Golden Iterations II (updated)

Write $x_n=\varphi+\varepsilon_n$ . (The " $\varepsilon$ "s and " $e$ "s from the question are the same in absolute value, but it's useful to work with the error term before taking the absolute value).

We have $x_{n+1}=\sqrt{1+x_n}=\sqrt{1+\varphi + \varepsilon_n}$

Since $\varphi^2=1+\varphi$ , this can be rewritten as $x_{n+1} = \varphi\sqrt{1 + \frac{\varepsilon_n}{1+\varphi}}$

Since for large $n$ , $|\varepsilon_n| \ll 1$ , we can apply the binomial theorem to get

$x_{n+1} = \varphi \left(1 + \frac{\varepsilon_n}{2(1+\varphi)} \right)+O(\varepsilon_n^2)$

Subtracting $\varphi$ from both sides we get

$\varepsilon_{n+1} = \frac{\varphi \varepsilon_n}{2(1+\varphi)} + O(\varepsilon_n^2)\\ =\frac{\varepsilon_n}{2\varphi} + O(\varepsilon_n^2)$

So in the limit, we have

$\frac{\varepsilon_{n+1}}{\varepsilon_n}=\frac{1}{2\varphi}=\frac{\varphi-1}{2}=\frac{\sqrt{5}-1}{4}$

and finally $(a,b,c,d)=(5,1,4,1)$ giving the answer $a+b+c+d=11$ .

notice that $\lim_{n\to \infty} x_n = \phi$ . let $f(x)= \sqrt{1+x}$ , then the limit in the problem is the absolute value of $\lim_{n \to \infty} \dfrac{f(x_n) - \phi}{(x_n-\phi)^d}= \lim_{x_n \to \phi} \dfrac{f(x_n) - \phi}{(x_n-\phi)^d}$ applying l'hopitals $\lim_{x_n \to \phi} \dfrac{f(x_n) - \phi}{(x_n-\phi)^d} = \lim_{x_n \to \phi} \dfrac{f'(x_n)}{d(x_n-\phi)^{d-1}}$ notice for $d>1$ this diverges and for $d<1$ it is $0$ . hence $d=1$ must be the solution, giving $\lim_{x_n \to \phi} \dfrac{f'(x_n)}{1(x_n-\phi)^{1-1}} = f'(\phi) =- \dfrac{1}{2\sqrt{1+\phi}} = -\dfrac{1}{2\phi}\\ = - \boxed{\dfrac{\sqrt{5}-1}{4}}$

Observe that $e_{n+1}=\sqrt{1+x_n}-φ=\sqrt{1+φ+e_n}-φ$ . So we can express the limit in terms of $e_n$ approaching 0.

$\lim_{e\to0} \frac{\sqrt{1+φ+e}-φ}{e^d}$

Both numerator and denominator evaluate to 0 for $e=0$ , so we use l'Hôpital's rule.

$\lim_{e\to0} \frac{\frac{d}{de}\sqrt{1+φ+e}-φ}{ \frac{d}{de}e^d}$ $=\lim_{e\to0}\frac{\frac{1}{2\sqrt{1+φ+e}}}{de^{d-1}}$ $=\lim_{e\to0}\frac{1}{2de^{d-1}\sqrt{1+φ+e}}$

This value is independent of e when we choose d=1. Our limit then evaluates to

$\lim_{e\to0} \frac{1}{2\sqrt{1+φ+e}} =\frac{1}{2φ}=\frac{\sqrt{5}-1}{4}$ .

So we have $a=1, b=5, c=4, d=1$ and $a+b+c+d=\boxed{11}$

It is clear that $\lim_{n \to \infty} x_n = \phi$ , so $|e_{n+1}|/|e_n|^d$ will diverge except if the numerator cancels the terms of the form of $x_n - \phi$ in the denominator.

We have

$\displaystyle{\frac{|e_{n+1}|}{|e_n|^d} = \frac{|x_{n+1} - \phi|}{|x_n- \phi|^d} }.$

We can rewrite the recursion of $x_n$ to find

$\displaystyle{ x_{n+1}^2 - 1 = x_n }.$

Inserting this in the previous equation

$\displaystyle{\frac{|e_{n+1}|}{|e_n|^d} = \frac{|x_{n+1} - \phi|}{|x_{n+1}^2 - 1 - \phi|^d} }.$

Using that $\phi^2 = \phi +1$ , we get

$\displaystyle{\frac{|e_{n+1}|}{|e_n|^d} = \frac{|x_{n+1} - \phi|}{|x_{n+1}^2 - \phi^2|^d} =\frac{|x_{n+1} - \phi|}{|(x_{n+1} - \phi) (x_{n+1} + \phi) |^d} }.$

Only if $d =1$ will we prevent the divergence in the limit. The result will then be

$\displaystyle{\lim_{n\to \infty} \frac{|e_{n+1}|}{|e_n|^d} = \lim_{n\to \infty} \frac{1}{| (x_{n+1} + \phi) |} = \frac{1}{2\phi} = \frac{\sqrt{5}-1}{4} }.$

And so the answer is $5+1+4+1 = \boxed{11}$ .