Is $e^{e^{2}}$ a relatively good approximation for $1000\phi$?

Question

Yesterday night, I found that $e^{e^{2}} \sim 1000\phi$, where $\phi$ is the golden ratio.

I believe that it is correct to four decimal places.

Would it be considered a relatively good approximation?

P.S. I know that an approximation is not needed, as it has a closed form, but, yeah...

$$\begin{align} e^{e^{2}} &\approx 1{,}618.17799191266 \\ 1000\phi &\approx 1{,}618.033988749895 \end{align}$$

Answer 1

The relative error is about $8.9 \cdot 10^{-5}$.

It is surprising that the relative error is somewhat small because $e$ and $\phi$ are apparently unrelated (*). It is probably a mathematical coincidence.

Whether this error implies a relatively good approximation depends on why you want this approximation or where you're going to use it.

(*) More precisely, $e$ and $\phi$ are algebraically independent because $e$ is transcendental and $\phi$ is algebraic.

Answer 2

I remember when I've found this approximation some years ago. Some other approximations of this type:

\begin{align} e^{e^{2}} &\approx 1000\phi \\ e^{e^{-2}} &\approx \ln\pi \approx \pi-2 \\ e^{e^{e^{-2}}} &\approx \pi \end{align} But these are matches just for few digits. You could find more accurate $\phi$ approximations at Golden Ratio Approximations MathWorld page. In this paper you could find relations between $e$ and $\phi$.

Answer 3

One can consider that an approximation formula is good if is achieves some information compression, i.e. it requires less bits of information than the decimal expansion of comparable accuracy.

Take the case of the well-known fractional approximations of $\pi$:

$$\frac{22}7=\color{green}{3.14}2857\cdots$$ $$\frac{335}{113}=\color{green}{3.141592}92\cdots$$

In my opinion they are virtually worthless as they don't require less digits, not counting the information contained in the shape of the formula, a fraction.

The case of $\dfrac{e^{e^2}}{1000}$ isn't so easy to evaluate, but it doesn't clearly improve on the four digits of accuracy it gives.

$\dfrac{\sqrt5+1}2$ isn't much more costly, is easier to evaluate, and isn't a bad approximation :)

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A four digits approximation of any constant isn't impressive. If you compute a million random simple expressions, you will quite probably end up finding several ones.