Summation Problem Please Help!!!

How can I do this summation formula in Matlab? The value of N = 0:2:32; Let P = (factorial(N)./(factorial(i).*factorial(N-i)))/(2.^N); Thanks.

Answer 1

[Roger Stafford]

Believe it or not, the following code will compute your H, and in fact is probably more accurate than using factorials. It uses a technique known as the "Pascal Triangle". It differs only in the division by 2 at each step in computing P.

 N = 32;
 P = zeros(N+1,N+2);
 P(1,2) = 1;
 for n = 1:N
   P(n+1,2:n+2) = (P(n,1:n+1)+P(n,2:n+2))/2;
 end
 h = zeros(N+1,1);
 for n = 0:N
   h(n+1) = -sum(P(n+1,2:n+2).*log2(P(n+1,2:n+2)));
 end

The array 'h' here is such that H(n) = h(n+1). It gets all H from H(0) to H(32). If you want it at 0:2:32 do:

 h(1:2:33)

[Edy]

What it would be like if I use factorial formula instead of Pascal Triangle?