In the first part of the code, plot the approximation of π as a function of N, the number of terms in the series, for N between 1 and 15. The function to be used is (pi^2-8)/16=Summation(n=1:N)(1/((2n-1)^2*(2n+1)^2))

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Varun 2013-10-26

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评论： Jie 2013-10-31

I basically do not have any idea how to start. any help would be appreciated.

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Walter Roberson 2013-10-26

How would you calculate the approximation of Pi if N is 1? How would you calculate it if N is 2? How would you plot the approximations?

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Jie 2013-10-26

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编辑：Jie 2013-10-26

在 MATLAB Online 中打开

wish this could help.

% the formula could be rewrote as pi=sqrt(sum(16/((2*n-1)^2*(2*n+1)^2)+8)
N=[1:15, 20, 30];
my_pi=[];
for n=N
    x=1:n;
    tmp=sqrt(sum(16./((2*x-1).^2.*(2*x+1).^2))+8);
    my_pi=[my_pi, tmp];
end
figure;h=axes('color',[.5,.5,.9],'fontangle','italic','fontname','Times New Roman','xcolor',[0,0,.7]);hold on; grid on,
plot(N,my_pi)
title('approcimation of \pi')