Approximating $\pi$ with least digits

Do you a digit efficient way to approximate $\pi$? I mean representing many digits of $\pi$ using only a few numeric digits and some sort of equation. Maybe mathematical operations also count as penalty.

For example the well known $\frac{355}{113}$ is an approximation, but it gives only 7 correct digits by using 6 digits (113355) in the approximation itself. Can you make a better digit ratio?

EDIT: to clarify the "game" let's assume that each mathematical operation (+, sqrt, power, ...) also counts as one digit. Otherwise one could of course make artifical infinitely nested structures of operations only. And preferably let's stick to basic arithmetics and powers/roots only.

EDIT: true. logarithm of imaginary numbers provides an easy way. let's not use complex numbers since that's what I had in mind. something you can present to non-mathematicians :)

Using continued fractions you may get as many ratios as you wish and all the ratios are as good as possible!) :

I'll steal part of what I wrote in this other thread and adapt it!

To use continued fractions for evaluation of $x=\pi$.

At each step :

• note the integer part $j\leftarrow \lfloor x\rfloor$ (illustrated in blue)
• compute the new fraction $f$ as illustrated :
  • write the previous fraction
  • multiply the numerator and denominator by $j$
  • add the numerator and denominator of the previous previous fraction (starting with $\frac 10$)
• evaluate the fractional part $x\leftarrow x-j$
• stop when $x$ becomes $0$ or at least very small (depending of the precision of your evaluation) or when you decide too...
• else compute $x$'s multiplicative inverse $x\leftarrow \frac 1x$ and repeat

$ \begin{array} {l|r|ccccc} x&j&&&&&f\\ \hline\\ 3.141592653589\cdots & \color{#0000ff}{3} & \color{#0000ff}{3}&=&\frac {\color{#0000ff}{3}}1&=&\frac 31\\ 1/0.141592\cdots=7.0625\cdots & \color{#0000ff}{7} &3+\cfrac 1{\color{#0000ff}{7}}&=& \frac {3\cdot \color{#0000ff}{7}+1}{1\cdot \color{#0000ff}{7} +0}&=&\frac {22}{7}\\ 1/7.0625\cdots=15.99659\cdots & \color{#0000ff}{15}&3+\cfrac 1{7+\cfrac 1{\color{#0000ff}{15}}}&=& \frac {22\cdot \color{#0000ff}{15}+3}{7\cdot \color{#0000ff}{15} +1}&=&\frac {333}{106}\\ 1/0.99659\cdots=1.0034172\cdots & \color{#0000ff}{1}&3+\cfrac 1{7+\cfrac 1{15+\cfrac 1{\color{#0000ff}{1}}}}&=& \frac {333\cdot \color{#0000ff}{1}+22}{106\cdot \color{#0000ff}{1} +7}&=&\frac {355}{113}\\ \cdots &\\ 1/0.003417231\cdots=292.63459\cdots & \color{#0000ff}{292}&3+\cfrac 1{7+\cfrac 1{15+\cfrac 1{1+\cfrac 1{\color{#0000ff}{292}}}}}&=& \frac {355\cdot \color{#0000ff}{292}+333}{113\cdot \color{#0000ff}{292} +106}&=&\frac {103993}{33102}\\ \cdots &\\ \end{array} $

without end since $\pi$ is transcendental!

With pure rational approximations, there are sharp limits (related to Roth's Theorem) in terms of how far you can go.

More generally, this will depend strongly on the operations allowed. For example, $$ \log(0-1)/\sqrt{0-1} $$ has 5 operations and 4 digits and is exact.