ln (phi) dalam Continued Fraction

Konstanta Euler (e) dan Golden Ratio (\phi) telah dikenal keduanya mempunyai bentuk yang sederhana dan menawan jika dinyatakan dalam bentuk continued fraction. Sedangkan \ln(\phi) — seperti halnya juga \pi — tidak begitu sederhana bila dinyatakan dalam bentuk continued fraction, yaitu partial denominator-nya tidak menunjukan pola yang jelas ketika semua partial numerator-nya adalah 1. Continued fraction demikian sering disebut simple continued fraction.

\ln(\phi) dalam simple continued fraction :

\ln(\phi) = \displaystyle \frac{1}{2+\displaystyle \frac{1}{12+\displaystyle \frac{1}{1+\displaystyle \frac{1}{4+\displaystyle \frac{1}{6+\displaystyle \frac{1}{4+\displaystyle \frac{1}{1+\displaystyle \frac{1}{3+\displaystyle \frac{1}{1+\displaystyle \frac{1}{314+\displaystyle \frac{1}{46+\displaystyle \frac{1}{2+\displaystyle \frac{1}{3+\displaystyle \frac{1}{3+\displaystyle \frac{1}{1+\displaystyle \frac{1}{1+\displaystyle \frac{1}{1+\displaystyle \frac{1}{6+\displaystyle \frac{1}{3+\displaystyle \frac{1}{...}}}}}}}}}}}}}}}}}}}}

\ln(\phi)=[0; 2, 12, 1, 4, 6, 4, 1, 3, 1, 314, 46, 2, 3, 3, 1, 1, 1, 6, 3, 3, 11, 23, 6, 1, 4, ...]

Seperti tampak di atas, dengan semua partial numerator-nya 1, partial denominator tidak menunjukan pola yang jelas (bandingkan dengan e dan \phi dalam simple continued fraction, lihat di bagian paling bawah).

Alternatif untuk menyatakan \ln(\phi) continued fraction di atas adalah melalui generalized continued fraction. Berikut diantarnya bentuk \ln(\phi) dalam generalized continued fraction yang bisa dibuktikan :

\ln(\phi)=\displaystyle \frac{\phi-1}{1+\displaystyle \frac{1^{2}(\phi-1)}{2-(\phi-1)+\displaystyle \frac{2^{2}(\phi-1)}{3-2(\phi-1)+\displaystyle \frac{3^{2}(\phi-1)}{4-3(\phi-1)+\displaystyle \frac{4^{2}(\phi-1)}{5-4(\phi-1)+\displaystyle \frac{5^{2}(\phi-1)}{6-5(\phi-1)+\displaystyle \frac{6^{2}(\phi-1)}{...}}}}}}}

Bila dinyatakan dengan notasi continued fraction \mathrm{K} dari Gauss :

\boxed{ \ln(\phi)=\displaystyle \frac{\phi-1}{1+\underset{n=1}{\overset{\infty}{\mathrm K}}\displaystyle\frac{n^{2}(\phi-1)}{(n+1)-n(\phi-1)}} }

Generalized continued fraction untuk \ln(\phi) lainnya yang menarik dibuktikan :

\boxed{ \ln(\phi)=\displaystyle \frac{\phi-1}{1+\underset{n=1}{\overset{\infty}{\mathrm K}}\displaystyle \frac{\left( \frac{n+1}{2}  \right )^{2}(\phi-1)}{n+1}} }

\boxed{ \ln(\phi)=\displaystyle \frac{\phi-1}{\phi+\underset{n=1}{\overset{\infty}{\mathrm K}}\displaystyle \frac{-n^{2}(\phi-1)\phi}{(n+1)+(2n+1)\phi}} }

Perluasan deret ini menarik dibuktikan :

\ln(\phi)=-\displaystyle \sum_{n=1}^{\infty}\displaystyle \frac{(-1)^{n}(\phi-1)^{n}}{n}=\displaystyle \frac{(\phi-1)}{1}-\displaystyle \frac{(\phi-1)^{2}}{2}+\displaystyle \frac{(\phi-1)^{3}}{3}-...

\boxed{ \ln(\phi)=-\displaystyle \sum_{n=1}^{\infty} \displaystyle\frac{(-1)^{n}(\sqrt{5}-1)^{n}}{2^{n}n} }

————————————————————–
Bandingkan :

e dalam simple continued fraction :

e = 2+ \displaystyle \frac{1}{1+\displaystyle \frac{1}{2+\displaystyle \frac{1}{1+\displaystyle \frac{1}{1+\displaystyle \frac{1}{4+\displaystyle \frac{1}{1+\displaystyle \frac{1}{1+\displaystyle \frac{1}{6+\displaystyle \frac{1}{1+\displaystyle \frac{1}{1+\displaystyle \frac{1}{8+\displaystyle \frac{1}{1+\displaystyle \frac{1}{1+\displaystyle \frac{1}{10+\displaystyle \frac{1}{1+\displaystyle \frac{1}{1+\displaystyle \frac{1}{12+\displaystyle \frac{1}{1+\displaystyle \frac{1}{1+\displaystyle \frac{1}{...}}}}}}}}}}}}}}}}}}}}

e = [2; 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, 1, 1, 10, 1, 1, 12, 1, 1, 14, 1, 1, 16, 1, 1, 18, 1, 1, 20, ...]

\phi dalam simple continued fraction :

\phi = 1+ \displaystyle \frac{1}{1+\displaystyle \frac{1}{1+\displaystyle \frac{1}{1+\displaystyle \frac{1}{1+\displaystyle \frac{1}{1+\displaystyle \frac{1}{1+\displaystyle \frac{1}{1+\displaystyle \frac{1}{1+\displaystyle \frac{1}{1+\displaystyle \frac{1}{1+\displaystyle \frac{1}{1+\displaystyle \frac{1}{1+\displaystyle \frac{1}{1+\displaystyle \frac{1}{1+\displaystyle \frac{1}{1+\displaystyle \frac{1}{1+\displaystyle \frac{1}{1+\displaystyle \frac{1}{1+\displaystyle \frac{1}{1+\displaystyle \frac{1}{...}}}}}}}}}}}}}}}}}}}}

\phi = [1; 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...]

2 thoughts on “ln (phi) dalam Continued Fraction

  1. Pingback: Prime in Golden Tree | matematika-ku

  2. Pingback: Pendekatan Rasional terhadap ln (phi) | matematika-ku

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