Golden Iterations I

Note that $x_{n+1}=1+\frac{1}{x_n}$ , and $\phi\ =\ 1+\frac{1}{\phi}$ . Hence $\frac{\left|e_{n+1}\right|}{\left|e_n\right|^d}$ $= \frac{\left|1+\frac{1}{x_n}-\left(1+\frac{1}{\phi}\right)\right|}{\left|x_n-\phi\right|^d}$ . Set $f(x) = 1+1/x$ and $d=1$ . Then the expression becomes $\left|\frac{f\left(x_n\right)-f\left(\phi\right)}{x_n-\phi}\right|$ Since $\lim_{n\to\infty}$ $x_n=\phi$ it becomes $\lim_{x\to\phi}$ $\left|\frac{f\left(x\right)\ -\ f\left(\phi\right)}{x-\phi}\right|$ which is $\left|f'\left(\phi\right)\right|$ , which is $\frac{1}{\phi^2}$ , which evaluates to $\frac{3-\sqrt{5}}{2}$ which gives the answer of 11.

We can easily see that $\displaystyle x_n=\frac{F_{n+2}}{F_{n+1}}$ where $\displaystyle F_n$ is the $n^{th}$ Fibonacci Number

So, $\displaystyle \left|e_n\right|=\frac{\phi F_{n+1}-F_{n+2}}{F_{n+1}}$

Now, using Binet's Formula we get

$e_n=\frac{1+\phi^2}{\phi^{2n+3}+\left(-1\right)^n\phi}$

Putting this expression in the limit, we can see that $\displaystyle d=1$ for the limit to exist and hence, evaluating the limit leads us to the answer which is $\displaystyle \frac{1}{\phi^2}$

So, $\displaystyle \frac{1}{\phi^2}=\frac{3-\sqrt{5}}{2}$ makes our answer $\displaystyle 11$

By definition, $x_n$ can be written as $1+\frac{1}{x_{n-1}}$ ; now, since $\phi$ is solution of $x=1+x^{-1}$ , we can rewrite it as $\phi=x_n=1+\frac{1}{x_{n}}$ . So, $|e_n|=|\frac{1}{x_{n-1}} - \frac{1}{x_{n}}|$ . Now we can calculate the limit, writing everything in function of $x_n$ : if you have not made mistakes, you will end up with $\frac{|1+\frac{1}{x_{n}}-x_n|^{1-d}}{|x_n+1|}$ . Note that $1+\frac{1}{x_{n}}-x_n$ , for $n \to \infty$ ,since $x_n=\phi$ , is equal to 0: putting $d=1$ , the limit exists and it's not 0, but tends to $\frac{1}{1+\phi}$ , which is equal to $\frac{3-\sqrt{5}}{2}$ . In the end, the answer is $3+5+2+1=11$ .