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im trying to compute value of "pd" for various values of "lnd"..... but it takes too much time without any results and when i terminate i get this message :

Operation terminated by user during sym/symsum (line 65)

In Untitled5 (line 12)

pd=(symsum(nchoosek(M,i)*(pd1)^(i)*(1-pd1)^(M-i), i, s, M));

equations i'm trying to implement are attached

M=8;

N=32;

K=14;

c=2;

s=1;

T=18.623390280430364437245197384232;

lnd=30*rand;

pd1=2*K*nchoosek(N/2,K)*((gamma(K+i)*gamma(N-K+i+1+T/(1+lnd)))/(gamma(N+1+T/(1+lnd)))-...

(symsum(nchoosek(N/2,i)*(gamma(K+i)*gamma(N-K-i+1+T/(1+lnd)))/(gamma(N+1+T/(1+lnd))), i, K, (N/2))));

pd=(symsum(nchoosek(M,i)*(pd1)^(i)*(1-pd1)^(M-i), i, s, M));

plot(lnd,pd,'b.')

Walter Roberson
on 31 May 2017

symsum is often the wrong function to call when you want to calculate a definite sum (that is, the upper and lower bound are numeric.) symsum() works by trying to calculate the general formula for the indefinite summation, and then substituting in the upper and lower bounds. Calculating the general formula can take a long time.

Instead, when you have numeric upper and lower bounds, you should typically generate all of the individual entries; vectorize if possible and otherwise use a for loop or arrayfun(). Then with those in hand, sum() the result.

I corrected some errors in your calculations of pd1

M = 8;

N = 32;

K = 14;

c = 2;

s = 1;

T = 18.623390280430364437245197384232;

nlambda = 50;

lambdavals = linspace(0,30,nlambda);

pd = zeros(1, nlambda);

for L = 1 : nlambda

lambda = lambdavals(L);

T1lambda = T./(1+lambda); %constant sub-expression factored out

pd1temp = arrayfun(@(i) nchoosek(N/2,i) * gamma(K+i) * gamma(N-K-i+1+T1lambda), K : N/2, 'Uniform', 0); %constant denominator removed from terms

pd1sum = sum( [pd1temp{:}] ) ./ gamma(N+1+T1lambda); %this is the first symsum

pd1 = 2 * K * nchoosek(N/2,K) * gamma(K) * gamma(N/2-K+1+T1lambda) / gamma(N/2+1+T1lambda) - pd1sum;

pdtemp = arrayfun(@(i) nchoosek(M,i)*(pd1)^(i)*(1-pd1)^(M-i), s:M, 'Uniform' ,0);

pdsum = sum( [pdtemp{:}] ); %this is the second symsum

pd(L) = pdsum;

end

plot(lambdavals, pd)

You will find that this will take less than a second to compute.

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