Continued Fraction Expansion of Golden Mean/Successive Convergents
Theorem
Consider the continued fraction expansion to the golden mean:
- $\phi = \sqbrk {1, 1, 1, 1, \ldots} = 1 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 {\ddots} } }$
The $n$th convergent is given by:
- $C_n = \dfrac {F_{n + 1} } {F_n}$
where $F_n$ denotes the $n$th Fibonacci number.
Proof
The proof proceeds by induction.
Listing the first few convergents, which can be calculated:
- $C_1 = \dfrac 1 1$
- $C_2 = \dfrac 2 1$
- $C_3 = \dfrac 3 2$
- $C_4 = \dfrac 5 3$
and so on.
For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:
- $C_n = \dfrac {F_{n + 1} } {F_n}$
$\map P 1$ is the case:
- $C_1 = \dfrac {F_{n + 1} } {F_n}$
\(\ds C_1\) | \(=\) | \(\ds \dfrac 1 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {F_2} {F_1}\) | Definition of Fibonacci Numbers |
Thus $\map P 1$ is seen to hold.
Basis for the Induction
$\map P 2$ is the case:
\(\ds C_2\) | \(=\) | \(\ds 1 + \cfrac 1 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 2 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {F_3} {F_2}\) | Definition of Fibonacci Numbers |
Thus $\map P 2$ is seen to hold.
This is the basis for the induction.
Induction Hypothesis
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
- $C_k = \dfrac {F_{k + 1} } {F_k}$
from which it is to be shown that:
- $C_{k + 1} = \dfrac {F_{k + 2} } {F_{k + 1} }$
Induction Step
This is the induction step:
Let $C_n$ be expressed as $\dfrac {p_n} {q_n}$ for any given $n$.
\(\ds C_{k + 1}\) | \(=\) | \(\ds \dfrac {p_{k + 1} } {q_{k + 1} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {p_k + p_{k - 1} } {q_k + q_{k - 1} }\) | Definition of Numerators and Denominators of Continued Fraction | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {F_{k + 1} + F_k} {F_k + F_{k - 1} }\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {F_{k + 2} } {F_{k + 1} }\) | Definition of Fibonacci Numbers |
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\forall n \in \Z_{>0}: C_n = \dfrac {F_{n + 1} } {F_n}$
$\blacksquare$
Sources
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $1 \cdotp 61803 \, 39887 \, 49894 \, 84820 \, 45868 \, 34365 \, 63811 \, 77203 \, 09179 \, 80576 \ldots$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $1 \cdotp 61803 \, 39887 \, 49894 \, 84820 \, 45868 \, 34365 \, 63811 \, 77203 \, 09179 \, 80576 \ldots$