How to plot an equation with two independent and one dependent variable?

 p= 0.5 * ∑ ( (-1)^i * (w!/(i!*(w-i)!))*(1-(2*f*i)/w)^k-1)

where ∑ is between i=0 to w, f= w^2/(4*l), l=any prime number greater than 53, k varies between 1 to 100, w varies between 5 to 40.

i want to plot a graph 'k vs p' with w=fixed value eg.w=9

l=53 w=9 f= w^2/(4*l); for k=1:100 for i=0:w temp1=(-1)^i*nchoosek(w,i)*(1-(1-(2*f*i)/w)^k-1); temp(1,i+1)=temp1; end; temp2=0.5 * sum(temp); temp3(1,k)=temp2; end; plot(k,temp3);

though this code is not showing error the plot generated is wrong. the p values must be in the order of 10^-4 to 10^-12 for k ranging from 1 to 100.

Answer 1

[Amit]

There is one difference between the equation and your code (however that does not fixes the issue though).

l=53; 
w=30;
f= w^2/(4*l);
temp3 = zeros(1,100);
for k=1:100
    temp = 0;
    for i=0:w 
        temp1=((-1)^i)*(nchoosek(w,i))*((1-(2*f*i)/w)^k-1); % This line was different from the equation 
        temp= temp + temp1; 
    end; 
    temp3(1,k)=0.5 * temp; 
end;
plot(1:100,temp3);

For small values of K, the order comes down to what you mentioned. Are you sure about the equation? or the values of K in which this order appear is 10 instead of 100?

[Pragathi]

Dear sir,

I think the difference you found between the code and the equation may be using nchoosek(w,i) line for (w!/(i!*(w-i)!)).If so, as far as i know i think both will give same answer.

The value of k varies between 1 to 100 but when i changed w=30 to w=9 and used semilogy in place of plot function in your answer i could get my plot (though it is not exact).

Thank you very much for the help.

Answer 2

[Paul]

it plots only points in vertical line because when you plot k has only a single value, namely 100. Add k=1:100; before the plot command, then it should plot what you want.