# 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!

Note by Tan Li Xuan
4 years, 1 month ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears 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} $$\boxed{123}$$

## Comments

Sort by:

Top Newest

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.

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 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 $$\frac{2549491779}{811528438}$$.

- 4 years, 1 month ago

Log in to reply

Thanks!

- 4 years, 1 month ago

Log in to reply

think of how it was invented.....

- 4 years, 1 month ago

Log in to reply

pi?or the method?

- 4 years, 1 month ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...