An Infinite Continued Fraction

Calculus Level 5

$\Large 1+\frac{2}{1+\frac{2}{ 1+\frac{2}{\ddots+\frac{2}{x}}}}=2$

The above infinite continued fraction can be viewed as a recurrence relation : $a_0=x$ , $a_{k+1}=1+\frac{2}{a_{k}}$ for $k\ge 0$ and $\displaystyle\lim_{k \to \infty}a_k=2$ .

What is the range of integers $x$ that satisfy this condition?

x \leq -2

x \in \mathbb{Z}

x \in \{-2, -1, 0, 1 \}

x=2

x \in \mathbb{Z} \setminus \{-2,-1,0\}

x \neq -1

x \geq -1

x \in \mathbb{Z} ^+

1 solution

Chan Lye Lee
May 9, 2016

Relevant wiki: Nested Functions

It is not difficult to see that if $x=0$ (or $-2$ respectively), then $a_1$ (or $a_2$ respectively) is not defined; if $x=-1$ , then $a_k=-1$ for all $k\ge 0$ . Here are the graphs showing iteration of $a_k$ for (i) $x<-2$ , (ii) $-2<x<-1$ , (iii) $-1<x<0$ and (iv) $x>0$ respectively.

We can get the idea from here that the $\displaystyle \lim_{k \to \infty} a_k=2$ for all $x \in \mathbb{R} \setminus \{-2,-1,0\}$ .

Any good algebraic proof?

There are other values that are not allowed. For $f(x) = 1 + \frac{2}{x}$ , $f^{-1}(x) = \frac{2}{x - 1}.$ Then the values $f^{-1}(-2) = -\frac{2}{3}$ , $f^{-1}(-\frac{2}{3}) = -\frac{6}{5}$ , $f^{-1}(-\frac{6}{5}) = -\frac{10}{11}$ , etc., are also not allowed.

Jon Haussmann - 5 years, 1 month ago

You are right. Thanks for pointing out the mistake.

Chan Lye Lee - 5 years, 1 month ago

I think for any other values of x, the value of 1 + 2/x will oscillate back and forth to 2. And at infinity, the limit is 1 + 1 = 2.

Rindell Mabunga - 5 years, 1 month ago

An Infinite Continued Fraction

1 solution

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