Clarifying the definition of big-oh I am a computer science student and I am familiar with big-oh notation in algorithm analysis books. But the book I am working for mathematical analysis is using somewhat different asymptotics definitions to analyze vanishing rate of the error term of the Taylor series. Here is the definition:
Defn: We say $f(t)$ vanishes at least to order p (at the origin), and write $f(t) = O(p)$, if there are positive constants $\delta$ and $C$ for which
$$\mid f(t) \mid \leq C \mid t \mid^p$$
when $\mid t \mid \lt p$.
I don't understand this "vanishes at least to order p" part. Later, the book says; if $f(t)$ vanishes to at least to order p, there is no reason to think $f(t)$ vanishes to higher order than p.
If it can not vanish to order higher than p, doesn't that mean it vanishes at most to order p, not at least? Can you clarify this definition and explain the last bold-faced part? Can a function $f(t) = O(p)$ vanish to order greater than p?
 A: If $|f(t)|\leq C|t|^p$, it might also be the case that $|f(t)|\leq D|t|^{p+1}$, because $|t|^{p+1}\leq |t|^p$ near $t=0$. But it must be the case that $|f(t)|\leq C|t|^k$ for $k<p$ because then $|t|^p\leq |t|^k$ near $t=0$. The statement "$f(t)$ is $O(p)$" does not say $p$ is the sharpest bound you can get, but it does imply a minimal degree of sharpness. As a result, it is best to read the statement as guaranteeing a minimal degree of vanishing, not specifying the best possible bound. That is why your book tells you to call this property vanishing to "at least" order $p$: they don't want you to make the mistake of thinking the order is exactly $p$.
Just look at a concrete example. Say you know $|f|$ is bounded by a constant times $t^2$. This means $|f|$ is also bounded by a constant times $|t|$, because $t^2<|t|$ for small nonzero $t$. But you don't know whether you can do better and bound $|f|$ by a higher-order power.
The general point is that if $a<b$ and you are interested in some $c<b$, it might be true that $a<c$, but it certainly doesn't have to be true.
