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

2 次查看(过去 30 天)
The following algorithm for computing ? is due to Archimedes:
1. Start with ? = 1 and ? = 6.
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
  9 个评论
Anastasia Kyriakou
Anastasia Kyriakou 2020-2-23
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')
while cond
%code to be repeated
end
end
end
% end
Like that?
Rik
Rik 2020-2-23
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.

请先登录,再进行评论。

回答(1 个)

Pravin Jagtap
Pravin Jagtap 2020-2-25
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

类别

Help CenterFile Exchange 中查找有关 Matrix Indexing 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by