What is the formula for pi used in the Python decimal library?
(Don't be alarmed by the title; this is a question about mathematics, not programming.)
In the documentation for the decimal
module in the Python Standard Library, an example is given for computing the digits of $\pi$ to a given precision:
def pi():
"""Compute Pi to the current precision.
>>> print(pi())
3.141592653589793238462643383
"""
getcontext().prec += 2 # extra digits for intermediate steps
three = Decimal(3) # substitute "three=3.0" for regular floats
lasts, t, s, n, na, d, da = 0, three, 3, 1, 0, 0, 24
while s != lasts:
lasts = s
n, na = n+na, na+8
d, da = d+da, da+32
t = (t * n) / d
s += t
getcontext().prec -= 2
return +s # unary plus applies the new precision
I was not able to find any reference for what formula or fact about $\pi$ this computation uses, hence this question.
Translating from code into more typical mathematical notation, and using some calculation and observation, this amounts to a formula for $\pi$ that begins like:
$$\begin{align}\pi &= 3+\frac{1}{8}+\frac{9}{640}+\frac{15}{7168}+\frac{35}{98304}+\frac{189}{2883584}+\frac{693}{54525952}+\frac{429}{167772160} + \dots\\ &= 3\left(1+\frac{1}{24}+\frac{1}{24}\frac{9}{80}+\frac{1}{24}\frac{9}{80}\frac{25}{168}+\frac{1}{24}\frac{9}{80}\frac{25}{168}\frac{49}{288}+\frac{1}{24}\frac{9}{80}\frac{25}{168}\frac{49}{288}\frac{81}{440}+\frac{1}{24}\frac{9}{80}\frac{25}{168}\frac{49}{288}\frac{81}{440}\frac{121}{624}+\frac{1}{24}\frac{9}{80}\frac{25}{168}\frac{49}{288}\frac{81}{440}\frac{121}{624}\frac{169}{840}+\dots\right) \end{align}$$
or, more compactly,
$$\pi = 3\left(1 + \sum_{n=1}^{\infty}\prod_{k=1}^{n}\frac{(2k-1)^2}{8k(2k+1)}\right)$$
Is this a well-known formula for $\pi$? How is it proved? How does it compare to other methods, in terms of how how quickly it converges, numerical stability issues, etc? At a glance I didn't see it on the Wikipedia page for List of formulae involving π or on the MathWorld page for Pi Formulas.
That is the Taylor series of $\arcsin(x)$ at $x=1/2$ (times 6).
This approximation for $\pi$ is attributed to Issac Newton:
- https://loresayer.com/2016/03/14/pi-infinite-sum-approximation/
- http://www.geom.uiuc.edu/~huberty/math5337/groupe/expresspi.html
- http://www.pi314.net/eng/newton.php
When I wrote that code shown in the Python docs, I got the formula came from p.53 in "The Joy of π". Of the many formulas listed, it was the first that:
- converged quickly,
- was short,
- was something I understood well-enough to derive by hand, and
- could be implemented using cheap operations: several additions with only a single multiply and single divide for each term. This allowed the estimate of $\pi$ to be easily be written as an efficient function using Python's floats, or with the decimal module, or with Python's multi-precision integers.
The formula solves for π in the equation $sin(\pi/6)=\frac{1}{2}$.
WolframAlpha gives the Maclaurin series for $6 \arcsin{(x)}$ as:
$$6 \arcsin{(x)} = 6 x + x^{3} + \frac{9 x^{5}}{20} + \frac{15 x^{7}}{56} + \frac{35 x^{9}}{192} + \dots $$
Evaluating the series at $x = \frac{1}{2}$ gives:
$$ \pi \approx 3+3 \frac{1}{24}+3 \frac{1}{24}\frac{9}{80}+3 \frac{1}{24}\frac{9}{80}\frac{25}{168}+\dots + \frac{(2k+1)^2}{16k^2+40k+24} + \dots\\ $$
From there, I used finite differences, to incrementally compute the numerators and denominators. The numerator differences were 8, 16, 24, ...
, hence the numerator adjustment na+8
in the code. The denominator differences were 56, 88, 120, ...
, hence the denominator adjustment da+32
in the code:
1 9 25 49 numerators
8 16 24 1st differences
8 8 2nd differences
24 80 168 288 denominator
56 88 120 1st differences
32 32 2nd differences
Here is the original code I wrote back in 1999 using Python's multi-precision integers (this predates the decimal module):
def pi(places=10):
"Computes pi to given number of decimal places"
# From p.53 in "The Joy of Pi". sin(pi/6) = 1/2
# 3 + 3*(1/24) + 3*(1/24)*(9/80) + 3*(1/24)*(9/80)*(25/168)
# The numerators 1, 9, 25, ... are given by (2x + 1) ^ 2
# The denominators 24, 80, 168 are given by 16x^2 +40x + 24
extra = 8
one = 10 ** (places+extra)
t, c, n, na, d, da = 3*one, 3*one, 1, 0, 0, 24
while t > 1:
n, na, d, da = n+na, na+8, d+da, da+32
t = t * n // d
c += t
return c // (10 ** extra)
It just computing $\pi = 6\sin^{-1}\left(\frac12\right)$ using the Taylor series expansion of arcsine. For reference,
$$6\sin^{-1}\frac{t}{2} = 3t+\frac{t^3}{8}+\frac{9t^5}{640}+\frac{15t^7}{7168}+\frac{35 t^9}{98304} + \cdots$$ and compare the coefficients with what you get.