Yet another new pi approximation - Printable Version +- HP Forums (https://www.hpmuseum.org/forum) +-- Forum: Not HP Calculators (/forum-7.html) +--- Forum: Not remotely HP Calculators (/forum-9.html) +--- Thread: Yet another new pi approximation (/thread-7576.html) |
Yet another new pi approximation - Gerson W. Barbosa - 01-14-2017 05:57 PM \(e+\frac{69-e^{-2^{2}\ln \ln \left ( e^{e^{2}}+\ln \ln \left ( 2 \right ) \right )}}{163}\) 'e+(69-e^(-2^2*LN(LN(e^e^2+LN(LN(2))))))/163' RE: Yet another new pi approximation - Maximilian Hohmann - 01-14-2017 08:27 PM 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. RE: Yet another new pi approximation - Gerson W. Barbosa - 01-14-2017 10:48 PM (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 :-) ----- 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' |