MATLAB: How to calculate pi

pi

question:
calculating pi:
pi/4= 1-(1/3)+(1/5)-(1/7)+(1/9)-... (a)
which comes from
arctanx= x-(x^3/3)+(x^5/5)-(x^7/7)+(x^9/9)-... (b)
write a function to cumpute pi using question a. you should find that this series converges slowly. Determine how many terms are required to calculate pi to a relative accuracy of 10^-5.
My function: 
function pi= calculating_pi
pi1=0
prompt='calculate pi';
n=input(100);
x=1
for i=1:n
    if mod(i,2)==1
        pi1=pi1+(1/x);
    else pi1=pi1-(1/x);
    end
    x=x+2
pi1 = 4 * pi1;
error = abs(pi - pi1);
fprintf('Calculated value of pi : %f\n', pi1);
fprintf('Actual value of pi : %f\n', pi);
fprintf('Error : %f\n', error);
end 
end
i keep getting error using input. I have no idea what I am doing wrong.

Best Answer

Easy. Type pi at the command line. Works like a charm.
Harder, since you did make a very credible effort at a solution, first, you need to recognize that you are using a pretty slowly convergent series.
Here is the sequence of estimates you should expect to see, for the first 50 terms.
n = 0:49;
cumsum(1./(2*n+1).*(-1).^n)*4
ans =
  Columns 1 through 11
            4       2.6667       3.4667       2.8952       3.3397        2.976       3.2837       3.0171       3.2524       3.0418       3.2323
  Columns 12 through 22
       3.0584       3.2184       3.0703       3.2082       3.0792       3.2004       3.0861       3.1942       3.0916       3.1892       3.0962
  Columns 23 through 33
       3.1851       3.0999       3.1816       3.1031       3.1786       3.1059       3.1761       3.1083       3.1738       3.1104       3.1719
  Columns 34 through 44
       3.1122       3.1702       3.1138       3.1686       3.1153       3.1672       3.1166        3.166       3.1178       3.1648       3.1189
  Columns 45 through 50
       3.1638       3.1199       3.1629       3.1208        3.162       3.1216
So, as I said, a slowly convergent series.
A very big part of the problem is in EVERY iteration of the loop, you multiply by 4.
pi1 = 4 * pi1;
That line is INSIDE your loop. So if I change your logic just slightly, multiplying by 4 only at the very end, it works nicely enough.
pi1 = 0;
x=1;
for i=1:n
    if mod(i,2)==1
        pi1=pi1+(1/x);
    else pi1=pi1-(1/x);
    end
    x=x+2;
end
pi1 = 4 * pi1;
error = abs(pi - pi1);
fprintf('Calculated value of pi : %f\n', pi1);
fprintf('Actual value of pi : %f\n', pi);
fprintf('Error : %f\n', error);
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