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accepted | What is the effect of adding 1/2 to a continued fraction? |
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comment |
What is the effect of adding 1/2 to a continued fraction?
Noam, thanks. I've been away from MO for a while, but this was actually exactly what I was looking for. |
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Apr
30 |
comment |
What is the effect of adding 1/2 to a continued fraction?
@Douglas: Could you contact me off list? |
Apr
30 |
comment |
What is the effect of adding 1/2 to a continued fraction?
@Douglas: Actually you might be right. The formula in my comment is a little cleaner though. |
Apr
30 |
comment |
What is the effect of adding 1/2 to a continued fraction?
@Douglas: The cases depending on whether a1 > 1 seems off. I think the (or rather a) formula is 1/2 x = [a0/2,2x'] or [(a0-1)/2,1,1,(x'-1)/2] (depending on parity of a0) and 2x=[2a0,x'/2]. This gives x/2 = [a0/2,2a1,x''/2] or [(a0-1)/2,1,1,(x'-1)/2]. Thanks again; this was exactly what I was looking for. I had dismissed finding a rule this simple for some reason. |
Apr
29 |
comment |
What is the effect of adding 1/2 to a continued fraction?
@Douglas: Check your formula for doubling. I think it is not quite correct. |
Apr
26 |
comment |
What is the effect of adding 1/2 to a continued fraction?
@Douglas: thanks, but this is not really what I'm asking. The 2[a0;2a1,a2,2a3,a4,...]=[2a0;a1,2a2,a3,2a4,...] part is in the spirit of what I want, but I want something analogous which works for any sequence. I was aware of this observation --- the hard part is of course in dealing with the remainders. |
Apr
26 |
comment |
What is the effect of adding 1/2 to a continued fraction?
Thanks, but this is not really what I'm asking. See the comments in the edit portion of the question. |
Apr
26 |
revised |
What is the effect of adding 1/2 to a continued fraction?
added the "edit:" part of the text |