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?
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
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.