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I have two real variables $\sigma_1$ and $\sigma_2$ where $0 < \sigma_1, \sigma_2 << 1$.

I am getting maximal errors of the form of products of linear combinations of the two variables. For example, I have a second order error of $$\sigma_1 (\sigma_1 - 2 \sigma_2).$$

Is this the same as writing $$\sigma_1 (\sigma_1 - 2 \sigma_2) = O((\sigma_1+\sigma_2)^2)?$$ How does it work with linear combinations?

Thanks!

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    If both sigmas are constants, the expression is O(1), constant – 2017-01-20
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    What if they're not constant? – 2017-01-20
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    Then it depends on their order. For exanple, if they are of the same order, the expression is O(\sigma_1^2) – 2017-01-20

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If $0 < \sigma_1,\sigma_2$ then it is indeed true that

$$ \sigma_1 (\sigma_1 - 2\sigma_2) = O((\sigma_1+\sigma_2)^2) $$

since

$$ -\frac{1}{3} \leq \frac{\sigma_1 (\sigma_1 - 2\sigma_2)}{(\sigma_1+\sigma_2)^2} \leq 1. $$

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    Does this method imply that $O(\sigma_1) = O(\sigma_1+ \sigma_2)$? Since $0 < \frac{\sigma_1}{\sigma_1+\sigma_2} < 1$. – 2017-01-24
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    @user202654 yes, presuming $\sigma_1,\sigma_2 > 0$, by the definition of $O(\cdots)$. – 2017-01-24