Different machine precision for scalars and vectors?
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I was prearing a demonstration for my students on the topic of numerical error, and after some trial and error devised the sequence:
for i = 1:8
a = sin(i*(pi/2))
end
which reveals errors due to machine precision (a ≠ 0) when i is even. However, these errors do not show up if I store the result in a vector:
for i = 1:8
a(i) = sin(i*(pi/2))
end
Can anyone explain why? I'm guessing the reason is that somehow fewer bits are being used to encode the fractional part of the number in the latter case, but it's not clear to me why this would be the case.
1 Commento
David Hill
il 31 Ago 2020
It is the same answer.
format long
a=sin(pi);
b=sin((1:8)*pi/2);
display(b(2));
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Più risposte (1)
the cyclist
il 31 Ago 2020
If you don't have a specific reason to be using a for loop here, why not also teach them canonical use of vectorized calculations?
i = 1:8;
a(i) = sin(i*(pi/2))
And either use the advice in David Hill's comment, or subtract the exact answer from a to show the error:
exact = [1 0 -1 0 1 0 -1 0];
floatingPointError = a - exact
1 Commento
Victoria Interrante
il 31 Ago 2020
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