Approximations for $ \pi $

I would like to know how are approximations for $\pi$ like $\frac{355}{113}$ computed.I tried to find some by guess and check and I came up with $\frac{65888736958667}{20973036362107}$ ( Yeah,it's very hard to memorize) but it's only accurate to 15 decimal digits.My method was to choose a random number for the denominator and multiply it by the value of $\pi$ up to 20 digits to get the numerator.So,are there more efficient methods to compute approximations for $\pi$ ?Thanks for any answers in advance!

#Pi #Approximation

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

Usually fractional approximations of irrational numbers( $e$ , $\sqrt(2)$ ..) can be easily computed from their Continued Fraction representation.

For instance the constant $e= [2;1,2,1,1,4,1,1,6,1,1,8,…]$ (AO03417) as you can see, follows a simple pattern in its continued fraction representation. By using this pattern we can easily compute more and more accurate convergents of $e$ .

Unfortunately the continued fraction representation of $\pi= [3;7,15,1,292,1,1,1,2,1,3,1,…]$ (AO01203) contains seemingly random digits.

However it is possible to compute such convergents by first finding the decimal representation of $\pi$ and then approximate the decimal using the continued fraction method .

To demonstrate, $\pi$ to seven decimal places is:

$\pi \approx 3.1415927$ Taking the reciprocal of $0.1415927$ $\pi \approx 3+ {1\over\displaystyle 7.0625133}$ And repeating this over and over we get.. $\pi \approx 3+ {1\over \displaystyle 7 + {1\over \displaystyle 15 + {1\over \displaystyle 1 + {1\over \displaystyle 293 + {1\over \displaystyle 10.320556}}}}}$ and so on.. If we truncate the fraction at $15$ it gives the approximation $\pi \approx 3+ {1\over \displaystyle 7 + {1\over \displaystyle 15 + 1}} = {3 + {16\over 113}} = {355\over 113}$

Since python already implements the rational approximation algorithm in its fractions module , all we have to do is find a large string of $\pi$ (I used the ‎The Spigot Algorithm here) and use the .limit_denominator(x) method to find a rational approximation.

from fractions import *
from decimal import *
#Spigot implementation http://stackoverflow.com/questions/9004789/1000-digits-of-pi-in-python
def make_pi():
    q, r, t, k, m, x = 1, 0, 1, 1, 3, 3
    for j in range(1000):
        if 4 * q + r - t < m * t:
            yield m
            q, r, t, k, m, x = 10*q, 10*(r-m*t), t, k, (10*(3*q+r))//t - 10*m, x
        else:
            q, r, t, k, m, x = q*k, (2*q+r)*x, t*x, k+1, (q*(7*k+2)+r*x)//(t*x), x+2


digits = make_pi()
pi_list = []
my_array = []

for i in make_pi():
    my_array.append(str(i))

my_array = my_array[:1] + ['.'] + my_array[1:]
big_float = Decimal("".join(my_array))
print Fraction(big_float).limit_denominator(10**9)

The code prints out a reasonable approximation $\frac{2549491779}{811528438}$ .

Thaddeus Abiy - 7 years, 1 month ago

Thanks!

Tan Li Xuan - 7 years, 1 month ago

think of how it was invented.....

Max B - 7 years, 1 month ago

pi?or the method?

Thaddeus Abiy - 7 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Approximations for \( \pi \)

Comments