# Why does $\frac{\partial}{\partial x}O\left(\left(\frac{1}{x}\right)^0\right)$ equal $O\left(\left(\frac{1}{x}\right)^0\right)$ in a series expansion?

When taking the derivative of a series expansion around a finite point, the $O(x^n)$ part is differentiated as expected. $O(x^n)$ becomes $O(x^{n-1})$ except $O(x^0)$ which stays $O(x^0)$.

When expanding around infinity, things do not work out that nicely. $O\left(\left(\frac{1}{x}\right)^n\right)$ in general becomes $O\left(\left(\frac{1}{x}\right)^{n+1}\right)$. So far so good but $O\left(\left(\frac{1}{x}\right)^0\right)$ stays the same as in the expansion around a point that is finite. This doesn't seem to make sense because the leading constant term should drop out and the second one should be differentiated so we should get $O\left(\left(\frac{1}{x}\right)^2\right)$.

Am I missing something? Is this behavior to be expected?

D[O[x, Infinity] x, x]
(* Out[1]= SeriesData[x, DirectedInfinity[1], {}, 0, 0, 1] *)

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Actually, your "as expected" comes as a surprise, because differentiating $O(x^n)$ should not yield anything definite, since (a) a function that grows like $O(x^n)$ need not be differentiable and (b) even when it is, its derivative is not necessarily $O(x^{n-1})$. For example, $f(x) = x^n\sin(x^2)$ is $O(x^n)$ but $f'(x)$ is $O(x^{n+1})$. – whuber Jan 26 '13 at 20:09
You are quite right, but when we assume analyticity there is a derivative and it grows like $O(x^{n-1})$. $x^n \mathrm{sin}(x^2)$ is actually $O(x^{n+2})$ around zero. – Friedrich Jan 26 '13 at 21:15