Computing π with Taylor Series

f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n \] \[ \arctan(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1}

Recall that the Taylor series for a function \(f(x)\) centered at 0 sums derivatives at the origin. For \(\arctan(x)\), working out the derivatives and substituting gives the Gregory series — valid for \(|x| \leq 1\).

\arctan(1) = \frac{\pi}{4} \] \[ \frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots

Since \(\arctan(1) = \frac{\pi}{4}\), plugging in \(x = 1\) gives the Leibniz formula. It's beautiful — but practically useless for computation. The series converges so slowly that you need millions of terms for just a handful of correct digits.

\frac{\pi}{4} = 4\arctan\!\left(\frac{1}{5}\right) - \arctan\!\left(\frac{1}{239}\right) \] \[ \arctan(a) + \arctan(b) = \arctan\!\left(\frac{a+b}{1-ab}\right)

In 1706, John Machin observed this identity, verifiable via the arctan addition formula. The key insight: both arguments are small fractions. Since each term in the arctan series is \(x^{2n+1}/(2n+1)\), small \(x\) means the series shrinks very rapidly — giving fast convergence instead of the glacial Leibniz pace.

\text{Leibniz: } \sim \frac{10^d}{2} \text{ terms for } d \text{ digits} \] \[ \text{Machin: } \sim \frac{d}{1.4} \text{ terms for } d \text{ digits}

For the Leibniz formula, the \(n\)-th term is \(1/(2n+1)\). To get error below \(10^{-d}\) you need roughly \(10^d / 2\) terms.

For Machin's formula, the dominant term shrinks by a factor of \(1/25\) each step, adding roughly \(\log_{10}(25) \approx 1.4\) new correct digits per iteration. For 314 digits we need around 224 terms — a dramatic improvement.

Terms: 1

The slider controls how many terms of Machin's formula are used. Watch how the approximation homes in on π, and compare the error to the naive Leibniz series at the same term count.

pi = 3.14159265358979323846264338327950288 419716939937510582097494459230781640 628620899862803482534211706798214808 651328230664709384460955058223172535 940812848111745028410270193852110555 964462294895493038196...

To go beyond floating-point precision (about 15 digits), we use Python's decimal module, which supports arbitrary-precision arithmetic. The algorithm is identical — we swap in Decimal types and set getcontext().prec high enough.

The loop runs until the current term drops below \(10^{-(d+10)}\), where \(d\) is the target digit count. Those extra 10 guard digits prevent rounding errors from creeping into the final result. Running with precision = 314 produces all 314 digits of π correctly.

Summary

The Taylor series for arctan gives us a direct route to π via \(\arctan(1) = \pi/4\), but naive evaluation converges too slowly to be useful. Machin's formula sidesteps this by decomposing \(\pi/4\) into arctan evaluations at small arguments, where the series converges geometrically. With arbitrary-precision arithmetic, a few hundred terms is all it takes to compute π to hundreds of digits.