# Big $\mathcal{O}$ notation

I'm learning about the big $\mathcal{O}$ notation and I'm a bit confused. Why is it that we can write things like $$\displaystyle x+\frac{x^3}{3}+\frac{x^5}{5}+\mathcal{O}(x^6)$$ when there's no $x^6$ term? Wouldn't it make sense to write $$\displaystyle x+\frac{x^3}{3}+\frac{x^5}{5}+\mathcal{O}(x^7)$$ instead? This is for the $\tanh^{-1}{x}$ series if it makes a difference.

• Because $O(x^7)\subset O(x^6)$, both statements are correct. Recall that the $O$-notation only states a domination, not an exact behaviour, for example $x^9=O(x^6)$ (when $x\to0$). – Did Jan 7 '16 at 17:47

The estimate $$\frac12\log\left(\frac{1+x}{1-x}\right)=x+\frac{x^3}3+\frac{x^5}5+O\left(x^7\right)\tag{1}$$ is only for small $x$. For $\left|x\right|\le a$, we have that $\left|x\right|^7\le a\left|x\right|^6$; therefore, $(1)$ implies $$\frac12\log\left(\frac{1+x}{1-x}\right)=x+\frac{x^3}3+\frac{x^5}5+O\left(x^6\right)\tag{2}$$ However, $(1)$ gives more information (that is, a closer approximation for small $x$) than $(2)$.
You're correct in saying that it would make more sense to say $\tanh^{-1}x=x+x^3/3+x^5/5+O(x^7)$ - this would be more useful. However, it's also still the case that $\tanh^{-1}x=x+x^3/3+x^5/5+O(x^6)$, it just happens to be a weaker condition.
$$\tanh^{-1}x=x+x^3/3+x^5/5+O((\sin^7 x)) (x \to 0)$$