Hypothetically you are put in math jail and the jailer says he will let you out only if you can give him 707 digits of pi. You can have a ream of paper and a couple pens, no computer, books, previous pi memorization or outside help.

What is the best formula to use? Where best will result in the least margin of error (most important) and is decently quick (second importance).

707 probably seems arbitrary, but I got it from here: http://en.wikipedia.org/wiki/William_Shanks

Also I don’t really understand how he could use that formula because I thought you need a table to get arctan values or it would probably take a long time to make values of arctan to use in the formula, but that isn’t too important.

I asked this question today because it celebrates the most accurate pi day for the next 100 years 🙂

**Answer**

As metioned in Wikipedia’s biography, Shanks used Machin’s formula

$$ \pi = 16\arctan(\frac15) – 4\arctan(\frac1{239}) $$

The standard way to use that (and the various Machin-*like* formulas found later) is to compute the arctangents using the power series

$$ \arctan x = x – \frac{x^3}3 + \frac{x^5}5 – \frac{x^7}7 + \frac{x^9}9 – \cdots $$

Getting $\arctan(\frac15)$ to 707 digits requires about 500 terms calculated to that precision. Each requires two long divisions — one to divide the previous numerator by 25, another to divide it by the denominator.

The series for $\arctan(\frac1{239})$ converges faster and only needs some 150 terms.

(You can know how many terms you need because the series is *alternating* (and absolutely decreasing) — so once you reach a term that is smaller than your desired precision, you can stop).

The point of Machin-like formulas is that the series for $\arctan x$ converges faster the smaller $x$ is. We could just compute $\pi$ as $4\arctan(1)$, but the series converges *hysterically slowly* when $x$ is as large as $1$ (and not at all if it is even larger). The trick embodied by Machin’s formula is to express a straight angle as a sum/difference of the corner angles of (a small number of different sizes of) long and thin right triangles with simple integer ratios between the cathetes.

The arctangent gets easier to compute the longer and thinner each triangle is, and especially if the neighboring side is an integer multiple of the opposite one, which corresponds to angles of the form $\arctan\frac{1}{\text{something}}$. Then going from one numerator in the series to the next costs only a division, rather than a division *and* a multiplication.

Machin observed that four copies of the $5$-$1$-$\sqrt{26}$ triangle makes the same angle as an $1$-$1$-$\sqrt2$ triangle (whose angle is $\pi/4$) plus one $239$-$1$-$\sqrt{239^2+1}$ triangle. These facts can be computed exactly using the techniques displayed here.

Later workers have found better variants of Machin’s idea, nut if you’re in prison without reference works, it’s probably easiest to rediscover Machin’s formula by remembering that some number of copies of $\arctan\frac1k$ for some fairly small $k$ adds up to something very close to 45°.

**Attribution***Source : Link , Question Author : Neil , Answer Author : Community*