Continued Fraction Expansion of Golden Mean

Theorem

The golden mean has the simplest possible continued fraction expansion, namely $\sqbrk {1, 1, 1, 1, \ldots}$:

$\phi = 1 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 {\ddots} } }$

This sequence is A000012 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).


Successive Convergents

The $n$th convergent is given by:

$C_n = \dfrac {F_{n + 1} } {F_n}$

where $F_n$ denotes the $n$th Fibonacci number.


Rate of Convergence

This continued fraction expansion has the slowest rate of convergence of all simple infinite continued fractions.


Proof

Let:

$x = 1 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 {\ddots} } }$

Then:

\(\ds x\) \(=\) \(\ds 1 + \frac 1 x\) substituting for $x$
\(\ds \leadsto \ \ \) \(\ds x^2\) \(=\) \(\ds x + 1\)
\(\ds \leadsto \ \ \) \(\ds x^2 - x - 1\) \(=\) \(\ds 0\)

The result follows from Golden Mean as Root of Quadratic.

$\blacksquare$


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): $1 \cdotp 61803 \, 39887 \, 49894 \, 84820 \, 45868 \, 34365 \, 63811 \, 77203 \, 09179 \, 80576 \ldots$
  • 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): golden section
  • 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): golden section