How do I approximate the value of pi

2015 11 8
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更新时间：2015 11 9
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Here is what I have so far:
function numberPI=(-1^n)*1/(2*n+1)*3^n
s=0;
n=input('enter value of n= ');
for i=1:n
s=s+i
numberPI
end
disp(['numberPI= ',num2str(numberPI)]);

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Dorothy Carter
Dorothy Carter 2015-11-8
function numberPI s=0; n=input('enter value of n= '); for i=1:n s=s+i; numberPI=sqrt(12)* 1./((2*n+1).*(-3).^n; end disp(['numberPI= ',num2str(numberPI)]); Still an error.

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Real Name
Real Name 2015-11-8

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Are you attempting to use the Leibniz series to approximate pi? That formula you have is not correct.

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Dorothy Carter
Dorothy Carter 2015-11-8
编辑：Dorothy Carter 2015-11-8
What is the formula then? That was what I was given Changed script: function numberPI s=0; n=input('enter value of n= '); for i=1:n s=s+i; numberPI=sqrt(12)* 1./((2*n+1).*(-3).^n; end disp(['numberPI= ',num2str(numberPI)]); Still an error.
Real Name
Real Name 2015-11-9
编辑：Real Name 2015-11-9
The formula is given here:
https://en.wikipedia.org/wiki/Leibniz_formula_for_%CF%80
SUM (-1)^n/(2n+1) = pi/4
So make sure to multiply the final result by 4. I'm not sure why you were given that formula. You should confirm with whomever your instructor is.

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