# Thread: Evaluate a recursion function.

1. ## Evaluate a recursion function.

2-$\displaystyle \frac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{...}}}}$ = ?

(A) 0
(B) 1
(C) 2
(D) 1/2
(E) -1/2

The answer booklet is separate from the book and I do not have it.

2. Hello, shailen.sobhee!

$\displaystyle 2-\frac{1}{2-\dfrac{1}{2-\dfrac{1}{2- \hdots}}} \;=\;?$

. . $\displaystyle (A)\;0 \qquad (B)\;1 \qquad (C)\;2 \qquad (D)\;\tfrac{1}{2} \qquad (E)\;-\tfrac{1}{2}$

$\displaystyle \text{Let: }\:x \;=\;2 - \frac{1}{2-\dfrac{1}{2-\dfrac{1}{2 - \hdots}}}$

$\displaystyle \text{Then: }\:x \;=\;2 - \frac{1}{x}$

Multiply by $\displaystyle x\!:\;\;x^2 \:=\:2x - 1 \quad\Rightarrow\quad x^2-2x+1\:=\:0 \quad\Rightarrow\quad(x-1)^2 \:=\:1$

Therefore: .$\displaystyle \boxed{x \:=\:1}$

3. That's brilliant.

There there exist a recursion in almost all expressions; take a very simple case:

y=$\displaystyle \frac{x}{y} = \frac{x}{\frac{x}{y}} = \frac{x}{\frac{x}{\frac{x}{y}}} = \frac{x}{\frac{x}{\frac{x}{\frac{x}{...}}}}$