How to make a function that calculate appoximate value of pi

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Muhammad Usman Saleem 2015-5-29

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/220586-how-to-make-a-function-that-calculate-appoximate-value-of-pi

评论： Muhammad Usman Saleem 2016-1-22

采纳的回答： Andrei Bobrov

Hi every one; I am going to make a function called a_pi that uses the following approximations of pi

http://www.mathworks.com/matlabcentral/answers/uploaded_files/31727/pi_appoximate.jpg

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..

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Muhammad Usman Saleem 2015-5-29

formaula is given in the image attached in the the post

James Tursa 2015-5-29

Please post what you have done so far so we can comment on it and make suggestions.

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Answer 1

Andrei Bobrov 2015-5-29

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/220586-how-to-make-a-function-that-calculate-appoximate-value-of-pi#answer_180920

编辑：Andrei Bobrov 2015-5-29

在 MATLAB Online 中打开

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

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Cedric 2016-1-18

在 MATLAB Online 中打开

 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.

Muhammad Usman Saleem 2016-1-22

Thanks, its lengthy approach but will be fine.

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