I have a function $f(x)$ that has convergent Taylor series expansion around $x=0$ in the following form:
$$f(0)-xg_1(0)+\frac{x^2}{2!}g_2(0)-\frac{x^3}{3!}g_3(0)+\frac{x^4}{4!}g_4(0)-\frac{x^5}{5!}g_5(0)+\ldots$$
where $(-1)^ig_i(x)$ denotes the $i$-th derivative of $f(x)$ with respect to $x$. I know that $g_i(x)>0$ for all $i$ at $x=0$. I am interested in bounding $f(x)$ around small positive $x$, say $0<x\leq\epsilon$.
I would like to make sure that, given all the facts that I described, I can make a claim that the terms $0$ through $i$ of Taylor series above form an upper bound on $f(x)$ for small positive $x$ if $i$ is even, and lower bound if $i$ is odd, or whether these facts are insufficient to make such a claim.
(I haven't worked with Taylor series in a while and I would rather make a fool of myself in front of the experts here than at work.)
Best Answer
Use Taylor's formula with remainder, $$ f(x) = f(0) + xf'(0) + \frac{x^2}{2} f^{(2)}(0) + \frac{x^3}{3!} f^{(3)}(0) + \cdots + \frac{x^m}{m!} f^{(m)}(0) + \frac{x^{m+1}}{(m+1)!} f^{(m+1)} (\xi), $$ for some $\xi \in (0,x)$. You know that $(-1)^i f^{(i)}(0) > 0$. Since $f^{(i)}$ is continuous, for some $\epsilon > 0$ it is the case that $(-1)^i f^{(i)}(\xi) > 0$ for all $\xi \in (0,\epsilon)$, and so for $x \in (0,\epsilon)$, $$ (-1)^{(m+1)} f(x) \geq (-1)^{(m+1)} \left[f(0) + xf'(0) + \frac{x^2}{2} f^{(2)}(0) + \frac{x^3}{3!} f^{(3)}(0) + \cdots + \frac{x^m}{m!} f^{(m)}(0)\right]. $$