The golden ratio is defined as:
$$\frac a b = \frac {a+b} a.$$
It comes down to:
$$\varphi = 1 +\frac 1 \varphi.$$
Is there a way to solve for $\varphi$ computationally/recursively? Is there any other way to solve it?
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3solve? it is $\frac{1+\sqrt 5}{2}$ – Will Jagy Oct 31 '22 at 04:18
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How to solve it? – Alexander Mills Oct 31 '22 at 04:19
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4What do you mean solve it? If $\phi = 1 +\frac 1\phi$ then $\phi^2 = \phi + 1$ and $\phi^2 -\phi -1 = 0$ so by quadratic formula $\phi = \frac {1\pm \sqrt 5}2$. But as $\phi > 0$ we have $\phi = \frac {1+\sqrt 5}2$. What's in doubt? – fleablood Oct 31 '22 at 04:27
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ahh I thought it was a recursive definition, didn't realize it was just a quadratic, thanks – Alexander Mills Oct 31 '22 at 04:30
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1You can define it recursively by looking at the quotient of two successive Fibonacci numbers. The quantity $F_n/F_{n-1}$ goes to $\phi$ as $n \to \infty$ – Mike Battaglia Oct 31 '22 at 06:10
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Aside from the quadratic construction provided by fleablood, there is also a way to solve for $\varphi$ recursively.
As $\varphi$ is the sole attractive stationary point of $f(x)=1 + \frac{1}{x}\,$, we can define the sequence $(x_n)_{n=0}^\infty$, where $x_{n+1} := 1+\frac{1}{x}$ and for any $x_0 \not \in \{1 - \varphi, 0\}$, we have
$$\lim_{n \to \infty}{x_n} = \varphi$$
(If $x_0 = 0$, then $x_1$ is undefined, and if $x_0= 1-\varphi$, then all $x_n = 1-\varphi$, as $1-\varphi$ is the other stationary point of $f$, but is unstable.)
EDIT: After further investigation, the limit of the sequence is undefined not just in the case of $x_0 = 0$, but for all $x_0$ in the sequence $(y_n)_{n=0}^\infty$ where $y_0:=0$ and $y_{n+1} := \frac{1}{y_n - 1}$. Interestingly, this sequence, defined by the inverse of the function defining $(x_n)_{n=0}^\infty$, converges to $1-\varphi$, and its defining function has an unstable stationary point at $\varphi$.
Lemmon
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Stability has nothing to do with which solution is the defined value of $\varphi$. – aschepler Oct 31 '22 at 13:53
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1I am aware, and do not say that it does. I was simply stating that for any $\varepsilon \not = 0$, $x_0 = 1 - \varphi + \varepsilon \not = 0$ gives a series that goes to $\varphi$, not $1 - \varphi$. – Lemmon Oct 31 '22 at 13:58