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Why does this process, when iterated, tend towards a certain number? (the golden ratio?)

Please post your favorit solution to the following

Compute $x=1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\ldots}}}$

Thank you

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    You might have heard of golden ratio.http://en.wikipedia.org/wiki/Golden_ratio This might be interesting for you .2012-06-09

3 Answers 3

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Here is another, more pedestrian, and sketchy, way of doing this:

Let the sequence $\frac{p_n}{q_n}$ be defined as follows:

$p_1=q_1=1$

$\frac{p_n}{q_n}=1+\frac 1{\frac{p_{n-1}}{q_{n-1}}} = \frac{p_{n-1}+q_{n-1}}{p_{n-1}}$

It is clear that this computes terminated versions of the continued fraction and we have $q_n=p_{n-1} \text{ and } p_n=p_{n-1}+q_{n-1}=p_{n-1}+p_{n-2}$

It follows that numerator and denominator are successive Fibonacci numbers, and the limit ratio is the golden ratio.

There is an efficient discussion of continued fractions in Hardy & Wright "An Introduction to the Theory of Numbers" which shows that successive estimates from terminated fractions lie either side of the limit and converge to it.

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    "terminated versions of the continued fraction" - *convergents* is the usual term of art. Basically, one is saying here that the $n$-th convergent of the continued fraction is $\dfrac{F_{n+1}}{F_n}$. For proving that the limit of that as $n\to\infty$ is $\phi$, see [this](http://math.stackexchange.com/questions/132305).2012-06-09
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Look at the equation $x^2-x-1=0$

It is clear is solution is not zero. Thus, write

$x^2=x+1$

This is equivalent to

$x=1+\frac 1 x$

Using this recursively

$x=1+\frac 1 {1+\frac 1 x}$

Thus, define $\{ x_n\} $ recursively as $x_0=1$ and $x_n=1+\dfrac{1}{x_{n-1}}$

Show the sequence is positive, increasing for $n>3$ and bounded, thus it converges to $\ell$. You can then show that $\lim x_n=\ell = x$, where $x$ is the positive solution of the first equation discussed, namely

$\phi = \frac{\sqrt 5 +1}{2}$

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Denote by $x=1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\ldots}}}$. Because the fraction goes on forever, the denominator of the second term in the RHS is identical to $x$. Thus, we have $x=1+1/x$ or the same as $x^2=1+x$. Thus, $x=\frac{\sqrt 5-1}{2}$.

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    This is definitely not a good explanation/solution of the problem. (-1). "goes on forever" is not a satisfactory term here.2012-06-11