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\(e+\frac{69-e^{-2^{2}\ln \ln \left ( e^{e^{2}}+\ln \ln \left ( 2 \right ) \right )}}{163}\)





'e+(69-e^(-2^2*LN(LN(e^e^2+LN(LN(2))))))/163'

I don't know... expressing one number with inifinite digits using another one? AFAIK, Euler once found a much more elegant way of linking Pi with e.

(01-14-2017 08:27 PM)Maximilian Hohmann Wrote: [ -> ]I don't know... expressing one number with infinite digits using another one? AFAIK, Euler once found a much more elegant way of linking Pi with e.

This is just an approximation starting from the fact that 163∙(π - e) ≃ 69, not an exact relationship in the realm of complex numbers like Euler's beautiful and wonderful formula. I think my correction term involving two levels of exponentiation and logarithm and the integer number 2 may have some aesthetic value, anyway beauty lies in the eyes of the beholder :-)

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Edited (22:Jan)

The following doesn't require a wider HP-50g screen to show the whole equation...



...but it does require a 16-digit calculator to show the accuracy of approximation.


'e+(69-e^(-8/(1+1/(163*401))))/163'