a simple formula linking the value of $e$ to the Golden Ratio $\phi$

Question

These last days, I was looking for an approximation formula to $\pi$. But, surprisingly, the formulas led to this other one:

$$ e = \left (\frac {\phi} {\phi - 1} \right)^{\frac {1} {2\text{Log}\phi}}\text{where }\phi\text { is the Golden Ratio} : \phi = \frac {1 + \sqrt {5}} {2} $$

Personally, I've never heard about something similar. Does someone know something about this result?

EDIT1: I would like to apology. The answer is so obvious that I didn't see it, because I got it by more complicated (than necessary) equations solving. I didn't really check my result and asked the question before really thinking to it. I've flagged this question to be deleted. sorry.

Is only a bit more surprising that $e^{\log x}=x$. The equation $\phi^2=\phi+1$ does the rest. — Martín-Blas Pérez Pinilla, Nov 19 '14 at 09:49
thank you @martin, I feel really bad about this question. See my EDIT1 on the answer. — Eddy Khemiri, Nov 19 '14 at 17:06
Don't feel bad. The question is simple but interesting. I've upvoted it. — Martín-Blas Pérez Pinilla, Nov 19 '14 at 19:22

score 4 · Accepted Answer · answered Nov 19 '14 at 09:43

4

Since $(\phi/(\phi-1))=\phi^2,$ after raising both sides to the $\log \phi$ this relation becomes an identity.

answered Nov 19 '14 at 09:43

coffeemath

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score 3 · Answer 2 · answered Nov 19 '14 at 09:42

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Here a simple one (smile):

$$\log{e}=\phi+1/\phi$$

answered Nov 19 '14 at 09:42

newzad

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score 0 · Answer 3 · answered Nov 19 '14 at 09:53

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$$e = \left (\frac {\phi} {\phi - 1} \right)^{\frac {1} {2\text{Log}\phi}}\iff \phi^2=(e^{\log\phi})^2=\frac{\phi}{\phi-1}=\cdots $$

answered Nov 19 '14 at 09:53

Martín-Blas Pérez Pinilla

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a simple formula linking the value of $e$ to the Golden Ratio $\phi$

3 Answers3