Golden ratio: Difference between revisions
imported>Joe Quick m (subpages) |
imported>Wlodzimierz Holsztynski m (→Properties: ort) |
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If <math>\scriptstyle \frac{a}{b}= \frac{1 + \sqrt{5}}{2}</math> it follows that <math>\scriptstyle \frac{a}{b}= \frac{a+b}{a} = 1 + \frac{b}{a}</math> | If <math>\scriptstyle \frac{a}{b}= \frac{1 + \sqrt{5}}{2}</math> it follows that <math>\scriptstyle \frac{a}{b}= \frac{a+b}{a} = 1 + \frac{b}{a}</math> | ||
With <math>\scriptstyle \Phi = 1 + \frac{1}{\Phi}</math> we could derive the | With <math>\scriptstyle \Phi = 1 + \frac{1}{\Phi}</math> we could derive the infinite [[continued fraction]] of the golden ratio: | ||
<math>\Phi = 1 + \frac{1}{\Phi} = 1 + \frac{1}{1 + \frac{1}{\Phi}} = 1 + \frac{1}{1 + \frac{1}{ 1 + \frac{1}{\Phi}}} = \dots = 1 + \frac{1}{1 + \frac{1}{ 1 + \frac{1}{1 + \dots}}}</math> | <math>\Phi = 1 + \frac{1}{\Phi} = 1 + \frac{1}{1 + \frac{1}{\Phi}} = 1 + \frac{1}{1 + \frac{1}{ 1 + \frac{1}{\Phi}}} = \dots = 1 + \frac{1}{1 + \frac{1}{ 1 + \frac{1}{1 + \dots}}}</math> | ||
Revision as of 21:49, 27 December 2007
If there is a longer line segment Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \scriptstyle a\ } and and a shorter line segment Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \scriptstyle b\ } , and if the ratio between Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \scriptstyle a + b\ } and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \scriptstyle a\ } is equal to the ratio between the line segment and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \scriptstyle b\ } , this ratio is called golden ratio. The value of the golden ratio is Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \scriptstyle \Phi = \frac{a}{b}= \frac{1 + \sqrt{5}}{2} = 1{,}618033988\dots}
Properties
If Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \scriptstyle \frac{a}{b}= \frac{1 + \sqrt{5}}{2}} it follows that
With we could derive the infinite continued fraction of the golden ratio: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Phi = 1 + \frac{1}{\Phi} = 1 + \frac{1}{1 + \frac{1}{\Phi}} = 1 + \frac{1}{1 + \frac{1}{ 1 + \frac{1}{\Phi}}} = \dots = 1 + \frac{1}{1 + \frac{1}{ 1 + \frac{1}{1 + \dots}}}}
