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approximation of pi with arctan

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Benjamin Trivers
Benjamin Trivers 2020-2-2
回答： David Hill 2020-2-3
term = @(x,n) (-1).^(n+1).*((x.^(2*n-1))./(2*n-1));
a1 = 0;
a2 = 0;
for n = 1:10
a1 = a1 + term(1/5,n);
a2 = a2 + term(1/239,n);
end
pi = (4*a1) - a2 % calculating pi/4
pi = 4*pi
this is my code
  1. Write a script that does the following: Approximate π using the formula from trigonom- etry (π = 4arctan1) and arctan values using the first 10 terms in the power series letting x=1.Then calculate the absolute error of your approximation using the above formula (hint: the error from k ∗ arctan x is k times the error from arctan x). Use fprintf to nicely display your results.
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John D'Errico
John D'Errico 2020-2-2
编辑：John D'Errico 2020-2-2
Using pi as a variable name is a bad idea. It will cause you many bugs in the futre, along with anguished questions like "Why does pi no longer work for me as pi?"
The question here is NOT to compute atan(1/5) or atan(1/239) at all. READ THE QUESTION. It is to compute pi using atan(1), using the indicate series. (Then you will worry about the factor of 4.) Read your assignment. It is very clear.
" using the first 10 terms in the power series letting x=1."
You need to use x=1 there, NOT 1/5 OR 1/239, even though OTHER approximations for pi use those values.
I will add that the series form you are using will be correct, IF you use the correct value of x inside the loop.
Benjamin Trivers
Benjamin Trivers 2020-2-2
Please don't get an attitude. This is a place to post questions. Not everyone gets how to get matlab to work right away. If you don't have something that will help, keep you comments to yourself

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回答（1 个）

David Hill
David Hill 2020-2-3
I provided this previous and you never commented.
Atan=0;
for k=1:10
Atan=Atan+(-1)^(k-1)/(2*k-1);
end
We tend to be blunt because we are all volunteers answering questions. You should listen to John. He is a matlab pro and well respected by all.

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