Asked by YI Hsu Lo
on 12 Dec 2012

i have 2 variables and 5 eqautions

i would like to present them in the form

X ( : , : , 1 ) = xx( : , : , 1 , 1 ) + xx ( : , : , 2 , 1 ) + xx ( : , : , 3 , 1 ) + xx ( : , : , 4 , 1 ) + xx ( : , : , 5 , 1 )

X ( : , : , 2 ) = xx ( : , : , 1 , 2 ) + xx ( : , : , 2 , 2 ) + xx ( : , : , 3 , 2 ) + xx ( : , : , 4 , 2 ) + xx (: , : , 5 , 2 )

. . . . . . .

and my method is really stupid:

for m=1:5;

xx (:,:,m,1)=sum(eh(:,:,1)./eh(:,:,m).*log((e(:,:,m+1)-z(:,:,1))./(e(:,:,m)-z(:,:,1))),3);

xx(:,:,m,2)=sum(eh(:,:,2)./eh(:,:,m).*log((e(:,:,m+1)-z(:,:,2))./(e(:,:,m)-z(:,:,2))),3);

xx(:,:,m,3)=sum(eh(:,:,3)./eh(:,:,m).*log((e(:,:,m+1)-z(:,:,3))./(e(:,:,m)-z(:,:,3))),3);

xx(:,:,m,4)=sum(eh(:,:,4)./eh(:,:,m).*log((e(:,:,m+1)-z(:,:,4))./(e(:,:,m)-z(:,:,4))),3);

xx(:,:,m,5)=sum(eh(:,:,5)./eh(:,:,m).*log((e(:,:,m+1)-z(:,:,5))./(e(:,:,m)-z(:,:,5))),3);

end;

for j=1:5;

X ( : , : , j ) = sum ( xx ( : , : , : , j ) , 3 ) ;

end;

how could i construct a for loop to shorten the code?

it seems very easy but somehow it confuse me! could someone answer me please!

and if there's some 'syms' in the equations, can i use the same method?

*No products are associated with this question.*

Answer by Azzi Abdelmalek
on 12 Dec 2012

Edited by Azzi Abdelmalek
on 12 Dec 2012

xx=rand(5,5,5,3); [n1,n2,n3,n4]=size(xx); X=zeros(n1,n2,n4); for k=1:n4 X(:,:,k)=0; for l=1:n3 X(:,:,k)=X(:,:,k)+xx(:,:,l,k) end end

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