Evaluating a function using the subs
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Why does the following fail to evaluate the function at x==6 and x==9?
p = 5E-3; L = 50E-3; x = 0:p:L; Q = x; y = Q(x==6*p)
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Star Strider
2017-11-15
If you run this:
Check = x - 6*p
you will see that it never evaluates to 0, so the condition is never met. This is called ‘floating-point approximation error’. The usual way of describing it is to note that in decimal notation, 1/3 evaluates to 0.3333... and multiplying that by 3 yields 0.9999..., not 1.
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2017-11-15
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