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