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# Golden ratio

The golden ratio (golden section, golden mean) is the positive root
 ϕ
${\displaystyle {\begin{array}{l}{\displaystyle x^{2}-x-1=0,}\end{array}}}$

which has roots

${\displaystyle {\begin{array}{l}{\displaystyle \phi ={\frac {1+{\sqrt {5}}}{2}},\ \varphi ={\frac {1-{\sqrt {5}}}{2}}.}\end{array}}}$

Note that

{\displaystyle {\begin{array}{l}{\displaystyle {\begin{aligned}\phi +\varphi &=1,\\\phi \,\varphi &=-1.\end{aligned}}}\end{array}}}

## Decimal expansion of the golden ratio

The decimal expansion of the golden ratio (A001622) is

${\displaystyle \phi =1.6180339887498948482045868343656381177203091798057628621\ldots }$

and the decimal expansion of the conjugate root of the golden ratio is

${\displaystyle \varphi =-0.6180339887498948482045868343656381177203091798057628621\ldots }$

Since

${\displaystyle x\,(x-1)=1,}$
the multiplicative inverse of the root
 x
is
 x  −  1
(same fractional part), and since
${\displaystyle x+[-(x-1)]=1,}$
the root
 x
 1
.

## Powers of ϕ and Fibonacci numbers

${\displaystyle \phi ^{n}={\bigg (}{\frac {1+{\sqrt {5}}}{2}}{\bigg )}^{n}=F_{n-1}+F_{n}\,\phi ,}$
where
 ϕ
is the golden ratio and
 Fn
is the
 n
th Fibonacci number.

Powers of
 ϕ

 n
 ϕ n = Fn  − 1 + Fn  ϕ
 ϕ  − n + ϕ n
6
 5 + 8 ϕ
18
5
 3 + 5 ϕ
4
 2 + 3 ϕ
7
3
 1 + 2 ϕ
2
 1 + 1 ϕ
3
1
 0 + 1 ϕ
0
 1 + 0 ϕ
2
−1
 − 1 + 1 ϕ
−2
 2 + ( − 1) ϕ
3
−3
 − 3 + 2 ϕ
−4
 5 + ( − 3) ϕ
7
−5
 − 8 + 5 ϕ
−6
 13 + ( − 8) ϕ
18

## Continued fraction and nested radicals expansions

The golden ratio has the simplest continued fraction expansion (the all ones sequence A000012)

${\displaystyle {\begin{array}{l}{\displaystyle \phi ={1+{\cfrac {1}{1+{\cfrac {1}{1+{\cfrac {1}{1+{\cfrac {1}{1+{\cfrac {1}{\ddots }}}}}}}}}}}=1+[1+[1+[1+[1+[1+\cdots \,]^{-1}]^{-1}]^{-1}]^{-1}]^{-1},}\end{array}}}$

since

${\displaystyle {\begin{array}{l}{\displaystyle \phi -1={\frac {1}{\phi }},}\end{array}}}$

and also the simplest nested radicals expansion (again, the all one's sequence)

${\displaystyle {\begin{array}{l}{\displaystyle \phi ={\sqrt {1\,+\,{\sqrt {1\,+\,{\sqrt {1\,+\,{\sqrt {1\,+\,{\sqrt {\cdots }}}}}}}}}}=[1+[1+[1+[1+[1+\cdots \,]^{\frac {1}{2}}]^{\frac {1}{2}}]^{\frac {1}{2}}]^{\frac {1}{2}}]^{\frac {1}{2}},}\end{array}}}$

since

${\displaystyle {\begin{array}{l}{\displaystyle \phi ^{2}-1=\phi .}\end{array}}}$

## Approximations

${\displaystyle e-{\frac {11}{10}}=1.61828182845904\ldots (1.000153173364\ldots \times \phi ),}$
where
 e
is Euler's number.
${\displaystyle {\sqrt {\frac {5\pi }{6}}}=1.6180215937964\ldots (0.999992339\ldots \times \phi ).}$

## As an infinite series

${\displaystyle {\begin{array}{l}{\displaystyle \phi =\sum _{k=0}^{\infty }\left({\frac {3-{\sqrt {5}}}{2}}\right)^{k}=\sum _{k=0}^{\infty }\left(1+\varphi \right)^{k}={\frac {1}{1-(1+\varphi )}}={\frac {-1}{\varphi }}=\phi .}\end{array}}}$

${\displaystyle {\begin{array}{l}{\displaystyle \varphi =\sum _{k=0}^{\infty }\phi ^{-2k}=\sum _{k=0}^{\infty }\left({\frac {1}{\phi ^{2}}}\right)^{k}=}\end{array}}}$ ${\displaystyle {\begin{array}{l}{\displaystyle \sum _{k=0}^{\infty }\left({\frac {1}{\phi +1}}\right)^{k}={\frac {1}{1-(\phi +1)}}={\frac {-1}{\phi }}=\varphi .}\end{array}}}$