Is it possible :$\pi \sim{\frac{-\gamma}{(\sqrt{2}e^2-\gamma)^4}}-e^{-13}+2(17^{\frac13}-1)-2\phi^{-16}-\sum_{n=1}^{\infty}(\frac{1}{17})^{2n+3}$?

54 Views Asked by At

I have done many attempts to give another approximation for $\pi$ I have got this

$$\pi \sim{\frac{-\gamma}{(\sqrt{2}e^2-\gamma)^4}}-e^{-13}+2(17^{\frac13}-1)-2\phi^{-16}-\sum_{n=1}^{\infty}(\frac{1}{17})^{2n+3}$$, My question here : Is it possible to accept this approximation if it is not known before , probably my computation with wolfram alpha not good however I have got an approximation with $10^{-7}$

Note:$\phi$ is the Golden ratio,$\gamma$ is the Euler -Mascheronni constant

1

There are 1 best solutions below

5
On

This too long for comments.

Be sure that I fully respect your work but when you look at the value of the different terms you introduced $$\left( \begin{array}{cc} \frac{-\gamma}{(\sqrt{2}e^2-\gamma)^4} & -0.0000607619504779 \\ -e^{-13} & -0.0000022603294070 \\ 2 \left(\sqrt[3]{17}-1\right) & +3.1425631813164707 \\ -2\phi^{-16} & -0.0009062077075697 \\ -\sum_{n=1}^{\infty}(\frac{1}{17})^{2n+3} & -0.0000007067417509 \\ \text{total} & +3.1415932445872653 \\ \pi & +3.1415926535897932 \end{array} \right)$$

So, the very interesting part is $ 2 \left(\sqrt[3]{17}-1\right)$.

Working $2 \left(\sqrt[3]{17}-1\right)-\pi$, we could find $$\pi \sim 2 \left(\sqrt[3]{17}-1\right)-\frac{81\ 3^{34/45} \log ^{\frac{283}{90}}(3)}{8000\ 2^{13/90} e^{39/10} \log^{\frac{13}{9}}(2)}=3.141592653589793238461944$$ which is correct for $22$ decimal places (but which does present much interest).

Currently, using CAS, we can make so good approximations !

For example, working your problem of $\pi$, I found that, using only whole numbers, a good approximation of it is the reciprocal of the real root of the simple cubic $$31352 x^3-28993 x^2+53597 x-15134=0$$ $$\pi\sim\frac{94056}{28993+2 \sqrt{4200525383} \sinh \left(\frac{1}{3} \sinh ^{-1}\left(\frac{5960683043029}{4200525383 \sqrt{4200525383}}\right)\right)}$$ which gives $20$ exact significant figures (does this show any interest ?).

In the same spirit $$\frac{248359849 \binom{e}{e!}+250062517 \binom{e!}{e}}{485577274}$$ gives $25$ significant digits.