How to make a function that calculate appoximate value of pi

Hi every one; I am going to make a function called a_pi that uses the following approximations of pi
but that function should have following specifications Instead of going to infinity, the function stops at the smallest k for which the approximation differs from pi (i.e., the value returned MATLAB’s built-­‐in function) by no more than the positive scalar, delta, which is the only input argument. The first output of the function is the approximate value of π, while the second is k. (Note: if your program or the grader takes a long time, you may have created an infinite loop and need to hit Ctrl-­‐C on your keyboard.) How to deal with that question.Thanks in advance for assistance..

2 Comments

formaula is given in the image attached in the the post
Please post what you have done so far so we can comment on it and make suggestions.

Sign in to comment.

 Accepted Answer

function [pi_here,k1] = a_pi(delta)
f = @(k)sqrt(12)* 1./( (2*k+1).*(-3).^k ) ;
k1 = ceil(fzero(@(k)abs(f(k)) - delta,1));
pi_here = sum(f(0:k1));
end
or
function [pi_here,k1] = a_pi(delta)
f = @(k)sqrt(12)* sum(1./( (2*k+1).*(-3).^k )) ;
k1 = ceil(fzero(@(k)abs(f(0:k) - pi) - delta,1));
pi_here = f(0:k1);
end
or
function [pi_here,k1] = a_pi(delta)
k = 0;
pi_here = 0;
while abs(pi_here - pi) > delta
pi_here = pi_here + sqrt(12)./( (2*k+1).*(-3).^k );
k = k + 1;
end
k1= k - 1;
end

3 Comments

@Andrei Bobrov thank you very much.Actually this and another question of my work was the only questions which i unable to understand or solve... thanks again for your assistance..
function main
tol = 1e-15 ;
n = 1e3 ;
tic ;
for k = 1 : n
a_pi1( tol ) ;
end
toc
tic ;
for k = 1 : n
a_pi2( tol ) ;
end
toc
tic ;
for k = 1 : n
a_pi3( tol ) ;
end
toc
end
function [pi_here,k1] = a_pi1(delta)
f = @(k)sqrt(12)* 1./( (2*k+1).*(-3).^k ) ;
k1 = ceil(fzero(@(k)abs(f(k)) - delta,1));
pi_here = sum(f(0:k1));
end
function [pi_here,k1] = a_pi2(delta)
f = @(k)sqrt(12)* sum(1./( (2*k+1).*(-3).^k )) ;
k1 = ceil(fzero(@(k)abs(f(0:k) - pi) - delta,1));
pi_here = f(0:k1);
end
function [pi_here,k1] = a_pi3(delta)
k = 0;
pi_here = 0;
while abs(pi_here - pi) > delta
pi_here = pi_here + sqrt(12)./( (2*k+1).*(-3).^k );
k = k + 1;
end
k1= k - 1;
end
Gives
>> main
Elapsed time is 0.731927 seconds.
Elapsed time is 2.314194 seconds.
Elapsed time is 0.007396 seconds.
Thanks, its lengthy approach but will be fine.

Sign in to comment.

More Answers (0)

Categories

Find more on MATLAB in Help Center and File Exchange

Community Treasure Hunt

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

Start Hunting!