Numerical Analysis best estimate on polynomial order I need to determine the best integer value of $k$ for the equation:
\begin{equation}
\arctan(x) = x + O(x^k) \text{    as $x\to 0$} 
\end{equation}
Taylor's Theorem with Lagrange Remainder would tell us:
\begin{equation}
\arctan(x) = \sum\limits_{k=0}^n(-1)^k\dfrac{x^{2k+1}}{2k+1} + \dfrac{f^{n+1}(\xi)}{(n+1)!}x^{n+1}
\end{equation}
Where $-1 < x < 1$ and $\xi$ is between $0$ and $x$. For this problem it seems that we want to approximate $\arctan(x)$ with $x$ and then let the $O(x^k)$ term by the Lagrange remainder (this is my best guess, let me know what you think). Anyway, let $n=2$ then we would get:
\begin{equation}arctan(x) = x + O(x^3)\end{equation}.
So my final answer would be $3$, what do you guys think?
Or it could be something like $\lim_{x\to 0} \dfrac{\arctan(x)-x}{x^k}$ where I need to choose $k$ to satisfy the $O$ condition?
By L'hoptial I could do:
\begin{align} =& \lim_{x\to 0} \dfrac{\arctan(x)-x}{x^k} \\
=& \lim_{x\to 0}\dfrac{\dfrac{1}{1+x^2}-1}{kx^{k-1}}. 
\end{align}
However, $k=-1$ if we want a constant, which doesn't even make sense. What can I do from here?
 A: When considering $x \to 0$, the "best" exponent in an $O(x^k)$ is the maximal $k$ such that $f(x) \in O(x^k)$ holds (pedants please excuse the notation(1)), as opposed to $x \to \pm\infty$, when "best" means minimal.
When you have a sufficiently smooth function $f$, and you approximate it with the Taylor polynomial of order $m$,
$$T_{f,m}(x) = \sum_{\nu = 0}^m \frac{f^{(\nu)}(0)}{\nu!}x^\nu,$$
you can see from the Lagrangian remainder $\dfrac{f^{(m+1)}(\xi)}{(m+1)!}x^{m+1}$ that $f(x) - T_{f,m}(x) \in O(x^{m+1})$. But that need not be the best exponent. From the fact that $\xi$ is between $0$ and $x$ in the Lagrangian remainder, it follows that
$$\lim\limits_{x \to 0} \frac{f(x) - T_{f,m}(x)}{x^{m+1}} = \frac{f^{(m+1)}(0)}{(m+1)!},$$
and thus if $f^{(m+1)}(0) = 0$, we know that $f(x) - T_{f,m}(x) \in o(x^{m+1})$ - if on the other hand $f^{(m+1)}(0) \neq 0$, the limit implies that $\dfrac{f(x) - T_{f,m}(x)}{x^{t}}$ is unbounded near $0$ for all $t > m+1$, so then $m+1$ is sharp.
Generally, if $f \in \mathscr{C}^{m+1}$ with $f^{(m+1)}(0) = 0$, $o(x^{m+1})$ is the best we can say.
But $f^{(m+1)}(0) = 0$ means $T_{f,m} \equiv T_{f,m+1}$, and hence if $f \in \mathscr{C}^{m+2}$, the Lagrangian remainder tells us that then $f(x) - T_{f,m}(x) \in O(x^{m+2})$.
Since $\arctan$ is analytic (in particular $\arctan \in \mathscr{C}^\infty$), the above tells us that the best exponent is $\min \lbrace k \colon k > m \land \arctan^{(k)}(0) \neq 0 \rbrace$ (since $\arctan$ is not a polynomial, the set is nonempty).
Since $\arctan$ is an odd function, all derivatives of even order at $0$ vanish, so the exponent is always an odd number. Since all derivatives of odd order at $0$ are nonzero, the exponent is the smallest odd number larger than the order of the Taylor polynomial, so

So my final answer would be 3, what do you guys think?

is correct.

(1) Formally, one would have to write $f \in O(g)$ for some function $g$, here $f \in O(x \mapsto x^k)$ for example.
