func=@(p)sum((a./(a-p)))
fzero for sum of functions?
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The actual equation is 0 = sum(ai/(ai-p)) where i is 1 to 4 and I have to look for p values.
This is what I put in the code,
a=[1.2 3.5 1.9 1];
func=@(p)sum((a/(a-p)))
fsolve(func,0)
and the answer it is giving me is -154.25. I checked with my calculator and it does not make sense. Is there something wrong with this code possibly? I am little bit confused about putting function inside the sum().
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