30:T41b,

Line segments in the golden ratio	A golden rectangle (in pink) with longer side a and shorter side b, when placed adjacent to a square with sides of length a, will produce a similar golden rectangle with longer side a + b and shorter side a. This illustrates the relationship ${tex`{\displaystyle {\frac {a+b}{a}}={\frac {a}{b}}\equiv \varphi } {\frac {a+b}{a}}={\frac {a}{b}}\equiv \varphi`}
Figure 1

31:[] 2e:{"title":"Golden ratio","cells":[{"id":0,"value":"# Golden ratio\n\nIn mathematics, two quantities are in the golden ratio if their ratio is the same as the ratio of their sum to the larger of the two quantities. The \"Figure 1\" illustrates the geometric relationship. Expressed algebraically, for quantities a and b with a > b > 0,","pinned":false,"hidden":false,"mode":"md"},{"id":2,"value":"tex.block `{\\frac {a+b}{a}}={\\frac {a}{b}}\\ {\\stackrel {\\text{def}}{=}}\\ \\varphi`","pinned":false,"hidden":false,"mode":"ojs"},{"id":7,"value":"md `where the Greek letter phi (${tex `\\varphi`} or ${tex `\\phi`} ) represents the golden ratio. It is an irrational number with a value of:`","pinned":false,"hidden":false,"mode":"ojs"},{"id":13,"value":"tex.block `\\varphi ={\\frac {1+{\\sqrt {5}}}{2}}=${ϕ}\\ldots`","pinned":false,"hidden":false,"mode":"ojs"},{"id":41,"value":"ϕ = (1 + Math.sqrt(5)) / 2","pinned":true,"hidden":false,"mode":"ojs"},{"id":16,"value":"$30","pinned":false,"hidden":false,"mode":"html"},{"id":239,"value":"md `## Fibonacci\n\nThe fibonacci series can be used to approximate the value of ${tex `\\varphi`}`","pinned":false,"hidden":false,"mode":"ojs"},{"id":52,"value":"function F(n) { // F = Fibonacci\n if (n == 0) {\n return 0;\n }\n else {\n if (n == 1) {\n return 1;\n }\n else {\n return F(n-2) + F(n-1);\n }\n }\n}","pinned":false,"hidden":false,"mode":"ojs"},{"id":54,"value":"viewof n = html ``","pinned":false,"hidden":false,"mode":"ojs"},{"id":61,"value":"tex `n = ${n}`","pinned":false,"hidden":false,"mode":"ojs"},{"id":63,"value":"tex `F(${n}) = ${F(n)}`","pinned":false,"hidden":false,"mode":"ojs"},{"id":102,"value":"tex.block `\\varphi = \\lim\\limits_{n\\to \\infty }{\\frac {F(n+1)}{F(n)}} \\approx {\\frac {F(${n+1})}{F(${n})}} = ${ϕ_fib}`","pinned":false,"hidden":false,"mode":"ojs"},{"id":93,"value":"ϕ_fib = F(n) / F(n-1)","pinned":false,"hidden":false,"mode":"ojs"},{"id":97,"value":"tex `Error = ${ϕ_fib - ϕ}`","pinned":false,"hidden":false,"mode":"ojs"},{"id":123,"value":"md `## Using The Golden Ratio to Calculate Fibonacci Numbers`","pinned":false,"hidden":false,"mode":"ojs"},{"id":140,"value":"tex.block `F(n) = \\Big(\\frac {\\varphi^n - (1-\\varphi)^n} { \\sqrt {5} }\\Big)`","pinned":false,"hidden":false,"mode":"ojs"},{"id":125,"value":"function F_from_ϕ(n) {\n return Math.round((Math.pow(ϕ,n) - Math.pow((1-ϕ),n)) / Math.sqrt(5))\n}","pinned":false,"hidden":false,"mode":"ojs"},{"id":131,"value":"md `Fibonacci(n) can be calculated non-recursively, using ${tex `\\varphi`} and ${tex `\\small \\sqrt(5)`}\n\n${tex `F(${n}) = ${F_from_ϕ(n)}`}`","pinned":false,"hidden":false,"mode":"ojs"},{"id":39,"value":"md `sources:\n* https://en.wikipedia.org/wiki/Golden_ratio\n* (es-ar) [Adrian Paenza, Grandes temas de la matemática: Capítulo 4: Fibonacci](https://youtu.be/0d4o57I3rn4)`","pinned":false,"hidden":false,"mode":"ojs"}],"files":"$W31","stdlib":"1","titling":"auto"}