alternating sign

Simple continued-fraction representations of constants are a source of much mathematical amusement. The picture is a scan of the cover of Carl Douglas Olds' 1963 classic Continued Fractions which served as my introduction to the topic around 1967.

The sequence [3; 7, 15, 1, 292, 1, 1, ...] is of course the start of the continued-fraction expansion for π. I wish C.D. Olds had used (and insisted on) that semicolon (instead of a comma) after the initial 3. The book shows us how to calculate from these numbers convergents to π, numerators and denominators of the convergents being derived separately. For numerators we pretend things start with [0, 1, ...] and for denominators, [1, 0, ...]. The numerator- and denominator-sequences are then calculated by taking an entry from the continued-fraction sequence (starting with 3), multiplying it by the previous number, and adding the previous-to-previous number.

So, for numerators: [0, 1, 3*1+0 = 3; 7*3+1 = 22, 15*22+3 = 333, ...]
For denominators: [1, 0, 3*0+1 = 1; 7*1+0 = 7, 15*7+1 = 106, ...]

Dropping our pretend prepends and dividing: [3/1; 22/7, 333/106, 355/113, ...]. These are the well-known convergents to π. Note that the convergents alternate in their relation to π: [<π; >π, <π, >π, ...]. Subtracting π: [<0; >0, <0, >0, ...]. Or, if you will, their sign: [-1; 1, -1, 1, ...].

One can do the same thing for τ (pronounced tau), π's less well-known parent (more commonly referred to as 2π). The continued-fraction expansion for τ is [6; 3, 1, 1, 7, 2, 146, ...]. The convergents are: [6; 19/3, 25/4, 44/7, ...]. Like π, these convergents alternate in their relation to τ: [<τ; >τ, <τ, >τ, ...], their sign likewise.

What happens when one divides the convergents to τ by the convergents to π? The resulting sequence is [2; 133/66, 1325/666, 4972/2485, ...], numbers very close or equal to 2. To generate their sign, subtract 2: [0; 1/66, -7/666, 2/2485, ...], hence [0; 1, -1, 1, ...]. The twist is that the sign sequence is now littered with zeros (in seemingly random — but decidedly clusteredlocations). How do those zeros affect the sign alternation? The surprising (to me) answer is: not at all. Remove all zeros from Convergents[τ]/Convergents[π]-2 and what remains will alternate sign, regardless of the number or placement of the previously intervening, single-or-consecutive zeros.