I need help with this function of estimate the value of PI

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Anthony Fuentes
Anthony Fuentes 2016-10-16
编辑： Walter Roberson 2016-10-16
I need to do the following:
Implement the equation pi/4= 1-1/3+1/5-1/7+1/9...- (Leibniz Equation) in a function that receives the relative error willing to accept and return the user a vector with all values of π (estimations) and other vector with relative errors associated with each value of π.
But, I have the following function (below) and I need to transform it in the desire function. I don't know is this is correct. Can you help me please? I'm a premier in this, and an university's student but my concentration is in Chemical Engineering and that is very difficult for me!. Thanks a lot.
%function:
function[p,err,i]=PI2function(m) % As you see m is the terms. But, I need that the function give me the terms and not to put them, only put the error that I want to accept (input).
p=zeros([1 m]); %This function is in Arrays
err=zeros([1 m]); %err is the relative error and p=is the estimated value of pi with that equation.
tolerance = 0.01; % the output of the function are the terms, the relative error associated with each pi value, and the estimated pi value (the last one in Arrays)
p(1) = 0;
err(1) = pi;
for i=1: m
p(i+1) = p(i) -4*((-1)^(i-1)/(2*i*-1+1));
err(i+1) = abs(pi-p(i+1));
if err(i+1) <= tolerance
break
end
end

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Walter Roberson
Walter Roberson 2016-10-16
编辑：Walter Roberson 2016-10-16
In this case, do not pre-allocate your arrays: let them grow with each iteration.
Your input parameter to the function should be tolerance rather than m. You do not need m in your code at all.
Change your for loop to be a while loop:
i = 1;
while true
...
if condition is true
break;
end
i = i + 1;
end
Keep in mind, though, that the tolerance they give might be more than pi itself, so you might not need any iterations.

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