Infinite series for pi

Hi I was wondering how I would go about writing a program for this infinite series of pi, so that the script works and I am would be able to plot a graph form it? http://upload.wikimedia.org/math/3/7/0/3706d96d8b2b4281a22700f2b642a5ad.png

 I am new to Matlab Student version and having a hard time with the series problems. than you all and have a great day

Answer 1

[Youssef Khmou]

hi, You can use loops to get the value pi=22/7, so it is an infinite serie, you specify in your software the format short to show 4 digits after the decimal point

example :

format short
f=2.6589

You implement the loop :

 % Infinite series for PI
 PI=0;

 %tolerance=1e-3;
 n=1;
 N=6351;
 v=zeros(N+1,1);
 for n=0:N
     PI=PI+4*(((-1)^n)/(2*n+1));
     v(n+1)=PI;
    %n=n+1;
 end
 PI=PI;
figure, plot(v)
figure, plot(v(1:40)), hold on, plot(0:0.1:40,pi,'r-')

[ChristianW]

Or using vector instead of for-loop.

n = 0:10000000;
PI = 4*sum((-1).^n./(2*n+1));

But the loop is needed to plot the evolution of PI:

N = 100;  % number of n-steps
nv = 0:N-1;  % n vector
PI = ones(1,N);  % preallocate PI
PI(1) = 4*1;  % initiate:  PI(1) = PI(n=0) = 4*1
for n = 1:N-1
    PI(n+1) = PI(n)+4*(-1)^n/(2*n+1);
end
plot(nv,PI,'k'); xlabel('n')

[Youssef Khmou]

right, also loops must be used to evaluate the tolerance like "while the quadratic error is > than tolerance"

[AngelsaAtWar]

How would I graph this to should it converging to Pi within 20 terms though?

[Youssef Khmou]

i edited the code, it plots two graphs : first for 6351 iterations, second only for 40 iterations,

[ChristianW]

Your x-axis is not n , its n+1 , the v-indices. (v(1) = v(n=0))

[Youssef Khmou]

@Christian : improve the code !