% Problem : First'for' loop will run four time. Each time z(n+1) will have % four different values.I want to plot this z(n+1) values for different % value of y. ( Y range can be 0.63095:3.98107) for y=0.63095:3.98107 B=0.5.*(1-​sqrt(y./(1​+y)));

Question relating to plotting

[Geoff Hayes]

naina - please show your code so that we can understand what the first for loop is. Are you trying to plot something on each iteration of the loop? Is there a second for loop?

[naina gupta]

% Problem : First'for' loop will run four time. Each time z(n+1) will have
% four different values.I want to plot this z(n+1) values for different
% value of y. ( Y range can be 0.63095:3.98107)
for y=0.63095:3.98107   
    B=0.5.*(1-sqrt(y./(1+y)));
    P=[ 0 0 0 0 0 0.0020 0.0134 0.0523 0.1498 0.3479 0.6496 0.9156 0.9968 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1];
    for n=1:32
        z(1)=0;
        z(n+1)=((factorial(32))./(factorial(n).*(factorial(32-n))).*(B.^(n)).*((1-B).^(32-n)).*(P(n)));
        z(n+1)=z(n)+z(n+1);
    end
    z(n+1) % This will have four different values as first 'for' loop will run four time

end

Answer 1

[Geoff Hayes]

naina - I suspect that you want to make z a 4x33 matrix

 z = zeros(4,33);
 k = 1;
 for y=0.63095:3.98107
     B=0.5.*(1-sqrt(y./(1+y)));
     P=[ 0 0 0 0 0 0.0020 0.0134 0.0523 0.1498 0.3479 0.6496 0.9156 0.9968 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1];
     z(k,1)=0;
     for n=1:32
        z(k,n+1)=((factorial(32))./(factorial(n).*(factorial(32-n))).*(B.^(n)).*((1-B).^(32-n)).*(P(n)));
        z(k,n+1)=z(k,n)+z(k,n+1);
    end
    k = k + 1;
 end