Approximating $\pi$ with least digits

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.

Perhaps you can use the Gauss Legendre Algorithm ? With 25 iterations, you will produce more than 45 million digits

For example, use bc with the desire scale, and do the algorithm. For a scale of 50, you obtain :

$$\pi_0=\color{red}{2.91421356237309504880168872420969807856967187537693}$$ $$\pi_1=3.14\color{red}{057925052216824831133126897582331177344023751285}$$ $$\pi_2=3.1415926\color{red}{4621354228214934443198269577431443722334535}$$ $$\pi_3=3.141592653589793238\color{red}{27951277480186397438122550483492}$$ $$\pi_4=3.1415926535897932384626433832795028841971\color{red}{1467828263}$$ $$\pi_5=3.14159265358979323846264338327950288419716939937\color{red}{240}$$

(the last 3 digits of error don't come from the algorithm but from the limited scale)

For a scale of 1000, you obtain :

$\pi_9=$3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679821480865132823066470938446095505822317253594081284811174502841027019385211055596446229489549303819644288109756659334461284756482337867831652712019091456485669234603486104543266482133936072602491412737245870066063155881748815209209628292540917153643678925903600113305305488204665213841469519415116094330572703657595919530921861173819326117931051185480744623799627495673518857527248912279381830119491298336733624406566430860213949463952247371907021798609437027705392171762931767523846748184676694051320005681271452635608277857713427577896091736371787214684409012249534301465495853710507922796892589235420199561121290219608640344181598136297747713099605187072113499999983729780499510597317328160963185950244594553469083026425223082533446850352619311881710100031378387528865875332083814206171776691473035982534904287554687311595628638823537875937519577818577805321712268066130019278766111959092164$\color{red}{199382}$

So you work on a fixed scale and stop when the algorithm is almost on a fixed point. Only the (very) few last digits will be wrong. And you only need $\log(n)$ iterations for a scale of $n$ ($n$ digits).