Pi Approximation Day

[Thomas Klemm]

(07-22-2022 11:47 PM)Gerson W. Barbosa Wrote:  \(
\begin{align}
e^{\sqrt[11_{3}]{\frac{2222_{4}-\frac{1}{22_{9}^{2}+22_{16}^{2}+\frac{1}{1111_{7}}}}{99_{10}}}}
\end{align}
\)

I tried to convert \(2222_{4}\) in my head:

\(
\begin{align}
2222_{4}
&= 2 \cdot 1111_{4} \\
&= 2 \cdot (4^3 + 4^2 + 4 + 1) \\
&= 2 \cdot \frac{4^4 - 1}{4 - 1} \\
&= 2 \cdot \frac{17 \cdot 15}{3} \\
&= 170 \\
\end{align}
\)

And similarly \(1111_{7}\):

\(
\begin{align}
1111_{7}
&= 7^3 + 7^2 + 7 + 1 \\
&= \frac{7^4 - 1}{7 - 1} \\
&= \frac{50 \cdot 48}{6} \\
&= 400 \\
\end{align}
\)

The other numbers were easy, so I ended up with:

3.141592653589793212117310511154447
-----------------------------------
3.141592653589793238462643383279503

You never cease to amaze me with your approximations of \(\pi\).

---

Today I learned how to enter the base of a number in the Wolfram language:

E^Surd[(4^^2222 - 1/(9^^22^2 + 16^^22^2 + 1/7^^1111))/99, 3^^11]