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real continued fraction has complex value?
let x be the continued fraction [1;-1]
1 - 1/( 1 - 1/( 1- 1/(1 - .... )))
if x = 1 - 1/x then x^2 = x - 1 so (x-1/2)^2 = -3/4
so x = (1+isqrt(3))/2
so 1 - 1/( 1 - 1/( 1- 1/(1 - .... ))) = (1+isqrt(3))/2
is there a fallacy in the working, or is that a genuine answer?
how do you test for convergence?
hi hotdog .. why do you have to assume that the tail is 1 so that the higher term is 1/(1-1) ?
i thikn my point is that once you know the solution (1+isqrt(3))/2 then it works propogating back to the next level .. so it has converged, it just hasn't converged to a real number. how is that when the explicit terms are all real?
3 Answers
- hotdogLv 41 decade agoFavorite Answer
If you cut the partial fraction off at some point you get infinities that you have to deal with.
1 - 1/(1 - 1)
1 - 1/(1 - 1/(1-1))
1 - 1/(1 - 1/(1-1/(1-1)))
etc.
Now, if we heuristically just say 1 - 1 = 0 and 1/0 = infinity and then work our way through the partial fraction, then you see that the values you get oscillate with a period of 3, not one. So it is not ok. to denote the partial fraction by x and then one level down say that it is x again.
New comment: Ok, I agree that you can assume that the tail has some arbitrary value and then demand that after one level the value is the same. You then get two complex solutions. This then implies that you cannot have a period of 1 with a real tail.
So, what you could do is demand that after two or three or more steps you get back to the same value. Maybe then there are real solutions....
- 5 years ago
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