Hello, I have a question about whether there is a way around the following problem. If I take different linear combinations of the same vector I should get the same vector as a result. However, applying the linear combination adds some almost-0 error which then implies that the two are not the same…

it’s expected, it’s Float number precision, you should use \approx

0.1, 0.2, 0.8, and 0.9 cannot be expressed exactly in floating point. If you change to x = rand(100); v1 = 0.25*x + (1-0.25)*x; v2 = 0.5*x + (1-0.5)*x; println(maximum(abs.(v1 .- v2))) you’ll get what you expect because 0.25 and 0.5 can be expressed exactly. This will be true in any language usin…

Machine Precision ... and 0 in Julia

General Usage

rdeits July 9, 2020, 4:59am 3

For a nice write-up of this, see PSA: floating-point arithmetic

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Machine Precision ... and 0 in Julia

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