Continued fraction question?

Question

If (x-1)^-1=2+(2+(2+....)^-1)^-1
Why does 2+(2+(2+....)^-1)^-1 equate to 1+x ??

e.g. (x-1)^-1=1+x

How is this done?

Answer 1

Usually these are done by "peeling off a layer" of the continued fraction. Let's see if that works here.

1/(x-1) = 2+(2+(2+....)^-1)^-1
1/(x-1) - 2 = (2+(2+....)^-1)^-1
(3 - 2x)/(x-1) = (2+(2+....)^-1)^-1
(x-1)/(3-2x) = (2+(2+....)^-1) = 2+(2+(2+....)^-1)^-1

Hmmm... I am not getting the same thing as you have above. Let me try one more thing. Let's find the value of 2+(2+(2+....)^-1)^-1

y = 2+(2+(2+....)^-1)^-1
y-2 = (2+(2+....)^-1)^-1
1/(y-2) = (2+(2+....)^-1) = 2+(2+(2+....)^-1)^-1 = y

So y = 1/(y-2)

Simplifying, we get
y^2 - 2y - 1 = 0

Using the quadratic formula, we get
y = 1 +/- sqrt(2)

Since y is positive it must be 1+sqrt(2). But if y is equal to 1+x, then x must be sqrt(2), but it is not the same as 1/(sqrt(2) - 1). So I do not believe the relation above is correct.

Answer 2

Clearly you didn't type the question properly because x appears on the LHS but not on the RHS.

Let y = 2 + [2 + (2 + ...)^(-1)]^(-1)
y - 2 = [2 + (2 + ...)^(-1)]^(-1)
1/(y-2) = [2 + (2 + ...)^(-1)]
The RHS is clearly the same expression we defined as y
therefore, 1/(y-2) = y
1 = y^2 - 2y = (y-1)^2 - 1
y = 1 +/- sqrt(2)
Clearly the expression is not negative, so
y = 1 + sqrt(2)
That's the value of the RHS of the original expression.