Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

From what I have read, it appears that the most efficient methods of calculating $ \pi $ are Machin-like formulae. And it is known that certain formulas are more efficient than others.

Are there any theoretical properties (rates of convergence, etc) of these formulas that make them more efficient than others?

Like, why is $$ \frac{\pi}{4} = 183\arctan\frac{1}{239} + 32\arctan\frac{1}{1023} - 68\arctan\frac{1}{5832} + 12\arctan\frac{1}{110443} - 12\arctan\frac{1}{4841182} - 100\arctan\frac{1}{6826318} $$ more efficient than $$ \frac{\pi}{4} = \arctan\frac{1}{2} + \arctan\frac{1}{3} $$ (as claimed by Wikipedia)

Or is this a purely empirical result?

I presume that the calculation of arctan is done using the Maclaurin series.

share|cite|improve this question
Many of the recent record computations use algorithms that are not at all Machin-like. In Machin-like methods, it is basically Maclaurin series, with speedup techniques. Note that if we have a "long" Machin-like formulas, like the $6$-term one you quote, the individual terms can be computed in parallel. – André Nicolas Jun 4 '12 at 22:27

With the Maclaurin series, after the first few summations, each successive sum will be slightly less than $1/4$ the value of the previous one when calculating $\arctan(1/2)$.

However, with $\arctan(1/239)$, each successive term in the series is less than $1/57,000$ the size as the previous term, so you are able to generate multiple additional decimal places with each additional term in the sequence.

However, to calculate millions of digits very quickly (using a computer, of course), it is better to use other algorithms.

See Chudnovsky algorithm (according to wikipedia, this was used to calculate 10 trillion digits of pi)

share|cite|improve this answer

In the Maclaurin series of $\arctan(x)$, the $n$'th term is $x^{2n-1}/(2n-1)$. When $|x|$ is small, the important part here is the $x^{2n-1}$. So to get accuracy $\epsilon$, you need $2n-1 \approx \log (\epsilon)/\log (x)$. Roughly speaking, in a Machin-like formula $\sum_j a_j \arctan(1/b_j)$ the number of terms you need to calculate is inversely proportional to the log of the smallest $b_j$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.