Estimating the value of pi using a summation through creation of an m.file by using a loop.

My script is missing something. Perhaps I do not understand the meaning of 'while' or 'for' correctly. But here is my code:

{% this script approximates the value of pi
n=0;
x=0;
while n<100000                    % loop ends at 10^6
n=n+2; % n will always be a multiple of 2        
while x<3.14
 x=4*((-1).^n/(2*n+1)) +x; % the summation formula
end 
disp(x)
end }

I want it such that my value of pi in this case 'x' is no greater than 1e-6 away from the actual value of pi. So when it satisfies this condition the script will cease. However my script gives me endless loop of 3.2 and it pissing me off, all day on this crap :S . Why is matlab so hard?????

Answer 1

[Roger Stafford]

Adam, your while-within-a-while construct is certainly not going to work for you.  The terms "4*(-1).^n/(2*n+1)" are supposed to be added once for each successive value of n, starting with n = 0, but, as it is, your code will add this term for an unchanging n a number of times before going on to the next n because of your inner while-loop arrangement.  Also the n = n+2 is wrong.  It should be for each successive n, n = n + 1, not every multiple of 2.

This approximation is based on the infinite series expansion of arctan:

 arctan(t) = t - 1/3*t^3 + 1/5*t^5 - 1/7*t^7 + ...

with t set to 1 giving

 pi/4 = arctan(1) = 1 - 1/3 + 1/5 - 1/7 + ...

With t equal to 1 this is a very, very slowly converging series, and you should have a single while-loop with the condition that 1/(2*n+1) < 1e-6 to achieve the accuracy you desire, which will require a great many loops.  You should start with n = 0 and compute your first x before adding 1 to n.  (Your present code erroneously uses n = 2 for its first term.)

Note that if you were to use t = 1/sqrt(2) instead of t = 1, your series would converge much, much faster, but the added terms would have to be changed to 6*(-1)^n/(2*n+1)*t^(2*n+1), since arctan(1/sqrt(2)) = pi/6.

Note 2: Sean is teasing you.  He starts with pi itself and his loop never executes.

Roger Stafford

[Roger Stafford]

My apologies.  I wrote down the wrong value for t in the last paragraph.  The paragraph should read:

Note that if you were to use t = 1/sqrt(3) instead of t = 1, your series would converge much, much faster, but the added terms would have to be changed to 6*(-1)^n/(2*n+1)*t^(2*n+1), since arctan(1/sqrt(3)) = pi/6.

Roger Stafford

[A K]

% this script approximates the value of pi
n=0;
x=0;
while (-1).^n/(2*n+1)<1e-6
 x=4*((-1).^n/(2*n+1)) +x;
 n=n+1; 
end 
x.

my question states that I have to use my summation formula with a single while loop. I don't think i can use the arctan thing

[bym]

@Adam - you might want to consider that

(-1).^n/(2*n+1)

can be positive or negative; but you are only interested in the magnitude. Also consider the comparison to 1e-6, I believe you mean greater than (>)

[A K]

%e
n=0;
x=0;
k=4*((-1)^n/(2*n+1));
while k>1e-6
n= n+1;
x=x+k;
k=4*((-1)^n/(2*n+1));
end
x
This doesn't work either and i only want n to increase by 2 each time through the loop. It still won't converge anywhere near pi... I don't get what i'm missing.

[A K]

yes!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

i got it!

 % if true
   n=0;
x=0;
k=4*((-1)^n/(2*n+1));
while abs(k)>1e-6
  n= n+1;
  x=x+k;
  k=4*((-1)^n/(2*n+1));
end
x

Answer 2

[Sean de Wolski]

function myPi = piEst()

    myPi = pi;
    for ii = 1:0
       myPi = myPi+ii;
    end
  end

[A K]

having difficulty comprehending what this means. you've equated myPi to pi, to it defeats the purpose of the estimation no?

[A K]

Your code is so simple! The sad thing is I have to use a loop.

[Sean de Wolski]

I used a loop!

3.141592653589793238466 is still an approximation to actual pi. So my code meets all of your constraints!

[A K]

Yes i had completely forgotten on the for loop because I felt so adamant on using the while loop.

Answer 3

[A K]

why did u create a vector 1:0