# algorithm for computing ? is due to Archimedes: how steps can be repeated

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Anastasia Kyriakou on 22 Feb 2020
Answered: Pravin Jagtap on 25 Feb 2020
The following algorithm for computing ? is due to Archimedes:
2. Replace ? by 2?.
3. Replace ? by sqrt(2 − sqrt(4 − ?))
4. Let ? = ??/2.
5. Let ? = na/2
6. Let ? = (? + ?)/2 (estimate of ?)
7. Let ? = (? − ?)/2 (estimate of error)
8. Repeat steps 2 – 7 until ? becomes smaller than a given tolerance ???.
9. Output ? and ?
Write a function that implements this algorithm. Use your function to determine ? and ? for ??? = 10^(=k)
, ? = 2, 3, … , 10.
I have written a code but i do not know how to write it so that if the value of e is larger than tol then steps 2-7 repeated. I have written this :
```
function [p,e] = algorithmPi(tol)
a = 1;
n = 6;
e = inf;
n = 2*n;
a = sqrt(2-sqrt(4-(a^2)));
l = (n*a)/2;
u = l/(sqrt((1-(a)^2)/2));
p = (u+l)/2; % estimate of pi
e = (u-l)/2; % estimate of error
if e < tol
done = true;
disp('Complete: Error below tolerance')
end
end
Rik on 23 Feb 2020
What code do you want to repeat? Make sure that is inside the while loop.
I would strongly urge you to read the documentation for the functions you're using if you don't understand what they do.

Pravin Jagtap on 25 Feb 2020
Hello Anastasia,
As mentioned in the above comments, I would recommend you to follow the documentation for understanding the loops. Refer to the following template which will help you to implement the algorithm:
function [p,e] = algorithmPi(tol)
% step 1 - Initialize a, n and e(to inf)
% step 2 - Iterative process from 2 -7
while (e > tol)
% step 3 - Replace n and a
% step 4 - Compute l, u, p and e
end
end

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