hi I want to solve this problem using posted script.
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Problem: Another formula for computing "pi" can be deduced from the identity pi/4 = 4arctan1/5— arctan1/239. Determine the number of terms that must be summed to ensure an approximation to pi to within 1e-3.![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/728314/image.png)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/728314/image.png)
Can some one tell me how to use a script to solve the problem.
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N/A
2021-9-3
编辑:N/A
2021-9-3
I understand I mistakenly deleted the equation too. I didn’t mean it. I just wanted make the question clear. But I re-edited that part and posted the equation. The only thing I wanted to delete was the Bisection script and the part that no one discussed nor has anything to do with matlab. I can leave the script if you think it’s important there. But before completely change something that I posted I think it would have been nice to comment and ask me to leave it as it was. Or ask me why I did that. Because being rude or hiding this question was not my intention at all. Thank you a lot for helping me out. I will leave the script if you think it’s not confusing there and i will only delete the non related second question. No one commented on that part anyway.
采纳的回答
Walter Roberson
2021-9-2
That script cannot be used to solve that problem. That script is for the case where a continuous variable x must be found such that f(x) is close to 0.
However, the current problem instead requires that you find the discrete variable n such that
4 * (approximating arctan 1/5 by n terms) - (approximating arctan 1/239 by n terms)
is within 1e-3 of pi/4
It is a completely different kind of problem.
8 个评论
Walter Roberson
2021-9-2
format long g
tol=1e-3;
arctan = @(x,n) (-1).^(n+1).*((x.^(2*n-1))./(2*n-1));
a1 = 0;
a2 = 0;
for n = 1:20
a1 = a1 + arctan(1/5,n);
a2 = a2 + arctan(1/239,n);
approximate_pi = 4*((4*a1) - a2)
if abs(pi - approximate_pi)<tol
fprintf('converges to %15.10f in %3d iterations\n', approximate_pi, n)
break
end
end
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