Subscript indices must either be real positive integers or logicals.
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function T=baryeval2(tau,s,c,belta,x)
u=zeros(size(x))
d=zeros(size(x))
T=zeros(size(x))
for k=1:5
u=u+s(k).*belta(k).*(x-tau(k)).^(-1);
d=d+c(k).*belta(k).*(x-tau(k)).^(-1);
end
T=u./d;
end
1 comentario
Azzi Abdelmalek
el 27 de Feb. de 2016
You can copy and paste your code instead of inserting an image.
Respuestas (4)
Azzi Abdelmalek
el 27 de Feb. de 2016
Editada: Azzi Abdelmalek
el 27 de Feb. de 2016
You cant use an index equal to zero, Matlab allows only positive integer indices, or logical indices
4 comentarios
the cyclist
el 27 de Feb. de 2016
MATLAB has 1-based indexing. When you enter
tau(k)
for k = 0, there is no such element.
Walter Roberson
el 27 de Feb. de 2016
You have, in part, (x-tau(k)).^(-1) . If there is any x equal to tau(k) then the subtraction would give 0 and raising that to power -1 would give infinity. Infinity times any non-zero value gives infinity (infinity times 0 gives nan), so you would be adding infinity to the value, resulting in infinity. And then since you are effectively totaling values, you are going to get an infinite result.
2 comentarios
Walter Roberson
el 27 de Feb. de 2016
If you do have an x equal to some tau, then inf or nan is the correct answer for the formulae you have written. The situation is exactly like asking to total 1/x when x is allowed to be 0.
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