Question 190809
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \frac{L}{1-L} = \frac{1}{L}]


is a delightful little rational equation, but unfortunately it has nothing whatever to do with the golden ratio.  Close, but no cigar I'm afraid.


The golden ratio is generally defined in terms of a golden section, that is a line segment divided into two parts <i><b>a</b></i> and <i><b>b</b></i> such that the following proportion holds:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{a + b}{a}=\frac{a}{b}=\varphi]


And <i><b>that</b></i> is the golden ratio.  The above has the unique positive solution for *[tex \LARGE \varphi = \frac{1+sqrt{5}}{2}] whereas your equation has the unique positive solution *[tex \LARGE L = \frac{-1+sqrt{5}}{2}] -- a very different number indeed.


However, I did say you were close.  You only missed it by a sign.  Your equation should have been:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \frac{L}{1+L} = \frac{1}{L}]


Where <i><b>L</b></i> is the golden ratio.  *[tex \LARGE {1 \over L}] is actually the reciprocal of the golden ratio, often named *[tex \LARGE \Phi]


Proof:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \frac{L}{1-L} = \frac{1}{L}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \frac{L^2}{(1-L)L} = \frac{1-L}{(1-L)L}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  L^2 = 1 - L]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  L^2 + L - 1 = 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{-1 \pm sqrt{1^2 - 4(1)(-1)}}{2(1)} = \frac{-1 \pm sqrt{5}}{2} ]


Exclude the negative root. *[tex \LARGE \frac{-1 + sqrt{5}}{2} ]


But:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \frac{L}{1+L} = \frac{1}{L}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \frac{L^2}{(1+L)L} = \frac{1+L}{(1+L)L}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  L^2 = 1 + L]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  L^2 - L - 1 = 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{-(-1) \pm sqrt{(-1)^2 - 4(1)(-1)}}{2(1)} = \frac{1 \pm sqrt{5}}{2} ]


Again, exclude the negative root. *[tex \LARGE \frac{1 + sqrt{5}}{2}]


But how do I know that *[tex \LARGE \frac{1 + sqrt{5}}{2}] is the correct value for the golden ratio?  Back to the original definition:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{a + b}{a}=\frac{a}{b}=\varphi]


From which we can see that *[tex \LARGE a = b\varphi]


Substituting:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{b\varphi + b}{b\varphi}=\frac{b\varphi}{b}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\varphi + 1}{\varphi}=\varphi]


And solving:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\varphi + 1}{\varphi}=\frac{\varphi^2}{\varphi}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \varphi^2 - \varphi - 1 = 0]


Which is the same as


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  L^2 - L - 1 = 0]


that we just solved above.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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