How do I approximate the value of pi

1 次查看(过去 30 天)
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)]);
  1 个评论
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.

请先登录,再进行评论。

采纳的回答

Real Name
Real Name 2015-11-8
Are you attempting to use the Leibniz series to approximate pi? That formula you have is not correct.
  2 个评论
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.

请先登录,再进行评论。

更多回答(0 个)

类别

Help CenterFile Exchange 中查找有关 Calculus 的更多信息

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by