Evaluating a function using the subs
1 view (last 30 days)
Show older comments
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)
0 Comments
Answers (1)
Star Strider
on 15 Nov 2017
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.
0 Comments
See Also
Categories
Find more on Elementary Math in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!