real continued fraction has complex value?

Question

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?

Answer 1

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....

Answer 2

Most things are paid for with credit, checks, and wire transfers not printed money, and if the government printed more money people would just put it in their bank accounts. The fed is increasing the amount of credit available for transactions but so far it is not being used. The problem is the government can not make people spend by easing credit restrictions, they can only slow down spending when people are spending to much and bidding up prices which produces inflation. It is much harder to prevent deflation which is what the problem is now.

Answer 3

The continued fraction doesn't converge.